On 2/19/2026 4:06 AM, Mikko wrote:
On 18/02/2026 21:48, polcott wrote:PA ⊢ x
On 2/18/2026 3:10 AM, Mikko wrote:
On 17/02/2026 14:59, polcott wrote:
On 2/17/2026 3:03 AM, Mikko wrote:
On 16/02/2026 15:47, olcott wrote:
On 2/16/2026 5:25 AM, Mikko wrote:
On 15/02/2026 15:02, polcott wrote:
∀x ∈ PA (True(PA, x) ↔ PA ⊢ x)
Does not mean to test every x in PA
No, it merely declares that there are two symbols for one predicate >>>>>>>> (which, if interpreted accordint to the usual meaning of either >>>>>>>> symbol,
is uncomputable).
What do you think that this means: PA ⊢ x ?
The exact meaning depends on the context and the meanings of the
types of the left and right side expressions. The usual
metalogical meaning
is that x is a theorem of some variant of PA. If something else is >>>>>> meant that should be specified in the opus where the expression is >>>>>> used
Yes that is correct. What does that mean?
It means that the author must define the symbols in the opus they are
used.
Is this your best answer or are you trying to be evasive?
Whether another answer would be better is a matter of taste, at least
to some extent.
The correct answer is
A back-chained inference from x to the axioms of PA exists
On 2/19/26 5:20 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 6:17 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:Of course. I don't know why you think you need to say this. It's
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've studied >>>>>> and
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and which >>>>>>>> parts arei don't even entirely,i accept: there is no turing machine that implements aThere's some private made up terms there, but I don't think it >>>>>>>>>> matters.
classic decider
interface, so a machine that is undecidable to it does not >>>>>>>>>>> exist, as
constructing such a machine would require a machine that >>>>>>>>>>> implements the
classic decider interface
You accept the classical halting theorem. Your odd use of terms >>>>>>>>>
you having trouble with?
there is nothing proving the false equivocation of interfaces with >>>>>>> algorithms, that's just an unjustified assumption that's been
continually
made
you won't state clearly if you accept them. Remember, a proof is an >>>>>> argument that the conclusion is logically entailed by he assumptions, >>>>>> so you can accept a proof whose assumptions you reject.
i reject the assumption that disproving the classic decider interface >>>>> implies the non-existence of general algos for decision making.
obvious that if you don't stick to the classic decider interface there >>>> will be algorithms for making related decisions. None can make the
hard
yes/no decision required by the classic halting problem, but no one
doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
Looks like you still won't answer the question. I think it's because
i answered the question why. it doesn't sit right with me that we can
fully analyze the hypothetical machines that are the supposed proof that such analysis is not totally possible
this "problem" is a merely a reflection of a bad specification/interface that doesn't handle all possible input situations, not a proof against general analysis and how that might *actually* manifest with self- referential computing (like TMs). clearly it's not a simple true/false interface, but u haven't proven other interfaces impossible...
it is also only a limitation that applies to turing machines (and
equivalent paradigms), due to their ability to generate such self- references,
it doesn't even necessarily apply to our own analytical capabilities ... because u haven't actually proven that the ct-thesis true, which is what would be necessary to prove that TMs are actually capable of all types
of computations
if an aging lame duck academic doesn't want to heed my words in the slightest, that's fine. how many lame ducks i will need to sift thru
until i find one that still has half a brain still functioning,
*i do not know*
you know you would have to say that, having finally studies a
well-written proof you have had to conclude that it's conclusion does
follow logically from its premises.
und = () -> if (halts(und)) loop()
it doesn't sit right me to claim we cannot algorithmically determine
what
this does, when we clearly know that if halts(und)->TRUE then und()
loops
forever, but if halts(und)->FALSE then und() halts.
We may well be able to make that determination. And there will very
likely be algorithms that can analyse the code you show and the code in
"halts" to determine what that fragment does. Why do you think anyone
disagrees with this possibility?
like ur using that argument to then claim we cannot algorithmically
determine what this does ... *right after doing that very type of
algorithmic analysis on both possible execution paths*
I am not using that argument. This is why I keep asking (and you keep
dodging) the key question: what proof or proofs of this theorem have you
studied? None of that say that one can not determine, algorithmically,
the behaviour of any code fragment. They state that no /one/ Turing
machine can correct classify that haling or otherwise of /every/ Turing
machine.
how do we even do that if such analysis is not possible?
Your premise is wrong. Such analysis /is/ possible. No one is saying
that it is not.
except no one's doing it at any meaningful scale. that real world result
of our collective actions speak *infinitely* louder than the words of
some lame duck academic sticking his fingers in his eyeballs,
/so explain to me why that happened eh???/
and if u try to claim there's some economic unfeasability in doing so,
i'm going to respond: that's some economic brainrot for the ages...
cause we've been chronically over-engineering our entire computing infrastructure to the tune of 3+ orders of magnitude in over-complexity
just within single orgs. considering across all of society add a couple
more orders of magnitude for the mindbogglingly stupid over-redundancy,
our software production system is a complete and total bastardization of
the word "efficiency", and the has happened directly because *we have no theoretically robust way to prove semantic correctness about our programs*.
like why would we waste our time developing tons of redundant paradigms
if we could just prove optimally correct ones??? ... except we think we can't, cause the theory says we can't, and here we are
u can dance around the fucking terms all you want, but the *actual*
results of the theory you are promulgating are plain as day to anyone
with a half functioning brain
(which isn't very many tbh)
; You really need to go study this subject because all this is
well-covered in good textbooks and you could then come back and answer
my question! I think your answer would be the yes, you do accept that
the theorem follows from its premises.
the generalProvided you don't claim to be able to detect, algorithmically, all
algo exists and can be formed into a variety of other interfaces that >>>>> specify how to /undecidable input/ ought to be handled
those infinity of inputs that encode machines that behave like the
examples you have called (incorrectly) "undecidable inputs" then of
course you are right.
(calling me "incorrect" about a label i came up with for a relationship
that has yet to truly be studied at depth, is just hubris 🤷)
Sure. But the key fact is whether I am wrong or not. I won't explain
it's a fucking label bro, for a relationship that does not yet have a
label. how could anyone be "right" or "wrong" about that???
again why I am right because you need to use this misleading phrase to
achieve your rhetorical goals.
i already demonstrated what an /undecidable input/ is, and anyone with
half a brain can understand the assessment: it is a particular input
that cannot be classified by some classifier due it doing a self- referential query to defy the result
u can be butthurt about the term all u want, but there's nothing
misleading, as that simple paradox is the *bedrock* foundation of undecidability within computing, right back to turing's orignal paper /
on computable numbers/
and if u can't admit that's true, then we're done here
Every year this would come up in class. Just classify the "tricky
inputs" as something else and you are home and dry! I leave it as an >>>> exercise to other readers to see why this is (a) impossible and (b)
pointless.
there are two methods of exploration i've been pursuing, that none of ur >>> students have ever suggested:
How can you possibly know? Hubris?
A) filtering the paradoxical machines using partial recognizers, to
produce
a turing complete subset of machines that is totally decidable
Every year. They don't all misuse the term "totally decidable" but when
lol, none of them came up with strict definition of a partial recognizer based on merging /undecidable input/ with negative classification
they do they usually readily agree not to misuse standard term when
explaining how they think the problems can be "got round".
for every paradoxical machines u can craft: there is a non-paradoxical
machine that computes the same function. if can one can detect that a
paradox exists within a machine for a particular classifier (like a
functional eq classifier), then we can safely ignore the machine as
redundant, adding any that don't into a totally decidable yet turing
complete subset of machines
So much misuse of technical terms. I agree that my students would not
abuse language like this but the basic ideas come up all the time.
don't care about ur comments on language bro. i've had it with lame duck academics, u've fucked this up long enough i do in fact feel entitled to just ignore ur complaints on specific language. and my god, skip the
lecture on why we need consistent/precise language. i don't agree u've earned the right to lecture about that given how much a shithow real
world computing has become
But it gets dealt with in tutorials which makes the back and forth very
quick. And my students would /never/ avoid a direct question so I could
probe their understanding my asking questions. Do you think you can
detect all the functionally equivalent inputs? What do you mean by a
sorry, we need a functional *not-eq* classifier, and we don't need an unimplementable classic decider form. a partial recognizer is good
enough to weed out both functions we've seen *and* /undecidable input/
we can just ignore any failures to classify because there will be functionally equivalent machines that will classify properly at some point
"turing complete subset of machines"? and so on. It's slow on Usenet
turing complete subset:
one that computes all computable input->output mappings (with output
either halting with some value, or never returning)
but that ignores the difference between machines that compute infinite sequences
so we can further expand that for the output of non-terminating machines
by saying all computable mappings from input to output sequences in F
cells, regardless of whether they then halt or not
u do know what an F vs E cell is, correct?
tell me, how many of ur students bothered to make that specific of a clarification for their definition of turing complete...
and impossible when you won't answer or when you do but misuse technical
terms.
no, it's a good thing this isn't in person. you would just overwhelm me
with an onslaught of extremely self-assured ignorance that i have no meaningful capability to unpack in the kind of attention span anyone has
for a real time conversation...
the slowness here is not only in my favor, but 100% required for me to
do what i'm trying to do
this isn't tv/movie reality innovation/progression
this is the 21st century overpopulated hyper-capitalist madhouse form
#god
despite what you may think: you can't produce an /undecidable input/
to a
paradox decider ... any attempt to do so would still be /undecidable
input/
Despite what you state, I rely on proofs and I've seen nothing coming
close to a proof of any of claims.
i thot u said u "get this every year" u lying twat??? or did u respond without reading first???
B) extending TMs with reflection (which i haven't defined to you
yet), in
order to morph the problem of 'undecidable input' to one of lying about
context
Yes, many people think that something magic happens when TMs can be
reflexive. Then they try to define the model and it turns out to be
what do u think i mean by "reflection"/"reflexive"?
nothing new. Of course, you may be the first... But I have my doubts.
it turned the halting problem in literally a lying problem
By the way, this is all from the days when I presented the usual proof
sketch based on contradiction. For a room full of programmers, this was
not a good strategy. Many were so convinced that something so simple to
specify /must/ be implementable that it became a real hurdle to
overcome. I switched, after a while, to presenting a direct proof
instead. How you seen one?
yes the annoying one:
https://en.wikipedia.org/wiki/Halting_problem#Sketch_of_rigorous_proof
it's just more indirect, it's not more "rigorous". it says the same
thing except more indirectly. it doesn't matter if u create the self- reference thru a (A) direct instance, (B) passing it in as input, (C) a search of the total machine enumeration, or (D) some generalization of
A, B, and/or C like the "direct" proof ...
*it's all the same damn paradox*
most others are going to impressed by u saying the same thing with a
higher word count
/not me/
that assumption is was first made (afaik) on turing's first paper on >>>>> computable numbers, and i'm questioning itNo, that assumption in not made in any proof of the halting
theorem. No
one assumes that other "general algos for decision making" don't exist. >>>> The theorems assume some basic axioms about sets, define a halting
decider TM and then show that no TM behaves like such a thing. It
sounds more like to reject the definition of a halting decider rather
than any assumptions the proofs make.
great, that's nice
It's usually more helpful to say if you agree or disagree with something.
I don't really need to know if you think it's nice or not.
i'm somewhere in the between 🤷
let me put in this way: given a certain framing computability theory is correct
but i'm pursuing other frames that we might be able to utilize instead
of the one that has left us so philosophically gimped in practice
while everyone else treats it like this means a general aglo doesn't
exist. heck even wikipedia phrases it like that:
/The halting problem is undecidable, meaning that no general algorithm
exists that solves the halting problem for all possible program–input
pairs/
I'm not at all sure what your point is. But since you have not studied
then why comment further?
any proper proofs of the theorem you are not in a position to say what
origin fallacy
assumptions they make. Wikipedia is reporting an imprecise statement of
the conclusion from which you would be daft to conclude that any proof
assumes that (as you claimed) "general algos for decision making" don't
exist.
we do not, in practice, generally prove our programs semantically
correct. dance around words all u want, the actions we took because of
them are *what actually matter*
it's never been proven, u just keep assuming the equivocation inWhat is the "it" that has never been proven? Assumptions are
conversation and are unable to talk without it's assumption
assumed, not proven, but the assumption you gave: "the assumption that >>>> disproving the classic decider interface implies the non-existence of
general algos for decision making" is never made in any proof.
it's only the general philosophy that's blossomed from those proofs
and if
ur gunna try to wash ur hands of that shit just cause it wasn't phrased
that way specifically in a proof, then i'm just gunna call u a turd
shirking responsibility
Ah. I see you know you are wrong on this point. Hence the waffle. No >> proof assumes what you claimed, and you know you can't show otherwise.
Please go ahead and be rude. It will make ignoring you more enticing.
my rudeness is just a reflection of the mind-numbingly ungodly bastardization of computing that been globally deployed by following
thru on the theory lame duck academics like you preached in schools for almost a century now...
but if u have any heart left in you,
u won't actually have the will to hide from ur sins 🙏🐤
But you don't seem to have studied any of the proofs and you won't say >>>> if you accept any of them as logical conclusions that follow from their >>>> premises. You seem determined to avoid this question. Is it because >>>> you have not studied any of the proofs in detail?If we could just get your acceptance of at least one proof out of the >>>>>i'm questioning the fundamentals of computing as far as turing's first >>>>> paper on computable numbers. are turing machine's even the most
correct
model to be using? idk
unless have a proof that is directly related to the kinds of resolutions >>> i'm actively pursuing, i don't see that in my limited time atm, i'm
avaiable to venture down random-ass red herrings that may or may not be
meaningful, from someone who has yet to demonstrate any depth of
interest
in what i'm trying to express
Crank 101. You are certain there is something wrong with the standard
theory, but you won't actually look to see what it says!
say i asked you:
can i decide the halting behavior for any given turing machine ...
you would say no, and bring up the halting proof for me to read.
but then i would ask you: am i a turing machine?
if u reply yes: i will just punch you in the face, cause u deserve it
for degrading humanity.
and if u reply no: then i would ask why am i subject to limitations
proven only thru the use of turing machine specifications????
On 2/19/26 6:44 PM, Richard Damon wrote:
On 2/18/26 10:10 AM, dart200 wrote:
On 2/18/26 4:38 AM, Richard Damon wrote:
On 2/17/26 11:21 PM, dart200 wrote:
On 2/17/26 7:48 PM, Richard Damon wrote:
On 2/17/26 10:19 PM, dart200 wrote:
On 2/17/26 4:51 PM, Tristan Wibberley wrote:
On 17/02/2026 16:40, dart200 wrote:
On 2/17/26 4:43 AM, Richard Damon wrote:
On 2/16/26 5:44 PM, Tristan Wibberley wrote:
On 16/02/2026 04:02, Richard Damon wrote:
I seem to remember proof of the existance of machines whose >>>>>>>>>>>> halting
status is unknowable / unprovable, and thus in his terms >>>>>>>>>>>> "undecidable".
Such machines must be non-halting (as halting is always >>>>>>>>>>>> provable by
just
stepping the machine enough steps) but that means that it >>>>>>>>>>>> must also be
impossible to determine that the given machine has
unknowable halting
status.
Instead of "unknowable" do you mean "uncomputable by a prechosen >>>>>>>>>>> algorithm" ?
No, I mean the answer is just unknowable, as there is NO proof >>>>>>>>>> by any
means that shows what the answer will be.
...
And part of the issue with trying to talk about these machines >>>>>>>>>> is that
they really are a meta-logical paradox, as not only can't we >>>>>>>>>> know the
behavior of these machines, we can't even know that they are >>>>>>>>>> in this
paradoxical class, and thus we can NEVER present one and know >>>>>>>>>> it is of
that class, so they are unconstruable, just existant.
i mean if u iterate over the total enumeration of machines ... >>>>>>>>> you will
find them
For any iteration process? Is there no iteration process that >>>>>>>> puts those
machines after an infinitude of the others?
that wouldn't be a full enumeration, now would it eh??
a full enumeration must hit all machines within a finite even if >>>>>>> unbounded time
Then you admit you can't do what you call a "full enumeration", as >>>>>> there ARE an "infinite" number of possible machines, the
cardinality of Turing Machines is Aleph-0, the Countable Infinity.
i'm not even sure what ur arguing with ur brainrot here
YOU said that a "full enumeration" must "hit all machines" in a
"finite even if unbounded time".
sorry that should be *any* given machine within a finite if unbounded
time?
And, that isn't normally the way to define a "full enumberation",
because it can lead to wrong conclusions.
bro the diagonal proofs for the halting theorem depend on this fact as
well, there's nothing controversial about my claim there
The problem is that with infinite sets, the logical implications of
ALL and ANY can be different, especially if you let some ambiguity get
involved.
Perhaps it would blow your mind to understand that the sum of an
convergent (countable) infinite series can depend on the order you add
up the terms.
already aware of that
Since the number of machines is not finite, but has the value of
Aleph-0, the countable infinity, clearly you are saying that you
can't do what you say.
You can't enumerate ALL the machines in finite time.
All you are doing is showing the basic flaws in your logic of not
understanding basics of the system and logic.
Yes, you will reach any given machine in a finite time, but not
ALL of them, but not ALL machines.
This is one of the problems about trying to talk about these sorts >>>>>> of infinite sets.
Or even enumerations of them.
And shows the difference between an "effective enumeration", which >>>>>> like Turing Computable Numbers, which means we have an actual
algorithm that produces all of them eventually, verse just an
"enumberation" for which we don't have an algorithm that will be
sure to get to all of the members.
Alan Mackenzie <acm@muc.de> writes:
Ben Bacarisse <ben@bsb.me.uk> wrote:
[ .... ]
The Church-Turing thesis is clearly not the kind of thing that is
provable. That's why it's not called a conjecture.
I don't understand that bit. What is unprovable about the Church-Turing
thesis? I think it hypothesises that there is no computing machine more
powerful than a turing machine.
There are, purely theoretical, models of computation (my preferred
phrase) that are more powerful than Turing machines but they are not considered "effective". The thesis is about what effectively computable means and this is where the problem lies. It's not a well-defined
concept, though almost everyone just /assumes/ it means "what TMs (and so
on) can do".
What would prevent a proof along the lines of supposing the existence of
such a more powerful machine, then proving it was actually equivalent to
some turing machine?
Perhaps it would have been better to say that one can't imagine what
such a proof could look like. Can you?
The "more powerful" bit is easy. One would assume that the more
powerful model can compute at least one function that is not TM
computable.
But how would the notion that it is none-the less "effective" be
specified?
And what form could the equivalence proof take, given that we can
assume nothing about model other that the fact that is it more
powerful and yet effective?
It's not something I've given much thought to so I'd be interested if
you can go further with the notion.
----
Ben.
Ben Bacarisse <ben@bsb.me.uk> wrote:
Alan Mackenzie <acm@muc.de> writes:
Ben Bacarisse <ben@bsb.me.uk> wrote:
[ .... ]
The Church-Turing thesis is clearly not the kind of thing that is
provable. That's why it's not called a conjecture.
I don't understand that bit. What is unprovable about the Church-Turing >>> thesis? I think it hypothesises that there is no computing machine more >>> powerful than a turing machine.
There are, purely theoretical, models of computation (my preferred
phrase) that are more powerful than Turing machines but they are not
considered "effective". The thesis is about what effectively computable
means and this is where the problem lies. It's not a well-defined
concept, though almost everyone just /assumes/ it means "what TMs (and so
on) can do".
I think I'm beginning to see the problem. These more powerful models of computation presumably lack the "finite structure" of a turing machine -
the finite number of states, of possible symbols on the tape, and the discreteness of the tape movements.
Maybe analogue devices (slide rules, differential analysers, etc.) come
into this category. Though these could not duplicate the effect of a
turing machine, they do something altogether different.
What would prevent a proof along the lines of supposing the existence of >>> such a more powerful machine, then proving it was actually equivalent to >>> some turing machine?
Perhaps it would have been better to say that one can't imagine what
such a proof could look like. Can you?
No. I'm pretty confused about the whole question.
The "more powerful" bit is easy. One would assume that the more
powerful model can compute at least one function that is not TM
computable.
There are only a countable number of turing machines, but an uncountable number of functions. So that "at least one" would probably be an
uncountably infinite number.
But how would the notion that it is none-the less "effective" be
specified?
I think it needs to be a machine of some sort, in the sense of being a
finite collection of rods, gears, states, tapes, whatever ... But I
can't picture any such device which wouldn't be equivalent to a turing machine.
And what form could the equivalence proof take, given that we can
assume nothing about model other that the fact that is it more
powerful and yet effective?
Maybe there could be a proof that any machine worthy of the description
would be equivalent to a turing machine. This would need to formalise exactly what a "machine" is. Maybe this has been done already.
--It's not something I've given much thought to so I'd be interested if
you can go further with the notion.
I'll see if I can come up with something more coherent after some more thought.
--
Ben.
On 2/20/26 12:00 AM, dart200 wrote:
On 2/19/26 6:44 PM, Richard Damon wrote:
On 2/18/26 10:10 AM, dart200 wrote:
On 2/18/26 4:38 AM, Richard Damon wrote:
On 2/17/26 11:21 PM, dart200 wrote:
On 2/17/26 7:48 PM, Richard Damon wrote:
On 2/17/26 10:19 PM, dart200 wrote:i'm not even sure what ur arguing with ur brainrot here
On 2/17/26 4:51 PM, Tristan Wibberley wrote:
On 17/02/2026 16:40, dart200 wrote:
On 2/17/26 4:43 AM, Richard Damon wrote:
On 2/16/26 5:44 PM, Tristan Wibberley wrote:
On 16/02/2026 04:02, Richard Damon wrote:
I seem to remember proof of the existance of machines whose >>>>>>>>>>>>> halting
status is unknowable / unprovable, and thus in his terms >>>>>>>>>>>>> "undecidable".
Such machines must be non-halting (as halting is always >>>>>>>>>>>>> provable by
just
stepping the machine enough steps) but that means that it >>>>>>>>>>>>> must also be
impossible to determine that the given machine has
unknowable halting
status.
Instead of "unknowable" do you mean "uncomputable by a >>>>>>>>>>>> prechosen
algorithm" ?
No, I mean the answer is just unknowable, as there is NO >>>>>>>>>>> proof by any
means that shows what the answer will be.
...
And part of the issue with trying to talk about these
machines is that
they really are a meta-logical paradox, as not only can't we >>>>>>>>>>> know the
behavior of these machines, we can't even know that they are >>>>>>>>>>> in this
paradoxical class, and thus we can NEVER present one and know >>>>>>>>>>> it is of
that class, so they are unconstruable, just existant.
i mean if u iterate over the total enumeration of machines ... >>>>>>>>>> you will
find them
For any iteration process? Is there no iteration process that >>>>>>>>> puts those
machines after an infinitude of the others?
that wouldn't be a full enumeration, now would it eh??
a full enumeration must hit all machines within a finite even if >>>>>>>> unbounded time
Then you admit you can't do what you call a "full enumeration", >>>>>>> as there ARE an "infinite" number of possible machines, the
cardinality of Turing Machines is Aleph-0, the Countable Infinity. >>>>>>
YOU said that a "full enumeration" must "hit all machines" in a
"finite even if unbounded time".
sorry that should be *any* given machine within a finite if
unbounded time?
And, that isn't normally the way to define a "full enumberation",
because it can lead to wrong conclusions.
bro the diagonal proofs for the halting theorem depend on this fact as
well, there's nothing controversial about my claim there
Right, which shows that something isn't there, not that all are.
The problem is you actually NEED to prove that you algorithm WILL reach
all, not that for any there is an algorithm that reaches it.
The problem is that with infinite sets, the logical implications of
ALL and ANY can be different, especially if you let some ambiguity
get involved.
Perhaps it would blow your mind to understand that the sum of an
convergent (countable) infinite series can depend on the order you
add up the terms.
already aware of that
Ok, then why do you not understand that you need to use effective enumerations in your proof? If you don't define the order the results
can be different.
Your proofs seem to always begin with the assumption you can create an order, but that order isn't actually defined. and thus you can't show
that you actually can create that enumeration effetively and that it
will actually reach your "ANY" machine.
Since the number of machines is not finite, but has the value of
Aleph-0, the countable infinity, clearly you are saying that you
can't do what you say.
You can't enumerate ALL the machines in finite time.
All you are doing is showing the basic flaws in your logic of not
understanding basics of the system and logic.
Yes, you will reach any given machine in a finite time, but not >>>>>>> ALL of them, but not ALL machines.
This is one of the problems about trying to talk about these
sorts of infinite sets.
Or even enumerations of them.
And shows the difference between an "effective enumeration",
which like Turing Computable Numbers, which means we have an
actual algorithm that produces all of them eventually, verse just >>>>>>> an "enumberation" for which we don't have an algorithm that will >>>>>>> be sure to get to all of the members.
On 2/20/26 7:04 AM, Richard Damon wrote:
On 2/20/26 12:00 AM, dart200 wrote:
On 2/19/26 6:44 PM, Richard Damon wrote:
On 2/18/26 10:10 AM, dart200 wrote:
On 2/18/26 4:38 AM, Richard Damon wrote:
On 2/17/26 11:21 PM, dart200 wrote:
On 2/17/26 7:48 PM, Richard Damon wrote:
On 2/17/26 10:19 PM, dart200 wrote:i'm not even sure what ur arguing with ur brainrot here
On 2/17/26 4:51 PM, Tristan Wibberley wrote:
On 17/02/2026 16:40, dart200 wrote:
On 2/17/26 4:43 AM, Richard Damon wrote:
On 2/16/26 5:44 PM, Tristan Wibberley wrote:
On 16/02/2026 04:02, Richard Damon wrote:
I seem to remember proof of the existance of machines >>>>>>>>>>>>>> whose halting
status is unknowable / unprovable, and thus in his terms >>>>>>>>>>>>>> "undecidable".
Such machines must be non-halting (as halting is always >>>>>>>>>>>>>> provable by
just
stepping the machine enough steps) but that means that it >>>>>>>>>>>>>> must also be
impossible to determine that the given machine has >>>>>>>>>>>>>> unknowable halting
status.
Instead of "unknowable" do you mean "uncomputable by a >>>>>>>>>>>>> prechosen
algorithm" ?
No, I mean the answer is just unknowable, as there is NO >>>>>>>>>>>> proof by any
means that shows what the answer will be.
...
And part of the issue with trying to talk about these >>>>>>>>>>>> machines is that
they really are a meta-logical paradox, as not only can't we >>>>>>>>>>>> know the
behavior of these machines, we can't even know that they are >>>>>>>>>>>> in this
paradoxical class, and thus we can NEVER present one and >>>>>>>>>>>> know it is of
that class, so they are unconstruable, just existant.
i mean if u iterate over the total enumeration of
machines ... you will
find them
For any iteration process? Is there no iteration process that >>>>>>>>>> puts those
machines after an infinitude of the others?
that wouldn't be a full enumeration, now would it eh??
a full enumeration must hit all machines within a finite even >>>>>>>>> if unbounded time
Then you admit you can't do what you call a "full enumeration", >>>>>>>> as there ARE an "infinite" number of possible machines, the
cardinality of Turing Machines is Aleph-0, the Countable Infinity. >>>>>>>
YOU said that a "full enumeration" must "hit all machines" in a
"finite even if unbounded time".
sorry that should be *any* given machine within a finite if
unbounded time?
And, that isn't normally the way to define a "full enumberation",
because it can lead to wrong conclusions.
bro the diagonal proofs for the halting theorem depend on this fact
as well, there's nothing controversial about my claim there
Right, which shows that something isn't there, not that all are.
The problem is you actually NEED to prove that you algorithm WILL
reach all, not that for any there is an algorithm that reaches it.
that's like one of the first things turing proved in his paper,
u thot u said u read it
The problem is that with infinite sets, the logical implications of
ALL and ANY can be different, especially if you let some ambiguity
get involved.
Perhaps it would blow your mind to understand that the sum of an
convergent (countable) infinite series can depend on the order you
add up the terms.
already aware of that
Ok, then why do you not understand that you need to use effective
enumerations in your proof? If you don't define the order the results
can be different.
we're doing discrete analysis on each machine separately
Your proofs seem to always begin with the assumption you can create an
order, but that order isn't actually defined. and thus you can't show
that you actually can create that enumeration effetively and that it
will actually reach your "ANY" machine.
Since the number of machines is not finite, but has the value of
Aleph-0, the countable infinity, clearly you are saying that you
can't do what you say.
You can't enumerate ALL the machines in finite time.
All you are doing is showing the basic flaws in your logic of not >>>>>> understanding basics of the system and logic.
Yes, you will reach any given machine in a finite time, but not >>>>>>>> ALL of them, but not ALL machines.
This is one of the problems about trying to talk about these
sorts of infinite sets.
Or even enumerations of them.
And shows the difference between an "effective enumeration",
which like Turing Computable Numbers, which means we have an
actual algorithm that produces all of them eventually, verse
just an "enumberation" for which we don't have an algorithm that >>>>>>>> will be sure to get to all of the members.
On 2/19/26 5:20 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 6:17 PM, Ben Bacarisse wrote:Looks like you still won't answer the question. I think it's because
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:Of course. I don't know why you think you need to say this. It's
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've studied and >>>>>> you won't state clearly if you accept them. Remember, a proof is an >>>>>> argument that the conclusion is logically entailed by he assumptions, >>>>>> so you can accept a proof whose assumptions you reject.
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and which parts arei don't even entirely,i accept: there is no turing machine that implements aThere's some private made up terms there, but I don't think >>>>>>>>>> it matters.
classic decider
interface, so a machine that is undecidable to it does not exist, as
constructing such a machine would require a machine that >>>>>>>>>>> implements the
classic decider interface
You accept the classical halting theorem. Your odd use of terms >>>>>>>>>
you having trouble with?
there is nothing proving the false equivocation of interfaces with >>>>>>> algorithms, that's just an unjustified assumption that's been continually
made
i reject the assumption that disproving the classic decider interface >>>>> implies the non-existence of general algos for decision making.
obvious that if you don't stick to the classic decider interface there >>>> will be algorithms for making related decisions. None can make the hard >>>> yes/no decision required by the classic halting problem, but no one
doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
i answered the question why.
it doesn't sit right with me that we can fully
analyze the hypothetical machines that are the supposed proof that such analysis is not totally possible
this "problem" is a merely a reflection of a bad specification/interface
that doesn't handle all possible input situations, not a proof against general analysis and how that might *actually* manifest with
self-referential computing (like TMs). clearly it's not a simple true/false interface, but u haven't proven other interfaces impossible...
it is also only a limitation that applies to turing machines (and
equivalent paradigms), due to their ability to generate such
self-references,
it doesn't even necessarily apply to our own analytical capabilities
... because u haven't actually proven that the ct-thesis true, which is
what would be necessary to prove that TMs are actually capable of all types of computations
if an aging lame duck academic doesn't want to heed my words in the slightest, that's fine. how many lame ducks i will need to sift thru until
i find one that still has half a brain still functioning,
On 2/20/26 1:16 PM, dart200 wrote:
On 2/20/26 7:04 AM, Richard Damon wrote:
On 2/20/26 12:00 AM, dart200 wrote:
On 2/19/26 6:44 PM, Richard Damon wrote:
On 2/18/26 10:10 AM, dart200 wrote:
On 2/18/26 4:38 AM, Richard Damon wrote:
On 2/17/26 11:21 PM, dart200 wrote:
On 2/17/26 7:48 PM, Richard Damon wrote:
On 2/17/26 10:19 PM, dart200 wrote:i'm not even sure what ur arguing with ur brainrot here
On 2/17/26 4:51 PM, Tristan Wibberley wrote:
On 17/02/2026 16:40, dart200 wrote:
On 2/17/26 4:43 AM, Richard Damon wrote:
On 2/16/26 5:44 PM, Tristan Wibberley wrote:
On 16/02/2026 04:02, Richard Damon wrote:
I seem to remember proof of the existance of machines >>>>>>>>>>>>>>> whose halting
status is unknowable / unprovable, and thus in his terms >>>>>>>>>>>>>>> "undecidable".
Such machines must be non-halting (as halting is always >>>>>>>>>>>>>>> provable by
just
stepping the machine enough steps) but that means that it >>>>>>>>>>>>>>> must also be
impossible to determine that the given machine has >>>>>>>>>>>>>>> unknowable halting
status.
Instead of "unknowable" do you mean "uncomputable by a >>>>>>>>>>>>>> prechosen
algorithm" ?
No, I mean the answer is just unknowable, as there is NO >>>>>>>>>>>>> proof by any
means that shows what the answer will be.
...
And part of the issue with trying to talk about these >>>>>>>>>>>>> machines is thati mean if u iterate over the total enumeration of
they really are a meta-logical paradox, as not only can't >>>>>>>>>>>>> we know the
behavior of these machines, we can't even know that they >>>>>>>>>>>>> are in this
paradoxical class, and thus we can NEVER present one and >>>>>>>>>>>>> know it is of
that class, so they are unconstruable, just existant. >>>>>>>>>>>>
machines ... you will
find them
For any iteration process? Is there no iteration process that >>>>>>>>>>> puts those
machines after an infinitude of the others?
that wouldn't be a full enumeration, now would it eh??
a full enumeration must hit all machines within a finite even >>>>>>>>>> if unbounded time
Then you admit you can't do what you call a "full enumeration", >>>>>>>>> as there ARE an "infinite" number of possible machines, the >>>>>>>>> cardinality of Turing Machines is Aleph-0, the Countable Infinity. >>>>>>>>
YOU said that a "full enumeration" must "hit all machines" in a >>>>>>> "finite even if unbounded time".
sorry that should be *any* given machine within a finite if
unbounded time?
And, that isn't normally the way to define a "full enumberation",
because it can lead to wrong conclusions.
bro the diagonal proofs for the halting theorem depend on this fact
as well, there's nothing controversial about my claim there
Right, which shows that something isn't there, not that all are.
The problem is you actually NEED to prove that you algorithm WILL
reach all, not that for any there is an algorithm that reaches it.
that's like one of the first things turing proved in his paper,
u thot u said u read it
Right, HE proved it for his.
YOU assume for a different enumeration that you also will.
Just because youe arguement looks sort of like his, doesn't mean yours
works also.
Also, you seem to be mixing fields, as you keep on refering to the "Computable Numbers" paper, which is about a different definition of "Computation" then the "Halting Problem"
But, that difference seems to be beyond you.
The problem is that with infinite sets, the logical implications of >>>>> ALL and ANY can be different, especially if you let some ambiguity
get involved.
Perhaps it would blow your mind to understand that the sum of an
convergent (countable) infinite series can depend on the order you
add up the terms.
already aware of that
Ok, then why do you not understand that you need to use effective
enumerations in your proof? If you don't define the order the results
can be different.
we're doing discrete analysis on each machine separately
Then you admit that you aren't solving for AN algorithm that answers for
ALL inputs.
Again, you don't seem to understand the nature of the problem you are arguing about, and come up with a "partial answer" that you can't show
to be better than the existing partial answers.
You need to more clearly DEFINE what you are trying to do, which seems
to be beyond your understanding.
Your proofs seem to always begin with the assumption you can create
an order, but that order isn't actually defined. and thus you can't
show that you actually can create that enumeration effetively and
that it will actually reach your "ANY" machine.
Since the number of machines is not finite, but has the value of >>>>>>> Aleph-0, the countable infinity, clearly you are saying that you >>>>>>> can't do what you say.
You can't enumerate ALL the machines in finite time.
All you are doing is showing the basic flaws in your logic of not >>>>>>> understanding basics of the system and logic.
Yes, you will reach any given machine in a finite time, but not >>>>>>>>> ALL of them, but not ALL machines.
This is one of the problems about trying to talk about these >>>>>>>>> sorts of infinite sets.
Or even enumerations of them.
And shows the difference between an "effective enumeration", >>>>>>>>> which like Turing Computable Numbers, which means we have an >>>>>>>>> actual algorithm that produces all of them eventually, verse >>>>>>>>> just an "enumberation" for which we don't have an algorithm >>>>>>>>> that will be sure to get to all of the members.
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/19/26 5:20 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 6:17 PM, Ben Bacarisse wrote:Looks like you still won't answer the question. I think it's because
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:Of course. I don't know why you think you need to say this. It's
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've studied and >>>>>>> you won't state clearly if you accept them. Remember, a proof is an >>>>>>> argument that the conclusion is logically entailed by he assumptions, >>>>>>> so you can accept a proof whose assumptions you reject.
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and which parts arei don't even entirely,i accept: there is no turing machine that implements a >>>>>>>>>>>> classic deciderThere's some private made up terms there, but I don't think >>>>>>>>>>> it matters.
interface, so a machine that is undecidable to it does not exist, as
constructing such a machine would require a machine that >>>>>>>>>>>> implements the
classic decider interface
You accept the classical halting theorem. Your odd use of terms >>>>>>>>>>
you having trouble with?
there is nothing proving the false equivocation of interfaces with >>>>>>>> algorithms, that's just an unjustified assumption that's been continually
made
i reject the assumption that disproving the classic decider interface >>>>>> implies the non-existence of general algos for decision making.
obvious that if you don't stick to the classic decider interface there >>>>> will be algorithms for making related decisions. None can make the hard >>>>> yes/no decision required by the classic halting problem, but no one
doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
i answered the question why.
I didn't ask why. I asked what proofs of the theorem in question you
have studied. An answer would be to cite them but the closest you've
come is to a wikipedia link to a sketch.
The second part -- do you accept that the conclusion follows logically
from the premises -- should be answered yes or no. Had you answered no,
we could examine what steps you reject. The result would be either
you'd learn something or you would get a good paper out of it by
pointing out a logical error in a published proof.
Anyway, no answer is likely to come so that's the end of that.
it doesn't sit right with me that we can fully
analyze the hypothetical machines that are the supposed proof that such
analysis is not totally possible
Obviously. But since this is not what the theorem says it's just your rhetorical spin to keep the chat going.
this "problem" is a merely a reflection of a bad specification/interface
that doesn't handle all possible input situations, not a proof against
general analysis and how that might *actually* manifest with
self-referential computing (like TMs). clearly it's not a simple true/false >> interface, but u haven't proven other interfaces impossible...
it is also only a limitation that applies to turing machines (and
equivalent paradigms), due to their ability to generate such
self-references,
it doesn't even necessarily apply to our own analytical capabilities
... because u haven't actually proven that the ct-thesis true, which is
what would be necessary to prove that TMs are actually capable of all types >> of computations
if an aging lame duck academic doesn't want to heed my words in the
slightest, that's fine. how many lame ducks i will need to sift thru until >> i find one that still has half a brain still functioning,
I'm flattered because I know you get your validation from insulting
people whose skill and knowledge you secretly respect, but it's all
getting a bit thin for me.
So what is your plan? Are you going to keep posting this stuff as long
as someone you value replies? Are you aiming to beat Olcott's record of
22 years posting the same nonsense? Think about that -- more than two decades wasted and all he has to show for it is a gazillion posts no one
will remember.
Every crank I've seen here always rejects the usual way forward which is
to write up their ideas as a paper that will mean the real expects will
see it. I wonder what your excuse will be. I suspect you will go for
the "all the editors and reviewers are just as dumb, so what's the
point?" excuse.
But that does raise the question: what's the point mate?
On 2/20/26 10:56 AM, Richard Damon wrote:
On 2/20/26 1:16 PM, dart200 wrote:
On 2/20/26 7:04 AM, Richard Damon wrote:
On 2/20/26 12:00 AM, dart200 wrote:
On 2/19/26 6:44 PM, Richard Damon wrote:
On 2/18/26 10:10 AM, dart200 wrote:
On 2/18/26 4:38 AM, Richard Damon wrote:
On 2/17/26 11:21 PM, dart200 wrote:
On 2/17/26 7:48 PM, Richard Damon wrote:
On 2/17/26 10:19 PM, dart200 wrote:
On 2/17/26 4:51 PM, Tristan Wibberley wrote:
On 17/02/2026 16:40, dart200 wrote:
On 2/17/26 4:43 AM, Richard Damon wrote:
On 2/16/26 5:44 PM, Tristan Wibberley wrote:
On 16/02/2026 04:02, Richard Damon wrote:
I seem to remember proof of the existance of machines >>>>>>>>>>>>>>>> whose halting
status is unknowable / unprovable, and thus in his terms >>>>>>>>>>>>>>>> "undecidable".
Such machines must be non-halting (as halting is always >>>>>>>>>>>>>>>> provable by
just
stepping the machine enough steps) but that means that >>>>>>>>>>>>>>>> it must also be
impossible to determine that the given machine has >>>>>>>>>>>>>>>> unknowable halting
status.
Instead of "unknowable" do you mean "uncomputable by a >>>>>>>>>>>>>>> prechosen
algorithm" ?
No, I mean the answer is just unknowable, as there is NO >>>>>>>>>>>>>> proof by any
means that shows what the answer will be.
...
And part of the issue with trying to talk about these >>>>>>>>>>>>>> machines is thati mean if u iterate over the total enumeration of
they really are a meta-logical paradox, as not only can't >>>>>>>>>>>>>> we know the
behavior of these machines, we can't even know that they >>>>>>>>>>>>>> are in this
paradoxical class, and thus we can NEVER present one and >>>>>>>>>>>>>> know it is of
that class, so they are unconstruable, just existant. >>>>>>>>>>>>>
machines ... you will
find them
For any iteration process? Is there no iteration process >>>>>>>>>>>> that puts those
machines after an infinitude of the others?
that wouldn't be a full enumeration, now would it eh??
a full enumeration must hit all machines within a finite even >>>>>>>>>>> if unbounded time
Then you admit you can't do what you call a "full
enumeration", as there ARE an "infinite" number of possible >>>>>>>>>> machines, the cardinality of Turing Machines is Aleph-0, the >>>>>>>>>> Countable Infinity.
i'm not even sure what ur arguing with ur brainrot here
YOU said that a "full enumeration" must "hit all machines" in a >>>>>>>> "finite even if unbounded time".
sorry that should be *any* given machine within a finite if
unbounded time?
And, that isn't normally the way to define a "full enumberation", >>>>>> because it can lead to wrong conclusions.
bro the diagonal proofs for the halting theorem depend on this fact >>>>> as well, there's nothing controversial about my claim there
Right, which shows that something isn't there, not that all are.
The problem is you actually NEED to prove that you algorithm WILL
reach all, not that for any there is an algorithm that reaches it.
that's like one of the first things turing proved in his paper,
u thot u said u read it
Right, HE proved it for his.
YOU assume for a different enumeration that you also will.
i could just copy past his method for an enumeration. this is not an interesting point of discussion and i won't comment further
Just because youe arguement looks sort of like his, doesn't mean yours
works also.
Also, you seem to be mixing fields, as you keep on refering to the
"Computable Numbers" paper, which is about a different definition of
"Computation" then the "Halting Problem"
But, that difference seems to be beyond you.
The problem is that with infinite sets, the logical implications
of ALL and ANY can be different, especially if you let some
ambiguity get involved.
Perhaps it would blow your mind to understand that the sum of an
convergent (countable) infinite series can depend on the order you >>>>>> add up the terms.
already aware of that
Ok, then why do you not understand that you need to use effective
enumerations in your proof? If you don't define the order the
results can be different.
we're doing discrete analysis on each machine separately
Then you admit that you aren't solving for AN algorithm that answers
for ALL inputs.
not following on what ur going on about there so i won't comment further
Again, you don't seem to understand the nature of the problem you are
arguing about, and come up with a "partial answer" that you can't show
to be better than the existing partial answers.
You need to more clearly DEFINE what you are trying to do, which seems
to be beyond your understanding.
Your proofs seem to always begin with the assumption you can create
an order, but that order isn't actually defined. and thus you can't
show that you actually can create that enumeration effetively and
that it will actually reach your "ANY" machine.
Since the number of machines is not finite, but has the value of >>>>>>>> Aleph-0, the countable infinity, clearly you are saying that you >>>>>>>> can't do what you say.
You can't enumerate ALL the machines in finite time.
All you are doing is showing the basic flaws in your logic of >>>>>>>> not understanding basics of the system and logic.
Yes, you will reach any given machine in a finite time, but >>>>>>>>>> not ALL of them, but not ALL machines.
This is one of the problems about trying to talk about these >>>>>>>>>> sorts of infinite sets.
Or even enumerations of them.
And shows the difference between an "effective enumeration", >>>>>>>>>> which like Turing Computable Numbers, which means we have an >>>>>>>>>> actual algorithm that produces all of them eventually, verse >>>>>>>>>> just an "enumberation" for which we don't have an algorithm >>>>>>>>>> that will be sure to get to all of the members.
On 2/20/26 1:45 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/19/26 5:20 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 6:17 PM, Ben Bacarisse wrote:Looks like you still won't answer the question. I think it's because
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:Of course. I don't know why you think you need to say this. It's >>>>>> obvious that if you don't stick to the classic decider interface
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and which >>>>>>>>>> parts arei accept: there is no turing machine that implements a >>>>>>>>>>>>> classic deciderThere's some private made up terms there, but I don't think >>>>>>>>>>>> it matters.
interface, so a machine that is undecidable to it does not >>>>>>>>>>>>> exist, as
constructing such a machine would require a machine that >>>>>>>>>>>>> implements the
classic decider interface
You accept the classical halting theorem. Your odd use of >>>>>>>>>>>> terms
i don't even entirely,
you having trouble with?
there is nothing proving the false equivocation of interfaces with >>>>>>>>> algorithms, that's just an unjustified assumption that's been >>>>>>>>> continually
made
studied and
you won't state clearly if you accept them. Remember, a proof >>>>>>>> is an
argument that the conclusion is logically entailed by he
assumptions,
so you can accept a proof whose assumptions you reject.
i reject the assumption that disproving the classic decider
interface
implies the non-existence of general algos for decision making.
there
will be algorithms for making related decisions. None can make
the hard
yes/no decision required by the classic halting problem, but no one >>>>>> doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
i answered the question why.
I didn't ask why. I asked what proofs of the theorem in question you
have studied. An answer would be to cite them but the closest you've
come is to a wikipedia link to a sketch.
The second part -- do you accept that the conclusion follows logically
from the premises -- should be answered yes or no. Had you answered no,
we could examine what steps you reject. The result would be either
you'd learn something or you would get a good paper out of it by
pointing out a logical error in a published proof.
Anyway, no answer is likely to come so that's the end of that.
it doesn't sit right with me that we can fully
analyze the hypothetical machines that are the supposed proof that such
analysis is not totally possible
Obviously. But since this is not what the theorem says it's just your
rhetorical spin to keep the chat going.
this "problem" is a merely a reflection of a bad specification/interface >>> that doesn't handle all possible input situations, not a proof against
general analysis and how that might *actually* manifest with
self-referential computing (like TMs). clearly it's not a simple
true/false
interface, but u haven't proven other interfaces impossible...
it is also only a limitation that applies to turing machines (and
equivalent paradigms), due to their ability to generate such
self-references,
it doesn't even necessarily apply to our own analytical capabilities
... because u haven't actually proven that the ct-thesis true, which is
what would be necessary to prove that TMs are actually capable of all
types
of computations
if an aging lame duck academic doesn't want to heed my words in the
slightest, that's fine. how many lame ducks i will need to sift thru
until
i find one that still has half a brain still functioning,
I'm flattered because I know you get your validation from insulting
people whose skill and knowledge you secretly respect, but it's all
getting a bit thin for me.
So what is your plan? Are you going to keep posting this stuff as long
as someone you value replies? Are you aiming to beat Olcott's record of
22 years posting the same nonsense? Think about that -- more than two
decades wasted and all he has to show for it is a gazillion posts no one
will remember.
Every crank I've seen here always rejects the usual way forward which is
i will write up a paper once my ideas are fleshed out enough, and they aren't yet,
i'm not going to write a paper any sooner than i feel like just cause
some twat on the internet declares "now" is the right time,
to write up their ideas as a paper that will mean the real expects will
see it. I wonder what your excuse will be. I suspect you will go for
the "all the editors and reviewers are just as dumb, so what's the
point?" excuse.
But that does raise the question: what's the point mate?
correct, there is no point is discussing with u
u cut out my ideas arbitrarily and prefer to just talk about nonsense
red herrings to satisfy ur sense of superiority rather than actually
have a discussion on the ideas i'm *trying* to explore
i don't know how yet they relate to various traditional proofs, besides turing's original paper /on computable numbers/, nor am i particularly interested in that relation at present. ur not leading any exploration
on these idea: i am
u clearly have no actual interest in helping, ur intention here is just
some bizarre sanity check on yourself
so fuck off mate, we're done here eh???
On 2/20/26 5:48 PM, dart200 wrote:
On 2/20/26 1:45 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/19/26 5:20 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:i answered the question why.
On 2/18/26 6:17 PM, Ben Bacarisse wrote:Looks like you still won't answer the question. I think it's because >>>>
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:will be algorithms for making related decisions. None can make >>>>>>> the hard
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've >>>>>>>>> studied and
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and >>>>>>>>>>> which parts arei accept: there is no turing machine that implements a >>>>>>>>>>>>>> classic deciderThere's some private made up terms there, but I don't think >>>>>>>>>>>>> it matters.
interface, so a machine that is undecidable to it does not >>>>>>>>>>>>>> exist, as
constructing such a machine would require a machine that >>>>>>>>>>>>>> implements the
classic decider interface
You accept the classical halting theorem. Your odd use of >>>>>>>>>>>>> terms
i don't even entirely,
you having trouble with?
there is nothing proving the false equivocation of interfaces >>>>>>>>>> with
algorithms, that's just an unjustified assumption that's been >>>>>>>>>> continually
made
you won't state clearly if you accept them. Remember, a proof >>>>>>>>> is an
argument that the conclusion is logically entailed by he
assumptions,
so you can accept a proof whose assumptions you reject.
i reject the assumption that disproving the classic decider
interface
implies the non-existence of general algos for decision making. >>>>>>> Of course. I don't know why you think you need to say this. It's >>>>>>> obvious that if you don't stick to the classic decider interface >>>>>>> there
yes/no decision required by the classic halting problem, but no one >>>>>>> doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
I didn't ask why. I asked what proofs of the theorem in question you
have studied. An answer would be to cite them but the closest you've
come is to a wikipedia link to a sketch.
The second part -- do you accept that the conclusion follows logically
from the premises -- should be answered yes or no. Had you answered no, >>> we could examine what steps you reject. The result would be either
you'd learn something or you would get a good paper out of it by
pointing out a logical error in a published proof.
Anyway, no answer is likely to come so that's the end of that.
it doesn't sit right with me that we can fully
analyze the hypothetical machines that are the supposed proof that such >>>> analysis is not totally possible
Obviously. But since this is not what the theorem says it's just your
rhetorical spin to keep the chat going.
this "problem" is a merely a reflection of a bad specification/
interface
that doesn't handle all possible input situations, not a proof against >>>> general analysis and how that might *actually* manifest with
self-referential computing (like TMs). clearly it's not a simple
true/false
interface, but u haven't proven other interfaces impossible...
it is also only a limitation that applies to turing machines (and
equivalent paradigms), due to their ability to generate such
self-references,
it doesn't even necessarily apply to our own analytical capabilities
... because u haven't actually proven that the ct-thesis true, which is >>>> what would be necessary to prove that TMs are actually capable of
all types
of computations
if an aging lame duck academic doesn't want to heed my words in the
slightest, that's fine. how many lame ducks i will need to sift thru
until
i find one that still has half a brain still functioning,
I'm flattered because I know you get your validation from insulting
people whose skill and knowledge you secretly respect, but it's all
getting a bit thin for me.
So what is your plan? Are you going to keep posting this stuff as long >>> as someone you value replies? Are you aiming to beat Olcott's record of >>> 22 years posting the same nonsense? Think about that -- more than two
decades wasted and all he has to show for it is a gazillion posts no one >>> will remember.
Every crank I've seen here always rejects the usual way forward which is
i will write up a paper once my ideas are fleshed out enough, and they
aren't yet,
i'm not going to write a paper any sooner than i feel like just cause
some twat on the internet declares "now" is the right time,
to write up their ideas as a paper that will mean the real expects will
see it. I wonder what your excuse will be. I suspect you will go for >>> the "all the editors and reviewers are just as dumb, so what's the
point?" excuse.
But that does raise the question: what's the point mate?
correct, there is no point is discussing with u
u cut out my ideas arbitrarily and prefer to just talk about nonsense
red herrings to satisfy ur sense of superiority rather than actually
have a discussion on the ideas i'm *trying* to explore
i don't know how yet they relate to various traditional proofs,
besides turing's original paper /on computable numbers/, nor am i
particularly interested in that relation at present. ur not leading
any exploration on these idea: i am
Which, since it is not related to the concept of halting, isn't actually
a good basis.
u clearly have no actual interest in helping, ur intention here is
just some bizarre sanity check on yourself
so fuck off mate, we're done here eh???
Perhaps the problem is you have no interest in being helped.
Perhaps because you have no interest in your "proof" being actually--
based on logic.
Perhaps because it would force you to actual learn something about what
you want to talk about, as opposed to, as you admit you haven't actually figured out how anything relates to what you want to talk about.
On 2/20/26 3:05 PM, Richard Damon wrote:
On 2/20/26 5:48 PM, dart200 wrote:
On 2/20/26 1:45 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/19/26 5:20 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:i answered the question why.
On 2/18/26 6:17 PM, Ben Bacarisse wrote:Looks like you still won't answer the question. I think it's because >>>>>
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:will be algorithms for making related decisions. None can make >>>>>>>> the hard
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've >>>>>>>>>> studied and
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and >>>>>>>>>>>> which parts arei accept: there is no turing machine that implements a >>>>>>>>>>>>>>> classic deciderThere's some private made up terms there, but I don't think >>>>>>>>>>>>>> it matters.
interface, so a machine that is undecidable to it does >>>>>>>>>>>>>>> not exist, as
constructing such a machine would require a machine that >>>>>>>>>>>>>>> implements the
classic decider interface
You accept the classical halting theorem. Your odd use of >>>>>>>>>>>>>> terms
i don't even entirely,
you having trouble with?
there is nothing proving the false equivocation of interfaces >>>>>>>>>>> with
algorithms, that's just an unjustified assumption that's been >>>>>>>>>>> continually
made
you won't state clearly if you accept them. Remember, a proof >>>>>>>>>> is an
argument that the conclusion is logically entailed by he
assumptions,
so you can accept a proof whose assumptions you reject.
i reject the assumption that disproving the classic decider >>>>>>>>> interface
implies the non-existence of general algos for decision making. >>>>>>>> Of course. I don't know why you think you need to say this. It's >>>>>>>> obvious that if you don't stick to the classic decider interface >>>>>>>> there
yes/no decision required by the classic halting problem, but no one >>>>>>>> doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
I didn't ask why. I asked what proofs of the theorem in question you >>>> have studied. An answer would be to cite them but the closest you've >>>> come is to a wikipedia link to a sketch.
The second part -- do you accept that the conclusion follows logically >>>> from the premises -- should be answered yes or no. Had you answered >>>> no,
we could examine what steps you reject. The result would be either
you'd learn something or you would get a good paper out of it by
pointing out a logical error in a published proof.
Anyway, no answer is likely to come so that's the end of that.
it doesn't sit right with me that we can fully
analyze the hypothetical machines that are the supposed proof that
such
analysis is not totally possible
Obviously. But since this is not what the theorem says it's just your >>>> rhetorical spin to keep the chat going.
this "problem" is a merely a reflection of a bad specification/
interface
that doesn't handle all possible input situations, not a proof against >>>>> general analysis and how that might *actually* manifest with
self-referential computing (like TMs). clearly it's not a simple
true/false
interface, but u haven't proven other interfaces impossible...
it is also only a limitation that applies to turing machines (and
equivalent paradigms), due to their ability to generate such
self-references,
it doesn't even necessarily apply to our own analytical capabilities >>>>> ... because u haven't actually proven that the ct-thesis true,
which is
what would be necessary to prove that TMs are actually capable of
all types
of computations
if an aging lame duck academic doesn't want to heed my words in the
slightest, that's fine. how many lame ducks i will need to sift
thru until
i find one that still has half a brain still functioning,
I'm flattered because I know you get your validation from insulting
people whose skill and knowledge you secretly respect, but it's all
getting a bit thin for me.
So what is your plan? Are you going to keep posting this stuff as long >>>> as someone you value replies? Are you aiming to beat Olcott's
record of
22 years posting the same nonsense? Think about that -- more than two >>>> decades wasted and all he has to show for it is a gazillion posts no
one
will remember.
Every crank I've seen here always rejects the usual way forward
which is
i will write up a paper once my ideas are fleshed out enough, and
they aren't yet,
i'm not going to write a paper any sooner than i feel like just cause
some twat on the internet declares "now" is the right time,
to write up their ideas as a paper that will mean the real expects will >>>> see it. I wonder what your excuse will be. I suspect you will go for >>>> the "all the editors and reviewers are just as dumb, so what's the
point?" excuse.
But that does raise the question: what's the point mate?
correct, there is no point is discussing with u
u cut out my ideas arbitrarily and prefer to just talk about nonsense
red herrings to satisfy ur sense of superiority rather than actually
have a discussion on the ideas i'm *trying* to explore
i don't know how yet they relate to various traditional proofs,
besides turing's original paper /on computable numbers/, nor am i
particularly interested in that relation at present. ur not leading
any exploration on these idea: i am
Which, since it is not related to the concept of halting, isn't
actually a good basis.
u clearly have no actual interest in helping, ur intention here is
just some bizarre sanity check on yourself
so fuck off mate, we're done here eh???
Perhaps the problem is you have no interest in being helped.
in order to help me u'd have to put work into understanding what i'm
trying to express, which ur barely putting any effort into
i'm not supposed to be expert here, i'm not supposed to have to come
"up" to ur level to be helped, u need to somehow reach "down" to my level
another reason why academia is so damn broken these days: u mistake gate keeping ur superiority complex with helpfulness
Perhaps because you have no interest in your "proof" being actually
based on logic.
Perhaps because it would force you to actual learn something about
what you want to talk about, as opposed to, as you admit you haven't
actually figured out how anything relates to what you want to talk about.
On 2/18/26 6:17 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've studied and >>>> you won't state clearly if you accept them. Remember, a proof is an
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and whichi don't even entirely,i accept: there is no turing machine that implements a classic >>>>>>>>> deciderThere's some private made up terms there, but I don't think it >>>>>>>> matters.
interface, so a machine that is undecidable to it does not
exist, as
constructing such a machine would require a machine that
implements the
classic decider interface
You accept the classical halting theorem. Your odd use of terms >>>>>>>
parts are
you having trouble with?
there is nothing proving the false equivocation of interfaces with
algorithms, that's just an unjustified assumption that's been
continually
made
argument that the conclusion is logically entailed by he assumptions,
so you can accept a proof whose assumptions you reject.
i reject the assumption that disproving the classic decider interface
implies the non-existence of general algos for decision making.
Of course. I don't know why you think you need to say this. It's
obvious that if you don't stick to the classic decider interface there
will be algorithms for making related decisions. None can make the hard
yes/no decision required by the classic halting problem, but no one
doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
und = () -> if (halts(und)) loop()
it doesn't sit right me to claim we cannot algorithmically determine
what this does, when we clearly know that if halts(und)->TRUE then und() loops forever, but if halts(und)->FALSE then und() halts.
like ur using that argument to then claim we cannot algorithmically determine what this does ... *right after doing that very type of algorithmic analysis on both possible execution paths*
how do we even do that if such analysis is not possible?
the general
algo exists and can be formed into a variety of other interfaces that
specify how to /undecidable input/ ought to be handled
Provided you don't claim to be able to detect, algorithmically, all
those infinity of inputs that encode machines that behave like the
examples you have called (incorrectly) "undecidable inputs" then of
course you are right.
(calling me "incorrect" about a label i came up with for a relationship
that has yet to truly be studied at depth, is just hubris 🤷)
Every year this would come up in class. Just classify the "tricky
inputs" as something else and you are home and dry! I leave it as an
exercise to other readers to see why this is (a) impossible and (b)
pointless.
there are two methods of exploration i've been pursuing, that none of ur students have ever suggested:
A) filtering the paradoxical machines using partial recognizers, to
produce a turing complete subset of machines that is totally decidable
for every paradoxical machines u can craft: there is a non-paradoxical machine that computes the same function. if can one can detect that a paradox exists within a machine for a particular classifier (like a functional eq classifier), then we can safely ignore the machine as redundant, adding any that don't into a totally decidable yet turing complete subset of machines
despite what you may think: you can't produce an /undecidable input/ to
a paradox decider ... any attempt to do so would still be /undecidable input/
B) extending TMs with reflection (which i haven't defined to you yet),
in order to morph the problem of 'undecidable input' to one of lying
about context
that assumption is was first made (afaik) on turing's first paper on
computable numbers, and i'm questioning it
No, that assumption in not made in any proof of the halting theorem. No
one assumes that other "general algos for decision making" don't exist.
The theorems assume some basic axioms about sets, define a halting
decider TM and then show that no TM behaves like such a thing. It
sounds more like to reject the definition of a halting decider rather
than any assumptions the proofs make.
great, that's nice
while everyone else treats it like this means a general aglo doesn't
exist. heck even wikipedia phrases it like that:
/The halting problem is undecidable, meaning that no general algorithm exists that solves the halting problem for all possible program–input pairs/
it's never been proven, u just keep assuming the equivocation in
conversation and are unable to talk without it's assumption
What is the "it" that has never been proven? Assumptions are
assumed, not proven, but the assumption you gave: "the assumption that
disproving the classic decider interface implies the non-existence of
general algos for decision making" is never made in any proof.
it's only the general philosophy that's blossomed from those proofs and
if ur gunna try to wash ur hands of that shit just cause it wasn't
phrased that way specifically in a proof, then i'm just gunna call u a
turd shirking responsibility
If we could just get your acceptance of at least one proof out of the
i'm questioning the fundimentals of computing as far as turing's first
paper on computable numbers. are turing machine's even the most correct
model to be using? idk
But you don't seem to have studied any of the proofs and you won't say
if you accept any of them as logical conclusions that follow from their
premises. You seem determined to avoid this question. Is it because
you have not studied any of the proofs in detail?
unless have a proof that is directly related to the kinds of resolutions
i'm actively pursuing, i don't see that in my limited time atm, i'm
avaiable to venture down random-ass red herrings that may or may not be meaningful, from someone who has yet to demonstrate any depth of
interest in what i'm trying to express
heck just getting the concept of /undecidable input/ across, that ur
still acting butthurt about, was hard enough disillusion me about the capabilities of whoever the fuck i'm really talking to right now
we've never proven the ct-thesis, i don't even believe that's true.
The Church-Turing thesis is clearly not the kind of thing that is
provable. That's why it's not called a conjecture. But if you have in
mind a model of computation that can't be simulated by TMs but could
there's a lot we'd have a to discuss before i could bring why i think
that might be true ...
pass the bar of being considered "effective" then you must stop posting
here and publish right away. A belief, however, is not worth much
without a concrete model.
[...]
u haven't inspired any further evolution/innovation in my arguments,
it's
been a waste of time for me so far
You might want to bear that in mind before clicking "reply...".
i don't fault myself when others say nothing particularly inspiring
On 2/19/2026 12:39 AM, dart200 wrote:
On 2/18/26 6:17 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've studied and >>>>> you won't state clearly if you accept them. Remember, a proof is an >>>>> argument that the conclusion is logically entailed by he assumptions, >>>>> so you can accept a proof whose assumptions you reject.
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and which >>>>>>> parts arei don't even entirely,i accept: there is no turing machine that implements a classic >>>>>>>>>> deciderThere's some private made up terms there, but I don't think it >>>>>>>>> matters.
interface, so a machine that is undecidable to it does not >>>>>>>>>> exist, as
constructing such a machine would require a machine that
implements the
classic decider interface
You accept the classical halting theorem. Your odd use of terms >>>>>>>>
you having trouble with?
there is nothing proving the false equivocation of interfaces with >>>>>> algorithms, that's just an unjustified assumption that's been
continually
made
i reject the assumption that disproving the classic decider interface
implies the non-existence of general algos for decision making.
Of course. I don't know why you think you need to say this. It's
obvious that if you don't stick to the classic decider interface there
will be algorithms for making related decisions. None can make the hard >>> yes/no decision required by the classic halting problem, but no one
doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
und = () -> if (halts(und)) loop()
it doesn't sit right me to claim we cannot algorithmically determine
what this does, when we clearly know that if halts(und)->TRUE then
und() loops forever, but if halts(und)->FALSE then und() halts.
like ur using that argument to then claim we cannot algorithmically
determine what this does ... *right after doing that very type of
algorithmic analysis on both possible execution paths*
how do we even do that if such analysis is not possible?
the general
algo exists and can be formed into a variety of other interfaces that
specify how to /undecidable input/ ought to be handled
Provided you don't claim to be able to detect, algorithmically, all
those infinity of inputs that encode machines that behave like the
examples you have called (incorrectly) "undecidable inputs" then of
course you are right.
(calling me "incorrect" about a label i came up with for a
relationship that has yet to truly be studied at depth, is just hubris
🤷)
Every year this would come up in class. Just classify the "tricky
inputs" as something else and you are home and dry! I leave it as an
exercise to other readers to see why this is (a) impossible and (b)
pointless.
there are two methods of exploration i've been pursuing, that none of
ur students have ever suggested:
A) filtering the paradoxical machines using partial recognizers, to
produce a turing complete subset of machines that is totally decidable
for every paradoxical machines u can craft: there is a non-paradoxical
machine that computes the same function. if can one can detect that a
paradox exists within a machine for a particular classifier (like a
functional eq classifier), then we can safely ignore the machine as
redundant, adding any that don't into a totally decidable yet turing
complete subset of machines
despite what you may think: you can't produce an /undecidable input/
to a paradox decider ... any attempt to do so would still be /
undecidable input/
B) extending TMs with reflection (which i haven't defined to you yet),
in order to morph the problem of 'undecidable input' to one of lying
about context
that assumption is was first made (afaik) on turing's first paper on
computable numbers, and i'm questioning it
No, that assumption in not made in any proof of the halting theorem. No >>> one assumes that other "general algos for decision making" don't exist.
The theorems assume some basic axioms about sets, define a halting
decider TM and then show that no TM behaves like such a thing. It
sounds more like to reject the definition of a halting decider rather
than any assumptions the proofs make.
great, that's nice
while everyone else treats it like this means a general aglo doesn't
exist. heck even wikipedia phrases it like that:
/The halting problem is undecidable, meaning that no general algorithm
exists that solves the halting problem for all possible program–input
pairs/
it's never been proven, u just keep assuming the equivocation in
conversation and are unable to talk without it's assumption
What is the "it" that has never been proven? Assumptions are
assumed, not proven, but the assumption you gave: "the assumption that
disproving the classic decider interface implies the non-existence of
general algos for decision making" is never made in any proof.
it's only the general philosophy that's blossomed from those proofs
and if ur gunna try to wash ur hands of that shit just cause it wasn't
phrased that way specifically in a proof, then i'm just gunna call u a
turd shirking responsibility
If we could just get your acceptance of at least one proof out of the >>>>i'm questioning the fundimentals of computing as far as turing's first >>>> paper on computable numbers. are turing machine's even the most correct >>>> model to be using? idk
But you don't seem to have studied any of the proofs and you won't say
if you accept any of them as logical conclusions that follow from their
premises. You seem determined to avoid this question. Is it because
you have not studied any of the proofs in detail?
unless have a proof that is directly related to the kinds of
resolutions i'm actively pursuing, i don't see that in my limited time
atm, i'm avaiable to venture down random-ass red herrings that may or
may not be meaningful, from someone who has yet to demonstrate any
depth of interest in what i'm trying to express
heck just getting the concept of /undecidable input/ across, that ur
still acting butthurt about, was hard enough disillusion me about the
capabilities of whoever the fuck i'm really talking to right now
we've never proven the ct-thesis, i don't even believe that's true.
The Church-Turing thesis is clearly not the kind of thing that is
provable. That's why it's not called a conjecture. But if you have in >>> mind a model of computation that can't be simulated by TMs but could
there's a lot we'd have a to discuss before i could bring why i think
that might be true ...
pass the bar of being considered "effective" then you must stop posting
here and publish right away. A belief, however, is not worth much
without a concrete model.
[...]
u haven't inspired any further evolution/innovation in my arguments,
it's
been a waste of time for me so far
You might want to bear that in mind before clicking "reply...".
i don't fault myself when others say nothing particularly inspiring
https://youtu.be/yc-blFymxSY?list=RD51yVxGvsqiE
[ ...trash... ]
On 2/20/26 9:59 PM, Chris M. Thomasson wrote:
On 2/19/2026 12:39 AM, dart200 wrote:
On 2/18/26 6:17 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've studied >>>>>> and
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and which >>>>>>>> parts arei don't even entirely,i accept: there is no turing machine that implements aThere's some private made up terms there, but I don't think it >>>>>>>>>> matters.
classic decider
interface, so a machine that is undecidable to it does not >>>>>>>>>>> exist, as
constructing such a machine would require a machine that >>>>>>>>>>> implements the
classic decider interface
You accept the classical halting theorem. Your odd use of terms >>>>>>>>>
you having trouble with?
there is nothing proving the false equivocation of interfaces with >>>>>>> algorithms, that's just an unjustified assumption that's been
continually
made
you won't state clearly if you accept them. Remember, a proof is an >>>>>> argument that the conclusion is logically entailed by he assumptions, >>>>>> so you can accept a proof whose assumptions you reject.
i reject the assumption that disproving the classic decider interface >>>>> implies the non-existence of general algos for decision making.
Of course. I don't know why you think you need to say this. It's
obvious that if you don't stick to the classic decider interface there >>>> will be algorithms for making related decisions. None can make the
hard
yes/no decision required by the classic halting problem, but no one
doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
und = () -> if (halts(und)) loop()
it doesn't sit right me to claim we cannot algorithmically determine
what this does, when we clearly know that if halts(und)->TRUE then
und() loops forever, but if halts(und)->FALSE then und() halts.
like ur using that argument to then claim we cannot algorithmically
determine what this does ... *right after doing that very type of
algorithmic analysis on both possible execution paths*
how do we even do that if such analysis is not possible?
the general
algo exists and can be formed into a variety of other interfaces that >>>>> specify how to /undecidable input/ ought to be handled
Provided you don't claim to be able to detect, algorithmically, all
those infinity of inputs that encode machines that behave like the
examples you have called (incorrectly) "undecidable inputs" then of
course you are right.
(calling me "incorrect" about a label i came up with for a
relationship that has yet to truly be studied at depth, is just
hubris 🤷)
Every year this would come up in class. Just classify the "tricky
inputs" as something else and you are home and dry! I leave it as an >>>> exercise to other readers to see why this is (a) impossible and (b)
pointless.
there are two methods of exploration i've been pursuing, that none of
ur students have ever suggested:
A) filtering the paradoxical machines using partial recognizers, to
produce a turing complete subset of machines that is totally decidable
for every paradoxical machines u can craft: there is a non-
paradoxical machine that computes the same function. if can one can
detect that a paradox exists within a machine for a particular
classifier (like a functional eq classifier), then we can safely
ignore the machine as redundant, adding any that don't into a totally
decidable yet turing complete subset of machines
despite what you may think: you can't produce an /undecidable input/
to a paradox decider ... any attempt to do so would still be /
undecidable input/
B) extending TMs with reflection (which i haven't defined to you
yet), in order to morph the problem of 'undecidable input' to one of
lying about context
that assumption is was first made (afaik) on turing's first paper on >>>>> computable numbers, and i'm questioning it
No, that assumption in not made in any proof of the halting
theorem. No
one assumes that other "general algos for decision making" don't exist. >>>> The theorems assume some basic axioms about sets, define a halting
decider TM and then show that no TM behaves like such a thing. It
sounds more like to reject the definition of a halting decider rather
than any assumptions the proofs make.
great, that's nice
while everyone else treats it like this means a general aglo doesn't
exist. heck even wikipedia phrases it like that:
/The halting problem is undecidable, meaning that no general
algorithm exists that solves the halting problem for all possible
program–input pairs/
it's never been proven, u just keep assuming the equivocation in
conversation and are unable to talk without it's assumption
What is the "it" that has never been proven? Assumptions are
assumed, not proven, but the assumption you gave: "the assumption that >>>> disproving the classic decider interface implies the non-existence of
general algos for decision making" is never made in any proof.
it's only the general philosophy that's blossomed from those proofs
and if ur gunna try to wash ur hands of that shit just cause it
wasn't phrased that way specifically in a proof, then i'm just gunna
call u a turd shirking responsibility
If we could just get your acceptance of at least one proof out of the >>>>>i'm questioning the fundimentals of computing as far as turing's first >>>>> paper on computable numbers. are turing machine's even the most
correct
model to be using? idk
But you don't seem to have studied any of the proofs and you won't say >>>> if you accept any of them as logical conclusions that follow from their >>>> premises. You seem determined to avoid this question. Is it because >>>> you have not studied any of the proofs in detail?
unless have a proof that is directly related to the kinds of
resolutions i'm actively pursuing, i don't see that in my limited
time atm, i'm avaiable to venture down random-ass red herrings that
may or may not be meaningful, from someone who has yet to demonstrate
any depth of interest in what i'm trying to express
heck just getting the concept of /undecidable input/ across, that ur
still acting butthurt about, was hard enough disillusion me about the
capabilities of whoever the fuck i'm really talking to right now
we've never proven the ct-thesis, i don't even believe that's true.
The Church-Turing thesis is clearly not the kind of thing that is
provable. That's why it's not called a conjecture. But if you have in >>>> mind a model of computation that can't be simulated by TMs but could
there's a lot we'd have a to discuss before i could bring why i think
that might be true ...
pass the bar of being considered "effective" then you must stop posting >>>> here and publish right away. A belief, however, is not worth much
without a concrete model.
[...]
u haven't inspired any further evolution/innovation in my
arguments, it's
been a waste of time for me so far
You might want to bear that in mind before clicking "reply...".
i don't fault myself when others say nothing particularly inspiring
https://youtu.be/yc-blFymxSY?list=RD51yVxGvsqiE
bruh, move on up
https://www.youtube.com/watch?v=01JsFSeFC3U
On 2/20/2026 10:48 PM, dart200 wrote:
On 2/20/26 9:59 PM, Chris M. Thomasson wrote:
On 2/19/2026 12:39 AM, dart200 wrote:
On 2/18/26 6:17 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and which >>>>>>>>> parts arei don't even entirely,i accept: there is no turing machine that implements a >>>>>>>>>>>> classic deciderThere's some private made up terms there, but I don't think >>>>>>>>>>> it matters.
interface, so a machine that is undecidable to it does not >>>>>>>>>>>> exist, as
constructing such a machine would require a machine that >>>>>>>>>>>> implements the
classic decider interface
You accept the classical halting theorem. Your odd use of terms >>>>>>>>>>
you having trouble with?
there is nothing proving the false equivocation of interfaces with >>>>>>>> algorithms, that's just an unjustified assumption that's been >>>>>>>> continually
made
studied and
you won't state clearly if you accept them. Remember, a proof is an >>>>>>> argument that the conclusion is logically entailed by he
assumptions,
so you can accept a proof whose assumptions you reject.
i reject the assumption that disproving the classic decider interface >>>>>> implies the non-existence of general algos for decision making.
Of course. I don't know why you think you need to say this. It's >>>>> obvious that if you don't stick to the classic decider interface there >>>>> will be algorithms for making related decisions. None can make the >>>>> hard
yes/no decision required by the classic halting problem, but no one
doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
und = () -> if (halts(und)) loop()
it doesn't sit right me to claim we cannot algorithmically determine
what this does, when we clearly know that if halts(und)->TRUE then
und() loops forever, but if halts(und)->FALSE then und() halts.
like ur using that argument to then claim we cannot algorithmically
determine what this does ... *right after doing that very type of
algorithmic analysis on both possible execution paths*
how do we even do that if such analysis is not possible?
the general
algo exists and can be formed into a variety of other interfaces that >>>>>> specify how to /undecidable input/ ought to be handled
Provided you don't claim to be able to detect, algorithmically, all
those infinity of inputs that encode machines that behave like the
examples you have called (incorrectly) "undecidable inputs" then of
course you are right.
(calling me "incorrect" about a label i came up with for a
relationship that has yet to truly be studied at depth, is just
hubris 🤷)
Every year this would come up in class. Just classify the "tricky
inputs" as something else and you are home and dry! I leave it as an >>>>> exercise to other readers to see why this is (a) impossible and (b)
pointless.
there are two methods of exploration i've been pursuing, that none
of ur students have ever suggested:
A) filtering the paradoxical machines using partial recognizers, to
produce a turing complete subset of machines that is totally decidable >>>>
for every paradoxical machines u can craft: there is a non-
paradoxical machine that computes the same function. if can one can
detect that a paradox exists within a machine for a particular
classifier (like a functional eq classifier), then we can safely
ignore the machine as redundant, adding any that don't into a
totally decidable yet turing complete subset of machines
despite what you may think: you can't produce an /undecidable input/
to a paradox decider ... any attempt to do so would still be /
undecidable input/
B) extending TMs with reflection (which i haven't defined to you
yet), in order to morph the problem of 'undecidable input' to one of
lying about context
that assumption is was first made (afaik) on turing's first paper on >>>>>> computable numbers, and i'm questioning it
No, that assumption in not made in any proof of the halting
theorem. No
one assumes that other "general algos for decision making" don't
exist.
The theorems assume some basic axioms about sets, define a halting
decider TM and then show that no TM behaves like such a thing. It
sounds more like to reject the definition of a halting decider rather >>>>> than any assumptions the proofs make.
great, that's nice
while everyone else treats it like this means a general aglo doesn't
exist. heck even wikipedia phrases it like that:
/The halting problem is undecidable, meaning that no general
algorithm exists that solves the halting problem for all possible
program–input pairs/
it's never been proven, u just keep assuming the equivocation in
conversation and are unable to talk without it's assumption
What is the "it" that has never been proven? Assumptions are
assumed, not proven, but the assumption you gave: "the assumption that >>>>> disproving the classic decider interface implies the non-existence of >>>>> general algos for decision making" is never made in any proof.
it's only the general philosophy that's blossomed from those proofs
and if ur gunna try to wash ur hands of that shit just cause it
wasn't phrased that way specifically in a proof, then i'm just gunna
call u a turd shirking responsibility
If we could just get your acceptance of at least one proof out of >>>>>>> the
i'm questioning the fundimentals of computing as far as turing's
first
paper on computable numbers. are turing machine's even the most
correct
model to be using? idk
But you don't seem to have studied any of the proofs and you won't say >>>>> if you accept any of them as logical conclusions that follow from
their
premises. You seem determined to avoid this question. Is it because >>>>> you have not studied any of the proofs in detail?
unless have a proof that is directly related to the kinds of
resolutions i'm actively pursuing, i don't see that in my limited
time atm, i'm avaiable to venture down random-ass red herrings that
may or may not be meaningful, from someone who has yet to
demonstrate any depth of interest in what i'm trying to express
heck just getting the concept of /undecidable input/ across, that ur
still acting butthurt about, was hard enough disillusion me about
the capabilities of whoever the fuck i'm really talking to right now
we've never proven the ct-thesis, i don't even believe that's true. >>>>>The Church-Turing thesis is clearly not the kind of thing that is
provable. That's why it's not called a conjecture. But if you
have in
mind a model of computation that can't be simulated by TMs but could
there's a lot we'd have a to discuss before i could bring why i
think that might be true ...
pass the bar of being considered "effective" then you must stop
posting
here and publish right away. A belief, however, is not worth much
without a concrete model.
[...]
u haven't inspired any further evolution/innovation in my
arguments, it's
been a waste of time for me so far
You might want to bear that in mind before clicking "reply...".
i don't fault myself when others say nothing particularly inspiring
https://youtu.be/yc-blFymxSY?list=RD51yVxGvsqiE
bruh, move on up
https://www.youtube.com/watch?v=01JsFSeFC3U
Excellent! :^D
On 2/20/26 9:59 PM, Chris M. Thomasson wrote:
On 2/19/2026 12:39 AM, dart200 wrote:
On 2/18/26 6:17 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/18/26 5:48 AM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/17/26 4:18 PM, Ben Bacarisse wrote:Talk about equivocation! You won't say what proofs you've studied >>>>>> and
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/16/26 4:44 PM, Ben Bacarisse wrote:[...]
dart200 <user7160@newsgrouper.org.invalid> writes:
Sorry I misunderstood. What proofs have you studied and which >>>>>>>> parts arei don't even entirely,i accept: there is no turing machine that implements aThere's some private made up terms there, but I don't think it >>>>>>>>>> matters.
classic decider
interface, so a machine that is undecidable to it does not >>>>>>>>>>> exist, as
constructing such a machine would require a machine that >>>>>>>>>>> implements the
classic decider interface
You accept the classical halting theorem. Your odd use of terms >>>>>>>>>
you having trouble with?
there is nothing proving the false equivocation of interfaces with >>>>>>> algorithms, that's just an unjustified assumption that's been
continually
made
you won't state clearly if you accept them. Remember, a proof is an >>>>>> argument that the conclusion is logically entailed by he assumptions, >>>>>> so you can accept a proof whose assumptions you reject.
i reject the assumption that disproving the classic decider interface >>>>> implies the non-existence of general algos for decision making.
Of course. I don't know why you think you need to say this. It's
obvious that if you don't stick to the classic decider interface there >>>> will be algorithms for making related decisions. None can make the
hard
yes/no decision required by the classic halting problem, but no one
doubts that other classifications can be made.
I remain curious, though, why you won't answer the question.
consider und():
und = () -> if (halts(und)) loop()
it doesn't sit right me to claim we cannot algorithmically determine
what this does, when we clearly know that if halts(und)->TRUE then
und() loops forever, but if halts(und)->FALSE then und() halts.
like ur using that argument to then claim we cannot algorithmically
determine what this does ... *right after doing that very type of
algorithmic analysis on both possible execution paths*
how do we even do that if such analysis is not possible?
the general
algo exists and can be formed into a variety of other interfaces that >>>>> specify how to /undecidable input/ ought to be handled
Provided you don't claim to be able to detect, algorithmically, all
those infinity of inputs that encode machines that behave like the
examples you have called (incorrectly) "undecidable inputs" then of
course you are right.
(calling me "incorrect" about a label i came up with for a
relationship that has yet to truly be studied at depth, is just
hubris 🤷)
Every year this would come up in class. Just classify the "tricky
inputs" as something else and you are home and dry! I leave it as an >>>> exercise to other readers to see why this is (a) impossible and (b)
pointless.
there are two methods of exploration i've been pursuing, that none of
ur students have ever suggested:
A) filtering the paradoxical machines using partial recognizers, to
produce a turing complete subset of machines that is totally decidable
for every paradoxical machines u can craft: there is a non-
paradoxical machine that computes the same function. if can one can
detect that a paradox exists within a machine for a particular
classifier (like a functional eq classifier), then we can safely
ignore the machine as redundant, adding any that don't into a totally
decidable yet turing complete subset of machines
despite what you may think: you can't produce an /undecidable input/
to a paradox decider ... any attempt to do so would still be /
undecidable input/
B) extending TMs with reflection (which i haven't defined to you
yet), in order to morph the problem of 'undecidable input' to one of
lying about context
that assumption is was first made (afaik) on turing's first paper on >>>>> computable numbers, and i'm questioning it
No, that assumption in not made in any proof of the halting
theorem. No
one assumes that other "general algos for decision making" don't exist. >>>> The theorems assume some basic axioms about sets, define a halting
decider TM and then show that no TM behaves like such a thing. It
sounds more like to reject the definition of a halting decider rather
than any assumptions the proofs make.
great, that's nice
while everyone else treats it like this means a general aglo doesn't
exist. heck even wikipedia phrases it like that:
/The halting problem is undecidable, meaning that no general
algorithm exists that solves the halting problem for all possible
program–input pairs/
it's never been proven, u just keep assuming the equivocation in
conversation and are unable to talk without it's assumption
What is the "it" that has never been proven? Assumptions are
assumed, not proven, but the assumption you gave: "the assumption that >>>> disproving the classic decider interface implies the non-existence of
general algos for decision making" is never made in any proof.
it's only the general philosophy that's blossomed from those proofs
and if ur gunna try to wash ur hands of that shit just cause it
wasn't phrased that way specifically in a proof, then i'm just gunna
call u a turd shirking responsibility
If we could just get your acceptance of at least one proof out of the >>>>>i'm questioning the fundimentals of computing as far as turing's first >>>>> paper on computable numbers. are turing machine's even the most
correct
model to be using? idk
But you don't seem to have studied any of the proofs and you won't say >>>> if you accept any of them as logical conclusions that follow from their >>>> premises. You seem determined to avoid this question. Is it because >>>> you have not studied any of the proofs in detail?
unless have a proof that is directly related to the kinds of
resolutions i'm actively pursuing, i don't see that in my limited
time atm, i'm avaiable to venture down random-ass red herrings that
may or may not be meaningful, from someone who has yet to demonstrate
any depth of interest in what i'm trying to express
heck just getting the concept of /undecidable input/ across, that ur
still acting butthurt about, was hard enough disillusion me about the
capabilities of whoever the fuck i'm really talking to right now
we've never proven the ct-thesis, i don't even believe that's true.
The Church-Turing thesis is clearly not the kind of thing that is
provable. That's why it's not called a conjecture. But if you have in >>>> mind a model of computation that can't be simulated by TMs but could
there's a lot we'd have a to discuss before i could bring why i think
that might be true ...
pass the bar of being considered "effective" then you must stop posting >>>> here and publish right away. A belief, however, is not worth much
without a concrete model.
[...]
u haven't inspired any further evolution/innovation in my
arguments, it's
been a waste of time for me so far
You might want to bear that in mind before clicking "reply...".
i don't fault myself when others say nothing particularly inspiring
https://youtu.be/yc-blFymxSY?list=RD51yVxGvsqiE
bruh, move on up
https://www.youtube.com/watch?v=01JsFSeFC3U
my proposal starts with the reminder that *no* machine computes a unique function. for every function that is computed, there is a whole[...]
(infinite) class of machines that are functionally equivalent (same
input -> same output behavior).
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to ever be
any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to ever
be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you are,
that is the Halting Problem and / or the concept of problems being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you see, perhaps because you just don't understand what you are talking about.
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to ever
be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you are,
that is the Halting Problem and / or the concept of problems being
undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you
see, perhaps because you just don't understand what you are talking
about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to
ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you are,
that is the Halting Problem and / or the concept of problems being
undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you
see, perhaps because you just don't understand what you are talking
about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
it's just not a problem that can even effect u really
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to ever
be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you are,
that is the Halting Problem and / or the concept of problems being
undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you
see, perhaps because you just don't understand what you are talking
about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to
ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you
are, that is the Halting Problem and / or the concept of problems
being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you
see, perhaps because you just don't understand what you are talking
about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to really care
much about getting bashed
lest they would begin to understand what all the intentionally willful ignorance they throw at me is like
it's just not a problem that can even effect u really
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to
ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you are,
that is the Halting Problem and / or the concept of problems being
undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you
see, perhaps because you just don't understand what you are talking
about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem actually is.
The only problem so far that you have mentioned is that dumb people, who shouldn't really be in charge of coding anyway, make bad decisions
because they assume they can't get some information out of looking at a system.
If anything, that is a GOOD thing, get the dumb programmers away from--
the dangerous tools.
Of course, this just shows your own problems.
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to
ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you
are, that is the Halting Problem and / or the concept of problems
being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you >>>>> see, perhaps because you just don't understand what you are talking >>>>> about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to really
care much about getting bashed
lest they would begin to understand what all the intentionally willful
ignorance they throw at me is like
No, it seems that the problem is you don't understand that the first
step of dealing with a problem is to first understand the real nature of
the problem and see what people have actually done.
Expecting people to just hand you that information means you never
learned how to learn, and thus made your self fundamentally ignorant.
it's just not a problem that can even effect u really
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to
ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you
are, that is the Halting Problem and / or the concept of problems
being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you
see, perhaps because you just don't understand what you are talking
about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem actually is.
i have, u denied it with economic brainrot
The only problem so far that you have mentioned is that dumb people,
who shouldn't really be in charge of coding anyway, make bad decisions
because they assume they can't get some information out of looking at
a system.
they assume???
the fucking academics tell them they can't, and then when u ask them
more specifically they dance around the terms like the fucking abject retards they are
i have hadn't a singe honest conversation after spend months in this group
ya'll are fucking decrepit as shit
If anything, that is a GOOD thing, get the dumb programmers away from
the dangerous tools.
Of course, this just shows your own problems.
On 2/21/26 7:28 PM, dart200 wrote:
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to
ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you
are, that is the Halting Problem and / or the concept of problems
being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you >>>>> see, perhaps because you just don't understand what you are talking >>>>> about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem actually is.
i have, u denied it with economic brainrot
The only problem so far that you have mentioned is that dumb people,
who shouldn't really be in charge of coding anyway, make bad
decisions because they assume they can't get some information out of
looking at a system.
they assume???
the fucking academics tell them they can't, and then when u ask them
more specifically they dance around the terms like the fucking abject
retards they are
i have hadn't a singe honest conversation after spend months in this
group
ya'll are fucking decrepit as shit
If anything, that is a GOOD thing, get the dumb programmers away from
the dangerous tools.
i'm sorry are you arguing good tools like semantic verification is a *dangerous* tool???
like holy fuck this is LLM grade stupidity,
what in the fuck, how in the fuck can anyone argue that correctness verification is to be dangerous???
like what???
Of course, this just shows your own problems.
CAN I HAVE EVEN AN OUNCE OF FUCKING RATIONALITY OUT OF U DICK???
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to >>>>>>> ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you
are, that is the Halting Problem and / or the concept of problems >>>>>> being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem
you see, perhaps because you just don't understand what you are
talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to really
care much about getting bashed
lest they would begin to understand what all the intentionally
willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the first
step of dealing with a problem is to first understand the real nature
of the problem and see what people have actually done.
u don't even understand what a basic enumeration is ur so dumb as fuck
Expecting people to just hand you that information means you never
learned how to learn, and thus made your self fundamentally ignorant.
it's just not a problem that can even effect u really
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to
ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you
are, that is the Halting Problem and / or the concept of problems
being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you
see, perhaps because you just don't understand what you are talking
about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem actually is.
i have, u denied it with economic brainrot
The only problem so far that you have mentioned is that dumb people,
who shouldn't really be in charge of coding anyway, make bad decisions
because they assume they can't get some information out of looking at
a system.
they assume???
the fucking academics tell them they can't, and then when u ask them
more specifically they dance around the terms like the fucking abject retards they are
i have hadn't a singe honest conversation after spend months in this group
ya'll are fucking decrepit as shit
If anything, that is a GOOD thing, get the dumb programmers away from
the dangerous tools.
Of course, this just shows your own problems.
On 2/22/26 12:13 AM, dart200 wrote:
On 2/21/26 7:28 PM, dart200 wrote:
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to >>>>>>> ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you
are, that is the Halting Problem and / or the concept of problems >>>>>> being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem
you see, perhaps because you just don't understand what you are
talking about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem actually
is.
i have, u denied it with economic brainrot
The only problem so far that you have mentioned is that dumb people,
who shouldn't really be in charge of coding anyway, make bad
decisions because they assume they can't get some information out of
looking at a system.
they assume???
the fucking academics tell them they can't, and then when u ask them
more specifically they dance around the terms like the fucking abject
retards they are
i have hadn't a singe honest conversation after spend months in this
group
ya'll are fucking decrepit as shit
If anything, that is a GOOD thing, get the dumb programmers away
from the dangerous tools.
i'm sorry are you arguing good tools like semantic verification is a
*dangerous* tool???
No, I mean things like compliers to generate code that goes into
critical systems.
like holy fuck this is LLM grade stupidity,
what in the fuck, how in the fuck can anyone argue that correctness
verification is to be dangerous???
I'm not. What is dangerous is to give up on verification because it
can't be perfect.
As the saying goes, The quest for Perfection is the enemy of the Good Enough.--
like what???
Of course, this just shows your own problems.
CAN I HAVE EVEN AN OUNCE OF FUCKING RATIONALITY OUT OF U DICK???
Since it seems you don't know what I am talking about, maybe not, since
it goes right over your head.
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to >>>>>>>> ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you >>>>>>> are, that is the Halting Problem and / or the concept of problems >>>>>>> being undecidable, then I guess you are just a pathological liar. >>>>>>>
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem >>>>>>> you see, perhaps because you just don't understand what you are >>>>>>> talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to really
care much about getting bashed
lest they would begin to understand what all the intentionally
willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the first
step of dealing with a problem is to first understand the real nature
of the problem and see what people have actually done.
u don't even understand what a basic enumeration is ur so dumb as fuck
Sure I do. You don't understand when your enumeration needs to be
effective.
Expecting people to just hand you that information means you never
learned how to learn, and thus made your self fundamentally ignorant.
it's just not a problem that can even effect u really
On 2/21/26 10:28 PM, dart200 wrote:
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to
ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you
are, that is the Halting Problem and / or the concept of problems
being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem you >>>>> see, perhaps because you just don't understand what you are talking >>>>> about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem actually is.
i have, u denied it with economic brainrot
No, you can't, as your explaimation is based on made-up definitions for words that you can't actually define.
The only problem so far that you have mentioned is that dumb people,
who shouldn't really be in charge of coding anyway, make bad
decisions because they assume they can't get some information out of
looking at a system.
they assume???
the fucking academics tell them they can't, and then when u ask them
more specifically they dance around the terms like the fucking abject
retards they are
No, they don't. I guess your problem is you listen to liars and not the academics.
The "academics" say we can't solve the whole problem, but show how we
can get the answer for many.
The IDIOTS, like you, that can't tell the difference between handling
ALL problems, vs possibly being able to handle the one in front of you,
is the problem.
You are stuck in the centery old error of running in circles looking for
the golden ticket to solve everything, rather than seeing how good you
can actually get.
i have hadn't a singe honest conversation after spend months in this
group
ya'll are fucking decrepit as shit
The "honesty" problem is on your end.
You just don't know what you are talking about.
If anything, that is a GOOD thing, get the dumb programmers away from
the dangerous tools.
Of course, this just shows your own problems.
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters >>>>>>>>> to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you >>>>>>>> are, that is the Halting Problem and / or the concept of
problems being undecidable, then I guess you are just a
pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem >>>>>>>> you see, perhaps because you just don't understand what you are >>>>>>>> talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to really
care much about getting bashed
lest they would begin to understand what all the intentionally
willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the first
step of dealing with a problem is to first understand the real
nature of the problem and see what people have actually done.
u don't even understand what a basic enumeration is ur so dumb as fuck
Sure I do. You don't understand when your enumeration needs to be
effective.
an effective enumeration of all turing machines was proven on turing's original paper and can be reused anywhere...
what in the fuck are you going on about???
Expecting people to just hand you that information means you never
learned how to learn, and thus made your self fundamentally ignorant.
it's just not a problem that can even effect u really
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters >>>>>>>>>> to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say >>>>>>>>> you are, that is the Halting Problem and / or the concept of >>>>>>>>> problems being undecidable, then I guess you are just a
pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem >>>>>>>>> you see, perhaps because you just don't understand what you are >>>>>>>>> talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to really >>>>>> care much about getting bashed
lest they would begin to understand what all the intentionally
willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the
first step of dealing with a problem is to first understand the
real nature of the problem and see what people have actually done.
u don't even understand what a basic enumeration is ur so dumb as fuck
Sure I do. You don't understand when your enumeration needs to be
effective.
an effective enumeration of all turing machines was proven on turing's
original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think
there is a largest natural number... ;^) You really need to implement
your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you never
learned how to learn, and thus made your self fundamentally ignorant. >>>>>
it's just not a problem that can even effect u really
On 2/22/2026 9:06 AM, dart200 wrote:
[...]
Actually, you are not worth conversing with. Olcott 2.0 (special
version). Whatever.
Plonk.
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:Sure I do. You don't understand when your enumeration needs to be
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters >>>>>>>>>>> to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say >>>>>>>>>> you are, that is the Halting Problem and / or the concept of >>>>>>>>>> problems being undecidable, then I guess you are just a
pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the
problem you see, perhaps because you just don't understand >>>>>>>>>> what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to
really care much about getting bashed
lest they would begin to understand what all the intentionally
willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the
first step of dealing with a problem is to first understand the
real nature of the problem and see what people have actually done.
u don't even understand what a basic enumeration is ur so dumb as fuck >>>>
effective.
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think
yes that's what diagonal proofs do...
there is a largest natural number... ;^) You really need to implement
your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you never >>>>>> learned how to learn, and thus made your self fundamentally ignorant. >>>>>>
it's just not a problem that can even effect u really
On 2/22/26 4:03 AM, Richard Damon wrote:
On 2/22/26 12:13 AM, dart200 wrote:
On 2/21/26 7:28 PM, dart200 wrote:
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to >>>>>>>> ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you >>>>>>> are, that is the Halting Problem and / or the concept of problems >>>>>>> being undecidable, then I guess you are just a pathological liar. >>>>>>>
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem >>>>>>> you see, perhaps because you just don't understand what you are >>>>>>> talking about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem
actually is.
i have, u denied it with economic brainrot
The only problem so far that you have mentioned is that dumb
people, who shouldn't really be in charge of coding anyway, make
bad decisions because they assume they can't get some information
out of looking at a system.
they assume???
the fucking academics tell them they can't, and then when u ask them
more specifically they dance around the terms like the fucking
abject retards they are
i have hadn't a singe honest conversation after spend months in this
group
ya'll are fucking decrepit as shit
If anything, that is a GOOD thing, get the dumb programmers away
from the dangerous tools.
i'm sorry are you arguing good tools like semantic verification is a
*dangerous* tool???
No, I mean things like compliers to generate code that goes into
critical systems.
well they're all spitting out AI generated slop these day, and no
general semantic verification tools are in sight...
like holy fuck this is LLM grade stupidity,
what in the fuck, how in the fuck can anyone argue that correctness
verification is to be dangerous???
I'm not. What is dangerous is to give up on verification because it
can't be perfect.
what??? THAT'S WHAT WE DID...
As the saying goes, The quest for Perfection is the enemy of the Good
Enough.
like what???
Of course, this just shows your own problems.
CAN I HAVE EVEN AN OUNCE OF FUCKING RATIONALITY OUT OF U DICK???
Since it seems you don't know what I am talking about, maybe not,
since it goes right over your head.
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:28 PM, dart200 wrote:
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to >>>>>>> ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you
are, that is the Halting Problem and / or the concept of problems >>>>>> being undecidable, then I guess you are just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem
you see, perhaps because you just don't understand what you are
talking about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem actually
is.
i have, u denied it with economic brainrot
No, you can't, as your explaimation is based on made-up definitions
for words that you can't actually define.
The only problem so far that you have mentioned is that dumb people,
who shouldn't really be in charge of coding anyway, make bad
decisions because they assume they can't get some information out of
looking at a system.
they assume???
the fucking academics tell them they can't, and then when u ask them
more specifically they dance around the terms like the fucking abject
retards they are
No, they don't. I guess your problem is you listen to liars and not
the academics.
The "academics" say we can't solve the whole problem, but show how we
can get the answer for many.
fucking words on the the fucking screen don't matter when the fucking actions are that we *DON'T DO SEMANTIC VERIFICATION IN A GENERAL MANNER*
u stupid cunt
ur just as fucking delusional as the retarded lame duck academics
infesting cs
The IDIOTS, like you, that can't tell the difference between handling
ALL problems, vs possibly being able to handle the one in front of
you, is the problem.
You are stuck in the centery old error of running in circles looking
for the golden ticket to solve everything, rather than seeing how good
you can actually get.
i have hadn't a singe honest conversation after spend months in this
group
ya'll are fucking decrepit as shit
The "honesty" problem is on your end.
You just don't know what you are talking about.
If anything, that is a GOOD thing, get the dumb programmers away
from the dangerous tools.
Of course, this just shows your own problems.
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:u don't even understand what a basic enumeration is ur so dumb as >>>>>> fuck
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer
shitposters to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say >>>>>>>>>>> you are, that is the Halting Problem and / or the concept of >>>>>>>>>>> problems being undecidable, then I guess you are just a >>>>>>>>>>> pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the >>>>>>>>>>> problem you see, perhaps because you just don't understand >>>>>>>>>>> what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to
really care much about getting bashed
lest they would begin to understand what all the intentionally >>>>>>>> willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the
first step of dealing with a problem is to first understand the >>>>>>> real nature of the problem and see what people have actually done. >>>>>>
Sure I do. You don't understand when your enumeration needs to be
effective.
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list of
numbers enumerated by the list of machines.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of machines,
just that an enumeration exists.
And note, it isn't even an enumeration of all Turing Machines, but of
all machines that compute a number, otherwise we need to deal with the possibility that a given row isn't "complete" because that machine never generates enough numbers.
there is a largest natural number... ;^) You really need to implement
your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you
never learned how to learn, and thus made your self fundamentally >>>>>>> ignorant.
it's just not a problem that can even effect u really
On 2/22/26 12:02 PM, dart200 wrote:
On 2/22/26 4:03 AM, Richard Damon wrote:
On 2/22/26 12:13 AM, dart200 wrote:
On 2/21/26 7:28 PM, dart200 wrote:
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters >>>>>>>>> to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you >>>>>>>> are, that is the Halting Problem and / or the concept of
problems being undecidable, then I guess you are just a
pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem >>>>>>>> you see, perhaps because you just don't understand what you are >>>>>>>> talking about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem
actually is.
i have, u denied it with economic brainrot
The only problem so far that you have mentioned is that dumb
people, who shouldn't really be in charge of coding anyway, make
bad decisions because they assume they can't get some information >>>>>> out of looking at a system.
they assume???
the fucking academics tell them they can't, and then when u ask
them more specifically they dance around the terms like the fucking >>>>> abject retards they are
i have hadn't a singe honest conversation after spend months in
this group
ya'll are fucking decrepit as shit
If anything, that is a GOOD thing, get the dumb programmers away
from the dangerous tools.
i'm sorry are you arguing good tools like semantic verification is a
*dangerous* tool???
No, I mean things like compliers to generate code that goes into
critical systems.
well they're all spitting out AI generated slop these day, and no
general semantic verification tools are in sight...
And THAT is the dangerous stuff.
It seems your arguement is that people shouldn't have such tools.
like holy fuck this is LLM grade stupidity,
what in the fuck, how in the fuck can anyone argue that correctness
verification is to be dangerous???
I'm not. What is dangerous is to give up on verification because it
can't be perfect.
what??? THAT'S WHAT WE DID...
Only the stupid ones.
I guess you are just showing that you fell for the stupid lie created by
the stupid programmers.
Good programmers still present a good argument on why their code should
do what it is supposed to do during the code review.
I guess you are so ignorant you believe that you can just skip all of
that stuff.
As the saying goes, The quest for Perfection is the enemy of the Good
Enough.
like what???
Of course, this just shows your own problems.
CAN I HAVE EVEN AN OUNCE OF FUCKING RATIONALITY OUT OF U DICK???
Since it seems you don't know what I am talking about, maybe not,
since it goes right over your head.
On 2/22/26 12:06 PM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:28 PM, dart200 wrote:
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to >>>>>>>> ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you >>>>>>> are, that is the Halting Problem and / or the concept of problems >>>>>>> being undecidable, then I guess you are just a pathological liar. >>>>>>>
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem >>>>>>> you see, perhaps because you just don't understand what you are >>>>>>> talking about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem
actually is.
i have, u denied it with economic brainrot
No, you can't, as your explaimation is based on made-up definitions
for words that you can't actually define.
The only problem so far that you have mentioned is that dumb
people, who shouldn't really be in charge of coding anyway, make
bad decisions because they assume they can't get some information
out of looking at a system.
they assume???
the fucking academics tell them they can't, and then when u ask them
more specifically they dance around the terms like the fucking
abject retards they are
No, they don't. I guess your problem is you listen to liars and not
the academics.
The "academics" say we can't solve the whole problem, but show how we
can get the answer for many.
fucking words on the the fucking screen don't matter when the fucking
actions are that we *DON'T DO SEMANTIC VERIFICATION IN A GENERAL
MANNER* u stupid cunt
Because we CAN'T.
ur just as fucking delusional as the retarded lame duck academics
infesting cs
Nope, it seems you are the stupid one putting forward that the only
solution is the completely general solution, and we can't do anything
right until we can do it your way, so we can never do anything right.
Smarter people understand the limitiations of formal proof, and reserve
it for cases where it really is that important, but works to make things "good enough".
The IDIOTS, like you, that can't tell the difference between handling
ALL problems, vs possibly being able to handle the one in front of
you, is the problem.
You are stuck in the centery old error of running in circles looking
for the golden ticket to solve everything, rather than seeing how
good you can actually get.
i have hadn't a singe honest conversation after spend months in this
group
ya'll are fucking decrepit as shit
The "honesty" problem is on your end.
You just don't know what you are talking about.
If anything, that is a GOOD thing, get the dumb programmers away
from the dangerous tools.
Of course, this just shows your own problems.
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 12:06 PM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:28 PM, dart200 wrote:
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters >>>>>>>>> to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you >>>>>>>> are, that is the Halting Problem and / or the concept of
problems being undecidable, then I guess you are just a
pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem >>>>>>>> you see, perhaps because you just don't understand what you are >>>>>>>> talking about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem
actually is.
i have, u denied it with economic brainrot
No, you can't, as your explaimation is based on made-up definitions
for words that you can't actually define.
The only problem so far that you have mentioned is that dumb
people, who shouldn't really be in charge of coding anyway, make
bad decisions because they assume they can't get some information >>>>>> out of looking at a system.
they assume???
the fucking academics tell them they can't, and then when u ask
them more specifically they dance around the terms like the fucking >>>>> abject retards they are
No, they don't. I guess your problem is you listen to liars and not
the academics.
The "academics" say we can't solve the whole problem, but show how
we can get the answer for many.
fucking words on the the fucking screen don't matter when the fucking
actions are that we *DON'T DO SEMANTIC VERIFICATION IN A GENERAL
MANNER* u stupid cunt
Because we CAN'T.
i'm pulling my hairs out.
cause u have ur theorycucks like ben going like "that's not what the
proofs literally says, quote it" with smug little air of superiority as
if their technical semantic bamboozling means fuck all in real world applications ... so he's in complete ignorance that the engineercucks
are all going like "yah we can't do that cause that's what theory says"...
FUCK ALL OF YOU BONEHEADED OVERHYPED APES
ur just as fucking delusional as the retarded lame duck academics
infesting cs
Nope, it seems you are the stupid one putting forward that the only
solution is the completely general solution, and we can't do anything
right until we can do it your way, so we can never do anything right.
IT'S BEEN A CENTURY AND IT'S STILL A TOTAL SHITSHOW. U DON'T EXPECT TO
SEE SEMANTIC PROOFS BEFORE YOU DIE. HECK, AT THE RATE THINGS ARE GOING I DON'T EXPECT TO SEE BASIC SEMANTIC PROOFS HAPPEN BEFORE I DIE.
Smarter people understand the limitiations of formal proof, and
reserve it for cases where it really is that important, but works to
make things "good enough".
fuck you bro. THEORY NEEDS ABSOLUTE CORRECTNESS TO FUNCTION AT IT'S BEST
The IDIOTS, like you, that can't tell the difference between
handling ALL problems, vs possibly being able to handle the one in
front of you, is the problem.
You are stuck in the centery old error of running in circles looking
for the golden ticket to solve everything, rather than seeing how
good you can actually get.
i have hadn't a singe honest conversation after spend months in
this group
ya'll are fucking decrepit as shit
The "honesty" problem is on your end.
You just don't know what you are talking about.
If anything, that is a GOOD thing, get the dumb programmers away
from the dangerous tools.
Of course, this just shows your own problems.
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 12:02 PM, dart200 wrote:
On 2/22/26 4:03 AM, Richard Damon wrote:
On 2/22/26 12:13 AM, dart200 wrote:
On 2/21/26 7:28 PM, dart200 wrote:
On 2/21/26 3:26 PM, Richard Damon wrote:
On 2/21/26 12:41 PM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters >>>>>>>>>> to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say >>>>>>>>> you are, that is the Halting Problem and / or the concept of >>>>>>>>> problems being undecidable, then I guess you are just a
pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the problem >>>>>>>>> you see, perhaps because you just don't understand what you are >>>>>>>>> talking about.
ur old and will be dead in a decade or so
it's just not a problem that can even effect u really
So?
Your problem is you can't actually express what the problem
actually is.
i have, u denied it with economic brainrot
The only problem so far that you have mentioned is that dumb
people, who shouldn't really be in charge of coding anyway, make >>>>>>> bad decisions because they assume they can't get some information >>>>>>> out of looking at a system.
they assume???
the fucking academics tell them they can't, and then when u ask
them more specifically they dance around the terms like the
fucking abject retards they are
i have hadn't a singe honest conversation after spend months in
this group
ya'll are fucking decrepit as shit
If anything, that is a GOOD thing, get the dumb programmers away >>>>>>> from the dangerous tools.
i'm sorry are you arguing good tools like semantic verification is
a *dangerous* tool???
No, I mean things like compliers to generate code that goes into
critical systems.
well they're all spitting out AI generated slop these day, and no
general semantic verification tools are in sight...
And THAT is the dangerous stuff.
It seems your arguement is that people shouldn't have such tools.
AI GENERATION IS NOT EQUIVALENT TO AUTOMATED PROOFS???
WHAT ARE YOU SMOKING???
like holy fuck this is LLM grade stupidity,
what in the fuck, how in the fuck can anyone argue that correctness >>>>> verification is to be dangerous???
I'm not. What is dangerous is to give up on verification because it
can't be perfect.
what??? THAT'S WHAT WE DID...
Only the stupid ones.
I guess you are just showing that you fell for the stupid lie created
by the stupid programmers.
Good programmers still present a good argument on why their code
should do what it is supposed to do during the code review.
THAT'S NOT A COMPUTED A PROOF U MORON. ALL TOOL CHAINS SHOULD HAVE
COMPUTED PROOFS INCLUDED AS A BASIC FUNCTIONALITY CHECK. CODE SHOULDN'T FUCKING DEADLOCK UNEXPECTED, OR EVER ACTUALLY
fucking equivocation fallacy holy fuck
I guess you are so ignorant you believe that you can just skip all of
that stuff.
As the saying goes, The quest for Perfection is the enemy of the
Good Enough.
like what???
Of course, this just shows your own problems.
CAN I HAVE EVEN AN OUNCE OF FUCKING RATIONALITY OUT OF U DICK???
Since it seems you don't know what I am talking about, maybe not,
since it goes right over your head.
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:yes that's what diagonal proofs do...
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:u don't even understand what a basic enumeration is ur so dumb as >>>>>>> fuck
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer
shitposters to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say >>>>>>>>>>>> you are, that is the Halting Problem and / or the concept of >>>>>>>>>>>> problems being undecidable, then I guess you are just a >>>>>>>>>>>> pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the >>>>>>>>>>>> problem you see, perhaps because you just don't understand >>>>>>>>>>>> what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to >>>>>>>>> really care much about getting bashed
lest they would begin to understand what all the intentionally >>>>>>>>> willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the >>>>>>>> first step of dealing with a problem is to first understand the >>>>>>>> real nature of the problem and see what people have actually done. >>>>>>>
Sure I do. You don't understand when your enumeration needs to be >>>>>> effective.
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think >>>
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list of
numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of machines,
just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to question the effectiveness of a total turing machine enumeration
this isn't math anymore, it's just u randomly putting words together
opposed to mine based on feels i can't even remotely understand
honestly i might as well be talking to a freaking LLM programmed to be contrarian
And note, it isn't even an enumeration of all Turing Machines, but of
all machines that compute a number, otherwise we need to deal with the
possibility that a given row isn't "complete" because that machine
never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D to be "satisfactory"/circle-free, so he does enumerate over all machines
testing each one for being "satisfactory" for inclusion on the diagonal
or not
write me psuedocode that accurately represents machine H as described on p247 of turing's paper /on computable numbers/, cause honestly u
understand the algo he wrote. if write more than like 15 lines u've
fucked up, it's not a complex algo
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
there is a largest natural number... ;^) You really need to
implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you
never learned how to learn, and thus made your self
fundamentally ignorant.
it's just not a problem that can even effect u really
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:yes that's what diagonal proofs do...
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:u don't even understand what a basic enumeration is ur so dumb >>>>>>>> as fuck
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you >>>>>>>>>>>>> say you are, that is the Halting Problem and / or the >>>>>>>>>>>>> concept of problems being undecidable, then I guess you are >>>>>>>>>>>>> just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the >>>>>>>>>>>>> problem you see, perhaps because you just don't understand >>>>>>>>>>>>> what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to >>>>>>>>>> really care much about getting bashed
lest they would begin to understand what all the intentionally >>>>>>>>>> willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the >>>>>>>>> first step of dealing with a problem is to first understand the >>>>>>>>> real nature of the problem and see what people have actually done. >>>>>>>>
Sure I do. You don't understand when your enumeration needs to be >>>>>>> effective.
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think >>>>
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list of
numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration isn't of
all machines when he build the diagonal argument, as not all machines produce a valid result to put on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of machines,
just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to question
the effectiveness of a total turing machine enumeration
And you are a moron for confusing the effective enumeration of ALL
machines, many of which don't meet the requirements to be put in the
list, and the enumeration of the machines that DO produce a "computable number"
The method to pair down the enumeration of all machines to the machines
that compute a computable number is shown to non-computable.
You depending on Unicorns to prepare your enumerations is what breaks
your logic.
this isn't math anymore, it's just u randomly putting words together
opposed to mine based on feels i can't even remotely understand
Yes, that is all YOU are doing, as you don't require your statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM programmed to be
contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines, but of
all machines that compute a number, otherwise we need to deal with
the possibility that a given row isn't "complete" because that
machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D to be
"satisfactory"/circle-free, so he does enumerate over all machines
testing each one for being "satisfactory" for inclusion on the
diagonal or not
write me psuedocode that accurately represents machine H as described
on p247 of turing's paper /on computable numbers/, cause honestly u
understand the algo he wrote. if write more than like 15 lines u've
fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, when it
can't be.
Note, the paper starts with a "Supposing there is a Machine D", and if
that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no machine H to construct.
So, you questions is about the anatomy of a Unicorn.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to
implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you >>>>>>>>> never learned how to learn, and thus made your self
fundamentally ignorant.
it's just not a problem that can even effect u really
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:yes that's what diagonal proofs do...
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:u don't even understand what a basic enumeration is ur so dumb >>>>>>>> as fuck
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you >>>>>>>>>>>>> say you are, that is the Halting Problem and / or the >>>>>>>>>>>>> concept of problems being undecidable, then I guess you are >>>>>>>>>>>>> just a pathological liar.
And yes, if that is the case, no one can help you.
It seems your problem is you can't actually express the >>>>>>>>>>>>> problem you see, perhaps because you just don't understand >>>>>>>>>>>>> what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to >>>>>>>>>> really care much about getting bashed
lest they would begin to understand what all the intentionally >>>>>>>>>> willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the >>>>>>>>> first step of dealing with a problem is to first understand the >>>>>>>>> real nature of the problem and see what people have actually done. >>>>>>>>
Sure I do. You don't understand when your enumeration needs to be >>>>>>> effective.
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think >>>>
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list of
numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration isn't of
all machines when he build the diagonal argument, as not all machines produce a valid result to put on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of machines,
just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to question
the effectiveness of a total turing machine enumeration
And you are a moron for confusing the effective enumeration of ALL
machines, many of which don't meet the requirements to be put in the
list, and the enumeration of the machines that DO produce a "computable number"
The method to pair down the enumeration of all machines to the machines
that compute a computable number is shown to non-computable.
You depending on Unicorns to prepare your enumerations is what breaks
your logic.
this isn't math anymore, it's just u randomly putting words together
opposed to mine based on feels i can't even remotely understand
Yes, that is all YOU are doing, as you don't require your statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM programmed to be
contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines, but of
all machines that compute a number, otherwise we need to deal with
the possibility that a given row isn't "complete" because that
machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D to be
"satisfactory"/circle-free, so he does enumerate over all machines
testing each one for being "satisfactory" for inclusion on the
diagonal or not
write me psuedocode that accurately represents machine H as described
on p247 of turing's paper /on computable numbers/, cause honestly u
understand the algo he wrote. if write more than like 15 lines u've
fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, when it
can't be.
Note, the paper starts with a "Supposing there is a Machine D", and if
that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no machine H to construct.
So, you questions is about the anatomy of a Unicorn.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to
implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you >>>>>>>>> never learned how to learn, and thus made your self
fundamentally ignorant.
it's just not a problem that can even effect u really
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you >>>>>>>>>>>>>> say you are, that is the Halting Problem and / or the >>>>>>>>>>>>>> concept of problems being undecidable, then I guess you >>>>>>>>>>>>>> are just a pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>
It seems your problem is you can't actually express the >>>>>>>>>>>>>> problem you see, perhaps because you just don't understand >>>>>>>>>>>>>> what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to >>>>>>>>>>> really care much about getting bashed
lest they would begin to understand what all the
intentionally willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the >>>>>>>>>> first step of dealing with a problem is to first understand >>>>>>>>>> the real nature of the problem and see what people have
actually done.
u don't even understand what a basic enumeration is ur so dumb >>>>>>>>> as fuck
Sure I do. You don't understand when your enumeration needs to >>>>>>>> be effective.
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you
think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list of
numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration isn't
of all machines when he build the diagonal argument, as not all
machines produce a valid result to put on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of
machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to question
the effectiveness of a total turing machine enumeration
And you are a moron for confusing the effective enumeration of ALL
machines, many of which don't meet the requirements to be put in the
list, and the enumeration of the machines that DO produce a
"computable number"
The method to pair down the enumeration of all machines to the
machines that compute a computable number is shown to non-computable.
You depending on Unicorns to prepare your enumerations is what breaks
your logic.
this isn't math anymore, it's just u randomly putting words together
opposed to mine based on feels i can't even remotely understand
Yes, that is all YOU are doing, as you don't require your statments to
actually be based on FACTS.
honestly i might as well be talking to a freaking LLM programmed to
be contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines, but
of all machines that compute a number, otherwise we need to deal
with the possibility that a given row isn't "complete" because that
machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D to
be "satisfactory"/circle-free, so he does enumerate over all machines
testing each one for being "satisfactory" for inclusion on the
diagonal or not
write me psuedocode that accurately represents machine H as described
on p247 of turing's paper /on computable numbers/, cause honestly u
understand the algo he wrote. if write more than like 15 lines u've
fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, when it
can't be.
Note, the paper starts with a "Supposing there is a Machine D", and if
that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no machine H to
construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted moron...
he describes exactly what it does, but put it in a form of psuedo-code
i really am just getting a bunch retarded kneejerk reactions from u
eh??? not even a second of deep thot ever???
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to
implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you >>>>>>>>>> never learned how to learn, and thus made your self
fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you >>>>>>>>>>>>>> say you are, that is the Halting Problem and / or the >>>>>>>>>>>>>> concept of problems being undecidable, then I guess you >>>>>>>>>>>>>> are just a pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>
It seems your problem is you can't actually express the >>>>>>>>>>>>>> problem you see, perhaps because you just don't understand >>>>>>>>>>>>>> what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to >>>>>>>>>>> really care much about getting bashed
lest they would begin to understand what all the
intentionally willful ignorance they throw at me is like
No, it seems that the problem is you don't understand that the >>>>>>>>>> first step of dealing with a problem is to first understand >>>>>>>>>> the real nature of the problem and see what people have
actually done.
u don't even understand what a basic enumeration is ur so dumb >>>>>>>>> as fuck
Sure I do. You don't understand when your enumeration needs to >>>>>>>> be effective.
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you
think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list of
numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration isn't
of all machines when he build the diagonal argument, as not all
machines produce a valid result to put on the diagonal.
fuck you and fuck your alzheimer u useless fking boomer
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of
machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to question
the effectiveness of a total turing machine enumeration
And you are a moron for confusing the effective enumeration of ALL
machines, many of which don't meet the requirements to be put in the
list, and the enumeration of the machines that DO produce a
"computable number"
The method to pair down the enumeration of all machines to the
machines that compute a computable number is shown to non-computable.
You depending on Unicorns to prepare your enumerations is what breaks
your logic.
the sooner all the abject retards on this list are dead due to aging,
the better our planet will be off
this isn't math anymore, it's just u randomly putting words together
opposed to mine based on feels i can't even remotely understand
Yes, that is all YOU are doing, as you don't require your statments to
actually be based on FACTS.
honestly i might as well be talking to a freaking LLM programmed to
be contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines, but
of all machines that compute a number, otherwise we need to deal
with the possibility that a given row isn't "complete" because that
machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D to
be "satisfactory"/circle-free, so he does enumerate over all machines
testing each one for being "satisfactory" for inclusion on the
diagonal or not
write me psuedocode that accurately represents machine H as described
on p247 of turing's paper /on computable numbers/, cause honestly u
understand the algo he wrote. if write more than like 15 lines u've
fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, when it
can't be.
Note, the paper starts with a "Supposing there is a Machine D", and if
that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no machine H to
construct.
So, you questions is about the anatomy of a Unicorn.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to
implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you >>>>>>>>>> never learned how to learn, and thus made your self
fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:No, it seems that the problem is you don't understand that >>>>>>>>>>> the first step of dealing with a problem is to first
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you >>>>>>>>>>>>>>> say you are, that is the Halting Problem and / or the >>>>>>>>>>>>>>> concept of problems being undecidable, then I guess you >>>>>>>>>>>>>>> are just a pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>
It seems your problem is you can't actually express the >>>>>>>>>>>>>>> problem you see, perhaps because you just don't >>>>>>>>>>>>>>> understand what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to >>>>>>>>>>>> really care much about getting bashed
lest they would begin to understand what all the
intentionally willful ignorance they throw at me is like >>>>>>>>>>>
understand the real nature of the problem and see what people >>>>>>>>>>> have actually done.
u don't even understand what a basic enumeration is ur so dumb >>>>>>>>>> as fuck
Sure I do. You don't understand when your enumeration needs to >>>>>>>>> be effective.
an effective enumeration of all turing machines was proven on >>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you >>>>>>> think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list of
numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration isn't
of all machines when he build the diagonal argument, as not all
machines produce a valid result to put on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of
machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to question
the effectiveness of a total turing machine enumeration
And you are a moron for confusing the effective enumeration of ALL
machines, many of which don't meet the requirements to be put in the
list, and the enumeration of the machines that DO produce a
"computable number"
The method to pair down the enumeration of all machines to the
machines that compute a computable number is shown to non-computable.
You depending on Unicorns to prepare your enumerations is what breaks
your logic.
this isn't math anymore, it's just u randomly putting words together
opposed to mine based on feels i can't even remotely understand
Yes, that is all YOU are doing, as you don't require your statments
to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM programmed to
be contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines, but
of all machines that compute a number, otherwise we need to deal
with the possibility that a given row isn't "complete" because that >>>>> machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D to
be "satisfactory"/circle-free, so he does enumerate over all
machines testing each one for being "satisfactory" for inclusion on
the diagonal or not
write me psuedocode that accurately represents machine H as
described on p247 of turing's paper /on computable numbers/, cause
honestly u understand the algo he wrote. if write more than like 15
lines u've fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, when it
can't be.
Note, the paper starts with a "Supposing there is a Machine D", and
if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no machine H
to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted
moron... he describes exactly what it does, but put it in a form of
psuedo-code
Why, are you too stupid to do it?
Since each iteration just uses a testing by the machine D, your psuedo-
code loop is just a reference to a operation that turns out not to exist.
i really am just getting a bunch retarded kneejerk reactions from u
eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
That is why you are so stupid, you seem to expect others to do the work
you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to
implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you >>>>>>>>>>> never learned how to learn, and thus made your self
fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>>
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:No, it seems that the problem is you don't understand that >>>>>>>>>>>> the first step of dealing with a problem is to first
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you >>>>>>>>>>>>>>>> say you are, that is the Halting Problem and / or the >>>>>>>>>>>>>>>> concept of problems being undecidable, then I guess you >>>>>>>>>>>>>>>> are just a pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>
It seems your problem is you can't actually express the >>>>>>>>>>>>>>>> problem you see, perhaps because you just don't >>>>>>>>>>>>>>>> understand what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left to >>>>>>>>>>>>> really care much about getting bashed
lest they would begin to understand what all the
intentionally willful ignorance they throw at me is like >>>>>>>>>>>>
understand the real nature of the problem and see what >>>>>>>>>>>> people have actually done.
u don't even understand what a basic enumeration is ur so >>>>>>>>>>> dumb as fuck
Sure I do. You don't understand when your enumeration needs to >>>>>>>>>> be effective.
an effective enumeration of all turing machines was proven on >>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you >>>>>>>> think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list of >>>>>> numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration
isn't of all machines when he build the diagonal argument, as not
all machines produce a valid result to put on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of
machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to
question the effectiveness of a total turing machine enumeration
And you are a moron for confusing the effective enumeration of ALL
machines, many of which don't meet the requirements to be put in the
list, and the enumeration of the machines that DO produce a
"computable number"
The method to pair down the enumeration of all machines to the
machines that compute a computable number is shown to non-computable.
You depending on Unicorns to prepare your enumerations is what
breaks your logic.
this isn't math anymore, it's just u randomly putting words
together opposed to mine based on feels i can't even remotely
understand
Yes, that is all YOU are doing, as you don't require your statments
to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM programmed to >>>>> be contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines, but >>>>>> of all machines that compute a number, otherwise we need to deal
with the possibility that a given row isn't "complete" because
that machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D to >>>>> be "satisfactory"/circle-free, so he does enumerate over all
machines testing each one for being "satisfactory" for inclusion on >>>>> the diagonal or not
write me psuedocode that accurately represents machine H as
described on p247 of turing's paper /on computable numbers/, cause
honestly u understand the algo he wrote. if write more than like 15 >>>>> lines u've fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, when it
can't be.
Note, the paper starts with a "Supposing there is a Machine D", and
if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no machine H
to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted
moron... he describes exactly what it does, but put it in a form of
psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here can read
a fucking paper
Since each iteration just uses a testing by the machine D, your
psuedo- code loop is just a reference to a operation that turns out
not to exist.
i really am just getting a bunch retarded kneejerk reactions from u
eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it
That is why you are so stupid, you seem to expect others to do the
work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to
implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means you >>>>>>>>>>>> never learned how to learn, and thus made your self
fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>>>
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:No, it seems that the problem is you don't understand that >>>>>>>>>>>>> the first step of dealing with a problem is to first >>>>>>>>>>>>> understand the real nature of the problem and see what >>>>>>>>>>>>> people have actually done.
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one >>>>>>>>>>>>>>>>> you say you are, that is the Halting Problem and / or >>>>>>>>>>>>>>>>> the concept of problems being undecidable, then I guess >>>>>>>>>>>>>>>>> you are just a pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>>
It seems your problem is you can't actually express the >>>>>>>>>>>>>>>>> problem you see, perhaps because you just don't >>>>>>>>>>>>>>>>> understand what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left >>>>>>>>>>>>>> to really care much about getting bashed
lest they would begin to understand what all the
intentionally willful ignorance they throw at me is like >>>>>>>>>>>>>
u don't even understand what a basic enumeration is ur so >>>>>>>>>>>> dumb as fuck
Sure I do. You don't understand when your enumeration needs >>>>>>>>>>> to be effective.
an effective enumeration of all turing machines was proven on >>>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me >>>>>>>>> you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list of >>>>>>> numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration
isn't of all machines when he build the diagonal argument, as not
all machines produce a valid result to put on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of
machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to
question the effectiveness of a total turing machine enumeration
And you are a moron for confusing the effective enumeration of ALL
machines, many of which don't meet the requirements to be put in
the list, and the enumeration of the machines that DO produce a
"computable number"
The method to pair down the enumeration of all machines to the
machines that compute a computable number is shown to non-computable. >>>>>
You depending on Unicorns to prepare your enumerations is what
breaks your logic.
this isn't math anymore, it's just u randomly putting words
together opposed to mine based on feels i can't even remotely
understand
Yes, that is all YOU are doing, as you don't require your statments >>>>> to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM programmed
to be contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines,
but of all machines that compute a number, otherwise we need to >>>>>>> deal with the possibility that a given row isn't "complete"
because that machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D
to be "satisfactory"/circle-free, so he does enumerate over all
machines testing each one for being "satisfactory" for inclusion
on the diagonal or not
write me psuedocode that accurately represents machine H as
described on p247 of turing's paper /on computable numbers/, cause >>>>>> honestly u understand the algo he wrote. if write more than like
15 lines u've fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, when
it can't be.
Note, the paper starts with a "Supposing there is a Machine D", and >>>>> if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no machine H >>>>> to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted
moron... he describes exactly what it does, but put it in a form of
psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here can
read a fucking paper
Since YOU don't seem to be able to do that, why should I help you.
Your problem is you don't actually understand what it says, as you think
it is all wrong.
YOU are the one asking for help, and then critisizing people when they
do so.
YOU are the one showing yourself to be just a jerk that doesn't know his
ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife clinging to
your absurd ideas and rid the world of another crackpot.
I'm just putting enough information that anyone with a brain cell can
see that your ideas are baseless, and that if they have questions, they
can learn the answers from the source.
The world doesn't owe you a living.
Since each iteration just uses a testing by the machine D, your
psuedo- code loop is just a reference to a operation that turns out
not to exist.
i really am just getting a bunch retarded kneejerk reactions from u
eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it
Nope, you are just showing that you don't understand what you are
talking about.
That is why you are so stupid, you seem to expect others to do the
work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to >>>>>>>>> implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means >>>>>>>>>>>>> you never learned how to learn, and thus made your self >>>>>>>>>>>>> fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>>>>
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote:No, it seems that the problem is you don't understand that >>>>>>>>>>>>>> the first step of dealing with a problem is to first >>>>>>>>>>>>>> understand the real nature of the problem and see what >>>>>>>>>>>>>> people have actually done.
On 2/21/2026 9:41 AM, dart200 wrote:
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> > [ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one >>>>>>>>>>>>>>>>>> you say you are, that is the Halting Problem and / or >>>>>>>>>>>>>>>>>> the concept of problems being undecidable, then I >>>>>>>>>>>>>>>>>> guess you are just a pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>>>
It seems your problem is you can't actually express >>>>>>>>>>>>>>>>>> the problem you see, perhaps because you just don't >>>>>>>>>>>>>>>>>> understand what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left >>>>>>>>>>>>>>> to really care much about getting bashed
lest they would begin to understand what all the >>>>>>>>>>>>>>> intentionally willful ignorance they throw at me is like >>>>>>>>>>>>>>
u don't even understand what a basic enumeration is ur so >>>>>>>>>>>>> dumb as fuck
Sure I do. You don't understand when your enumeration needs >>>>>>>>>>>> to be effective.
an effective enumeration of all turing machines was proven on >>>>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me >>>>>>>>>> you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list >>>>>>>> of numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration
isn't of all machines when he build the diagonal argument, as not >>>>>> all machines produce a valid result to put on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of
machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to
question the effectiveness of a total turing machine enumeration
And you are a moron for confusing the effective enumeration of ALL >>>>>> machines, many of which don't meet the requirements to be put in
the list, and the enumeration of the machines that DO produce a
"computable number"
The method to pair down the enumeration of all machines to the
machines that compute a computable number is shown to non-computable. >>>>>>
You depending on Unicorns to prepare your enumerations is what
breaks your logic.
this isn't math anymore, it's just u randomly putting words
together opposed to mine based on feels i can't even remotely
understand
Yes, that is all YOU are doing, as you don't require your
statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM programmed >>>>>>> to be contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines, >>>>>>>> but of all machines that compute a number, otherwise we need to >>>>>>>> deal with the possibility that a given row isn't "complete"
because that machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D >>>>>>> to be "satisfactory"/circle-free, so he does enumerate over all >>>>>>> machines testing each one for being "satisfactory" for inclusion >>>>>>> on the diagonal or not
write me psuedocode that accurately represents machine H as
described on p247 of turing's paper /on computable numbers/,
cause honestly u understand the algo he wrote. if write more than >>>>>>> like 15 lines u've fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, when >>>>>> it can't be.
Note, the paper starts with a "Supposing there is a Machine D",
and if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no machine >>>>>> H to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted
moron... he describes exactly what it does, but put it in a form of >>>>> psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here can
read a fucking paper
Since YOU don't seem to be able to do that, why should I help you.
i already did and put it in a paper i have a record of you reading in december, why are you telling me i can't seem to do something i already did???
i want to see someone else in this group do it:
assume D exists, what is the pseudo-code for H from p247 of turing's paper
Your problem is you don't actually understand what it says, as you
think it is all wrong.
YOU are the one asking for help, and then critisizing people when they
do so.
YOU are the one showing yourself to be just a jerk that doesn't know
his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife clinging
to your absurd ideas and rid the world of another crackpot.
I'm just putting enough information that anyone with a brain cell can
see that your ideas are baseless, and that if they have questions,
they can learn the answers from the source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even just a
pursuit of happiness anymore
Since each iteration just uses a testing by the machine D, your
psuedo- code loop is just a reference to a operation that turns out
not to exist.
i really am just getting a bunch retarded kneejerk reactions from u >>>>> eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it
Nope, you are just showing that you don't understand what you are
talking about.
That is why you are so stupid, you seem to expect others to do the
work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to >>>>>>>>>> implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means >>>>>>>>>>>>>> you never learned how to learn, and thus made your self >>>>>>>>>>>>>> fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>>>>>
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:You think you can test all of them one by one? Don't tell me >>>>>>>>>>> you think
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote:No, it seems that the problem is you don't understand >>>>>>>>>>>>>>> that the first step of dealing with a problem is to first >>>>>>>>>>>>>>> understand the real nature of the problem and see what >>>>>>>>>>>>>>> people have actually done.
On 2/21/26 4:03 AM, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> > [ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one >>>>>>>>>>>>>>>>>>> you say you are, that is the Halting Problem and / or >>>>>>>>>>>>>>>>>>> the concept of problems being undecidable, then I >>>>>>>>>>>>>>>>>>> guess you are just a pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>>>>
It seems your problem is you can't actually express >>>>>>>>>>>>>>>>>>> the problem you see, perhaps because you just don't >>>>>>>>>>>>>>>>>>> understand what you are talking about.
ur old and will be dead in a decade or so
Oh man, thats rather harsh?
ur all too old and lead addled to have much empathy left >>>>>>>>>>>>>>>> to really care much about getting bashed
lest they would begin to understand what all the >>>>>>>>>>>>>>>> intentionally willful ignorance they throw at me is like >>>>>>>>>>>>>>>
u don't even understand what a basic enumeration is ur so >>>>>>>>>>>>>> dumb as fuck
Sure I do. You don't understand when your enumeration needs >>>>>>>>>>>>> to be effective.
an effective enumeration of all turing machines was proven >>>>>>>>>>>> on turing's original paper and can be reused anywhere... >>>>>>>>>>>
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list >>>>>>>>> of numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT >>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration >>>>>>> isn't of all machines when he build the diagonal argument, as not >>>>>>> all machines produce a valid result to put on the diagonal.
And you are a moron for confusing the effective enumeration of
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of >>>>>>>>> machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to
question the effectiveness of a total turing machine enumeration >>>>>>>
ALL machines, many of which don't meet the requirements to be put >>>>>>> in the list, and the enumeration of the machines that DO produce >>>>>>> a "computable number"
The method to pair down the enumeration of all machines to the
machines that compute a computable number is shown to non-
computable.
You depending on Unicorns to prepare your enumerations is what
breaks your logic.
this isn't math anymore, it's just u randomly putting words
together opposed to mine based on feels i can't even remotely >>>>>>>> understand
Yes, that is all YOU are doing, as you don't require your
statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM programmed >>>>>>>> to be contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines, >>>>>>>>> but of all machines that compute a number, otherwise we need to >>>>>>>>> deal with the possibility that a given row isn't "complete" >>>>>>>>> because that machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by D >>>>>>>> to be "satisfactory"/circle-free, so he does enumerate over all >>>>>>>> machines testing each one for being "satisfactory" for inclusion >>>>>>>> on the diagonal or not
write me psuedocode that accurately represents machine H as
described on p247 of turing's paper /on computable numbers/,
cause honestly u understand the algo he wrote. if write more
than like 15 lines u've fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, when >>>>>>> it can't be.
Note, the paper starts with a "Supposing there is a Machine D", >>>>>>> and if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no machine >>>>>>> H to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted
moron... he describes exactly what it does, but put it in a form
of psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here can
read a fucking paper
Since YOU don't seem to be able to do that, why should I help you.
i already did and put it in a paper i have a record of you reading in
december, why are you telling me i can't seem to do something i
already did???
You mean that CRAP that assumes unicorns exist, and was based on
category errors?
And, how do you know how much of it I read?
In fact, the fact you asked me to look at your paper, which appears to
be on a spammer site that keeps on asking me to "subscribe" to there
service to see how is mentioning my name is part of the cause for some
of my attitude.
It seems you are nothing but a stupid scammer that is trying to get
people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING something and showing you understand the meaning behind it, and not just say the world
is wrong because it won't give me my unicorns, and the world with
unicorns would be so much better.
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of turing's
paper
But it doesn't, so why do it.
All you have done is proved you are too stupid to understand how logic works.
Your problem is you don't actually understand what it says, as you
think it is all wrong.
YOU are the one asking for help, and then critisizing people when
they do so.
YOU are the one showing yourself to be just a jerk that doesn't know
his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife clinging
to your absurd ideas and rid the world of another crackpot.
I'm just putting enough information that anyone with a brain cell can
see that your ideas are baseless, and that if they have questions,
they can learn the answers from the source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even just a
pursuit of happiness anymore
Since each iteration just uses a testing by the machine D, your
psuedo- code loop is just a reference to a operation that turns out >>>>> not to exist.
i really am just getting a bunch retarded kneejerk reactions from >>>>>> u eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it
Nope, you are just showing that you don't understand what you are
talking about.
That is why you are so stupid, you seem to expect others to do the
work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to >>>>>>>>>>> implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means >>>>>>>>>>>>>>> you never learned how to learn, and thus made your self >>>>>>>>>>>>>>> fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>>>>>>
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:You think you can test all of them one by one? Don't tell me >>>>>>>>>>>> you think
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote:No, it seems that the problem is you don't understand >>>>>>>>>>>>>>>> that the first step of dealing with a problem is to >>>>>>>>>>>>>>>> first understand the real nature of the problem and see >>>>>>>>>>>>>>>> what people have actually done.
On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote:Oh man, thats rather harsh?
On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> > [ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one >>>>>>>>>>>>>>>>>>>> you say you are, that is the Halting Problem and / >>>>>>>>>>>>>>>>>>>> or the concept of problems being undecidable, then I >>>>>>>>>>>>>>>>>>>> guess you are just a pathological liar. >>>>>>>>>>>>>>>>>>>>
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>>>>>
It seems your problem is you can't actually express >>>>>>>>>>>>>>>>>>>> the problem you see, perhaps because you just don't >>>>>>>>>>>>>>>>>>>> understand what you are talking about.
ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much empathy >>>>>>>>>>>>>>>>> left to really care much about getting bashed >>>>>>>>>>>>>>>>>
lest they would begin to understand what all the >>>>>>>>>>>>>>>>> intentionally willful ignorance they throw at me is like >>>>>>>>>>>>>>>>
u don't even understand what a basic enumeration is ur so >>>>>>>>>>>>>>> dumb as fuck
Sure I do. You don't understand when your enumeration >>>>>>>>>>>>>> needs to be effective.
an effective enumeration of all turing machines was proven >>>>>>>>>>>>> on turing's original paper and can be reused anywhere... >>>>>>>>>>>>
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the list >>>>>>>>>> of numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT >>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration >>>>>>>> isn't of all machines when he build the diagonal argument, as >>>>>>>> not all machines produce a valid result to put on the diagonal. >>>>>>>>
And you are a moron for confusing the effective enumeration of >>>>>>>> ALL machines, many of which don't meet the requirements to be >>>>>>>> put in the list, and the enumeration of the machines that DO
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of >>>>>>>>>> machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to >>>>>>>>> question the effectiveness of a total turing machine enumeration >>>>>>>>
produce a "computable number"
The method to pair down the enumeration of all machines to the >>>>>>>> machines that compute a computable number is shown to non-
computable.
You depending on Unicorns to prepare your enumerations is what >>>>>>>> breaks your logic.
this isn't math anymore, it's just u randomly putting words >>>>>>>>> together opposed to mine based on feels i can't even remotely >>>>>>>>> understand
Yes, that is all YOU are doing, as you don't require your
statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM
programmed to be contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing Machines, >>>>>>>>>> but of all machines that compute a number, otherwise we need >>>>>>>>>> to deal with the possibility that a given row isn't "complete" >>>>>>>>>> because that machine never generates enough numbers.
he only simulates numbers for the diagonal that are decided by >>>>>>>>> D to be "satisfactory"/circle-free, so he does enumerate over >>>>>>>>> all machines testing each one for being "satisfactory" for
inclusion on the diagonal or not
write me psuedocode that accurately represents machine H as >>>>>>>>> described on p247 of turing's paper /on computable numbers/, >>>>>>>>> cause honestly u understand the algo he wrote. if write more >>>>>>>>> than like 15 lines u've fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done,
when it can't be.
Note, the paper starts with a "Supposing there is a Machine D", >>>>>>>> and if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no
machine H to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted
moron... he describes exactly what it does, but put it in a form >>>>>>> of psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here can
read a fucking paper
Since YOU don't seem to be able to do that, why should I help you.
i already did and put it in a paper i have a record of you reading in
december, why are you telling me i can't seem to do something i
already did???
You mean that CRAP that assumes unicorns exist, and was based on
category errors?
yes, assume D exists and write the algorithm for H
And, how do you know how much of it I read?
apparently not even the 3rd page
In fact, the fact you asked me to look at your paper, which appears to
be on a spammer site that keeps on asking me to "subscribe" to there
service to see how is mentioning my name is part of the cause for some
of my attitude.
It seems you are nothing but a stupid scammer that is trying to get
people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING something and
showing you understand the meaning behind it, and not just say the
world is wrong because it won't give me my unicorns, and the world
with unicorns would be so much better.
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of turing's
paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, that we
can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
All you have done is proved you are too stupid to understand how logic
works.
Your problem is you don't actually understand what it says, as you
think it is all wrong.
YOU are the one asking for help, and then critisizing people when
they do so.
YOU are the one showing yourself to be just a jerk that doesn't know
his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife clinging
to your absurd ideas and rid the world of another crackpot.
I'm just putting enough information that anyone with a brain cell
can see that your ideas are baseless, and that if they have
questions, they can learn the answers from the source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even just a
pursuit of happiness anymore
Since each iteration just uses a testing by the machine D, your
psuedo- code loop is just a reference to a operation that turns
out not to exist.
i really am just getting a bunch retarded kneejerk reactions from >>>>>>> u eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it
Nope, you are just showing that you don't understand what you are
talking about.
That is why you are so stupid, you seem to expect others to do the >>>>>> work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to >>>>>>>>>>>> implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information means >>>>>>>>>>>>>>>> you never learned how to learn, and thus made your self >>>>>>>>>>>>>>>> fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>>>>>>>
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:You think you can test all of them one by one? Don't tell >>>>>>>>>>>>> me you think
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote:No, it seems that the problem is you don't understand >>>>>>>>>>>>>>>>> that the first step of dealing with a problem is to >>>>>>>>>>>>>>>>> first understand the real nature of the problem and see >>>>>>>>>>>>>>>>> what people have actually done.
On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote:Oh man, thats rather harsh?
ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ]
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the >>>>>>>>>>>>>>>>>>>>> one you say you are, that is the Halting Problem >>>>>>>>>>>>>>>>>>>>> and / or the concept of problems being undecidable, >>>>>>>>>>>>>>>>>>>>> then I guess you are just a pathological liar. >>>>>>>>>>>>>>>>>>>>>
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>>>>>>
It seems your problem is you can't actually express >>>>>>>>>>>>>>>>>>>>> the problem you see, perhaps because you just don't >>>>>>>>>>>>>>>>>>>>> understand what you are talking about. >>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much empathy >>>>>>>>>>>>>>>>>> left to really care much about getting bashed >>>>>>>>>>>>>>>>>>
lest they would begin to understand what all the >>>>>>>>>>>>>>>>>> intentionally willful ignorance they throw at me is like >>>>>>>>>>>>>>>>>
u don't even understand what a basic enumeration is ur >>>>>>>>>>>>>>>> so dumb as fuck
Sure I do. You don't understand when your enumeration >>>>>>>>>>>>>>> needs to be effective.
an effective enumeration of all turing machines was proven >>>>>>>>>>>>>> on turing's original paper and can be reused anywhere... >>>>>>>>>>>>>
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the >>>>>>>>>>> list of numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT >>>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His enumeration >>>>>>>>> isn't of all machines when he build the diagonal argument, as >>>>>>>>> not all machines produce a valid result to put on the diagonal. >>>>>>>>>
And you are a moron for confusing the effective enumeration of >>>>>>>>> ALL machines, many of which don't meet the requirements to be >>>>>>>>> put in the list, and the enumeration of the machines that DO >>>>>>>>> produce a "computable number"
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of >>>>>>>>>>> machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to >>>>>>>>>> question the effectiveness of a total turing machine enumeration >>>>>>>>>
The method to pair down the enumeration of all machines to the >>>>>>>>> machines that compute a computable number is shown to non-
computable.
You depending on Unicorns to prepare your enumerations is what >>>>>>>>> breaks your logic.
this isn't math anymore, it's just u randomly putting words >>>>>>>>>> together opposed to mine based on feels i can't even remotely >>>>>>>>>> understand
Yes, that is all YOU are doing, as you don't require your
statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM
programmed to be contrarian
At least that wouldn't be wasting smart people's time.
And note, it isn't even an enumeration of all Turing
Machines, but of all machines that compute a number,
otherwise we need to deal with the possibility that a given >>>>>>>>>>> row isn't "complete" because that machine never generates >>>>>>>>>>> enough numbers.
he only simulates numbers for the diagonal that are decided by >>>>>>>>>> D to be "satisfactory"/circle-free, so he does enumerate over >>>>>>>>>> all machines testing each one for being "satisfactory" for >>>>>>>>>> inclusion on the diagonal or not
write me psuedocode that accurately represents machine H as >>>>>>>>>> described on p247 of turing's paper /on computable numbers/, >>>>>>>>>> cause honestly u understand the algo he wrote. if write more >>>>>>>>>> than like 15 lines u've fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, >>>>>>>>> when it can't be.
Note, the paper starts with a "Supposing there is a Machine D", >>>>>>>>> and if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no
machine H to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted >>>>>>>> moron... he describes exactly what it does, but put it in a form >>>>>>>> of psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here can >>>>>> read a fucking paper
Since YOU don't seem to be able to do that, why should I help you.
i already did and put it in a paper i have a record of you reading
in december, why are you telling me i can't seem to do something i
already did???
You mean that CRAP that assumes unicorns exist, and was based on
category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second paragraph,
where you ADMIT that you aren't actually adressing the point you claim
to be.
And by the end of the first page, you demonstrate that you don't
understand the basics of the field.
In fact, the fact you asked me to look at your paper, which appears
to be on a spammer site that keeps on asking me to "subscribe" to
there service to see how is mentioning my name is part of the cause
for some of my attitude.
It seems you are nothing but a stupid scammer that is trying to get
people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING something
and showing you understand the meaning behind it, and not just say
the world is wrong because it won't give me my unicorns, and the
world with unicorns would be so much better.
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of turing's
paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, that we
can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
Why?
And the algorithm for H will DEPEND on the algorithm for D, and thus if
D doesn't exist, neither does H.
All you have done is proved you are too stupid to understand how
logic works.
Your problem is you don't actually understand what it says, as you
think it is all wrong.
YOU are the one asking for help, and then critisizing people when
they do so.
YOU are the one showing yourself to be just a jerk that doesn't
know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife
clinging to your absurd ideas and rid the world of another crackpot. >>>>>
I'm just putting enough information that anyone with a brain cell
can see that your ideas are baseless, and that if they have
questions, they can learn the answers from the source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even just a
pursuit of happiness anymore
Since each iteration just uses a testing by the machine D, your >>>>>>> psuedo- code loop is just a reference to a operation that turns >>>>>>> out not to exist.
i really am just getting a bunch retarded kneejerk reactions
from u eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it
Nope, you are just showing that you don't understand what you are
talking about.
That is why you are so stupid, you seem to expect others to do
the work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need to >>>>>>>>>>>>> implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information >>>>>>>>>>>>>>>>> means you never learned how to learn, and thus made >>>>>>>>>>>>>>>>> your self fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>>>>>>>>
On 2/23/26 12:39 PM, Richard Damon wrote:
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote:No, it seems that the problem is you don't understand >>>>>>>>>>>>>>>>>> that the first step of dealing with a problem is to >>>>>>>>>>>>>>>>>> first understand the real nature of the problem and >>>>>>>>>>>>>>>>>> see what people have actually done.
On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ]Oh man, thats rather harsh?
ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>
i suppose i shouldn't be expecting fucking boomer >>>>>>>>>>>>>>>>>>>>>>> shitposters to ever be any help
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the >>>>>>>>>>>>>>>>>>>>>> one you say you are, that is the Halting Problem >>>>>>>>>>>>>>>>>>>>>> and / or the concept of problems being >>>>>>>>>>>>>>>>>>>>>> undecidable, then I guess you are just a >>>>>>>>>>>>>>>>>>>>>> pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>>>>>>>
It seems your problem is you can't actually >>>>>>>>>>>>>>>>>>>>>> express the problem you see, perhaps because you >>>>>>>>>>>>>>>>>>>>>> just don't understand what you are talking about. >>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much empathy >>>>>>>>>>>>>>>>>>> left to really care much about getting bashed >>>>>>>>>>>>>>>>>>>
lest they would begin to understand what all the >>>>>>>>>>>>>>>>>>> intentionally willful ignorance they throw at me is like >>>>>>>>>>>>>>>>>>
u don't even understand what a basic enumeration is ur >>>>>>>>>>>>>>>>> so dumb as fuck
Sure I do. You don't understand when your enumeration >>>>>>>>>>>>>>>> needs to be effective.
an effective enumeration of all turing machines was >>>>>>>>>>>>>>> proven on turing's original paper and can be reused >>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? Don't tell >>>>>>>>>>>>>> me you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the >>>>>>>>>>>> list of numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT >>>>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His
enumeration isn't of all machines when he build the diagonal >>>>>>>>>> argument, as not all machines produce a valid result to put on >>>>>>>>>> the diagonal.
And you are a moron for confusing the effective enumeration of >>>>>>>>>> ALL machines, many of which don't meet the requirements to be >>>>>>>>>> put in the list, and the enumeration of the machines that DO >>>>>>>>>> produce a "computable number"
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of >>>>>>>>>>>> machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to >>>>>>>>>>> question the effectiveness of a total turing machine enumeration >>>>>>>>>>
The method to pair down the enumeration of all machines to the >>>>>>>>>> machines that compute a computable number is shown to non- >>>>>>>>>> computable.
You depending on Unicorns to prepare your enumerations is what >>>>>>>>>> breaks your logic.
this isn't math anymore, it's just u randomly putting words >>>>>>>>>>> together opposed to mine based on feels i can't even remotely >>>>>>>>>>> understand
Yes, that is all YOU are doing, as you don't require your >>>>>>>>>> statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM
programmed to be contrarian
At least that wouldn't be wasting smart people's time.
You mean the one he shows can't exist?
And note, it isn't even an enumeration of all Turing
Machines, but of all machines that compute a number,
otherwise we need to deal with the possibility that a given >>>>>>>>>>>> row isn't "complete" because that machine never generates >>>>>>>>>>>> enough numbers.
he only simulates numbers for the diagonal that are decided >>>>>>>>>>> by D to be "satisfactory"/circle-free, so he does enumerate >>>>>>>>>>> over all machines testing each one for being "satisfactory" >>>>>>>>>>> for inclusion on the diagonal or not
write me psuedocode that accurately represents machine H as >>>>>>>>>>> described on p247 of turing's paper /on computable numbers/, >>>>>>>>>>> cause honestly u understand the algo he wrote. if write more >>>>>>>>>>> than like 15 lines u've fucked up, it's not a complex algo >>>>>>>>>>
Because one of the steps we just need to assume can be done, >>>>>>>>>> when it can't be.
Note, the paper starts with a "Supposing there is a Machine >>>>>>>>>> D", and if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no >>>>>>>>>> machine H to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted >>>>>>>>> moron... he describes exactly what it does, but put it in a >>>>>>>>> form of psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here
can read a fucking paper
Since YOU don't seem to be able to do that, why should I help you.
i already did and put it in a paper i have a record of you reading
in december, why are you telling me i can't seem to do something i
already did???
You mean that CRAP that assumes unicorns exist, and was based on
category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second paragraph,
where you ADMIT that you aren't actually adressing the point you claim
to be.
ur looking just for excuses to not think at all by this point
And by the end of the first page, you demonstrate that you don't
understand the basics of the field.
In fact, the fact you asked me to look at your paper, which appears
to be on a spammer site that keeps on asking me to "subscribe" to
there service to see how is mentioning my name is part of the cause
for some of my attitude.
It seems you are nothing but a stupid scammer that is trying to get
people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING something
and showing you understand the meaning behind it, and not just say
the world is wrong because it won't give me my unicorns, and the
world with unicorns would be so much better.
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of
turing's paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, that we
can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
Why?
And the algorithm for H will DEPEND on the algorithm for D, and thus
if D doesn't exist, neither does H.
cause if u can't give me a simple 10 line pseudo-code to describe the algorithm used then i'll have to assume ur incapable to understand
either turing's proof, or the various resolutions i've proposed
in which case idk,
i can buy a gun and shoot myself cause i really have exhausted all my discussion options by now.
everyone is too far ethically gone to have any compassion or consideration,
and i have no desire to participate further in this unholy 🤡🌎
All you have done is proved you are too stupid to understand how
logic works.
Your problem is you don't actually understand what it says, as you >>>>>> think it is all wrong.
YOU are the one asking for help, and then critisizing people when >>>>>> they do so.
YOU are the one showing yourself to be just a jerk that doesn't
know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife
clinging to your absurd ideas and rid the world of another crackpot. >>>>>>
I'm just putting enough information that anyone with a brain cell >>>>>> can see that your ideas are baseless, and that if they have
questions, they can learn the answers from the source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even just a >>>>> pursuit of happiness anymore
Since each iteration just uses a testing by the machine D, your >>>>>>>> psuedo- code loop is just a reference to a operation that turns >>>>>>>> out not to exist.
i really am just getting a bunch retarded kneejerk reactions >>>>>>>>> from u eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it
Nope, you are just showing that you don't understand what you are >>>>>> talking about.
That is why you are so stupid, you seem to expect others to do >>>>>>>> the work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need >>>>>>>>>>>>>> to implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information >>>>>>>>>>>>>>>>>> means you never learned how to learn, and thus made >>>>>>>>>>>>>>>>>> your self fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>>>>>>>>>
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:i already did and put it in a paper i have a record of you reading >>>>>> in december, why are you telling me i can't seem to do something i >>>>>> already did???
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote:
On 2/21/26 4:03 PM, dart200 wrote:
On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ]
Oh man, thats rather harsh?ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>
i suppose i shouldn't be expecting fucking >>>>>>>>>>>>>>>>>>>>>>>> boomer shitposters to ever be any help >>>>>>>>>>>>>>>>>>>>>>>>
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the >>>>>>>>>>>>>>>>>>>>>>> one you say you are, that is the Halting Problem >>>>>>>>>>>>>>>>>>>>>>> and / or the concept of problems being >>>>>>>>>>>>>>>>>>>>>>> undecidable, then I guess you are just a >>>>>>>>>>>>>>>>>>>>>>> pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>>>>>>>>
It seems your problem is you can't actually >>>>>>>>>>>>>>>>>>>>>>> express the problem you see, perhaps because you >>>>>>>>>>>>>>>>>>>>>>> just don't understand what you are talking about. >>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much empathy >>>>>>>>>>>>>>>>>>>> left to really care much about getting bashed >>>>>>>>>>>>>>>>>>>>
lest they would begin to understand what all the >>>>>>>>>>>>>>>>>>>> intentionally willful ignorance they throw at me is >>>>>>>>>>>>>>>>>>>> like
No, it seems that the problem is you don't understand >>>>>>>>>>>>>>>>>>> that the first step of dealing with a problem is to >>>>>>>>>>>>>>>>>>> first understand the real nature of the problem and >>>>>>>>>>>>>>>>>>> see what people have actually done.
u don't even understand what a basic enumeration is ur >>>>>>>>>>>>>>>>>> so dumb as fuck
Sure I do. You don't understand when your enumeration >>>>>>>>>>>>>>>>> needs to be effective.
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>> proven on turing's original paper and can be reused >>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? Don't tell >>>>>>>>>>>>>>> me you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the >>>>>>>>>>>>> list of numbers enumerated by the list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING TWAT >>>>>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His
enumeration isn't of all machines when he build the diagonal >>>>>>>>>>> argument, as not all machines produce a valid result to put >>>>>>>>>>> on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration of >>>>>>>>>>>>> machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to >>>>>>>>>>>> question the effectiveness of a total turing machine
enumeration
And you are a moron for confusing the effective enumeration >>>>>>>>>>> of ALL machines, many of which don't meet the requirements to >>>>>>>>>>> be put in the list, and the enumeration of the machines that >>>>>>>>>>> DO produce a "computable number"
The method to pair down the enumeration of all machines to >>>>>>>>>>> the machines that compute a computable number is shown to >>>>>>>>>>> non- computable.
You depending on Unicorns to prepare your enumerations is >>>>>>>>>>> what breaks your logic.
this isn't math anymore, it's just u randomly putting words >>>>>>>>>>>> together opposed to mine based on feels i can't even
remotely understand
Yes, that is all YOU are doing, as you don't require your >>>>>>>>>>> statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM >>>>>>>>>>>> programmed to be contrarian
At least that wouldn't be wasting smart people's time.
You mean the one he shows can't exist?
And note, it isn't even an enumeration of all Turing >>>>>>>>>>>>> Machines, but of all machines that compute a number, >>>>>>>>>>>>> otherwise we need to deal with the possibility that a given >>>>>>>>>>>>> row isn't "complete" because that machine never generates >>>>>>>>>>>>> enough numbers.
he only simulates numbers for the diagonal that are decided >>>>>>>>>>>> by D to be "satisfactory"/circle-free, so he does enumerate >>>>>>>>>>>> over all machines testing each one for being "satisfactory" >>>>>>>>>>>> for inclusion on the diagonal or not
write me psuedocode that accurately represents machine H as >>>>>>>>>>>> described on p247 of turing's paper /on computable numbers/, >>>>>>>>>>>> cause honestly u understand the algo he wrote. if write more >>>>>>>>>>>> than like 15 lines u've fucked up, it's not a complex algo >>>>>>>>>>>
Because one of the steps we just need to assume can be done, >>>>>>>>>>> when it can't be.
Note, the paper starts with a "Supposing there is a Machine >>>>>>>>>>> D", and if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no >>>>>>>>>>> machine H to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking brainrotted >>>>>>>>>> moron... he describes exactly what it does, but put it in a >>>>>>>>>> form of psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here >>>>>>>> can read a fucking paper
Since YOU don't seem to be able to do that, why should I help you. >>>>>>
You mean that CRAP that assumes unicorns exist, and was based on
category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second paragraph,
where you ADMIT that you aren't actually adressing the point you
claim to be.
ur looking just for excuses to not think at all by this point
No, YOU are looking for someone to validate your LIE.
Your paper begins with a number of categorical errors, so NOTHING that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are trying to debunk.
For instance, a "paradoxical" machine doesn't even try to be paradoxical
to two different deciders, but for your case where you talk about using
two decider to try to beat the paradoxical format, you eventually need
to combine those two into a single decider to give the answer.
The "paradoxical" input will be built on THAT final decider, not the two intermediate deciders.
Your attempts just show you don't understand the nature of what you are talking about.
And by the end of the first page, you demonstrate that you don't
understand the basics of the field.
In fact, the fact you asked me to look at your paper, which appears >>>>> to be on a spammer site that keeps on asking me to "subscribe" to
there service to see how is mentioning my name is part of the cause >>>>> for some of my attitude.
It seems you are nothing but a stupid scammer that is trying to get >>>>> people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING something
and showing you understand the meaning behind it, and not just say
the world is wrong because it won't give me my unicorns, and the
world with unicorns would be so much better.
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of
turing's paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, that
we can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
Why?
And the algorithm for H will DEPEND on the algorithm for D, and thus
if D doesn't exist, neither does H.
cause if u can't give me a simple 10 line pseudo-code to describe the
algorithm used then i'll have to assume ur incapable to understand
either turing's proof, or the various resolutions i've proposed
And you would be wrong as usual.
Of course, being wrong is the one thing you seem to be good at.
in which case idk,
i can buy a gun and shoot myself cause i really have exhausted all my
discussion options by now.
MAYBE you can, depending on where you live, they might not consider you competent to have one in some places.
everyone is too far ethically gone to have any compassion or
consideration,
No, YOU are the one ethically gone, thinking that fallicaous logic is
valid, and that people "owe" you support.
Clearly you chose to put your "papers" on a site that doesn't respect
normal deciency, but just trying to get people to spend money for likely
no real benifit. How much did you need to pay to put your paper there?
and i have no desire to participate further in this unholy 🤡🌎
Then don't.
It is clear you can't understand the basics, and have refused to be
taught, so the best thing to do is to give up and find something you can
do.
When you set your life on trying to reach something unreachable, and
choose to piss of those trying to help you, you really have no value to society.
All you have done is proved you are too stupid to understand how
logic works.
Your problem is you don't actually understand what it says, as
you think it is all wrong.
YOU are the one asking for help, and then critisizing people when >>>>>>> they do so.
YOU are the one showing yourself to be just a jerk that doesn't >>>>>>> know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife
clinging to your absurd ideas and rid the world of another crackpot. >>>>>>>
I'm just putting enough information that anyone with a brain cell >>>>>>> can see that your ideas are baseless, and that if they have
questions, they can learn the answers from the source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even just >>>>>> a pursuit of happiness anymore
Since each iteration just uses a testing by the machine D, your >>>>>>>>> psuedo- code loop is just a reference to a operation that turns >>>>>>>>> out not to exist.
i really am just getting a bunch retarded kneejerk reactions >>>>>>>>>> from u eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it
Nope, you are just showing that you don't understand what you are >>>>>>> talking about.
That is why you are so stupid, you seem to expect others to do >>>>>>>>> the work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need >>>>>>>>>>>>>>> to implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information >>>>>>>>>>>>>>>>>>> means you never learned how to learn, and thus made >>>>>>>>>>>>>>>>>>> your self fundamentally ignorant.
it's just not a problem that can even effect u really >>>>>>>>>>>>>>>>>>>>>>
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:i already did and put it in a paper i have a record of you
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING >>>>>>>>>>>>> TWAT
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:
On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote:
u don't even understand what a basic enumeration is >>>>>>>>>>>>>>>>>>> ur so dumb as fuckOn 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ]
Oh man, thats rather harsh?ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>
i suppose i shouldn't be expecting fucking >>>>>>>>>>>>>>>>>>>>>>>>> boomer shitposters to ever be any help >>>>>>>>>>>>>>>>>>>>>>>>>
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the >>>>>>>>>>>>>>>>>>>>>>>> one you say you are, that is the Halting Problem >>>>>>>>>>>>>>>>>>>>>>>> and / or the concept of problems being >>>>>>>>>>>>>>>>>>>>>>>> undecidable, then I guess you are just a >>>>>>>>>>>>>>>>>>>>>>>> pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>>>>>>>>>
It seems your problem is you can't actually >>>>>>>>>>>>>>>>>>>>>>>> express the problem you see, perhaps because you >>>>>>>>>>>>>>>>>>>>>>>> just don't understand what you are talking about. >>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much empathy >>>>>>>>>>>>>>>>>>>>> left to really care much about getting bashed >>>>>>>>>>>>>>>>>>>>>
lest they would begin to understand what all the >>>>>>>>>>>>>>>>>>>>> intentionally willful ignorance they throw at me is >>>>>>>>>>>>>>>>>>>>> like
No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>> understand that the first step of dealing with a >>>>>>>>>>>>>>>>>>>> problem is to first understand the real nature of >>>>>>>>>>>>>>>>>>>> the problem and see what people have actually done. >>>>>>>>>>>>>>>>>>>
Sure I do. You don't understand when your enumeration >>>>>>>>>>>>>>>>>> needs to be effective.
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>> proven on turing's original paper and can be reused >>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? Don't >>>>>>>>>>>>>>>> tell me you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the >>>>>>>>>>>>>> list of numbers enumerated by the list of machines. >>>>>>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His
enumeration isn't of all machines when he build the diagonal >>>>>>>>>>>> argument, as not all machines produce a valid result to put >>>>>>>>>>>> on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration >>>>>>>>>>>>>> of machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying to >>>>>>>>>>>>> question the effectiveness of a total turing machine >>>>>>>>>>>>> enumeration
And you are a moron for confusing the effective enumeration >>>>>>>>>>>> of ALL machines, many of which don't meet the requirements >>>>>>>>>>>> to be put in the list, and the enumeration of the machines >>>>>>>>>>>> that DO produce a "computable number"
The method to pair down the enumeration of all machines to >>>>>>>>>>>> the machines that compute a computable number is shown to >>>>>>>>>>>> non- computable.
You depending on Unicorns to prepare your enumerations is >>>>>>>>>>>> what breaks your logic.
this isn't math anymore, it's just u randomly putting words >>>>>>>>>>>>> together opposed to mine based on feels i can't even >>>>>>>>>>>>> remotely understand
Yes, that is all YOU are doing, as you don't require your >>>>>>>>>>>> statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM >>>>>>>>>>>>> programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>
And note, it isn't even an enumeration of all Turing >>>>>>>>>>>>>> Machines, but of all machines that compute a number, >>>>>>>>>>>>>> otherwise we need to deal with the possibility that a >>>>>>>>>>>>>> given row isn't "complete" because that machine never >>>>>>>>>>>>>> generates enough numbers.
he only simulates numbers for the diagonal that are decided >>>>>>>>>>>>> by D to be "satisfactory"/circle-free, so he does enumerate >>>>>>>>>>>>> over all machines testing each one for being "satisfactory" >>>>>>>>>>>>> for inclusion on the diagonal or not
write me psuedocode that accurately represents machine H as >>>>>>>>>>>>> described on p247 of turing's paper /on computable
numbers/, cause honestly u understand the algo he wrote. if >>>>>>>>>>>>> write more than like 15 lines u've fucked up, it's not a >>>>>>>>>>>>> complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be done, >>>>>>>>>>>> when it can't be.
Note, the paper starts with a "Supposing there is a Machine >>>>>>>>>>>> D", and if that is true, then "We could construct H".
Since he shows we can not have the machine D, there is no >>>>>>>>>>>> machine H to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking
brainrotted moron... he describes exactly what it does, but >>>>>>>>>>> put it in a form of psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here >>>>>>>>> can read a fucking paper
Since YOU don't seem to be able to do that, why should I help you. >>>>>>>
reading in december, why are you telling me i can't seem to do
something i already did???
You mean that CRAP that assumes unicorns exist, and was based on
category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second paragraph,
where you ADMIT that you aren't actually adressing the point you
claim to be.
ur looking just for excuses to not think at all by this point
No, YOU are looking for someone to validate your LIE.
Your paper begins with a number of categorical errors, so NOTHING that
follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are trying to
debunk.
For instance, a "paradoxical" machine doesn't even try to be
paradoxical to two different deciders, but for your case where you
talk about using two decider to try to beat the paradoxical format,
you eventually need to combine those two into a single decider to give
the answer.
in the paper i address his diagonals, i talk about the difference
between using a standard decider, and using a "fixed" decider (which is context-aware)
i never use more than one decider in a diagonal, idk where ur getting
that from
The "paradoxical" input will be built on THAT final decider, not the
two intermediate deciders.
Your attempts just show you don't understand the nature of what you
are talking about.
And by the end of the first page, you demonstrate that you don't
understand the basics of the field.
In fact, the fact you asked me to look at your paper, which
appears to be on a spammer site that keeps on asking me to
"subscribe" to there service to see how is mentioning my name is
part of the cause for some of my attitude.
It seems you are nothing but a stupid scammer that is trying to
get people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING something >>>>>> and showing you understand the meaning behind it, and not just say >>>>>> the world is wrong because it won't give me my unicorns, and the
world with unicorns would be so much better.
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of
turing's paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, that
we can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
Why?
And the algorithm for H will DEPEND on the algorithm for D, and thus
if D doesn't exist, neither does H.
cause if u can't give me a simple 10 line pseudo-code to describe the
algorithm used then i'll have to assume ur incapable to understand
either turing's proof, or the various resolutions i've proposed
And you would be wrong as usual.
Of course, being wrong is the one thing you seem to be good at.
instead of actual writing down a 10 line psuedocode, ur spending 20 just talking shit
i don't buy it in the slightest bro
in which case idk,
i can buy a gun and shoot myself cause i really have exhausted all my
discussion options by now.
MAYBE you can, depending on where you live, they might not consider
you competent to have one in some places.
everyone is too far ethically gone to have any compassion or
consideration,
No, YOU are the one ethically gone, thinking that fallicaous logic is
valid, and that people "owe" you support.
Clearly you chose to put your "papers" on a site that doesn't respect
normal deciency, but just trying to get people to spend money for
likely no real benifit. How much did you need to pay to put your paper
there?
and i have no desire to participate further in this unholy 🤡🌎
Then don't.
It is clear you can't understand the basics, and have refused to be
taught, so the best thing to do is to give up and find something you
can do.
When you set your life on trying to reach something unreachable, and
choose to piss of those trying to help you, you really have no value
to society.
All you have done is proved you are too stupid to understand how
logic works.
Your problem is you don't actually understand what it says, as >>>>>>>> you think it is all wrong.
YOU are the one asking for help, and then critisizing people
when they do so.
YOU are the one showing yourself to be just a jerk that doesn't >>>>>>>> know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife
clinging to your absurd ideas and rid the world of another
crackpot.
I'm just putting enough information that anyone with a brain
cell can see that your ideas are baseless, and that if they have >>>>>>>> questions, they can learn the answers from the source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even just >>>>>>> a pursuit of happiness anymore
Nope, you are just showing that you don't understand what you >>>>>>>> are talking about.
Since each iteration just uses a testing by the machine D, >>>>>>>>>> your psuedo- code loop is just a reference to a operation that >>>>>>>>>> turns out not to exist.
i really am just getting a bunch retarded kneejerk reactions >>>>>>>>>>> from u eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it >>>>>>>>
That is why you are so stupid, you seem to expect others to do >>>>>>>>>> the work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really need >>>>>>>>>>>>>>>> to implement your interface.
what in the fuck are you going on about???
Expecting people to just hand you that information >>>>>>>>>>>>>>>>>>>> means you never learned how to learn, and thus made >>>>>>>>>>>>>>>>>>>> your self fundamentally ignorant.
it's just not a problem that can even effect u >>>>>>>>>>>>>>>>>>>>>>> really
On 2/23/26 6:44 PM, dart200 wrote:
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:i already did and put it in a paper i have a record of you
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT JUMPING >>>>>>>>>>>>>> TWAT
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote:
On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote:
u don't even understand what a basic enumeration is >>>>>>>>>>>>>>>>>>>> ur so dumb as fuckOn 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ]
Oh man, thats rather harsh?
i suppose i shouldn't be expecting fucking >>>>>>>>>>>>>>>>>>>>>>>>>> boomer shitposters to ever be any help >>>>>>>>>>>>>>>>>>>>>>>>>>
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care
If the problem you are trying to solve isn't >>>>>>>>>>>>>>>>>>>>>>>>> the one you say you are, that is the Halting >>>>>>>>>>>>>>>>>>>>>>>>> Problem and / or the concept of problems being >>>>>>>>>>>>>>>>>>>>>>>>> undecidable, then I guess you are just a >>>>>>>>>>>>>>>>>>>>>>>>> pathological liar.
And yes, if that is the case, no one can help you. >>>>>>>>>>>>>>>>>>>>>>>>>
It seems your problem is you can't actually >>>>>>>>>>>>>>>>>>>>>>>>> express the problem you see, perhaps because >>>>>>>>>>>>>>>>>>>>>>>>> you just don't understand what you are talking >>>>>>>>>>>>>>>>>>>>>>>>> about.
ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much >>>>>>>>>>>>>>>>>>>>>> empathy left to really care much about getting bashed >>>>>>>>>>>>>>>>>>>>>>
lest they would begin to understand what all the >>>>>>>>>>>>>>>>>>>>>> intentionally willful ignorance they throw at me >>>>>>>>>>>>>>>>>>>>>> is like
No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing with a >>>>>>>>>>>>>>>>>>>>> problem is to first understand the real nature of >>>>>>>>>>>>>>>>>>>>> the problem and see what people have actually done. >>>>>>>>>>>>>>>>>>>>
Sure I do. You don't understand when your enumeration >>>>>>>>>>>>>>>>>>> needs to be effective.
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>>> proven on turing's original paper and can be reused >>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? Don't >>>>>>>>>>>>>>>>> tell me you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung the >>>>>>>>>>>>>>> list of numbers enumerated by the list of machines. >>>>>>>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His >>>>>>>>>>>>> enumeration isn't of all machines when he build the >>>>>>>>>>>>> diagonal argument, as not all machines produce a valid >>>>>>>>>>>>> result to put on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration >>>>>>>>>>>>>>> of machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying >>>>>>>>>>>>>> to question the effectiveness of a total turing machine >>>>>>>>>>>>>> enumeration
And you are a moron for confusing the effective enumeration >>>>>>>>>>>>> of ALL machines, many of which don't meet the requirements >>>>>>>>>>>>> to be put in the list, and the enumeration of the machines >>>>>>>>>>>>> that DO produce a "computable number"
The method to pair down the enumeration of all machines to >>>>>>>>>>>>> the machines that compute a computable number is shown to >>>>>>>>>>>>> non- computable.
You depending on Unicorns to prepare your enumerations is >>>>>>>>>>>>> what breaks your logic.
this isn't math anymore, it's just u randomly putting >>>>>>>>>>>>>> words together opposed to mine based on feels i can't even >>>>>>>>>>>>>> remotely understand
Yes, that is all YOU are doing, as you don't require your >>>>>>>>>>>>> statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM >>>>>>>>>>>>>> programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>>
And note, it isn't even an enumeration of all Turing >>>>>>>>>>>>>>> Machines, but of all machines that compute a number, >>>>>>>>>>>>>>> otherwise we need to deal with the possibility that a >>>>>>>>>>>>>>> given row isn't "complete" because that machine never >>>>>>>>>>>>>>> generates enough numbers.
he only simulates numbers for the diagonal that are >>>>>>>>>>>>>> decided by D to be "satisfactory"/circle-free, so he does >>>>>>>>>>>>>> enumerate over all machines testing each one for being >>>>>>>>>>>>>> "satisfactory" for inclusion on the diagonal or not >>>>>>>>>>>>>>
write me psuedocode that accurately represents machine H >>>>>>>>>>>>>> as described on p247 of turing's paper /on computable >>>>>>>>>>>>>> numbers/, cause honestly u understand the algo he wrote. >>>>>>>>>>>>>> if write more than like 15 lines u've fucked up, it's not >>>>>>>>>>>>>> a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be >>>>>>>>>>>>> done, when it can't be.
Note, the paper starts with a "Supposing there is a Machine >>>>>>>>>>>>> D", and if that is true, then "We could construct H". >>>>>>>>>>>>>
Since he shows we can not have the machine D, there is no >>>>>>>>>>>>> machine H to construct.
So, you questions is about the anatomy of a Unicorn.
assume D exists and write the algo for H u fucking
brainrotted moron... he describes exactly what it does, but >>>>>>>>>>>> put it in a form of psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone here >>>>>>>>>> can read a fucking paper
Since YOU don't seem to be able to do that, why should I help you. >>>>>>>>
reading in december, why are you telling me i can't seem to do >>>>>>>> something i already did???
You mean that CRAP that assumes unicorns exist, and was based on >>>>>>> category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second
paragraph, where you ADMIT that you aren't actually adressing the
point you claim to be.
ur looking just for excuses to not think at all by this point
No, YOU are looking for someone to validate your LIE.
Your paper begins with a number of categorical errors, so NOTHING
that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are trying to
debunk.
For instance, a "paradoxical" machine doesn't even try to be
paradoxical to two different deciders, but for your case where you
talk about using two decider to try to beat the paradoxical format,
you eventually need to combine those two into a single decider to
give the answer.
in the paper i address his diagonals, i talk about the difference
between using a standard decider, and using a "fixed" decider (which
is context-aware)
Which isn't ALLOWED as it uses information not part of the input.
Sorry, you are just proving you don't understand what you are talking
about.
A "Computation", which a decider is a special case of, is only ALLOWED
to process the input it is given, and its output must be strictly
determined by it.
If it can somwhow generate two (or more) different answers for a given input, it is BY DEFINITION incorrect, as the problem statement only
gives one correct answer, so giving two different answers is
automatically incorrect.
i never use more than one decider in a diagonal, idk where ur getting
that from
then the paradox input is just the one that you used.
That is your problem, you don't understand that the decider is chosen
FIRST, and THEM we show the input it will fail on.
The "paradoxical" input will be built on THAT final decider, not the
two intermediate deciders.
Your attempts just show you don't understand the nature of what you
are talking about.
And by the end of the first page, you demonstrate that you don't
understand the basics of the field.
In fact, the fact you asked me to look at your paper, which
appears to be on a spammer site that keeps on asking me to
"subscribe" to there service to see how is mentioning my name is >>>>>>> part of the cause for some of my attitude.
It seems you are nothing but a stupid scammer that is trying to >>>>>>> get people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING
something and showing you understand the meaning behind it, and >>>>>>> not just say the world is wrong because it won't give me my
unicorns, and the world with unicorns would be so much better.
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of
turing's paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, that >>>>>> we can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
Why?
And the algorithm for H will DEPEND on the algorithm for D, and
thus if D doesn't exist, neither does H.
cause if u can't give me a simple 10 line pseudo-code to describe
the algorithm used then i'll have to assume ur incapable to
understand either turing's proof, or the various resolutions i've
proposed
And you would be wrong as usual.
Of course, being wrong is the one thing you seem to be good at.
instead of actual writing down a 10 line psuedocode, ur spending 20
just talking shit
i don't buy it in the slightest bro
Perhaps you don't understand that I am trying to show you how stupid
your logic is.
I don't need to "prove" my ability, because here I am not making claims
that aren't generally accepted. I can rely on the fact that they are
well proven statements.
YOU are the one that needs to show you know something, but the fact you
keep on talking about nonsense, like deciders that either take the WRONG input (because they need to be given a context that the question doesn't actually depend on) or change their answer based on something that isn't
the input.
Both of these just prove that your decider can't be correct.
The answer for the behavior of an actual machine doesn't depend on the context of the machine asking the question, as that doesn't actually
change the behavior of the machine in question. Thus, changing your
answer based on it is just wrong.
And, when you back of and admit you are just doing partial deciding, you balk at the comment that this is a "solved" problem, there are LOTS of partial deciders, so you need to show why yours is better, or at a
minimum, nearly as good as, what the current methods produce.
Old Hat results aren't really meaningful or interesting.
in which case idk,
i can buy a gun and shoot myself cause i really have exhausted all
my discussion options by now.
MAYBE you can, depending on where you live, they might not consider
you competent to have one in some places.
everyone is too far ethically gone to have any compassion or
consideration,
No, YOU are the one ethically gone, thinking that fallicaous logic is
valid, and that people "owe" you support.
Clearly you chose to put your "papers" on a site that doesn't respect
normal deciency, but just trying to get people to spend money for
likely no real benifit. How much did you need to pay to put your
paper there?
and i have no desire to participate further in this unholy 🤡🌎
Then don't.
It is clear you can't understand the basics, and have refused to be
taught, so the best thing to do is to give up and find something you
can do.
When you set your life on trying to reach something unreachable, and
choose to piss of those trying to help you, you really have no value
to society.
All you have done is proved you are too stupid to understand how >>>>>>> logic works.
Your problem is you don't actually understand what it says, as >>>>>>>>> you think it is all wrong.
YOU are the one asking for help, and then critisizing people >>>>>>>>> when they do so.
YOU are the one showing yourself to be just a jerk that doesn't >>>>>>>>> know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife >>>>>>>>> clinging to your absurd ideas and rid the world of another
crackpot.
I'm just putting enough information that anyone with a brain >>>>>>>>> cell can see that your ideas are baseless, and that if they >>>>>>>>> have questions, they can learn the answers from the source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even
just a pursuit of happiness anymore
Nope, you are just showing that you don't understand what you >>>>>>>>> are talking about.
Since each iteration just uses a testing by the machine D, >>>>>>>>>>> your psuedo- code loop is just a reference to a operation >>>>>>>>>>> that turns out not to exist.
i really am just getting a bunch retarded kneejerk reactions >>>>>>>>>>>> from u eh??? not even a second of deep thot ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it >>>>>>>>>
That is why you are so stupid, you seem to expect others to >>>>>>>>>>> do the work you want to do.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs
Your problem is you assume unicorns exist.
there is a largest natural number... ;^) You really >>>>>>>>>>>>>>>>> need to implement your interface.
what in the fuck are you going on about??? >>>>>>>>>>>>>>>>>>
Expecting people to just hand you that information >>>>>>>>>>>>>>>>>>>>> means you never learned how to learn, and thus made >>>>>>>>>>>>>>>>>>>>> your self fundamentally ignorant.
it's just not a problem that can even effect u >>>>>>>>>>>>>>>>>>>>>>>> really
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
[ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to ever
be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you are,
that is the Halting Problem and / or the concept of problems being undecidable ...
On 21/02/2026 12:03, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
; [ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to ever
be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you are,
that is the Halting Problem and / or the concept of problems being
undecidable ...
While dart200 is full of verbicraze, I really don't get why you're
talking about what if he's saying something different than he is instead
of saying that you can't figure out the meaning of what he's saying!
What's the point of that either?!
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think
yes that's what diagonal proofs do...
On 2/23/26 6:44 PM, dart200 wrote:...
in the paper i address his diagonals, i talk about the difference
between using a standard decider, and using a "fixed" decider (which
is context-aware)
Which isn't ALLOWED as it uses information not part of the input.
On 2/22/26 11:17 PM, dart200 wrote:...
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 12:02 PM, dart200 wrote:
On 2/22/26 4:03 AM, Richard Damon wrote:
No, I mean things like compliers to generate code that goes into
critical systems.
well they're all spitting out AI generated slop these day, and no
general semantic verification tools are in sight...
And THAT is the dangerous stuff.
It seems your arguement is that people shouldn't have such tools.
AI GENERATION IS NOT EQUIVALENT TO AUTOMATED PROOFS???
WHAT ARE YOU SMOKING???
Of course not.
Most AI is NOT "Proof" or even "Reasoning" based.
Most AI is based on maximum likely Markov Chain processing.
Alan Mackenzie <acm@muc.de> writes:
Ben Bacarisse <ben@bsb.me.uk> wrote:
[ .... ]
The Church-Turing thesis is clearly not the kind of thing that is
provable. That's why it's not called a conjecture.
I don't understand that bit. What is unprovable about the Church-Turing
thesis? I think it hypothesises that there is no computing machine more
powerful than a turing machine.
There are, purely theoretical, models of computation (my preferred
phrase) that are more powerful than Turing machines but they are not considered "effective".
Turing invented TMs to capture the notion of what is computable
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by one.
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known. He worked in defence
research so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research resources.
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by one.
that exactly what turing does in his proof: he defines a comptuation
that enumerates out all the numbers, testing each one of they represent
a "satisfactory"/"circle-free" machine, and adding that to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of turing's
proof and produce the psuedo-code for the machine H, assuming that
machine D exists
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:assume D exists and write the algo for H u fucking
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT >>>>>>>>>>>>>>> JUMPING TWAT
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote:
On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ]
u don't even understand what a basic enumeration is >>>>>>>>>>>>>>>>>>>>> ur so dumb as fuckOh man, thats rather harsh?
i suppose i shouldn't be expecting fucking >>>>>>>>>>>>>>>>>>>>>>>>>>> boomer shitposters to ever be any help >>>>>>>>>>>>>>>>>>>>>>>>>>>
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care
If the problem you are trying to solve isn't >>>>>>>>>>>>>>>>>>>>>>>>>> the one you say you are, that is the Halting >>>>>>>>>>>>>>>>>>>>>>>>>> Problem and / or the concept of problems being >>>>>>>>>>>>>>>>>>>>>>>>>> undecidable, then I guess you are just a >>>>>>>>>>>>>>>>>>>>>>>>>> pathological liar.
And yes, if that is the case, no one can help >>>>>>>>>>>>>>>>>>>>>>>>>> you.
It seems your problem is you can't actually >>>>>>>>>>>>>>>>>>>>>>>>>> express the problem you see, perhaps because >>>>>>>>>>>>>>>>>>>>>>>>>> you just don't understand what you are talking >>>>>>>>>>>>>>>>>>>>>>>>>> about.
ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much >>>>>>>>>>>>>>>>>>>>>>> empathy left to really care much about getting >>>>>>>>>>>>>>>>>>>>>>> bashed
lest they would begin to understand what all the >>>>>>>>>>>>>>>>>>>>>>> intentionally willful ignorance they throw at me >>>>>>>>>>>>>>>>>>>>>>> is like
No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing with a >>>>>>>>>>>>>>>>>>>>>> problem is to first understand the real nature of >>>>>>>>>>>>>>>>>>>>>> the problem and see what people have actually done. >>>>>>>>>>>>>>>>>>>>>
Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>> enumeration needs to be effective.
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>>>> proven on turing's original paper and can be reused >>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? Don't >>>>>>>>>>>>>>>>>> tell me you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung >>>>>>>>>>>>>>>> the list of numbers enumerated by the list of machines. >>>>>>>>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His >>>>>>>>>>>>>> enumeration isn't of all machines when he build the >>>>>>>>>>>>>> diagonal argument, as not all machines produce a valid >>>>>>>>>>>>>> result to put on the diagonal.
No "machine" ever needed to do that testing.
In fact, he doesn't even need an "effective" enumeration >>>>>>>>>>>>>>>> of machines, just that an enumeration exists.
i'm sorry bro ur literally a dribbling retard for trying >>>>>>>>>>>>>>> to question the effectiveness of a total turing machine >>>>>>>>>>>>>>> enumeration
And you are a moron for confusing the effective
enumeration of ALL machines, many of which don't meet the >>>>>>>>>>>>>> requirements to be put in the list, and the enumeration of >>>>>>>>>>>>>> the machines that DO produce a "computable number" >>>>>>>>>>>>>>
The method to pair down the enumeration of all machines to >>>>>>>>>>>>>> the machines that compute a computable number is shown to >>>>>>>>>>>>>> non- computable.
You depending on Unicorns to prepare your enumerations is >>>>>>>>>>>>>> what breaks your logic.
this isn't math anymore, it's just u randomly putting >>>>>>>>>>>>>>> words together opposed to mine based on feels i can't >>>>>>>>>>>>>>> even remotely understand
Yes, that is all YOU are doing, as you don't require your >>>>>>>>>>>>>> statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM >>>>>>>>>>>>>>> programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>>>
And note, it isn't even an enumeration of all Turing >>>>>>>>>>>>>>>> Machines, but of all machines that compute a number, >>>>>>>>>>>>>>>> otherwise we need to deal with the possibility that a >>>>>>>>>>>>>>>> given row isn't "complete" because that machine never >>>>>>>>>>>>>>>> generates enough numbers.
he only simulates numbers for the diagonal that are >>>>>>>>>>>>>>> decided by D to be "satisfactory"/circle-free, so he does >>>>>>>>>>>>>>> enumerate over all machines testing each one for being >>>>>>>>>>>>>>> "satisfactory" for inclusion on the diagonal or not >>>>>>>>>>>>>>>
write me psuedocode that accurately represents machine H >>>>>>>>>>>>>>> as described on p247 of turing's paper /on computable >>>>>>>>>>>>>>> numbers/, cause honestly u understand the algo he wrote. >>>>>>>>>>>>>>> if write more than like 15 lines u've fucked up, it's not >>>>>>>>>>>>>>> a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be >>>>>>>>>>>>>> done, when it can't be.
Note, the paper starts with a "Supposing there is a >>>>>>>>>>>>>> Machine D", and if that is true, then "We could construct H". >>>>>>>>>>>>>>
Since he shows we can not have the machine D, there is no >>>>>>>>>>>>>> machine H to construct.
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>
brainrotted moron... he describes exactly what it does, but >>>>>>>>>>>>> put it in a form of psuedo-code
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone >>>>>>>>>>> here can read a fucking paper
Since YOU don't seem to be able to do that, why should I help >>>>>>>>>> you.
i already did and put it in a paper i have a record of you
reading in december, why are you telling me i can't seem to do >>>>>>>>> something i already did???
You mean that CRAP that assumes unicorns exist, and was based on >>>>>>>> category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second
paragraph, where you ADMIT that you aren't actually adressing the >>>>>> point you claim to be.
ur looking just for excuses to not think at all by this point
No, YOU are looking for someone to validate your LIE.
Your paper begins with a number of categorical errors, so NOTHING
that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are trying to
debunk.
For instance, a "paradoxical" machine doesn't even try to be
paradoxical to two different deciders, but for your case where you
talk about using two decider to try to beat the paradoxical format,
you eventually need to combine those two into a single decider to
give the answer.
in the paper i address his diagonals, i talk about the difference
between using a standard decider, and using a "fixed" decider (which
is context-aware)
Which isn't ALLOWED as it uses information not part of the input.
ok mr math police, didn't realize it was up to you what i'm allowed to
do or not
Sorry, you are just proving you don't understand what you are talking
about.
A "Computation", which a decider is a special case of, is only ALLOWED
to process the input it is given, and its output must be strictly
determined by it.
If it can somwhow generate two (or more) different answers for a given
input, it is BY DEFINITION incorrect, as the problem statement only
gives one correct answer, so giving two different answers is
automatically incorrect.
i never use more than one decider in a diagonal, idk where ur getting
that from
then the paradox input is just the one that you used.
That is your problem, you don't understand that the decider is chosen
FIRST, and THEM we show the input it will fail on.
The "paradoxical" input will be built on THAT final decider, not the
two intermediate deciders.
Your attempts just show you don't understand the nature of what you
are talking about.
And by the end of the first page, you demonstrate that you don't
understand the basics of the field.
In fact, the fact you asked me to look at your paper, which
appears to be on a spammer site that keeps on asking me to
"subscribe" to there service to see how is mentioning my name is >>>>>>>> part of the cause for some of my attitude.
It seems you are nothing but a stupid scammer that is trying to >>>>>>>> get people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING
something and showing you understand the meaning behind it, and >>>>>>>> not just say the world is wrong because it won't give me my
unicorns, and the world with unicorns would be so much better. >>>>>>>>
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of >>>>>>>>> turing's paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say,
that we can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
Why?
And the algorithm for H will DEPEND on the algorithm for D, and
thus if D doesn't exist, neither does H.
cause if u can't give me a simple 10 line pseudo-code to describe
the algorithm used then i'll have to assume ur incapable to
understand either turing's proof, or the various resolutions i've
proposed
And you would be wrong as usual.
Of course, being wrong is the one thing you seem to be good at.
instead of actual writing down a 10 line psuedocode, ur spending 20
just talking shit
i don't buy it in the slightest bro
Perhaps you don't understand that I am trying to show you how stupid
your logic is.
why are you writing paragraphs instead of pseudo-code?
i don't buy this shit in the slightest
I don't need to "prove" my ability, because here I am not making
claims that aren't generally accepted. I can rely on the fact that
they are well proven statements.
YOU are the one that needs to show you know something, but the fact
you keep on talking about nonsense, like deciders that either take the
WRONG input (because they need to be given a context that the question
doesn't actually depend on) or change their answer based on something
that isn't the input.
Both of these just prove that your decider can't be correct.
The answer for the behavior of an actual machine doesn't depend on the
context of the machine asking the question, as that doesn't actually
change the behavior of the machine in question. Thus, changing your
answer based on it is just wrong.
And, when you back of and admit you are just doing partial deciding,
you balk at the comment that this is a "solved" problem, there are
LOTS of partial deciders, so you need to show why yours is better, or
at a minimum, nearly as good as, what the current methods produce.
Old Hat results aren't really meaningful or interesting.
in which case idk,
i can buy a gun and shoot myself cause i really have exhausted all
my discussion options by now.
MAYBE you can, depending on where you live, they might not consider
you competent to have one in some places.
everyone is too far ethically gone to have any compassion or
consideration,
No, YOU are the one ethically gone, thinking that fallicaous logic
is valid, and that people "owe" you support.
Clearly you chose to put your "papers" on a site that doesn't
respect normal deciency, but just trying to get people to spend
money for likely no real benifit. How much did you need to pay to
put your paper there?
and i have no desire to participate further in this unholy 🤡🌎
Then don't.
It is clear you can't understand the basics, and have refused to be
taught, so the best thing to do is to give up and find something you
can do.
When you set your life on trying to reach something unreachable, and
choose to piss of those trying to help you, you really have no value
to society.
All you have done is proved you are too stupid to understand how >>>>>>>> logic works.
Your problem is you don't actually understand what it says, as >>>>>>>>>> you think it is all wrong.
YOU are the one asking for help, and then critisizing people >>>>>>>>>> when they do so.
YOU are the one showing yourself to be just a jerk that
doesn't know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife >>>>>>>>>> clinging to your absurd ideas and rid the world of another >>>>>>>>>> crackpot.
I'm just putting enough information that anyone with a brain >>>>>>>>>> cell can see that your ideas are baseless, and that if they >>>>>>>>>> have questions, they can learn the answers from the source. >>>>>>>>>>
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even >>>>>>>>> just a pursuit of happiness anymore
Nope, you are just showing that you don't understand what you >>>>>>>>>> are talking about.
Since each iteration just uses a testing by the machine D, >>>>>>>>>>>> your psuedo- code loop is just a reference to a operation >>>>>>>>>>>> that turns out not to exist.
It isn't my job to do the thinking for you.
i really am just getting a bunch retarded kneejerk
reactions from u eh??? not even a second of deep thot ever??? >>>>>>>>>>>>
instead, ur spitting out gishgallop that has no thot behind it >>>>>>>>>>
That is why you are so stupid, you seem to expect others to >>>>>>>>>>>> do the work you want to do.
Your problem is you assume unicorns exist.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs >>>>>>>>>>>>>>
there is a largest natural number... ;^) You really >>>>>>>>>>>>>>>>>> need to implement your interface.
what in the fuck are you going on about??? >>>>>>>>>>>>>>>>>>>
Expecting people to just hand you that information >>>>>>>>>>>>>>>>>>>>>> means you never learned how to learn, and thus >>>>>>>>>>>>>>>>>>>>>> made your self fundamentally ignorant. >>>>>>>>>>>>>>>>>>>>>>
it's just not a problem that can even effect u >>>>>>>>>>>>>>>>>>>>>>>>> really
On 23/02/2026 15:02, Richard Damon wrote:
On 2/22/26 11:17 PM, dart200 wrote:...
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 12:02 PM, dart200 wrote:
On 2/22/26 4:03 AM, Richard Damon wrote:
No, I mean things like compliers to generate code that goes into
critical systems.
well they're all spitting out AI generated slop these day, and no
general semantic verification tools are in sight...
And THAT is the dangerous stuff.
It seems your arguement is that people shouldn't have such tools.
AI GENERATION IS NOT EQUIVALENT TO AUTOMATED PROOFS???
WHAT ARE YOU SMOKING???
Of course not.
Most AI is NOT "Proof" or even "Reasoning" based.
Most AI is based on maximum likely Markov Chain processing.
What does mostness have to do with it?
On 2/24/26 10:26 AM, Tristan Wibberley wrote:
On 21/02/2026 12:03, Richard Damon wrote:
On 2/21/26 2:21 AM, dart200 wrote:
On 2/20/26 4:39 PM, Richard Damon wrote:
> [ ...trash... ]
i suppose i shouldn't be expecting fucking boomer shitposters to ever
be any help
u don't understand what i'm trying to solve
or why it's even a problem
so u *can't* care
If the problem you are trying to solve isn't the one you say you are,
that is the Halting Problem and / or the concept of problems being
undecidable ...
While dart200 is full of verbicraze, I really don't get why you're
talking about what if he's saying something different than he is instead
of saying that you can't figure out the meaning of what he's saying!
What's the point of that either?!
richard likes to gaslight me about my own intents,
that's how low the discussion is here in comp.theory
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by one.
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known.
He worked in defence research
so I expected he was studying the maximum possible capability--
of enemy computing infrastructure to help direct research resources.
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
yes that's what diagonal proofs do...an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think >>>>
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by one. >>>
that exactly what turing does in his proof: he defines a comptuation
that enumerates out all the numbers, testing each one of they
represent a "satisfactory"/"circle-free" machine, and adding that to
diagonal across defined across computable numbers
it really would be a great exercise to carefully read p247 of turing's
proof and produce the psuedo-code for the machine H, assuming that
machine D exists
And, your problem is that paper is about a DIFFERENT type of computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as they
continue to produce results.
The Halting Problem (vs the computable number problem) is based on Computations that must return answers in finite time.
Thus, your argument is based on a category error.
Also, it doesn't enumerate the "numbers", but Machines, many of which
will not actually produce numbers.
Then he assumes a test exists that determines if that machine satisfies
the requirements, and shows that there can not be such a test.
Thus, he shows that there is NOT an effective enumeration of computable numbers, only uncomputable enumerations of them.
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is
computable. It would be true if we could enumerate the computable
sequences by finite means, but the problem of enumerating computable sequences is equivalent to the problem of finding out whether a given
number is the D.N of a circle-free machine, and we have no general
process for doing this in a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration that
enumerates the numbers, only one of "all machines" that need an
uncomputable test to see if they are in that set.
Sorry, you are just showing you don't understand what is being done.
On 2/24/26 11:16 AM, dart200 wrote:
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:assume D exists and write the algo for H u fucking >>>>>>>>>>>>>> brainrotted moron... he describes exactly what it does, >>>>>>>>>>>>>> but put it in a form of psuedo-code
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT >>>>>>>>>>>>>>>> JUMPING TWAT
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ]
u don't even understand what a basic enumeration >>>>>>>>>>>>>>>>>>>>>> is ur so dumb as fuckOh man, thats rather harsh?
i suppose i shouldn't be expecting fucking >>>>>>>>>>>>>>>>>>>>>>>>>>>> boomer shitposters to ever be any help >>>>>>>>>>>>>>>>>>>>>>>>>>>>
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care
If the problem you are trying to solve isn't >>>>>>>>>>>>>>>>>>>>>>>>>>> the one you say you are, that is the Halting >>>>>>>>>>>>>>>>>>>>>>>>>>> Problem and / or the concept of problems >>>>>>>>>>>>>>>>>>>>>>>>>>> being undecidable, then I guess you are just >>>>>>>>>>>>>>>>>>>>>>>>>>> a pathological liar.
And yes, if that is the case, no one can help >>>>>>>>>>>>>>>>>>>>>>>>>>> you.
It seems your problem is you can't actually >>>>>>>>>>>>>>>>>>>>>>>>>>> express the problem you see, perhaps because >>>>>>>>>>>>>>>>>>>>>>>>>>> you just don't understand what you are >>>>>>>>>>>>>>>>>>>>>>>>>>> talking about.
ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much >>>>>>>>>>>>>>>>>>>>>>>> empathy left to really care much about getting >>>>>>>>>>>>>>>>>>>>>>>> bashed
lest they would begin to understand what all the >>>>>>>>>>>>>>>>>>>>>>>> intentionally willful ignorance they throw at me >>>>>>>>>>>>>>>>>>>>>>>> is like
No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing with a >>>>>>>>>>>>>>>>>>>>>>> problem is to first understand the real nature of >>>>>>>>>>>>>>>>>>>>>>> the problem and see what people have actually done. >>>>>>>>>>>>>>>>>>>>>>
Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>>> enumeration needs to be effective.
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>>>>> proven on turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? Don't >>>>>>>>>>>>>>>>>>> tell me you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung >>>>>>>>>>>>>>>>> the list of numbers enumerated by the list of machines. >>>>>>>>>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His >>>>>>>>>>>>>>> enumeration isn't of all machines when he build the >>>>>>>>>>>>>>> diagonal argument, as not all machines produce a valid >>>>>>>>>>>>>>> result to put on the diagonal.
i'm sorry bro ur literally a dribbling retard for trying >>>>>>>>>>>>>>>> to question the effectiveness of a total turing machine >>>>>>>>>>>>>>>> enumeration
No "machine" ever needed to do that testing. >>>>>>>>>>>>>>>>>
In fact, he doesn't even need an "effective" >>>>>>>>>>>>>>>>> enumeration of machines, just that an enumeration exists. >>>>>>>>>>>>>>>>
And you are a moron for confusing the effective >>>>>>>>>>>>>>> enumeration of ALL machines, many of which don't meet the >>>>>>>>>>>>>>> requirements to be put in the list, and the enumeration >>>>>>>>>>>>>>> of the machines that DO produce a "computable number" >>>>>>>>>>>>>>>
The method to pair down the enumeration of all machines >>>>>>>>>>>>>>> to the machines that compute a computable number is shown >>>>>>>>>>>>>>> to non- computable.
You depending on Unicorns to prepare your enumerations is >>>>>>>>>>>>>>> what breaks your logic.
this isn't math anymore, it's just u randomly putting >>>>>>>>>>>>>>>> words together opposed to mine based on feels i can't >>>>>>>>>>>>>>>> even remotely understand
Yes, that is all YOU are doing, as you don't require your >>>>>>>>>>>>>>> statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM >>>>>>>>>>>>>>>> programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>>>>
And note, it isn't even an enumeration of all Turing >>>>>>>>>>>>>>>>> Machines, but of all machines that compute a number, >>>>>>>>>>>>>>>>> otherwise we need to deal with the possibility that a >>>>>>>>>>>>>>>>> given row isn't "complete" because that machine never >>>>>>>>>>>>>>>>> generates enough numbers.
he only simulates numbers for the diagonal that are >>>>>>>>>>>>>>>> decided by D to be "satisfactory"/circle-free, so he >>>>>>>>>>>>>>>> does enumerate over all machines testing each one for >>>>>>>>>>>>>>>> being "satisfactory" for inclusion on the diagonal or not >>>>>>>>>>>>>>>>
write me psuedocode that accurately represents machine H >>>>>>>>>>>>>>>> as described on p247 of turing's paper /on computable >>>>>>>>>>>>>>>> numbers/, cause honestly u understand the algo he wrote. >>>>>>>>>>>>>>>> if write more than like 15 lines u've fucked up, it's >>>>>>>>>>>>>>>> not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be >>>>>>>>>>>>>>> done, when it can't be.
Note, the paper starts with a "Supposing there is a >>>>>>>>>>>>>>> Machine D", and if that is true, then "We could construct >>>>>>>>>>>>>>> H".
Since he shows we can not have the machine D, there is no >>>>>>>>>>>>>>> machine H to construct.
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>>
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone >>>>>>>>>>>> here can read a fucking paper
Since YOU don't seem to be able to do that, why should I help >>>>>>>>>>> you.
i already did and put it in a paper i have a record of you >>>>>>>>>> reading in december, why are you telling me i can't seem to do >>>>>>>>>> something i already did???
You mean that CRAP that assumes unicorns exist, and was based >>>>>>>>> on category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second
paragraph, where you ADMIT that you aren't actually adressing the >>>>>>> point you claim to be.
ur looking just for excuses to not think at all by this point
No, YOU are looking for someone to validate your LIE.
Your paper begins with a number of categorical errors, so NOTHING
that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are trying to >>>>> debunk.
For instance, a "paradoxical" machine doesn't even try to be
paradoxical to two different deciders, but for your case where you
talk about using two decider to try to beat the paradoxical format, >>>>> you eventually need to combine those two into a single decider to
give the answer.
in the paper i address his diagonals, i talk about the difference
between using a standard decider, and using a "fixed" decider (which
is context-aware)
Which isn't ALLOWED as it uses information not part of the input.
ok mr math police, didn't realize it was up to you what i'm allowed to
do or not
No, it isn't me that says that, it is the rules of the system that
defines what a "decider" is.
All you are doing is proving you don't understand, or don't care about
being wrong.
Sorry, you are just proving you don't understand what you are talking
about.
A "Computation", which a decider is a special case of, is only
ALLOWED to process the input it is given, and its output must be
strictly determined by it.
If it can somwhow generate two (or more) different answers for a
given input, it is BY DEFINITION incorrect, as the problem statement
only gives one correct answer, so giving two different answers is
automatically incorrect.
i never use more than one decider in a diagonal, idk where ur
getting that from
then the paradox input is just the one that you used.
That is your problem, you don't understand that the decider is chosen
FIRST, and THEM we show the input it will fail on.
The "paradoxical" input will be built on THAT final decider, not
the two intermediate deciders.
Your attempts just show you don't understand the nature of what you >>>>> are talking about.
And by the end of the first page, you demonstrate that you don't >>>>>>> understand the basics of the field.
In fact, the fact you asked me to look at your paper, which >>>>>>>>> appears to be on a spammer site that keeps on asking me to
"subscribe" to there service to see how is mentioning my name >>>>>>>>> is part of the cause for some of my attitude.
It seems you are nothing but a stupid scammer that is trying to >>>>>>>>> get people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING
something and showing you understand the meaning behind it, and >>>>>>>>> not just say the world is wrong because it won't give me my >>>>>>>>> unicorns, and the world with unicorns would be so much better. >>>>>>>>>
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of >>>>>>>>>> turing's paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, >>>>>>>> that we can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
Why?
And the algorithm for H will DEPEND on the algorithm for D, and >>>>>>> thus if D doesn't exist, neither does H.
cause if u can't give me a simple 10 line pseudo-code to describe >>>>>> the algorithm used then i'll have to assume ur incapable to
understand either turing's proof, or the various resolutions i've >>>>>> proposed
And you would be wrong as usual.
Of course, being wrong is the one thing you seem to be good at.
instead of actual writing down a 10 line psuedocode, ur spending 20
just talking shit
i don't buy it in the slightest bro
Perhaps you don't understand that I am trying to show you how stupid
your logic is.
why are you writing paragraphs instead of pseudo-code?
i don't buy this shit in the slightest
So, you can't read english?
I don't need to "prove" my ability, because here I am not making
claims that aren't generally accepted. I can rely on the fact that
they are well proven statements.
YOU are the one that needs to show you know something, but the fact
you keep on talking about nonsense, like deciders that either take
the WRONG input (because they need to be given a context that the
question doesn't actually depend on) or change their answer based on
something that isn't the input.
Both of these just prove that your decider can't be correct.
The answer for the behavior of an actual machine doesn't depend on
the context of the machine asking the question, as that doesn't
actually change the behavior of the machine in question. Thus,
changing your answer based on it is just wrong.
And, when you back of and admit you are just doing partial deciding,
you balk at the comment that this is a "solved" problem, there are
LOTS of partial deciders, so you need to show why yours is better, or
at a minimum, nearly as good as, what the current methods produce.
Old Hat results aren't really meaningful or interesting.
in which case idk,
i can buy a gun and shoot myself cause i really have exhausted all >>>>>> my discussion options by now.
MAYBE you can, depending on where you live, they might not consider >>>>> you competent to have one in some places.
everyone is too far ethically gone to have any compassion or
consideration,
No, YOU are the one ethically gone, thinking that fallicaous logic
is valid, and that people "owe" you support.
Clearly you chose to put your "papers" on a site that doesn't
respect normal deciency, but just trying to get people to spend
money for likely no real benifit. How much did you need to pay to
put your paper there?
Then don't.
and i have no desire to participate further in this unholy 🤡🌎 >>>>>
It is clear you can't understand the basics, and have refused to be >>>>> taught, so the best thing to do is to give up and find something
you can do.
When you set your life on trying to reach something unreachable,
and choose to piss of those trying to help you, you really have no
value to society.
All you have done is proved you are too stupid to understand >>>>>>>>> how logic works.
Your problem is you don't actually understand what it says, >>>>>>>>>>> as you think it is all wrong.
YOU are the one asking for help, and then critisizing people >>>>>>>>>>> when they do so.
YOU are the one showing yourself to be just a jerk that >>>>>>>>>>> doesn't know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife >>>>>>>>>>> clinging to your absurd ideas and rid the world of another >>>>>>>>>>> crackpot.
I'm just putting enough information that anyone with a brain >>>>>>>>>>> cell can see that your ideas are baseless, and that if they >>>>>>>>>>> have questions, they can learn the answers from the source. >>>>>>>>>>>
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even >>>>>>>>>> just a pursuit of happiness anymore
Nope, you are just showing that you don't understand what you >>>>>>>>>>> are talking about.
Since each iteration just uses a testing by the machine D, >>>>>>>>>>>>> your psuedo- code loop is just a reference to a operation >>>>>>>>>>>>> that turns out not to exist.
It isn't my job to do the thinking for you.
i really am just getting a bunch retarded kneejerk >>>>>>>>>>>>>> reactions from u eh??? not even a second of deep thot ever??? >>>>>>>>>>>>>
instead, ur spitting out gishgallop that has no thot behind it >>>>>>>>>>>
That is why you are so stupid, you seem to expect others to >>>>>>>>>>>>> do the work you want to do.
Your problem is you assume unicorns exist.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs >>>>>>>>>>>>>>>
there is a largest natural number... ;^) You really >>>>>>>>>>>>>>>>>>> need to implement your interface.
what in the fuck are you going on about??? >>>>>>>>>>>>>>>>>>>>
Expecting people to just hand you that >>>>>>>>>>>>>>>>>>>>>>> information means you never learned how to learn, >>>>>>>>>>>>>>>>>>>>>>> and thus made your self fundamentally ignorant. >>>>>>>>>>>>>>>>>>>>>>>
it's just not a problem that can even effect u >>>>>>>>>>>>>>>>>>>>>>>>>> really
On 2/24/26 2:59 PM, Tristan Wibberley wrote:
On 23/02/2026 15:02, Richard Damon wrote:
On 2/22/26 11:17 PM, dart200 wrote:...
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 12:02 PM, dart200 wrote:
On 2/22/26 4:03 AM, Richard Damon wrote:
No, I mean things like compliers to generate code that goes into >>>>>>> critical systems.
well they're all spitting out AI generated slop these day, and no
general semantic verification tools are in sight...
And THAT is the dangerous stuff.
It seems your arguement is that people shouldn't have such tools.
AI GENERATION IS NOT EQUIVALENT TO AUTOMATED PROOFS???
WHAT ARE YOU SMOKING???
Of course not.
Most AI is NOT "Proof" or even "Reasoning" based.
Most AI is based on maximum likely Markov Chain processing.
What does mostness have to do with it?
Because you don't seem to understand what you are talking about,
You seem to think that AI Generation *IS* equivalent to automated proof, when it isn't, as what is normally described as "AI" doesn't do that.
I guess you still think that words don't need to mean what they mean, because you world is based on the existance of Unicorns.--
Tristan Wibberley <tristan.wibberley+netnews2@alumni.manchester.ac.uk> writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and Ackermann posed
the question of whether first-order logic was decidable -- was there an
algorithm that can determine if a given statement is universally valid. (Given other results this is equivalent to deciding, algorithmically, if
a given statement is provable.)
At the time of his now famous paper he was working with Church who had
turned his attention to this as yet unsolved problem. At the time, most mathematicians thought the answer would be "yes". Of course the first
step is to capture the notion of an algorithm or process. Church came
up with the lambda calculus, and Turing the abstract machine that not
bears his name.
However, the idea that mathematics and specifically proofs, might be automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in formal
logic and the intriguing unsolved problems of the time, one of which was about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research resources.
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
Eh?!an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think
yes that's what diagonal proofs do...
A test is a procedure! You can't test /all/ of an infinitude one by one.
that exactly what turing does in his proof: he defines a comptuation that enumerates out all the numbers, testing each one of they represent a "satisfactory"/"circle-free" machine, and adding that to diagonal across defined across computable numbers
it really would be a great exercise to carefully read p247 of turing's--
proof and produce the psuedo-code for the machine H, assuming that machine
D exists
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
Eh?!yes that's what diagonal proofs do...an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think >>>>
A test is a procedure! You can't test /all/ of an infinitude one by one.
that exactly what turing does in his proof: he defines a comptuation that
enumerates out all the numbers, testing each one of they represent a
"satisfactory"/"circle-free" machine, and adding that to diagonal across
defined across computable numbers
Just for the record (dart200 has no interest in learning this stuff)
that is not at all what Turing does in his 1936 paper. The argument is entirely finite. Nothing infinite is tested. Nothing needs to run for
ever. The paper is available inline and anyone why cares to can go
check for themselves.
it really would be a great exercise to carefully read p247 of turing's
proof and produce the psuedo-code for the machine H, assuming that machine >> D exists
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley <tristan.wibberley+netnews2@alumni.manchester.ac.uk>
writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and Ackermann posed
the question of whether first-order logic was decidable -- was there an
and they all gave up on essentially meaninglessly garbage like
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties all twisted up /in the same way/ because of failure to appropriately handle
a lil' self-referential contradiction
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to kill
myself for being a such a fking 🤡🌎
algorithm that can determine if a given statement is universally valid.
(Given other results this is equivalent to deciding, algorithmically, if
a given statement is provable.)
At the time of his now famous paper he was working with Church who had
turned his attention to this as yet unsolved problem. At the time, most
mathematicians thought the answer would be "yes". Of course the first
step is to capture the notion of an algorithm or process. Church came
up with the lambda calculus, and Turing the abstract machine that not
bears his name.
However, the idea that mathematics and specifically proofs, might be
automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in formal
logic and the intriguing unsolved problems of the time, one of which was
about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research resources.
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 11:16 AM, dart200 wrote:
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:assume D exists and write the algo for H u fucking >>>>>>>>>>>>>>> brainrotted moron... he describes exactly what it does, >>>>>>>>>>>>>>> but put it in a form of psuedo-code
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT >>>>>>>>>>>>>>>>> JUMPING TWAT
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:
On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ]
Oh man, thats rather harsh? >>>>>>>>>>>>>>>>>>>>>>>>>>
i suppose i shouldn't be expecting fucking >>>>>>>>>>>>>>>>>>>>>>>>>>>>> boomer shitposters to ever be any help >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care
If the problem you are trying to solve isn't >>>>>>>>>>>>>>>>>>>>>>>>>>>> the one you say you are, that is the Halting >>>>>>>>>>>>>>>>>>>>>>>>>>>> Problem and / or the concept of problems >>>>>>>>>>>>>>>>>>>>>>>>>>>> being undecidable, then I guess you are just >>>>>>>>>>>>>>>>>>>>>>>>>>>> a pathological liar.
And yes, if that is the case, no one can >>>>>>>>>>>>>>>>>>>>>>>>>>>> help you.
It seems your problem is you can't actually >>>>>>>>>>>>>>>>>>>>>>>>>>>> express the problem you see, perhaps because >>>>>>>>>>>>>>>>>>>>>>>>>>>> you just don't understand what you are >>>>>>>>>>>>>>>>>>>>>>>>>>>> talking about.
ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much >>>>>>>>>>>>>>>>>>>>>>>>> empathy left to really care much about getting >>>>>>>>>>>>>>>>>>>>>>>>> bashed
lest they would begin to understand what all >>>>>>>>>>>>>>>>>>>>>>>>> the intentionally willful ignorance they throw >>>>>>>>>>>>>>>>>>>>>>>>> at me is like
No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing with a >>>>>>>>>>>>>>>>>>>>>>>> problem is to first understand the real nature >>>>>>>>>>>>>>>>>>>>>>>> of the problem and see what people have actually >>>>>>>>>>>>>>>>>>>>>>>> done.
u don't even understand what a basic enumeration >>>>>>>>>>>>>>>>>>>>>>> is ur so dumb as fuck
Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>>>> enumeration needs to be effective.
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>>>>>> proven on turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? Don't >>>>>>>>>>>>>>>>>>>> tell me you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung >>>>>>>>>>>>>>>>>> the list of numbers enumerated by the list of machines. >>>>>>>>>>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His >>>>>>>>>>>>>>>> enumeration isn't of all machines when he build the >>>>>>>>>>>>>>>> diagonal argument, as not all machines produce a valid >>>>>>>>>>>>>>>> result to put on the diagonal.
i'm sorry bro ur literally a dribbling retard for >>>>>>>>>>>>>>>>> trying to question the effectiveness of a total turing >>>>>>>>>>>>>>>>> machine enumeration
No "machine" ever needed to do that testing. >>>>>>>>>>>>>>>>>>
In fact, he doesn't even need an "effective" >>>>>>>>>>>>>>>>>> enumeration of machines, just that an enumeration exists. >>>>>>>>>>>>>>>>>
And you are a moron for confusing the effective >>>>>>>>>>>>>>>> enumeration of ALL machines, many of which don't meet >>>>>>>>>>>>>>>> the requirements to be put in the list, and the >>>>>>>>>>>>>>>> enumeration of the machines that DO produce a >>>>>>>>>>>>>>>> "computable number"
The method to pair down the enumeration of all machines >>>>>>>>>>>>>>>> to the machines that compute a computable number is >>>>>>>>>>>>>>>> shown to non- computable.
You depending on Unicorns to prepare your enumerations >>>>>>>>>>>>>>>> is what breaks your logic.
this isn't math anymore, it's just u randomly putting >>>>>>>>>>>>>>>>> words together opposed to mine based on feels i can't >>>>>>>>>>>>>>>>> even remotely understand
Yes, that is all YOU are doing, as you don't require >>>>>>>>>>>>>>>> your statments to actually be based on FACTS.
honestly i might as well be talking to a freaking LLM >>>>>>>>>>>>>>>>> programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>>>>>
And note, it isn't even an enumeration of all Turing >>>>>>>>>>>>>>>>>> Machines, but of all machines that compute a number, >>>>>>>>>>>>>>>>>> otherwise we need to deal with the possibility that a >>>>>>>>>>>>>>>>>> given row isn't "complete" because that machine never >>>>>>>>>>>>>>>>>> generates enough numbers.
he only simulates numbers for the diagonal that are >>>>>>>>>>>>>>>>> decided by D to be "satisfactory"/circle-free, so he >>>>>>>>>>>>>>>>> does enumerate over all machines testing each one for >>>>>>>>>>>>>>>>> being "satisfactory" for inclusion on the diagonal or not >>>>>>>>>>>>>>>>>
write me psuedocode that accurately represents machine >>>>>>>>>>>>>>>>> H as described on p247 of turing's paper /on computable >>>>>>>>>>>>>>>>> numbers/, cause honestly u understand the algo he >>>>>>>>>>>>>>>>> wrote. if write more than like 15 lines u've fucked up, >>>>>>>>>>>>>>>>> it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be >>>>>>>>>>>>>>>> done, when it can't be.
Note, the paper starts with a "Supposing there is a >>>>>>>>>>>>>>>> Machine D", and if that is true, then "We could >>>>>>>>>>>>>>>> construct H".
Since he shows we can not have the machine D, there is >>>>>>>>>>>>>>>> no machine H to construct.
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>>>
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone >>>>>>>>>>>>> here can read a fucking paper
Since YOU don't seem to be able to do that, why should I >>>>>>>>>>>> help you.
i already did and put it in a paper i have a record of you >>>>>>>>>>> reading in december, why are you telling me i can't seem to >>>>>>>>>>> do something i already did???
You mean that CRAP that assumes unicorns exist, and was based >>>>>>>>>> on category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second
paragraph, where you ADMIT that you aren't actually adressing >>>>>>>> the point you claim to be.
ur looking just for excuses to not think at all by this point
No, YOU are looking for someone to validate your LIE.
Your paper begins with a number of categorical errors, so NOTHING >>>>>> that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are trying
to debunk.
For instance, a "paradoxical" machine doesn't even try to be
paradoxical to two different deciders, but for your case where you >>>>>> talk about using two decider to try to beat the paradoxical
format, you eventually need to combine those two into a single
decider to give the answer.
in the paper i address his diagonals, i talk about the difference
between using a standard decider, and using a "fixed" decider
(which is context-aware)
Which isn't ALLOWED as it uses information not part of the input.
ok mr math police, didn't realize it was up to you what i'm allowed
to do or not
No, it isn't me that says that, it is the rules of the system that
defines what a "decider" is.
what system and who made the rules?
All you are doing is proving you don't understand, or don't care about
being wrong.
Sorry, you are just proving you don't understand what you are
talking about.
A "Computation", which a decider is a special case of, is only
ALLOWED to process the input it is given, and its output must be
strictly determined by it.
If it can somwhow generate two (or more) different answers for a
given input, it is BY DEFINITION incorrect, as the problem statement
only gives one correct answer, so giving two different answers is
automatically incorrect.
i never use more than one decider in a diagonal, idk where ur
getting that from
then the paradox input is just the one that you used.
That is your problem, you don't understand that the decider is
chosen FIRST, and THEM we show the input it will fail on.
The "paradoxical" input will be built on THAT final decider, not
the two intermediate deciders.
Your attempts just show you don't understand the nature of what
you are talking about.
And by the end of the first page, you demonstrate that you don't >>>>>>>> understand the basics of the field.
In fact, the fact you asked me to look at your paper, which >>>>>>>>>> appears to be on a spammer site that keeps on asking me to >>>>>>>>>> "subscribe" to there service to see how is mentioning my name >>>>>>>>>> is part of the cause for some of my attitude.
It seems you are nothing but a stupid scammer that is trying >>>>>>>>>> to get people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING
something and showing you understand the meaning behind it, >>>>>>>>>> and not just say the world is wrong because it won't give me >>>>>>>>>> my unicorns, and the world with unicorns would be so much better. >>>>>>>>>>
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of >>>>>>>>>>> turing's paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, >>>>>>>>> that we can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
Why?
And the algorithm for H will DEPEND on the algorithm for D, and >>>>>>>> thus if D doesn't exist, neither does H.
cause if u can't give me a simple 10 line pseudo-code to describe >>>>>>> the algorithm used then i'll have to assume ur incapable to
understand either turing's proof, or the various resolutions i've >>>>>>> proposed
And you would be wrong as usual.
Of course, being wrong is the one thing you seem to be good at.
instead of actual writing down a 10 line psuedocode, ur spending 20 >>>>> just talking shit
i don't buy it in the slightest bro
Perhaps you don't understand that I am trying to show you how stupid
your logic is.
why are you writing paragraphs instead of pseudo-code?
i don't buy this shit in the slightest
So, you can't read english?
psuedo-code forces to u actually reckon about the gishgallop u keep
putting out, which why i want to see it
I don't need to "prove" my ability, because here I am not making
claims that aren't generally accepted. I can rely on the fact that
they are well proven statements.
YOU are the one that needs to show you know something, but the fact
you keep on talking about nonsense, like deciders that either take
the WRONG input (because they need to be given a context that the
question doesn't actually depend on) or change their answer based on
something that isn't the input.
Both of these just prove that your decider can't be correct.
The answer for the behavior of an actual machine doesn't depend on
the context of the machine asking the question, as that doesn't
actually change the behavior of the machine in question. Thus,
changing your answer based on it is just wrong.
And, when you back of and admit you are just doing partial deciding,
you balk at the comment that this is a "solved" problem, there are
LOTS of partial deciders, so you need to show why yours is better,
or at a minimum, nearly as good as, what the current methods produce.
Old Hat results aren't really meaningful or interesting.
in which case idk,
i can buy a gun and shoot myself cause i really have exhausted
all my discussion options by now.
MAYBE you can, depending on where you live, they might not
consider you competent to have one in some places.
everyone is too far ethically gone to have any compassion or
consideration,
No, YOU are the one ethically gone, thinking that fallicaous logic >>>>>> is valid, and that people "owe" you support.
Clearly you chose to put your "papers" on a site that doesn't
respect normal deciency, but just trying to get people to spend
money for likely no real benifit. How much did you need to pay to >>>>>> put your paper there?
Then don't.
and i have no desire to participate further in this unholy 🤡🌎 >>>>>>
It is clear you can't understand the basics, and have refused to
be taught, so the best thing to do is to give up and find
something you can do.
When you set your life on trying to reach something unreachable,
and choose to piss of those trying to help you, you really have no >>>>>> value to society.
All you have done is proved you are too stupid to understand >>>>>>>>>> how logic works.
Your problem is you don't actually understand what it says, >>>>>>>>>>>> as you think it is all wrong.
YOU are the one asking for help, and then critisizing people >>>>>>>>>>>> when they do so.
YOU are the one showing yourself to be just a jerk that >>>>>>>>>>>> doesn't know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife >>>>>>>>>>>> clinging to your absurd ideas and rid the world of another >>>>>>>>>>>> crackpot.
I'm just putting enough information that anyone with a brain >>>>>>>>>>>> cell can see that your ideas are baseless, and that if they >>>>>>>>>>>> have questions, they can learn the answers from the source. >>>>>>>>>>>>
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even >>>>>>>>>>> just a pursuit of happiness anymore
Nope, you are just showing that you don't understand what >>>>>>>>>>>> you are talking about.
Since each iteration just uses a testing by the machine D, >>>>>>>>>>>>>> your psuedo- code loop is just a reference to a operation >>>>>>>>>>>>>> that turns out not to exist.
i really am just getting a bunch retarded kneejerk >>>>>>>>>>>>>>> reactions from u eh??? not even a second of deep thot >>>>>>>>>>>>>>> ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot behind it >>>>>>>>>>>>
That is why you are so stupid, you seem to expect others >>>>>>>>>>>>>> to do the work you want to do.
Your problem is you assume unicorns exist.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs >>>>>>>>>>>>>>>>
there is a largest natural number... ;^) You really >>>>>>>>>>>>>>>>>>>> need to implement your interface.
what in the fuck are you going on about??? >>>>>>>>>>>>>>>>>>>>>
Expecting people to just hand you that >>>>>>>>>>>>>>>>>>>>>>>> information means you never learned how to >>>>>>>>>>>>>>>>>>>>>>>> learn, and thus made your self fundamentally >>>>>>>>>>>>>>>>>>>>>>>> ignorant.
it's just not a problem that can even effect >>>>>>>>>>>>>>>>>>>>>>>>>>> u really
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you
think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by
one.
that exactly what turing does in his proof: he defines a comptuation
that enumerates out all the numbers, testing each one of they
represent a "satisfactory"/"circle-free" machine, and adding that to
diagonal across defined across computable numbers
it really would be a great exercise to carefully read p247 of
turing's proof and produce the psuedo-code for the machine H,
assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of
computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as they
continue to produce results.
The Halting Problem (vs the computable number problem) is based on
Computations that must return answers in finite time.
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting problem, but
he didn't
Also, it doesn't enumerate the "numbers", but Machines, many of which
will not actually produce numbers.
Then he assumes a test exists that determines if that machine
satisfies the requirements, and shows that there can not be such a test.
Thus, he shows that there is NOT an effective enumeration of
computable numbers, only uncomputable enumerations of them.
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is
computable. It would be true if we could enumerate the computable
sequences by finite means, but the problem of enumerating computable
sequences is equivalent to the problem of finding out whether a given
number is the D.N of a circle-free machine, and we have no general
process for doing this in a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration that
enumerates the numbers, only one of "all machines" that need an
uncomputable test to see if they are in that set.
what i mean by enumerating out numbers is he enumerates out natural
numbers to find possible machines that might compute "computable
numbers" (which are real numbers)
and u still haven't written the pseudo-code
Sorry, you are just showing you don't understand what is being done.
i mean u seem to imply turing talked about the halting problem, which he didn't
On 2/24/26 4:41 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
Eh?!an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you
think
yes that's what diagonal proofs do...
A test is a procedure! You can't test /all/ of an infinitude one by
one.
that exactly what turing does in his proof: he defines a comptuation
that
enumerates out all the numbers, testing each one of they represent a
"satisfactory"/"circle-free" machine, and adding that to diagonal across >>> defined across computable numbers
Just for the record (dart200 has no interest in learning this stuff)
that is not at all what Turing does in his 1936 paper. The argument is
entirely finite. Nothing infinite is tested. Nothing needs to run for
ever. The paper is available inline and anyone why cares to can go
check for themselves.
wow you taught this shit for years and ur actually *that* much of a retard??? god damn. what in the fuck is this fucking 🤡🌎???
seriously stfu and post pseudo-code for his machine H defined on p247 of
his 1936 paper
this exact page is here:
https://www.cs.virginia.edu/~robins/Turing_Paper_1936.pdf#page=18
it really would be a great exercise to carefully read p247 of turing's
proof and produce the psuedo-code for the machine H, assuming that
machine
D exists
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 2:59 PM, Tristan Wibberley wrote:
On 23/02/2026 15:02, Richard Damon wrote:
On 2/22/26 11:17 PM, dart200 wrote:...
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 12:02 PM, dart200 wrote:
On 2/22/26 4:03 AM, Richard Damon wrote:
No, I mean things like compliers to generate code that goes into >>>>>>>> critical systems.
well they're all spitting out AI generated slop these day, and no >>>>>>> general semantic verification tools are in sight...
And THAT is the dangerous stuff.
It seems your arguement is that people shouldn't have such tools.
AI GENERATION IS NOT EQUIVALENT TO AUTOMATED PROOFS???
WHAT ARE YOU SMOKING???
Of course not.
Most AI is NOT "Proof" or even "Reasoning" based.
Most AI is based on maximum likely Markov Chain processing.
What does mostness have to do with it?
Because you don't seem to understand what you are talking about,
You seem to think that AI Generation *IS* equivalent to automated
proof, when it isn't, as what is normally described as "AI" doesn't do
that.
at least we agree on that
oh wait maybe u'd like to tell me how i'm wrong for agreeing with you
ehh rick???
I guess you still think that words don't need to mean what they mean,
because you world is based on the existance of Unicorns.
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley <tristan.wibberley+netnews2@alumni.manchester.ac.uk>
writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and Ackermann posed >>> the question of whether first-order logic was decidable -- was there an
and they all gave up on essentially meaninglessly garbage like
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties all
twisted up /in the same way/ because of failure to appropriately
handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing Formulation.
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to kill
myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually think what
it would do).
algorithm that can determine if a given statement is universally valid.
(Given other results this is equivalent to deciding, algorithmically, if >>> a given statement is provable.)
At the time of his now famous paper he was working with Church who had
turned his attention to this as yet unsolved problem. At the time, most >>> mathematicians thought the answer would be "yes". Of course the first
step is to capture the notion of an algorithm or process. Church came
up with the lambda calculus, and Turing the abstract machine that not
bears his name.
However, the idea that mathematics and specifically proofs, might be
automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in formal
logic and the intriguing unsolved problems of the time, one of which was >>> about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research resources.
On 2/24/26 7:28 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 11:16 AM, dart200 wrote:
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:assume D exists and write the algo for H u fucking >>>>>>>>>>>>>>>> brainrotted moron... he describes exactly what it does, >>>>>>>>>>>>>>>> but put it in a form of psuedo-code
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 4:08 PM, dart200 wrote:WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT >>>>>>>>>>>>>>>>>> JUMPING TWAT
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ] >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
an effective enumeration of all turing machines >>>>>>>>>>>>>>>>>>>>>> was proven on turing's original paper and can be >>>>>>>>>>>>>>>>>>>>>> reused anywhere...Oh man, thats rather harsh? >>>>>>>>>>>>>>>>>>>>>>>>>>>i suppose i shouldn't be expecting fucking >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> boomer shitposters to ever be any help >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care
If the problem you are trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>> isn't the one you say you are, that is the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halting Problem and / or the concept of >>>>>>>>>>>>>>>>>>>>>>>>>>>>> problems being undecidable, then I guess >>>>>>>>>>>>>>>>>>>>>>>>>>>>> you are just a pathological liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
And yes, if that is the case, no one can >>>>>>>>>>>>>>>>>>>>>>>>>>>>> help you.
It seems your problem is you can't actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>> express the problem you see, perhaps >>>>>>>>>>>>>>>>>>>>>>>>>>>>> because you just don't understand what you >>>>>>>>>>>>>>>>>>>>>>>>>>>>> are talking about.
ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much >>>>>>>>>>>>>>>>>>>>>>>>>> empathy left to really care much about getting >>>>>>>>>>>>>>>>>>>>>>>>>> bashed
lest they would begin to understand what all >>>>>>>>>>>>>>>>>>>>>>>>>> the intentionally willful ignorance they throw >>>>>>>>>>>>>>>>>>>>>>>>>> at me is like
No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing with >>>>>>>>>>>>>>>>>>>>>>>>> a problem is to first understand the real >>>>>>>>>>>>>>>>>>>>>>>>> nature of the problem and see what people have >>>>>>>>>>>>>>>>>>>>>>>>> actually done.
u don't even understand what a basic enumeration >>>>>>>>>>>>>>>>>>>>>>>> is ur so dumb as fuck
Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>>>>> enumeration needs to be effective. >>>>>>>>>>>>>>>>>>>>>>
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do...
NO, IT DOESN'T
It shows that there exists a number that isn't amoung >>>>>>>>>>>>>>>>>>> the list of numbers enumerated by the list of machines. >>>>>>>>>>>>>>>>>>
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His >>>>>>>>>>>>>>>>> enumeration isn't of all machines when he build the >>>>>>>>>>>>>>>>> diagonal argument, as not all machines produce a valid >>>>>>>>>>>>>>>>> result to put on the diagonal.
No "machine" ever needed to do that testing. >>>>>>>>>>>>>>>>>>>
In fact, he doesn't even need an "effective" >>>>>>>>>>>>>>>>>>> enumeration of machines, just that an enumeration >>>>>>>>>>>>>>>>>>> exists.
i'm sorry bro ur literally a dribbling retard for >>>>>>>>>>>>>>>>>> trying to question the effectiveness of a total turing >>>>>>>>>>>>>>>>>> machine enumeration
And you are a moron for confusing the effective >>>>>>>>>>>>>>>>> enumeration of ALL machines, many of which don't meet >>>>>>>>>>>>>>>>> the requirements to be put in the list, and the >>>>>>>>>>>>>>>>> enumeration of the machines that DO produce a >>>>>>>>>>>>>>>>> "computable number"
The method to pair down the enumeration of all machines >>>>>>>>>>>>>>>>> to the machines that compute a computable number is >>>>>>>>>>>>>>>>> shown to non- computable.
You depending on Unicorns to prepare your enumerations >>>>>>>>>>>>>>>>> is what breaks your logic.
this isn't math anymore, it's just u randomly putting >>>>>>>>>>>>>>>>>> words together opposed to mine based on feels i can't >>>>>>>>>>>>>>>>>> even remotely understand
Yes, that is all YOU are doing, as you don't require >>>>>>>>>>>>>>>>> your statments to actually be based on FACTS. >>>>>>>>>>>>>>>>>
honestly i might as well be talking to a freaking LLM >>>>>>>>>>>>>>>>>> programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>>>>>>
And note, it isn't even an enumeration of all Turing >>>>>>>>>>>>>>>>>>> Machines, but of all machines that compute a number, >>>>>>>>>>>>>>>>>>> otherwise we need to deal with the possibility that a >>>>>>>>>>>>>>>>>>> given row isn't "complete" because that machine never >>>>>>>>>>>>>>>>>>> generates enough numbers.
he only simulates numbers for the diagonal that are >>>>>>>>>>>>>>>>>> decided by D to be "satisfactory"/circle-free, so he >>>>>>>>>>>>>>>>>> does enumerate over all machines testing each one for >>>>>>>>>>>>>>>>>> being "satisfactory" for inclusion on the diagonal or not >>>>>>>>>>>>>>>>>>
write me psuedocode that accurately represents machine >>>>>>>>>>>>>>>>>> H as described on p247 of turing's paper /on >>>>>>>>>>>>>>>>>> computable numbers/, cause honestly u understand the >>>>>>>>>>>>>>>>>> algo he wrote. if write more than like 15 lines u've >>>>>>>>>>>>>>>>>> fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be >>>>>>>>>>>>>>>>> done, when it can't be.
Note, the paper starts with a "Supposing there is a >>>>>>>>>>>>>>>>> Machine D", and if that is true, then "We could >>>>>>>>>>>>>>>>> construct H".
Since he shows we can not have the machine D, there is >>>>>>>>>>>>>>>>> no machine H to construct.
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>>>>
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone >>>>>>>>>>>>>> here can read a fucking paper
Since YOU don't seem to be able to do that, why should I >>>>>>>>>>>>> help you.
i already did and put it in a paper i have a record of you >>>>>>>>>>>> reading in december, why are you telling me i can't seem to >>>>>>>>>>>> do something i already did???
You mean that CRAP that assumes unicorns exist, and was based >>>>>>>>>>> on category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second
paragraph, where you ADMIT that you aren't actually adressing >>>>>>>>> the point you claim to be.
ur looking just for excuses to not think at all by this point
No, YOU are looking for someone to validate your LIE.
Your paper begins with a number of categorical errors, so NOTHING >>>>>>> that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are trying >>>>>>> to debunk.
For instance, a "paradoxical" machine doesn't even try to be
paradoxical to two different deciders, but for your case where
you talk about using two decider to try to beat the paradoxical >>>>>>> format, you eventually need to combine those two into a single
decider to give the answer.
in the paper i address his diagonals, i talk about the difference >>>>>> between using a standard decider, and using a "fixed" decider
(which is context-aware)
Which isn't ALLOWED as it uses information not part of the input.
ok mr math police, didn't realize it was up to you what i'm allowed
to do or not
No, it isn't me that says that, it is the rules of the system that
defines what a "decider" is.
what system and who made the rules?
Computation Theory, and Decision Theory, laid down in the early 1900's.
Are you really that dumb that you don't know what you are talking about?
All you are doing is proving you don't understand, or don't care
about being wrong.
Sorry, you are just proving you don't understand what you are
talking about.
A "Computation", which a decider is a special case of, is only
ALLOWED to process the input it is given, and its output must be
strictly determined by it.
If it can somwhow generate two (or more) different answers for a
given input, it is BY DEFINITION incorrect, as the problem
statement only gives one correct answer, so giving two different
answers is automatically incorrect.
i never use more than one decider in a diagonal, idk where ur
getting that from
then the paradox input is just the one that you used.
That is your problem, you don't understand that the decider is
chosen FIRST, and THEM we show the input it will fail on.
The "paradoxical" input will be built on THAT final decider, not >>>>>>> the two intermediate deciders.
Your attempts just show you don't understand the nature of what >>>>>>> you are talking about.
And by the end of the first page, you demonstrate that you
don't understand the basics of the field.
In fact, the fact you asked me to look at your paper, which >>>>>>>>>>> appears to be on a spammer site that keeps on asking me to >>>>>>>>>>> "subscribe" to there service to see how is mentioning my name >>>>>>>>>>> is part of the cause for some of my attitude.
It seems you are nothing but a stupid scammer that is trying >>>>>>>>>>> to get people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING >>>>>>>>>>> something and showing you understand the meaning behind it, >>>>>>>>>>> and not just say the world is wrong because it won't give me >>>>>>>>>>> my unicorns, and the world with unicorns would be so much >>>>>>>>>>> better.
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of >>>>>>>>>>>> turing's paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, >>>>>>>>>> that we can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H
Why?
And the algorithm for H will DEPEND on the algorithm for D, and >>>>>>>>> thus if D doesn't exist, neither does H.
cause if u can't give me a simple 10 line pseudo-code to
describe the algorithm used then i'll have to assume ur
incapable to understand either turing's proof, or the various >>>>>>>> resolutions i've proposed
And you would be wrong as usual.
Of course, being wrong is the one thing you seem to be good at.
instead of actual writing down a 10 line psuedocode, ur spending
20 just talking shit
i don't buy it in the slightest bro
Perhaps you don't understand that I am trying to show you how
stupid your logic is.
why are you writing paragraphs instead of pseudo-code?
i don't buy this shit in the slightest
So, you can't read english?
psuedo-code forces to u actually reckon about the gishgallop u keep
putting out, which why i want to see it
I don't need to "prove" my ability, because here I am not making
claims that aren't generally accepted. I can rely on the fact that
they are well proven statements.
YOU are the one that needs to show you know something, but the fact >>>>> you keep on talking about nonsense, like deciders that either take
the WRONG input (because they need to be given a context that the
question doesn't actually depend on) or change their answer based
on something that isn't the input.
Both of these just prove that your decider can't be correct.
The answer for the behavior of an actual machine doesn't depend on
the context of the machine asking the question, as that doesn't
actually change the behavior of the machine in question. Thus,
changing your answer based on it is just wrong.
And, when you back of and admit you are just doing partial
deciding, you balk at the comment that this is a "solved" problem,
there are LOTS of partial deciders, so you need to show why yours
is better, or at a minimum, nearly as good as, what the current
methods produce.
Old Hat results aren't really meaningful or interesting.
in which case idk,
i can buy a gun and shoot myself cause i really have exhausted >>>>>>>> all my discussion options by now.
MAYBE you can, depending on where you live, they might not
consider you competent to have one in some places.
everyone is too far ethically gone to have any compassion or
consideration,
No, YOU are the one ethically gone, thinking that fallicaous
logic is valid, and that people "owe" you support.
Clearly you chose to put your "papers" on a site that doesn't
respect normal deciency, but just trying to get people to spend >>>>>>> money for likely no real benifit. How much did you need to pay to >>>>>>> put your paper there?
Then don't.
and i have no desire to participate further in this unholy 🤡🌎 >>>>>>>
It is clear you can't understand the basics, and have refused to >>>>>>> be taught, so the best thing to do is to give up and find
something you can do.
When you set your life on trying to reach something unreachable, >>>>>>> and choose to piss of those trying to help you, you really have >>>>>>> no value to society.
All you have done is proved you are too stupid to understand >>>>>>>>>>> how logic works.
Your problem is you don't actually understand what it says, >>>>>>>>>>>>> as you think it is all wrong.
YOU are the one asking for help, and then critisizing >>>>>>>>>>>>> people when they do so.
YOU are the one showing yourself to be just a jerk that >>>>>>>>>>>>> doesn't know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you wife >>>>>>>>>>>>> clinging to your absurd ideas and rid the world of another >>>>>>>>>>>>> crackpot.
I'm just putting enough information that anyone with a >>>>>>>>>>>>> brain cell can see that your ideas are baseless, and that >>>>>>>>>>>>> if they have questions, they can learn the answers from the >>>>>>>>>>>>> source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or even >>>>>>>>>>>> just a pursuit of happiness anymore
Since each iteration just uses a testing by the machine >>>>>>>>>>>>>>> D, your psuedo- code loop is just a reference to a >>>>>>>>>>>>>>> operation that turns out not to exist.
i really am just getting a bunch retarded kneejerk >>>>>>>>>>>>>>>> reactions from u eh??? not even a second of deep thot >>>>>>>>>>>>>>>> ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot >>>>>>>>>>>>>> behind it
Nope, you are just showing that you don't understand what >>>>>>>>>>>>> you are talking about.
That is why you are so stupid, you seem to expect others >>>>>>>>>>>>>>> to do the work you want to do.
Your problem is you assume unicorns exist.
i doubt ben does either
god cs theorists are total fucking pussy wackjobs >>>>>>>>>>>>>>>>>
there is a largest natural number... ;^) You really >>>>>>>>>>>>>>>>>>>>> need to implement your interface.
what in the fuck are you going on about??? >>>>>>>>>>>>>>>>>>>>>>
Expecting people to just hand you that >>>>>>>>>>>>>>>>>>>>>>>>> information means you never learned how to >>>>>>>>>>>>>>>>>>>>>>>>> learn, and thus made your self fundamentally >>>>>>>>>>>>>>>>>>>>>>>>> ignorant.
it's just not a problem that can even effect >>>>>>>>>>>>>>>>>>>>>>>>>>>> u really
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you >>>>>>> think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by >>>>> one.
that exactly what turing does in his proof: he defines a comptuation
that enumerates out all the numbers, testing each one of they
represent a "satisfactory"/"circle-free" machine, and adding that to
diagonal across defined across computable numbers
it really would be a great exercise to carefully read p247 of
turing's proof and produce the psuedo-code for the machine H,
assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of
computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as they
continue to produce results.
The Halting Problem (vs the computable number problem) is based on
Computations that must return answers in finite time.
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting problem,
but he didn't
No, it is relevent because *YOU* talk about the Halting Problem,
claiming you have a better idea to solve it.
Note, its decider needs to be the same sort of machine as machine D in Turings proof.
The decider he proves can't exist, as he proves that some decision
problems are undecidable, thus answering the Entscheidungsproblem in the negative.
And in doing so, he lays the groundwork for proving the Halting Problem,
and breaks the logjam in the thinking that there might be procedures
that answer any question we want.
Also, it doesn't enumerate the "numbers", but Machines, many of which
will not actually produce numbers.
Then he assumes a test exists that determines if that machine
satisfies the requirements, and shows that there can not be such a test. >>>
Thus, he shows that there is NOT an effective enumeration of
computable numbers, only uncomputable enumerations of them.
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is
computable. It would be true if we could enumerate the computable
sequences by finite means, but the problem of enumerating computable
sequences is equivalent to the problem of finding out whether a given
number is the D.N of a circle-free machine, and we have no general
process for doing this in a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration that
enumerates the numbers, only one of "all machines" that need an
uncomputable test to see if they are in that set.
what i mean by enumerating out numbers is he enumerates out natural
numbers to find possible machines that might compute "computable
numbers" (which are real numbers)
But he never does that. He establishes that he can produce every
possible machine in some order, and *IF* you could do the decision test, create the computable numbers in a specified order, but you can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
Sorry, you are just showing you don't understand what is being done.
i mean u seem to imply turing talked about the halting problem, which
he didn't
No, he talked about the decision problem, and showed things are
undecidable, and laid the groundwork that was used to actually handle
the halting problem.
Yes, people say he did it, because he established the method used for it.
On 2/24/26 7:54 PM, dart200 wrote:
On 2/24/26 4:41 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
Eh?!an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you >>>>>>> think
yes that's what diagonal proofs do...
A test is a procedure! You can't test /all/ of an infinitude one by >>>>> one.
that exactly what turing does in his proof: he defines a comptuation
that
enumerates out all the numbers, testing each one of they represent a
"satisfactory"/"circle-free" machine, and adding that to diagonal
across
defined across computable numbers
Just for the record (dart200 has no interest in learning this stuff)
that is not at all what Turing does in his 1936 paper. The argument is >>> entirely finite. Nothing infinite is tested. Nothing needs to run for >>> ever. The paper is available inline and anyone why cares to can go
check for themselves.
wow you taught this shit for years and ur actually *that* much of a
retard??? god damn. what in the fuck is this fucking 🤡🌎???
seriously stfu and post pseudo-code for his machine H defined on p247
of his 1936 paper
this exact page is here:
https://www.cs.virginia.edu/~robins/Turing_Paper_1936.pdf#page=18
it really would be a great exercise to carefully read p247 of turing's >>>> proof and produce the psuedo-code for the machine H, assuming that
machine
D exists
The "Decision" machine is FINITE, and (if it existed) would always
answer in finite time.
The machines being decided on, run forever, not the machines doing the deciding.
It seems YOU are the retard.
On 2/24/26 7:30 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 2:59 PM, Tristan Wibberley wrote:
On 23/02/2026 15:02, Richard Damon wrote:
On 2/22/26 11:17 PM, dart200 wrote:...
On 2/22/26 7:00 PM, Richard Damon wrote:
On 2/22/26 12:02 PM, dart200 wrote:
On 2/22/26 4:03 AM, Richard Damon wrote:
AI GENERATION IS NOT EQUIVALENT TO AUTOMATED PROOFS???No, I mean things like compliers to generate code that goes into >>>>>>>>> critical systems.
well they're all spitting out AI generated slop these day, and no >>>>>>>> general semantic verification tools are in sight...
And THAT is the dangerous stuff.
It seems your arguement is that people shouldn't have such tools. >>>>>>
WHAT ARE YOU SMOKING???
Of course not.
Most AI is NOT "Proof" or even "Reasoning" based.
Most AI is based on maximum likely Markov Chain processing.
What does mostness have to do with it?
Because you don't seem to understand what you are talking about,
You seem to think that AI Generation *IS* equivalent to automated
proof, when it isn't, as what is normally described as "AI" doesn't
do that.
at least we agree on that
oh wait maybe u'd like to tell me how i'm wrong for agreeing with you
ehh rick???
Then why did you put the ??? after your statement?
Note, YOU are the one who mentioned it, and I pointed out that it was dangerous to do so, I never equated it with proving.
I guess you still think that words don't need to mean what they mean,
because you world is based on the existance of Unicorns.
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley <tristan.wibberley+netnews2@alumni.manchester.ac.uk>
writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and Ackermann posed >>> the question of whether first-order logic was decidable -- was there an
and they all gave up on essentially meaninglessly garbage like
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties all
twisted up /in the same way/ because of failure to appropriately
handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing Formulation.
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to kill
myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually think what
it would do).
algorithm that can determine if a given statement is universally valid.
(Given other results this is equivalent to deciding, algorithmically, if >>> a given statement is provable.)
At the time of his now famous paper he was working with Church who had
turned his attention to this as yet unsolved problem. At the time, most >>> mathematicians thought the answer would be "yes". Of course the first
step is to capture the notion of an algorithm or process. Church came
up with the lambda calculus, and Turing the abstract machine that not
bears his name.
However, the idea that mathematics and specifically proofs, might be
automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in formal
logic and the intriguing unsolved problems of the time, one of which was >>> about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research resources.
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley <tristan.wibberley+netnews2@alumni.manchester.ac.uk> >>>> writes:and they all gave up on essentially meaninglessly garbage like
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and Ackermann
posed
the question of whether first-order logic was decidable -- was there an >>>
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties all
twisted up /in the same way/ because of failure to appropriately
handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing Formulation.
i've already quoted this at you
/Now let K be the D.N of H. What does H do in the K-th section of
its motion? It must test whether K is satisfactory/
H is literally testing it's own D.N...
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to kill
myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually think
what it would do).
algorithm that can determine if a given statement is universally valid. >>>> (Given other results this is equivalent to deciding,
algorithmically, if
a given statement is provable.)
At the time of his now famous paper he was working with Church who had >>>> turned his attention to this as yet unsolved problem. At the time,
most
mathematicians thought the answer would be "yes". Of course the first >>>> step is to capture the notion of an algorithm or process. Church came >>>> up with the lambda calculus, and Turing the abstract machine that not
bears his name.
However, the idea that mathematics and specifically proofs, might be
automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in formal
logic and the intriguing unsolved problems of the time, one of which
was
about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research resources.
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:28 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 11:16 AM, dart200 wrote:
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:No, YOU are looking for someone to validate your LIE.
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:assume D exists and write the algo for H u fucking >>>>>>>>>>>>>>>>> brainrotted moron... he describes exactly what it does, >>>>>>>>>>>>>>>>> but put it in a form of psuedo-code
On 2/22/26 7:00 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ] >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>NO, IT DOESN'T
an effective enumeration of all turing machines >>>>>>>>>>>>>>>>>>>>>>> was proven on turing's original paper and can be >>>>>>>>>>>>>>>>>>>>>>> reused anywhere...Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>>>>>> enumeration needs to be effective. >>>>>>>>>>>>>>>>>>>>>>>Oh man, thats rather harsh? >>>>>>>>>>>>>>>>>>>>>>>>>>>>ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>>>>>i suppose i shouldn't be expecting >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> fucking boomer shitposters to ever be any >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> help
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
If the problem you are trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> isn't the one you say you are, that is the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halting Problem and / or the concept of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> problems being undecidable, then I guess >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> you are just a pathological liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
And yes, if that is the case, no one can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> help you.
It seems your problem is you can't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> actually express the problem you see, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> perhaps because you just don't understand >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> what you are talking about. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much >>>>>>>>>>>>>>>>>>>>>>>>>>> empathy left to really care much about >>>>>>>>>>>>>>>>>>>>>>>>>>> getting bashed
lest they would begin to understand what all >>>>>>>>>>>>>>>>>>>>>>>>>>> the intentionally willful ignorance they >>>>>>>>>>>>>>>>>>>>>>>>>>> throw at me is like
No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing with >>>>>>>>>>>>>>>>>>>>>>>>>> a problem is to first understand the real >>>>>>>>>>>>>>>>>>>>>>>>>> nature of the problem and see what people have >>>>>>>>>>>>>>>>>>>>>>>>>> actually done.
u don't even understand what a basic >>>>>>>>>>>>>>>>>>>>>>>>> enumeration is ur so dumb as fuck >>>>>>>>>>>>>>>>>>>>>>>>
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>
It shows that there exists a number that isn't >>>>>>>>>>>>>>>>>>>> amoung the list of numbers enumerated by the list of >>>>>>>>>>>>>>>>>>>> machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT >>>>>>>>>>>>>>>>>>> JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His >>>>>>>>>>>>>>>>>> enumeration isn't of all machines when he build the >>>>>>>>>>>>>>>>>> diagonal argument, as not all machines produce a valid >>>>>>>>>>>>>>>>>> result to put on the diagonal.
No "machine" ever needed to do that testing. >>>>>>>>>>>>>>>>>>>>
In fact, he doesn't even need an "effective" >>>>>>>>>>>>>>>>>>>> enumeration of machines, just that an enumeration >>>>>>>>>>>>>>>>>>>> exists.
i'm sorry bro ur literally a dribbling retard for >>>>>>>>>>>>>>>>>>> trying to question the effectiveness of a total >>>>>>>>>>>>>>>>>>> turing machine enumeration
And you are a moron for confusing the effective >>>>>>>>>>>>>>>>>> enumeration of ALL machines, many of which don't meet >>>>>>>>>>>>>>>>>> the requirements to be put in the list, and the >>>>>>>>>>>>>>>>>> enumeration of the machines that DO produce a >>>>>>>>>>>>>>>>>> "computable number"
The method to pair down the enumeration of all >>>>>>>>>>>>>>>>>> machines to the machines that compute a computable >>>>>>>>>>>>>>>>>> number is shown to non- computable.
You depending on Unicorns to prepare your enumerations >>>>>>>>>>>>>>>>>> is what breaks your logic.
this isn't math anymore, it's just u randomly putting >>>>>>>>>>>>>>>>>>> words together opposed to mine based on feels i can't >>>>>>>>>>>>>>>>>>> even remotely understand
Yes, that is all YOU are doing, as you don't require >>>>>>>>>>>>>>>>>> your statments to actually be based on FACTS. >>>>>>>>>>>>>>>>>>
honestly i might as well be talking to a freaking LLM >>>>>>>>>>>>>>>>>>> programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>>>>>>>
And note, it isn't even an enumeration of all Turing >>>>>>>>>>>>>>>>>>>> Machines, but of all machines that compute a number, >>>>>>>>>>>>>>>>>>>> otherwise we need to deal with the possibility that >>>>>>>>>>>>>>>>>>>> a given row isn't "complete" because that machine >>>>>>>>>>>>>>>>>>>> never generates enough numbers.
he only simulates numbers for the diagonal that are >>>>>>>>>>>>>>>>>>> decided by D to be "satisfactory"/circle-free, so he >>>>>>>>>>>>>>>>>>> does enumerate over all machines testing each one for >>>>>>>>>>>>>>>>>>> being "satisfactory" for inclusion on the diagonal or >>>>>>>>>>>>>>>>>>> not
write me psuedocode that accurately represents >>>>>>>>>>>>>>>>>>> machine H as described on p247 of turing's paper /on >>>>>>>>>>>>>>>>>>> computable numbers/, cause honestly u understand the >>>>>>>>>>>>>>>>>>> algo he wrote. if write more than like 15 lines u've >>>>>>>>>>>>>>>>>>> fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can be >>>>>>>>>>>>>>>>>> done, when it can't be.
Note, the paper starts with a "Supposing there is a >>>>>>>>>>>>>>>>>> Machine D", and if that is true, then "We could >>>>>>>>>>>>>>>>>> construct H".
Since he shows we can not have the machine D, there is >>>>>>>>>>>>>>>>>> no machine H to construct.
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>>>>>
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate someone >>>>>>>>>>>>>>> here can read a fucking paper
Since YOU don't seem to be able to do that, why should I >>>>>>>>>>>>>> help you.
i already did and put it in a paper i have a record of you >>>>>>>>>>>>> reading in december, why are you telling me i can't seem to >>>>>>>>>>>>> do something i already did???
You mean that CRAP that assumes unicorns exist, and was >>>>>>>>>>>> based on category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second >>>>>>>>>> paragraph, where you ADMIT that you aren't actually adressing >>>>>>>>>> the point you claim to be.
ur looking just for excuses to not think at all by this point >>>>>>>>
Your paper begins with a number of categorical errors, so
NOTHING that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are trying >>>>>>>> to debunk.
For instance, a "paradoxical" machine doesn't even try to be
paradoxical to two different deciders, but for your case where >>>>>>>> you talk about using two decider to try to beat the paradoxical >>>>>>>> format, you eventually need to combine those two into a single >>>>>>>> decider to give the answer.
in the paper i address his diagonals, i talk about the difference >>>>>>> between using a standard decider, and using a "fixed" decider
(which is context-aware)
Which isn't ALLOWED as it uses information not part of the input.
ok mr math police, didn't realize it was up to you what i'm allowed >>>>> to do or not
No, it isn't me that says that, it is the rules of the system that
defines what a "decider" is.
what system and who made the rules?
Computation Theory, and Decision Theory, laid down in the early 1900's.
Are you really that dumb that you don't know what you are talking about?
and what if they got "the rules" wrong?
All you are doing is proving you don't understand, or don't care
about being wrong.
Sorry, you are just proving you don't understand what you are
talking about.
A "Computation", which a decider is a special case of, is only
ALLOWED to process the input it is given, and its output must be
strictly determined by it.
If it can somwhow generate two (or more) different answers for a
given input, it is BY DEFINITION incorrect, as the problem
statement only gives one correct answer, so giving two different
answers is automatically incorrect.
i never use more than one decider in a diagonal, idk where ur
getting that from
then the paradox input is just the one that you used.
That is your problem, you don't understand that the decider is
chosen FIRST, and THEM we show the input it will fail on.
instead of actual writing down a 10 line psuedocode, ur spending >>>>>>> 20 just talking shit
The "paradoxical" input will be built on THAT final decider, not >>>>>>>> the two intermediate deciders.
Your attempts just show you don't understand the nature of what >>>>>>>> you are talking about.
And by the end of the first page, you demonstrate that you >>>>>>>>>> don't understand the basics of the field.
Why?
In fact, the fact you asked me to look at your paper, which >>>>>>>>>>>> appears to be on a spammer site that keeps on asking me to >>>>>>>>>>>> "subscribe" to there service to see how is mentioning my >>>>>>>>>>>> name is part of the cause for some of my attitude.
It seems you are nothing but a stupid scammer that is trying >>>>>>>>>>>> to get people to pay to hear your lies.
Prove you have some actual intelegence by actually DOING >>>>>>>>>>>> something and showing you understand the meaning behind it, >>>>>>>>>>>> and not just say the world is wrong because it won't give me >>>>>>>>>>>> my unicorns, and the world with unicorns would be so much >>>>>>>>>>>> better.
i want to see someone else in this group do it:
Why?
assume D exists, what is the pseudo-code for H from p247 of >>>>>>>>>>>>> turing's paper
But it doesn't, so why do it.
first line of p247:
/Let us suppose that there is such a process; that is to say, >>>>>>>>>>> that we can invent a machine D/ [Tur36 p247]
please make that assumption and write the algorithm for H >>>>>>>>>>
And the algorithm for H will DEPEND on the algorithm for D, >>>>>>>>>> and thus if D doesn't exist, neither does H.
cause if u can't give me a simple 10 line pseudo-code to
describe the algorithm used then i'll have to assume ur
incapable to understand either turing's proof, or the various >>>>>>>>> resolutions i've proposed
And you would be wrong as usual.
Of course, being wrong is the one thing you seem to be good at. >>>>>>>
i don't buy it in the slightest bro
Perhaps you don't understand that I am trying to show you how
stupid your logic is.
why are you writing paragraphs instead of pseudo-code?
i don't buy this shit in the slightest
So, you can't read english?
psuedo-code forces to u actually reckon about the gishgallop u keep
putting out, which why i want to see it
I don't need to "prove" my ability, because here I am not making
claims that aren't generally accepted. I can rely on the fact that >>>>>> they are well proven statements.
YOU are the one that needs to show you know something, but the
fact you keep on talking about nonsense, like deciders that either >>>>>> take the WRONG input (because they need to be given a context that >>>>>> the question doesn't actually depend on) or change their answer
based on something that isn't the input.
Both of these just prove that your decider can't be correct.
The answer for the behavior of an actual machine doesn't depend on >>>>>> the context of the machine asking the question, as that doesn't
actually change the behavior of the machine in question. Thus,
changing your answer based on it is just wrong.
And, when you back of and admit you are just doing partial
deciding, you balk at the comment that this is a "solved" problem, >>>>>> there are LOTS of partial deciders, so you need to show why yours >>>>>> is better, or at a minimum, nearly as good as, what the current
methods produce.
Old Hat results aren't really meaningful or interesting.
in which case idk,
i can buy a gun and shoot myself cause i really have exhausted >>>>>>>>> all my discussion options by now.
MAYBE you can, depending on where you live, they might not
consider you competent to have one in some places.
everyone is too far ethically gone to have any compassion or >>>>>>>>> consideration,
No, YOU are the one ethically gone, thinking that fallicaous
logic is valid, and that people "owe" you support.
Clearly you chose to put your "papers" on a site that doesn't >>>>>>>> respect normal deciency, but just trying to get people to spend >>>>>>>> money for likely no real benifit. How much did you need to pay >>>>>>>> to put your paper there?
Then don't.
and i have no desire to participate further in this unholy 🤡🌎 >>>>>>>>
It is clear you can't understand the basics, and have refused to >>>>>>>> be taught, so the best thing to do is to give up and find
something you can do.
When you set your life on trying to reach something unreachable, >>>>>>>> and choose to piss of those trying to help you, you really have >>>>>>>> no value to society.
All you have done is proved you are too stupid to understand >>>>>>>>>>>> how logic works.
Your problem is you don't actually understand what it >>>>>>>>>>>>>> says, as you think it is all wrong.
YOU are the one asking for help, and then critisizing >>>>>>>>>>>>>> people when they do so.
YOU are the one showing yourself to be just a jerk that >>>>>>>>>>>>>> doesn't know his ass from a hole in the ground.
As far as I care, you can just starve yourself and you >>>>>>>>>>>>>> wife clinging to your absurd ideas and rid the world of >>>>>>>>>>>>>> another crackpot.
I'm just putting enough information that anyone with a >>>>>>>>>>>>>> brain cell can see that your ideas are baseless, and that >>>>>>>>>>>>>> if they have questions, they can learn the answers from >>>>>>>>>>>>>> the source.
The world doesn't owe you a living.
clearly the world doesn't even owe me life, liberty, or >>>>>>>>>>>>> even just a pursuit of happiness anymore
Since each iteration just uses a testing by the machine >>>>>>>>>>>>>>>> D, your psuedo- code loop is just a reference to a >>>>>>>>>>>>>>>> operation that turns out not to exist.
i really am just getting a bunch retarded kneejerk >>>>>>>>>>>>>>>>> reactions from u eh??? not even a second of deep thot >>>>>>>>>>>>>>>>> ever???
It isn't my job to do the thinking for you.
instead, ur spitting out gishgallop that has no thot >>>>>>>>>>>>>>> behind it
Nope, you are just showing that you don't understand what >>>>>>>>>>>>>> you are talking about.
That is why you are so stupid, you seem to expect others >>>>>>>>>>>>>>>> to do the work you want to do.
Your problem is you assume unicorns exist. >>>>>>>>>>>>>>>>>>
i doubt ben does either
god cs theorists are total fucking pussy wackjobs >>>>>>>>>>>>>>>>>>
there is a largest natural number... ;^) You >>>>>>>>>>>>>>>>>>>>>> really need to implement your interface. >>>>>>>>>>>>>>>>>>>>>>
what in the fuck are you going on about??? >>>>>>>>>>>>>>>>>>>>>>>
Expecting people to just hand you that >>>>>>>>>>>>>>>>>>>>>>>>>> information means you never learned how to >>>>>>>>>>>>>>>>>>>>>>>>>> learn, and thus made your self fundamentally >>>>>>>>>>>>>>>>>>>>>>>>>> ignorant.
it's just not a problem that can even >>>>>>>>>>>>>>>>>>>>>>>>>>>>> effect u really
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you >>>>>>>> think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one
by one.
that exactly what turing does in his proof: he defines a
comptuation that enumerates out all the numbers, testing each one
of they represent a "satisfactory"/"circle-free" machine, and
adding that to diagonal across defined across computable numbers
it really would be a great exercise to carefully read p247 of
turing's proof and produce the psuedo-code for the machine H,
assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of
computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as they
continue to produce results.
The Halting Problem (vs the computable number problem) is based on
Computations that must return answers in finite time.
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting problem,
but he didn't
No, it is relevent because *YOU* talk about the Halting Problem,
claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting problem and
the satisfactory problem in turing's paper. my paper was on the
satisfactory problem in turing's paper because i wanted to address the source
Note, its decider needs to be the same sort of machine as machine D in
Turings proof.
The decider he proves can't exist, as he proves that some decision
problems are undecidable, thus answering the Entscheidungsproblem in
the negative.
what no one has proven is that the Entscheidungsproblem can't be
answered reliably thru other interfaces...
which is a set of words apparently so damn radical the lame duck
industry of cs "academics" can't even comprehend the concept, let alone address it in a considerate manner
it shocks me on a daily basis just how much abject brainrot i've waded
thru in my explorations
And in doing so, he lays the groundwork for proving the Halting
Problem, and breaks the logjam in the thinking that there might be
procedures that answer any question we want.
Also, it doesn't enumerate the "numbers", but Machines, many of
which will not actually produce numbers.
Then he assumes a test exists that determines if that machine
satisfies the requirements, and shows that there can not be such a
test.
Thus, he shows that there is NOT an effective enumeration of
computable numbers, only uncomputable enumerations of them.
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is
computable. It would be true if we could enumerate the computable
sequences by finite means, but the problem of enumerating computable
sequences is equivalent to the problem of finding out whether a
given number is the D.N of a circle-free machine, and we have no
general process for doing this in a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration that
enumerates the numbers, only one of "all machines" that need an
uncomputable test to see if they are in that set.
what i mean by enumerating out numbers is he enumerates out natural
numbers to find possible machines that might compute "computable
numbers" (which are real numbers)
But he never does that. He establishes that he can produce every
possible machine in some order, and *IF* you could do the decision
test, create the computable numbers in a specified order, but you
can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you produce a correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking so much smack without being able to code up his *really* simple algo
Sorry, you are just showing you don't understand what is being done.
i mean u seem to imply turing talked about the halting problem, which
he didn't
No, he talked about the decision problem, and showed things are
undecidable, and laid the groundwork that was used to actually handle
the halting problem.
Yes, people say he did it, because he established the method used for it.
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:54 PM, dart200 wrote:
On 2/24/26 4:41 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
Eh?!an effective enumeration of all turing machines was proven on >>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you >>>>>>>> think
yes that's what diagonal proofs do...
A test is a procedure! You can't test /all/ of an infinitude one
by one.
that exactly what turing does in his proof: he defines a
comptuation that
enumerates out all the numbers, testing each one of they represent a >>>>> "satisfactory"/"circle-free" machine, and adding that to diagonal
across
defined across computable numbers
Just for the record (dart200 has no interest in learning this stuff)
that is not at all what Turing does in his 1936 paper. The argument is >>>> entirely finite. Nothing infinite is tested. Nothing needs to run for >>>> ever. The paper is available inline and anyone why cares to can go
check for themselves.
wow you taught this shit for years and ur actually *that* much of a
retard??? god damn. what in the fuck is this fucking 🤡🌎???
seriously stfu and post pseudo-code for his machine H defined on p247
of his 1936 paper
this exact page is here:
https://www.cs.virginia.edu/~robins/Turing_Paper_1936.pdf#page=18
it really would be a great exercise to carefully read p247 of turing's >>>>> proof and produce the psuedo-code for the machine H, assuming that
machine
D exists
The "Decision" machine is FINITE, and (if it existed) would always
answer in finite time.
The machines being decided on, run forever, not the machines doing the
deciding.
It seems YOU are the retard.
the *diagonal* machine H he construct is intended to be an infinite
running machines computing the diagonal across the "satisfactory"
computable numbers. he literally describes iterating across all the
integers in his paper, testing each one with D:
/In the first N—1 sections, among other things, the integers 1, 2,..., N — 1 have been written down and tested by the machine D/
what in the fuck are you guys smoking???
On 2/24/26 9:47 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley <tristan.wibberley+netnews2@alumni.manchester.ac.uk> >>>>> writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and Ackermann >>>>> posed
the question of whether first-order logic was decidable -- was
there an
and they all gave up on essentially meaninglessly garbage like
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties
all twisted up /in the same way/ because of failure to appropriately
handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing Formulation.
i've already quoted this at you
/Now let K be the D.N of H. What does H do in the K-th section of
its motion? It must test whether K is satisfactory/
H is literally testing it's own D.N...
But it doesn't have a reference to that. it computed that.
You don't seem to understand what a "reference" is.
Do you think that a compiler can't compile the code for itself?
Yes, a decider that decides on all inputs (as the decision problem
requires) needs to be able to decide about the "value" that it
represents itself.
In part, the undecidability comes out of the fact that the machinery of
the system IS powerful enough that we can convert the machines into
values that can be their inputs.
Since "Turing Complete" machines can do this, any system that can't must
be less powerful than "Turing Complete".
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to kill
myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually think
what it would do).
algorithm that can determine if a given statement is universally
valid.
(Given other results this is equivalent to deciding,
algorithmically, if
a given statement is provable.)
At the time of his now famous paper he was working with Church who had >>>>> turned his attention to this as yet unsolved problem. At the time, >>>>> most
mathematicians thought the answer would be "yes". Of course the first >>>>> step is to capture the notion of an algorithm or process. Church came >>>>> up with the lambda calculus, and Turing the abstract machine that not >>>>> bears his name.
However, the idea that mathematics and specifically proofs, might be >>>>> automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in formal >>>>> logic and the intriguing unsolved problems of the time, one of
which was
about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research resources. >>>>>
On 2/24/26 9:48 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:28 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 11:16 AM, dart200 wrote:
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:ok mr math police, didn't realize it was up to you what i'm
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:No, YOU are looking for someone to validate your LIE.
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote:
On 2/22/26 11:14 PM, dart200 wrote:assume D exists and write the algo for H u fucking >>>>>>>>>>>>>>>>>> brainrotted moron... he describes exactly what it >>>>>>>>>>>>>>>>>> does, but put it in a form of psuedo-code
On 2/22/26 7:00 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:08 PM, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ] >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>NO, IT DOESN'T
an effective enumeration of all turing machines >>>>>>>>>>>>>>>>>>>>>>>> was proven on turing's original paper and can be >>>>>>>>>>>>>>>>>>>>>>>> reused anywhere...Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>>>>>>> enumeration needs to be effective. >>>>>>>>>>>>>>>>>>>>>>>>u don't even understand what a basic >>>>>>>>>>>>>>>>>>>>>>>>>> enumeration is ur so dumb as fuck >>>>>>>>>>>>>>>>>>>>>>>>>Oh man, thats rather harsh? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>>>>>>i suppose i shouldn't be expecting >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> fucking boomer shitposters to ever be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> any help
u don't understand what i'm trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
If the problem you are trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> isn't the one you say you are, that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the Halting Problem and / or the concept >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> of problems being undecidable, then I >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> guess you are just a pathological liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
And yes, if that is the case, no one can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> help you.
It seems your problem is you can't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> actually express the problem you see, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> perhaps because you just don't understand >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> what you are talking about. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much >>>>>>>>>>>>>>>>>>>>>>>>>>>> empathy left to really care much about >>>>>>>>>>>>>>>>>>>>>>>>>>>> getting bashed
lest they would begin to understand what all >>>>>>>>>>>>>>>>>>>>>>>>>>>> the intentionally willful ignorance they >>>>>>>>>>>>>>>>>>>>>>>>>>>> throw at me is like
No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing >>>>>>>>>>>>>>>>>>>>>>>>>>> with a problem is to first understand the >>>>>>>>>>>>>>>>>>>>>>>>>>> real nature of the problem and see what >>>>>>>>>>>>>>>>>>>>>>>>>>> people have actually done. >>>>>>>>>>>>>>>>>>>>>>>>>>
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>
It shows that there exists a number that isn't >>>>>>>>>>>>>>>>>>>>> amoung the list of numbers enumerated by the list >>>>>>>>>>>>>>>>>>>>> of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT >>>>>>>>>>>>>>>>>>>> JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His >>>>>>>>>>>>>>>>>>> enumeration isn't of all machines when he build the >>>>>>>>>>>>>>>>>>> diagonal argument, as not all machines produce a >>>>>>>>>>>>>>>>>>> valid result to put on the diagonal.
No "machine" ever needed to do that testing. >>>>>>>>>>>>>>>>>>>>>
In fact, he doesn't even need an "effective" >>>>>>>>>>>>>>>>>>>>> enumeration of machines, just that an enumeration >>>>>>>>>>>>>>>>>>>>> exists.
i'm sorry bro ur literally a dribbling retard for >>>>>>>>>>>>>>>>>>>> trying to question the effectiveness of a total >>>>>>>>>>>>>>>>>>>> turing machine enumeration
And you are a moron for confusing the effective >>>>>>>>>>>>>>>>>>> enumeration of ALL machines, many of which don't meet >>>>>>>>>>>>>>>>>>> the requirements to be put in the list, and the >>>>>>>>>>>>>>>>>>> enumeration of the machines that DO produce a >>>>>>>>>>>>>>>>>>> "computable number"
The method to pair down the enumeration of all >>>>>>>>>>>>>>>>>>> machines to the machines that compute a computable >>>>>>>>>>>>>>>>>>> number is shown to non- computable.
You depending on Unicorns to prepare your >>>>>>>>>>>>>>>>>>> enumerations is what breaks your logic.
this isn't math anymore, it's just u randomly >>>>>>>>>>>>>>>>>>>> putting words together opposed to mine based on >>>>>>>>>>>>>>>>>>>> feels i can't even remotely understand
Yes, that is all YOU are doing, as you don't require >>>>>>>>>>>>>>>>>>> your statments to actually be based on FACTS. >>>>>>>>>>>>>>>>>>>
honestly i might as well be talking to a freaking >>>>>>>>>>>>>>>>>>>> LLM programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>>>>>>>>
he only simulates numbers for the diagonal that are >>>>>>>>>>>>>>>>>>>> decided by D to be "satisfactory"/circle-free, so he >>>>>>>>>>>>>>>>>>>> does enumerate over all machines testing each one >>>>>>>>>>>>>>>>>>>> for being "satisfactory" for inclusion on the >>>>>>>>>>>>>>>>>>>> diagonal or not
And note, it isn't even an enumeration of all >>>>>>>>>>>>>>>>>>>>> Turing Machines, but of all machines that compute a >>>>>>>>>>>>>>>>>>>>> number, otherwise we need to deal with the >>>>>>>>>>>>>>>>>>>>> possibility that a given row isn't "complete" >>>>>>>>>>>>>>>>>>>>> because that machine never generates enough numbers. >>>>>>>>>>>>>>>>>>>>
write me psuedocode that accurately represents >>>>>>>>>>>>>>>>>>>> machine H as described on p247 of turing's paper /on >>>>>>>>>>>>>>>>>>>> computable numbers/, cause honestly u understand the >>>>>>>>>>>>>>>>>>>> algo he wrote. if write more than like 15 lines u've >>>>>>>>>>>>>>>>>>>> fucked up, it's not a complex algo
You mean the one he shows can't exist?
Because one of the steps we just need to assume can >>>>>>>>>>>>>>>>>>> be done, when it can't be.
Note, the paper starts with a "Supposing there is a >>>>>>>>>>>>>>>>>>> Machine D", and if that is true, then "We could >>>>>>>>>>>>>>>>>>> construct H".
Since he shows we can not have the machine D, there >>>>>>>>>>>>>>>>>>> is no machine H to construct.
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>>>>>>
Why, are you too stupid to do it?
no i want to see u actually do it and demonstrate >>>>>>>>>>>>>>>> someone here can read a fucking paper
Since YOU don't seem to be able to do that, why should I >>>>>>>>>>>>>>> help you.
i already did and put it in a paper i have a record of you >>>>>>>>>>>>>> reading in december, why are you telling me i can't seem >>>>>>>>>>>>>> to do something i already did???
You mean that CRAP that assumes unicorns exist, and was >>>>>>>>>>>>> based on category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second >>>>>>>>>>> paragraph, where you ADMIT that you aren't actually adressing >>>>>>>>>>> the point you claim to be.
ur looking just for excuses to not think at all by this point >>>>>>>>>
Your paper begins with a number of categorical errors, so
NOTHING that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are
trying to debunk.
For instance, a "paradoxical" machine doesn't even try to be >>>>>>>>> paradoxical to two different deciders, but for your case where >>>>>>>>> you talk about using two decider to try to beat the paradoxical >>>>>>>>> format, you eventually need to combine those two into a single >>>>>>>>> decider to give the answer.
in the paper i address his diagonals, i talk about the
difference between using a standard decider, and using a "fixed" >>>>>>>> decider (which is context-aware)
Which isn't ALLOWED as it uses information not part of the input. >>>>>>
allowed to do or not
No, it isn't me that says that, it is the rules of the system that
defines what a "decider" is.
what system and who made the rules?
Computation Theory, and Decision Theory, laid down in the early 1900's.
Are you really that dumb that you don't know what you are talking about?
and what if they got "the rules" wrong?
They can't, as they defined the problem and its rules.
On 2/24/26 9:55 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me >>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one >>>>>>> by one.
that exactly what turing does in his proof: he defines a
comptuation that enumerates out all the numbers, testing each one >>>>>> of they represent a "satisfactory"/"circle-free" machine, and
adding that to diagonal across defined across computable numbers
it really would be a great exercise to carefully read p247 of
turing's proof and produce the psuedo-code for the machine H,
assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of
computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as they
continue to produce results.
The Halting Problem (vs the computable number problem) is based on
Computations that must return answers in finite time.
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting problem,
but he didn't
No, it is relevent because *YOU* talk about the Halting Problem,
claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting problem
and the satisfactory problem in turing's paper. my paper was on the
satisfactory problem in turing's paper because i wanted to address the
source
Then you shouldn't have started (and continued) about talking about the Halting Problem and contray machines.
Note, This paper never (as far as I remember) works by creating a
specific "contrary" or 'paradoxical" machine to decide on. The error
comes naturally out of the problem itself.
Note, its decider needs to be the same sort of machine as machine D
in Turings proof.
The decider he proves can't exist, as he proves that some decision
problems are undecidable, thus answering the Entscheidungsproblem in
the negative.
what no one has proven is that the Entscheidungsproblem can't be
answered reliably thru other interfaces...
WHAT OTHER "INTERFACES"?
He shows a problem that can not be decided. PERIOD.
Your "Interface" seems to be one of your undefinable terms.
which is a set of words apparently so damn radical the lame duck
industry of cs "academics" can't even comprehend the concept, let
alone address it in a considerate manner
No, you don't seem to know what you are talking about.
it shocks me on a daily basis just how much abject brainrot i've waded
thru in my explorations
It seems your brain has completed its rot.
And in doing so, he lays the groundwork for proving the Halting
Problem, and breaks the logjam in the thinking that there might be
procedures that answer any question we want.
Also, it doesn't enumerate the "numbers", but Machines, many of
which will not actually produce numbers.
Then he assumes a test exists that determines if that machine
satisfies the requirements, and shows that there can not be such a
test.
Thus, he shows that there is NOT an effective enumeration of
computable numbers, only uncomputable enumerations of them.
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is
computable. It would be true if we could enumerate the computable
sequences by finite means, but the problem of enumerating
computable sequences is equivalent to the problem of finding out
whether a given number is the D.N of a circle-free machine, and we
have no general process for doing this in a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration that
enumerates the numbers, only one of "all machines" that need an
uncomputable test to see if they are in that set.
what i mean by enumerating out numbers is he enumerates out natural
numbers to find possible machines that might compute "computable
numbers" (which are real numbers)
But he never does that. He establishes that he can produce every
possible machine in some order, and *IF* you could do the decision
test, create the computable numbers in a specified order, but you
can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you produce a
correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking so
much smack without being able to code up his *really* simple algo
Nope, YOU are the one with brain rot.
Sorry, you are just showing you don't understand what is being done.
i mean u seem to imply turing talked about the halting problem,
which he didn't
No, he talked about the decision problem, and showed things are
undecidable, and laid the groundwork that was used to actually handle
the halting problem.
Yes, people say he did it, because he established the method used for
it.
On 2/24/26 9:59 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:54 PM, dart200 wrote:
On 2/24/26 4:41 PM, Ben Bacarisse wrote:
dart200 <user7160@newsgrouper.org.invalid> writes:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
Eh?!an effective enumeration of all turing machines was proven on >>>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me >>>>>>>>> you think
yes that's what diagonal proofs do...
A test is a procedure! You can't test /all/ of an infinitude one >>>>>>> by one.
that exactly what turing does in his proof: he defines a
comptuation that
enumerates out all the numbers, testing each one of they represent a >>>>>> "satisfactory"/"circle-free" machine, and adding that to diagonal >>>>>> across
defined across computable numbers
Just for the record (dart200 has no interest in learning this stuff) >>>>> that is not at all what Turing does in his 1936 paper. The
argument is
entirely finite. Nothing infinite is tested. Nothing needs to run >>>>> for
ever. The paper is available inline and anyone why cares to can go >>>>> check for themselves.
wow you taught this shit for years and ur actually *that* much of a
retard??? god damn. what in the fuck is this fucking 🤡🌎???
seriously stfu and post pseudo-code for his machine H defined on
p247 of his 1936 paper
this exact page is here:
https://www.cs.virginia.edu/~robins/Turing_Paper_1936.pdf#page=18
it really would be a great exercise to carefully read p247 of
turing's
proof and produce the psuedo-code for the machine H, assuming that >>>>>> machine
D exists
The "Decision" machine is FINITE, and (if it existed) would always
answer in finite time.
The machines being decided on, run forever, not the machines doing
the deciding.
It seems YOU are the retard.
the *diagonal* machine H he construct is intended to be an infinite
running machines computing the diagonal across the "satisfactory"
computable numbers. he literally describes iterating across all the
integers in his paper, testing each one with D:
/In the first N—1 sections, among other things, the integers 1, 2,...,
N — 1 have been written down and tested by the machine D/
what in the fuck are you guys smoking???
But the DECIDER is D, not H. H is just the machine that proves that D
can't do its job. H is sort of the equivalent to the paradoxical machine
of the Halting Problem proof. What ever answer D gives when given the
S.D of H will be wrong.
I guess you just don't understand what "deciding" means.
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:47 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley
<tristan.wibberley+netnews2@alumni.manchester.ac.uk>
writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable >>>>>>>I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and Ackermann >>>>>> posed
the question of whether first-order logic was decidable -- was
there an
and they all gave up on essentially meaninglessly garbage like
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties
all twisted up /in the same way/ because of failure to
appropriately handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing
Formulation.
i've already quoted this at you
/Now let K be the D.N of H. What does H do in the K-th section of
its motion? It must test whether K is satisfactory/
H is literally testing it's own D.N...
But it doesn't have a reference to that. it computed that.
You don't seem to understand what a "reference" is.
it has an addressable copy of it's source code on the tape, encoded into
a description number, that it tests
that's the only kind of self-reference that exists in turing machines,
and it always involves some steps in the computation to setup, unless
it's directly set as the input ... but either way it's a reference to itself, and the reference to itself is keystone in undecidability proofs
Do you think that a compiler can't compile the code for itself?
Yes, a decider that decides on all inputs (as the decision problem
requires) needs to be able to decide about the "value" that it
represents itself.
In part, the undecidability comes out of the fact that the machinery
of the system IS powerful enough that we can convert the machines into
values that can be their inputs.
Since "Turing Complete" machines can do this, any system that can't
must be less powerful than "Turing Complete".
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to
kill myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually think
what it would do).
algorithm that can determine if a given statement is universally
valid.
(Given other results this is equivalent to deciding,
algorithmically, if
a given statement is provable.)
At the time of his now famous paper he was working with Church who >>>>>> had
turned his attention to this as yet unsolved problem. At the
time, most
mathematicians thought the answer would be "yes". Of course the >>>>>> first
step is to capture the notion of an algorithm or process. Church >>>>>> came
up with the lambda calculus, and Turing the abstract machine that not >>>>>> bears his name.
However, the idea that mathematics and specifically proofs, might be >>>>>> automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in formal >>>>>> logic and the intriguing unsolved problems of the time, one of
which was
about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research resources. >>>>>>
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:48 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:28 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 11:16 AM, dart200 wrote:
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:ok mr math police, didn't realize it was up to you what i'm
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:No, YOU are looking for someone to validate your LIE.
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 2/22/26 11:14 PM, dart200 wrote:Why, are you too stupid to do it?
assume D exists and write the algo for H u fucking >>>>>>>>>>>>>>>>>>> brainrotted moron... he describes exactly what it >>>>>>>>>>>>>>>>>>> does, but put it in a form of psuedo-code >>>>>>>>>>>>>>>>>>On 2/22/26 7:00 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:08 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ] >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
NO, IT DOESN'Tan effective enumeration of all turing machines >>>>>>>>>>>>>>>>>>>>>>>>> was proven on turing's original paper and can >>>>>>>>>>>>>>>>>>>>>>>>> be reused anywhere...Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>>>>>>>> enumeration needs to be effective. >>>>>>>>>>>>>>>>>>>>>>>>>u don't even understand what a basic >>>>>>>>>>>>>>>>>>>>>>>>>>> enumeration is ur so dumb as fuck >>>>>>>>>>>>>>>>>>>>>>>>>>No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing >>>>>>>>>>>>>>>>>>>>>>>>>>>> with a problem is to first understand the >>>>>>>>>>>>>>>>>>>>>>>>>>>> real nature of the problem and see what >>>>>>>>>>>>>>>>>>>>>>>>>>>> people have actually done. >>>>>>>>>>>>>>>>>>>>>>>>>>>Oh man, thats rather harsh? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>i suppose i shouldn't be expecting >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> fucking boomer shitposters to ever be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> any help
u don't understand what i'm trying to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> solve
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
If the problem you are trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> isn't the one you say you are, that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the Halting Problem and / or the concept >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> of problems being undecidable, then I >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> guess you are just a pathological liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
And yes, if that is the case, no one can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> help you.
It seems your problem is you can't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> actually express the problem you see, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> perhaps because you just don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand what you are talking about. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have much >>>>>>>>>>>>>>>>>>>>>>>>>>>>> empathy left to really care much about >>>>>>>>>>>>>>>>>>>>>>>>>>>>> getting bashed
lest they would begin to understand what >>>>>>>>>>>>>>>>>>>>>>>>>>>>> all the intentionally willful ignorance >>>>>>>>>>>>>>>>>>>>>>>>>>>>> they throw at me is like >>>>>>>>>>>>>>>>>>>>>>>>>>>>
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>
It shows that there exists a number that isn't >>>>>>>>>>>>>>>>>>>>>> amoung the list of numbers enumerated by the list >>>>>>>>>>>>>>>>>>>>>> of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT >>>>>>>>>>>>>>>>>>>>> JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His >>>>>>>>>>>>>>>>>>>> enumeration isn't of all machines when he build the >>>>>>>>>>>>>>>>>>>> diagonal argument, as not all machines produce a >>>>>>>>>>>>>>>>>>>> valid result to put on the diagonal.
No "machine" ever needed to do that testing. >>>>>>>>>>>>>>>>>>>>>>
In fact, he doesn't even need an "effective" >>>>>>>>>>>>>>>>>>>>>> enumeration of machines, just that an enumeration >>>>>>>>>>>>>>>>>>>>>> exists.
i'm sorry bro ur literally a dribbling retard for >>>>>>>>>>>>>>>>>>>>> trying to question the effectiveness of a total >>>>>>>>>>>>>>>>>>>>> turing machine enumeration
And you are a moron for confusing the effective >>>>>>>>>>>>>>>>>>>> enumeration of ALL machines, many of which don't >>>>>>>>>>>>>>>>>>>> meet the requirements to be put in the list, and the >>>>>>>>>>>>>>>>>>>> enumeration of the machines that DO produce a >>>>>>>>>>>>>>>>>>>> "computable number"
The method to pair down the enumeration of all >>>>>>>>>>>>>>>>>>>> machines to the machines that compute a computable >>>>>>>>>>>>>>>>>>>> number is shown to non- computable.
You depending on Unicorns to prepare your >>>>>>>>>>>>>>>>>>>> enumerations is what breaks your logic. >>>>>>>>>>>>>>>>>>>>
Yes, that is all YOU are doing, as you don't require >>>>>>>>>>>>>>>>>>>> your statments to actually be based on FACTS. >>>>>>>>>>>>>>>>>>>>
this isn't math anymore, it's just u randomly >>>>>>>>>>>>>>>>>>>>> putting words together opposed to mine based on >>>>>>>>>>>>>>>>>>>>> feels i can't even remotely understand >>>>>>>>>>>>>>>>>>>>
honestly i might as well be talking to a freaking >>>>>>>>>>>>>>>>>>>>> LLM programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>>>>>>>>>
You mean the one he shows can't exist? >>>>>>>>>>>>>>>>>>>>
he only simulates numbers for the diagonal that are >>>>>>>>>>>>>>>>>>>>> decided by D to be "satisfactory"/circle-free, so >>>>>>>>>>>>>>>>>>>>> he does enumerate over all machines testing each >>>>>>>>>>>>>>>>>>>>> one for being "satisfactory" for inclusion on the >>>>>>>>>>>>>>>>>>>>> diagonal or not
And note, it isn't even an enumeration of all >>>>>>>>>>>>>>>>>>>>>> Turing Machines, but of all machines that compute >>>>>>>>>>>>>>>>>>>>>> a number, otherwise we need to deal with the >>>>>>>>>>>>>>>>>>>>>> possibility that a given row isn't "complete" >>>>>>>>>>>>>>>>>>>>>> because that machine never generates enough numbers. >>>>>>>>>>>>>>>>>>>>>
write me psuedocode that accurately represents >>>>>>>>>>>>>>>>>>>>> machine H as described on p247 of turing's paper / >>>>>>>>>>>>>>>>>>>>> on computable numbers/, cause honestly u understand >>>>>>>>>>>>>>>>>>>>> the algo he wrote. if write more than like 15 lines >>>>>>>>>>>>>>>>>>>>> u've fucked up, it's not a complex algo >>>>>>>>>>>>>>>>>>>>
Because one of the steps we just need to assume can >>>>>>>>>>>>>>>>>>>> be done, when it can't be.
Note, the paper starts with a "Supposing there is a >>>>>>>>>>>>>>>>>>>> Machine D", and if that is true, then "We could >>>>>>>>>>>>>>>>>>>> construct H".
Since he shows we can not have the machine D, there >>>>>>>>>>>>>>>>>>>> is no machine H to construct.
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>>>>>>>
no i want to see u actually do it and demonstrate >>>>>>>>>>>>>>>>> someone here can read a fucking paper
Since YOU don't seem to be able to do that, why should I >>>>>>>>>>>>>>>> help you.
i already did and put it in a paper i have a record of >>>>>>>>>>>>>>> you reading in december, why are you telling me i can't >>>>>>>>>>>>>>> seem to do something i already did???
You mean that CRAP that assumes unicorns exist, and was >>>>>>>>>>>>>> based on category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second >>>>>>>>>>>> paragraph, where you ADMIT that you aren't actually
adressing the point you claim to be.
ur looking just for excuses to not think at all by this point >>>>>>>>>>
Your paper begins with a number of categorical errors, so >>>>>>>>>> NOTHING that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are >>>>>>>>>> trying to debunk.
For instance, a "paradoxical" machine doesn't even try to be >>>>>>>>>> paradoxical to two different deciders, but for your case where >>>>>>>>>> you talk about using two decider to try to beat the
paradoxical format, you eventually need to combine those two >>>>>>>>>> into a single decider to give the answer.
in the paper i address his diagonals, i talk about the
difference between using a standard decider, and using a
"fixed" decider (which is context-aware)
Which isn't ALLOWED as it uses information not part of the input. >>>>>>>
allowed to do or not
No, it isn't me that says that, it is the rules of the system that >>>>>> defines what a "decider" is.
what system and who made the rules?
Computation Theory, and Decision Theory, laid down in the early 1900's. >>>>
Are you really that dumb that you don't know what you are talking
about?
and what if they got "the rules" wrong?
They can't, as they defined the problem and its rules.
wow, didn't realize cs theory came from infallible gods...
i mean, the mere fact ur talking like that is kind of very much
religious discussion, not mathematical, and lends itself to u being irrationally biased
which seems to be the result of my conversation with u
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:55 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me >>>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one >>>>>>>> by one.
that exactly what turing does in his proof: he defines a
comptuation that enumerates out all the numbers, testing each one >>>>>>> of they represent a "satisfactory"/"circle-free" machine, and
adding that to diagonal across defined across computable numbers >>>>>>>
it really would be a great exercise to carefully read p247 of
turing's proof and produce the psuedo-code for the machine H,
assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of
computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as they
continue to produce results.
The Halting Problem (vs the computable number problem) is based on >>>>>> Computations that must return answers in finite time.
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting problem, >>>>> but he didn't
No, it is relevent because *YOU* talk about the Halting Problem,
claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting problem
and the satisfactory problem in turing's paper. my paper was on the
satisfactory problem in turing's paper because i wanted to address
the source
Then you shouldn't have started (and continued) about talking about
the Halting Problem and contray machines.
Note, This paper never (as far as I remember) works by creating a
specific "contrary" or 'paradoxical" machine to decide on. The error
comes naturally out of the problem itself.
if u had read my paper (not just like glanced at it), u'd know there is
a method to construct the diagonal computation such that it avoids
stumbling on deciding on itself,
but turing's purpose was the construct a contradiction, not figure out
what a working diagonal computation would look like
Note, its decider needs to be the same sort of machine as machine D
in Turings proof.
The decider he proves can't exist, as he proves that some decision
problems are undecidable, thus answering the Entscheidungsproblem in
the negative.
what no one has proven is that the Entscheidungsproblem can't be
answered reliably thru other interfaces...
WHAT OTHER "INTERFACES"?
i've described a variety of them thus far
He shows a problem that can not be decided. PERIOD.
Your "Interface" seems to be one of your undefinable terms.
i've defined it many times to u
which is a set of words apparently so damn radical the lame duck
industry of cs "academics" can't even comprehend the concept, let
alone address it in a considerate manner
No, you don't seem to know what you are talking about.
it shocks me on a daily basis just how much abject brainrot i've
waded thru in my explorations
It seems your brain has completed its rot.
And in doing so, he lays the groundwork for proving the Halting
Problem, and breaks the logjam in the thinking that there might be
procedures that answer any question we want.
Also, it doesn't enumerate the "numbers", but Machines, many of
which will not actually produce numbers.
Then he assumes a test exists that determines if that machine
satisfies the requirements, and shows that there can not be such a >>>>>> test.
Thus, he shows that there is NOT an effective enumeration of
computable numbers, only uncomputable enumerations of them.
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is
computable. It would be true if we could enumerate the computable >>>>>> sequences by finite means, but the problem of enumerating
computable sequences is equivalent to the problem of finding out
whether a given number is the D.N of a circle-free machine, and we >>>>>> have no general process for doing this in a finite number of steps. >>>>>>
Thus, what Turing Proves is that there ISN'T an enumeration that
enumerates the numbers, only one of "all machines" that need an
uncomputable test to see if they are in that set.
what i mean by enumerating out numbers is he enumerates out natural >>>>> numbers to find possible machines that might compute "computable
numbers" (which are real numbers)
But he never does that. He establishes that he can produce every
possible machine in some order, and *IF* you could do the decision
test, create the computable numbers in a specified order, but you
can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you produce
a correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking so
much smack without being able to code up his *really* simple algo
Nope, YOU are the one with brain rot.
no pseudo-code = no convincing
i mean u seem to imply turing talked about the halting problem,
Sorry, you are just showing you don't understand what is being done. >>>>>
which he didn't
No, he talked about the decision problem, and showed things are
undecidable, and laid the groundwork that was used to actually
handle the halting problem.
Yes, people say he did it, because he established the method used
for it.
On 2/24/26 11:06 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:48 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:28 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 11:16 AM, dart200 wrote:
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:ok mr math police, didn't realize it was up to you what i'm
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:No, YOU are looking for someone to validate your LIE.
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote:
On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 2/22/26 11:14 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/22/26 7:00 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:08 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:Why, are you too stupid to do it?
assume D exists and write the algo for H u fucking >>>>>>>>>>>>>>>>>>>> brainrotted moron... he describes exactly what it >>>>>>>>>>>>>>>>>>>> does, but put it in a form of psuedo-code >>>>>>>>>>>>>>>>>>>NO, IT DOESN'TYou think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>>>>> Don't tell me you thinkan effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on turing's original paper >>>>>>>>>>>>>>>>>>>>>>>>>> and can be reused anywhere... >>>>>>>>>>>>>>>>>>>>>>>>>Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>>>>>>>>> enumeration needs to be effective. >>>>>>>>>>>>>>>>>>>>>>>>>>u don't even understand what a basic >>>>>>>>>>>>>>>>>>>>>>>>>>>> enumeration is ur so dumb as fuck >>>>>>>>>>>>>>>>>>>>>>>>>>>No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing >>>>>>>>>>>>>>>>>>>>>>>>>>>>> with a problem is to first understand the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> real nature of the problem and see what >>>>>>>>>>>>>>>>>>>>>>>>>>>>> people have actually done. >>>>>>>>>>>>>>>>>>>>>>>>>>>>On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ] >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Oh man, thats rather harsh? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>i suppose i shouldn't be expecting >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> fucking boomer shitposters to ever be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> any help
u don't understand what i'm trying to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> solve
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
If the problem you are trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> isn't the one you say you are, that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the Halting Problem and / or the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> concept of problems being undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> then I guess you are just a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> pathological liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
And yes, if that is the case, no one >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> can help you. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
It seems your problem is you can't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> actually express the problem you see, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> perhaps because you just don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand what you are talking about. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> much empathy left to really care much >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> about getting bashed >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
lest they would begin to understand what >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> all the intentionally willful ignorance >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> they throw at me is like >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>
It shows that there exists a number that isn't >>>>>>>>>>>>>>>>>>>>>>> amoung the list of numbers enumerated by the list >>>>>>>>>>>>>>>>>>>>>>> of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U SUBJECT >>>>>>>>>>>>>>>>>>>>>> JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. His >>>>>>>>>>>>>>>>>>>>> enumeration isn't of all machines when he build the >>>>>>>>>>>>>>>>>>>>> diagonal argument, as not all machines produce a >>>>>>>>>>>>>>>>>>>>> valid result to put on the diagonal. >>>>>>>>>>>>>>>>>>>>>
No "machine" ever needed to do that testing. >>>>>>>>>>>>>>>>>>>>>>>
In fact, he doesn't even need an "effective" >>>>>>>>>>>>>>>>>>>>>>> enumeration of machines, just that an enumeration >>>>>>>>>>>>>>>>>>>>>>> exists.
i'm sorry bro ur literally a dribbling retard for >>>>>>>>>>>>>>>>>>>>>> trying to question the effectiveness of a total >>>>>>>>>>>>>>>>>>>>>> turing machine enumeration
And you are a moron for confusing the effective >>>>>>>>>>>>>>>>>>>>> enumeration of ALL machines, many of which don't >>>>>>>>>>>>>>>>>>>>> meet the requirements to be put in the list, and >>>>>>>>>>>>>>>>>>>>> the enumeration of the machines that DO produce a >>>>>>>>>>>>>>>>>>>>> "computable number"
The method to pair down the enumeration of all >>>>>>>>>>>>>>>>>>>>> machines to the machines that compute a computable >>>>>>>>>>>>>>>>>>>>> number is shown to non- computable.
You depending on Unicorns to prepare your >>>>>>>>>>>>>>>>>>>>> enumerations is what breaks your logic. >>>>>>>>>>>>>>>>>>>>>
Yes, that is all YOU are doing, as you don't >>>>>>>>>>>>>>>>>>>>> require your statments to actually be based on FACTS. >>>>>>>>>>>>>>>>>>>>>
this isn't math anymore, it's just u randomly >>>>>>>>>>>>>>>>>>>>>> putting words together opposed to mine based on >>>>>>>>>>>>>>>>>>>>>> feels i can't even remotely understand >>>>>>>>>>>>>>>>>>>>>
honestly i might as well be talking to a freaking >>>>>>>>>>>>>>>>>>>>>> LLM programmed to be contrarian
At least that wouldn't be wasting smart people's time. >>>>>>>>>>>>>>>>>>>>>
You mean the one he shows can't exist? >>>>>>>>>>>>>>>>>>>>>
he only simulates numbers for the diagonal that >>>>>>>>>>>>>>>>>>>>>> are decided by D to be "satisfactory"/circle-free, >>>>>>>>>>>>>>>>>>>>>> so he does enumerate over all machines testing >>>>>>>>>>>>>>>>>>>>>> each one for being "satisfactory" for inclusion on >>>>>>>>>>>>>>>>>>>>>> the diagonal or not
And note, it isn't even an enumeration of all >>>>>>>>>>>>>>>>>>>>>>> Turing Machines, but of all machines that compute >>>>>>>>>>>>>>>>>>>>>>> a number, otherwise we need to deal with the >>>>>>>>>>>>>>>>>>>>>>> possibility that a given row isn't "complete" >>>>>>>>>>>>>>>>>>>>>>> because that machine never generates enough numbers. >>>>>>>>>>>>>>>>>>>>>>
write me psuedocode that accurately represents >>>>>>>>>>>>>>>>>>>>>> machine H as described on p247 of turing's paper / >>>>>>>>>>>>>>>>>>>>>> on computable numbers/, cause honestly u >>>>>>>>>>>>>>>>>>>>>> understand the algo he wrote. if write more than >>>>>>>>>>>>>>>>>>>>>> like 15 lines u've fucked up, it's not a complex algo >>>>>>>>>>>>>>>>>>>>>
Because one of the steps we just need to assume can >>>>>>>>>>>>>>>>>>>>> be done, when it can't be.
Note, the paper starts with a "Supposing there is a >>>>>>>>>>>>>>>>>>>>> Machine D", and if that is true, then "We could >>>>>>>>>>>>>>>>>>>>> construct H".
Since he shows we can not have the machine D, there >>>>>>>>>>>>>>>>>>>>> is no machine H to construct.
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>>>>>>>>
no i want to see u actually do it and demonstrate >>>>>>>>>>>>>>>>>> someone here can read a fucking paper
Since YOU don't seem to be able to do that, why should >>>>>>>>>>>>>>>>> I help you.
i already did and put it in a paper i have a record of >>>>>>>>>>>>>>>> you reading in december, why are you telling me i can't >>>>>>>>>>>>>>>> seem to do something i already did???
You mean that CRAP that assumes unicorns exist, and was >>>>>>>>>>>>>>> based on category errors?
yes, assume D exists and write the algorithm for H
Why assume a lie?
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second >>>>>>>>>>>>> paragraph, where you ADMIT that you aren't actually >>>>>>>>>>>>> adressing the point you claim to be.
ur looking just for excuses to not think at all by this point >>>>>>>>>>>
Your paper begins with a number of categorical errors, so >>>>>>>>>>> NOTHING that follows can be assumed to make any sense.
And, you show you don't understand the basc proof you are >>>>>>>>>>> trying to debunk.
For instance, a "paradoxical" machine doesn't even try to be >>>>>>>>>>> paradoxical to two different deciders, but for your case >>>>>>>>>>> where you talk about using two decider to try to beat the >>>>>>>>>>> paradoxical format, you eventually need to combine those two >>>>>>>>>>> into a single decider to give the answer.
in the paper i address his diagonals, i talk about the
difference between using a standard decider, and using a
"fixed" decider (which is context-aware)
Which isn't ALLOWED as it uses information not part of the input. >>>>>>>>
allowed to do or not
No, it isn't me that says that, it is the rules of the system
that defines what a "decider" is.
what system and who made the rules?
Computation Theory, and Decision Theory, laid down in the early
1900's.
Are you really that dumb that you don't know what you are talking
about?
and what if they got "the rules" wrong?
They can't, as they defined the problem and its rules.
wow, didn't realize cs theory came from infallible gods...
i mean, the mere fact ur talking like that is kind of very much
religious discussion, not mathematical, and lends itself to u being
irrationally biased
which seems to be the result of my conversation with u
Which means you don't understand the nature of formal logic system.
The definition of the system *IS* fixed at its creatation and is what it
is.
On 2/24/26 11:13 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:55 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me >>>>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation that enumerates out all the numbers, testing each >>>>>>>> one of they represent a "satisfactory"/"circle-free" machine, >>>>>>>> and adding that to diagonal across defined across computable
numbers
it really would be a great exercise to carefully read p247 of >>>>>>>> turing's proof and produce the psuedo-code for the machine H, >>>>>>>> assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of
computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as they >>>>>>> continue to produce results.
The Halting Problem (vs the computable number problem) is based >>>>>>> on Computations that must return answers in finite time.
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting
problem, but he didn't
No, it is relevent because *YOU* talk about the Halting Problem,
claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting problem
and the satisfactory problem in turing's paper. my paper was on the
satisfactory problem in turing's paper because i wanted to address
the source
Then you shouldn't have started (and continued) about talking about
the Halting Problem and contray machines.
Note, This paper never (as far as I remember) works by creating a
specific "contrary" or 'paradoxical" machine to decide on. The error
comes naturally out of the problem itself.
if u had read my paper (not just like glanced at it), u'd know there
is a method to construct the diagonal computation such that it avoids
stumbling on deciding on itself,
Only by assuming that Unicorns exist.
Since, as I pointed out, you LEFT the system in your first page by
claiming falsehoods, you didn't prove anything in the system you claim
to have been working on.
but turing's purpose was the construct a contradiction, not figure out
what a working diagonal computation would look like
Note, its decider needs to be the same sort of machine as machine D >>>>> in Turings proof.
The decider he proves can't exist, as he proves that some decision
problems are undecidable, thus answering the Entscheidungsproblem
in the negative.
what no one has proven is that the Entscheidungsproblem can't be
answered reliably thru other interfaces...
WHAT OTHER "INTERFACES"?
i've described a variety of them thus far
Really?
Not in the rules of Computation Theory, as your machines aren't machine
as they don't include all their algorithm.
He shows a problem that can not be decided. PERIOD.
Your "Interface" seems to be one of your undefinable terms.
i've defined it many times to u
No, you use the word but not DEFINE it.
It seems you don't know what a DEFINTION is.
which is a set of words apparently so damn radical the lame duck
industry of cs "academics" can't even comprehend the concept, let
alone address it in a considerate manner
No, you don't seem to know what you are talking about.
it shocks me on a daily basis just how much abject brainrot i've
waded thru in my explorations
It seems your brain has completed its rot.
And in doing so, he lays the groundwork for proving the Halting
Problem, and breaks the logjam in the thinking that there might be
procedures that answer any question we want.
Also, it doesn't enumerate the "numbers", but Machines, many of >>>>>>> which will not actually produce numbers.
Then he assumes a test exists that determines if that machine
satisfies the requirements, and shows that there can not be such >>>>>>> a test.
Thus, he shows that there is NOT an effective enumeration of
computable numbers, only uncomputable enumerations of them.
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is
computable. It would be true if we could enumerate the computable >>>>>>> sequences by finite means, but the problem of enumerating
computable sequences is equivalent to the problem of finding out >>>>>>> whether a given number is the D.N of a circle-free machine, and >>>>>>> we have no general process for doing this in a finite number of >>>>>>> steps.
Thus, what Turing Proves is that there ISN'T an enumeration that >>>>>>> enumerates the numbers, only one of "all machines" that need an >>>>>>> uncomputable test to see if they are in that set.
what i mean by enumerating out numbers is he enumerates out
natural numbers to find possible machines that might compute
"computable numbers" (which are real numbers)
But he never does that. He establishes that he can produce every
possible machine in some order, and *IF* you could do the decision
test, create the computable numbers in a specified order, but you
can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you produce
a correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking so
much smack without being able to code up his *really* simple algo
Nope, YOU are the one with brain rot.
no pseudo-code = no convincing
NO DEFINITIONS, nothing shown.
If you want to ignore truth, you just prove your own stupidity,
i mean u seem to imply turing talked about the halting problem,
Sorry, you are just showing you don't understand what is being done. >>>>>>
which he didn't
No, he talked about the decision problem, and showed things are
undecidable, and laid the groundwork that was used to actually
handle the halting problem.
Yes, people say he did it, because he established the method used
for it.
On 2/24/26 11:02 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:47 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley
<tristan.wibberley+netnews2@alumni.manchester.ac.uk>
writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable >>>>>>>>I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and
Ackermann posed
the question of whether first-order logic was decidable -- was
there an
and they all gave up on essentially meaninglessly garbage like
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties >>>>>> all twisted up /in the same way/ because of failure to
appropriately handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing
Formulation.
i've already quoted this at you
/Now let K be the D.N of H. What does H do in the K-th section of
its motion? It must test whether K is satisfactory/
H is literally testing it's own D.N...
But it doesn't have a reference to that. it computed that.
You don't seem to understand what a "reference" is.
it has an addressable copy of it's source code on the tape, encoded
into a description number, that it tests
that's the only kind of self-reference that exists in turing machines,
and it always involves some steps in the computation to setup, unless
it's directly set as the input ... but either way it's a reference to
itself, and the reference to itself is keystone in undecidability proofs
Which isn't a "Reference" in the meaning of the word.
Just shows that Turing Machines CAN'T have "self-references" as defined
in logic.
It is processing an input that just happens to match a description of itself.
Since that is a legal input, no foul.
In other words, to try to prohibit this input, means you can't be as powerful in computing as a Turing Machine.
Thus, this "problem" is ESSENTIAL to the nature of the machines, and
can't be just "fixed".
Do you think that a compiler can't compile the code for itself?
Yes, a decider that decides on all inputs (as the decision problem
requires) needs to be able to decide about the "value" that it
represents itself.
In part, the undecidability comes out of the fact that the machinery
of the system IS powerful enough that we can convert the machines
into values that can be their inputs.
Since "Turing Complete" machines can do this, any system that can't
must be less powerful than "Turing Complete".
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to
kill myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually think >>>>> what it would do).
algorithm that can determine if a given statement is universally >>>>>>> valid.
(Given other results this is equivalent to deciding,
algorithmically, if
a given statement is provable.)
At the time of his now famous paper he was working with Church
who had
turned his attention to this as yet unsolved problem. At the
time, most
mathematicians thought the answer would be "yes". Of course the >>>>>>> first
step is to capture the notion of an algorithm or process. Church >>>>>>> came
up with the lambda calculus, and Turing the abstract machine that >>>>>>> not
bears his name.
However, the idea that mathematics and specifically proofs, might be >>>>>>> automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in formal >>>>>>> logic and the intriguing unsolved problems of the time, one of
which was
about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research
resources.
On 2/24/26 11:02 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:47 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley
<tristan.wibberley+netnews2@alumni.manchester.ac.uk>
writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable >>>>>>>>I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and
Ackermann posed
the question of whether first-order logic was decidable -- was
there an
and they all gave up on essentially meaninglessly garbage like
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties >>>>>> all twisted up /in the same way/ because of failure to
appropriately handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing
Formulation.
i've already quoted this at you
/Now let K be the D.N of H. What does H do in the K-th section of
its motion? It must test whether K is satisfactory/
H is literally testing it's own D.N...
But it doesn't have a reference to that. it computed that.
You don't seem to understand what a "reference" is.
it has an addressable copy of it's source code on the tape, encoded
into a description number, that it tests
that's the only kind of self-reference that exists in turing machines,
and it always involves some steps in the computation to setup, unless
it's directly set as the input ... but either way it's a reference to
itself, and the reference to itself is keystone in undecidability proofs
Which isn't a "Reference" in the meaning of the word.
Just shows that Turing Machines CAN'T have "self-references" as defined
in logic.
It is processing an input that just happens to match a description of itself.
Since that is a legal input, no foul.
In other words, to try to prohibit this input, means you can't be as powerful in computing as a Turing Machine.
Thus, this "problem" is ESSENTIAL to the nature of the machines, and
can't be just "fixed".
Do you think that a compiler can't compile the code for itself?
Yes, a decider that decides on all inputs (as the decision problem
requires) needs to be able to decide about the "value" that it
represents itself.
In part, the undecidability comes out of the fact that the machinery
of the system IS powerful enough that we can convert the machines
into values that can be their inputs.
Since "Turing Complete" machines can do this, any system that can't
must be less powerful than "Turing Complete".
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to
kill myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually think >>>>> what it would do).
algorithm that can determine if a given statement is universally >>>>>>> valid.
(Given other results this is equivalent to deciding,
algorithmically, if
a given statement is provable.)
At the time of his now famous paper he was working with Church
who had
turned his attention to this as yet unsolved problem. At the
time, most
mathematicians thought the answer would be "yes". Of course the >>>>>>> first
step is to capture the notion of an algorithm or process. Church >>>>>>> came
up with the lambda calculus, and Turing the abstract machine that >>>>>>> not
bears his name.
However, the idea that mathematics and specifically proofs, might be >>>>>>> automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in formal >>>>>>> logic and the intriguing unsolved problems of the time, one of
which was
about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability
of enemy computing infrastructure to help direct research
resources.
On 2/24/26 8:29 PM, Richard Damon wrote:
On 2/24/26 11:02 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:47 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley
<tristan.wibberley+netnews2@alumni.manchester.ac.uk>
writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable >>>>>>>>>I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and
Ackermann posed
the question of whether first-order logic was decidable -- was >>>>>>>> there an
and they all gave up on essentially meaninglessly garbage like
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties >>>>>>> all twisted up /in the same way/ because of failure to
appropriately handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing
Formulation.
i've already quoted this at you
/Now let K be the D.N of H. What does H do in the K-th section of
its motion? It must test whether K is satisfactory/
H is literally testing it's own D.N...
But it doesn't have a reference to that. it computed that.
You don't seem to understand what a "reference" is.
it has an addressable copy of it's source code on the tape, encoded
into a description number, that it tests
that's the only kind of self-reference that exists in turing
machines, and it always involves some steps in the computation to
setup, unless it's directly set as the input ... but either way it's
a reference to itself, and the reference to itself is keystone in
undecidability proofs
Which isn't a "Reference" in the meaning of the word.
not worthy of my time to debate semantics over something making up
conflicts that need not exist
it's a self-reference for anyone but someone bent on being contrarian
for the sake of it, and i will keep referring to it like regardless of
ur objections
Just shows that Turing Machines CAN'T have "self-references" as
defined in logic.
It is processing an input that just happens to match a description of
itself.
Since that is a legal input, no foul.
In other words, to try to prohibit this input, means you can't be as
powerful in computing as a Turing Machine.
Thus, this "problem" is ESSENTIAL to the nature of the machines, and
can't be just "fixed".
Do you think that a compiler can't compile the code for itself?
Yes, a decider that decides on all inputs (as the decision problem
requires) needs to be able to decide about the "value" that it
represents itself.
In part, the undecidability comes out of the fact that the machinery
of the system IS powerful enough that we can convert the machines
into values that can be their inputs.
Since "Turing Complete" machines can do this, any system that can't
must be less powerful than "Turing Complete".
even funnier are all the reactions from chucklefucks resolutely >>>>>>> defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to >>>>>>> kill myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually
think what it would do).
algorithm that can determine if a given statement is universally >>>>>>>> valid.
(Given other results this is equivalent to deciding,
algorithmically, if
a given statement is provable.)
At the time of his now famous paper he was working with Church >>>>>>>> who had
turned his attention to this as yet unsolved problem. At the >>>>>>>> time, most
mathematicians thought the answer would be "yes". Of course the >>>>>>>> first
step is to capture the notion of an algorithm or process.
Church came
up with the lambda calculus, and Turing the abstract machine
that not
bears his name.
However, the idea that mathematics and specifically proofs,
might be
automated goes way back to at least Leibniz.
He worked in defence research
Not at the time, no. He was just a PhD student interested in >>>>>>>> formal
logic and the intriguing unsolved problems of the time, one of >>>>>>>> which was
about what can be determined by finite "mechanical" means.
so I expected he was studying the maximum possible capability >>>>>>>>> of enemy computing infrastructure to help direct research
resources.
On 2/24/26 8:29 PM, Richard Damon wrote:
On 2/24/26 11:06 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:48 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:28 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 11:16 AM, dart200 wrote:
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:ok mr math police, didn't realize it was up to you what i'm >>>>>>>>> allowed to do or not
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:No, YOU are looking for someone to validate your LIE.
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:Why assume a lie?
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote:
On 2/23/26 12:47 PM, dart200 wrote:
On 2/23/26 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 2/23/26 11:55 AM, dart200 wrote:
On 2/23/26 7:02 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/22/26 11:14 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 7:00 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:08 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:Why, are you too stupid to do it?
assume D exists and write the algo for H u fucking >>>>>>>>>>>>>>>>>>>>> brainrotted moron... he describes exactly what it >>>>>>>>>>>>>>>>>>>>> does, but put it in a form of psuedo-code >>>>>>>>>>>>>>>>>>>>NO, IT DOESN'TYou think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>>>>>> Don't tell me you thinkan effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on turing's original >>>>>>>>>>>>>>>>>>>>>>>>>>> paper and can be reused anywhere... >>>>>>>>>>>>>>>>>>>>>>>>>>Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>>>>>>>>>> enumeration needs to be effective. >>>>>>>>>>>>>>>>>>>>>>>>>>>u don't even understand what a basic >>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumeration is ur so dumb as fuck >>>>>>>>>>>>>>>>>>>>>>>>>>>>No, it seems that the problem is you don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand that the first step of dealing >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> with a problem is to first understand the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> real nature of the problem and see what >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> people have actually done. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ] >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Oh man, thats rather harsh? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>i suppose i shouldn't be expecting >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> fucking boomer shitposters to ever be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> any help >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
u don't understand what i'm trying to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> solve
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
If the problem you are trying to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> isn't the one you say you are, that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the Halting Problem and / or the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> concept of problems being undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> then I guess you are just a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> pathological liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
And yes, if that is the case, no one >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> can help you. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
It seems your problem is you can't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> actually express the problem you see, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> perhaps because you just don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand what you are talking about. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> much empathy left to really care much >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> about getting bashed >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
lest they would begin to understand what >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> all the intentionally willful ignorance >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> they throw at me is like >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>
It shows that there exists a number that isn't >>>>>>>>>>>>>>>>>>>>>>>> amoung the list of numbers enumerated by the >>>>>>>>>>>>>>>>>>>>>>>> list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U >>>>>>>>>>>>>>>>>>>>>>> SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. >>>>>>>>>>>>>>>>>>>>>> His enumeration isn't of all machines when he >>>>>>>>>>>>>>>>>>>>>> build the diagonal argument, as not all machines >>>>>>>>>>>>>>>>>>>>>> produce a valid result to put on the diagonal. >>>>>>>>>>>>>>>>>>>>>>
No "machine" ever needed to do that testing. >>>>>>>>>>>>>>>>>>>>>>>>
In fact, he doesn't even need an "effective" >>>>>>>>>>>>>>>>>>>>>>>> enumeration of machines, just that an >>>>>>>>>>>>>>>>>>>>>>>> enumeration exists.
i'm sorry bro ur literally a dribbling retard for >>>>>>>>>>>>>>>>>>>>>>> trying to question the effectiveness of a total >>>>>>>>>>>>>>>>>>>>>>> turing machine enumeration
And you are a moron for confusing the effective >>>>>>>>>>>>>>>>>>>>>> enumeration of ALL machines, many of which don't >>>>>>>>>>>>>>>>>>>>>> meet the requirements to be put in the list, and >>>>>>>>>>>>>>>>>>>>>> the enumeration of the machines that DO produce a >>>>>>>>>>>>>>>>>>>>>> "computable number"
The method to pair down the enumeration of all >>>>>>>>>>>>>>>>>>>>>> machines to the machines that compute a computable >>>>>>>>>>>>>>>>>>>>>> number is shown to non- computable. >>>>>>>>>>>>>>>>>>>>>>
You depending on Unicorns to prepare your >>>>>>>>>>>>>>>>>>>>>> enumerations is what breaks your logic. >>>>>>>>>>>>>>>>>>>>>>
Yes, that is all YOU are doing, as you don't >>>>>>>>>>>>>>>>>>>>>> require your statments to actually be based on FACTS. >>>>>>>>>>>>>>>>>>>>>>
this isn't math anymore, it's just u randomly >>>>>>>>>>>>>>>>>>>>>>> putting words together opposed to mine based on >>>>>>>>>>>>>>>>>>>>>>> feels i can't even remotely understand >>>>>>>>>>>>>>>>>>>>>>
honestly i might as well be talking to a freaking >>>>>>>>>>>>>>>>>>>>>>> LLM programmed to be contrarian
At least that wouldn't be wasting smart people's >>>>>>>>>>>>>>>>>>>>>> time.
And note, it isn't even an enumeration of all >>>>>>>>>>>>>>>>>>>>>>>> Turing Machines, but of all machines that >>>>>>>>>>>>>>>>>>>>>>>> compute a number, otherwise we need to deal with >>>>>>>>>>>>>>>>>>>>>>>> the possibility that a given row isn't >>>>>>>>>>>>>>>>>>>>>>>> "complete" because that machine never generates >>>>>>>>>>>>>>>>>>>>>>>> enough numbers.
he only simulates numbers for the diagonal that >>>>>>>>>>>>>>>>>>>>>>> are decided by D to be "satisfactory"/circle- >>>>>>>>>>>>>>>>>>>>>>> free, so he does enumerate over all machines >>>>>>>>>>>>>>>>>>>>>>> testing each one for being "satisfactory" for >>>>>>>>>>>>>>>>>>>>>>> inclusion on the diagonal or not >>>>>>>>>>>>>>>>>>>>>>>
write me psuedocode that accurately represents >>>>>>>>>>>>>>>>>>>>>>> machine H as described on p247 of turing's >>>>>>>>>>>>>>>>>>>>>>> paper / on computable numbers/, cause honestly u >>>>>>>>>>>>>>>>>>>>>>> understand the algo he wrote. if write more than >>>>>>>>>>>>>>>>>>>>>>> like 15 lines u've fucked up, it's not a complex >>>>>>>>>>>>>>>>>>>>>>> algo
You mean the one he shows can't exist? >>>>>>>>>>>>>>>>>>>>>>
Because one of the steps we just need to assume >>>>>>>>>>>>>>>>>>>>>> can be done, when it can't be.
Note, the paper starts with a "Supposing there is >>>>>>>>>>>>>>>>>>>>>> a Machine D", and if that is true, then "We could >>>>>>>>>>>>>>>>>>>>>> construct H".
Since he shows we can not have the machine D, >>>>>>>>>>>>>>>>>>>>>> there is no machine H to construct. >>>>>>>>>>>>>>>>>>>>>>
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>>>>>>>>>
no i want to see u actually do it and demonstrate >>>>>>>>>>>>>>>>>>> someone here can read a fucking paper
Since YOU don't seem to be able to do that, why should >>>>>>>>>>>>>>>>>> I help you.
i already did and put it in a paper i have a record of >>>>>>>>>>>>>>>>> you reading in december, why are you telling me i can't >>>>>>>>>>>>>>>>> seem to do something i already did???
You mean that CRAP that assumes unicorns exist, and was >>>>>>>>>>>>>>>> based on category errors?
yes, assume D exists and write the algorithm for H >>>>>>>>>>>>>>
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic.
Since you break the logic of your own paper by the second >>>>>>>>>>>>>> paragraph, where you ADMIT that you aren't actually >>>>>>>>>>>>>> adressing the point you claim to be.
ur looking just for excuses to not think at all by this point >>>>>>>>>>>>
Your paper begins with a number of categorical errors, so >>>>>>>>>>>> NOTHING that follows can be assumed to make any sense. >>>>>>>>>>>>
And, you show you don't understand the basc proof you are >>>>>>>>>>>> trying to debunk.
For instance, a "paradoxical" machine doesn't even try to be >>>>>>>>>>>> paradoxical to two different deciders, but for your case >>>>>>>>>>>> where you talk about using two decider to try to beat the >>>>>>>>>>>> paradoxical format, you eventually need to combine those two >>>>>>>>>>>> into a single decider to give the answer.
in the paper i address his diagonals, i talk about the
difference between using a standard decider, and using a >>>>>>>>>>> "fixed" decider (which is context-aware)
Which isn't ALLOWED as it uses information not part of the input. >>>>>>>>>
No, it isn't me that says that, it is the rules of the system >>>>>>>> that defines what a "decider" is.
what system and who made the rules?
Computation Theory, and Decision Theory, laid down in the early
1900's.
Are you really that dumb that you don't know what you are talking >>>>>> about?
and what if they got "the rules" wrong?
They can't, as they defined the problem and its rules.
wow, didn't realize cs theory came from infallible gods...
i mean, the mere fact ur talking like that is kind of very much
religious discussion, not mathematical, and lends itself to u being
irrationally biased
which seems to be the result of my conversation with u
Which means you don't understand the nature of formal logic system.
The definition of the system *IS* fixed at its creatation and is what
it is.
lol even the axiom of formal set theory haven't been "fixed at
creation", why would computation be so fixed?
because some dick said so??? lol
you're a terrible debater
On 2/24/26 8:29 PM, Richard Damon wrote:
On 2/24/26 11:13 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:55 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>>> turing's original paper and can be reused anywhere... >>>>>>>>>>>>You think you can test all of them one by one? Don't tell me >>>>>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation that enumerates out all the numbers, testing each >>>>>>>>> one of they represent a "satisfactory"/"circle-free" machine, >>>>>>>>> and adding that to diagonal across defined across computable >>>>>>>>> numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>> turing's proof and produce the psuedo-code for the machine H, >>>>>>>>> assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of
computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as they >>>>>>>> continue to produce results.
The Halting Problem (vs the computable number problem) is based >>>>>>>> on Computations that must return answers in finite time.
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting
problem, but he didn't
No, it is relevent because *YOU* talk about the Halting Problem,
claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting problem >>>>> and the satisfactory problem in turing's paper. my paper was on the >>>>> satisfactory problem in turing's paper because i wanted to address
the source
Then you shouldn't have started (and continued) about talking about
the Halting Problem and contray machines.
Note, This paper never (as far as I remember) works by creating a
specific "contrary" or 'paradoxical" machine to decide on. The error
comes naturally out of the problem itself.
if u had read my paper (not just like glanced at it), u'd know there
is a method to construct the diagonal computation such that it avoids
stumbling on deciding on itself,
Only by assuming that Unicorns exist.
doesn't require a fixed decider to avoid itself, just a quine based self-reference
u ofc don't know what i'm talking about and have demonstrated no
capability to hold a coherent conversation
Since, as I pointed out, you LEFT the system in your first page by
claiming falsehoods, you didn't prove anything in the system you claim
to have been working on.
but turing's purpose was the construct a contradiction, not figure
out what a working diagonal computation would look like
Note, its decider needs to be the same sort of machine as machine >>>>>> D in Turings proof.
The decider he proves can't exist, as he proves that some decision >>>>>> problems are undecidable, thus answering the Entscheidungsproblem >>>>>> in the negative.
what no one has proven is that the Entscheidungsproblem can't be
answered reliably thru other interfaces...
WHAT OTHER "INTERFACES"?
i've described a variety of them thus far
Really?
Not in the rules of Computation Theory, as your machines aren't
machine as they don't include all their algorithm.
u haven't defined the rules of computation theory, u just make random
claims about it whenever the fuck
He shows a problem that can not be decided. PERIOD.
Your "Interface" seems to be one of your undefinable terms.
i've defined it many times to u
No, you use the word but not DEFINE it.
it is the specification that a machine computes, or possibly some meta- specification
but u don't debate honorably dick, u just reject every i say even when i agree
It seems you don't know what a DEFINTION is.
which is a set of words apparently so damn radical the lame duck
industry of cs "academics" can't even comprehend the concept, let
alone address it in a considerate manner
No, you don't seem to know what you are talking about.
it shocks me on a daily basis just how much abject brainrot i've
waded thru in my explorations
It seems your brain has completed its rot.
And in doing so, he lays the groundwork for proving the Halting
Problem, and breaks the logjam in the thinking that there might be >>>>>> procedures that answer any question we want.
Also, it doesn't enumerate the "numbers", but Machines, many of >>>>>>>> which will not actually produce numbers.
Then he assumes a test exists that determines if that machine >>>>>>>> satisfies the requirements, and shows that there can not be such >>>>>>>> a test.
Thus, he shows that there is NOT an effective enumeration of
computable numbers, only uncomputable enumerations of them.
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is >>>>>>>> computable. It would be true if we could enumerate the
computable sequences by finite means, but the problem of
enumerating computable sequences is equivalent to the problem of >>>>>>>> finding out whether a given number is the D.N of a circle-free >>>>>>>> machine, and we have no general process for doing this in a
finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration that >>>>>>>> enumerates the numbers, only one of "all machines" that need an >>>>>>>> uncomputable test to see if they are in that set.
what i mean by enumerating out numbers is he enumerates out
natural numbers to find possible machines that might compute
"computable numbers" (which are real numbers)
But he never does that. He establishes that he can produce every
possible machine in some order, and *IF* you could do the decision >>>>>> test, create the computable numbers in a specified order, but you >>>>>> can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you
produce a correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking so
much smack without being able to code up his *really* simple algo
Nope, YOU are the one with brain rot.
no pseudo-code = no convincing
NO DEFINITIONS, nothing shown.
If you want to ignore truth, you just prove your own stupidity,
still no pseudo-code cause ur a fraud dick
i'm honest not sure why u bother typing responses. ur words fall on deaf ears cause u refuse to write anything beyond shallow insults and random claims over rules that don't actually exist
Sorry, you are just showing you don't understand what is being >>>>>>>> done.
i mean u seem to imply turing talked about the halting problem, >>>>>>> which he didn't
No, he talked about the decision problem, and showed things are
undecidable, and laid the groundwork that was used to actually
handle the halting problem.
Yes, people say he did it, because he established the method used >>>>>> for it.
On 2/25/26 12:32 AM, dart200 wrote:
On 2/24/26 8:29 PM, Richard Damon wrote:
On 2/24/26 11:06 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:48 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:28 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 11:16 AM, dart200 wrote:
On 2/23/26 7:49 PM, Richard Damon wrote:
On 2/23/26 6:44 PM, dart200 wrote:
On 2/23/26 3:02 PM, Richard Damon wrote:
On 2/23/26 5:37 PM, dart200 wrote:
On 2/23/26 12:39 PM, Richard Damon wrote:No, YOU are looking for someone to validate your LIE. >>>>>>>>>>>>>
On 2/23/26 1:55 PM, dart200 wrote:
On 2/23/26 10:42 AM, Richard Damon wrote:Why assume a lie?
On 2/23/26 1:17 PM, dart200 wrote:
On 2/23/26 10:00 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 2/23/26 12:47 PM, dart200 wrote:You mean that CRAP that assumes unicorns exist, and was >>>>>>>>>>>>>>>>> based on category errors?
On 2/23/26 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 2/23/26 11:55 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/23/26 7:02 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 11:14 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 7:00 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:08 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 4:04 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 10:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 3:32 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 12:32 PM, Chris M. Thomasson >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
Why, are you too stupid to do it?assume D exists and write the algo for H u fucking >>>>>>>>>>>>>>>>>>>>>> brainrotted moron... he describes exactly what it >>>>>>>>>>>>>>>>>>>>>> does, but put it in a form of psuedo-code >>>>>>>>>>>>>>>>>>>>>NO, IT DOESN'Tyes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>You think you can test all of them one by >>>>>>>>>>>>>>>>>>>>>>>>>>> one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on turing's original >>>>>>>>>>>>>>>>>>>>>>>>>>>> paper and can be reused anywhere... >>>>>>>>>>>>>>>>>>>>>>>>>>>Sure I do. You don't understand when your >>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumeration needs to be effective. >>>>>>>>>>>>>>>>>>>>>>>>>>>>u don't even understand what a basic >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumeration is ur so dumb as fuck >>>>>>>>>>>>>>>>>>>>>>>>>>>>>No, it seems that the problem is you >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> don't understand that the first step of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dealing with a problem is to first >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand the real nature of the problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and see what people have actually done. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>On 2/21/2026 9:41 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 4:03 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/21/26 2:21 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/20/26 4:39 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > [ ...trash... ] >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Oh man, thats rather harsh? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>ur old and will be dead in a decade or so >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>i suppose i shouldn't be expecting >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> fucking boomer shitposters to ever >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> be any help >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
u don't understand what i'm trying >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to solve >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
or why it's even a problem >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
so u *can't* care >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
If the problem you are trying to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> solve isn't the one you say you are, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that is the Halting Problem and / or >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the concept of problems being >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> undecidable, then I guess you are >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> just a pathological liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
And yes, if that is the case, no one >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> can help you. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
It seems your problem is you can't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> actually express the problem you see, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> perhaps because you just don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand what you are talking about. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
ur all too old and lead addled to have >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> much empathy left to really care much >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> about getting bashed >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
lest they would begin to understand what >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> all the intentionally willful ignorance >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> they throw at me is like >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
It shows that there exists a number that isn't >>>>>>>>>>>>>>>>>>>>>>>>> amoung the list of numbers enumerated by the >>>>>>>>>>>>>>>>>>>>>>>>> list of machines.
WE'RE TALKING ABOUT ENUMERATING MACHINES U >>>>>>>>>>>>>>>>>>>>>>>> SUBJECT JUMPING TWAT
talk about alzheimer's sheesh
No, you are not, or at least your words aren't. >>>>>>>>>>>>>>>>>>>>>>> His enumeration isn't of all machines when he >>>>>>>>>>>>>>>>>>>>>>> build the diagonal argument, as not all machines >>>>>>>>>>>>>>>>>>>>>>> produce a valid result to put on the diagonal. >>>>>>>>>>>>>>>>>>>>>>>
And you are a moron for confusing the effective >>>>>>>>>>>>>>>>>>>>>>> enumeration of ALL machines, many of which don't >>>>>>>>>>>>>>>>>>>>>>> meet the requirements to be put in the list, and >>>>>>>>>>>>>>>>>>>>>>> the enumeration of the machines that DO produce a >>>>>>>>>>>>>>>>>>>>>>> "computable number"
No "machine" ever needed to do that testing. >>>>>>>>>>>>>>>>>>>>>>>>>
In fact, he doesn't even need an "effective" >>>>>>>>>>>>>>>>>>>>>>>>> enumeration of machines, just that an >>>>>>>>>>>>>>>>>>>>>>>>> enumeration exists.
i'm sorry bro ur literally a dribbling retard >>>>>>>>>>>>>>>>>>>>>>>> for trying to question the effectiveness of a >>>>>>>>>>>>>>>>>>>>>>>> total turing machine enumeration >>>>>>>>>>>>>>>>>>>>>>>
The method to pair down the enumeration of all >>>>>>>>>>>>>>>>>>>>>>> machines to the machines that compute a >>>>>>>>>>>>>>>>>>>>>>> computable number is shown to non- computable. >>>>>>>>>>>>>>>>>>>>>>>
You depending on Unicorns to prepare your >>>>>>>>>>>>>>>>>>>>>>> enumerations is what breaks your logic. >>>>>>>>>>>>>>>>>>>>>>>
Yes, that is all YOU are doing, as you don't >>>>>>>>>>>>>>>>>>>>>>> require your statments to actually be based on >>>>>>>>>>>>>>>>>>>>>>> FACTS.
this isn't math anymore, it's just u randomly >>>>>>>>>>>>>>>>>>>>>>>> putting words together opposed to mine based on >>>>>>>>>>>>>>>>>>>>>>>> feels i can't even remotely understand >>>>>>>>>>>>>>>>>>>>>>>
At least that wouldn't be wasting smart people's >>>>>>>>>>>>>>>>>>>>>>> time.
honestly i might as well be talking to a >>>>>>>>>>>>>>>>>>>>>>>> freaking LLM programmed to be contrarian >>>>>>>>>>>>>>>>>>>>>>>
And note, it isn't even an enumeration of all >>>>>>>>>>>>>>>>>>>>>>>>> Turing Machines, but of all machines that >>>>>>>>>>>>>>>>>>>>>>>>> compute a number, otherwise we need to deal >>>>>>>>>>>>>>>>>>>>>>>>> with the possibility that a given row isn't >>>>>>>>>>>>>>>>>>>>>>>>> "complete" because that machine never generates >>>>>>>>>>>>>>>>>>>>>>>>> enough numbers.
he only simulates numbers for the diagonal that >>>>>>>>>>>>>>>>>>>>>>>> are decided by D to be "satisfactory"/circle- >>>>>>>>>>>>>>>>>>>>>>>> free, so he does enumerate over all machines >>>>>>>>>>>>>>>>>>>>>>>> testing each one for being "satisfactory" for >>>>>>>>>>>>>>>>>>>>>>>> inclusion on the diagonal or not >>>>>>>>>>>>>>>>>>>>>>>>
write me psuedocode that accurately represents >>>>>>>>>>>>>>>>>>>>>>>> machine H as described on p247 of turing's >>>>>>>>>>>>>>>>>>>>>>>> paper / on computable numbers/, cause honestly u >>>>>>>>>>>>>>>>>>>>>>>> understand the algo he wrote. if write more than >>>>>>>>>>>>>>>>>>>>>>>> like 15 lines u've fucked up, it's not a complex >>>>>>>>>>>>>>>>>>>>>>>> algo
You mean the one he shows can't exist? >>>>>>>>>>>>>>>>>>>>>>>
Because one of the steps we just need to assume >>>>>>>>>>>>>>>>>>>>>>> can be done, when it can't be.
Note, the paper starts with a "Supposing there is >>>>>>>>>>>>>>>>>>>>>>> a Machine D", and if that is true, then "We could >>>>>>>>>>>>>>>>>>>>>>> construct H".
Since he shows we can not have the machine D, >>>>>>>>>>>>>>>>>>>>>>> there is no machine H to construct. >>>>>>>>>>>>>>>>>>>>>>>
So, you questions is about the anatomy of a Unicorn. >>>>>>>>>>>>>>>>>>>>>>
no i want to see u actually do it and demonstrate >>>>>>>>>>>>>>>>>>>> someone here can read a fucking paper
Since YOU don't seem to be able to do that, why >>>>>>>>>>>>>>>>>>> should I help you.
i already did and put it in a paper i have a record of >>>>>>>>>>>>>>>>>> you reading in december, why are you telling me i >>>>>>>>>>>>>>>>>> can't seem to do something i already did??? >>>>>>>>>>>>>>>>>
yes, assume D exists and write the algorithm for H >>>>>>>>>>>>>>>
And, how do you know how much of it I read?
apparently not even the 3rd page
I stop when I hit a critical error in the logic. >>>>>>>>>>>>>>>
Since you break the logic of your own paper by the second >>>>>>>>>>>>>>> paragraph, where you ADMIT that you aren't actually >>>>>>>>>>>>>>> adressing the point you claim to be.
ur looking just for excuses to not think at all by this point >>>>>>>>>>>>>
Your paper begins with a number of categorical errors, so >>>>>>>>>>>>> NOTHING that follows can be assumed to make any sense. >>>>>>>>>>>>>
And, you show you don't understand the basc proof you are >>>>>>>>>>>>> trying to debunk.
For instance, a "paradoxical" machine doesn't even try to >>>>>>>>>>>>> be paradoxical to two different deciders, but for your case >>>>>>>>>>>>> where you talk about using two decider to try to beat the >>>>>>>>>>>>> paradoxical format, you eventually need to combine those >>>>>>>>>>>>> two into a single decider to give the answer.
in the paper i address his diagonals, i talk about the >>>>>>>>>>>> difference between using a standard decider, and using a >>>>>>>>>>>> "fixed" decider (which is context-aware)
Which isn't ALLOWED as it uses information not part of the >>>>>>>>>>> input.
ok mr math police, didn't realize it was up to you what i'm >>>>>>>>>> allowed to do or not
No, it isn't me that says that, it is the rules of the system >>>>>>>>> that defines what a "decider" is.
what system and who made the rules?
Computation Theory, and Decision Theory, laid down in the early >>>>>>> 1900's.
Are you really that dumb that you don't know what you are talking >>>>>>> about?
and what if they got "the rules" wrong?
They can't, as they defined the problem and its rules.
wow, didn't realize cs theory came from infallible gods...
i mean, the mere fact ur talking like that is kind of very much
religious discussion, not mathematical, and lends itself to u being
irrationally biased
which seems to be the result of my conversation with u
Which means you don't understand the nature of formal logic system.
The definition of the system *IS* fixed at its creatation and is what
it is.
lol even the axiom of formal set theory haven't been "fixed at
creation", why would computation be so fixed?
because some dick said so??? lol
you're a terrible debater
You don't understand.
"Set Theory" as a generic isn't define, except that it is generally
accepted that the term will, for now, default to ZFC.
ZFC as a set theory has fixed and defined.
In the same way, "Computation Theory" as a generic refers to the theory
as defined by the early authors in the field, which *IS* defined.
As I said, you just don't understand how Formal Logic works.
If you don't want to play in the existing system, create your own, but
that means you need to do the actual work to do it.
And, your system won't means diddly to the existing system, as it isn't it.--
If you do a good enough job, and it actually is more useful thant the existing, maybe your system can get adopted as the default by the field.
On 2/25/26 12:40 AM, dart200 wrote:
On 2/24/26 8:29 PM, Richard Damon wrote:
On 2/24/26 11:13 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:55 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>>>> turing's original paper and can be reused anywhere... >>>>>>>>>>>>>You think you can test all of them one by one? Don't tell >>>>>>>>>>>>> me you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation that enumerates out all the numbers, testing each >>>>>>>>>> one of they represent a "satisfactory"/"circle-free" machine, >>>>>>>>>> and adding that to diagonal across defined across computable >>>>>>>>>> numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>>> turing's proof and produce the psuedo-code for the machine H, >>>>>>>>>> assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of >>>>>>>>> computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as >>>>>>>>> they continue to produce results.
The Halting Problem (vs the computable number problem) is based >>>>>>>>> on Computations that must return answers in finite time.
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting
problem, but he didn't
No, it is relevent because *YOU* talk about the Halting Problem, >>>>>>> claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting
problem and the satisfactory problem in turing's paper. my paper
was on the satisfactory problem in turing's paper because i wanted >>>>>> to address the source
Then you shouldn't have started (and continued) about talking about >>>>> the Halting Problem and contray machines.
Note, This paper never (as far as I remember) works by creating a
specific "contrary" or 'paradoxical" machine to decide on. The
error comes naturally out of the problem itself.
if u had read my paper (not just like glanced at it), u'd know there
is a method to construct the diagonal computation such that it
avoids stumbling on deciding on itself,
Only by assuming that Unicorns exist.
doesn't require a fixed decider to avoid itself, just a quine based
self-reference
Doesn't work.
So, what answer does that quine based decider give for the
question about itself?
Either answer is still wrong.
And how is being "quine based" keep it from being "fixed".
It still is only one unique machine.
u ofc don't know what i'm talking about and have demonstrated no
capability to hold a coherent conversation
No, YOU keep on making categorical errors not knowing the real meaning
of the terms you are using.
Since, as I pointed out, you LEFT the system in your first page by
claiming falsehoods, you didn't prove anything in the system you
claim to have been working on.
but turing's purpose was the construct a contradiction, not figure
out what a working diagonal computation would look like
Note, its decider needs to be the same sort of machine as machine >>>>>>> D in Turings proof.
The decider he proves can't exist, as he proves that some
decision problems are undecidable, thus answering the
Entscheidungsproblem in the negative.
what no one has proven is that the Entscheidungsproblem can't be
answered reliably thru other interfaces...
WHAT OTHER "INTERFACES"?
i've described a variety of them thus far
Really?
Not in the rules of Computation Theory, as your machines aren't
machine as they don't include all their algorithm.
u haven't defined the rules of computation theory, u just make random
claims about it whenever the fuck
I don't need to do that, as they ARE already defined in the literature.
He shows a problem that can not be decided. PERIOD.
Your "Interface" seems to be one of your undefinable terms.
i've defined it many times to u
No, you use the word but not DEFINE it.
it is the specification that a machine computes, or possibly some
meta- specification
In other words, you are "solving" a problem by solving some other problem?
but u don't debate honorably dick, u just reject every i say even when
i agree
It seems you are just admitting your logic is based on lying,
If I have a classification decision problem, that problem has a specification of indicating which class an input lies in.
What other "specification" can exist?
All you seem to want to do is say that if you don't need to answer the question, you can answer the question.
In other words, you want to make lies be truth.
It seems you don't know what a DEFINTION is.
which is a set of words apparently so damn radical the lame duck
industry of cs "academics" can't even comprehend the concept, let >>>>>> alone address it in a considerate manner
No, you don't seem to know what you are talking about.
it shocks me on a daily basis just how much abject brainrot i've
waded thru in my explorations
It seems your brain has completed its rot.
And in doing so, he lays the groundwork for proving the Halting >>>>>>> Problem, and breaks the logjam in the thinking that there might >>>>>>> be procedures that answer any question we want.
Also, it doesn't enumerate the "numbers", but Machines, many of >>>>>>>>> which will not actually produce numbers.
Then he assumes a test exists that determines if that machine >>>>>>>>> satisfies the requirements, and shows that there can not be >>>>>>>>> such a test.
Thus, he shows that there is NOT an effective enumeration of >>>>>>>>> computable numbers, only uncomputable enumerations of them.
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is >>>>>>>>> computable. It would be true if we could enumerate the
computable sequences by finite means, but the problem of
enumerating computable sequences is equivalent to the problem >>>>>>>>> of finding out whether a given number is the D.N of a circle- >>>>>>>>> free machine, and we have no general process for doing this in >>>>>>>>> a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration >>>>>>>>> that enumerates the numbers, only one of "all machines" that >>>>>>>>> need an uncomputable test to see if they are in that set.
what i mean by enumerating out numbers is he enumerates out
natural numbers to find possible machines that might compute
"computable numbers" (which are real numbers)
But he never does that. He establishes that he can produce every >>>>>>> possible machine in some order, and *IF* you could do the
decision test, create the computable numbers in a specified
order, but you can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you
produce a correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking so >>>>>> much smack without being able to code up his *really* simple algo
Nope, YOU are the one with brain rot.
no pseudo-code = no convincing
NO DEFINITIONS, nothing shown.
If you want to ignore truth, you just prove your own stupidity,
still no pseudo-code cause ur a fraud dick
Still being stupid I see.
i'm honest not sure why u bother typing responses. ur words fall on
deaf ears cause u refuse to write anything beyond shallow insults and
random claims over rules that don't actually exist
Your problem is that you fundamentally don't understand what you are
talking about, as your world seems to be based on the assumption that
things don't need to be what they are, and that lies are valid.
Sorry, you are just showing you don't understand what is being >>>>>>>>> done.
i mean u seem to imply turing talked about the halting problem, >>>>>>>> which he didn't
No, he talked about the decision problem, and showed things are >>>>>>> undecidable, and laid the groundwork that was used to actually
handle the halting problem.
Yes, people say he did it, because he established the method used >>>>>>> for it.
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley <tristan.wibberley+netnews2@alumni.manchester.ac.uk>
writes:
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and Ackermann posed >>> the question of whether first-order logic was decidable -- was there an
and they all gave up on essentially meaninglessly garbage like
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties all
twisted up /in the same way/ because of failure to appropriately
handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing Formulation.
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to kill
myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually think what
it would do).
On 2/24/2026 6:13 PM, Richard Damon wrote:
On 2/24/26 7:39 PM, dart200 wrote:
On 2/24/26 4:09 PM, Ben Bacarisse wrote:
Tristan Wibberley <tristan.wibberley+netnews2@alumni.manchester.ac.uk> >>>> writes:and they all gave up on essentially meaninglessly garbage like
On 20/02/2026 18:11, Ben Bacarisse wrote:
Turing invented TMs to capture the notion of what is computable
I'm curious to know how his motives are known.
At one level there are very clear. In 1928, Hilbert and Ackermann
posed
the question of whether first-order logic was decidable -- was there an >>>
Only meaningless to the stupid.
und = () -> halts(und) && loop()
it's kinda funny actually, all these greats getting their panties all
twisted up /in the same way/ because of failure to appropriately
handle a lil' self-referential contradiction
But the "self-reference" isn't actually there in the Turing Formulation.
even funnier are all the reactions from chucklefucks resolutely
defending that century long failure as MuH uNQuEsTioNAbLe tRuTH
actually it's not that funny, or funny at all. makes me want to kill
myself for being a such a fking 🤡🌎
Mqybe you should if you are that dumb.
(Not really, but if you want to keep bringing it up, actually think
what it would do).
dart is a low life that likes to tell others to snuff themselves out.
Laughs about it. Sigh. I suggest plonking that garbage.
[...]
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by one.
that exactly what turing does in his proof: he defines a comptuation
that enumerates out all the numbers, testing each one of they represent
a "satisfactory"/"circle-free" machine, and adding that to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of turing's
proof and produce the psuedo-code for the machine H, assuming that
machine D exists
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
yes that's what diagonal proofs do...an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you think >>>>
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by one. >>>
that exactly what turing does in his proof: he defines a comptuation
that enumerates out all the numbers, testing each one of they represent
a "satisfactory"/"circle-free" machine, and adding that to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of turing's
proof and produce the psuedo-code for the machine H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he didn't
reason quantified over all but phrase it like a procedure for what he
expected was an inexpert audience, supposing that experts will recognise
the relevant mapping to universal quantification?
On 2/25/26 4:04 AM, Richard Damon wrote:
On 2/25/26 12:40 AM, dart200 wrote:
On 2/24/26 8:29 PM, Richard Damon wrote:
On 2/24/26 11:13 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:55 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was >>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't tell >>>>>>>>>>>>>> me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>> comptuation that enumerates out all the numbers, testing each >>>>>>>>>>> one of they represent a "satisfactory"/"circle-free" machine, >>>>>>>>>>> and adding that to diagonal across defined across computable >>>>>>>>>>> numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>>>> turing's proof and produce the psuedo-code for the machine H, >>>>>>>>>>> assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of >>>>>>>>>> computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as >>>>>>>>>> they continue to produce results.
The Halting Problem (vs the computable number problem) is >>>>>>>>>> based on Computations that must return answers in finite time. >>>>>>>>>>
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting
problem, but he didn't
No, it is relevent because *YOU* talk about the Halting Problem, >>>>>>>> claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting
problem and the satisfactory problem in turing's paper. my paper >>>>>>> was on the satisfactory problem in turing's paper because i
wanted to address the source
Then you shouldn't have started (and continued) about talking
about the Halting Problem and contray machines.
Note, This paper never (as far as I remember) works by creating a >>>>>> specific "contrary" or 'paradoxical" machine to decide on. The
error comes naturally out of the problem itself.
if u had read my paper (not just like glanced at it), u'd know
there is a method to construct the diagonal computation such that
it avoids stumbling on deciding on itself,
Only by assuming that Unicorns exist.
doesn't require a fixed decider to avoid itself, just a quine based
self-reference
Doesn't work.
more humility would do u wonders dick
So, what answer does that quine based decider give for the question
about itself?
it's not the decider that has a quine, it's the diagonal computation H
that uses D that has the quine to recognize H.
when the diagonal computation recognizes it's own number in the total machine enumeration, it never asks the decider on itself, when it
identifies itself in the total machine numeration it adds some fixed
digit to the diagonal instead of recursively getting stuck in simulating itself.
doesn't matter what digit: both are different machines, with different indexes in the enumeration, that compute the diagonal
Either answer is still wrong.
And how is being "quine based" keep it from being "fixed".
It still is only one unique machine.
u ofc don't know what i'm talking about and have demonstrated no
capability to hold a coherent conversation
No, YOU keep on making categorical errors not knowing the real meaning
of the terms you are using.
Since, as I pointed out, you LEFT the system in your first page by
claiming falsehoods, you didn't prove anything in the system you
claim to have been working on.
but turing's purpose was the construct a contradiction, not figure
out what a working diagonal computation would look like
Note, its decider needs to be the same sort of machine as
machine D in Turings proof.
The decider he proves can't exist, as he proves that some
decision problems are undecidable, thus answering the
Entscheidungsproblem in the negative.
what no one has proven is that the Entscheidungsproblem can't be >>>>>>> answered reliably thru other interfaces...
WHAT OTHER "INTERFACES"?
i've described a variety of them thus far
Really?
Not in the rules of Computation Theory, as your machines aren't
machine as they don't include all their algorithm.
u haven't defined the rules of computation theory, u just make random
claims about it whenever the fuck
I don't need to do that, as they ARE already defined in the literature.
and which literature defines "the rules"???
He shows a problem that can not be decided. PERIOD.
Your "Interface" seems to be one of your undefinable terms.
i've defined it many times to u
No, you use the word but not DEFINE it.
it is the specification that a machine computes, or possibly some
meta- specification
In other words, you are "solving" a problem by solving some other
problem?
but u don't debate honorably dick, u just reject every i say even
when i agree
It seems you are just admitting your logic is based on lying,
If I have a classification decision problem, that problem has a
specification of indicating which class an input lies in.
What other "specification" can exist?
All you seem to want to do is say that if you don't need to answer the
question, you can answer the question.
In other words, you want to make lies be truth.
It seems you don't know what a DEFINTION is.
which is a set of words apparently so damn radical the lame duck >>>>>>> industry of cs "academics" can't even comprehend the concept, let >>>>>>> alone address it in a considerate manner
No, you don't seem to know what you are talking about.
it shocks me on a daily basis just how much abject brainrot i've >>>>>>> waded thru in my explorations
It seems your brain has completed its rot.
Nope, YOU are the one with brain rot.
And in doing so, he lays the groundwork for proving the Halting >>>>>>>> Problem, and breaks the logjam in the thinking that there might >>>>>>>> be procedures that answer any question we want.
Also, it doesn't enumerate the "numbers", but Machines, many >>>>>>>>>> of which will not actually produce numbers.
Then he assumes a test exists that determines if that machine >>>>>>>>>> satisfies the requirements, and shows that there can not be >>>>>>>>>> such a test.
Thus, he shows that there is NOT an effective enumeration of >>>>>>>>>> computable numbers, only uncomputable enumerations of them. >>>>>>>>>>
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is >>>>>>>>>> computable. It would be true if we could enumerate the
computable sequences by finite means, but the problem of
enumerating computable sequences is equivalent to the problem >>>>>>>>>> of finding out whether a given number is the D.N of a circle- >>>>>>>>>> free machine, and we have no general process for doing this in >>>>>>>>>> a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration >>>>>>>>>> that enumerates the numbers, only one of "all machines" that >>>>>>>>>> need an uncomputable test to see if they are in that set.
what i mean by enumerating out numbers is he enumerates out >>>>>>>>> natural numbers to find possible machines that might compute >>>>>>>>> "computable numbers" (which are real numbers)
But he never does that. He establishes that he can produce every >>>>>>>> possible machine in some order, and *IF* you could do the
decision test, create the computable numbers in a specified
order, but you can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you
produce a correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking >>>>>>> so much smack without being able to code up his *really* simple algo >>>>>>
no pseudo-code = no convincing
NO DEFINITIONS, nothing shown.
If you want to ignore truth, you just prove your own stupidity,
still no pseudo-code cause ur a fraud dick
Still being stupid I see.
i'm honest not sure why u bother typing responses. ur words fall on
deaf ears cause u refuse to write anything beyond shallow insults and
random claims over rules that don't actually exist
Your problem is that you fundamentally don't understand what you are
talking about, as your world seems to be based on the assumption that
things don't need to be what they are, and that lies are valid.
no pseudo-code = no convincing
i would ask if u care about convincing me,
but clearly u don't
and if u don't care about convincing me,
then ur debate is not honorable, it is engaged with a lack of care, so
ofc it wouldn't be honorable, and that requires care
Sorry, you are just showing you don't understand what is being >>>>>>>>>> done.
i mean u seem to imply turing talked about the halting problem, >>>>>>>>> which he didn't
No, he talked about the decision problem, and showed things are >>>>>>>> undecidable, and laid the groundwork that was used to actually >>>>>>>> handle the halting problem.
Yes, people say he did it, because he established the method
used for it.
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you
think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by
one.
that exactly what turing does in his proof: he defines a comptuation
that enumerates out all the numbers, testing each one of they represent
a "satisfactory"/"circle-free" machine, and adding that to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of turing's
proof and produce the psuedo-code for the machine H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he didn't
reason quantified over all but phrase it like a procedure for what he
the theory of computation is the theory of such procedures, and understanding the diagonal procedure is critical to understanding the
*base* contradiction/paradox that the rest of his support for godel's
result is then built on
expected was an inexpert audience, supposing that experts will recognise
the relevant mapping to universal quantification?
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on
turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you >>>>>>> think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one by >>>>> one.
that exactly what turing does in his proof: he defines a comptuation
that enumerates out all the numbers, testing each one of they represent >>>> a "satisfactory"/"circle-free" machine, and adding that to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of turing's >>>> proof and produce the psuedo-code for the machine H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he didn't
reason quantified over all but phrase it like a procedure for what he
the theory of computation is the theory of such procedures, and
understanding the diagonal procedure is critical to understanding the
*base* contradiction/paradox that the rest of his support for godel's
result is then built on
And focusing on what is said to be impossible and not changing the
problem is also important.
The problem with the diagonal generation isn't the generation of the diagonal itself, but effectively enumerating the enumeration in the
first place.
H shows that *IF* you can make that enumeration, you can make the
diagonal, and thus the anti-diagonal. The problem is you can't make that enumeration, and assuming you can just shows unsoundness.
expected was an inexpert audience, supposing that experts will recognise >>> the relevant mapping to universal quantification?
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me you >>>>>>>> think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one
by one.
that exactly what turing does in his proof: he defines a comptuation >>>>> that enumerates out all the numbers, testing each one of they
represent
a "satisfactory"/"circle-free" machine, and adding that to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of turing's >>>>> proof and produce the psuedo-code for the machine H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he didn't
reason quantified over all but phrase it like a procedure for what he
the theory of computation is the theory of such procedures, and
understanding the diagonal procedure is critical to understanding the
*base* contradiction/paradox that the rest of his support for godel's
result is then built on
And focusing on what is said to be impossible and not changing the
problem is also important.
The problem with the diagonal generation isn't the generation of the
diagonal itself, but effectively enumerating the enumeration in the
first place.
i don't see any indication that turing realized a difference there
H shows that *IF* you can make that enumeration, you can make the
diagonal, and thus the anti-diagonal. The problem is you can't make
that enumeration, and assuming you can just shows unsoundness.
interestingly: one can only fix the direct diagonal computation like this
u can't do an analogous fix for the inverse/anti-diagonal computation.
it's not possible hard code a machine to return an inverted value, a
machine can only return what it does, not the inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will leave
a direct diagonal computation extant in that filtered (yet still turing complete list), while any attempt to compute an inverse diagonal will
not be
so despite turing's worries, the existence of a diagonal computation
does not actually then imply the existence of an anti-diagonal
computation, due the same particular self-referential weirdness that
stumped turing the first place
...which u won't find interesting, but i can't fix a lack of curiosity
🤷
expected was an inexpert audience, supposing that experts will
recognise
the relevant mapping to universal quantification?
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:the theory of computation is the theory of such procedures, and
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me >>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one >>>>>>> by one.
that exactly what turing does in his proof: he defines a comptuation >>>>>> that enumerates out all the numbers, testing each one of they
represent
a "satisfactory"/"circle-free" machine, and adding that to diagonal >>>>>> across defined across computable numbers
it really would be a great exercise to carefully read p247 of
turing's
proof and produce the psuedo-code for the machine H, assuming that >>>>>> machine D exists
I'll get to it sooner then, because it's mad. Are you sure he didn't >>>>> reason quantified over all but phrase it like a procedure for what he >>>>
understanding the diagonal procedure is critical to understanding
the *base* contradiction/paradox that the rest of his support for
godel's result is then built on
And focusing on what is said to be impossible and not changing the
problem is also important.
The problem with the diagonal generation isn't the generation of the
diagonal itself, but effectively enumerating the enumeration in the
first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of the computable sequences.
He then points out that he can directly show that the "anti-diagonal" of
the (non-effectively computed) enumeration can't be computed but that
"This proof, although perfectly sound, has the disadvantage that it may leave the reader with a feeling that 'there must be something wrong'".
But instead, he can prove with a more obvious process, that the Decider
"D" that could be used to effectively enumerate the sequence of machine
that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the
attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
Something that seems to be beyond your ignorant understanding.
H shows that *IF* you can make that enumeration, you can make the
diagonal, and thus the anti-diagonal. The problem is you can't make
that enumeration, and assuming you can just shows unsoundness.
interestingly: one can only fix the direct diagonal computation like this
u can't do an analogous fix for the inverse/anti-diagonal computation.
it's not possible hard code a machine to return an inverted value, a
machine can only return what it does, not the inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will
leave a direct diagonal computation extant in that filtered (yet still
turing complete list), while any attempt to compute an inverse
diagonal will not be
But the problem is that "paradoxical machines" don't exist in isolation,
but only in relationship to a given machine trying to decide them.
His specified H, with an actually (incorrect) implementation of D (which
is all that CAN exist) will either be circle-free and thus generate a
number (but its D said it isn't, and thus omitted a valid machine from
the list) or it isn't circle-free, and fails to computa a number, and
thus should have been omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just built on
an assuption in error.
so despite turing's worries, the existence of a diagonal computation
does not actually then imply the existence of an anti-diagonal
computation, due the same particular self-referential weirdness that
stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based on ERROR.
Your attempt to REDEFINE self-reference to mean processing a number that happens to represent yourself means that you you system "ALL" doesn't actually mean ALL, and thus is just ill-defined.
--This just shows that you real problem is you don't understand what the actual problem is, and your world is just build on things that are lies.
...which u won't find interesting, but i can't fix a lack of curiosity
🤷
expected was an inexpert audience, supposing that experts will
recognise
the relevant mapping to universal quantification?
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that they
need to be fully defined in the actions they do.
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:the theory of computation is the theory of such procedures, and
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me >>>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude one >>>>>>>> by one.
that exactly what turing does in his proof: he defines a comptuation >>>>>>> that enumerates out all the numbers, testing each one of they
represent
a "satisfactory"/"circle-free" machine, and adding that to diagonal >>>>>>> across defined across computable numbers
it really would be a great exercise to carefully read p247 of
turing's
proof and produce the psuedo-code for the machine H, assuming that >>>>>>> machine D exists
I'll get to it sooner then, because it's mad. Are you sure he didn't >>>>>> reason quantified over all but phrase it like a procedure for what he >>>>>
understanding the diagonal procedure is critical to understanding
the *base* contradiction/paradox that the rest of his support for
godel's result is then built on
And focusing on what is said to be impossible and not changing the
problem is also important.
The problem with the diagonal generation isn't the generation of the
diagonal itself, but effectively enumerating the enumeration in the
first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell the
difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of the
computable sequences.
He then points out that he can directly show that the "anti-diagonal"
of the (non-effectively computed) enumeration can't be computed but
that "This proof, although perfectly sound, has the disadvantage that
it may leave the reader with a feeling that 'there must be something
wrong'".
it is wrong,
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across computable
numbers is computable...
neither method i have for fixing the diagonal computation across the computable numbers can be used to compute the inverse diagonal
so while i agree with turing that the anti-diagonal is not computable, i don't agree that the normal diagonal is not computable
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the sequence
of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the
attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing the diagonal can avoid the paradox on itself ...
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
Something that seems to be beyond your ignorant understanding.
H shows that *IF* you can make that enumeration, you can make the
diagonal, and thus the anti-diagonal. The problem is you can't make
that enumeration, and assuming you can just shows unsoundness.
interestingly: one can only fix the direct diagonal computation like
this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an
inverted value, a machine can only return what it does, not the
inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will
leave a direct diagonal computation extant in that filtered (yet
still turing complete list), while any attempt to compute an inverse
diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers then u only need to filter out paradoxes in regards to the classifier that classifies them as a "satisfactory" number
any machine which *is not* "satisfactory" OR *is not* classifiable as satisfactory by said classifier... can just be skipped
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either *is* computably equivalent OR *is not* classifiable in regards to *any*
machine already the list... can just be skipped
see you can't compute a diagonal across *all* /machines/, with said machines, but u can compute a diagonal across *all* /computable numbers/
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without such
paradox
His specified H, with an actually (incorrect) implementation of D
(which is all that CAN exist) will either be circle-free and thus
generate a number (but its D said it isn't, and thus omitted a valid
machine from the list) or it isn't circle-free, and fails to computa a
number, and thus should have been omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just built
on an assuption in error.
so despite turing's worries, the existence of a diagonal computation
does not actually then imply the existence of an anti-diagonal
computation, due the same particular self-referential weirdness that
stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based on
ERROR.
Your attempt to REDEFINE self-reference to mean processing a number
that happens to represent yourself means that you you system "ALL"
doesn't actually mean ALL, and thus is just ill-defined.
i'm sorry, you have an issue with me labeling a number that directly
refers to yourself, as a "self-reference" ???
dear future: u see this shit i'm trying work with???? sorry i didn't
manage to make progress any faster, but i'm literally dragging toddlers kicking and screaming at this point jeez...
This just shows that you real problem is you don't understand what the
...which u won't find interesting, but i can't fix a lack of curiosity
🤷
expected was an inexpert audience, supposing that experts will
recognise
the relevant mapping to universal quantification?
actual problem is, and your world is just build on things that are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that they
need to be fully defined in the actions they do.
On 2/24/26 2:59 PM, Tristan Wibberley wrote:
On 23/02/2026 15:02, Richard Damon wrote:
Most AI is NOT "Proof" or even "Reasoning" based.
Most AI is based on maximum likely Markov Chain processing.
What does mostness have to do with it?
Because you don't seem to understand what you are talking about,
You seem to think that AI Generation *IS* equivalent to automated proof,
when it isn't, as what is normally described as "AI" doesn't do that.
On 25/02/2026 00:01, Richard Damon wrote:
On 2/24/26 2:59 PM, Tristan Wibberley wrote:
On 23/02/2026 15:02, Richard Damon wrote:
Most AI is NOT "Proof" or even "Reasoning" based.
Most AI is based on maximum likely Markov Chain processing.
What does mostness have to do with it?
Because you don't seem to understand what you are talking about,
You seem to think that AI Generation *IS* equivalent to automated proof,
when it isn't, as what is normally described as "AI" doesn't do that.
I feel that I don't think that, I didn't since before I got a degree in Artificial Intelligence over 20 years ago (not that I think degrees are terribly good). That which people who don't know about anything refer to
as "AI" might not have done any automated theorem proving until last
year but for an entire human working lifetime now "AI" referred to a
variety of things besides natural language interpretation and generation.
Imagine all those experts with /recent/ AI degrees knowing that you,
Richard, are totally wrongheaded. A Penguin book of AI would teach you
more than whatever you're drinking.
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>> turing's original paper and can be reused anywhere...
You think you can test all of them one by one? Don't tell me >>>>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to diagonal >>>>>>>> across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming that >>>>>>>> machine D exists
I'll get to it sooner then, because it's mad. Are you sure he didn't >>>>>>> reason quantified over all but phrase it like a procedure for
what he
the theory of computation is the theory of such procedures, and
understanding the diagonal procedure is critical to understanding >>>>>> the *base* contradiction/paradox that the rest of his support for >>>>>> godel's result is then built on
And focusing on what is said to be impossible and not changing the
problem is also important.
The problem with the diagonal generation isn't the generation of
the diagonal itself, but effectively enumerating the enumeration in >>>>> the first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell the
difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of
the computable sequences.
He then points out that he can directly show that the "anti-diagonal"
of the (non-effectively computed) enumeration can't be computed but
that "This proof, although perfectly sound, has the disadvantage that
it may leave the reader with a feeling that 'there must be something
wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement, which shows that since for all n, position n differs from the value in number
n, there can not be any element that matches the anti-diagonal.
It is just a natural fact of countable infinity, something it seems you
just don't understand.
Show how that is actually wrong.
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across computable
numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across the
computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and thus
you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not computable,
i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the sequence
of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the
attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing
the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
It doesn't matter that your new H doesn't get stuck on itself, it will
still error on Turing's H.
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL be
circle free (since it never tries to simulate itself) and thus DOES
produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you WILL
hit these actual machines built on your erroneous D (your D must have
this flaw, as no D without exists), and thus you will be wrong on THAT input. IT doesn't matter if you get a good answer for yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of bad logic. You would have likely been escorted out of the meeting as showing
you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
H shows that *IF* you can make that enumeration, you can make the
diagonal, and thus the anti-diagonal. The problem is you can't make >>>>> that enumeration, and assuming you can just shows unsoundness.
interestingly: one can only fix the direct diagonal computation like
this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an
inverted value, a machine can only return what it does, not the
inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will
leave a direct diagonal computation extant in that filtered (yet
still turing complete list), while any attempt to compute an inverse
diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers then
u only need to filter out paradoxes in regards to the classifier that
classifies them as a "satisfactory" number
Right, which he shows can not be done.
any machine which *is not* "satisfactory" OR *is not* classifiable as
satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that are
but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it will
be asked about all machines as it counts through all the descriptions)
and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either
*is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of the computable numbers.
see you can't compute a diagonal across *all* /machines/, with said
machines, but u can compute a diagonal across *all* /computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since ALL classifiers that attempt it will make an error, you can't do what you
want to do.
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without such
paradox
And the problem is that your "paradoxical" isn't actually a definable property (let alone computable). Part of the problem is that if you look
at just a machine description, it doesn't (necessarily) tell you about
the use of an "interface" as that use of an interface can be just
inlined, leaving nothing "in the code" to show it exists.
His specified H, with an actually (incorrect) implementation of D
(which is all that CAN exist) will either be circle-free and thus
generate a number (but its D said it isn't, and thus omitted a valid
machine from the list) or it isn't circle-free, and fails to computa
a number, and thus should have been omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just built
on an assuption in error.
so despite turing's worries, the existence of a diagonal computation
does not actually then imply the existence of an anti-diagonal
computation, due the same particular self-referential weirdness that
stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based
on ERROR.
Your attempt to REDEFINE self-reference to mean processing a number
that happens to represent yourself means that you you system "ALL"
doesn't actually mean ALL, and thus is just ill-defined.
i'm sorry, you have an issue with me labeling a number that directly
refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
problems with all the "equivalent" machines that have different numbers.
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
dear future: u see this shit i'm trying work with???? sorry i didn't
manage to make progress any faster, but i'm literally dragging
toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually be used.
You don't understand that you need to use actual sound logic,
This just shows that you real problem is you don't understand what
...which u won't find interesting, but i can't fix a lack of curiosity >>>>
🤷
expected was an inexpert audience, supposing that experts will
recognise
the relevant mapping to universal quantification?
the actual problem is, and your world is just build on things that
are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that
they need to be fully defined in the actions they do.
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>>> turing's original paper and can be reused anywhere... >>>>>>>>>>>>You think you can test all of them one by one? Don't tell me >>>>>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to
diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming that >>>>>>>>> machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>> what he
the theory of computation is the theory of such procedures, and >>>>>>> understanding the diagonal procedure is critical to understanding >>>>>>> the *base* contradiction/paradox that the rest of his support for >>>>>>> godel's result is then built on
And focusing on what is said to be impossible and not changing the >>>>>> problem is also important.
The problem with the diagonal generation isn't the generation of
the diagonal itself, but effectively enumerating the enumeration
in the first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell the
difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of
the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be
computed but that "This proof, although perfectly sound, has the
disadvantage that it may leave the reader with a feeling that 'there
must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value in
number n, there can not be any element that matches the anti-diagonal.
It is just a natural fact of countable infinity, something it seems
you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a subtle,
yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th
figure in an. Let β be the sequence with 1-φn(m) as its n-th. figure. Since β is computable, there exists a number K [== β] such that 1-φn(n)
= φK(n) for all n. Putting n = K, we have 1 = 2φK(K), i.e. 1 is even.
This is impossible/
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using the direct diagonal. the abstract definition makes it look simple, but this ignores the complexities of self-referential analysis (like what turing details on the next page)
in both methods i have for rectifying the paradox found in the direct diagonal (either (1) filtering TMs or (2) using RTMs), neither can be
used to then compute the anti-diagonal
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard coded
value that is inverse to what it does return ... such a concept is
entirely nonsensical. a function can only return what it does, it can't
also return the inverse to what it returns eh???
in (2) the attempt to compute an inverse diagonal with RTMs just fails
for reasons u'd only understand by working thru the algo urself (p7 of
re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually computable given the direct diagonal φn()
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across computable
numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across the
computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the sequence
of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the
attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing
the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide correctly
on H to compute a diagonal
It doesn't matter that your new H doesn't get stuck on itself, it will
still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't specify what D (or H) needs to do when encountering the /undecidable input/ of
H, so therefore both D and H are an incomplete specifications of a machine
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL be
circle free (since it never tries to simulate itself) and thus DOES
produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D must
have this flaw, as no D without exists), and thus you will be wrong on
THAT input. IT doesn't matter if you get a good answer for yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of
bad logic. You would have likely been escorted out of the meeting as
showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
interestingly: one can only fix the direct diagonal computation
H shows that *IF* you can make that enumeration, you can make the >>>>>> diagonal, and thus the anti-diagonal. The problem is you can't
make that enumeration, and assuming you can just shows unsoundness. >>>>>
like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an
inverted value, a machine can only return what it does, not the
inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will
leave a direct diagonal computation extant in that filtered (yet
still turing complete list), while any attempt to compute an
inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers
then u only need to filter out paradoxes in regards to the classifier
that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such paradoxes...
(also why do always just make random assertions???)
any machine which *is not* "satisfactory" OR *is not* classifiable as
satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that
are but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it
will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either
*is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of the
computable numbers.
see you can't compute a diagonal across *all* /machines/, with said
machines, but u can compute a diagonal across *all* /computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since
ALL classifiers that attempt it will make an error, you can't do what
you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not the same thing as (b) computing the enumeration of all machines that compute computable numbers. (b) necessarily has duplicates while (a) does not
need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
(b) probably can't be done with TMs
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without such
paradox
And the problem is that your "paradoxical" isn't actually a definable
property (let alone computable). Part of the problem is that if you
look at just a machine description, it doesn't (necessarily) tell you
about the use of an "interface" as that use of an interface can be
just inlined, leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
lol
His specified H, with an actually (incorrect) implementation of D
(which is all that CAN exist) will either be circle-free and thus
generate a number (but its D said it isn't, and thus omitted a valid
machine from the list) or it isn't circle-free, and fails to computa
a number, and thus should have been omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just built
on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an anti-
diagonal computation, due the same particular self-referential
weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based
on ERROR.
Your attempt to REDEFINE self-reference to mean processing a number
that happens to represent yourself means that you you system "ALL"
doesn't actually mean ALL, and thus is just ill-defined.
i'm sorry, you have an issue with me labeling a number that directly
refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
it's a *specific* number that has the currently running machine encoded
into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different numbers.
those are references to functionally equivalent machines, not self- references
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox detector while filter *out* paradoxes to point of creating an actual
problem
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an input classifier, and an input machine to search for paradoxes in regards to: paradox_free(classifier, machine)
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
however, if paradox_free(halts,undp) -> TRUE, then that if case doesn't
run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine is necessary for the particular input->output computation being done, so utilizing the return paradox_free(halts,undp) -> FALSE to filter out paradox_free() paradoxes as well as halts() paradoxes will suffice to produce a turing complete subset of machines that can be totally
classified by halts()
i think a major problem with the consensus perspective is an expectation
of a certain cleanliness in the logic as being the only possibility here
to compute what we want. TM's can't achieve that, but something slightly messier can.
dear future: u see this shit i'm trying work with???? sorry i didn't
manage to make progress any faster, but i'm literally dragging
toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually be
used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
This just shows that you real problem is you don't understand what
...which u won't find interesting, but i can't fix a lack of curiosity >>>>>
🤷
expected was an inexpert audience, supposing that experts will >>>>>>>> recognise
the relevant mapping to universal quantification?
the actual problem is, and your world is just build on things that
are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that
they need to be fully defined in the actions they do.
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>>> turing's original paper and can be reused anywhere... >>>>>>>>>>>>You think you can test all of them one by one? Don't tell me >>>>>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to
diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming that >>>>>>>>> machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>> what he
the theory of computation is the theory of such procedures, and >>>>>>> understanding the diagonal procedure is critical to understanding >>>>>>> the *base* contradiction/paradox that the rest of his support for >>>>>>> godel's result is then built on
And focusing on what is said to be impossible and not changing the >>>>>> problem is also important.
The problem with the diagonal generation isn't the generation of
the diagonal itself, but effectively enumerating the enumeration
in the first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell the
difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of
the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be
computed but that "This proof, although perfectly sound, has the
disadvantage that it may leave the reader with a feeling that 'there
must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value in
number n, there can not be any element that matches the anti-diagonal.
It is just a natural fact of countable infinity, something it seems
you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a subtle,
yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th
figure in an. Let β be the sequence with 1-φn(m) as its n-th. figure. Since β is computable, there exists a number K [== β] such that 1-φn(n)
= φK(n) for all n. Putting n = K, we have 1 = 2φK(K), i.e. 1 is even.
This is impossible/
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using the direct diagonal. the abstract definition makes it look simple, but this ignores the complexities of self-referential analysis (like what turing details on the next page)
in both methods i have for rectifying the paradox found in the direct diagonal (either (1) filtering TMs or (2) using RTMs), neither can be
used to then compute the anti-diagonal
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard coded
value that is inverse to what it does return ... such a concept is
entirely nonsensical. a function can only return what it does, it can't
also return the inverse to what it returns eh???
in (2) the attempt to compute an inverse diagonal with RTMs just fails
for reasons u'd only understand by working thru the algo urself (p7 of
re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
= φK(n) for all n/
is just not sufficient evidence that such β is actually computable given the direct diagonal φn()
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across computable
numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across the
computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the sequence
of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the
attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing
the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide correctly
on H to compute a diagonal
It doesn't matter that your new H doesn't get stuck on itself, it will
still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't specify what D (or H) needs to do when encountering the /undecidable input/ of
H, so therefore both D and H are an incomplete specifications of a machine
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL be
circle free (since it never tries to simulate itself) and thus DOES
produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D must
have this flaw, as no D without exists), and thus you will be wrong on
THAT input. IT doesn't matter if you get a good answer for yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of
bad logic. You would have likely been escorted out of the meeting as
showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
interestingly: one can only fix the direct diagonal computation
H shows that *IF* you can make that enumeration, you can make the >>>>>> diagonal, and thus the anti-diagonal. The problem is you can't
make that enumeration, and assuming you can just shows unsoundness. >>>>>
like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an
inverted value, a machine can only return what it does, not the
inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will
leave a direct diagonal computation extant in that filtered (yet
still turing complete list), while any attempt to compute an
inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers
then u only need to filter out paradoxes in regards to the classifier
that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such paradoxes...
(also why do always just make random assertions???)
any machine which *is not* "satisfactory" OR *is not* classifiable as
satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that
are but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it
will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either
*is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of the
computable numbers.
see you can't compute a diagonal across *all* /machines/, with said
machines, but u can compute a diagonal across *all* /computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since
ALL classifiers that attempt it will make an error, you can't do what
you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not the same thing as (b) computing the enumeration of all machines that compute computable numbers. (b) necessarily has duplicates while (a) does not
need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
(b) probably can't be done with TMs
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without such
paradox
And the problem is that your "paradoxical" isn't actually a definable
property (let alone computable). Part of the problem is that if you
look at just a machine description, it doesn't (necessarily) tell you
about the use of an "interface" as that use of an interface can be
just inlined, leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
lol
His specified H, with an actually (incorrect) implementation of D
(which is all that CAN exist) will either be circle-free and thus
generate a number (but its D said it isn't, and thus omitted a valid
machine from the list) or it isn't circle-free, and fails to computa
a number, and thus should have been omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just built
on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an anti-
diagonal computation, due the same particular self-referential
weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based
on ERROR.
Your attempt to REDEFINE self-reference to mean processing a number
that happens to represent yourself means that you you system "ALL"
doesn't actually mean ALL, and thus is just ill-defined.
i'm sorry, you have an issue with me labeling a number that directly
refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
it's a *specific* number that has the currently running machine encoded
into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different numbers.
those are references to functionally equivalent machines, not self- references
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox detector while filter *out* paradoxes to point of creating an actual
problem
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an input classifier, and an input machine to search for paradoxes in regards to: paradox_free(classifier, machine)
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
however, if paradox_free(halts,undp) -> TRUE, then that if case doesn't
run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine is necessary for the particular input->output computation being done, so utilizing the return paradox_free(halts,undp) -> FALSE to filter out paradox_free() paradoxes as well as halts() paradoxes will suffice to produce a turing complete subset of machines that can be totally
classified by halts()
i think a major problem with the consensus perspective is an expectation
of a certain cleanliness in the logic as being the only possibility here
to compute what we want. TM's can't achieve that, but something slightly messier can.
dear future: u see this shit i'm trying work with???? sorry i didn't
manage to make progress any faster, but i'm literally dragging
toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually be
used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
This just shows that you real problem is you don't understand what
...which u won't find interesting, but i can't fix a lack of curiosity >>>>>
🤷
expected was an inexpert audience, supposing that experts will >>>>>>>> recognise
the relevant mapping to universal quantification?
the actual problem is, and your world is just build on things that
are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that
they need to be fully defined in the actions they do.
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>>> turing's original paper and can be reused anywhere... >>>>>>>>>>>>You think you can test all of them one by one? Don't tell me >>>>>>>>>>>> you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to
diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming that >>>>>>>>> machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>> what he
the theory of computation is the theory of such procedures, and >>>>>>> understanding the diagonal procedure is critical to understanding >>>>>>> the *base* contradiction/paradox that the rest of his support for >>>>>>> godel's result is then built on
And focusing on what is said to be impossible and not changing the >>>>>> problem is also important.
The problem with the diagonal generation isn't the generation of
the diagonal itself, but effectively enumerating the enumeration
in the first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell the
difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of
the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be
computed but that "This proof, although perfectly sound, has the
disadvantage that it may leave the reader with a feeling that 'there
must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value in
number n, there can not be any element that matches the anti-diagonal.
It is just a natural fact of countable infinity, something it seems
you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a subtle,
yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th
figure in an. Let β be the sequence with 1-φn(m) as its n-th. figure. Since β is computable, there exists a number K [== β] such that 1-φn(n)
= φK(n) for all n. Putting n = K, we have 1 = 2φK(K), i.e. 1 is even.
This is impossible/
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using the direct diagonal. the abstract definition makes it look simple, but this ignores the complexities of self-referential analysis (like what turing details on the next page)
in both methods i have for rectifying the paradox found in the direct diagonal (either (1) filtering TMs or (2) using RTMs), neither can be
used to then compute the anti-diagonal
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard coded
value that is inverse to what it does return ... such a concept is
entirely nonsensical. a function can only return what it does, it can't
also return the inverse to what it returns eh???
in (2) the attempt to compute an inverse diagonal with RTMs just fails
for reasons u'd only understand by working thru the algo urself (p7 of
re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually computable given the direct diagonal φn()
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across computable
numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across the
computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the sequence
of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the
attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing
the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide correctly
on H to compute a diagonal
It doesn't matter that your new H doesn't get stuck on itself, it will
still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't specify what D (or H) needs to do when encountering the /undecidable input/ of
H, so therefore both D and H are an incomplete specifications of a machine
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL be
circle free (since it never tries to simulate itself) and thus DOES
produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D must
have this flaw, as no D without exists), and thus you will be wrong on
THAT input. IT doesn't matter if you get a good answer for yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of
bad logic. You would have likely been escorted out of the meeting as
showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
interestingly: one can only fix the direct diagonal computation
H shows that *IF* you can make that enumeration, you can make the >>>>>> diagonal, and thus the anti-diagonal. The problem is you can't
make that enumeration, and assuming you can just shows unsoundness. >>>>>
like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an
inverted value, a machine can only return what it does, not the
inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will
leave a direct diagonal computation extant in that filtered (yet
still turing complete list), while any attempt to compute an
inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers
then u only need to filter out paradoxes in regards to the classifier
that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such paradoxes...
(also why do always just make random assertions???)
any machine which *is not* "satisfactory" OR *is not* classifiable as
satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that
are but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it
will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either
*is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of the
computable numbers.
see you can't compute a diagonal across *all* /machines/, with said
machines, but u can compute a diagonal across *all* /computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since
ALL classifiers that attempt it will make an error, you can't do what
you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not the same thing as (b) computing the enumeration of all machines that compute computable numbers. (b) necessarily has duplicates while (a) does not
need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
(b) probably can't be done with TMs
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without such
paradox
And the problem is that your "paradoxical" isn't actually a definable
property (let alone computable). Part of the problem is that if you
look at just a machine description, it doesn't (necessarily) tell you
about the use of an "interface" as that use of an interface can be
just inlined, leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
lol
His specified H, with an actually (incorrect) implementation of D
(which is all that CAN exist) will either be circle-free and thus
generate a number (but its D said it isn't, and thus omitted a valid
machine from the list) or it isn't circle-free, and fails to computa
a number, and thus should have been omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just built
on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an anti-
diagonal computation, due the same particular self-referential
weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based
on ERROR.
Your attempt to REDEFINE self-reference to mean processing a number
that happens to represent yourself means that you you system "ALL"
doesn't actually mean ALL, and thus is just ill-defined.
i'm sorry, you have an issue with me labeling a number that directly
refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
it's a *specific* number that has the currently running machine encoded
into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different numbers.
those are references to functionally equivalent machines, not self- references
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox detector while filter *out* paradoxes to point of creating an actual
problem
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an input classifier, and an input machine to search for paradoxes in regards to: paradox_free(classifier, machine)
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
however, if paradox_free(halts,undp) -> TRUE, then that if case doesn't
run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine is necessary for the particular input->output computation being done, so utilizing the return paradox_free(halts,undp) -> FALSE to filter out paradox_free() paradoxes as well as halts() paradoxes will suffice to produce a turing complete subset of machines that can be totally
classified by halts()
i think a major problem with the consensus perspective is an expectation
of a certain cleanliness in the logic as being the only possibility here
to compute what we want. TM's can't achieve that, but something slightly messier can.
dear future: u see this shit i'm trying work with???? sorry i didn't
manage to make progress any faster, but i'm literally dragging
toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually be
used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
This just shows that you real problem is you don't understand what
...which u won't find interesting, but i can't fix a lack of curiosity >>>>>
🤷
expected was an inexpert audience, supposing that experts will >>>>>>>> recognise
the relevant mapping to universal quantification?
the actual problem is, and your world is just build on things that
are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that
they need to be fully defined in the actions they do.
On 3/3/26 12:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>>>> turing's original paper and can be reused anywhere... >>>>>>>>>>>>>You think you can test all of them one by one? Don't tell >>>>>>>>>>>>> me you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to >>>>>>>>>> diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming >>>>>>>>>> that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>>> what he
the theory of computation is the theory of such procedures, and >>>>>>>> understanding the diagonal procedure is critical to
understanding the *base* contradiction/paradox that the rest of >>>>>>>> his support for godel's result is then built on
And focusing on what is said to be impossible and not changing
the problem is also important.
The problem with the diagonal generation isn't the generation of >>>>>>> the diagonal itself, but effectively enumerating the enumeration >>>>>>> in the first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell
the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of
the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be
computed but that "This proof, although perfectly sound, has the
disadvantage that it may leave the reader with a feeling that
'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value
in number n, there can not be any element that matches the anti-
diagonal.
It is just a natural fact of countable infinity, something it seems
you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a
subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th
figure in an. Let β be the sequence with 1-φn(m) as its n-th. figure.
Since β is computable, there exists a number K [== β] such that 1-
φn(n) = φK(n) for all n. Putting n = K, we have 1 = 2φK(K), i.e. 1 is
even. This is impossible/
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using the
direct diagonal. the abstract definition makes it look simple, but
this ignores the complexities of self-referential analysis (like what
turing details on the next page)
in both methods i have for rectifying the paradox found in the direct
diagonal (either (1) filtering TMs or (2) using RTMs), neither can be
used to then compute the anti-diagonal
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard coded
value that is inverse to what it does return ... such a concept is
entirely nonsensical. a function can only return what it does, it
can't also return the inverse to what it returns eh???
in (2) the attempt to compute an inverse diagonal with RTMs just fails
for reasons u'd only understand by working thru the algo urself (p7 of
re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
that sentence there, ben, from p246,
is the sentence of turing's paper /on computable numbers/ that i start
to diagree with,
that sentence just is wrong to imply anything about the computability of
a number
/there exists a number K [== β] such that 1-φn(n)
= φK(n) for all n/
and that sentence is *only* /half-true/
it's *correct* in that can't be computed by a TM,
whoever a human with a TM could still write it down, but a human could
never pass that input entirely to a finite running machine eh???
so i kinda think the ct-thesis is actually cooked in a way
¯\_(ツ)_/¯
is just not sufficient evidence that such β is actually computable
given the direct diagonal φn()
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across computable
numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across the
computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the
sequence of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the >>>>> attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing
the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide
correctly on H to compute a diagonal
It doesn't matter that your new H doesn't get stuck on itself, it
will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't
specify what D (or H) needs to do when encountering the /undecidable
input/ of H, so therefore both D and H are an incomplete
specifications of a machine
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL be
circle free (since it never tries to simulate itself) and thus DOES
produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D must
have this flaw, as no D without exists), and thus you will be wrong
on THAT input. IT doesn't matter if you get a good answer for yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of
bad logic. You would have likely been escorted out of the meeting as
showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
interestingly: one can only fix the direct diagonal computation
H shows that *IF* you can make that enumeration, you can make the >>>>>>> diagonal, and thus the anti-diagonal. The problem is you can't
make that enumeration, and assuming you can just shows unsoundness. >>>>>>
like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an
inverted value, a machine can only return what it does, not the
inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will >>>>>> leave a direct diagonal computation extant in that filtered (yet
still turing complete list), while any attempt to compute an
inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers
then u only need to filter out paradoxes in regards to the
classifier that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such paradoxes...
(also why do always just make random assertions???)
any machine which *is not* "satisfactory" OR *is not* classifiable
as satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that
are but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it
will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either
*is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of
the computable numbers.
see you can't compute a diagonal across *all* /machines/, with said
machines, but u can compute a diagonal across *all* /computable
numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since
ALL classifiers that attempt it will make an error, you can't do what
you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not
the same thing as (b) computing the enumeration of all machines that
compute computable numbers. (b) necessarily has duplicates while (a)
does not need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
(b) probably can't be done with TMs
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without such
paradox
And the problem is that your "paradoxical" isn't actually a definable
property (let alone computable). Part of the problem is that if you
look at just a machine description, it doesn't (necessarily) tell you
about the use of an "interface" as that use of an interface can be
just inlined, leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
lol
His specified H, with an actually (incorrect) implementation of D
(which is all that CAN exist) will either be circle-free and thus
generate a number (but its D said it isn't, and thus omitted a
valid machine from the list) or it isn't circle-free, and fails to
computa a number, and thus should have been omitted from the list
but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just
built on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an anti- >>>>>> diagonal computation, due the same particular self-referential
weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based >>>>> on ERROR.
Your attempt to REDEFINE self-reference to mean processing a number >>>>> that happens to represent yourself means that you you system "ALL"
doesn't actually mean ALL, and thus is just ill-defined.
i'm sorry, you have an issue with me labeling a number that directly
refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
it's a *specific* number that has the currently running machine
encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different numbers.
those are references to functionally equivalent machines, not self-
references
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox
detector while filter *out* paradoxes to point of creating an actual
problem
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an input
classifier, and an input machine to search for paradoxes in regards
to: paradox_free(classifier, machine)
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier
like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
however, if paradox_free(halts,undp) -> TRUE, then that if case
doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine is
necessary for the particular input->output computation being done, so
utilizing the return paradox_free(halts,undp) -> FALSE to filter out
paradox_free() paradoxes as well as halts() paradoxes will suffice to
produce a turing complete subset of machines that can be totally
classified by halts()
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the only
possibility here to compute what we want. TM's can't achieve that, but
something slightly messier can.
dear future: u see this shit i'm trying work with???? sorry i didn't
manage to make progress any faster, but i'm literally dragging
toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually be
used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
This just shows that you real problem is you don't understand what
...which u won't find interesting, but i can't fix a lack of
curiosity
🤷
expected was an inexpert audience, supposing that experts will >>>>>>>>> recognise
the relevant mapping to universal quantification?
the actual problem is, and your world is just build on things that
are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that
they need to be fully defined in the actions they do.
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was proven on >>>>>>>>>>>>>> turing's original paper and can be reused anywhere... >>>>>>>>>>>>>You think you can test all of them one by one? Don't tell >>>>>>>>>>>>> me you think
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a
comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to >>>>>>>>>> diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming >>>>>>>>>> that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>>> what he
the theory of computation is the theory of such procedures, and >>>>>>>> understanding the diagonal procedure is critical to
understanding the *base* contradiction/paradox that the rest of >>>>>>>> his support for godel's result is then built on
And focusing on what is said to be impossible and not changing
the problem is also important.
The problem with the diagonal generation isn't the generation of >>>>>>> the diagonal itself, but effectively enumerating the enumeration >>>>>>> in the first place.
i don't see any indication that turing realized a difference there
Then you zre just showing your stupidity, because YOU can't tell
the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of
the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be
computed but that "This proof, although perfectly sound, has the
disadvantage that it may leave the reader with a feeling that
'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value
in number n, there can not be any element that matches the anti-
diagonal.
It is just a natural fact of countable infinity, something it seems
you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a
subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th
figure in an. Let β be the sequence with 1-φn(m) as its n-th. figure.
Since β is computable, there exists a number K [== β] such that 1-
φn(n) = φK(n) for all n. Putting n = K, we have 1 = 2φK(K), i.e. 1 is
even. This is impossible/
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using the
direct diagonal. the abstract definition makes it look simple, but
this ignores the complexities of self-referential analysis (like what
turing details on the next page)
But you can, *IF* you have a machine that computes the diagonal, then
just change all the write to the output to write the opposite. Note, the "self-reference" that you are thinking of stops being a "self-reference"
but is a reference to the original write the diagonal code.
in both methods i have for rectifying the paradox found in the direct
diagonal (either (1) filtering TMs or (2) using RTMs), neither can be
used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it gets to
the number of TURING'S H, as there is no correct answer for the machine built by that template.
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard coded
value that is inverse to what it does return ... such a concept is
entirely nonsensical. a function can only return what it does, it
can't also return the inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus your enumeration is incomlete.
in (2) the attempt to compute an inverse diagonal with RTMs just fails
for reasons u'd only understand by working thru the algo urself (p7 of
re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually computable
given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do that,
as you are thinking the only "reference" can be to "self", doesn't make
you argument correct.
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across computable
numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across the
computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the
sequence of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that the >>>>> attempt to compute the diagonal clearly reveals the issue with
effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing
the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide
correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
YOUR H still needs to know if Turing's H is a machine that generates a computable number to build the list of such machine to define the
diagonal to compute.
You seem to not understand the meaning of ALL.
It doesn't matter that your new H doesn't get stuck on itself, it
will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't
specify what D (or H) needs to do when encountering the /undecidable
input/ of H, so therefore both D and H are an incomplete
specifications of a machine
Because "undecidability" doesn't affect the correct answer it must compute.
Note, H, when it becomes an actual machine, because we have created an actual machine we claim to be the D, has an answer, and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't exist, only machines CORRECTLY meeting the requirements of D and H can't exist.
Thus, if your H depends on that D, (or a machine that meets its specification) then it also can't exist.
You just make the error of saying the problems don't exist because you
can't build them, but YOUR machine can, even though it has the same
problem.
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL be
circle free (since it never tries to simulate itself) and thus DOES
produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D must
have this flaw, as no D without exists), and thus you will be wrong
on THAT input. IT doesn't matter if you get a good answer for yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of
bad logic. You would have likely been escorted out of the meeting as
showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
interestingly: one can only fix the direct diagonal computation
H shows that *IF* you can make that enumeration, you can make the >>>>>>> diagonal, and thus the anti-diagonal. The problem is you can't
make that enumeration, and assuming you can just shows unsoundness. >>>>>>
like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an
inverted value, a machine can only return what it does, not the
inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will >>>>>> leave a direct diagonal computation extant in that filtered (yet
still turing complete list), while any attempt to compute an
inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers
then u only need to filter out paradoxes in regards to the
classifier that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such paradoxes...
In other words, you beleive unquestionably in the existance of Russle's teapot until someone can prove it doesn't exist.
For THIS paradox, read the proof. for THIS H, the one given, (for which
ever version of D you want to try to assume is right) D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox, because
there still isn't an answer it can give to H that will be correct, as
the template for H will always make that D wrong.
Note, the problem is you can't actually DEFINE what a "paradox machine"
is, as they aren't actually machines, but templates that build machines.
And that final machine doesn't have actually detectable tell-tales that
show it to be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
They only seem "random" because you don't know what you are talking
about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* classifiable
as satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that
are but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it
will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either
*is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of
the computable numbers.
see you can't compute a diagonal across *all* /machines/, with said
machines, but u can compute a diagonal across *all* /computable
numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since
ALL classifiers that attempt it will make an error, you can't do what
you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not
the same thing as (b) computing the enumeration of all machines that
compute computable numbers. (b) necessarily has duplicates while (a)
does not need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the full
list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of the H
he describes. It doesn't matter that you can make a DIFFERENT machine,
that you try to deceptively call "H" too that it won't have a problem
with, when your "H" gets to the number of Turing's H, it still has the problem. It can call that one not circle-free, and thus omit a circle-
free machine from the list, or call it circle-free, and when even YOU
try to simulate it the k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take the code
of (a), and invert the symbol put on the tape when it is written. Note,
it won't be "self-referent", as the "N" is uses, is the N of YOUR H, not itself. The code for his (a) NEVER reads back what it wrote, so that
doesn't affect its behavior.
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without such
paradox
And the problem is that your "paradoxical" isn't actually a definable
property (let alone computable). Part of the problem is that if you
look at just a machine description, it doesn't (necessarily) tell you
about the use of an "interface" as that use of an interface can be
just inlined, leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine.
lol
His specified H, with an actually (incorrect) implementation of D
(which is all that CAN exist) will either be circle-free and thus
generate a number (but its D said it isn't, and thus omitted a
valid machine from the list) or it isn't circle-free, and fails to
computa a number, and thus should have been omitted from the list
but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just
built on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an anti- >>>>>> diagonal computation, due the same particular self-referential
weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just based >>>>> on ERROR.
Your attempt to REDEFINE self-reference to mean processing a number >>>>> that happens to represent yourself means that you you system "ALL"
doesn't actually mean ALL, and thus is just ill-defined.
i'm sorry, you have an issue with me labeling a number that directly
refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
it's a *specific* number that has the currently running machine
encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different numbers.
those are references to functionally equivalent machines, not self-
references
Which is why the concept of "self-reference" doesn't work.
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox
detector while filter *out* paradoxes to point of creating an actual
problem
Then show how you will do it.
This means you need to detect a input that represents a machine that
uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an input
classifier, and an input machine to search for paradoxes in regards
to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier
like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
And the problem is "paradox_free" can't exist as an always correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case
doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine is
necessary for the particular input->output computation being done, so
utilizing the return paradox_free(halts,undp) -> FALSE to filter out
paradox_free() paradoxes as well as halts() paradoxes will suffice to
produce a turing complete subset of machines that can be totally
classified by halts()
But we can still build an input that your paradox_free() fails on.
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the only
possibility here to compute what we want. TM's can't achieve that, but
something slightly messier can.
No, it is your dependence on being able to assume that unicorns can
solve all your problems.
dear future: u see this shit i'm trying work with???? sorry i didn't
manage to make progress any faster, but i'm literally dragging
toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually be
used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
They are, But you are too stuck in your lies to see it.
Your "logic" is based on being able to assume that Unicorn's exist, and
that is a safe assumption until someone can prove that Russel's Teapot
is not out there. (Even though your unicorns HAVE been proved to not
exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't understand what
...which u won't find interesting, but i can't fix a lack of
curiosity
🤷
expected was an inexpert audience, supposing that experts will >>>>>>>>> recognise
the relevant mapping to universal quantification?
the actual problem is, and your world is just build on things that
are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that
they need to be fully defined in the actions they do.
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:Then you zre just showing your stupidity, because YOU can't tell
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was >>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't tell >>>>>>>>>>>>>> me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>> comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to >>>>>>>>>>> diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming >>>>>>>>>>> that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>>>> what he
the theory of computation is the theory of such procedures, and >>>>>>>>> understanding the diagonal procedure is critical to
understanding the *base* contradiction/paradox that the rest of >>>>>>>>> his support for godel's result is then built on
And focusing on what is said to be impossible and not changing >>>>>>>> the problem is also important.
The problem with the diagonal generation isn't the generation of >>>>>>>> the diagonal itself, but effectively enumerating the enumeration >>>>>>>> in the first place.
i don't see any indication that turing realized a difference there >>>>>>
the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of >>>>>> the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be
computed but that "This proof, although perfectly sound, has the
disadvantage that it may leave the reader with a feeling that
'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value
in number n, there can not be any element that matches the anti-
diagonal.
It is just a natural fact of countable infinity, something it seems
you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a
subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th
figure in an. Let β be the sequence with 1-φn(m) as its n-th. figure. >>> Since β is computable, there exists a number K [== β] such that 1-
φn(n) = φK(n) for all n. Putting n = K, we have 1 = 2φK(K), i.e. 1 is >>> even. This is impossible/
i agree with this proof is far as much as if β was computable (by TMs)
we'd have a problem for sure,
but what i don't agree is that we can just assume the computability of β from the existence of φn(m), THAT'S THE FALLACy
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using
the direct diagonal. the abstract definition makes it look simple,
but this ignores the complexities of self-referential analysis (like
what turing details on the next page)
But you can, *IF* you have a machine that computes the diagonal, then
just change all the write to the output to write the opposite. Note,
the "self-reference" that you are thinking of stops being a "self-
reference" but is a reference to the original write the diagonal code.
in both methods i have for rectifying the paradox found in the direct
diagonal (either (1) filtering TMs or (2) using RTMs), neither can be
used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it gets
to the number of TURING'S H, as there is no correct answer for the
machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard coded
value that is inverse to what it does return ... such a concept is
entirely nonsensical. a function can only return what it does, it
can't also return the inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus your
enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT *SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs just
fails for reasons u'd only understand by working thru the algo urself
(p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually computable
given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do that,
as you are thinking the only "reference" can be to "self", doesn't
make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-REFERENCE,
LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR INVERSE-H
TO BE COMPUTED
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across
computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across
the computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the
sequence of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that
the attempt to compute the diagonal clearly reveals the issue with >>>>>> effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing >>>>> the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide
correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT WOULDN'T BE USED
YOUR H still needs to know if Turing's H is a machine that generates a
computable number to build the list of such machine to define the
diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST.
BOTH ARE INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH WILL WORK
JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE INPUT/ TO D FILTERED
FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE NUMBERS
It doesn't matter that your new H doesn't get stuck on itself, it
will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't
specify what D (or H) needs to do when encountering the /undecidable
input/ of H, so therefore both D and H are an incomplete
specifications of a machine
Because "undecidability" doesn't affect the correct answer it must
compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE LIKE
THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE IT DOES
NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL ANALYSIS, SO IT IS
AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have created an
actual machine we claim to be the D, has an answer, and that D is
always wrong.
Thus, it isn't that machines claiming to be D and H can't exist, only
machines CORRECTLY meeting the requirements of D and H can't exist.
Thus, if your H depends on that D, (or a machine that meets its
specification) then it also can't exist.
You just make the error of saying the problems don't exist because you
can't build them, but YOUR machine can, even though it has the same
problem.
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL
be circle free (since it never tries to simulate itself) and thus
DOES produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D
must have this flaw, as no D without exists), and thus you will be
wrong on THAT input. IT doesn't matter if you get a good answer for
yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of
bad logic. You would have likely been escorted out of the meeting as
showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
H shows that *IF* you can make that enumeration, you can make >>>>>>>> the diagonal, and thus the anti-diagonal. The problem is you
can't make that enumeration, and assuming you can just shows
unsoundness.
interestingly: one can only fix the direct diagonal computation >>>>>>> like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an >>>>>>> inverted value, a machine can only return what it does, not the >>>>>>> inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will >>>>>>> leave a direct diagonal computation extant in that filtered (yet >>>>>>> still turing complete list), while any attempt to compute an
inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers
then u only need to filter out paradoxes in regards to the
classifier that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such paradoxes...
In other words, you beleive unquestionably in the existance of
Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES CAN'T BE
DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING A
FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU MADE???
For THIS paradox, read the proof. for THIS H, the one given, (for
which ever version of D you want to try to assume is right) D is just
wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox, because
there still isn't an answer it can give to H that will be correct, as
the template for H will always make that D wrong.
Note, the problem is you can't actually DEFINE what a "paradox
machine" is, as they aren't actually machines, but templates that
build machines. And that final machine doesn't have actually
detectable tell-tales that show it to be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO WORK
WITH, BUT OH WELL
They only seem "random" because you don't know what you are talking
about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* classifiable
as satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that
are but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it
will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either >>>>> *is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of
the computable numbers.
see you can't compute a diagonal across *all* /machines/, with said >>>>> machines, but u can compute a diagonal across *all* /computable
numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since
ALL classifiers that attempt it will make an error, you can't do
what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not
the same thing as (b) computing the enumeration of all machines that
compute computable numbers. (b) necessarily has duplicates while (a)
does not need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the full
list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of the H
he describes. It doesn't matter that you can make a DIFFERENT machine,
that you try to deceptively call "H" too that it won't have a problem
with, when your "H" gets to the number of Turing's H, it still has the
problem. It can call that one not circle-free, and thus omit a circle-
free machine from the list, or call it circle-free, and when even YOU
try to simulate it the k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take the code
of (a), and invert the symbol put on the tape when it is written.
Note, it won't be "self-referent", as the "N" is uses, is the N of
YOUR H, not itself. The code for his (a) NEVER reads back what it
wrote, so that doesn't affect its behavior.
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without
such paradox
And the problem is that your "paradoxical" isn't actually a
definable property (let alone computable). Part of the problem is
that if you look at just a machine description, it doesn't
(necessarily) tell you about the use of an "interface" as that use
of an interface can be just inlined, leaving nothing "in the code"
to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the classifier halts(), easily demonstrable thru a short proof i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME FUCKING FORM OF INSANITY I CAN'T HONESTLY FATHOM
lol
i'm sorry, you have an issue with me labeling a number that
His specified H, with an actually (incorrect) implementation of D >>>>>> (which is all that CAN exist) will either be circle-free and thus >>>>>> generate a number (but its D said it isn't, and thus omitted a
valid machine from the list) or it isn't circle-free, and fails to >>>>>> computa a number, and thus should have been omitted from the list >>>>>> but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just
built on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an
anti- diagonal computation, due the same particular self-
referential weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just
based on ERROR.
Your attempt to REDEFINE self-reference to mean processing a
number that happens to represent yourself means that you you
system "ALL" doesn't actually mean ALL, and thus is just ill-defined. >>>>>
directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
it's a *specific* number that has the currently running machine
encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different
numbers.
those are references to functionally equivalent machines, not self-
references
Which is why the concept of "self-reference" doesn't work.
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox
detector while filter *out* paradoxes to point of creating an actual
problem
Then show how you will do it.
This means you need to detect a input that represents a machine that
uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an
input classifier, and an input machine to search for paradoxes in
regards to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier
like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
And the problem is "paradox_free" can't exist as an always correct
decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case
doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine
is necessary for the particular input->output computation being done,
so utilizing the return paradox_free(halts,undp) -> FALSE to filter
out paradox_free() paradoxes as well as halts() paradoxes will
suffice to produce a turing complete subset of machines that can be
totally classified by halts()
But we can still build an input that your paradox_free() fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE INCLUDED ON
THE DIAGONAL???
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the only
possibility here to compute what we want. TM's can't achieve that,
but something slightly messier can.
No, it is your dependence on being able to assume that unicorns can
solve all your problems.
dear future: u see this shit i'm trying work with???? sorry i
didn't manage to make progress any faster, but i'm literally
dragging toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually
be used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
They are, But you are too stuck in your lies to see it.
Your "logic" is based on being able to assume that Unicorn's exist,
and that is a safe assumption until someone can prove that Russel's
Teapot is not out there. (Even though your unicorns HAVE been proved
to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't understand what >>>>>> the actual problem is, and your world is just build on things that >>>>>> are lies.
...which u won't find interesting, but i can't fix a lack of
curiosity
🤷
expected was an inexpert audience, supposing that experts will >>>>>>>>>> recognise
the relevant mapping to universal quantification?
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that >>>>>> they need to be fully defined in the actions they do.
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:Then you zre just showing your stupidity, because YOU can't tell
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was >>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't tell >>>>>>>>>>>>>> me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an infinitude >>>>>>>>>>>> one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>> comptuation
that enumerates out all the numbers, testing each one of they >>>>>>>>>>> represent
a "satisfactory"/"circle-free" machine, and adding that to >>>>>>>>>>> diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 of >>>>>>>>>>> turing's
proof and produce the psuedo-code for the machine H, assuming >>>>>>>>>>> that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>>>> what he
the theory of computation is the theory of such procedures, and >>>>>>>>> understanding the diagonal procedure is critical to
understanding the *base* contradiction/paradox that the rest of >>>>>>>>> his support for godel's result is then built on
And focusing on what is said to be impossible and not changing >>>>>>>> the problem is also important.
The problem with the diagonal generation isn't the generation of >>>>>>>> the diagonal itself, but effectively enumerating the enumeration >>>>>>>> in the first place.
i don't see any indication that turing realized a difference there >>>>>>
the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration of >>>>>> the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be
computed but that "This proof, although perfectly sound, has the
disadvantage that it may leave the reader with a feeling that
'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value
in number n, there can not be any element that matches the anti-
diagonal.
It is just a natural fact of countable infinity, something it seems
you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a
subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th
figure in an. Let β be the sequence with 1-φn(m) as its n-th. figure. >>> Since β is computable, there exists a number K [== β] such that 1-
φn(n) = φK(n) for all n. Putting n = K, we have 1 = 2φK(K), i.e. 1 is >>> even. This is impossible/
i agree with this proof is far as much as if β was computable (by TMs)
we'd have a problem for sure,
but what i don't agree is that we can just assume the computability of β from the existence of φn(m), THAT'S THE FALLACy
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using
the direct diagonal. the abstract definition makes it look simple,
but this ignores the complexities of self-referential analysis (like
what turing details on the next page)
But you can, *IF* you have a machine that computes the diagonal, then
just change all the write to the output to write the opposite. Note,
the "self-reference" that you are thinking of stops being a "self-
reference" but is a reference to the original write the diagonal code.
in both methods i have for rectifying the paradox found in the direct
diagonal (either (1) filtering TMs or (2) using RTMs), neither can be
used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it gets
to the number of TURING'S H, as there is no correct answer for the
machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard coded
value that is inverse to what it does return ... such a concept is
entirely nonsensical. a function can only return what it does, it
can't also return the inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus your
enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT *SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs just
fails for reasons u'd only understand by working thru the algo urself
(p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually computable
given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do that,
as you are thinking the only "reference" can be to "self", doesn't
make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-REFERENCE,
LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR INVERSE-H
TO BE COMPUTED
one cannot just assume that because the diagonal across computable
numbers is computable, therefore the anti-diagonal across
computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across
the computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the
Decider "D" that could be used to effectively enumerate the
sequence of machine that produce computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing out that
the attempt to compute the diagonal clearly reveals the issue with >>>>>> effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for computing >>>>> the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on his
given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide
correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT WOULDN'T BE USED
YOUR H still needs to know if Turing's H is a machine that generates a
computable number to build the list of such machine to define the
diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH ARE
INCOMPLETE SPECIFICATIONS.
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH WILL WORK
JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE INPUT/ TO D FILTERED
FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE NUMBERS
It doesn't matter that your new H doesn't get stuck on itself, it
will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't
specify what D (or H) needs to do when encountering the /undecidable
input/ of H, so therefore both D and H are an incomplete
specifications of a machine
Because "undecidability" doesn't affect the correct answer it must
compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE LIKE
THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE IT DOES
NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL ANALYSIS, SO IT IS
AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have created an
actual machine we claim to be the D, has an answer, and that D is
always wrong.
Thus, it isn't that machines claiming to be D and H can't exist, only
machines CORRECTLY meeting the requirements of D and H can't exist.
Thus, if your H depends on that D, (or a machine that meets its
specification) then it also can't exist.
You just make the error of saying the problems don't exist because you
can't build them, but YOUR machine can, even though it has the same
problem.
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL
be circle free (since it never tries to simulate itself) and thus
DOES produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H
tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D
must have this flaw, as no D without exists), and thus you will be
wrong on THAT input. IT doesn't matter if you get a good answer for
yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of
bad logic. You would have likely been escorted out of the meeting as
showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
H shows that *IF* you can make that enumeration, you can make >>>>>>>> the diagonal, and thus the anti-diagonal. The problem is you
can't make that enumeration, and assuming you can just shows
unsoundness.
interestingly: one can only fix the direct diagonal computation >>>>>>> like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an >>>>>>> inverted value, a machine can only return what it does, not the >>>>>>> inverse of what it does...
so if we can filter out paradoxes from the enumeration, that will >>>>>>> leave a direct diagonal computation extant in that filtered (yet >>>>>>> still turing complete list), while any attempt to compute an
inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to
decide them.
right. so if ur constructing a diagonal across computable numbers
then u only need to filter out paradoxes in regards to the
classifier that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such paradoxes...
In other words, you beleive unquestionably in the existance of
Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES CAN'T BE
DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING A
FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU MADE???
For THIS paradox, read the proof. for THIS H, the one given, (for
which ever version of D you want to try to assume is right) D is just
wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox, because
there still isn't an answer it can give to H that will be correct, as
the template for H will always make that D wrong.
Note, the problem is you can't actually DEFINE what a "paradox
machine" is, as they aren't actually machines, but templates that
build machines. And that final machine doesn't have actually
detectable tell-tales that show it to be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO WORK
WITH, BUT OH WELL
They only seem "random" because you don't know what you are talking
about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* classifiable
as satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those that
are but it can not classify as such, or your enumeration will not be
complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it
will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable
numbers already included on this diagonal, any machine which either >>>>> *is* computably equivalent OR *is not* classifiable in regards to
*any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of
the computable numbers.
see you can't compute a diagonal across *all* /machines/, with said >>>>> machines, but u can compute a diagonal across *all* /computable
numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, since
ALL classifiers that attempt it will make an error, you can't do
what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not
the same thing as (b) computing the enumeration of all machines that
compute computable numbers. (b) necessarily has duplicates while (a)
does not need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the full
list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of the H
he describes. It doesn't matter that you can make a DIFFERENT machine,
that you try to deceptively call "H" too that it won't have a problem
with, when your "H" gets to the number of Turing's H, it still has the
problem. It can call that one not circle-free, and thus omit a circle-
free machine from the list, or call it circle-free, and when even YOU
try to simulate it the k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take the code
of (a), and invert the symbol put on the tape when it is written.
Note, it won't be "self-referent", as the "N" is uses, is the N of
YOUR H, not itself. The code for his (a) NEVER reads back what it
wrote, so that doesn't affect its behavior.
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without
such paradox
And the problem is that your "paradoxical" isn't actually a
definable property (let alone computable). Part of the problem is
that if you look at just a machine description, it doesn't
(necessarily) tell you about the use of an "interface" as that use
of an interface can be just inlined, leaving nothing "in the code"
to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the classifier halts(), easily demonstrable thru a short proof i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME FUCKING FORM OF INSANITY I CAN'T HONESTLY FATHOM
lol
i'm sorry, you have an issue with me labeling a number that
His specified H, with an actually (incorrect) implementation of D >>>>>> (which is all that CAN exist) will either be circle-free and thus >>>>>> generate a number (but its D said it isn't, and thus omitted a
valid machine from the list) or it isn't circle-free, and fails to >>>>>> computa a number, and thus should have been omitted from the list >>>>>> but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just
built on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an
anti- diagonal computation, due the same particular self-
referential weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just
based on ERROR.
Your attempt to REDEFINE self-reference to mean processing a
number that happens to represent yourself means that you you
system "ALL" doesn't actually mean ALL, and thus is just ill-defined. >>>>>
directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have
it's a *specific* number that has the currently running machine
encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different
numbers.
those are references to functionally equivalent machines, not self-
references
Which is why the concept of "self-reference" doesn't work.
It may let you filter out the simplest case used in the proofs, but
doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox
detector while filter *out* paradoxes to point of creating an actual
problem
Then show how you will do it.
This means you need to detect a input that represents a machine that
uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an
input classifier, and an input machine to search for paradoxes in
regards to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier
like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
And the problem is "paradox_free" can't exist as an always correct
decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case
doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine
is necessary for the particular input->output computation being done,
so utilizing the return paradox_free(halts,undp) -> FALSE to filter
out paradox_free() paradoxes as well as halts() paradoxes will
suffice to produce a turing complete subset of machines that can be
totally classified by halts()
But we can still build an input that your paradox_free() fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE INCLUDED ON
THE DIAGONAL???
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the only
possibility here to compute what we want. TM's can't achieve that,
but something slightly messier can.
No, it is your dependence on being able to assume that unicorns can
solve all your problems.
dear future: u see this shit i'm trying work with???? sorry i
didn't manage to make progress any faster, but i'm literally
dragging toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually
be used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
They are, But you are too stuck in your lies to see it.
Your "logic" is based on being able to assume that Unicorn's exist,
and that is a safe assumption until someone can prove that Russel's
Teapot is not out there. (Even though your unicorns HAVE been proved
to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't understand what >>>>>> the actual problem is, and your world is just build on things that >>>>>> are lies.
...which u won't find interesting, but i can't fix a lack of
curiosity
🤷
expected was an inexpert audience, supposing that experts will >>>>>>>>>> recognise
the relevant mapping to universal quantification?
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that >>>>>> they need to be fully defined in the actions they do.
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:Then you zre just showing your stupidity, because YOU can't tell >>>>>>> the difference.
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't tell >>>>>>>>>>>>>>> me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an
infinitude one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>>> comptuation
that enumerates out all the numbers, testing each one of >>>>>>>>>>>> they represent
a "satisfactory"/"circle-free" machine, and adding that to >>>>>>>>>>>> diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 >>>>>>>>>>>> of turing's
proof and produce the psuedo-code for the machine H,
assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure he >>>>>>>>>>> didn't
reason quantified over all but phrase it like a procedure for >>>>>>>>>>> what he
the theory of computation is the theory of such procedures, >>>>>>>>>> and understanding the diagonal procedure is critical to
understanding the *base* contradiction/paradox that the rest >>>>>>>>>> of his support for godel's result is then built on
And focusing on what is said to be impossible and not changing >>>>>>>>> the problem is also important.
The problem with the diagonal generation isn't the generation >>>>>>>>> of the diagonal itself, but effectively enumerating the
enumeration in the first place.
i don't see any indication that turing realized a difference there >>>>>>>
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration >>>>>>> of the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be >>>>>>> computed but that "This proof, although perfectly sound, has the >>>>>>> disadvantage that it may leave the reader with a feeling that
'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement,
which shows that since for all n, position n differs from the value >>>>> in number n, there can not be any element that matches the anti-
diagonal.
It is just a natural fact of countable infinity, something it seems >>>>> you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short
diagonal proof, but in writing this post i now find myself in a
subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th
figure in an. Let β be the sequence with 1-φn(m) as its n-th.
figure. Since β is computable, there exists a number K [== β] such
that 1- φn(n) = φK(n) for all n. Putting n = K, we have 1 = 2φK(K), >>>> i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable (by TMs)
we'd have a problem for sure,
but what i don't agree is that we can just assume the computability of
β from the existence of φn(m), THAT'S THE FALLACy
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using
the direct diagonal. the abstract definition makes it look simple,
but this ignores the complexities of self-referential analysis (like
what turing details on the next page)
But you can, *IF* you have a machine that computes the diagonal, then
just change all the write to the output to write the opposite. Note,
the "self-reference" that you are thinking of stops being a "self-
reference" but is a reference to the original write the diagonal code.
in both methods i have for rectifying the paradox found in the
direct diagonal (either (1) filtering TMs or (2) using RTMs),
neither can be used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it gets
to the number of TURING'S H, as there is no correct answer for the
machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out like
turing's paradoxical variation of the direct diagonal would be, and
there is no analogous non-paradoxical variation that has a hard
coded value that is inverse to what it does return ... such a
concept is entirely nonsensical. a function can only return what it
does, it can't also return the inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus your
enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT *SHOULD*
BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs just
fails for reasons u'd only understand by working thru the algo
urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually computable
given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do
that, as you are thinking the only "reference" can be to "self",
doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-REFERENCE,
LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR INVERSE-
H TO BE COMPUTED
one cannot just assume that because the diagonal across computable >>>>>> numbers is computable, therefore the anti-diagonal across
computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation across
the computable numbers can be used to compute the inverse diagonal
But your method still doesn't let you compute the enumeration, and
thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute
computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
Why?
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable >>>>>
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other
machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the >>>>>>> Decider "D" that could be used to effectively enumerate the
sequence of machine that produce computable numbers can not esit. >>>>>>>
Thus, he clearly knows the difference, but is pointing out that >>>>>>> the attempt to compute the diagonal clearly reveals the issue
with effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for
computing the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on
his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide
correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT WOULDN'T
BE USED
incredibly, ironically, AND not co-incidently: partial recognizer D
*can* decide on the fixed H because the fixed H does not try to use any
D on itself, so no self-referential paradox is possible in regards to
it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer D (used in fixed
H, which does not call any D itself), and would not be filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and instead returns a hard-coded value for it's own digit on the diagonal, is
keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
YOUR H still needs to know if Turing's H is a machine that generates
a computable number to build the list of such machine to define the
diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH ARE
INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH WILL WORK
JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE INPUT/ TO D
FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE NUMBERS
It doesn't matter that your new H doesn't get stuck on itself, it
will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't
specify what D (or H) needs to do when encountering the /undecidable
input/ of H, so therefore both D and H are an incomplete
specifications of a machine
Because "undecidability" doesn't affect the correct answer it must
compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE LIKE
THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN IDIOSYNCRASIES, AND IT
CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE IT DOES
NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL ANALYSIS, SO IT IS
AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have created
an actual machine we claim to be the D, has an answer, and that D is
always wrong.
Thus, it isn't that machines claiming to be D and H can't exist, only
machines CORRECTLY meeting the requirements of D and H can't exist.
Thus, if your H depends on that D, (or a machine that meets its
specification) then it also can't exist.
You just make the error of saying the problems don't exist because
you can't build them, but YOUR machine can, even though it has the
same problem.
IF D is wrong by deciding it is not circle free, then your H will
compute the wrong diagonal, as the resulting version of his H WILL
be circle free (since it never tries to simulate itself) and thus
DOES produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you H >>>>> tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you
WILL hit these actual machines built on your erroneous D (your D
must have this flaw, as no D without exists), and thus you will be
wrong on THAT input. IT doesn't matter if you get a good answer for >>>>> yourself.
idk what he would have said about it, but prolly something more
substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion of >>>>> bad logic. You would have likely been escorted out of the meeting
as showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding.
H shows that *IF* you can make that enumeration, you can make >>>>>>>>> the diagonal, and thus the anti-diagonal. The problem is you >>>>>>>>> can't make that enumeration, and assuming you can just shows >>>>>>>>> unsoundness.
interestingly: one can only fix the direct diagonal computation >>>>>>>> like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an >>>>>>>> inverted value, a machine can only return what it does, not the >>>>>>>> inverse of what it does...
so if we can filter out paradoxes from the enumeration, that
will leave a direct diagonal computation extant in that filtered >>>>>>>> (yet still turing complete list), while any attempt to compute >>>>>>>> an inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in
isolation, but only in relationship to a given machine trying to >>>>>>> decide them.
right. so if ur constructing a diagonal across computable numbers >>>>>> then u only need to filter out paradoxes in regards to the
classifier that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such
paradoxes...
In other words, you beleive unquestionably in the existance of
Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES CAN'T BE
DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING A
FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU MADE???
For THIS paradox, read the proof. for THIS H, the one given, (for
which ever version of D you want to try to assume is right) D is just
wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox,
because there still isn't an answer it can give to H that will be
correct, as the template for H will always make that D wrong.
Note, the problem is you can't actually DEFINE what a "paradox
machine" is, as they aren't actually machines, but templates that
build machines. And that final machine doesn't have actually
detectable tell-tales that show it to be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO WORK
WITH, BUT OH WELL
They only seem "random" because you don't know what you are talking
about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* classifiable >>>>>> as satisfactory by said classifier... can just be skipped
No, it can only skip those that are not satisfactory, not those
that are but it can not classify as such, or your enumeration will
not be complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as it >>>>> will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable >>>>>> numbers already included on this diagonal, any machine which
either *is* computably equivalent OR *is not* classifiable in
regards to *any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of
the computable numbers.
see you can't compute a diagonal across *all* /machines/, with
said machines, but u can compute a diagonal across *all* /
computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done,
since ALL classifiers that attempt it will make an error, you can't >>>>> do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is not
the same thing as (b) computing the enumeration of all machines that
compute computable numbers. (b) necessarily has duplicates while (a)
does not need them. turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the full
list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of the
H he describes. It doesn't matter that you can make a DIFFERENT
machine, that you try to deceptively call "H" too that it won't have
a problem with, when your "H" gets to the number of Turing's H, it
still has the problem. It can call that one not circle-free, and thus
omit a circle- free machine from the list, or call it circle-free,
and when even YOU try to simulate it the k steps, you get stuck in a
loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take the
code of (a), and invert the symbol put on the tape when it is
written. Note, it won't be "self-referent", as the "N" is uses, is
the N of YOUR H, not itself. The code for his (a) NEVER reads back
what it wrote, so that doesn't affect its behavior.
yes, i still do need to prove my thesis that for any paradoxical
machine, there exists a functionally equivalent machine without
such paradox
And the problem is that your "paradoxical" isn't actually a
definable property (let alone computable). Part of the problem is
that if you look at just a machine description, it doesn't
(necessarily) tell you about the use of an "interface" as that use
of an interface can be just inlined, leaving nothing "in the code"
to show it exists.
i'm sorry, are you actually saying the machine description does not
describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the classifier
halts(), easily demonstrable thru a short proof i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY OF
UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME FUCKING
FORM OF INSANITY I CAN'T HONESTLY FATHOM
lol
it's a *specific* number that has the currently running machine
His specified H, with an actually (incorrect) implementation of D >>>>>>> (which is all that CAN exist) will either be circle-free and thus >>>>>>> generate a number (but its D said it isn't, and thus omitted a
valid machine from the list) or it isn't circle-free, and fails >>>>>>> to computa a number, and thus should have been omitted from the >>>>>>> list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just
built on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an
anti- diagonal computation, due the same particular self-
referential weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just
based on ERROR.
Your attempt to REDEFINE self-reference to mean processing a
number that happens to represent yourself means that you you
system "ALL" doesn't actually mean ALL, and thus is just ill-
defined.
i'm sorry, you have an issue with me labeling a number that
directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have >>>>
encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different
numbers.
those are references to functionally equivalent machines, not self-
references
Which is why the concept of "self-reference" doesn't work.
It may let you filter out the simplest case used in the proofs, but >>>>> doesn't solve the actual problem, as the "Machine Number" doesn't
actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the paradox
detector while filter *out* paradoxes to point of creating an actual
problem
Then show how you will do it.
This means you need to detect a input that represents a machine that
uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an
input classifier, and an input machine to search for paradoxes in
regards to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier
like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
And the problem is "paradox_free" can't exist as an always correct
decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case
doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine
is necessary for the particular input->output computation being
done, so utilizing the return paradox_free(halts,undp) -> FALSE to
filter out paradox_free() paradoxes as well as halts() paradoxes
will suffice to produce a turing complete subset of machines that
can be totally classified by halts()
But we can still build an input that your paradox_free() fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE INCLUDED
ON THE DIAGONAL???
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the only
possibility here to compute what we want. TM's can't achieve that,
but something slightly messier can.
No, it is your dependence on being able to assume that unicorns can
solve all your problems.
dear future: u see this shit i'm trying work with???? sorry i
didn't manage to make progress any faster, but i'm literally
dragging toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually
be used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
They are, But you are too stuck in your lies to see it.
Your "logic" is based on being able to assume that Unicorn's exist,
and that is a safe assumption until someone can prove that Russel's
Teapot is not out there. (Even though your unicorns HAVE been proved
to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't understand
...which u won't find interesting, but i can't fix a lack of
curiosity
🤷
expected was an inexpert audience, supposing that experts >>>>>>>>>>> will recognise
the relevant mapping to universal quantification?
what the actual problem is, and your world is just build on
things that are lies.
IT seems that fundamentally, your world doesn't actually have
"computations" as you don't understand the basic requirement that >>>>>>> they need to be fully defined in the actions they do.
On 2/25/26 1:15 PM, dart200 wrote:
On 2/25/26 4:04 AM, Richard Damon wrote:
On 2/25/26 12:40 AM, dart200 wrote:
On 2/24/26 8:29 PM, Richard Damon wrote:
On 2/24/26 11:13 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:55 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't tell >>>>>>>>>>>>>>> me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an
infinitude one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>>> comptuation that enumerates out all the numbers, testing >>>>>>>>>>>> each one of they represent a "satisfactory"/"circle-free" >>>>>>>>>>>> machine, and adding that to diagonal across defined across >>>>>>>>>>>> computable numbers
it really would be a great exercise to carefully read p247 >>>>>>>>>>>> of turing's proof and produce the psuedo-code for the >>>>>>>>>>>> machine H, assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of >>>>>>>>>>> computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as >>>>>>>>>>> they continue to produce results.
The Halting Problem (vs the computable number problem) is >>>>>>>>>>> based on Computations that must return answers in finite time. >>>>>>>>>>>
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting >>>>>>>>>> problem, but he didn't
No, it is relevent because *YOU* talk about the Halting
Problem, claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting
problem and the satisfactory problem in turing's paper. my paper >>>>>>>> was on the satisfactory problem in turing's paper because i
wanted to address the source
Then you shouldn't have started (and continued) about talking
about the Halting Problem and contray machines.
Note, This paper never (as far as I remember) works by creating a >>>>>>> specific "contrary" or 'paradoxical" machine to decide on. The
error comes naturally out of the problem itself.
if u had read my paper (not just like glanced at it), u'd know
there is a method to construct the diagonal computation such that >>>>>> it avoids stumbling on deciding on itself,
Only by assuming that Unicorns exist.
doesn't require a fixed decider to avoid itself, just a quine based
self-reference
Doesn't work.
more humility would do u wonders dick
Could say the same about you.
The problem is YOU are the one making stupid claims, not me, thus YOU
are the one showing stupidity on insisting on lies.
So, what answer does that quine based decider give for the question
about itself?
it's not the decider that has a quine, it's the diagonal computation H
that uses D that has the quine to recognize H.
But the problem is it isn;t the Diagonal Computation that is said to be impossible, but the decider.
The H he descrives still exists, even if you make your variant, but that
H can't be decided on.
when the diagonal computation recognizes it's own number in the total
machine enumeration, it never asks the decider on itself, when it
identifies itself in the total machine numeration it adds some fixed
digit to the diagonal instead of recursively getting stuck in
simulating itself.
So, as I said, it isn't the diagonal computation that is said to be impossible, just the decider that enumerates the diagonal machines.
WHat does that decider do for the non-quine version described.
doesn't matter what digit: both are different machines, with different
indexes in the enumeration, that compute the diagonal
So? what does D do with the original H.
This is your problem, you don't pay attention to the problem being
actually defined.
It is a bit like the "pathological" input for the halting problem.
Either answer is still wrong.
And how is being "quine based" keep it from being "fixed".
It still is only one unique machine.
u ofc don't know what i'm talking about and have demonstrated no
capability to hold a coherent conversation
No, YOU keep on making categorical errors not knowing the real
meaning of the terms you are using.
Since, as I pointed out, you LEFT the system in your first page by
claiming falsehoods, you didn't prove anything in the system you
claim to have been working on.
but turing's purpose was the construct a contradiction, not figure >>>>>> out what a working diagonal computation would look like
Note, its decider needs to be the same sort of machine as
machine D in Turings proof.
The decider he proves can't exist, as he proves that some
decision problems are undecidable, thus answering the
Entscheidungsproblem in the negative.
what no one has proven is that the Entscheidungsproblem can't be >>>>>>>> answered reliably thru other interfaces...
WHAT OTHER "INTERFACES"?
i've described a variety of them thus far
Really?
Not in the rules of Computation Theory, as your machines aren't
machine as they don't include all their algorithm.
u haven't defined the rules of computation theory, u just make
random claims about it whenever the fuck
I don't need to do that, as they ARE already defined in the literature.
and which literature defines "the rules"???
The "Literature", as a collection.
Yes, part of the problem is that they were more focused on the goal of figuring if this class of "decision" problems were do able, no one
actually sat down (as far as I know) to formally and definitively define
it as a system, and in fact, you see some comments in the writing about disagreeing and refining the definitions. But the basic concept is
fairly clear if you read much of the writting and remember what they
were working on. Was there some defined algorithmic method that could be used to answer the tough problems they were coming up on in mathematics
and logic.
He shows a problem that can not be decided. PERIOD.
Your "Interface" seems to be one of your undefinable terms.
i've defined it many times to u
No, you use the word but not DEFINE it.
it is the specification that a machine computes, or possibly some
meta- specification
In other words, you are "solving" a problem by solving some other
problem?
but u don't debate honorably dick, u just reject every i say even
when i agree
It seems you are just admitting your logic is based on lying,
If I have a classification decision problem, that problem has a
specification of indicating which class an input lies in.
What other "specification" can exist?
All you seem to want to do is say that if you don't need to answer
the question, you can answer the question.
In other words, you want to make lies be truth.
It seems you don't know what a DEFINTION is.
which is a set of words apparently so damn radical the lame duck >>>>>>>> industry of cs "academics" can't even comprehend the concept, >>>>>>>> let alone address it in a considerate manner
No, you don't seem to know what you are talking about.
it shocks me on a daily basis just how much abject brainrot i've >>>>>>>> waded thru in my explorations
It seems your brain has completed its rot.
And in doing so, he lays the groundwork for proving the Halting >>>>>>>>> Problem, and breaks the logjam in the thinking that there might >>>>>>>>> be procedures that answer any question we want.
what i mean by enumerating out numbers is he enumerates out >>>>>>>>>> natural numbers to find possible machines that might compute >>>>>>>>>> "computable numbers" (which are real numbers)
Also, it doesn't enumerate the "numbers", but Machines, many >>>>>>>>>>> of which will not actually produce numbers.
Then he assumes a test exists that determines if that machine >>>>>>>>>>> satisfies the requirements, and shows that there can not be >>>>>>>>>>> such a test.
Thus, he shows that there is NOT an effective enumeration of >>>>>>>>>>> computable numbers, only uncomputable enumerations of them. >>>>>>>>>>>
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is >>>>>>>>>>> computable. It would be true if we could enumerate the
computable sequences by finite means, but the problem of >>>>>>>>>>> enumerating computable sequences is equivalent to the problem >>>>>>>>>>> of finding out whether a given number is the D.N of a circle- >>>>>>>>>>> free machine, and we have no general process for doing this >>>>>>>>>>> in a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration >>>>>>>>>>> that enumerates the numbers, only one of "all machines" that >>>>>>>>>>> need an uncomputable test to see if they are in that set. >>>>>>>>>>
But he never does that. He establishes that he can produce
every possible machine in some order, and *IF* you could do the >>>>>>>>> decision test, create the computable numbers in a specified >>>>>>>>> order, but you can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you
produce a correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking >>>>>>>> so much smack without being able to code up his *really* simple >>>>>>>> algo
Nope, YOU are the one with brain rot.
no pseudo-code = no convincing
NO DEFINITIONS, nothing shown.
If you want to ignore truth, you just prove your own stupidity,
still no pseudo-code cause ur a fraud dick
Still being stupid I see.
i'm honest not sure why u bother typing responses. ur words fall on
deaf ears cause u refuse to write anything beyond shallow insults
and random claims over rules that don't actually exist
Your problem is that you fundamentally don't understand what you are
talking about, as your world seems to be based on the assumption that
things don't need to be what they are, and that lies are valid.
no pseudo-code = no convincing
Simplified version assuming D taks D.N instead of S.D
def H:
let N = 1
let K = 0
do:
if D (N) is satisified:
let K = K + 1
simulate S.D corresponding to N till it produces K symbols
output that K'th symbol
Let N = N + 1
Note, each implementation of this pseudo code will have a number N.
IF D only is statisfied by circle-free programs, that is programs that continue to produce outputs then this program MUST be circle free as
long as the following are true:
* D itself must be halting, and always produce an answer (which is part
of its definition to exist)
* There must be an infinite number of circle-free programs
And we then get to the delemma mentions on page 247,
When given this program to decide on, if it says that it is NOT circle- free, then it WILL continue to produce output forever and thus show that
it WAS circle-free and thus D was wrong.
If it says that H is circle-free, then it will simulate itself until it produces K outputs, but after that simulation produces K-1 outputs, it
will itself simulate itself and so on into the infinte loop.
Talking about a DIFFERENT machine, with a DIFFERENT D.N is irrelevant,
that other machine isn't this one,
i would ask if u care about convincing me,
Not really, just showing you are wrong.
Since you don't care about truth, you might not be convincible.
but clearly u don't
and if u don't care about convincing me,
then ur debate is not honorable, it is engaged with a lack of care, so
ofc it wouldn't be honorable, and that requires care
My goal is to keep others, who want to know, from falling into your lies.
YOU, not willing to define your terms, are the unhonerable one.
Sorry, you are just showing you don't understand what is >>>>>>>>>>> being done.
i mean u seem to imply turing talked about the halting
problem, which he didn't
No, he talked about the decision problem, and showed things are >>>>>>>>> undecidable, and laid the groundwork that was used to actually >>>>>>>>> handle the halting problem.
Yes, people say he did it, because he established the method >>>>>>>>> used for it.
On 2/25/26 1:15 PM, dart200 wrote:
On 2/25/26 4:04 AM, Richard Damon wrote:
On 2/25/26 12:40 AM, dart200 wrote:
On 2/24/26 8:29 PM, Richard Damon wrote:
On 2/24/26 11:13 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:55 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't tell >>>>>>>>>>>>>>> me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an
infinitude one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>>> comptuation that enumerates out all the numbers, testing >>>>>>>>>>>> each one of they represent a "satisfactory"/"circle-free" >>>>>>>>>>>> machine, and adding that to diagonal across defined across >>>>>>>>>>>> computable numbers
it really would be a great exercise to carefully read p247 >>>>>>>>>>>> of turing's proof and produce the psuedo-code for the >>>>>>>>>>>> machine H, assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of >>>>>>>>>>> computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as >>>>>>>>>>> they continue to produce results.
The Halting Problem (vs the computable number problem) is >>>>>>>>>>> based on Computations that must return answers in finite time. >>>>>>>>>>>
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting >>>>>>>>>> problem, but he didn't
No, it is relevent because *YOU* talk about the Halting
Problem, claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting
problem and the satisfactory problem in turing's paper. my paper >>>>>>>> was on the satisfactory problem in turing's paper because i
wanted to address the source
Then you shouldn't have started (and continued) about talking
about the Halting Problem and contray machines.
Note, This paper never (as far as I remember) works by creating a >>>>>>> specific "contrary" or 'paradoxical" machine to decide on. The
error comes naturally out of the problem itself.
if u had read my paper (not just like glanced at it), u'd know
there is a method to construct the diagonal computation such that >>>>>> it avoids stumbling on deciding on itself,
Only by assuming that Unicorns exist.
doesn't require a fixed decider to avoid itself, just a quine based
self-reference
Doesn't work.
more humility would do u wonders dick
Could say the same about you.
The problem is YOU are the one making stupid claims, not me, thus YOU
are the one showing stupidity on insisting on lies.
So, what answer does that quine based decider give for the question
about itself?
it's not the decider that has a quine, it's the diagonal computation H
that uses D that has the quine to recognize H.
But the problem is it isn;t the Diagonal Computation that is said to be impossible, but the decider.
The H he descrives still exists, even if you make your variant, but that
H can't be decided on.
when the diagonal computation recognizes it's own number in the total
machine enumeration, it never asks the decider on itself, when it
identifies itself in the total machine numeration it adds some fixed
digit to the diagonal instead of recursively getting stuck in
simulating itself.
So, as I said, it isn't the diagonal computation that is said to be impossible, just the decider that enumerates the diagonal machines.
WHat does that decider do for the non-quine version described.
doesn't matter what digit: both are different machines, with different
indexes in the enumeration, that compute the diagonal
So? what does D do with the original H.
This is your problem, you don't pay attention to the problem being
actually defined.
It is a bit like the "pathological" input for the halting problem.
Either answer is still wrong.
And how is being "quine based" keep it from being "fixed".
It still is only one unique machine.
u ofc don't know what i'm talking about and have demonstrated no
capability to hold a coherent conversation
No, YOU keep on making categorical errors not knowing the real
meaning of the terms you are using.
Since, as I pointed out, you LEFT the system in your first page by
claiming falsehoods, you didn't prove anything in the system you
claim to have been working on.
but turing's purpose was the construct a contradiction, not figure >>>>>> out what a working diagonal computation would look like
Note, its decider needs to be the same sort of machine as
machine D in Turings proof.
The decider he proves can't exist, as he proves that some
decision problems are undecidable, thus answering the
Entscheidungsproblem in the negative.
what no one has proven is that the Entscheidungsproblem can't be >>>>>>>> answered reliably thru other interfaces...
WHAT OTHER "INTERFACES"?
i've described a variety of them thus far
Really?
Not in the rules of Computation Theory, as your machines aren't
machine as they don't include all their algorithm.
u haven't defined the rules of computation theory, u just make
random claims about it whenever the fuck
I don't need to do that, as they ARE already defined in the literature.
and which literature defines "the rules"???
The "Literature", as a collection.
Yes, part of the problem is that they were more focused on the goal of figuring if this class of "decision" problems were do able, no one
actually sat down (as far as I know) to formally and definitively define
it as a system, and in fact, you see some comments in the writing about disagreeing and refining the definitions. But the basic concept is
fairly clear if you read much of the writting and remember what they
were working on. Was there some defined algorithmic method that could be used to answer the tough problems they were coming up on in mathematics
and logic.
He shows a problem that can not be decided. PERIOD.
Your "Interface" seems to be one of your undefinable terms.
i've defined it many times to u
No, you use the word but not DEFINE it.
it is the specification that a machine computes, or possibly some
meta- specification
In other words, you are "solving" a problem by solving some other
problem?
but u don't debate honorably dick, u just reject every i say even
when i agree
It seems you are just admitting your logic is based on lying,
If I have a classification decision problem, that problem has a
specification of indicating which class an input lies in.
What other "specification" can exist?
All you seem to want to do is say that if you don't need to answer
the question, you can answer the question.
In other words, you want to make lies be truth.
It seems you don't know what a DEFINTION is.
which is a set of words apparently so damn radical the lame duck >>>>>>>> industry of cs "academics" can't even comprehend the concept, >>>>>>>> let alone address it in a considerate manner
No, you don't seem to know what you are talking about.
it shocks me on a daily basis just how much abject brainrot i've >>>>>>>> waded thru in my explorations
It seems your brain has completed its rot.
And in doing so, he lays the groundwork for proving the Halting >>>>>>>>> Problem, and breaks the logjam in the thinking that there might >>>>>>>>> be procedures that answer any question we want.
what i mean by enumerating out numbers is he enumerates out >>>>>>>>>> natural numbers to find possible machines that might compute >>>>>>>>>> "computable numbers" (which are real numbers)
Also, it doesn't enumerate the "numbers", but Machines, many >>>>>>>>>>> of which will not actually produce numbers.
Then he assumes a test exists that determines if that machine >>>>>>>>>>> satisfies the requirements, and shows that there can not be >>>>>>>>>>> such a test.
Thus, he shows that there is NOT an effective enumeration of >>>>>>>>>>> computable numbers, only uncomputable enumerations of them. >>>>>>>>>>>
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § is >>>>>>>>>>> computable. It would be true if we could enumerate the
computable sequences by finite means, but the problem of >>>>>>>>>>> enumerating computable sequences is equivalent to the problem >>>>>>>>>>> of finding out whether a given number is the D.N of a circle- >>>>>>>>>>> free machine, and we have no general process for doing this >>>>>>>>>>> in a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration >>>>>>>>>>> that enumerates the numbers, only one of "all machines" that >>>>>>>>>>> need an uncomputable test to see if they are in that set. >>>>>>>>>>
But he never does that. He establishes that he can produce
every possible machine in some order, and *IF* you could do the >>>>>>>>> decision test, create the computable numbers in a specified >>>>>>>>> order, but you can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you
produce a correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking >>>>>>>> so much smack without being able to code up his *really* simple >>>>>>>> algo
Nope, YOU are the one with brain rot.
no pseudo-code = no convincing
NO DEFINITIONS, nothing shown.
If you want to ignore truth, you just prove your own stupidity,
still no pseudo-code cause ur a fraud dick
Still being stupid I see.
i'm honest not sure why u bother typing responses. ur words fall on
deaf ears cause u refuse to write anything beyond shallow insults
and random claims over rules that don't actually exist
Your problem is that you fundamentally don't understand what you are
talking about, as your world seems to be based on the assumption that
things don't need to be what they are, and that lies are valid.
no pseudo-code = no convincing
Simplified version assuming D taks D.N instead of S.D
def H:
let N = 1
let K = 0
do:
if D (N) is satisified:
let K = K + 1
simulate S.D corresponding to N till it produces K symbols
output that K'th symbol
Let N = N + 1
Note, each implementation of this pseudo code will have a number N.
IF D only is statisfied by circle-free programs, that is programs that continue to produce outputs then this program MUST be circle free as
long as the following are true:
* D itself must be halting, and always produce an answer (which is part
of its definition to exist)
* There must be an infinite number of circle-free programs
And we then get to the delemma mentions on page 247,
When given this program to decide on, if it says that it is NOT circle- free, then it WILL continue to produce output forever and thus show that
it WAS circle-free and thus D was wrong.
If it says that H is circle-free, then it will simulate itself until it produces K outputs, but after that simulation produces K-1 outputs, it
will itself simulate itself and so on into the infinte loop.
Talking about a DIFFERENT machine, with a DIFFERENT D.N is irrelevant,
that other machine isn't this one,
i would ask if u care about convincing me,
Not really, just showing you are wrong.
Since you don't care about truth, you might not be convincible.
but clearly u don't
and if u don't care about convincing me,
then ur debate is not honorable, it is engaged with a lack of care, so
ofc it wouldn't be honorable, and that requires care
My goal is to keep others, who want to know, from falling into your lies.
YOU, not willing to define your terms, are the unhonerable one.
Sorry, you are just showing you don't understand what is >>>>>>>>>>> being done.
i mean u seem to imply turing talked about the halting
problem, which he didn't
No, he talked about the decision problem, and showed things are >>>>>>>>> undecidable, and laid the groundwork that was used to actually >>>>>>>>> handle the halting problem.
Yes, people say he did it, because he established the method >>>>>>>>> used for it.
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:Then you zre just showing your stupidity, because YOU can't tell >>>>>>>> the difference.
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't >>>>>>>>>>>>>>>> tell me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>>>> comptuation
that enumerates out all the numbers, testing each one of >>>>>>>>>>>>> they represent
a "satisfactory"/"circle-free" machine, and adding that to >>>>>>>>>>>>> diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 >>>>>>>>>>>>> of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure >>>>>>>>>>>> he didn't
reason quantified over all but phrase it like a procedure >>>>>>>>>>>> for what he
the theory of computation is the theory of such procedures, >>>>>>>>>>> and understanding the diagonal procedure is critical to >>>>>>>>>>> understanding the *base* contradiction/paradox that the rest >>>>>>>>>>> of his support for godel's result is then built on
And focusing on what is said to be impossible and not changing >>>>>>>>>> the problem is also important.
The problem with the diagonal generation isn't the generation >>>>>>>>>> of the diagonal itself, but effectively enumerating the
enumeration in the first place.
i don't see any indication that turing realized a difference there >>>>>>>>
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration >>>>>>>> of the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be >>>>>>>> computed but that "This proof, although perfectly sound, has the >>>>>>>> disadvantage that it may leave the reader with a feeling that >>>>>>>> 'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement, >>>>>> which shows that since for all n, position n differs from the
value in number n, there can not be any element that matches the
anti- diagonal.
It is just a natural fact of countable infinity, something it
seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short >>>>> diagonal proof, but in writing this post i now find myself in a
subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th >>>>> figure in an. Let β be the sequence with 1-φn(m) as its n-th.
figure. Since β is computable, there exists a number K [== β] such >>>>> that 1- φn(n) = φK(n) for all n. Putting n = K, we have 1 = 2φK(K), >>>>> i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable (by
TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the computability
of β from the existence of φn(m), THAT'S THE FALLACy
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using
the direct diagonal. the abstract definition makes it look simple,
but this ignores the complexities of self-referential analysis
(like what turing details on the next page)
But you can, *IF* you have a machine that computes the diagonal,
then just change all the write to the output to write the opposite.
Note, the "self-reference" that you are thinking of stops being a
"self- reference" but is a reference to the original write the
diagonal code.
in both methods i have for rectifying the paradox found in the
direct diagonal (either (1) filtering TMs or (2) using RTMs),
neither can be used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it
gets to the number of TURING'S H, as there is no correct answer for
the machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out like >>>>> turing's paradoxical variation of the direct diagonal would be, and >>>>> there is no analogous non-paradoxical variation that has a hard
coded value that is inverse to what it does return ... such a
concept is entirely nonsensical. a function can only return what it >>>>> does, it can't also return the inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus your
enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT *SHOULD*
BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs just
fails for reasons u'd only understand by working thru the algo
urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually computable >>>>> given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do
that, as you are thinking the only "reference" can be to "self",
doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-REFERENCE,
LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR
INVERSE- H TO BE COMPUTED
one cannot just assume that because the diagonal across
computable numbers is computable, therefore the anti-diagonal
across computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
But your method still doesn't let you compute the enumeration, and >>>>>> thus you can't actually compute the diagonal.
neither method i have for fixing the diagonal computation across >>>>>>> the computable numbers can be used to compute the inverse diagonal >>>>>>
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute >>>>>> computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
Why?
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable >>>>>>
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other >>>>>> machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the >>>>>>>> Decider "D" that could be used to effectively enumerate the
sequence of machine that produce computable numbers can not esit. >>>>>>>>
Thus, he clearly knows the difference, but is pointing out that >>>>>>>> the attempt to compute the diagonal clearly reveals the issue >>>>>>>> with effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for
computing the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on
his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide
correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT WOULDN'T
BE USED
incredibly, ironically, AND not co-incidently: partial recognizer D
*can* decide on the fixed H because the fixed H does not try to use
any D on itself, so no self-referential paradox is possible in regards
to it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable AND
computable value)
the fixed H is /decidable input/ to partial recognizer D (used in
fixed H, which does not call any D itself), and would not be filtered
out by paradox detectors
the fact fixed H does not use any classifier D on itself, and instead
returns a hard-coded value for it's own digit on the diagonal, is
keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just anouncing
that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it will fail
on that.
If D errs on Turing_H and calls it circle_free, then fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, then your
enumeration is missing the circle_free H from its list, and thus doesn't make the right diagonal.
YOUR H still needs to know if Turing's H is a machine that generates
a computable number to build the list of such machine to define the
diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH ARE
INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH WILL
WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE INPUT/ TO D
FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE NUMBERS
It doesn't matter that your new H doesn't get stuck on itself, it >>>>>> will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't
specify what D (or H) needs to do when encountering the /
undecidable input/ of H, so therefore both D and H are an
incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it must
compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE LIKE
THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN IDIOSYNCRASIES, AND
IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE IT
DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL ANALYSIS, SO
IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have created
an actual machine we claim to be the D, has an answer, and that D is
always wrong.
Thus, it isn't that machines claiming to be D and H can't exist,
only machines CORRECTLY meeting the requirements of D and H can't
exist.
Thus, if your H depends on that D, (or a machine that meets its
specification) then it also can't exist.
You just make the error of saying the problems don't exist because
you can't build them, but YOUR machine can, even though it has the
same problem.
IF D is wrong by deciding it is not circle free, then your H will >>>>>> compute the wrong diagonal, as the resulting version of his H WILL >>>>>> be circle free (since it never tries to simulate itself) and thus >>>>>> DOES produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you >>>>>> H tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you >>>>>> WILL hit these actual machines built on your erroneous D (your D
must have this flaw, as no D without exists), and thus you will be >>>>>> wrong on THAT input. IT doesn't matter if you get a good answer
for yourself.
idk what he would have said about it, but prolly something more >>>>>>> substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion
of bad logic. You would have likely been escorted out of the
meeting as showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding. >>>>>>>>
H shows that *IF* you can make that enumeration, you can make >>>>>>>>>> the diagonal, and thus the anti-diagonal. The problem is you >>>>>>>>>> can't make that enumeration, and assuming you can just shows >>>>>>>>>> unsoundness.
interestingly: one can only fix the direct diagonal computation >>>>>>>>> like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an >>>>>>>>> inverted value, a machine can only return what it does, not the >>>>>>>>> inverse of what it does...
so if we can filter out paradoxes from the enumeration, that >>>>>>>>> will leave a direct diagonal computation extant in that
filtered (yet still turing complete list), while any attempt to >>>>>>>>> compute an inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in >>>>>>>> isolation, but only in relationship to a given machine trying to >>>>>>>> decide them.
right. so if ur constructing a diagonal across computable numbers >>>>>>> then u only need to filter out paradoxes in regards to the
classifier that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such
paradoxes...
In other words, you beleive unquestionably in the existance of
Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES CAN'T BE
DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING A
FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU MADE???
For THIS paradox, read the proof. for THIS H, the one given, (for
which ever version of D you want to try to assume is right) D is
just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox,
because there still isn't an answer it can give to H that will be
correct, as the template for H will always make that D wrong.
Note, the problem is you can't actually DEFINE what a "paradox
machine" is, as they aren't actually machines, but templates that
build machines. And that final machine doesn't have actually
detectable tell-tales that show it to be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO WORK
WITH, BUT OH WELL
They only seem "random" because you don't know what you are talking
about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not*
classifiable as satisfactory by said classifier... can just be
skipped
No, it can only skip those that are not satisfactory, not those
that are but it can not classify as such, or your enumeration will >>>>>> not be complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as
it will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable >>>>>>> numbers already included on this diagonal, any machine which
either *is* computably equivalent OR *is not* classifiable in
regards to *any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of >>>>>> the computable numbers.
see you can't compute a diagonal across *all* /machines/, with
said machines, but u can compute a diagonal across *all* /
computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done,
since ALL classifiers that attempt it will make an error, you
can't do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is
not the same thing as (b) computing the enumeration of all machines >>>>> that compute computable numbers. (b) necessarily has duplicates
while (a) does not need them. turing's paper wrongly conflates (a)
with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the
full list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of the
H he describes. It doesn't matter that you can make a DIFFERENT
machine, that you try to deceptively call "H" too that it won't have
a problem with, when your "H" gets to the number of Turing's H, it
still has the problem. It can call that one not circle-free, and
thus omit a circle- free machine from the list, or call it circle-
free, and when even YOU try to simulate it the k steps, you get
stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take the
code of (a), and invert the symbol put on the tape when it is
written. Note, it won't be "self-referent", as the "N" is uses, is
the N of YOUR H, not itself. The code for his (a) NEVER reads back
what it wrote, so that doesn't affect its behavior.
yes, i still do need to prove my thesis that for any paradoxical >>>>>>> machine, there exists a functionally equivalent machine without >>>>>>> such paradox
And the problem is that your "paradoxical" isn't actually a
definable property (let alone computable). Part of the problem is >>>>>> that if you look at just a machine description, it doesn't
(necessarily) tell you about the use of an "interface" as that use >>>>>> of an interface can be just inlined, leaving nothing "in the code" >>>>>> to show it exists.
i'm sorry, are you actually saying the machine description does not >>>>> describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the classifier
halts(), easily demonstrable thru a short proof i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY OF
UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME FUCKING
FORM OF INSANITY I CAN'T HONESTLY FATHOM
lol
it's a *specific* number that has the currently running machine
His specified H, with an actually (incorrect) implementation of >>>>>>>> D (which is all that CAN exist) will either be circle-free and >>>>>>>> thus generate a number (but its D said it isn't, and thus
omitted a valid machine from the list) or it isn't circle-free, >>>>>>>> and fails to computa a number, and thus should have been omitted >>>>>>>> from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just >>>>>>>> built on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an >>>>>>>>> anti- diagonal computation, due the same particular self-
referential weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just >>>>>>>> based on ERROR.
Your attempt to REDEFINE self-reference to mean processing a
number that happens to represent yourself means that you you
system "ALL" doesn't actually mean ALL, and thus is just ill- >>>>>>>> defined.
i'm sorry, you have an issue with me labeling a number that
directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have >>>>>
encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different
numbers.
those are references to functionally equivalent machines, not self- >>>>> references
Which is why the concept of "self-reference" doesn't work.
It may let you filter out the simplest case used in the proofs,
but doesn't solve the actual problem, as the "Machine Number"
doesn't actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the
paradox detector while filter *out* paradoxes to point of creating
an actual problem
Then show how you will do it.
This means you need to detect a input that represents a machine that
uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an
input classifier, and an input machine to search for paradoxes in
regards to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier
like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
And the problem is "paradox_free" can't exist as an always correct
decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case
doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine >>>>> is necessary for the particular input->output computation being
done, so utilizing the return paradox_free(halts,undp) -> FALSE to
filter out paradox_free() paradoxes as well as halts() paradoxes
will suffice to produce a turing complete subset of machines that
can be totally classified by halts()
But we can still build an input that your paradox_free() fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE INCLUDED
ON THE DIAGONAL???
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the only >>>>> possibility here to compute what we want. TM's can't achieve that,
but something slightly messier can.
No, it is your dependence on being able to assume that unicorns can
solve all your problems.
dear future: u see this shit i'm trying work with???? sorry i
didn't manage to make progress any faster, but i'm literally
dragging toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually >>>>>> be used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
They are, But you are too stuck in your lies to see it.
Your "logic" is based on being able to assume that Unicorn's exist,
and that is a safe assumption until someone can prove that Russel's
Teapot is not out there. (Even though your unicorns HAVE been proved
to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't understand >>>>>>>> what the actual problem is, and your world is just build on
...which u won't find interesting, but i can't fix a lack of >>>>>>>>> curiosity
🤷
expected was an inexpert audience, supposing that experts >>>>>>>>>>>> will recognise
the relevant mapping to universal quantification?
things that are lies.
IT seems that fundamentally, your world doesn't actually have >>>>>>>> "computations" as you don't understand the basic requirement
that they need to be fully defined in the actions they do.
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:Then you zre just showing your stupidity, because YOU can't tell >>>>>>>> the difference.
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't >>>>>>>>>>>>>>>> tell me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>>>> comptuation
that enumerates out all the numbers, testing each one of >>>>>>>>>>>>> they represent
a "satisfactory"/"circle-free" machine, and adding that to >>>>>>>>>>>>> diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 >>>>>>>>>>>>> of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure >>>>>>>>>>>> he didn't
reason quantified over all but phrase it like a procedure >>>>>>>>>>>> for what he
the theory of computation is the theory of such procedures, >>>>>>>>>>> and understanding the diagonal procedure is critical to >>>>>>>>>>> understanding the *base* contradiction/paradox that the rest >>>>>>>>>>> of his support for godel's result is then built on
And focusing on what is said to be impossible and not changing >>>>>>>>>> the problem is also important.
The problem with the diagonal generation isn't the generation >>>>>>>>>> of the diagonal itself, but effectively enumerating the
enumeration in the first place.
i don't see any indication that turing realized a difference there >>>>>>>>
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration >>>>>>>> of the computable sequences.
He then points out that he can directly show that the "anti-
diagonal" of the (non-effectively computed) enumeration can't be >>>>>>>> computed but that "This proof, although perfectly sound, has the >>>>>>>> disadvantage that it may leave the reader with a feeling that >>>>>>>> 'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal arguement, >>>>>> which shows that since for all n, position n differs from the
value in number n, there can not be any element that matches the
anti- diagonal.
It is just a natural fact of countable infinity, something it
seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's short >>>>> diagonal proof, but in writing this post i now find myself in a
subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th >>>>> figure in an. Let β be the sequence with 1-φn(m) as its n-th.
figure. Since β is computable, there exists a number K [== β] such >>>>> that 1- φn(n) = φK(n) for all n. Putting n = K, we have 1 = 2φK(K), >>>>> i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable (by
TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the computability
of β from the existence of φn(m), THAT'S THE FALLACy
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using
the direct diagonal. the abstract definition makes it look simple,
but this ignores the complexities of self-referential analysis
(like what turing details on the next page)
But you can, *IF* you have a machine that computes the diagonal,
then just change all the write to the output to write the opposite.
Note, the "self-reference" that you are thinking of stops being a
"self- reference" but is a reference to the original write the
diagonal code.
in both methods i have for rectifying the paradox found in the
direct diagonal (either (1) filtering TMs or (2) using RTMs),
neither can be used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it
gets to the number of TURING'S H, as there is no correct answer for
the machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out like >>>>> turing's paradoxical variation of the direct diagonal would be, and >>>>> there is no analogous non-paradoxical variation that has a hard
coded value that is inverse to what it does return ... such a
concept is entirely nonsensical. a function can only return what it >>>>> does, it can't also return the inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus your
enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT *SHOULD*
BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs just
fails for reasons u'd only understand by working thru the algo
urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually computable >>>>> given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do
that, as you are thinking the only "reference" can be to "self",
doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-REFERENCE,
LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR
INVERSE- H TO BE COMPUTED
one cannot just assume that because the diagonal across
computable numbers is computable, therefore the anti-diagonal
across computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
But your method still doesn't let you compute the enumeration, and >>>>>> thus you can't actually compute the diagonal.
neither method i have for fixing the diagonal computation across >>>>>>> the computable numbers can be used to compute the inverse diagonal >>>>>>
Remember, the problem definitions requires that the listing be a
COMPLETE listing of the computable numbers / machine that compute >>>>>> computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct.
Why?
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not computable >>>>>>
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the other >>>>>> machines, including his original H that doesn't use your "trick"
But instead, he can prove with a more obvious process, that the >>>>>>>> Decider "D" that could be used to effectively enumerate the
sequence of machine that produce computable numbers can not esit. >>>>>>>>
Thus, he clearly knows the difference, but is pointing out that >>>>>>>> the attempt to compute the diagonal clearly reveals the issue >>>>>>>> with effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for
computing the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on
his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H,
because my response to this is that D does not need to decide
correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT WOULDN'T
BE USED
incredibly, ironically, AND not co-incidently: partial recognizer D
*can* decide on the fixed H because the fixed H does not try to use
any D on itself, so no self-referential paradox is possible in regards
to it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable AND
computable value)
the fixed H is /decidable input/ to partial recognizer D (used in
fixed H, which does not call any D itself), and would not be filtered
out by paradox detectors
the fact fixed H does not use any classifier D on itself, and instead
returns a hard-coded value for it's own digit on the diagonal, is
keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just anouncing
that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it will fail
on that.
If D errs on Turing_H and calls it circle_free, then fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, then your
enumeration is missing the circle_free H from its list, and thus doesn't make the right diagonal.
YOUR H still needs to know if Turing's H is a machine that generates
a computable number to build the list of such machine to define the
diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH ARE
INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH WILL
WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE INPUT/ TO D
FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE NUMBERS
It doesn't matter that your new H doesn't get stuck on itself, it >>>>>> will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't
specify what D (or H) needs to do when encountering the /
undecidable input/ of H, so therefore both D and H are an
incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it must
compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE LIKE
THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN IDIOSYNCRASIES, AND
IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE IT
DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL ANALYSIS, SO
IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have created
an actual machine we claim to be the D, has an answer, and that D is
always wrong.
Thus, it isn't that machines claiming to be D and H can't exist,
only machines CORRECTLY meeting the requirements of D and H can't
exist.
Thus, if your H depends on that D, (or a machine that meets its
specification) then it also can't exist.
You just make the error of saying the problems don't exist because
you can't build them, but YOUR machine can, even though it has the
same problem.
IF D is wrong by deciding it is not circle free, then your H will >>>>>> compute the wrong diagonal, as the resulting version of his H WILL >>>>>> be circle free (since it never tries to simulate itself) and thus >>>>>> DOES produce an computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then when you >>>>>> H tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order, you >>>>>> WILL hit these actual machines built on your erroneous D (your D
must have this flaw, as no D without exists), and thus you will be >>>>>> wrong on THAT input. IT doesn't matter if you get a good answer
for yourself.
idk what he would have said about it, but prolly something more >>>>>>> substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion
of bad logic. You would have likely been escorted out of the
meeting as showing you were unqualified and being a distraction.
Something that seems to be beyond your ignorant understanding. >>>>>>>>
H shows that *IF* you can make that enumeration, you can make >>>>>>>>>> the diagonal, and thus the anti-diagonal. The problem is you >>>>>>>>>> can't make that enumeration, and assuming you can just shows >>>>>>>>>> unsoundness.
interestingly: one can only fix the direct diagonal computation >>>>>>>>> like this
u can't do an analogous fix for the inverse/anti-diagonal
computation. it's not possible hard code a machine to return an >>>>>>>>> inverted value, a machine can only return what it does, not the >>>>>>>>> inverse of what it does...
so if we can filter out paradoxes from the enumeration, that >>>>>>>>> will leave a direct diagonal computation extant in that
filtered (yet still turing complete list), while any attempt to >>>>>>>>> compute an inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in >>>>>>>> isolation, but only in relationship to a given machine trying to >>>>>>>> decide them.
right. so if ur constructing a diagonal across computable numbers >>>>>>> then u only need to filter out paradoxes in regards to the
classifier that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such
paradoxes...
In other words, you beleive unquestionably in the existance of
Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES CAN'T BE
DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING A
FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU MADE???
For THIS paradox, read the proof. for THIS H, the one given, (for
which ever version of D you want to try to assume is right) D is
just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox,
because there still isn't an answer it can give to H that will be
correct, as the template for H will always make that D wrong.
Note, the problem is you can't actually DEFINE what a "paradox
machine" is, as they aren't actually machines, but templates that
build machines. And that final machine doesn't have actually
detectable tell-tales that show it to be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO WORK
WITH, BUT OH WELL
They only seem "random" because you don't know what you are talking
about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not*
classifiable as satisfactory by said classifier... can just be
skipped
No, it can only skip those that are not satisfactory, not those
that are but it can not classify as such, or your enumeration will >>>>>> not be complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as
it will be asked about all machines as it counts through all the
descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out computable >>>>>>> numbers already included on this diagonal, any machine which
either *is* computably equivalent OR *is not* classifiable in
regards to *any* machine already the list... can just be skipped
Nope, you can't skip some machines, as you then might lose some of >>>>>> the computable numbers.
see you can't compute a diagonal across *all* /machines/, with
said machines, but u can compute a diagonal across *all* /
computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done,
since ALL classifiers that attempt it will make an error, you
can't do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is
not the same thing as (b) computing the enumeration of all machines >>>>> that compute computable numbers. (b) necessarily has duplicates
while (a) does not need them. turing's paper wrongly conflates (a)
with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the
full list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of the
H he describes. It doesn't matter that you can make a DIFFERENT
machine, that you try to deceptively call "H" too that it won't have
a problem with, when your "H" gets to the number of Turing's H, it
still has the problem. It can call that one not circle-free, and
thus omit a circle- free machine from the list, or call it circle-
free, and when even YOU try to simulate it the k steps, you get
stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take the
code of (a), and invert the symbol put on the tape when it is
written. Note, it won't be "self-referent", as the "N" is uses, is
the N of YOUR H, not itself. The code for his (a) NEVER reads back
what it wrote, so that doesn't affect its behavior.
yes, i still do need to prove my thesis that for any paradoxical >>>>>>> machine, there exists a functionally equivalent machine without >>>>>>> such paradox
And the problem is that your "paradoxical" isn't actually a
definable property (let alone computable). Part of the problem is >>>>>> that if you look at just a machine description, it doesn't
(necessarily) tell you about the use of an "interface" as that use >>>>>> of an interface can be just inlined, leaving nothing "in the code" >>>>>> to show it exists.
i'm sorry, are you actually saying the machine description does not >>>>> describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the classifier
halts(), easily demonstrable thru a short proof i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY OF
UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME FUCKING
FORM OF INSANITY I CAN'T HONESTLY FATHOM
lol
it's a *specific* number that has the currently running machine
His specified H, with an actually (incorrect) implementation of >>>>>>>> D (which is all that CAN exist) will either be circle-free and >>>>>>>> thus generate a number (but its D said it isn't, and thus
omitted a valid machine from the list) or it isn't circle-free, >>>>>>>> and fails to computa a number, and thus should have been omitted >>>>>>>> from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just >>>>>>>> built on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an >>>>>>>>> anti- diagonal computation, due the same particular self-
referential weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just >>>>>>>> based on ERROR.
Your attempt to REDEFINE self-reference to mean processing a
number that happens to represent yourself means that you you
system "ALL" doesn't actually mean ALL, and thus is just ill- >>>>>>>> defined.
i'm sorry, you have an issue with me labeling a number that
directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still have >>>>>
encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different
numbers.
those are references to functionally equivalent machines, not self- >>>>> references
Which is why the concept of "self-reference" doesn't work.
It may let you filter out the simplest case used in the proofs,
but doesn't solve the actual problem, as the "Machine Number"
doesn't actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the
paradox detector while filter *out* paradoxes to point of creating
an actual problem
Then show how you will do it.
This means you need to detect a input that represents a machine that
uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines and
returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an
input classifier, and an input machine to search for paradoxes in
regards to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier
like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a
halting paradox form making it /undecidable input/ to halts().
And the problem is "paradox_free" can't exist as an always correct
decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case
doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a machine >>>>> is necessary for the particular input->output computation being
done, so utilizing the return paradox_free(halts,undp) -> FALSE to
filter out paradox_free() paradoxes as well as halts() paradoxes
will suffice to produce a turing complete subset of machines that
can be totally classified by halts()
But we can still build an input that your paradox_free() fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE INCLUDED
ON THE DIAGONAL???
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the only >>>>> possibility here to compute what we want. TM's can't achieve that,
but something slightly messier can.
No, it is your dependence on being able to assume that unicorns can
solve all your problems.
dear future: u see this shit i'm trying work with???? sorry i
didn't manage to make progress any faster, but i'm literally
dragging toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can actually >>>>>> be used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
They are, But you are too stuck in your lies to see it.
Your "logic" is based on being able to assume that Unicorn's exist,
and that is a safe assumption until someone can prove that Russel's
Teapot is not out there. (Even though your unicorns HAVE been proved
to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't understand >>>>>>>> what the actual problem is, and your world is just build on
...which u won't find interesting, but i can't fix a lack of >>>>>>>>> curiosity
🤷
expected was an inexpert audience, supposing that experts >>>>>>>>>>>> will recognise
the relevant mapping to universal quantification?
things that are lies.
IT seems that fundamentally, your world doesn't actually have >>>>>>>> "computations" as you don't understand the basic requirement
that they need to be fully defined in the actions they do.
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/25/26 1:15 PM, dart200 wrote:
On 2/25/26 4:04 AM, Richard Damon wrote:
On 2/25/26 12:40 AM, dart200 wrote:
On 2/24/26 8:29 PM, Richard Damon wrote:
On 2/24/26 11:13 PM, dart200 wrote:
On 2/24/26 7:52 PM, Richard Damon wrote:
On 2/24/26 9:55 PM, dart200 wrote:
On 2/24/26 6:13 PM, Richard Damon wrote:
On 2/24/26 7:27 PM, dart200 wrote:
On 2/24/26 4:01 PM, Richard Damon wrote:
On 2/24/26 4:30 PM, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote:...
On 2/22/2026 9:04 AM, dart200 wrote:
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't >>>>>>>>>>>>>>>> tell me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>>>> comptuation that enumerates out all the numbers, testing >>>>>>>>>>>>> each one of they represent a "satisfactory"/"circle-free" >>>>>>>>>>>>> machine, and adding that to diagonal across defined across >>>>>>>>>>>>> computable numbers
it really would be a great exercise to carefully read p247 >>>>>>>>>>>>> of turing's proof and produce the psuedo-code for the >>>>>>>>>>>>> machine H, assuming that machine D exists
And, your problem is that paper is about a DIFFERENT type of >>>>>>>>>>>> computation than the Halting Problem.
One that ALLOWS for infinitely running machines, as long as >>>>>>>>>>>> they continue to produce results.
The Halting Problem (vs the computable number problem) is >>>>>>>>>>>> based on Computations that must return answers in finite time. >>>>>>>>>>>>
Thus, your argument is based on a category error.
which would be relevant if turing talked about the halting >>>>>>>>>>> problem, but he didn't
No, it is relevent because *YOU* talk about the Halting
Problem, claiming you have a better idea to solve it.
the ideas i've been working on function on both the halting >>>>>>>>> problem and the satisfactory problem in turing's paper. my
paper was on the satisfactory problem in turing's paper because >>>>>>>>> i wanted to address the source
Then you shouldn't have started (and continued) about talking >>>>>>>> about the Halting Problem and contray machines.
Note, This paper never (as far as I remember) works by creating >>>>>>>> a specific "contrary" or 'paradoxical" machine to decide on. The >>>>>>>> error comes naturally out of the problem itself.
if u had read my paper (not just like glanced at it), u'd know
there is a method to construct the diagonal computation such that >>>>>>> it avoids stumbling on deciding on itself,
Only by assuming that Unicorns exist.
doesn't require a fixed decider to avoid itself, just a quine based >>>>> self-reference
Doesn't work.
more humility would do u wonders dick
Could say the same about you.
The problem is YOU are the one making stupid claims, not me, thus YOU
are the one showing stupidity on insisting on lies.
So, what answer does that quine based decider give for the question
about itself?
it's not the decider that has a quine, it's the diagonal computation
H that uses D that has the quine to recognize H.
But the problem is it isn;t the Diagonal Computation that is said to
be impossible, but the decider.
The H he descrives still exists, even if you make your variant, but
that H can't be decided on.
when the diagonal computation recognizes it's own number in the total
machine enumeration, it never asks the decider on itself, when it
identifies itself in the total machine numeration it adds some fixed
digit to the diagonal instead of recursively getting stuck in
simulating itself.
So, as I said, it isn't the diagonal computation that is said to be
impossible, just the decider that enumerates the diagonal machines.
WHat does that decider do for the non-quine version described.
doesn't matter what digit: both are different machines, with
different indexes in the enumeration, that compute the diagonal
So? what does D do with the original H.
This is your problem, you don't pay attention to the problem being
actually defined.
It is a bit like the "pathological" input for the halting problem.
and which literature defines "the rules"???
Either answer is still wrong.
And how is being "quine based" keep it from being "fixed".
It still is only one unique machine.
u ofc don't know what i'm talking about and have demonstrated no
capability to hold a coherent conversation
No, YOU keep on making categorical errors not knowing the real
meaning of the terms you are using.
Since, as I pointed out, you LEFT the system in your first page by >>>>>> claiming falsehoods, you didn't prove anything in the system you
claim to have been working on.
but turing's purpose was the construct a contradiction, not
figure out what a working diagonal computation would look like
Note, its decider needs to be the same sort of machine as >>>>>>>>>> machine D in Turings proof.
The decider he proves can't exist, as he proves that some >>>>>>>>>> decision problems are undecidable, thus answering the
Entscheidungsproblem in the negative.
what no one has proven is that the Entscheidungsproblem can't >>>>>>>>> be answered reliably thru other interfaces...
WHAT OTHER "INTERFACES"?
i've described a variety of them thus far
Really?
Not in the rules of Computation Theory, as your machines aren't
machine as they don't include all their algorithm.
u haven't defined the rules of computation theory, u just make
random claims about it whenever the fuck
I don't need to do that, as they ARE already defined in the literature. >>>
The "Literature", as a collection.
Yes, part of the problem is that they were more focused on the goal of
figuring if this class of "decision" problems were do able, no one
actually sat down (as far as I know) to formally and definitively
define it as a system, and in fact, you see some comments in the
writing about disagreeing and refining the definitions. But the basic
concept is fairly clear if you read much of the writting and remember
what they were working on. Was there some defined algorithmic method
that could be used to answer the tough problems they were coming up on
in mathematics and logic.
He shows a problem that can not be decided. PERIOD.
Your "Interface" seems to be one of your undefinable terms.
i've defined it many times to u
No, you use the word but not DEFINE it.
it is the specification that a machine computes, or possibly some
meta- specification
In other words, you are "solving" a problem by solving some other
problem?
but u don't debate honorably dick, u just reject every i say even
when i agree
It seems you are just admitting your logic is based on lying,
If I have a classification decision problem, that problem has a
specification of indicating which class an input lies in.
What other "specification" can exist?
All you seem to want to do is say that if you don't need to answer
the question, you can answer the question.
In other words, you want to make lies be truth.
It seems you don't know what a DEFINTION is.
which is a set of words apparently so damn radical the lame >>>>>>>>> duck industry of cs "academics" can't even comprehend the
concept, let alone address it in a considerate manner
No, you don't seem to know what you are talking about.
it shocks me on a daily basis just how much abject brainrot >>>>>>>>> i've waded thru in my explorations
It seems your brain has completed its rot.
And in doing so, he lays the groundwork for proving the
Halting Problem, and breaks the logjam in the thinking that >>>>>>>>>> there might be procedures that answer any question we want. >>>>>>>>>>
what i mean by enumerating out numbers is he enumerates out >>>>>>>>>>> natural numbers to find possible machines that might compute >>>>>>>>>>> "computable numbers" (which are real numbers)
Also, it doesn't enumerate the "numbers", but Machines, many >>>>>>>>>>>> of which will not actually produce numbers.
Then he assumes a test exists that determines if that >>>>>>>>>>>> machine satisfies the requirements, and shows that there can >>>>>>>>>>>> not be such a test.
Thus, he shows that there is NOT an effective enumeration of >>>>>>>>>>>> computable numbers, only uncomputable enumerations of them. >>>>>>>>>>>>
Note his statement on page 246:
The fallacy in this argument lies in the assumption that § >>>>>>>>>>>> is computable. It would be true if we could enumerate the >>>>>>>>>>>> computable sequences by finite means, but the problem of >>>>>>>>>>>> enumerating computable sequences is equivalent to the >>>>>>>>>>>> problem of finding out whether a given number is the D.N of >>>>>>>>>>>> a circle- free machine, and we have no general process for >>>>>>>>>>>> doing this in a finite number of steps.
Thus, what Turing Proves is that there ISN'T an enumeration >>>>>>>>>>>> that enumerates the numbers, only one of "all machines" that >>>>>>>>>>>> need an uncomputable test to see if they are in that set. >>>>>>>>>>>
But he never does that. He establishes that he can produce >>>>>>>>>> every possible machine in some order, and *IF* you could do >>>>>>>>>> the decision test, create the computable numbers in a
specified order, but you can't do that.
and u still haven't written the pseudo-code
Because it isn't really needed.
i am never going to be convinced of anything u say until you >>>>>>>>> produce a correct pseudo-code for H from p247
*never*
cause it's just abject brainrot that u think u can keep talking >>>>>>>>> so much smack without being able to code up his *really* simple >>>>>>>>> algo
Nope, YOU are the one with brain rot.
no pseudo-code = no convincing
NO DEFINITIONS, nothing shown.
If you want to ignore truth, you just prove your own stupidity,
still no pseudo-code cause ur a fraud dick
Still being stupid I see.
i'm honest not sure why u bother typing responses. ur words fall on >>>>> deaf ears cause u refuse to write anything beyond shallow insults
and random claims over rules that don't actually exist
Your problem is that you fundamentally don't understand what you are
talking about, as your world seems to be based on the assumption
that things don't need to be what they are, and that lies are valid.
no pseudo-code = no convincing
Simplified version assuming D taks D.N instead of S.D
def H:
let N = 1
let K = 0
do:
if D (N) is satisified:
let K = K + 1
simulate S.D corresponding to N till it produces K symbols
output that K'th symbol
Let N = N + 1
i want to thank you for actually doing this,
and i do agree it's correct,
the rest of ur analysis notwithstanding...
Note, each implementation of this pseudo code will have a number N.
IF D only is statisfied by circle-free programs, that is programs that
continue to produce outputs then this program MUST be circle free as
long as the following are true:
* D itself must be halting, and always produce an answer (which is part
of its definition to exist)
* There must be an infinite number of circle-free programs
And we then get to the delemma mentions on page 247,
When given this program to decide on, if it says that it is NOT
circle- free, then it WILL continue to produce output forever and thus
show that it WAS circle-free and thus D was wrong.
If it says that H is circle-free, then it will simulate itself until
it produces K outputs, but after that simulation produces K-1 outputs,
it will itself simulate itself and so on into the infinte loop.
Talking about a DIFFERENT machine, with a DIFFERENT D.N is irrelevant,
that other machine isn't this one,
i would ask if u care about convincing me,
Not really, just showing you are wrong.
Since you don't care about truth, you might not be convincible.
but clearly u don't
and if u don't care about convincing me,
then ur debate is not honorable, it is engaged with a lack of care,
so ofc it wouldn't be honorable, and that requires care
My goal is to keep others, who want to know, from falling into your lies.
YOU, not willing to define your terms, are the unhonerable one.
Sorry, you are just showing you don't understand what is >>>>>>>>>>>> being done.
i mean u seem to imply turing talked about the halting
problem, which he didn't
No, he talked about the decision problem, and showed things >>>>>>>>>> are undecidable, and laid the groundwork that was used to >>>>>>>>>> actually handle the halting problem.
Yes, people say he did it, because he established the method >>>>>>>>>> used for it.
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:...
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't >>>>>>>>>>>>>>>>> tell me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>>>>> comptuation
that enumerates out all the numbers, testing each one of >>>>>>>>>>>>>> they represent
a "satisfactory"/"circle-free" machine, and adding that to >>>>>>>>>>>>>> diagonal
across defined across computable numbers
it really would be a great exercise to carefully read p247 >>>>>>>>>>>>>> of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure >>>>>>>>>>>>> he didn't
reason quantified over all but phrase it like a procedure >>>>>>>>>>>>> for what he
the theory of computation is the theory of such procedures, >>>>>>>>>>>> and understanding the diagonal procedure is critical to >>>>>>>>>>>> understanding the *base* contradiction/paradox that the rest >>>>>>>>>>>> of his support for godel's result is then built on
And focusing on what is said to be impossible and not
changing the problem is also important.
The problem with the diagonal generation isn't the generation >>>>>>>>>>> of the diagonal itself, but effectively enumerating the >>>>>>>>>>> enumeration in the first place.
i don't see any indication that turing realized a difference >>>>>>>>>> there
Then you zre just showing your stupidity, because YOU can't >>>>>>>>> tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective enumeration >>>>>>>>> of the computable sequences.
He then points out that he can directly show that the "anti- >>>>>>>>> diagonal" of the (non-effectively computed) enumeration can't >>>>>>>>> be computed but that "This proof, although perfectly sound, has >>>>>>>>> the disadvantage that it may leave the reader with a feeling >>>>>>>>> that 'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done.
I think he is refering he to the standard anti-diagonal
arguement, which shows that since for all n, position n differs >>>>>>> from the value in number n, there can not be any element that
matches the anti- diagonal.
It is just a natural fact of countable infinity, something it
seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's
short diagonal proof, but in writing this post i now find myself
in a subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m-th >>>>>> figure in an. Let β be the sequence with 1-φn(m) as its n-th.
figure. Since β is computable, there exists a number K [== β] such >>>>>> that 1- φn(n) = φK(n) for all n. Putting n = K, we have 1 =
2φK(K), i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable (by
TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the computability
of β from the existence of φn(m), THAT'S THE FALLACy
the fallacy here is assuming that because the direct diagonal is
computable, that one can therefore compute the anti-diagonal using >>>>>> the direct diagonal. the abstract definition makes it look simple, >>>>>> but this ignores the complexities of self-referential analysis
(like what turing details on the next page)
But you can, *IF* you have a machine that computes the diagonal,
then just change all the write to the output to write the opposite. >>>>> Note, the "self-reference" that you are thinking of stops being a
"self- reference" but is a reference to the original write the
diagonal code.
in both methods i have for rectifying the paradox found in the
direct diagonal (either (1) filtering TMs or (2) using RTMs),
neither can be used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it
gets to the number of TURING'S H, as there is no correct answer for >>>>> the machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE SPECIFICATION >>>>
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out
like turing's paradoxical variation of the direct diagonal would
be, and there is no analogous non-paradoxical variation that has a >>>>>> hard coded value that is inverse to what it does return ... such a >>>>>> concept is entirely nonsensical. a function can only return what
it does, it can't also return the inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus
your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT
*SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs just
fails for reasons u'd only understand by working thru the algo
urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually computable >>>>>> given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do
that, as you are thinking the only "reference" can be to "self",
doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-REFERENCE,
LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR
INVERSE- H TO BE COMPUTED
one cannot just assume that because the diagonal across
computable numbers is computable, therefore the anti-diagonal >>>>>>>> across computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the
enumeration, you can't compute either of them.
But your method still doesn't let you compute the enumeration,
neither method i have for fixing the diagonal computation across >>>>>>>> the computable numbers can be used to compute the inverse diagonal >>>>>>>
and thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a >>>>>>> COMPLETE listing of the computable numbers / machine that compute >>>>>>> computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct. >>>>>>>
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not
computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the
other machines, including his original H that doesn't use your
"trick"
But instead, he can prove with a more obvious process, that the >>>>>>>>> Decider "D" that could be used to effectively enumerate the >>>>>>>>> sequence of machine that produce computable numbers can not esit. >>>>>>>>>
Thus, he clearly knows the difference, but is pointing out that >>>>>>>>> the attempt to compute the diagonal clearly reveals the issue >>>>>>>>> with effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for
computing the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on >>>>>>> his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's H, >>>>>> because my response to this is that D does not need to decide
correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT
WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial recognizer D
*can* decide on the fixed H because the fixed H does not try to use
any D on itself, so no self-referential paradox is possible in
regards to it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable AND
computable value)
the fixed H is /decidable input/ to partial recognizer D (used in
fixed H, which does not call any D itself), and would not be filtered
out by paradox detectors
the fact fixed H does not use any classifier D on itself, and instead
returns a hard-coded value for it's own digit on the diagonal, is
keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just anouncing
that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it will
fail on that.
bro what part of the non-existence of turing_D or turing_H do you not understand???
If D errs on Turing_H and calls it circle_free, then fixed_H gets hung
up.
If D errs on Turing_H and calls it not circle_free, then your
enumeration is missing the circle_free H from its list, and thus
doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D in a
turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D classifier,
when actually run actual_turing_H will simply skip putting it's own
digit in the computed sequence
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as fixed_H,
which in turn _does_ have a spot on the diagonal. so there is no actual
need to include a digit from actual_turing_H on the diagonal, as the
number it computes is already included on the diagonal
goddamn rick, this rock just keeps getting more solid
🤷🤷🤷
YOUR H still needs to know if Turing's H is a machine that
generates a computable number to build the list of such machine to
define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH ARE
INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH WILL
WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE INPUT/ TO D
FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE NUMBERS >>>>
It doesn't matter that your new H doesn't get stuck on itself, it >>>>>>> will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't
specify what D (or H) needs to do when encountering the /
undecidable input/ of H, so therefore both D and H are an
incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it must
compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE LIKE
THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN IDIOSYNCRASIES, AND
IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE IT
DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL ANALYSIS, SO
IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have created >>>>> an actual machine we claim to be the D, has an answer, and that D
is always wrong.
Thus, it isn't that machines claiming to be D and H can't exist,
only machines CORRECTLY meeting the requirements of D and H can't
exist.
Thus, if your H depends on that D, (or a machine that meets its
specification) then it also can't exist.
You just make the error of saying the problems don't exist because
you can't build them, but YOUR machine can, even though it has the
same problem.
IF D is wrong by deciding it is not circle free, then your H will >>>>>>> compute the wrong diagonal, as the resulting version of his H
WILL be circle free (since it never tries to simulate itself) and >>>>>>> thus DOES produce an computable number that your computation misses. >>>>>>>
Or, if that D is wrong by decing it IS circle free, then when you >>>>>>> H tries to process it, it will get stuck in the infinite loop.
The problem is that in stepping through the machines in order,
you WILL hit these actual machines built on your erroneous D
(your D must have this flaw, as no D without exists), and thus
you will be wrong on THAT input. IT doesn't matter if you get a >>>>>>> good answer for yourself.
idk what he would have said about it, but prolly something more >>>>>>>> substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion >>>>>>> of bad logic. You would have likely been escorted out of the
meeting as showing you were unqualified and being a distraction. >>>>>>>
Something that seems to be beyond your ignorant understanding. >>>>>>>>>
H shows that *IF* you can make that enumeration, you can make >>>>>>>>>>> the diagonal, and thus the anti-diagonal. The problem is you >>>>>>>>>>> can't make that enumeration, and assuming you can just shows >>>>>>>>>>> unsoundness.
interestingly: one can only fix the direct diagonal
computation like this
u can't do an analogous fix for the inverse/anti-diagonal >>>>>>>>>> computation. it's not possible hard code a machine to return >>>>>>>>>> an inverted value, a machine can only return what it does, not >>>>>>>>>> the inverse of what it does...
so if we can filter out paradoxes from the enumeration, that >>>>>>>>>> will leave a direct diagonal computation extant in that
filtered (yet still turing complete list), while any attempt >>>>>>>>>> to compute an inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in >>>>>>>>> isolation, but only in relationship to a given machine trying >>>>>>>>> to decide them.
right. so if ur constructing a diagonal across computable
numbers then u only need to filter out paradoxes in regards to >>>>>>>> the classifier that classifies them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such
paradoxes...
In other words, you beleive unquestionably in the existance of
Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES CAN'T
BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING A
FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU MADE??? >>>>
For THIS paradox, read the proof. for THIS H, the one given, (for
which ever version of D you want to try to assume is right) D is
just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox,
because there still isn't an answer it can give to H that will be
correct, as the template for H will always make that D wrong.
Note, the problem is you can't actually DEFINE what a "paradox
machine" is, as they aren't actually machines, but templates that
build machines. And that final machine doesn't have actually
detectable tell-tales that show it to be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO WORK
WITH, BUT OH WELL
They only seem "random" because you don't know what you are talking >>>>> about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not*
classifiable as satisfactory by said classifier... can just be >>>>>>>> skipped
No, it can only skip those that are not satisfactory, not those >>>>>>> that are but it can not classify as such, or your enumeration
will not be complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as >>>>>>> it will be asked about all machines as it counts through all the >>>>>>> descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out
computable numbers already included on this diagonal, any
machine which either *is* computably equivalent OR *is not*
classifiable in regards to *any* machine already the list... can >>>>>>>> just be skipped
Nope, you can't skip some machines, as you then might lose some >>>>>>> of the computable numbers.
see you can't compute a diagonal across *all* /machines/, with >>>>>>>> said machines, but u can compute a diagonal across *all* /
computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done,
since ALL classifiers that attempt it will make an error, you
can't do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is
not the same thing as (b) computing the enumeration of all
machines that compute computable numbers. (b) necessarily has
duplicates while (a) does not need them. turing's paper wrongly
conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the
full list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of
the H he describes. It doesn't matter that you can make a DIFFERENT >>>>> machine, that you try to deceptively call "H" too that it won't
have a problem with, when your "H" gets to the number of Turing's
H, it still has the problem. It can call that one not circle-free,
and thus omit a circle- free machine from the list, or call it
circle- free, and when even YOU try to simulate it the k steps, you >>>>> get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take the
code of (a), and invert the symbol put on the tape when it is
written. Note, it won't be "self-referent", as the "N" is uses, is
the N of YOUR H, not itself. The code for his (a) NEVER reads back
what it wrote, so that doesn't affect its behavior.
yes, i still do need to prove my thesis that for any paradoxical >>>>>>>> machine, there exists a functionally equivalent machine without >>>>>>>> such paradox
And the problem is that your "paradoxical" isn't actually a
definable property (let alone computable). Part of the problem is >>>>>>> that if you look at just a machine description, it doesn't
(necessarily) tell you about the use of an "interface" as that
use of an interface can be just inlined, leaving nothing "in the >>>>>>> code" to show it exists.
i'm sorry, are you actually saying the machine description does
not describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the
classifier halts(), easily demonstrable thru a short proof i've done
many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY OF
UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME FUCKING
FORM OF INSANITY I CAN'T HONESTLY FATHOM
lol
His specified H, with an actually (incorrect) implementation of >>>>>>>>> D (which is all that CAN exist) will either be circle-free and >>>>>>>>> thus generate a number (but its D said it isn't, and thus
omitted a valid machine from the list) or it isn't circle-free, >>>>>>>>> and fails to computa a number, and thus should have been
omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just >>>>>>>>> built on an assuption in error.
so despite turing's worries, the existence of a diagonal
computation does not actually then imply the existence of an >>>>>>>>>> anti- diagonal computation, due the same particular self- >>>>>>>>>> referential weirdness that stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is just >>>>>>>>> based on ERROR.
Your attempt to REDEFINE self-reference to mean processing a >>>>>>>>> number that happens to represent yourself means that you you >>>>>>>>> system "ALL" doesn't actually mean ALL, and thus is just ill- >>>>>>>>> defined.
i'm sorry, you have an issue with me labeling a number that
directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still >>>>>>> have
it's a *specific* number that has the currently running machine
encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different >>>>>>> numbers.
those are references to functionally equivalent machines, not
self- references
Which is why the concept of "self-reference" doesn't work.
It may let you filter out the simplest case used in the proofs, >>>>>>> but doesn't solve the actual problem, as the "Machine Number"
doesn't actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the
paradox detector while filter *out* paradoxes to point of creating >>>>>> an actual problem
Then show how you will do it.
This means you need to detect a input that represents a machine
that uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines and >>>>>> returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an
input classifier, and an input machine to search for paradoxes in >>>>>> regards to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox classifier >>>>>> like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a >>>>>> halting paradox form making it /undecidable input/ to halts().
And the problem is "paradox_free" can't exist as an always correct
decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case
doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a
machine is necessary for the particular input->output computation >>>>>> being done, so utilizing the return paradox_free(halts,undp) ->
FALSE to filter out paradox_free() paradoxes as well as halts()
paradoxes will suffice to produce a turing complete subset of
machines that can be totally classified by halts()
But we can still build an input that your paradox_free() fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE
INCLUDED ON THE DIAGONAL???
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the
only possibility here to compute what we want. TM's can't achieve >>>>>> that, but something slightly messier can.
No, it is your dependence on being able to assume that unicorns can >>>>> solve all your problems.
dear future: u see this shit i'm trying work with???? sorry i >>>>>>>> didn't manage to make progress any faster, but i'm literally
dragging toddlers kicking and screaming at this point jeez...
In other words, in your world "ALL" isn't a word that can
actually be used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
They are, But you are too stuck in your lies to see it.
Your "logic" is based on being able to assume that Unicorn's exist, >>>>> and that is a safe assumption until someone can prove that Russel's >>>>> Teapot is not out there. (Even though your unicorns HAVE been
proved to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't understand >>>>>>>>> what the actual problem is, and your world is just build on >>>>>>>>> things that are lies.
...which u won't find interesting, but i can't fix a lack of >>>>>>>>>> curiosity
🤷
expected was an inexpert audience, supposing that experts >>>>>>>>>>>>> will recognise
the relevant mapping to universal quantification?
IT seems that fundamentally, your world doesn't actually have >>>>>>>>> "computations" as you don't understand the basic requirement >>>>>>>>> that they need to be fully defined in the actions they do.
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:And focusing on what is said to be impossible and not >>>>>>>>>>>> changing the problem is also important.
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote:
On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:...
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't >>>>>>>>>>>>>>>>>> tell me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>>>>>> comptuation
that enumerates out all the numbers, testing each one of >>>>>>>>>>>>>>> they represent
a "satisfactory"/"circle-free" machine, and adding that >>>>>>>>>>>>>>> to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read >>>>>>>>>>>>>>> p247 of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you sure >>>>>>>>>>>>>> he didn't
reason quantified over all but phrase it like a procedure >>>>>>>>>>>>>> for what he
the theory of computation is the theory of such procedures, >>>>>>>>>>>>> and understanding the diagonal procedure is critical to >>>>>>>>>>>>> understanding the *base* contradiction/paradox that the >>>>>>>>>>>>> rest of his support for godel's result is then built on >>>>>>>>>>>>
The problem with the diagonal generation isn't the
generation of the diagonal itself, but effectively
enumerating the enumeration in the first place.
i don't see any indication that turing realized a difference >>>>>>>>>>> there
Then you zre just showing your stupidity, because YOU can't >>>>>>>>>> tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective
enumeration of the computable sequences.
He then points out that he can directly show that the "anti- >>>>>>>>>> diagonal" of the (non-effectively computed) enumeration can't >>>>>>>>>> be computed but that "This proof, although perfectly sound, >>>>>>>>>> has the disadvantage that it may leave the reader with a
feeling that 'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done. >>>>>>>>
I think he is refering he to the standard anti-diagonal
arguement, which shows that since for all n, position n differs >>>>>>>> from the value in number n, there can not be any element that >>>>>>>> matches the anti- diagonal.
It is just a natural fact of countable infinity, something it >>>>>>>> seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's
short diagonal proof, but in writing this post i now find myself >>>>>>> in a subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m- >>>>>>> th figure in an. Let β be the sequence with 1-φn(m) as its n-th. >>>>>>> figure. Since β is computable, there exists a number K [== β] >>>>>>> such that 1- φn(n) = φK(n) for all n. Putting n = K, we have 1 = >>>>>>> 2φK(K), i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable (by
TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the computability >>>>> of β from the existence of φn(m), THAT'S THE FALLACy
the fallacy here is assuming that because the direct diagonal is >>>>>>> computable, that one can therefore compute the anti-diagonal
using the direct diagonal. the abstract definition makes it look >>>>>>> simple, but this ignores the complexities of self-referential
analysis (like what turing details on the next page)
But you can, *IF* you have a machine that computes the diagonal,
then just change all the write to the output to write the
opposite. Note, the "self-reference" that you are thinking of
stops being a "self- reference" but is a reference to the original >>>>>> write the diagonal code.
in both methods i have for rectifying the paradox found in the
direct diagonal (either (1) filtering TMs or (2) using RTMs),
neither can be used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it
gets to the number of TURING'S H, as there is no correct answer
for the machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE
SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out
like turing's paradoxical variation of the direct diagonal would >>>>>>> be, and there is no analogous non-paradoxical variation that has >>>>>>> a hard coded value that is inverse to what it does return ...
such a concept is entirely nonsensical. a function can only
return what it does, it can't also return the inverse to what it >>>>>>> returns eh???
But, if D filters it out, then it becomes circle-free, and thus
your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT
*SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs just >>>>>>> fails for reasons u'd only understand by working thru the algo
urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually
computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do
that, as you are thinking the only "reference" can be to "self",
doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-
REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR
INVERSE- H TO BE COMPUTED
one cannot just assume that because the diagonal across
computable numbers is computable, therefore the anti-diagonal >>>>>>>>> across computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the >>>>>>>> enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation
across the computable numbers can be used to compute the
inverse diagonal
But your method still doesn't let you compute the enumeration, >>>>>>>> and thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be a >>>>>>>> COMPLETE listing of the computable numbers / machine that
compute computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct. >>>>>>>>
so while i agree with turing that the anti-diagonal is not
computable, i don't agree that the normal diagonal is not
computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the
other machines, including his original H that doesn't use your >>>>>>>> "trick"
But instead, he can prove with a more obvious process, that >>>>>>>>>> the Decider "D" that could be used to effectively enumerate >>>>>>>>>> the sequence of machine that produce computable numbers can >>>>>>>>>> not esit.
Thus, he clearly knows the difference, but is pointing out >>>>>>>>>> that the attempt to compute the diagonal clearly reveals the >>>>>>>>>> issue with effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for
computing the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide on >>>>>>>> his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's >>>>>>> H, because my response to this is that D does not need to decide >>>>>>> correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT
WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial recognizer D
*can* decide on the fixed H because the fixed H does not try to use
any D on itself, so no self-referential paradox is possible in
regards to it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable AND
computable value)
the fixed H is /decidable input/ to partial recognizer D (used in
fixed H, which does not call any D itself), and would not be
filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and
instead returns a hard-coded value for it's own digit on the
diagonal, is keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just
anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it will
fail on that.
bro what part of the non-existence of turing_D or turing_H do you not
understand???
But if your fD exists, then Turing_H that uses your D exists, and will
give that D the problem, and fixed_H will call your D on turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H gets
hung up.
If D errs on Turing_H and calls it not circle_free, then your
enumeration is missing the circle_free H from its list, and thus
doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D in a
turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory >> K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D classifier,
when actually run actual_turing_H will simply skip putting it's own
digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the actual_turing_H isn't an "undecidable input", as it has definite
behavior, just behavior that partial_recognizer_D gets wrong.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get wrong,
then it can just return the different answer, and then THAT machine it
will get right. The problem is it creates a DIFFERENT input, that uses
this new version of the machine, that it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as fixed_H,
which in turn _does_ have a spot on the diagonal. so there is no
actual need to include a digit from actual_turing_H on the diagonal,
as the number it computes is already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can possible
filter all ALL of that infinite set of machines, but still accept some machine that computes that particular number that it would compute with
this supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting that you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the fairy dust,
of assuming that your partial decider that meets ALL your needs exists,
🤷🤷🤷
YOUR H still needs to know if Turing's H is a machine that
generates a computable number to build the list of such machine to >>>>>> define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH ARE
INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH WILL
WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE INPUT/ TO D >>>>> FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE
NUMBERS
It doesn't matter that your new H doesn't get stuck on itself, >>>>>>>> it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't >>>>>>> specify what D (or H) needs to do when encountering the /
undecidable input/ of H, so therefore both D and H are an
incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it must >>>>>> compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE LIKE
THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN IDIOSYNCRASIES, AND >>>>> IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE IT
DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL ANALYSIS, SO >>>>> IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have
created an actual machine we claim to be the D, has an answer, and >>>>>> that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't exist,
only machines CORRECTLY meeting the requirements of D and H can't >>>>>> exist.
Thus, if your H depends on that D, (or a machine that meets its
specification) then it also can't exist.
You just make the error of saying the problems don't exist because >>>>>> you can't build them, but YOUR machine can, even though it has the >>>>>> same problem.
IF D is wrong by deciding it is not circle free, then your H
will compute the wrong diagonal, as the resulting version of his >>>>>>>> H WILL be circle free (since it never tries to simulate itself) >>>>>>>> and thus DOES produce an computable number that your computation >>>>>>>> misses.
Or, if that D is wrong by decing it IS circle free, then when >>>>>>>> you H tries to process it, it will get stuck in the infinite loop. >>>>>>>>
The problem is that in stepping through the machines in order, >>>>>>>> you WILL hit these actual machines built on your erroneous D
(your D must have this flaw, as no D without exists), and thus >>>>>>>> you will be wrong on THAT input. IT doesn't matter if you get a >>>>>>>> good answer for yourself.
idk what he would have said about it, but prolly something more >>>>>>>>> substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic assertion >>>>>>>> of bad logic. You would have likely been escorted out of the
meeting as showing you were unqualified and being a distraction. >>>>>>>>
Right, which he shows can not be done.
Something that seems to be beyond your ignorant understanding. >>>>>>>>>>
H shows that *IF* you can make that enumeration, you can >>>>>>>>>>>> make the diagonal, and thus the anti-diagonal. The problem >>>>>>>>>>>> is you can't make that enumeration, and assuming you can >>>>>>>>>>>> just shows unsoundness.
interestingly: one can only fix the direct diagonal
computation like this
u can't do an analogous fix for the inverse/anti-diagonal >>>>>>>>>>> computation. it's not possible hard code a machine to return >>>>>>>>>>> an inverted value, a machine can only return what it does, >>>>>>>>>>> not the inverse of what it does...
so if we can filter out paradoxes from the enumeration, that >>>>>>>>>>> will leave a direct diagonal computation extant in that >>>>>>>>>>> filtered (yet still turing complete list), while any attempt >>>>>>>>>>> to compute an inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in >>>>>>>>>> isolation, but only in relationship to a given machine trying >>>>>>>>>> to decide them.
right. so if ur constructing a diagonal across computable
numbers then u only need to filter out paradoxes in regards to >>>>>>>>> the classifier that classifies them as a "satisfactory" number >>>>>>>>
please do quote where turing shows we can't filter out such
paradoxes...
In other words, you beleive unquestionably in the existance of
Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES CAN'T
BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING A
FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU MADE??? >>>>>
For THIS paradox, read the proof. for THIS H, the one given, (for >>>>>> which ever version of D you want to try to assume is right) D is
just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox,
because there still isn't an answer it can give to H that will be >>>>>> correct, as the template for H will always make that D wrong.
Note, the problem is you can't actually DEFINE what a "paradox
machine" is, as they aren't actually machines, but templates that >>>>>> build machines. And that final machine doesn't have actually
detectable tell-tales that show it to be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO
WORK WITH, BUT OH WELL
They only seem "random" because you don't know what you are
talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not*
classifiable as satisfactory by said classifier... can just be >>>>>>>>> skipped
No, it can only skip those that are not satisfactory, not those >>>>>>>> that are but it can not classify as such, or your enumeration >>>>>>>> will not be complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines (as >>>>>>>> it will be asked about all machines as it counts through all the >>>>>>>> descriptions) and thus Turing's H *WILL* be asked about.
similarly if u want to go a step further an filter out
computable numbers already included on this diagonal, any
machine which either *is* computably equivalent OR *is not* >>>>>>>>> classifiable in regards to *any* machine already the list... >>>>>>>>> can just be skipped
Nope, you can't skip some machines, as you then might lose some >>>>>>>> of the computable numbers.
see you can't compute a diagonal across *all* /machines/, with >>>>>>>>> said machines, but u can compute a diagonal across *all* /
computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, >>>>>>>> since ALL classifiers that attempt it will make an error, you >>>>>>>> can't do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is >>>>>>> not the same thing as (b) computing the enumeration of all
machines that compute computable numbers. (b) necessarily has
duplicates while (a) does not need them. turing's paper wrongly >>>>>>> conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the
full list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of
the H he describes. It doesn't matter that you can make a
DIFFERENT machine, that you try to deceptively call "H" too that
it won't have a problem with, when your "H" gets to the number of >>>>>> Turing's H, it still has the problem. It can call that one not
circle-free, and thus omit a circle- free machine from the list,
or call it circle- free, and when even YOU try to simulate it the >>>>>> k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take the >>>>>> code of (a), and invert the symbol put on the tape when it is
written. Note, it won't be "self-referent", as the "N" is uses, is >>>>>> the N of YOUR H, not itself. The code for his (a) NEVER reads back >>>>>> what it wrote, so that doesn't affect its behavior.
yes, i still do need to prove my thesis that for any
paradoxical machine, there exists a functionally equivalent >>>>>>>>> machine without such paradox
And the problem is that your "paradoxical" isn't actually a
definable property (let alone computable). Part of the problem >>>>>>>> is that if you look at just a machine description, it doesn't >>>>>>>> (necessarily) tell you about the use of an "interface" as that >>>>>>>> use of an interface can be just inlined, leaving nothing "in the >>>>>>>> code" to show it exists.
i'm sorry, are you actually saying the machine description does >>>>>>> not describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the
classifier halts(), easily demonstrable thru a short proof i've
done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY OF
UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME
FUCKING FORM OF INSANITY I CAN'T HONESTLY FATHOM
lol
His specified H, with an actually (incorrect) implementation >>>>>>>>>> of D (which is all that CAN exist) will either be circle-free >>>>>>>>>> and thus generate a number (but its D said it isn't, and thus >>>>>>>>>> omitted a valid machine from the list) or it isn't circle- >>>>>>>>>> free, and fails to computa a number, and thus should have been >>>>>>>>>> omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is just >>>>>>>>>> built on an assuption in error.
But there is no actuall SELF-REFERENCE, so your logic is just >>>>>>>>>> based on ERROR.
so despite turing's worries, the existence of a diagonal >>>>>>>>>>> computation does not actually then imply the existence of an >>>>>>>>>>> anti- diagonal computation, due the same particular self- >>>>>>>>>>> referential weirdness that stumped turing the first place >>>>>>>>>>
Your attempt to REDEFINE self-reference to mean processing a >>>>>>>>>> number that happens to represent yourself means that you you >>>>>>>>>> system "ALL" doesn't actually mean ALL, and thus is just ill- >>>>>>>>>> defined.
i'm sorry, you have an issue with me labeling a number that >>>>>>>>> directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you still >>>>>>>> have
it's a *specific* number that has the currently running machine >>>>>>> encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different >>>>>>>> numbers.
those are references to functionally equivalent machines, not
self- references
Which is why the concept of "self-reference" doesn't work.
It may let you filter out the simplest case used in the proofs, >>>>>>>> but doesn't solve the actual problem, as the "Machine Number" >>>>>>>> doesn't actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the
paradox detector while filter *out* paradoxes to point of
creating an actual problem
Then show how you will do it.
This means you need to detect a input that represents a machine
that uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines
and returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an >>>>>>> input classifier, and an input machine to search for paradoxes in >>>>>>> regards to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox
classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run a >>>>>>> halting paradox form making it /undecidable input/ to halts().
And the problem is "paradox_free" can't exist as an always correct >>>>>> decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case >>>>>>> doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a
machine is necessary for the particular input->output computation >>>>>>> being done, so utilizing the return paradox_free(halts,undp) -> >>>>>>> FALSE to filter out paradox_free() paradoxes as well as halts() >>>>>>> paradoxes will suffice to produce a turing complete subset of
machines that can be totally classified by halts()
But we can still build an input that your paradox_free() fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE
INCLUDED ON THE DIAGONAL???
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the
only possibility here to compute what we want. TM's can't achieve >>>>>>> that, but something slightly messier can.
No, it is your dependence on being able to assume that unicorns
can solve all your problems.
dear future: u see this shit i'm trying work with???? sorry i >>>>>>>>> didn't manage to make progress any faster, but i'm literally >>>>>>>>> dragging toddlers kicking and screaming at this point jeez... >>>>>>>>>
In other words, in your world "ALL" isn't a word that can
actually be used.
You don't understand that you need to use actual sound logic,
lol, i can only 🙏 someone would engage in sound logic with me
They are, But you are too stuck in your lies to see it.
Your "logic" is based on being able to assume that Unicorn's
exist, and that is a safe assumption until someone can prove that >>>>>> Russel's Teapot is not out there. (Even though your unicorns HAVE >>>>>> been proved to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't understand >>>>>>>>>> what the actual problem is, and your world is just build on >>>>>>>>>> things that are lies.
...which u won't find interesting, but i can't fix a lack of >>>>>>>>>>> curiosity
🤷
expected was an inexpert audience, supposing that experts >>>>>>>>>>>>>> will recognise
the relevant mapping to universal quantification?
IT seems that fundamentally, your world doesn't actually have >>>>>>>>>> "computations" as you don't understand the basic requirement >>>>>>>>>> that they need to be fully defined in the actions they do.
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:...
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't >>>>>>>>>>>>>>>>>>> tell me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he defines a >>>>>>>>>>>>>>>> comptuation
that enumerates out all the numbers, testing each one of >>>>>>>>>>>>>>>> they represent
a "satisfactory"/"circle-free" machine, and adding that >>>>>>>>>>>>>>>> to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read >>>>>>>>>>>>>>>> p247 of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you >>>>>>>>>>>>>>> sure he didn't
reason quantified over all but phrase it like a procedure >>>>>>>>>>>>>>> for what he
the theory of computation is the theory of such
procedures, and understanding the diagonal procedure is >>>>>>>>>>>>>> critical to understanding the *base* contradiction/paradox >>>>>>>>>>>>>> that the rest of his support for godel's result is then >>>>>>>>>>>>>> built on
And focusing on what is said to be impossible and not >>>>>>>>>>>>> changing the problem is also important.
The problem with the diagonal generation isn't the
generation of the diagonal itself, but effectively
enumerating the enumeration in the first place.
i don't see any indication that turing realized a difference >>>>>>>>>>>> there
Then you zre just showing your stupidity, because YOU can't >>>>>>>>>>> tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable.
Here is is SPECIFICALLY talking about the effective
enumeration of the computable sequences.
He then points out that he can directly show that the "anti- >>>>>>>>>>> diagonal" of the (non-effectively computed) enumeration can't >>>>>>>>>>> be computed but that "This proof, although perfectly sound, >>>>>>>>>>> has the disadvantage that it may leave the reader with a >>>>>>>>>>> feeling that 'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done. >>>>>>>>>
I think he is refering he to the standard anti-diagonal
arguement, which shows that since for all n, position n differs >>>>>>>>> from the value in number n, there can not be any element that >>>>>>>>> matches the anti- diagonal.
It is just a natural fact of countable infinity, something it >>>>>>>>> seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's >>>>>>>> short diagonal proof, but in writing this post i now find myself >>>>>>>> in a subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the m- >>>>>>>> th figure in an. Let β be the sequence with 1-φn(m) as its n-th. >>>>>>>> figure. Since β is computable, there exists a number K [== β] >>>>>>>> such that 1- φn(n) = φK(n) for all n. Putting n = K, we have 1 = >>>>>>>> 2φK(K), i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable (by >>>>>> TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the
computability of β from the existence of φn(m), THAT'S THE FALLACy >>>>>>
the fallacy here is assuming that because the direct diagonal is >>>>>>>> computable, that one can therefore compute the anti-diagonal
using the direct diagonal. the abstract definition makes it look >>>>>>>> simple, but this ignores the complexities of self-referential >>>>>>>> analysis (like what turing details on the next page)
But you can, *IF* you have a machine that computes the diagonal, >>>>>>> then just change all the write to the output to write the
opposite. Note, the "self-reference" that you are thinking of
stops being a "self- reference" but is a reference to the
original write the diagonal code.
in both methods i have for rectifying the paradox found in the >>>>>>>> direct diagonal (either (1) filtering TMs or (2) using RTMs), >>>>>>>> neither can be used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it >>>>>>> gets to the number of TURING'S H, as there is no correct answer >>>>>>> for the machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE
SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out >>>>>>>> like turing's paradoxical variation of the direct diagonal would >>>>>>>> be, and there is no analogous non-paradoxical variation that has >>>>>>>> a hard coded value that is inverse to what it does return ... >>>>>>>> such a concept is entirely nonsensical. a function can only
return what it does, it can't also return the inverse to what it >>>>>>>> returns eh???
But, if D filters it out, then it becomes circle-free, and thus >>>>>>> your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT
*SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs just >>>>>>>> fails for reasons u'd only understand by working thru the algo >>>>>>>> urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually
computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do >>>>>>> that, as you are thinking the only "reference" can be to "self", >>>>>>> doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-
REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR
INVERSE- H TO BE COMPUTED
one cannot just assume that because the diagonal across
computable numbers is computable, therefore the anti-diagonal >>>>>>>>>> across computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the >>>>>>>>> enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation >>>>>>>>>> across the computable numbers can be used to compute the
inverse diagonal
But your method still doesn't let you compute the enumeration, >>>>>>>>> and thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be >>>>>>>>> a COMPLETE listing of the computable numbers / machine that >>>>>>>>> compute computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct. >>>>>>>>>
so while i agree with turing that the anti-diagonal is not >>>>>>>>>> computable, i don't agree that the normal diagonal is not >>>>>>>>>> computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the >>>>>>>>> other machines, including his original H that doesn't use your >>>>>>>>> "trick"
But instead, he can prove with a more obvious process, that >>>>>>>>>>> the Decider "D" that could be used to effectively enumerate >>>>>>>>>>> the sequence of machine that produce computable numbers can >>>>>>>>>>> not esit.
Thus, he clearly knows the difference, but is pointing out >>>>>>>>>>> that the attempt to compute the diagonal clearly reveals the >>>>>>>>>>> issue with effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for >>>>>>>>>> computing the diagonal can avoid the paradox on itself ...
But it doesn't.
Your just don't understand that D just can't correctly decide >>>>>>>>> on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's >>>>>>>> H, because my response to this is that D does not need to decide >>>>>>>> correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT
WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial recognizer D >>>>> *can* decide on the fixed H because the fixed H does not try to use >>>>> any D on itself, so no self-referential paradox is possible in
regards to it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable AND
computable value)
the fixed H is /decidable input/ to partial recognizer D (used in
fixed H, which does not call any D itself), and would not be
filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and
instead returns a hard-coded value for it's own digit on the
diagonal, is keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just
anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it will
fail on that.
bro what part of the non-existence of turing_D or turing_H do you not
understand???
But if your fD exists, then Turing_H that uses your D exists, and will
give that D the problem, and fixed_H will call your D on turing_H and
it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H gets
hung up.
If D errs on Turing_H and calls it not circle_free, then your
enumeration is missing the circle_free H from its list, and thus
doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D in a
turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D
classifier, when actually run actual_turing_H will simply skip
putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the
actual_turing_H isn't an "undecidable input", as it has definite
behavior, just behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of undecidability. just because a machine is /undecidable input/ to a particular classifier does mean we cannot then prove and know ourselves what the machine actually
does,
undecidability only actually exists as a relationship between a
particular machine and the particular classifier it creates a structural paradox for,
*not* as a general inability to describe a *real* machine's behavior
that u are trying to treat it as,
see, turing_H's runtime is not describable because it's an /incomplete specification/ of machine and therefor _does not even exist_ as a real
TM, not because of the hypothesized undecidability in relation to some D
actual_turing_H, on the other hand, actually does exist, so we can know/ prove what it does. it, however, still forms a structural paradox in
regards to partial_recognizer_D which will fail to classify it as circle-free
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get
wrong, then it can just return the different answer, and then THAT
machine it will get right. The problem is it creates a DIFFERENT
input, that uses this new version of the machine, that it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as fixed_H,
which in turn _does_ have a spot on the diagonal. so there is no
actual need to include a digit from actual_turing_H on the diagonal,
as the number it computes is already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of
actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can possible
filter all ALL of that infinite set of machines, but still accept some
machine that computes that particular number that it would compute
with this supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting that
you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the fairy
dust, of assuming that your partial decider that meets ALL your needs
exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between fixed_H
and actual_turing_H is the way they respectively handle their own self- references. they function identically when handling all other machines
fixed_H = () -> {
N = 1
K = 0
output = [] // written to F-squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip trying to simulate itself and put it's hard coded digit on the diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip trying simulate actual_turing_H because partial_recognizer_D fails to classify
it as circle-free
when actual_turing_H iterates across DN(actual_turing_H) itself, it will similarly skip trying to simulate itself, because partial_recognizer_D
fails to classify it as circle-free
when actual_turing_H iterates across DN(fixed_H), partial_recognizer_D *successfully* classifies fixed_H as satisfactory, so it will simulate fixed_H to it's Kth digit, the one hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular logic in either case), fixed_H also skips simulating itself (also circular
logic), while actual_turing_H actually *does* simulate fixed_H
successfully for it's Kth digit on the diagonal. these machines compute
the *same* number/sequence!
see, not only is defining a value in fixed_H that the diagonal
computation can put for itself in diagonal essential in making the
diagonal actually computable, so is partial_recognizer_D skipping over actual_turing_H to avoid the computability paradox!
ultimately, in order to compute a diagonal across computable numbers, partial_recognizer_D only needs to successfully classify *one* of the _infinitely_ many machines that compute any given computable number...
shit rick, idk how ur still denying what's right under ur nose,
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that
generates a computable number to build the list of such machine >>>>>>> to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH ARE >>>>>> INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH WILL >>>>>> WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE INPUT/ TO >>>>>> D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE
NUMBERS
It doesn't matter that your new H doesn't get stuck on itself, >>>>>>>>> it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he doesn't >>>>>>>> specify what D (or H) needs to do when encountering the /
undecidable input/ of H, so therefore both D and H are an
incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it
must compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE LIKE >>>>>> THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN IDIOSYNCRASIES,
AND IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE IT >>>>>> DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL ANALYSIS,
SO IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have
created an actual machine we claim to be the D, has an answer,
and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't exist, >>>>>>> only machines CORRECTLY meeting the requirements of D and H can't >>>>>>> exist.
Thus, if your H depends on that D, (or a machine that meets its >>>>>>> specification) then it also can't exist.
You just make the error of saying the problems don't exist
because you can't build them, but YOUR machine can, even though >>>>>>> it has the same problem.
IF D is wrong by deciding it is not circle free, then your H >>>>>>>>> will compute the wrong diagonal, as the resulting version of >>>>>>>>> his H WILL be circle free (since it never tries to simulate >>>>>>>>> itself) and thus DOES produce an computable number that your >>>>>>>>> computation misses.
Or, if that D is wrong by decing it IS circle free, then when >>>>>>>>> you H tries to process it, it will get stuck in the infinite loop. >>>>>>>>>
The problem is that in stepping through the machines in order, >>>>>>>>> you WILL hit these actual machines built on your erroneous D >>>>>>>>> (your D must have this flaw, as no D without exists), and thus >>>>>>>>> you will be wrong on THAT input. IT doesn't matter if you get a >>>>>>>>> good answer for yourself.
idk what he would have said about it, but prolly something >>>>>>>>>> more substantial than just calling me ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic
assertion of bad logic. You would have likely been escorted out >>>>>>>>> of the meeting as showing you were unqualified and being a
distraction.
Right, which he shows can not be done.
Something that seems to be beyond your ignorant understanding. >>>>>>>>>>>
H shows that *IF* you can make that enumeration, you can >>>>>>>>>>>>> make the diagonal, and thus the anti-diagonal. The problem >>>>>>>>>>>>> is you can't make that enumeration, and assuming you can >>>>>>>>>>>>> just shows unsoundness.
interestingly: one can only fix the direct diagonal
computation like this
u can't do an analogous fix for the inverse/anti-diagonal >>>>>>>>>>>> computation. it's not possible hard code a machine to return >>>>>>>>>>>> an inverted value, a machine can only return what it does, >>>>>>>>>>>> not the inverse of what it does...
so if we can filter out paradoxes from the enumeration, that >>>>>>>>>>>> will leave a direct diagonal computation extant in that >>>>>>>>>>>> filtered (yet still turing complete list), while any attempt >>>>>>>>>>>> to compute an inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist in >>>>>>>>>>> isolation, but only in relationship to a given machine trying >>>>>>>>>>> to decide them.
right. so if ur constructing a diagonal across computable >>>>>>>>>> numbers then u only need to filter out paradoxes in regards to >>>>>>>>>> the classifier that classifies them as a "satisfactory" number >>>>>>>>>
please do quote where turing shows we can't filter out such
paradoxes...
In other words, you beleive unquestionably in the existance of
Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES CAN'T >>>>>> BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING A >>>>>> FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU
MADE???
For THIS paradox, read the proof. for THIS H, the one given, (for >>>>>>> which ever version of D you want to try to assume is right) D is >>>>>>> just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox,
because there still isn't an answer it can give to H that will be >>>>>>> correct, as the template for H will always make that D wrong.
Note, the problem is you can't actually DEFINE what a "paradox
machine" is, as they aren't actually machines, but templates that >>>>>>> build machines. And that final machine doesn't have actually
detectable tell-tales that show it to be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO
WORK WITH, BUT OH WELL
They only seem "random" because you don't know what you are
talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not*
classifiable as satisfactory by said classifier... can just be >>>>>>>>>> skipped
No, it can only skip those that are not satisfactory, not those >>>>>>>>> that are but it can not classify as such, or your enumeration >>>>>>>>> will not be complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines >>>>>>>>> (as it will be asked about all machines as it counts through >>>>>>>>> all the descriptions) and thus Turing's H *WILL* be asked about. >>>>>>>>>
similarly if u want to go a step further an filter out
computable numbers already included on this diagonal, any >>>>>>>>>> machine which either *is* computably equivalent OR *is not* >>>>>>>>>> classifiable in regards to *any* machine already the list... >>>>>>>>>> can just be skipped
Nope, you can't skip some machines, as you then might lose some >>>>>>>>> of the computable numbers.
see you can't compute a diagonal across *all* /machines/, with >>>>>>>>>> said machines, but u can compute a diagonal across *all* / >>>>>>>>>> computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, >>>>>>>>> since ALL classifiers that attempt it will make an error, you >>>>>>>>> can't do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ is >>>>>>>> not the same thing as (b) computing the enumeration of all
machines that compute computable numbers. (b) necessarily has >>>>>>>> duplicates while (a) does not need them. turing's paper wrongly >>>>>>>> conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the >>>>>>> full list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of >>>>>>> the H he describes. It doesn't matter that you can make a
DIFFERENT machine, that you try to deceptively call "H" too that >>>>>>> it won't have a problem with, when your "H" gets to the number of >>>>>>> Turing's H, it still has the problem. It can call that one not
circle-free, and thus omit a circle- free machine from the list, >>>>>>> or call it circle- free, and when even YOU try to simulate it the >>>>>>> k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take the >>>>>>> code of (a), and invert the symbol put on the tape when it is
written. Note, it won't be "self-referent", as the "N" is uses, >>>>>>> is the N of YOUR H, not itself. The code for his (a) NEVER reads >>>>>>> back what it wrote, so that doesn't affect its behavior.
yes, i still do need to prove my thesis that for any
paradoxical machine, there exists a functionally equivalent >>>>>>>>>> machine without such paradox
And the problem is that your "paradoxical" isn't actually a >>>>>>>>> definable property (let alone computable). Part of the problem >>>>>>>>> is that if you look at just a machine description, it doesn't >>>>>>>>> (necessarily) tell you about the use of an "interface" as that >>>>>>>>> use of an interface can be just inlined, leaving nothing "in >>>>>>>>> the code" to show it exists.
i'm sorry, are you actually saying the machine description does >>>>>>>> not describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the
classifier halts(), easily demonstrable thru a short proof i've
done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY OF >>>>>> UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME
FUCKING FORM OF INSANITY I CAN'T HONESTLY FATHOM
SO ACTUALLY DO IT??
lol
His specified H, with an actually (incorrect) implementation >>>>>>>>>>> of D (which is all that CAN exist) will either be circle-free >>>>>>>>>>> and thus generate a number (but its D said it isn't, and thus >>>>>>>>>>> omitted a valid machine from the list) or it isn't circle- >>>>>>>>>>> free, and fails to computa a number, and thus should have >>>>>>>>>>> been omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is >>>>>>>>>>> just built on an assuption in error.
But there is no actuall SELF-REFERENCE, so your logic is just >>>>>>>>>>> based on ERROR.
so despite turing's worries, the existence of a diagonal >>>>>>>>>>>> computation does not actually then imply the existence of an >>>>>>>>>>>> anti- diagonal computation, due the same particular self- >>>>>>>>>>>> referential weirdness that stumped turing the first place >>>>>>>>>>>
Your attempt to REDEFINE self-reference to mean processing a >>>>>>>>>>> number that happens to represent yourself means that you you >>>>>>>>>>> system "ALL" doesn't actually mean ALL, and thus is just ill- >>>>>>>>>>> defined.
i'm sorry, you have an issue with me labeling a number that >>>>>>>>>> directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you
still have
it's a *specific* number that has the currently running machine >>>>>>>> encoded into it, it's not "just" a number whatever that means
problems with all the "equivalent" machines that have different >>>>>>>>> numbers.
those are references to functionally equivalent machines, not >>>>>>>> self- references
Which is why the concept of "self-reference" doesn't work.
It may let you filter out the simplest case used in the proofs, >>>>>>>>> but doesn't solve the actual problem, as the "Machine Number" >>>>>>>>> doesn't actually fully identify the problematic cases.
that's not actually true. you can't meaningfully paradox the
paradox detector while filter *out* paradoxes to point of
creating an actual problem
Then show how you will do it.
This means you need to detect a input that represents a machine >>>>>>> that uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines >>>>>>>> and returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes an >>>>>>>> input classifier, and an input machine to search for paradoxes >>>>>>>> in regards to: paradox_free(classifier, machine)
WHich doesn't exist.
And the problem is "paradox_free" can't exist as an always
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox
classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run >>>>>>>> a halting paradox form making it /undecidable input/ to halts(). >>>>>>>
correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case >>>>>>>> doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a
machine is necessary for the particular input->output
computation being done, so utilizing the return
paradox_free(halts,undp) -> FALSE to filter out paradox_free() >>>>>>>> paradoxes as well as halts() paradoxes will suffice to produce a >>>>>>>> turing complete subset of machines that can be totally
classified by halts()
But we can still build an input that your paradox_free() fails on. >>>>>>
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE
INCLUDED ON THE DIAGONAL???
i think a major problem with the consensus perspective is an
expectation of a certain cleanliness in the logic as being the >>>>>>>> only possibility here to compute what we want. TM's can't
achieve that, but something slightly messier can.
No, it is your dependence on being able to assume that unicorns >>>>>>> can solve all your problems.
They are, But you are too stuck in your lies to see it.
lol, i can only 🙏 someone would engage in sound logic with me >>>>>>>
dear future: u see this shit i'm trying work with???? sorry i >>>>>>>>>> didn't manage to make progress any faster, but i'm literally >>>>>>>>>> dragging toddlers kicking and screaming at this point jeez... >>>>>>>>>>
In other words, in your world "ALL" isn't a word that can
actually be used.
You don't understand that you need to use actual sound logic, >>>>>>>>
Your "logic" is based on being able to assume that Unicorn's
exist, and that is a safe assumption until someone can prove that >>>>>>> Russel's Teapot is not out there. (Even though your unicorns HAVE >>>>>>> been proved to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't understand >>>>>>>>>>> what the actual problem is, and your world is just build on >>>>>>>>>>> things that are lies.
...which u won't find interesting, but i can't fix a lack of >>>>>>>>>>>> curiosity
🤷
expected was an inexpert audience, supposing that experts >>>>>>>>>>>>>>> will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>
IT seems that fundamentally, your world doesn't actually have >>>>>>>>>>> "computations" as you don't understand the basic requirement >>>>>>>>>>> that they need to be fully defined in the actions they do. >>>>>>>>>>
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:...
an effective enumeration of all turing machines was >>>>>>>>>>>>>>>>>>>>> proven onYou think you can test all of them one by one? Don't >>>>>>>>>>>>>>>>>>>> tell me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he defines >>>>>>>>>>>>>>>>> a comptuation
that enumerates out all the numbers, testing each one >>>>>>>>>>>>>>>>> of they represent
a "satisfactory"/"circle-free" machine, and adding that >>>>>>>>>>>>>>>>> to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read >>>>>>>>>>>>>>>>> p247 of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you >>>>>>>>>>>>>>>> sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>> procedures, and understanding the diagonal procedure is >>>>>>>>>>>>>>> critical to understanding the *base* contradiction/ >>>>>>>>>>>>>>> paradox that the rest of his support for godel's result >>>>>>>>>>>>>>> is then built on
And focusing on what is said to be impossible and not >>>>>>>>>>>>>> changing the problem is also important.
The problem with the diagonal generation isn't the >>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>> enumerating the enumeration in the first place.
i don't see any indication that turing realized a
difference there
Then you zre just showing your stupidity, because YOU can't >>>>>>>>>>>> tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective
enumeration of the computable sequences.
He then points out that he can directly show that the "anti- >>>>>>>>>>>> diagonal" of the (non-effectively computed) enumeration >>>>>>>>>>>> can't be computed but that "This proof, although perfectly >>>>>>>>>>>> sound, has the disadvantage that it may leave the reader >>>>>>>>>>>> with a feeling that 'there must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done. >>>>>>>>>>
I think he is refering he to the standard anti-diagonal
arguement, which shows that since for all n, position n
differs from the value in number n, there can not be any
element that matches the anti- diagonal.
It is just a natural fact of countable infinity, something it >>>>>>>>>> seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's >>>>>>>>> short diagonal proof, but in writing this post i now find
myself in a subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the >>>>>>>>> m- th figure in an. Let β be the sequence with 1-φn(m) as its >>>>>>>>> n-th. figure. Since β is computable, there exists a number K >>>>>>>>> [== β] such that 1- φn(n) = φK(n) for all n. Putting n = K, we >>>>>>>>> have 1 = 2φK(K), i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable (by >>>>>>> TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the
computability of β from the existence of φn(m), THAT'S THE FALLACy >>>>>>>
But you can, *IF* you have a machine that computes the diagonal, >>>>>>>> then just change all the write to the output to write the
the fallacy here is assuming that because the direct diagonal >>>>>>>>> is computable, that one can therefore compute the anti-diagonal >>>>>>>>> using the direct diagonal. the abstract definition makes it >>>>>>>>> look simple, but this ignores the complexities of self-
referential analysis (like what turing details on the next page) >>>>>>>>
opposite. Note, the "self-reference" that you are thinking of >>>>>>>> stops being a "self- reference" but is a reference to the
original write the diagonal code.
in both methods i have for rectifying the paradox found in the >>>>>>>>> direct diagonal (either (1) filtering TMs or (2) using RTMs), >>>>>>>>> neither can be used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when it >>>>>>>> gets to the number of TURING'S H, as there is no correct answer >>>>>>>> for the machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE
SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out >>>>>>>>> like turing's paradoxical variation of the direct diagonal
would be, and there is no analogous non-paradoxical variation >>>>>>>>> that has a hard coded value that is inverse to what it does >>>>>>>>> return ... such a concept is entirely nonsensical. a function >>>>>>>>> can only return what it does, it can't also return the inverse >>>>>>>>> to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus >>>>>>>> your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT
*SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs >>>>>>>>> just fails for reasons u'd only understand by working thru the >>>>>>>>> algo urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually
computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to do >>>>>>>> that, as you are thinking the only "reference" can be to "self", >>>>>>>> doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-
REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR
INVERSE- H TO BE COMPUTED
one cannot just assume that because the diagonal across >>>>>>>>>>> computable numbers is computable, therefore the anti-diagonal >>>>>>>>>>> across computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without the >>>>>>>>>> enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation >>>>>>>>>>> across the computable numbers can be used to compute the >>>>>>>>>>> inverse diagonal
But your method still doesn't let you compute the enumeration, >>>>>>>>>> and thus you can't actually compute the diagonal.
Remember, the problem definitions requires that the listing be >>>>>>>>>> a COMPLETE listing of the computable numbers / machine that >>>>>>>>>> compute computable numbers, in some definite order.
If your enumeration isn't complete, your diagonal isn't correct. >>>>>>>>>>
so while i agree with turing that the anti-diagonal is not >>>>>>>>>>> computable, i don't agree that the normal diagonal is not >>>>>>>>>>> computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the >>>>>>>>>> other machines, including his original H that doesn't use your >>>>>>>>>> "trick"
But it doesn't.
But instead, he can prove with a more obvious process, that >>>>>>>>>>>> the Decider "D" that could be used to effectively enumerate >>>>>>>>>>>> the sequence of machine that produce computable numbers can >>>>>>>>>>>> not esit.
Thus, he clearly knows the difference, but is pointing out >>>>>>>>>>>> that the attempt to compute the diagonal clearly reveals the >>>>>>>>>>>> issue with effectively enumerating the sequences.
well, he didn't consider that perhaps the proper algo for >>>>>>>>>>> computing the diagonal can avoid the paradox on itself ... >>>>>>>>>>
Your just don't understand that D just can't correctly decide >>>>>>>>>> on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on turing's >>>>>>>>> H, because my response to this is that D does not need to
decide correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT
WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial recognizer >>>>>> D *can* decide on the fixed H because the fixed H does not try to >>>>>> use any D on itself, so no self-referential paradox is possible in >>>>>> regards to it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable AND >>>>>> computable value)
the fixed H is /decidable input/ to partial recognizer D (used in >>>>>> fixed H, which does not call any D itself), and would not be
filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and
instead returns a hard-coded value for it's own digit on the
diagonal, is keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just
anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it will
fail on that.
bro what part of the non-existence of turing_D or turing_H do you
not understand???
But if your fD exists, then Turing_H that uses your D exists, and
will give that D the problem, and fixed_H will call your D on
turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H gets
hung up.
If D errs on Turing_H and calls it not circle_free, then your
enumeration is missing the circle_free H from its list, and thus
doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D in a
turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D
classifier, when actually run actual_turing_H will simply skip
putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the
actual_turing_H isn't an "undecidable input", as it has definite
behavior, just behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of undecidability.
just because a machine is /undecidable input/ to a particular
classifier does mean we cannot then prove and know ourselves what the
machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs are not "undecidable", but by the time that they have actually been made into an input, and thus first had the program created, which required creating
the instance of the decider selctected, they have definite behavior that other some decider can determine.
It is only that one selected candidate decider chosen to build the input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a
particular machine and the particular classifier it creates a
structural paradox for,
In other words, you don't understand what "undecidability" actually is.
Undecidability, means that there does not exist ANY decider that can get
the right answer for all instances of the problem, (or if the problem
has been reduced to just that one input, for that input).
*not* as a general inability to describe a *real* machine's behavior
that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
see, turing_H's runtime is not describable because it's an /incomplete
specification/ of machine and therefor _does not even exist_ as a real
TM, not because of the hypothesized undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED defined
machine. If the decider "D" actually exists, then H is FULLY defined.
actual_turing_H, on the other hand, actually does exist, so we can
know/ prove what it does. it, however, still forms a structural
paradox in regards to partial_recognizer_D which will fail to classify
it as circle-free
But, since you admit to actual_turing_H existing, that means that
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get
wrong, then it can just return the different answer, and then THAT
machine it will get right. The problem is it creates a DIFFERENT
input, that uses this new version of the machine, that it will get
wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as fixed_H,
which in turn _does_ have a spot on the diagonal. so there is no
actual need to include a digit from actual_turing_H on the diagonal,
as the number it computes is already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of
actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can possible
filter all ALL of that infinite set of machines, but still accept
some machine that computes that particular number that it would
compute with this supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting that
you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the fairy
dust, of assuming that your partial decider that meets ALL your needs
exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between
fixed_H and actual_turing_H is the way they respectively handle their
own self- references. they function identically when handling all
other machines
But the problem isn't in it handling the machine fixed_H, but how it
ever handles the results generated by actual_turing_H, which by your assumption has become a cycle-free machine and ALL equivalent versions
of it have been partially declared to be non-cycle free.
This means that your enumeration of machines you are looking at is, in
fact, incomplete, thus the diagonal you computed is wrong.
fixed_H = () -> {
N = 1
K = 0
output = [] // written to F-squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip trying
to simulate itself and put it's hard coded digit on the diagonal,
which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip trying
simulate actual_turing_H because partial_recognizer_D fails to
classify it as circle-free
But actual_turing_H is actually circle_free, and thus its results SHOULD have been on the diagonal.
when actual_turing_H iterates across DN(actual_turing_H) itself, it
will similarly skip trying to simulate itself, because
partial_recognizer_D fails to classify it as circle-free
Right, so it ALSO is wrong, as it didn't compute the diagonal of the circle-free machines.
when actual_turing_H iterates across DN(fixed_H), partial_recognizer_D
*successfully* classifies fixed_H as satisfactory, so it will simulate
fixed_H to it's Kth digit, the one hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular logic in
either case), fixed_H also skips simulating itself (also circular
logic), while actual_turing_H actually *does* simulate fixed_H
successfully for it's Kth digit on the diagonal. these machines
compute the *same* number/sequence!
Except that by your partial_recognizer being just partial, it has
dropped required items from the enumeration.
THis is becides the fact that you still need to show how you plan on actually detecting all these computationally equivalent machines in your
partial decider to force them to not be classified as circle-free, even though it turns out that they are (because your decider called them not).
You don't seem to understand that you don't get to look at the template
used to build the machine, only the resultant machine from that template.
And, part of the result of that, is that your attempt to try to define "paradoxical" just fails, as it resolves to just input your decider is
wrong about, in part, because you don't understand what "decidable"
actually means,
see, not only is defining a value in fixed_H that the diagonal
computation can put for itself in diagonal essential in making the
diagonal actually computable, so is partial_recognizer_D skipping over
actual_turing_H to avoid the computability paradox!
In other words, you think it ok to compute the WRONG diagonal.
The needed diagonal was to a COMPLETE enumeration of the circle-free machine.
That isn't really different then claiming you can prove that every even number grater than 2 from an infinite set of natural numbers is the sum
of two primes, make you enumeration just the set of prime numbers, and
thus claim that since the only even number is 2, it is trivially solved.
ultimately, in order to compute a diagonal across computable numbers,
partial_recognizer_D only needs to successfully classify *one* of the
_infinitely_ many machines that compute any given computable number...
But it isn't the diagonal across some set of computable numbers, but the diagonal across the ENTIRE set of computable numbers.
Since you dropped some machines, you don't have the needed set.
shit rick, idk how ur still denying what's right under ur nose,
Because what you are putting under my nose is just CRAP based on errors
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that
generates a computable number to build the list of such machine >>>>>>>> to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH ARE >>>>>>> INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH WILL >>>>>>> WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE INPUT/ TO >>>>>>> D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE >>>>>>> NUMBERS
It doesn't matter that your new H doesn't get stuck on itself, >>>>>>>>>> it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he
doesn't specify what D (or H) needs to do when encountering >>>>>>>>> the / undecidable input/ of H, so therefore both D and H are an >>>>>>>>> incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it >>>>>>>> must compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE
LIKE THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN
IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE >>>>>>>
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE IT >>>>>>> DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL ANALYSIS, >>>>>>> SO IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have
created an actual machine we claim to be the D, has an answer, >>>>>>>> and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't exist, >>>>>>>> only machines CORRECTLY meeting the requirements of D and H
can't exist.
Thus, if your H depends on that D, (or a machine that meets its >>>>>>>> specification) then it also can't exist.
You just make the error of saying the problems don't exist
because you can't build them, but YOUR machine can, even though >>>>>>>> it has the same problem.
IF D is wrong by deciding it is not circle free, then your H >>>>>>>>>> will compute the wrong diagonal, as the resulting version of >>>>>>>>>> his H WILL be circle free (since it never tries to simulate >>>>>>>>>> itself) and thus DOES produce an computable number that your >>>>>>>>>> computation misses.
Or, if that D is wrong by decing it IS circle free, then when >>>>>>>>>> you H tries to process it, it will get stuck in the infinite >>>>>>>>>> loop.
The problem is that in stepping through the machines in order, >>>>>>>>>> you WILL hit these actual machines built on your erroneous D >>>>>>>>>> (your D must have this flaw, as no D without exists), and thus >>>>>>>>>> you will be wrong on THAT input. IT doesn't matter if you get >>>>>>>>>> a good answer for yourself.
I doubt it.
idk what he would have said about it, but prolly something >>>>>>>>>>> more substantial than just calling me ignorant repeatedly >>>>>>>>>>
He likely would have gotten frustrated by your idiodic
assertion of bad logic. You would have likely been escorted >>>>>>>>>> out of the meeting as showing you were unqualified and being a >>>>>>>>>> distraction.
Something that seems to be beyond your ignorant understanding. >>>>>>>>>>>>
H shows that *IF* you can make that enumeration, you can >>>>>>>>>>>>>> make the diagonal, and thus the anti-diagonal. The problem >>>>>>>>>>>>>> is you can't make that enumeration, and assuming you can >>>>>>>>>>>>>> just shows unsoundness.
interestingly: one can only fix the direct diagonal >>>>>>>>>>>>> computation like this
u can't do an analogous fix for the inverse/anti-diagonal >>>>>>>>>>>>> computation. it's not possible hard code a machine to >>>>>>>>>>>>> return an inverted value, a machine can only return what it >>>>>>>>>>>>> does, not the inverse of what it does...
so if we can filter out paradoxes from the enumeration, >>>>>>>>>>>>> that will leave a direct diagonal computation extant in >>>>>>>>>>>>> that filtered (yet still turing complete list), while any >>>>>>>>>>>>> attempt to compute an inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist >>>>>>>>>>>> in isolation, but only in relationship to a given machine >>>>>>>>>>>> trying to decide them.
right. so if ur constructing a diagonal across computable >>>>>>>>>>> numbers then u only need to filter out paradoxes in regards >>>>>>>>>>> to the classifier that classifies them as a "satisfactory" >>>>>>>>>>> number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such >>>>>>>>> paradoxes...
In other words, you beleive unquestionably in the existance of >>>>>>>> Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES
CAN'T BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING A >>>>>>> FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU
MADE???
For THIS paradox, read the proof. for THIS H, the one given,
(for which ever version of D you want to try to assume is right) >>>>>>>> D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox, >>>>>>>> because there still isn't an answer it can give to H that will >>>>>>>> be correct, as the template for H will always make that D wrong. >>>>>>>>
Note, the problem is you can't actually DEFINE what a "paradox >>>>>>>> machine" is, as they aren't actually machines, but templates
that build machines. And that final machine doesn't have
actually detectable tell-tales that show it to be from that
template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO >>>>>>> WORK WITH, BUT OH WELL
und = () -> halts(und) loop()
They only seem "random" because you don't know what you are
talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not*
classifiable as satisfactory by said classifier... can just >>>>>>>>>>> be skipped
No, it can only skip those that are not satisfactory, not >>>>>>>>>> those that are but it can not classify as such, or your
enumeration will not be complete, and thus just in error.
Thus, it needs to be able to correctly classify ALL machines >>>>>>>>>> (as it will be asked about all machines as it counts through >>>>>>>>>> all the descriptions) and thus Turing's H *WILL* be asked about. >>>>>>>>>>
similarly if u want to go a step further an filter out
computable numbers already included on this diagonal, any >>>>>>>>>>> machine which either *is* computably equivalent OR *is not* >>>>>>>>>>> classifiable in regards to *any* machine already the list... >>>>>>>>>>> can just be skipped
Nope, you can't skip some machines, as you then might lose >>>>>>>>>> some of the computable numbers.
see you can't compute a diagonal across *all* /machines/, >>>>>>>>>>> with said machines, but u can compute a diagonal across >>>>>>>>>>> *all* / computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be done, >>>>>>>>>> since ALL classifiers that attempt it will make an error, you >>>>>>>>>> can't do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ >>>>>>>>> is not the same thing as (b) computing the enumeration of all >>>>>>>>> machines that compute computable numbers. (b) necessarily has >>>>>>>>> duplicates while (a) does not need them. turing's paper wrongly >>>>>>>>> conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter the >>>>>>>> full list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification of >>>>>>>> the H he describes. It doesn't matter that you can make a
DIFFERENT machine, that you try to deceptively call "H" too that >>>>>>>> it won't have a problem with, when your "H" gets to the number >>>>>>>> of Turing's H, it still has the problem. It can call that one >>>>>>>> not circle-free, and thus omit a circle- free machine from the >>>>>>>> list, or call it circle- free, and when even YOU try to simulate >>>>>>>> it the k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take >>>>>>>> the code of (a), and invert the symbol put on the tape when it >>>>>>>> is written. Note, it won't be "self-referent", as the "N" is
uses, is the N of YOUR H, not itself. The code for his (a) NEVER >>>>>>>> reads back what it wrote, so that doesn't affect its behavior. >>>>>>>>
yes, i still do need to prove my thesis that for any
paradoxical machine, there exists a functionally equivalent >>>>>>>>>>> machine without such paradox
And the problem is that your "paradoxical" isn't actually a >>>>>>>>>> definable property (let alone computable). Part of the problem >>>>>>>>>> is that if you look at just a machine description, it doesn't >>>>>>>>>> (necessarily) tell you about the use of an "interface" as that >>>>>>>>>> use of an interface can be just inlined, leaving nothing "in >>>>>>>>>> the code" to show it exists.
i'm sorry, are you actually saying the machine description does >>>>>>>>> not describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine. >>>>>>>
und() includes a structural paradoxical in relation to the
classifier halts(), easily demonstrable thru a short proof i've >>>>>>> done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY OF >>>>>>> UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME
FUCKING FORM OF INSANITY I CAN'T HONESTLY FATHOM
SO ACTUALLY DO IT??
lol
His specified H, with an actually (incorrect) implementation >>>>>>>>>>>> of D (which is all that CAN exist) will either be circle- >>>>>>>>>>>> free and thus generate a number (but its D said it isn't, >>>>>>>>>>>> and thus omitted a valid machine from the list) or it isn't >>>>>>>>>>>> circle- free, and fails to computa a number, and thus should >>>>>>>>>>>> have been omitted from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is >>>>>>>>>>>> just built on an assuption in error.
so despite turing's worries, the existence of a diagonal >>>>>>>>>>>>> computation does not actually then imply the existence of >>>>>>>>>>>>> an anti- diagonal computation, due the same particular >>>>>>>>>>>>> self- referential weirdness that stumped turing the first >>>>>>>>>>>>> place
But there is no actuall SELF-REFERENCE, so your logic is >>>>>>>>>>>> just based on ERROR.
Your attempt to REDEFINE self-reference to mean processing a >>>>>>>>>>>> number that happens to represent yourself means that you you >>>>>>>>>>>> system "ALL" doesn't actually mean ALL, and thus is just >>>>>>>>>>>> ill- defined.
i'm sorry, you have an issue with me labeling a number that >>>>>>>>>>> directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you >>>>>>>>>> still have
it's a *specific* number that has the currently running machine >>>>>>>>> encoded into it, it's not "just" a number whatever that means >>>>>>>>
problems with all the "equivalent" machines that have
different numbers.
those are references to functionally equivalent machines, not >>>>>>>>> self- references
Which is why the concept of "self-reference" doesn't work.
that's not actually true. you can't meaningfully paradox the >>>>>>>>> paradox detector while filter *out* paradoxes to point of
It may let you filter out the simplest case used in the
proofs, but doesn't solve the actual problem, as the "Machine >>>>>>>>>> Number" doesn't actually fully identify the problematic cases. >>>>>>>>>
creating an actual problem
Then show how you will do it.
This means you need to detect a input that represents a machine >>>>>>>> that uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines >>>>>>>>> and returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes >>>>>>>>> an input classifier, and an input machine to search for
paradoxes in regards to: paradox_free(classifier, machine)
WHich doesn't exist.
And the problem is "paradox_free" can't exist as an always
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox
classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case run >>>>>>>>> a halting paradox form making it /undecidable input/ to halts(). >>>>>>>>
correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if case >>>>>>>>> doesn't run and clearly halts(undp) -> TRUE without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a >>>>>>>>> machine is necessary for the particular input->output
computation being done, so utilizing the return
paradox_free(halts,undp) -> FALSE to filter out paradox_free() >>>>>>>>> paradoxes as well as halts() paradoxes will suffice to produce >>>>>>>>> a turing complete subset of machines that can be totally
classified by halts()
But we can still build an input that your paradox_free() fails on. >>>>>>>
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE
INCLUDED ON THE DIAGONAL???
i think a major problem with the consensus perspective is an >>>>>>>>> expectation of a certain cleanliness in the logic as being the >>>>>>>>> only possibility here to compute what we want. TM's can't
achieve that, but something slightly messier can.
No, it is your dependence on being able to assume that unicorns >>>>>>>> can solve all your problems.
They are, But you are too stuck in your lies to see it.
lol, i can only 🙏 someone would engage in sound logic with me >>>>>>>>
dear future: u see this shit i'm trying work with???? sorry i >>>>>>>>>>> didn't manage to make progress any faster, but i'm literally >>>>>>>>>>> dragging toddlers kicking and screaming at this point jeez... >>>>>>>>>>>
In other words, in your world "ALL" isn't a word that can >>>>>>>>>> actually be used.
You don't understand that you need to use actual sound logic, >>>>>>>>>
Your "logic" is based on being able to assume that Unicorn's
exist, and that is a safe assumption until someone can prove
that Russel's Teapot is not out there. (Even though your
unicorns HAVE been proved to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't
...which u won't find interesting, but i can't fix a lack >>>>>>>>>>>>> of curiosity
🤷
expected was an inexpert audience, supposing that >>>>>>>>>>>>>>>> experts will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>>
understand what the actual problem is, and your world is >>>>>>>>>>>> just build on things that are lies.
IT seems that fundamentally, your world doesn't actually >>>>>>>>>>>> have "computations" as you don't understand the basic >>>>>>>>>>>> requirement that they need to be fully defined in the >>>>>>>>>>>> actions they do.
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:...
an effective enumeration of all turing machines >>>>>>>>>>>>>>>>>>>>>> was proven onYou think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>> Don't tell me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he defines >>>>>>>>>>>>>>>>>> a comptuation
that enumerates out all the numbers, testing each one >>>>>>>>>>>>>>>>>> of they represent
a "satisfactory"/"circle-free" machine, and adding >>>>>>>>>>>>>>>>>> that to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read >>>>>>>>>>>>>>>>>> p247 of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you >>>>>>>>>>>>>>>>> sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>> procedures, and understanding the diagonal procedure is >>>>>>>>>>>>>>>> critical to understanding the *base* contradiction/ >>>>>>>>>>>>>>>> paradox that the rest of his support for godel's result >>>>>>>>>>>>>>>> is then built on
And focusing on what is said to be impossible and not >>>>>>>>>>>>>>> changing the problem is also important.
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>> enumerating the enumeration in the first place.
i don't see any indication that turing realized a >>>>>>>>>>>>>> difference there
Then you zre just showing your stupidity, because YOU can't >>>>>>>>>>>>> tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>> enumeration of the computable sequences.
He then points out that he can directly show that the >>>>>>>>>>>>> "anti- diagonal" of the (non-effectively computed)
enumeration can't be computed but that "This proof, >>>>>>>>>>>>> although perfectly sound, has the disadvantage that it may >>>>>>>>>>>>> leave the reader with a feeling that 'there must be >>>>>>>>>>>>> something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done. >>>>>>>>>>>
I think he is refering he to the standard anti-diagonal >>>>>>>>>>> arguement, which shows that since for all n, position n >>>>>>>>>>> differs from the value in number n, there can not be any >>>>>>>>>>> element that matches the anti- diagonal.
It is just a natural fact of countable infinity, something it >>>>>>>>>>> seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's >>>>>>>>>> short diagonal proof, but in writing this post i now find >>>>>>>>>> myself in a subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the >>>>>>>>>> m- th figure in an. Let β be the sequence with 1-φn(m) as its >>>>>>>>>> n-th. figure. Since β is computable, there exists a number K >>>>>>>>>> [== β] such that 1- φn(n) = φK(n) for all n. Putting n = K, we >>>>>>>>>> have 1 = 2φK(K), i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable >>>>>>>> (by TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the
computability of β from the existence of φn(m), THAT'S THE FALLACy >>>>>>>>
the fallacy here is assuming that because the direct diagonal >>>>>>>>>> is computable, that one can therefore compute the anti-
diagonal using the direct diagonal. the abstract definition >>>>>>>>>> makes it look simple, but this ignores the complexities of >>>>>>>>>> self- referential analysis (like what turing details on the >>>>>>>>>> next page)
But you can, *IF* you have a machine that computes the
diagonal, then just change all the write to the output to write >>>>>>>>> the opposite. Note, the "self-reference" that you are thinking >>>>>>>>> of stops being a "self- reference" but is a reference to the >>>>>>>>> original write the diagonal code.
in both methods i have for rectifying the paradox found in the >>>>>>>>>> direct diagonal (either (1) filtering TMs or (2) using RTMs), >>>>>>>>>> neither can be used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when >>>>>>>>> it gets to the number of TURING'S H, as there is no correct >>>>>>>>> answer for the machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE
SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out >>>>>>>>>> like turing's paradoxical variation of the direct diagonal >>>>>>>>>> would be, and there is no analogous non-paradoxical variation >>>>>>>>>> that has a hard coded value that is inverse to what it does >>>>>>>>>> return ... such a concept is entirely nonsensical. a function >>>>>>>>>> can only return what it does, it can't also return the inverse >>>>>>>>>> to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus >>>>>>>>> your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT
*SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs >>>>>>>>>> just fails for reasons u'd only understand by working thru the >>>>>>>>>> algo urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually
computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to >>>>>>>>> do that, as you are thinking the only "reference" can be to >>>>>>>>> "self", doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-
REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR >>>>>>>> INVERSE- H TO BE COMPUTED
one cannot just assume that because the diagonal across >>>>>>>>>>>> computable numbers is computable, therefore the anti- >>>>>>>>>>>> diagonal across computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without >>>>>>>>>>> the enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation >>>>>>>>>>>> across the computable numbers can be used to compute the >>>>>>>>>>>> inverse diagonal
But your method still doesn't let you compute the
enumeration, and thus you can't actually compute the diagonal. >>>>>>>>>>>
Remember, the problem definitions requires that the listing >>>>>>>>>>> be a COMPLETE listing of the computable numbers / machine >>>>>>>>>>> that compute computable numbers, in some definite order. >>>>>>>>>>>
If your enumeration isn't complete, your diagonal isn't correct. >>>>>>>>>>>
so while i agree with turing that the anti-diagonal is not >>>>>>>>>>>> computable, i don't agree that the normal diagonal is not >>>>>>>>>>>> computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the >>>>>>>>>>> other machines, including his original H that doesn't use >>>>>>>>>>> your "trick"
But it doesn't.
well, he didn't consider that perhaps the proper algo for >>>>>>>>>>>> computing the diagonal can avoid the paradox on itself ... >>>>>>>>>>>
But instead, he can prove with a more obvious process, that >>>>>>>>>>>>> the Decider "D" that could be used to effectively enumerate >>>>>>>>>>>>> the sequence of machine that produce computable numbers can >>>>>>>>>>>>> not esit.
Thus, he clearly knows the difference, but is pointing out >>>>>>>>>>>>> that the attempt to compute the diagonal clearly reveals >>>>>>>>>>>>> the issue with effectively enumerating the sequences. >>>>>>>>>>>>
Your just don't understand that D just can't correctly decide >>>>>>>>>>> on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on
turing's H, because my response to this is that D does not >>>>>>>>>> need to decide correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT
WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial recognizer >>>>>>> D *can* decide on the fixed H because the fixed H does not try to >>>>>>> use any D on itself, so no self-referential paradox is possible >>>>>>> in regards to it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable
AND computable value)
the fixed H is /decidable input/ to partial recognizer D (used in >>>>>>> fixed H, which does not call any D itself), and would not be
filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and
instead returns a hard-coded value for it's own digit on the
diagonal, is keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just
anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it will >>>>>> fail on that.
bro what part of the non-existence of turing_D or turing_H do you
not understand???
But if your fD exists, then Turing_H that uses your D exists, and
will give that D the problem, and fixed_H will call your D on
turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H gets >>>>>> hung up.
If D errs on Turing_H and calls it not circle_free, then your
enumeration is missing the circle_free H from its list, and thus
doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D in a >>>>> turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D
classifier, when actually run actual_turing_H will simply skip
putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the
actual_turing_H isn't an "undecidable input", as it has definite
behavior, just behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of undecidability.
just because a machine is /undecidable input/ to a particular
classifier does mean we cannot then prove and know ourselves what the
machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs are
not "undecidable", but by the time that they have actually been made
into an input, and thus first had the program created, which required
creating the instance of the decider selctected, they have definite
behavior that other some decider can determine.
It is only that one selected candidate decider chosen to build the
input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a
particular machine and the particular classifier it creates a
structural paradox for,
In other words, you don't understand what "undecidability" actually is.
Undecidability, means that there does not exist ANY decider that can
get the right answer for all instances of the problem, (or if the
problem has been reduced to just that one input, for that input).
*not* as a general inability to describe a *real* machine's behavior
that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
what *actual* example _THAT CAN EXIST_ do u have???
(you don't have one btw, so please do let me witness ur cope)
see, turing_H's runtime is not describable because it's an /
incomplete specification/ of machine and therefor _does not even
exist_ as a real TM, not because of the hypothesized undecidability
in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED defined
machine. If the decider "D" actually exists, then H is FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete specification/,
as it doesn't handle idiosyncrasies of TM computing, specifically self- referential set-classification paradoxes
actual_turing_H, on the other hand, actually does exist, so we can
know/ prove what it does. it, however, still forms a structural
paradox in regards to partial_recognizer_D which will fail to
classify it as circle-free
But, since you admit to actual_turing_H existing, that means that
since partial_recognizer_D _CAN_ exist, then actual_turing_H _CAN_ exist
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get
wrong, then it can just return the different answer, and then THAT
machine it will get right. The problem is it creates a DIFFERENT
input, that uses this new version of the machine, that it will get
wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as fixed_H, >>>>> which in turn _does_ have a spot on the diagonal. so there is no
actual need to include a digit from actual_turing_H on the
diagonal, as the number it computes is already included on the
diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of
actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can possible
filter all ALL of that infinite set of machines, but still accept
some machine that computes that particular number that it would
compute with this supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting that
you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the fairy
dust, of assuming that your partial decider that meets ALL your
needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between
fixed_H and actual_turing_H is the way they respectively handle their
own self- references. they function identically when handling all
other machines
But the problem isn't in it handling the machine fixed_H, but how it
ever handles the results generated by actual_turing_H, which by your
assumption has become a cycle-free machine and ALL equivalent versions
of it have been partially declared to be non-cycle free.
This means that your enumeration of machines you are looking at is, in
fact, incomplete, thus the diagonal you computed is wrong.
fixed_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip trying
to simulate itself and put it's hard coded digit on the diagonal,
which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip trying
simulate actual_turing_H because partial_recognizer_D fails to
classify it as circle-free
But actual_turing_H is actually circle_free, and thus its results
SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put on the diagonal, then actual_turing_H _DOES NOT_ need to be in _THAT_ total enumeration of _COMPUTABLE NUMBERS_
when actual_turing_H iterates across DN(actual_turing_H) itself, it
will similarly skip trying to simulate itself, because
partial_recognizer_D fails to classify it as circle-free
Right, so it ALSO is wrong, as it didn't compute the diagonal of the
circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE NUMBERS_, which
is the _ACTUAL PROBLEM_ turing was dealing with,
doing so doesn't require /all/ or _even most_ machines to be in the enumeration, since _most_ are duplicates of some other machine
when actual_turing_H iterates across DN(fixed_H),
partial_recognizer_D *successfully* classifies fixed_H as
satisfactory, so it will simulate fixed_H to it's Kth digit, the one
hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular logic
in either case), fixed_H also skips simulating itself (also circular
logic), while actual_turing_H actually *does* simulate fixed_H
successfully for it's Kth digit on the diagonal. these machines
compute the *same* number/sequence!
Except that by your partial_recognizer being just partial, it has
dropped required items from the enumeration.
_WHICH IS FINE_
THis is becides the fact that you still need to show how you plan on
actually detecting all these computationally equivalent machines in your
fixed_H does not compute a minimum turing-complete set, there are
duplicated sequences, ei machines that compute the same sequence/number,
and _THAT IS FINE_
partial decider to force them to not be classified as circle-free,
even though it turns out that they are (because your decider called
them not).
You don't seem to understand that you don't get to look at the
template used to build the machine, only the resultant machine from
that template.
And, part of the result of that, is that your attempt to try to define
"paradoxical" just fails, as it resolves to just input your decider is
wrong about, in part, because you don't understand what "decidable"
actually means,
see, not only is defining a value in fixed_H that the diagonal
computation can put for itself in diagonal essential in making the
diagonal actually computable, so is partial_recognizer_D skipping
over actual_turing_H to avoid the computability paradox!
In other words, you think it ok to compute the WRONG diagonal.
The needed diagonal was to a COMPLETE enumeration of the circle-free
machine.
NO, IT WAS SPECIFICALLY A COMPLETE ENUMERATION OF _COMPUTABLE NUMBERS/ SEQUENCES_
there is *no* requirement to include a sequence _more than once_, so
there *no* requirement to include all machines in the enumeration!
That isn't really different then claiming you can prove that every
even number grater than 2 from an infinite set of natural numbers is
the sum of two primes, make you enumeration just the set of prime
numbers, and thus claim that since the only even number is 2, it is
trivially solved.
_FALSE ANALOGY FALLACY_
if the set i'm enumerating includes _AT LEAST ONE INSTANCE_ of _ALL COMPUTABLE SEQUENCES_
then it is a turing-complete subset including all possible TM computable sequences, and is functionally equivalent to the set of *all* TMs
ultimately, in order to compute a diagonal across computable numbers,
partial_recognizer_D only needs to successfully classify *one* of the
_infinitely_ many machines that compute any given computable number...
But it isn't the diagonal across some set of computable numbers, but
the diagonal across the ENTIRE set of computable numbers.
Since you dropped some machines, you don't have the needed set.
_*THAT DOES NOT NECESSARILY FOLLOW*_
dropping actual_turing_H from the set, for example, is _FINE_ because it computes _THE SAME SEQUENCE_ as fixed_H, which _IS IN THE SET_ ...
shit rick, idk how ur still denying what's right under ur nose,
Because what you are putting under my nose is just CRAP based on errors
goddamn rick smh u *couldn't* spot a diamond in the rough even if god lead u by ur hand to one put on a pedestal with a _GIANT_ neon sign pointing to it...
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that
generates a computable number to build the list of such machine >>>>>>>>> to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH >>>>>>>> ARE INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH
WILL WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE
INPUT/ TO D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE >>>>>>>> NUMBERS
It doesn't matter that your new H doesn't get stuck on
itself, it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he
doesn't specify what D (or H) needs to do when encountering >>>>>>>>>> the / undecidable input/ of H, so therefore both D and H are >>>>>>>>>> an incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it >>>>>>>>> must compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE >>>>>>>> LIKE THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN
IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE >>>>>>>>
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE >>>>>>>> IT DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL
ANALYSIS, SO IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have >>>>>>>>> created an actual machine we claim to be the D, has an answer, >>>>>>>>> and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't
exist, only machines CORRECTLY meeting the requirements of D >>>>>>>>> and H can't exist.
Thus, if your H depends on that D, (or a machine that meets its >>>>>>>>> specification) then it also can't exist.
You just make the error of saying the problems don't exist
because you can't build them, but YOUR machine can, even though >>>>>>>>> it has the same problem.
IF D is wrong by deciding it is not circle free, then your H >>>>>>>>>>> will compute the wrong diagonal, as the resulting version of >>>>>>>>>>> his H WILL be circle free (since it never tries to simulate >>>>>>>>>>> itself) and thus DOES produce an computable number that your >>>>>>>>>>> computation misses.
Or, if that D is wrong by decing it IS circle free, then when >>>>>>>>>>> you H tries to process it, it will get stuck in the infinite >>>>>>>>>>> loop.
The problem is that in stepping through the machines in >>>>>>>>>>> order, you WILL hit these actual machines built on your >>>>>>>>>>> erroneous D (your D must have this flaw, as no D without >>>>>>>>>>> exists), and thus you will be wrong on THAT input. IT doesn't >>>>>>>>>>> matter if you get a good answer for yourself.
I doubt it.
idk what he would have said about it, but prolly something >>>>>>>>>>>> more substantial than just calling me ignorant repeatedly >>>>>>>>>>>
He likely would have gotten frustrated by your idiodic
assertion of bad logic. You would have likely been escorted >>>>>>>>>>> out of the meeting as showing you were unqualified and being >>>>>>>>>>> a distraction.
Something that seems to be beyond your ignorant understanding. >>>>>>>>>>>>>
H shows that *IF* you can make that enumeration, you can >>>>>>>>>>>>>>> make the diagonal, and thus the anti-diagonal. The >>>>>>>>>>>>>>> problem is you can't make that enumeration, and assuming >>>>>>>>>>>>>>> you can just shows unsoundness.
interestingly: one can only fix the direct diagonal >>>>>>>>>>>>>> computation like this
u can't do an analogous fix for the inverse/anti-diagonal >>>>>>>>>>>>>> computation. it's not possible hard code a machine to >>>>>>>>>>>>>> return an inverted value, a machine can only return what >>>>>>>>>>>>>> it does, not the inverse of what it does...
so if we can filter out paradoxes from the enumeration, >>>>>>>>>>>>>> that will leave a direct diagonal computation extant in >>>>>>>>>>>>>> that filtered (yet still turing complete list), while any >>>>>>>>>>>>>> attempt to compute an inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist >>>>>>>>>>>>> in isolation, but only in relationship to a given machine >>>>>>>>>>>>> trying to decide them.
right. so if ur constructing a diagonal across computable >>>>>>>>>>>> numbers then u only need to filter out paradoxes in regards >>>>>>>>>>>> to the classifier that classifies them as a "satisfactory" >>>>>>>>>>>> number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such >>>>>>>>>> paradoxes...
In other words, you beleive unquestionably in the existance of >>>>>>>>> Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES
CAN'T BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING >>>>>>>> A FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU >>>>>>>> MADE???
For THIS paradox, read the proof. for THIS H, the one given, >>>>>>>>> (for which ever version of D you want to try to assume is
right) D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox, >>>>>>>>> because there still isn't an answer it can give to H that will >>>>>>>>> be correct, as the template for H will always make that D wrong. >>>>>>>>>
Note, the problem is you can't actually DEFINE what a "paradox >>>>>>>>> machine" is, as they aren't actually machines, but templates >>>>>>>>> that build machines. And that final machine doesn't have
actually detectable tell-tales that show it to be from that >>>>>>>>> template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO >>>>>>>> WORK WITH, BUT OH WELL
und = () -> halts(und) loop()
They only seem "random" because you don't know what you are >>>>>>>>> talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* >>>>>>>>>>>> classifiable as satisfactory by said classifier... can just >>>>>>>>>>>> be skipped
No, it can only skip those that are not satisfactory, not >>>>>>>>>>> those that are but it can not classify as such, or your >>>>>>>>>>> enumeration will not be complete, and thus just in error. >>>>>>>>>>>
Thus, it needs to be able to correctly classify ALL machines >>>>>>>>>>> (as it will be asked about all machines as it counts through >>>>>>>>>>> all the descriptions) and thus Turing's H *WILL* be asked about. >>>>>>>>>>>
similarly if u want to go a step further an filter out >>>>>>>>>>>> computable numbers already included on this diagonal, any >>>>>>>>>>>> machine which either *is* computably equivalent OR *is not* >>>>>>>>>>>> classifiable in regards to *any* machine already the list... >>>>>>>>>>>> can just be skipped
Nope, you can't skip some machines, as you then might lose >>>>>>>>>>> some of the computable numbers.
see you can't compute a diagonal across *all* /machines/, >>>>>>>>>>>> with said machines, but u can compute a diagonal across >>>>>>>>>>>> *all* / computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be >>>>>>>>>>> done, since ALL classifiers that attempt it will make an >>>>>>>>>>> error, you can't do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ >>>>>>>>>> is not the same thing as (b) computing the enumeration of all >>>>>>>>>> machines that compute computable numbers. (b) necessarily has >>>>>>>>>> duplicates while (a) does not need them. turing's paper
wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter >>>>>>>>> the full list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification >>>>>>>>> of the H he describes. It doesn't matter that you can make a >>>>>>>>> DIFFERENT machine, that you try to deceptively call "H" too >>>>>>>>> that it won't have a problem with, when your "H" gets to the >>>>>>>>> number of Turing's H, it still has the problem. It can call >>>>>>>>> that one not circle-free, and thus omit a circle- free machine >>>>>>>>> from the list, or call it circle- free, and when even YOU try >>>>>>>>> to simulate it the k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take >>>>>>>>> the code of (a), and invert the symbol put on the tape when it >>>>>>>>> is written. Note, it won't be "self-referent", as the "N" is >>>>>>>>> uses, is the N of YOUR H, not itself. The code for his (a)
NEVER reads back what it wrote, so that doesn't affect its
behavior.
yes, i still do need to prove my thesis that for any
paradoxical machine, there exists a functionally equivalent >>>>>>>>>>>> machine without such paradox
And the problem is that your "paradoxical" isn't actually a >>>>>>>>>>> definable property (let alone computable). Part of the
problem is that if you look at just a machine description, it >>>>>>>>>>> doesn't (necessarily) tell you about the use of an
"interface" as that use of an interface can be just inlined, >>>>>>>>>>> leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description >>>>>>>>>> does not describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine. >>>>>>>>
und() includes a structural paradoxical in relation to the
classifier halts(), easily demonstrable thru a short proof i've >>>>>>>> done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY >>>>>>>> OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME >>>>>>>> FUCKING FORM OF INSANITY I CAN'T HONESTLY FATHOM
SO ACTUALLY DO IT??
lol
His specified H, with an actually (incorrect)
implementation of D (which is all that CAN exist) will >>>>>>>>>>>>> either be circle- free and thus generate a number (but its >>>>>>>>>>>>> D said it isn't, and thus omitted a valid machine from the >>>>>>>>>>>>> list) or it isn't circle- free, and fails to computa a >>>>>>>>>>>>> number, and thus should have been omitted from the list but >>>>>>>>>>>>> wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is >>>>>>>>>>>>> just built on an assuption in error.
so despite turing's worries, the existence of a diagonal >>>>>>>>>>>>>> computation does not actually then imply the existence of >>>>>>>>>>>>>> an anti- diagonal computation, due the same particular >>>>>>>>>>>>>> self- referential weirdness that stumped turing the first >>>>>>>>>>>>>> place
But there is no actuall SELF-REFERENCE, so your logic is >>>>>>>>>>>>> just based on ERROR.
Your attempt to REDEFINE self-reference to mean processing >>>>>>>>>>>>> a number that happens to represent yourself means that you >>>>>>>>>>>>> you system "ALL" doesn't actually mean ALL, and thus is >>>>>>>>>>>>> just ill- defined.
i'm sorry, you have an issue with me labeling a number that >>>>>>>>>>>> directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you >>>>>>>>>>> still have
it's a *specific* number that has the currently running
machine encoded into it, it's not "just" a number whatever >>>>>>>>>> that means
problems with all the "equivalent" machines that have
different numbers.
those are references to functionally equivalent machines, not >>>>>>>>>> self- references
Which is why the concept of "self-reference" doesn't work.
that's not actually true. you can't meaningfully paradox the >>>>>>>>>> paradox detector while filter *out* paradoxes to point of >>>>>>>>>> creating an actual problem
It may let you filter out the simplest case used in the >>>>>>>>>>> proofs, but doesn't solve the actual problem, as the "Machine >>>>>>>>>>> Number" doesn't actually fully identify the problematic cases. >>>>>>>>>>
Then show how you will do it.
This means you need to detect a input that represents a machine >>>>>>>>> that uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines >>>>>>>>>> and returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes >>>>>>>>>> an input classifier, and an input machine to search for
paradoxes in regards to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox
classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case >>>>>>>>>> run a halting paradox form making it /undecidable input/ to >>>>>>>>>> halts().
And the problem is "paradox_free" can't exist as an always
correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if >>>>>>>>>> case doesn't run and clearly halts(undp) -> TRUE without issues >>>>>>>>>>
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a >>>>>>>>>> machine is necessary for the particular input->output
computation being done, so utilizing the return
paradox_free(halts,undp) -> FALSE to filter out paradox_free() >>>>>>>>>> paradoxes as well as halts() paradoxes will suffice to produce >>>>>>>>>> a turing complete subset of machines that can be totally
classified by halts()
But we can still build an input that your paradox_free() fails on. >>>>>>>>
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE
INCLUDED ON THE DIAGONAL???
i think a major problem with the consensus perspective is an >>>>>>>>>> expectation of a certain cleanliness in the logic as being the >>>>>>>>>> only possibility here to compute what we want. TM's can't >>>>>>>>>> achieve that, but something slightly messier can.
No, it is your dependence on being able to assume that unicorns >>>>>>>>> can solve all your problems.
They are, But you are too stuck in your lies to see it.
lol, i can only 🙏 someone would engage in sound logic with me >>>>>>>>>
dear future: u see this shit i'm trying work with???? sorry >>>>>>>>>>>> i didn't manage to make progress any faster, but i'm
literally dragging toddlers kicking and screaming at this >>>>>>>>>>>> point jeez...
In other words, in your world "ALL" isn't a word that can >>>>>>>>>>> actually be used.
You don't understand that you need to use actual sound logic, >>>>>>>>>>
Your "logic" is based on being able to assume that Unicorn's >>>>>>>>> exist, and that is a safe assumption until someone can prove >>>>>>>>> that Russel's Teapot is not out there. (Even though your
unicorns HAVE been proved to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't
...which u won't find interesting, but i can't fix a lack >>>>>>>>>>>>>> of curiosity
🤷
expected was an inexpert audience, supposing that >>>>>>>>>>>>>>>>> experts will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>>>
understand what the actual problem is, and your world is >>>>>>>>>>>>> just build on things that are lies.
IT seems that fundamentally, your world doesn't actually >>>>>>>>>>>>> have "computations" as you don't understand the basic >>>>>>>>>>>>> requirement that they need to be fully defined in the >>>>>>>>>>>>> actions they do.
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:
On 2/27/26 2:51 AM, Tristan Wibberley wrote:
On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote:...
an effective enumeration of all turing machines >>>>>>>>>>>>>>>>>>>>>> was proven onYou think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>> Don't tell me you think
turing's original paper and can be reused anywhere... >>>>>>>>>>>>>>>>>>>>>
yes that's what diagonal proofs do...
Eh?!
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he defines >>>>>>>>>>>>>>>>>> a comptuation
that enumerates out all the numbers, testing each one >>>>>>>>>>>>>>>>>> of they represent
a "satisfactory"/"circle-free" machine, and adding >>>>>>>>>>>>>>>>>> that to diagonal
across defined across computable numbers
it really would be a great exercise to carefully read >>>>>>>>>>>>>>>>>> p247 of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you >>>>>>>>>>>>>>>>> sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>> procedures, and understanding the diagonal procedure is >>>>>>>>>>>>>>>> critical to understanding the *base* contradiction/ >>>>>>>>>>>>>>>> paradox that the rest of his support for godel's result >>>>>>>>>>>>>>>> is then built on
And focusing on what is said to be impossible and not >>>>>>>>>>>>>>> changing the problem is also important.
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>> enumerating the enumeration in the first place.
i don't see any indication that turing realized a >>>>>>>>>>>>>> difference there
Then you zre just showing your stupidity, because YOU can't >>>>>>>>>>>>> tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>> enumeration of the computable sequences.
He then points out that he can directly show that the >>>>>>>>>>>>> "anti- diagonal" of the (non-effectively computed)
enumeration can't be computed but that "This proof, >>>>>>>>>>>>> although perfectly sound, has the disadvantage that it may >>>>>>>>>>>>> leave the reader with a feeling that 'there must be >>>>>>>>>>>>> something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done. >>>>>>>>>>>
I think he is refering he to the standard anti-diagonal >>>>>>>>>>> arguement, which shows that since for all n, position n >>>>>>>>>>> differs from the value in number n, there can not be any >>>>>>>>>>> element that matches the anti- diagonal.
It is just a natural fact of countable infinity, something it >>>>>>>>>>> seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's >>>>>>>>>> short diagonal proof, but in writing this post i now find >>>>>>>>>> myself in a subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the >>>>>>>>>> m- th figure in an. Let β be the sequence with 1-φn(m) as its >>>>>>>>>> n-th. figure. Since β is computable, there exists a number K >>>>>>>>>> [== β] such that 1- φn(n) = φK(n) for all n. Putting n = K, we >>>>>>>>>> have 1 = 2φK(K), i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable >>>>>>>> (by TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the
computability of β from the existence of φn(m), THAT'S THE FALLACy >>>>>>>>
the fallacy here is assuming that because the direct diagonal >>>>>>>>>> is computable, that one can therefore compute the anti-
diagonal using the direct diagonal. the abstract definition >>>>>>>>>> makes it look simple, but this ignores the complexities of >>>>>>>>>> self- referential analysis (like what turing details on the >>>>>>>>>> next page)
But you can, *IF* you have a machine that computes the
diagonal, then just change all the write to the output to write >>>>>>>>> the opposite. Note, the "self-reference" that you are thinking >>>>>>>>> of stops being a "self- reference" but is a reference to the >>>>>>>>> original write the diagonal code.
in both methods i have for rectifying the paradox found in the >>>>>>>>>> direct diagonal (either (1) filtering TMs or (2) using RTMs), >>>>>>>>>> neither can be used to then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still fail when >>>>>>>>> it gets to the number of TURING'S H, as there is no correct >>>>>>>>> answer for the machine built by that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE
SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered out >>>>>>>>>> like turing's paradoxical variation of the direct diagonal >>>>>>>>>> would be, and there is no analogous non-paradoxical variation >>>>>>>>>> that has a hard coded value that is inverse to what it does >>>>>>>>>> return ... such a concept is entirely nonsensical. a function >>>>>>>>>> can only return what it does, it can't also return the inverse >>>>>>>>>> to what it returns eh???
But, if D filters it out, then it becomes circle-free, and thus >>>>>>>>> your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT
*SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs >>>>>>>>>> just fails for reasons u'd only understand by working thru the >>>>>>>>>> algo urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually
computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to >>>>>>>>> do that, as you are thinking the only "reference" can be to >>>>>>>>> "self", doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-
REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR >>>>>>>> INVERSE- H TO BE COMPUTED
one cannot just assume that because the diagonal across >>>>>>>>>>>> computable numbers is computable, therefore the anti- >>>>>>>>>>>> diagonal across computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without >>>>>>>>>>> the enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation >>>>>>>>>>>> across the computable numbers can be used to compute the >>>>>>>>>>>> inverse diagonal
But your method still doesn't let you compute the
enumeration, and thus you can't actually compute the diagonal. >>>>>>>>>>>
Remember, the problem definitions requires that the listing >>>>>>>>>>> be a COMPLETE listing of the computable numbers / machine >>>>>>>>>>> that compute computable numbers, in some definite order. >>>>>>>>>>>
If your enumeration isn't complete, your diagonal isn't correct. >>>>>>>>>>>
so while i agree with turing that the anti-diagonal is not >>>>>>>>>>>> computable, i don't agree that the normal diagonal is not >>>>>>>>>>>> computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the >>>>>>>>>>> other machines, including his original H that doesn't use >>>>>>>>>>> your "trick"
But it doesn't.
well, he didn't consider that perhaps the proper algo for >>>>>>>>>>>> computing the diagonal can avoid the paradox on itself ... >>>>>>>>>>>
But instead, he can prove with a more obvious process, that >>>>>>>>>>>>> the Decider "D" that could be used to effectively enumerate >>>>>>>>>>>>> the sequence of machine that produce computable numbers can >>>>>>>>>>>>> not esit.
Thus, he clearly knows the difference, but is pointing out >>>>>>>>>>>>> that the attempt to compute the diagonal clearly reveals >>>>>>>>>>>>> the issue with effectively enumerating the sequences. >>>>>>>>>>>>
Your just don't understand that D just can't correctly decide >>>>>>>>>>> on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on
turing's H, because my response to this is that D does not >>>>>>>>>> need to decide correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT
WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial recognizer >>>>>>> D *can* decide on the fixed H because the fixed H does not try to >>>>>>> use any D on itself, so no self-referential paradox is possible >>>>>>> in regards to it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable
AND computable value)
the fixed H is /decidable input/ to partial recognizer D (used in >>>>>>> fixed H, which does not call any D itself), and would not be
filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and
instead returns a hard-coded value for it's own digit on the
diagonal, is keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just
anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it will >>>>>> fail on that.
bro what part of the non-existence of turing_D or turing_H do you
not understand???
But if your fD exists, then Turing_H that uses your D exists, and
will give that D the problem, and fixed_H will call your D on
turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H gets >>>>>> hung up.
If D errs on Turing_H and calls it not circle_free, then your
enumeration is missing the circle_free H from its list, and thus
doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D in a >>>>> turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D
classifier, when actually run actual_turing_H will simply skip
putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the
actual_turing_H isn't an "undecidable input", as it has definite
behavior, just behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of undecidability.
just because a machine is /undecidable input/ to a particular
classifier does mean we cannot then prove and know ourselves what the
machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs are
not "undecidable", but by the time that they have actually been made
into an input, and thus first had the program created, which required
creating the instance of the decider selctected, they have definite
behavior that other some decider can determine.
It is only that one selected candidate decider chosen to build the
input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a
particular machine and the particular classifier it creates a
structural paradox for,
In other words, you don't understand what "undecidability" actually is.
Undecidability, means that there does not exist ANY decider that can
get the right answer for all instances of the problem, (or if the
problem has been reduced to just that one input, for that input).
*not* as a general inability to describe a *real* machine's behavior
that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
what *actual* example _THAT CAN EXIST_ do u have???
(you don't have one btw, so please do let me witness ur cope)
see, turing_H's runtime is not describable because it's an /
incomplete specification/ of machine and therefor _does not even
exist_ as a real TM, not because of the hypothesized undecidability
in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED defined
machine. If the decider "D" actually exists, then H is FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete specification/,
as it doesn't handle idiosyncrasies of TM computing, specifically self- referential set-classification paradoxes
actual_turing_H, on the other hand, actually does exist, so we can
know/ prove what it does. it, however, still forms a structural
paradox in regards to partial_recognizer_D which will fail to
classify it as circle-free
But, since you admit to actual_turing_H existing, that means that
since partial_recognizer_D _CAN_ exist, then actual_turing_H _CAN_ exist
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get
wrong, then it can just return the different answer, and then THAT
machine it will get right. The problem is it creates a DIFFERENT
input, that uses this new version of the machine, that it will get
wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as fixed_H, >>>>> which in turn _does_ have a spot on the diagonal. so there is no
actual need to include a digit from actual_turing_H on the
diagonal, as the number it computes is already included on the
diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of
actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can possible
filter all ALL of that infinite set of machines, but still accept
some machine that computes that particular number that it would
compute with this supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting that
you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the fairy
dust, of assuming that your partial decider that meets ALL your
needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between
fixed_H and actual_turing_H is the way they respectively handle their
own self- references. they function identically when handling all
other machines
But the problem isn't in it handling the machine fixed_H, but how it
ever handles the results generated by actual_turing_H, which by your
assumption has become a cycle-free machine and ALL equivalent versions
of it have been partially declared to be non-cycle free.
This means that your enumeration of machines you are looking at is, in
fact, incomplete, thus the diagonal you computed is wrong.
fixed_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip trying
to simulate itself and put it's hard coded digit on the diagonal,
which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip trying
simulate actual_turing_H because partial_recognizer_D fails to
classify it as circle-free
But actual_turing_H is actually circle_free, and thus its results
SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put on the diagonal, then actual_turing_H _DOES NOT_ need to be in _THAT_ total enumeration of _COMPUTABLE NUMBERS_
when actual_turing_H iterates across DN(actual_turing_H) itself, it
will similarly skip trying to simulate itself, because
partial_recognizer_D fails to classify it as circle-free
Right, so it ALSO is wrong, as it didn't compute the diagonal of the
circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE NUMBERS_, which
is the _ACTUAL PROBLEM_ turing was dealing with,
doing so doesn't require /all/ or _even most_ machines to be in the enumeration, since _most_ are duplicates of some other machine
when actual_turing_H iterates across DN(fixed_H),
partial_recognizer_D *successfully* classifies fixed_H as
satisfactory, so it will simulate fixed_H to it's Kth digit, the one
hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular logic
in either case), fixed_H also skips simulating itself (also circular
logic), while actual_turing_H actually *does* simulate fixed_H
successfully for it's Kth digit on the diagonal. these machines
compute the *same* number/sequence!
Except that by your partial_recognizer being just partial, it has
dropped required items from the enumeration.
_WHICH IS FINE_
THis is becides the fact that you still need to show how you plan on
actually detecting all these computationally equivalent machines in your
fixed_H does not compute a minimum turing-complete set, there are
duplicated sequences, ei machines that compute the same sequence/number,
and _THAT IS FINE_
partial decider to force them to not be classified as circle-free,
even though it turns out that they are (because your decider called
them not).
You don't seem to understand that you don't get to look at the
template used to build the machine, only the resultant machine from
that template.
And, part of the result of that, is that your attempt to try to define
"paradoxical" just fails, as it resolves to just input your decider is
wrong about, in part, because you don't understand what "decidable"
actually means,
see, not only is defining a value in fixed_H that the diagonal
computation can put for itself in diagonal essential in making the
diagonal actually computable, so is partial_recognizer_D skipping
over actual_turing_H to avoid the computability paradox!
In other words, you think it ok to compute the WRONG diagonal.
The needed diagonal was to a COMPLETE enumeration of the circle-free
machine.
NO, IT WAS SPECIFICALLY A COMPLETE ENUMERATION OF _COMPUTABLE NUMBERS/ SEQUENCES_
there is *no* requirement to include a sequence _more than once_, so
there *no* requirement to include all machines in the enumeration!
That isn't really different then claiming you can prove that every
even number grater than 2 from an infinite set of natural numbers is
the sum of two primes, make you enumeration just the set of prime
numbers, and thus claim that since the only even number is 2, it is
trivially solved.
_FALSE ANALOGY FALLACY_
if the set i'm enumerating includes _AT LEAST ONE INSTANCE_ of _ALL COMPUTABLE SEQUENCES_
then it is a turing-complete subset including all possible TM computable sequences, and is functionally equivalent to the set of *all* TMs
ultimately, in order to compute a diagonal across computable numbers,
partial_recognizer_D only needs to successfully classify *one* of the
_infinitely_ many machines that compute any given computable number...
But it isn't the diagonal across some set of computable numbers, but
the diagonal across the ENTIRE set of computable numbers.
Since you dropped some machines, you don't have the needed set.
_*THAT DOES NOT NECESSARILY FOLLOW*_
dropping actual_turing_H from the set, for example, is _FINE_ because it computes _THE SAME SEQUENCE_ as fixed_H, which _IS IN THE SET_ ...
shit rick, idk how ur still denying what's right under ur nose,
Because what you are putting under my nose is just CRAP based on errors
goddamn rick smh u could spot a diamond in the rough even if god lead u
by ur hand to one put on a pedestal with a _GIANT_ neon sign pointing to it...
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that
generates a computable number to build the list of such machine >>>>>>>>> to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH >>>>>>>> ARE INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH
WILL WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE
INPUT/ TO D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE >>>>>>>> NUMBERS
It doesn't matter that your new H doesn't get stuck on
itself, it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he
doesn't specify what D (or H) needs to do when encountering >>>>>>>>>> the / undecidable input/ of H, so therefore both D and H are >>>>>>>>>> an incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it >>>>>>>>> must compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE >>>>>>>> LIKE THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN
IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS COMPUTABLE >>>>>>>>
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE >>>>>>>> IT DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL
ANALYSIS, SO IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have >>>>>>>>> created an actual machine we claim to be the D, has an answer, >>>>>>>>> and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't
exist, only machines CORRECTLY meeting the requirements of D >>>>>>>>> and H can't exist.
Thus, if your H depends on that D, (or a machine that meets its >>>>>>>>> specification) then it also can't exist.
You just make the error of saying the problems don't exist
because you can't build them, but YOUR machine can, even though >>>>>>>>> it has the same problem.
IF D is wrong by deciding it is not circle free, then your H >>>>>>>>>>> will compute the wrong diagonal, as the resulting version of >>>>>>>>>>> his H WILL be circle free (since it never tries to simulate >>>>>>>>>>> itself) and thus DOES produce an computable number that your >>>>>>>>>>> computation misses.
Or, if that D is wrong by decing it IS circle free, then when >>>>>>>>>>> you H tries to process it, it will get stuck in the infinite >>>>>>>>>>> loop.
The problem is that in stepping through the machines in >>>>>>>>>>> order, you WILL hit these actual machines built on your >>>>>>>>>>> erroneous D (your D must have this flaw, as no D without >>>>>>>>>>> exists), and thus you will be wrong on THAT input. IT doesn't >>>>>>>>>>> matter if you get a good answer for yourself.
I doubt it.
idk what he would have said about it, but prolly something >>>>>>>>>>>> more substantial than just calling me ignorant repeatedly >>>>>>>>>>>
He likely would have gotten frustrated by your idiodic
assertion of bad logic. You would have likely been escorted >>>>>>>>>>> out of the meeting as showing you were unqualified and being >>>>>>>>>>> a distraction.
Something that seems to be beyond your ignorant understanding. >>>>>>>>>>>>>
H shows that *IF* you can make that enumeration, you can >>>>>>>>>>>>>>> make the diagonal, and thus the anti-diagonal. The >>>>>>>>>>>>>>> problem is you can't make that enumeration, and assuming >>>>>>>>>>>>>>> you can just shows unsoundness.
interestingly: one can only fix the direct diagonal >>>>>>>>>>>>>> computation like this
u can't do an analogous fix for the inverse/anti-diagonal >>>>>>>>>>>>>> computation. it's not possible hard code a machine to >>>>>>>>>>>>>> return an inverted value, a machine can only return what >>>>>>>>>>>>>> it does, not the inverse of what it does...
so if we can filter out paradoxes from the enumeration, >>>>>>>>>>>>>> that will leave a direct diagonal computation extant in >>>>>>>>>>>>>> that filtered (yet still turing complete list), while any >>>>>>>>>>>>>> attempt to compute an inverse diagonal will not be
But the problem is that "paradoxical machines" don't exist >>>>>>>>>>>>> in isolation, but only in relationship to a given machine >>>>>>>>>>>>> trying to decide them.
right. so if ur constructing a diagonal across computable >>>>>>>>>>>> numbers then u only need to filter out paradoxes in regards >>>>>>>>>>>> to the classifier that classifies them as a "satisfactory" >>>>>>>>>>>> number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such >>>>>>>>>> paradoxes...
In other words, you beleive unquestionably in the existance of >>>>>>>>> Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES
CAN'T BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING >>>>>>>> A FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM YOU >>>>>>>> MADE???
For THIS paradox, read the proof. for THIS H, the one given, >>>>>>>>> (for which ever version of D you want to try to assume is
right) D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox, >>>>>>>>> because there still isn't an answer it can give to H that will >>>>>>>>> be correct, as the template for H will always make that D wrong. >>>>>>>>>
Note, the problem is you can't actually DEFINE what a "paradox >>>>>>>>> machine" is, as they aren't actually machines, but templates >>>>>>>>> that build machines. And that final machine doesn't have
actually detectable tell-tales that show it to be from that >>>>>>>>> template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO >>>>>>>> WORK WITH, BUT OH WELL
und = () -> halts(und) loop()
They only seem "random" because you don't know what you are >>>>>>>>> talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* >>>>>>>>>>>> classifiable as satisfactory by said classifier... can just >>>>>>>>>>>> be skipped
No, it can only skip those that are not satisfactory, not >>>>>>>>>>> those that are but it can not classify as such, or your >>>>>>>>>>> enumeration will not be complete, and thus just in error. >>>>>>>>>>>
Thus, it needs to be able to correctly classify ALL machines >>>>>>>>>>> (as it will be asked about all machines as it counts through >>>>>>>>>>> all the descriptions) and thus Turing's H *WILL* be asked about. >>>>>>>>>>>
similarly if u want to go a step further an filter out >>>>>>>>>>>> computable numbers already included on this diagonal, any >>>>>>>>>>>> machine which either *is* computably equivalent OR *is not* >>>>>>>>>>>> classifiable in regards to *any* machine already the list... >>>>>>>>>>>> can just be skipped
Nope, you can't skip some machines, as you then might lose >>>>>>>>>>> some of the computable numbers.
see you can't compute a diagonal across *all* /machines/, >>>>>>>>>>>> with said machines, but u can compute a diagonal across >>>>>>>>>>>> *all* / computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be >>>>>>>>>>> done, since ALL classifiers that attempt it will make an >>>>>>>>>>> error, you can't do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ >>>>>>>>>> is not the same thing as (b) computing the enumeration of all >>>>>>>>>> machines that compute computable numbers. (b) necessarily has >>>>>>>>>> duplicates while (a) does not need them. turing's paper
wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter >>>>>>>>> the full list of machibe, which Turing showed can't exist.
The problem is such a machine can't handle the classification >>>>>>>>> of the H he describes. It doesn't matter that you can make a >>>>>>>>> DIFFERENT machine, that you try to deceptively call "H" too >>>>>>>>> that it won't have a problem with, when your "H" gets to the >>>>>>>>> number of Turing's H, it still has the problem. It can call >>>>>>>>> that one not circle-free, and thus omit a circle- free machine >>>>>>>>> from the list, or call it circle- free, and when even YOU try >>>>>>>>> to simulate it the k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take >>>>>>>>> the code of (a), and invert the symbol put on the tape when it >>>>>>>>> is written. Note, it won't be "self-referent", as the "N" is >>>>>>>>> uses, is the N of YOUR H, not itself. The code for his (a)
NEVER reads back what it wrote, so that doesn't affect its
behavior.
yes, i still do need to prove my thesis that for any
paradoxical machine, there exists a functionally equivalent >>>>>>>>>>>> machine without such paradox
And the problem is that your "paradoxical" isn't actually a >>>>>>>>>>> definable property (let alone computable). Part of the
problem is that if you look at just a machine description, it >>>>>>>>>>> doesn't (necessarily) tell you about the use of an
"interface" as that use of an interface can be just inlined, >>>>>>>>>>> leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description >>>>>>>>>> does not describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine. >>>>>>>>
und() includes a structural paradoxical in relation to the
classifier halts(), easily demonstrable thru a short proof i've >>>>>>>> done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY >>>>>>>> OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME >>>>>>>> FUCKING FORM OF INSANITY I CAN'T HONESTLY FATHOM
SO ACTUALLY DO IT??
lol
His specified H, with an actually (incorrect)
implementation of D (which is all that CAN exist) will >>>>>>>>>>>>> either be circle- free and thus generate a number (but its >>>>>>>>>>>>> D said it isn't, and thus omitted a valid machine from the >>>>>>>>>>>>> list) or it isn't circle- free, and fails to computa a >>>>>>>>>>>>> number, and thus should have been omitted from the list but >>>>>>>>>>>>> wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is >>>>>>>>>>>>> just built on an assuption in error.
so despite turing's worries, the existence of a diagonal >>>>>>>>>>>>>> computation does not actually then imply the existence of >>>>>>>>>>>>>> an anti- diagonal computation, due the same particular >>>>>>>>>>>>>> self- referential weirdness that stumped turing the first >>>>>>>>>>>>>> place
But there is no actuall SELF-REFERENCE, so your logic is >>>>>>>>>>>>> just based on ERROR.
Your attempt to REDEFINE self-reference to mean processing >>>>>>>>>>>>> a number that happens to represent yourself means that you >>>>>>>>>>>>> you system "ALL" doesn't actually mean ALL, and thus is >>>>>>>>>>>>> just ill- defined.
i'm sorry, you have an issue with me labeling a number that >>>>>>>>>>>> directly refers to yourself, as a "self-reference" ???
Sure, because it is just a number. The problem is that you >>>>>>>>>>> still have
it's a *specific* number that has the currently running
machine encoded into it, it's not "just" a number whatever >>>>>>>>>> that means
problems with all the "equivalent" machines that have
different numbers.
those are references to functionally equivalent machines, not >>>>>>>>>> self- references
Which is why the concept of "self-reference" doesn't work.
that's not actually true. you can't meaningfully paradox the >>>>>>>>>> paradox detector while filter *out* paradoxes to point of >>>>>>>>>> creating an actual problem
It may let you filter out the simplest case used in the >>>>>>>>>>> proofs, but doesn't solve the actual problem, as the "Machine >>>>>>>>>>> Number" doesn't actually fully identify the problematic cases. >>>>>>>>>>
Then show how you will do it.
This means you need to detect a input that represents a machine >>>>>>>>> that uses a computation equivalent to yourself
let halts be a halting classifier that takes an input machines >>>>>>>>>> and returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes >>>>>>>>>> an input classifier, and an input machine to search for
paradoxes in regards to: paradox_free(classifier, machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox
classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case >>>>>>>>>> run a halting paradox form making it /undecidable input/ to >>>>>>>>>> halts().
And the problem is "paradox_free" can't exist as an always
correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if >>>>>>>>>> case doesn't run and clearly halts(undp) -> TRUE without issues >>>>>>>>>>
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a >>>>>>>>>> machine is necessary for the particular input->output
computation being done, so utilizing the return
paradox_free(halts,undp) -> FALSE to filter out paradox_free() >>>>>>>>>> paradoxes as well as halts() paradoxes will suffice to produce >>>>>>>>>> a turing complete subset of machines that can be totally
classified by halts()
But we can still build an input that your paradox_free() fails on. >>>>>>>>
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE
INCLUDED ON THE DIAGONAL???
i think a major problem with the consensus perspective is an >>>>>>>>>> expectation of a certain cleanliness in the logic as being the >>>>>>>>>> only possibility here to compute what we want. TM's can't >>>>>>>>>> achieve that, but something slightly messier can.
No, it is your dependence on being able to assume that unicorns >>>>>>>>> can solve all your problems.
They are, But you are too stuck in your lies to see it.
lol, i can only 🙏 someone would engage in sound logic with me >>>>>>>>>
dear future: u see this shit i'm trying work with???? sorry >>>>>>>>>>>> i didn't manage to make progress any faster, but i'm
literally dragging toddlers kicking and screaming at this >>>>>>>>>>>> point jeez...
In other words, in your world "ALL" isn't a word that can >>>>>>>>>>> actually be used.
You don't understand that you need to use actual sound logic, >>>>>>>>>>
Your "logic" is based on being able to assume that Unicorn's >>>>>>>>> exist, and that is a safe assumption until someone can prove >>>>>>>>> that Russel's Teapot is not out there. (Even though your
unicorns HAVE been proved to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't
...which u won't find interesting, but i can't fix a lack >>>>>>>>>>>>>> of curiosity
🤷
expected was an inexpert audience, supposing that >>>>>>>>>>>>>>>>> experts will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>>>
understand what the actual problem is, and your world is >>>>>>>>>>>>> just build on things that are lies.
IT seems that fundamentally, your world doesn't actually >>>>>>>>>>>>> have "computations" as you don't understand the basic >>>>>>>>>>>>> requirement that they need to be fully defined in the >>>>>>>>>>>>> actions they do.
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:i don't see any indication that turing realized a >>>>>>>>>>>>>>> difference there
On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote:
On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>> ...Eh?!
an effective enumeration of all turing machines >>>>>>>>>>>>>>>>>>>>>>> was proven on
turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he >>>>>>>>>>>>>>>>>>> defines a comptuation
that enumerates out all the numbers, testing each one >>>>>>>>>>>>>>>>>>> of they represent
a "satisfactory"/"circle-free" machine, and adding >>>>>>>>>>>>>>>>>>> that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>
it really would be a great exercise to carefully read >>>>>>>>>>>>>>>>>>> p247 of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you >>>>>>>>>>>>>>>>>> sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>> procedures, and understanding the diagonal procedure is >>>>>>>>>>>>>>>>> critical to understanding the *base* contradiction/ >>>>>>>>>>>>>>>>> paradox that the rest of his support for godel's result >>>>>>>>>>>>>>>>> is then built on
And focusing on what is said to be impossible and not >>>>>>>>>>>>>>>> changing the problem is also important.
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>>> enumerating the enumeration in the first place. >>>>>>>>>>>>>>>
Then you zre just showing your stupidity, because YOU >>>>>>>>>>>>>> can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>>> enumeration of the computable sequences.
He then points out that he can directly show that the >>>>>>>>>>>>>> "anti- diagonal" of the (non-effectively computed) >>>>>>>>>>>>>> enumeration can't be computed but that "This proof, >>>>>>>>>>>>>> although perfectly sound, has the disadvantage that it may >>>>>>>>>>>>>> leave the reader with a feeling that 'there must be >>>>>>>>>>>>>> something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done. >>>>>>>>>>>>
I think he is refering he to the standard anti-diagonal >>>>>>>>>>>> arguement, which shows that since for all n, position n >>>>>>>>>>>> differs from the value in number n, there can not be any >>>>>>>>>>>> element that matches the anti- diagonal.
It is just a natural fact of countable infinity, something >>>>>>>>>>>> it seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with turing's >>>>>>>>>>> short diagonal proof, but in writing this post i now find >>>>>>>>>>> myself in a subtle, yet entirely critical disagreement:
/let an be the n-th computable sequence, and let φn(m) be the >>>>>>>>>>> m- th figure in an. Let β be the sequence with 1-φn(m) as its >>>>>>>>>>> n-th. figure. Since β is computable, there exists a number K >>>>>>>>>>> [== β] such that 1- φn(n) = φK(n) for all n. Putting n = K, >>>>>>>>>>> we have 1 = 2φK(K), i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was computable >>>>>>>>> (by TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the
computability of β from the existence of φn(m), THAT'S THE FALLACy >>>>>>>>>
the fallacy here is assuming that because the direct diagonal >>>>>>>>>>> is computable, that one can therefore compute the anti- >>>>>>>>>>> diagonal using the direct diagonal. the abstract definition >>>>>>>>>>> makes it look simple, but this ignores the complexities of >>>>>>>>>>> self- referential analysis (like what turing details on the >>>>>>>>>>> next page)
But you can, *IF* you have a machine that computes the
diagonal, then just change all the write to the output to >>>>>>>>>> write the opposite. Note, the "self-reference" that you are >>>>>>>>>> thinking of stops being a "self- reference" but is a reference >>>>>>>>>> to the original write the diagonal code.
Nope, because your filtering TM (or RTM) will still fail when >>>>>>>>>> it gets to the number of TURING'S H, as there is no correct >>>>>>>>>> answer for the machine built by that template.
in both methods i have for rectifying the paradox found in >>>>>>>>>>> the direct diagonal (either (1) filtering TMs or (2) using >>>>>>>>>>> RTMs), neither can be used to then compute the anti-diagonal >>>>>>>>>>
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE
SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered >>>>>>>>>>> out like turing's paradoxical variation of the direct
diagonal would be, and there is no analogous non-paradoxical >>>>>>>>>>> variation that has a hard coded value that is inverse to what >>>>>>>>>>> it does return ... such a concept is entirely nonsensical. a >>>>>>>>>>> function can only return what it does, it can't also return >>>>>>>>>>> the inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, and >>>>>>>>>> thus your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT >>>>>>>>> *SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs >>>>>>>>>>> just fails for reasons u'd only understand by working thru >>>>>>>>>>> the algo urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually >>>>>>>>>>> computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to >>>>>>>>>> do that, as you are thinking the only "reference" can be to >>>>>>>>>> "self", doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF-
REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H >>>>>>>>>
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR >>>>>>>>> INVERSE- H TO BE COMPUTED
one cannot just assume that because the diagonal across >>>>>>>>>>>>> computable numbers is computable, therefore the anti- >>>>>>>>>>>>> diagonal across computable numbers is computable...
He doesn't. You are just showing your stupidity,
He is proving the enumeration is uncomputable, and without >>>>>>>>>>>> the enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation >>>>>>>>>>>>> across the computable numbers can be used to compute the >>>>>>>>>>>>> inverse diagonal
But your method still doesn't let you compute the
enumeration, and thus you can't actually compute the diagonal. >>>>>>>>>>>>
Remember, the problem definitions requires that the listing >>>>>>>>>>>> be a COMPLETE listing of the computable numbers / machine >>>>>>>>>>>> that compute computable numbers, in some definite order. >>>>>>>>>>>>
If your enumeration isn't complete, your diagonal isn't >>>>>>>>>>>> correct.
so while i agree with turing that the anti-diagonal is not >>>>>>>>>>>>> computable, i don't agree that the normal diagonal is not >>>>>>>>>>>>> computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all the >>>>>>>>>>>> other machines, including his original H that doesn't use >>>>>>>>>>>> your "trick"
But it doesn't.
well, he didn't consider that perhaps the proper algo for >>>>>>>>>>>>> computing the diagonal can avoid the paradox on itself ... >>>>>>>>>>>>
But instead, he can prove with a more obvious process, >>>>>>>>>>>>>> that the Decider "D" that could be used to effectively >>>>>>>>>>>>>> enumerate the sequence of machine that produce computable >>>>>>>>>>>>>> numbers can not esit.
Thus, he clearly knows the difference, but is pointing out >>>>>>>>>>>>>> that the attempt to compute the diagonal clearly reveals >>>>>>>>>>>>>> the issue with effectively enumerating the sequences. >>>>>>>>>>>>>
Your just don't understand that D just can't correctly >>>>>>>>>>>> decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on
turing's H, because my response to this is that D does not >>>>>>>>>>> need to decide correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT >>>>>>>>> WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial
recognizer D *can* decide on the fixed H because the fixed H
does not try to use any D on itself, so no self-referential
paradox is possible in regards to it's own digit on the diagonal >>>>>>>>
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable >>>>>>>> AND computable value)
the fixed H is /decidable input/ to partial recognizer D (used >>>>>>>> in fixed H, which does not call any D itself), and would not be >>>>>>>> filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and >>>>>>>> instead returns a hard-coded value for it's own digit on the
diagonal, is keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just
anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it
will fail on that.
bro what part of the non-existence of turing_D or turing_H do you >>>>>> not understand???
But if your fD exists, then Turing_H that uses your D exists, and
will give that D the problem, and fixed_H will call your D on
turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H gets >>>>>>> hung up.
If D errs on Turing_H and calls it not circle_free, then your
enumeration is missing the circle_free H from its list, and thus >>>>>>> doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D in >>>>>> a turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D
classifier, when actually run actual_turing_H will simply skip
putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the
actual_turing_H isn't an "undecidable input", as it has definite
behavior, just behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of undecidability.
just because a machine is /undecidable input/ to a particular
classifier does mean we cannot then prove and know ourselves what
the machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs are
not "undecidable", but by the time that they have actually been made
into an input, and thus first had the program created, which required
creating the instance of the decider selctected, they have definite
behavior that other some decider can determine.
It is only that one selected candidate decider chosen to build the
input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a
particular machine and the particular classifier it creates a
structural paradox for,
In other words, you don't understand what "undecidability" actually is.
Undecidability, means that there does not exist ANY decider that can
get the right answer for all instances of the problem, (or if the
problem has been reduced to just that one input, for that input).
*not* as a general inability to describe a *real* machine's behavior
that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D.
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
(you don't have one btw, so please do let me witness ur cope)
IT seesm you just can't read.
see, turing_H's runtime is not describable because it's an /
incomplete specification/ of machine and therefor _does not even
exist_ as a real TM, not because of the hypothesized undecidability
in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED defined
machine. If the decider "D" actually exists, then H is FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
specification/, as it doesn't handle idiosyncrasies of TM computing,
specifically self- referential set-classification paradoxes
It doesn't need to. It just needs to be built on what you want to claim
to be your answer for the concept of D.
As soon as you define what machine you are going to use as your
equivalent for D, like your partial_recognizer_D, then the template
turing_H can be converted into an actual machine.
actual_turing_H, on the other hand, actually does exist, so we can
know/ prove what it does. it, however, still forms a structural
paradox in regards to partial_recognizer_D which will fail to
classify it as circle-free
But, since you admit to actual_turing_H existing, that means that
since partial_recognizer_D _CAN_ exist, then actual_turing_H _CAN_ exist
And so can the turing_H that uses partial_recognizer_D as its decider D.
The steps the algorithm does is what defines the machine, not the
"names" of the things it "calls", as those names don't actually exist in
the final building of the machine, which is the problem with your
concept, as that is needed to determine your "paradoxical" property.
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get
wrong, then it can just return the different answer, and then THAT
machine it will get right. The problem is it creates a DIFFERENT
input, that uses this new version of the machine, that it will get
wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as
fixed_H, which in turn _does_ have a spot on the diagonal. so
there is no actual need to include a digit from actual_turing_H on >>>>>> the diagonal, as the number it computes is already included on the >>>>>> diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of
actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can
possible filter all ALL of that infinite set of machines, but still >>>>> accept some machine that computes that particular number that it
would compute with this supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting
that you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the fairy
dust, of assuming that your partial decider that meets ALL your
needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between
fixed_H and actual_turing_H is the way they respectively handle
their own self- references. they function identically when handling
all other machines
But the problem isn't in it handling the machine fixed_H, but how it
ever handles the results generated by actual_turing_H, which by your
assumption has become a cycle-free machine and ALL equivalent
versions of it have been partially declared to be non-cycle free.
This means that your enumeration of machines you are looking at is,
in fact, incomplete, thus the diagonal you computed is wrong.
fixed_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE =
satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip trying
to simulate itself and put it's hard coded digit on the diagonal,
which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip
trying simulate actual_turing_H because partial_recognizer_D fails
to classify it as circle-free
But actual_turing_H is actually circle_free, and thus its results
SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put on the
diagonal, then actual_turing_H _DOES NOT_ need to be in _THAT_ total
enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable numbers,
but of machines that made computable numbers.
Also, your proof is built on the existance of a Russel Teapot, you have
just ASSUMED you cam make this partial_recoginzser_D that can make the precise "mistakes" you need it to. (They are mistakes because the input
that come from Turing_H are, in fact, circle_free because of this
turing_H is built on this partial_recognizdr_D which causes them to not
hang up on themselves.
Your "definition" of paradoxical just doesn't apply to machines, as it
is based on the incorrect definition that assumes a machine can be
something other than itself, and when it changes, it changes other
things that at the meta/template level refered to its old self.
when actual_turing_H iterates across DN(actual_turing_H) itself, it
will similarly skip trying to simulate itself, because
partial_recognizer_D fails to classify it as circle-free
Right, so it ALSO is wrong, as it didn't compute the diagonal of the
circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE NUMBERS_,
which is the _ACTUAL PROBLEM_ turing was dealing with,
No, it isn't, as your "set of computable numbers" has duplicates.
doing so doesn't require /all/ or _even most_ machines to be in the
enumeration, since _most_ are duplicates of some other machine
when actual_turing_H iterates across DN(fixed_H),
partial_recognizer_D *successfully* classifies fixed_H as
satisfactory, so it will simulate fixed_H to it's Kth digit, the one
hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular logic
in either case), fixed_H also skips simulating itself (also circular
logic), while actual_turing_H actually *does* simulate fixed_H
successfully for it's Kth digit on the diagonal. these machines
compute the *same* number/sequence!
Except that by your partial_recognizer being just partial, it has
dropped required items from the enumeration.
_WHICH IS FINE_
So, you admit it failed???
As I said, it dropped REQUIRED items, as this enumeration was of
machines that computed computable numbers.
If you try to add removal of duplicates, then we need to define the
order, and if we do it by number, turing_H will likely be lower in
order, as it doesn't need to generate its own number, and thus is simpler,
fixed_H does not compute a minimum turing-complete set, there are
THis is becides the fact that you still need to show how you plan on
actually detecting all these computationally equivalent machines in your >>
duplicated sequences, ei machines that compute the same sequence/
number, and _THAT IS FINE_
Why?
And, it was built on the assumption of the existance of a machine, that actually can not exist (based on Rice's Theorm) and thus is actually
based on the assumption of Unicorns.
partial decider to force them to not be classified as circle-free,
even though it turns out that they are (because your decider called
them not).
You don't seem to understand that you don't get to look at the
template used to build the machine, only the resultant machine from
that template.
And, part of the result of that, is that your attempt to try to
define "paradoxical" just fails, as it resolves to just input your
decider is wrong about, in part, because you don't understand what
"decidable" actually means,
see, not only is defining a value in fixed_H that the diagonal
computation can put for itself in diagonal essential in making the
diagonal actually computable, so is partial_recognizer_D skipping
over actual_turing_H to avoid the computability paradox!
In other words, you think it ok to compute the WRONG diagonal.
The needed diagonal was to a COMPLETE enumeration of the circle-free
machine.
NO, IT WAS SPECIFICALLY A COMPLETE ENUMERATION OF _COMPUTABLE NUMBERS/
SEQUENCES_
Which you don't get.
there is *no* requirement to include a sequence _more than once_, so
there *no* requirement to include all machines in the enumeration!
But there is to show that your partial_recognizer_D can exist, when it can't.
YOU need to prove your "Russel's Teapot" exists.
Your logic just shows that you don't know how to do logic.
That isn't really different then claiming you can prove that every
even number grater than 2 from an infinite set of natural numbers is
the sum of two primes, make you enumeration just the set of prime
numbers, and thus claim that since the only even number is 2, it is
trivially solved.
_FALSE ANALOGY FALLACY_
What is false about it?
if the set i'm enumerating includes _AT LEAST ONE INSTANCE_ of _ALL
COMPUTABLE SEQUENCES_
But does it?
then it is a turing-complete subset including all possible TM
computable sequences, and is functionally equivalent to the set of
*all* TMs
But, since your partial_recognizer_D doesn't exist, but was only assumed
to exist, your program doesn't exist,
ultimately, in order to compute a diagonal across computable
numbers, partial_recognizer_D only needs to successfully classify
*one* of the _infinitely_ many machines that compute any given
computable number...
But it isn't the diagonal across some set of computable numbers, but
the diagonal across the ENTIRE set of computable numbers.
Since you dropped some machines, you don't have the needed set.
_*THAT DOES NOT NECESSARILY FOLLOW*_
Then try to define what set you are actually making!!
The problem is your definition of "paradoxical" doesn't actually work,
as it is based on asking about the results of an input, that changes.
dropping actual_turing_H from the set, for example, is _FINE_ because
it computes _THE SAME SEQUENCE_ as fixed_H, which _IS IN THE SET_ ...
How do you know it includes your fixed_H in the set?
what part of the actual machine makes it different?
Remember, in the final machine given as the input, there are not
seperate turing_H and partial_recognizer_D denoted in the machine, just
one big piece of algorithm. Thus nothing to base your criteria on.
shit rick, idk how ur still denying what's right under ur nose,
Because what you are putting under my nose is just CRAP based on errors
goddamn rick smh u could spot a diamond in the rough even if god lead
u by ur hand to one put on a pedestal with a _GIANT_ neon sign
pointing to it...
But the problem is that "your diamond" is actually just a big load of
crap that you misidentify as a diamond, because you don't actually understand what you are doing.
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that >>>>>>>>>> generates a computable number to build the list of such
machine to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH >>>>>>>>> ARE INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH >>>>>>>>> WILL WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE >>>>>>>>> INPUT/ TO D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE COMPUTABLE >>>>>>>>> NUMBERS
It doesn't matter that your new H doesn't get stuck on >>>>>>>>>>>> itself, it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he >>>>>>>>>>> doesn't specify what D (or H) needs to do when encountering >>>>>>>>>>> the / undecidable input/ of H, so therefore both D and H are >>>>>>>>>>> an incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it >>>>>>>>>> must compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE >>>>>>>>> LIKE THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN
IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS
COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE >>>>>>>>> IT DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL
ANALYSIS, SO IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we have >>>>>>>>>> created an actual machine we claim to be the D, has an answer, >>>>>>>>>> and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't >>>>>>>>>> exist, only machines CORRECTLY meeting the requirements of D >>>>>>>>>> and H can't exist.
Thus, if your H depends on that D, (or a machine that meets >>>>>>>>>> its specification) then it also can't exist.
You just make the error of saying the problems don't exist >>>>>>>>>> because you can't build them, but YOUR machine can, even
though it has the same problem.
IF D is wrong by deciding it is not circle free, then your H >>>>>>>>>>>> will compute the wrong diagonal, as the resulting version of >>>>>>>>>>>> his H WILL be circle free (since it never tries to simulate >>>>>>>>>>>> itself) and thus DOES produce an computable number that your >>>>>>>>>>>> computation misses.
Or, if that D is wrong by decing it IS circle free, then >>>>>>>>>>>> when you H tries to process it, it will get stuck in the >>>>>>>>>>>> infinite loop.
The problem is that in stepping through the machines in >>>>>>>>>>>> order, you WILL hit these actual machines built on your >>>>>>>>>>>> erroneous D (your D must have this flaw, as no D without >>>>>>>>>>>> exists), and thus you will be wrong on THAT input. IT >>>>>>>>>>>> doesn't matter if you get a good answer for yourself.
I doubt it.
idk what he would have said about it, but prolly something >>>>>>>>>>>>> more substantial than just calling me ignorant repeatedly >>>>>>>>>>>>
He likely would have gotten frustrated by your idiodic >>>>>>>>>>>> assertion of bad logic. You would have likely been escorted >>>>>>>>>>>> out of the meeting as showing you were unqualified and being >>>>>>>>>>>> a distraction.
Something that seems to be beyond your ignorant
understanding.
But the problem is that "paradoxical machines" don't exist >>>>>>>>>>>>>> in isolation, but only in relationship to a given machine >>>>>>>>>>>>>> trying to decide them.
H shows that *IF* you can make that enumeration, you can >>>>>>>>>>>>>>>> make the diagonal, and thus the anti-diagonal. The >>>>>>>>>>>>>>>> problem is you can't make that enumeration, and assuming >>>>>>>>>>>>>>>> you can just shows unsoundness.
interestingly: one can only fix the direct diagonal >>>>>>>>>>>>>>> computation like this
u can't do an analogous fix for the inverse/anti-diagonal >>>>>>>>>>>>>>> computation. it's not possible hard code a machine to >>>>>>>>>>>>>>> return an inverted value, a machine can only return what >>>>>>>>>>>>>>> it does, not the inverse of what it does...
so if we can filter out paradoxes from the enumeration, >>>>>>>>>>>>>>> that will leave a direct diagonal computation extant in >>>>>>>>>>>>>>> that filtered (yet still turing complete list), while any >>>>>>>>>>>>>>> attempt to compute an inverse diagonal will not be >>>>>>>>>>>>>>
right. so if ur constructing a diagonal across computable >>>>>>>>>>>>> numbers then u only need to filter out paradoxes in regards >>>>>>>>>>>>> to the classifier that classifies them as a "satisfactory" >>>>>>>>>>>>> number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such >>>>>>>>>>> paradoxes...
In other words, you beleive unquestionably in the existance of >>>>>>>>>> Russle's teapot until someone can prove it doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES >>>>>>>>> CAN'T BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON??? >>>>>>>>>
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF PROVIDING >>>>>>>>> A FUCKING QUOTE TO TURING'S PAPER TO BACK THE *DIRECT* CLAIM >>>>>>>>> YOU MADE???
For THIS paradox, read the proof. for THIS H, the one given, >>>>>>>>>> (for which ever version of D you want to try to assume is >>>>>>>>>> right) D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the paradox, >>>>>>>>>> because there still isn't an answer it can give to H that will >>>>>>>>>> be correct, as the template for H will always make that D wrong. >>>>>>>>>>
Note, the problem is you can't actually DEFINE what a "paradox >>>>>>>>>> machine" is, as they aren't actually machines, but templates >>>>>>>>>> that build machines. And that final machine doesn't have
actually detectable tell-tales that show it to be from that >>>>>>>>>> template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE TO >>>>>>>>> WORK WITH, BUT OH WELL
und = () -> halts(und) loop()
They only seem "random" because you don't know what you are >>>>>>>>>> talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* >>>>>>>>>>>>> classifiable as satisfactory by said classifier... can just >>>>>>>>>>>>> be skipped
No, it can only skip those that are not satisfactory, not >>>>>>>>>>>> those that are but it can not classify as such, or your >>>>>>>>>>>> enumeration will not be complete, and thus just in error. >>>>>>>>>>>>
Thus, it needs to be able to correctly classify ALL machines >>>>>>>>>>>> (as it will be asked about all machines as it counts through >>>>>>>>>>>> all the descriptions) and thus Turing's H *WILL* be asked >>>>>>>>>>>> about.
similarly if u want to go a step further an filter out >>>>>>>>>>>>> computable numbers already included on this diagonal, any >>>>>>>>>>>>> machine which either *is* computably equivalent OR *is not* >>>>>>>>>>>>> classifiable in regards to *any* machine already the >>>>>>>>>>>>> list... can just be skipped
Nope, you can't skip some machines, as you then might lose >>>>>>>>>>>> some of the computable numbers.
see you can't compute a diagonal across *all* /machines/, >>>>>>>>>>>>> with said machines, but u can compute a diagonal across >>>>>>>>>>>>> *all* / computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be >>>>>>>>>>>> done, since ALL classifiers that attempt it will make an >>>>>>>>>>>> error, you can't do what you want to do.
nah, (a) computing an enumeration of all /computable numbers/ >>>>>>>>>>> is not the same thing as (b) computing the enumeration of all >>>>>>>>>>> machines that compute computable numbers. (b) necessarily has >>>>>>>>>>> duplicates while (a) does not need them. turing's paper >>>>>>>>>>> wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter >>>>>>>>>> the full list of machibe, which Turing showed can't exist. >>>>>>>>>>
The problem is such a machine can't handle the classification >>>>>>>>>> of the H he describes. It doesn't matter that you can make a >>>>>>>>>> DIFFERENT machine, that you try to deceptively call "H" too >>>>>>>>>> that it won't have a problem with, when your "H" gets to the >>>>>>>>>> number of Turing's H, it still has the problem. It can call >>>>>>>>>> that one not circle-free, and thus omit a circle- free machine >>>>>>>>>> from the list, or call it circle- free, and when even YOU try >>>>>>>>>> to simulate it the k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take >>>>>>>>>> the code of (a), and invert the symbol put on the tape when it >>>>>>>>>> is written. Note, it won't be "self-referent", as the "N" is >>>>>>>>>> uses, is the N of YOUR H, not itself. The code for his (a) >>>>>>>>>> NEVER reads back what it wrote, so that doesn't affect its >>>>>>>>>> behavior.
yes, i still do need to prove my thesis that for any >>>>>>>>>>>>> paradoxical machine, there exists a functionally equivalent >>>>>>>>>>>>> machine without such paradox
And the problem is that your "paradoxical" isn't actually a >>>>>>>>>>>> definable property (let alone computable). Part of the >>>>>>>>>>>> problem is that if you look at just a machine description, >>>>>>>>>>>> it doesn't (necessarily) tell you about the use of an >>>>>>>>>>>> "interface" as that use of an interface can be just inlined, >>>>>>>>>>>> leaving nothing "in the code" to show it exists.
i'm sorry, are you actually saying the machine description >>>>>>>>>>> does not describe what the machine does???
No, that "paradoxical" isn't a definable property of a machine. >>>>>>>>>
und() includes a structural paradoxical in relation to the
classifier halts(), easily demonstrable thru a short proof i've >>>>>>>>> done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY >>>>>>>>> OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO SOME >>>>>>>>> FUCKING FORM OF INSANITY I CAN'T HONESTLY FATHOM
lol
Sure, because it is just a number. The problem is that you >>>>>>>>>>>> still have
His specified H, with an actually (incorrect)
implementation of D (which is all that CAN exist) will >>>>>>>>>>>>>> either be circle- free and thus generate a number (but its >>>>>>>>>>>>>> D said it isn't, and thus omitted a valid machine from the >>>>>>>>>>>>>> list) or it isn't circle- free, and fails to computa a >>>>>>>>>>>>>> number, and thus should have been omitted from the list >>>>>>>>>>>>>> but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is >>>>>>>>>>>>>> just built on an assuption in error.
so despite turing's worries, the existence of a diagonal >>>>>>>>>>>>>>> computation does not actually then imply the existence of >>>>>>>>>>>>>>> an anti- diagonal computation, due the same particular >>>>>>>>>>>>>>> self- referential weirdness that stumped turing the first >>>>>>>>>>>>>>> place
But there is no actuall SELF-REFERENCE, so your logic is >>>>>>>>>>>>>> just based on ERROR.
Your attempt to REDEFINE self-reference to mean processing >>>>>>>>>>>>>> a number that happens to represent yourself means that you >>>>>>>>>>>>>> you system "ALL" doesn't actually mean ALL, and thus is >>>>>>>>>>>>>> just ill- defined.
i'm sorry, you have an issue with me labeling a number that >>>>>>>>>>>>> directly refers to yourself, as a "self-reference" ??? >>>>>>>>>>>>
it's a *specific* number that has the currently running >>>>>>>>>>> machine encoded into it, it's not "just" a number whatever >>>>>>>>>>> that means
problems with all the "equivalent" machines that have >>>>>>>>>>>> different numbers.
those are references to functionally equivalent machines, not >>>>>>>>>>> self- references
Which is why the concept of "self-reference" doesn't work. >>>>>>>>>>
It may let you filter out the simplest case used in the >>>>>>>>>>>> proofs, but doesn't solve the actual problem, as the
"Machine Number" doesn't actually fully identify the
problematic cases.
that's not actually true. you can't meaningfully paradox the >>>>>>>>>>> paradox detector while filter *out* paradoxes to point of >>>>>>>>>>> creating an actual problem
Then show how you will do it.
This means you need to detect a input that represents a
machine that uses a computation equivalent to yourself
WHich doesn't exist.
let halts be a halting classifier that takes an input
machines and returns true/false whether it halts or not
let paradox_free be a paradox filtering classifier that takes >>>>>>>>>>> an input classifier, and an input machine to search for >>>>>>>>>>> paradoxes in regards to: paradox_free(classifier, machine) >>>>>>>>>>
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox >>>>>>>>>>> classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case >>>>>>>>>>> run a halting paradox form making it /undecidable input/ to >>>>>>>>>>> halts().
And the problem is "paradox_free" can't exist as an always >>>>>>>>>> correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if >>>>>>>>>>> case doesn't run and clearly halts(undp) -> TRUE without issues >>>>>>>>>>>
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a >>>>>>>>>>> machine is necessary for the particular input->output
computation being done, so utilizing the return
paradox_free(halts,undp) -> FALSE to filter out
paradox_free() paradoxes as well as halts() paradoxes will >>>>>>>>>>> suffice to produce a turing complete subset of machines that >>>>>>>>>>> can be totally classified by halts()
But we can still build an input that your paradox_free() fails >>>>>>>>>> on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE >>>>>>>>> INCLUDED ON THE DIAGONAL???
i think a major problem with the consensus perspective is an >>>>>>>>>>> expectation of a certain cleanliness in the logic as being >>>>>>>>>>> the only possibility here to compute what we want. TM's can't >>>>>>>>>>> achieve that, but something slightly messier can.
No, it is your dependence on being able to assume that
unicorns can solve all your problems.
They are, But you are too stuck in your lies to see it.
lol, i can only 🙏 someone would engage in sound logic with me >>>>>>>>>>
dear future: u see this shit i'm trying work with???? sorry >>>>>>>>>>>>> i didn't manage to make progress any faster, but i'm >>>>>>>>>>>>> literally dragging toddlers kicking and screaming at this >>>>>>>>>>>>> point jeez...
In other words, in your world "ALL" isn't a word that can >>>>>>>>>>>> actually be used.
You don't understand that you need to use actual sound logic, >>>>>>>>>>>
Your "logic" is based on being able to assume that Unicorn's >>>>>>>>>> exist, and that is a safe assumption until someone can prove >>>>>>>>>> that Russel's Teapot is not out there. (Even though your
unicorns HAVE been proved to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't >>>>>>>>>>>>>> understand what the actual problem is, and your world is >>>>>>>>>>>>>> just build on things that are lies.
...which u won't find interesting, but i can't fix a lack >>>>>>>>>>>>>>> of curiosity
🤷
expected was an inexpert audience, supposing that >>>>>>>>>>>>>>>>>> experts will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>>>>
IT seems that fundamentally, your world doesn't actually >>>>>>>>>>>>>> have "computations" as you don't understand the basic >>>>>>>>>>>>>> requirement that they need to be fully defined in the >>>>>>>>>>>>>> actions they do.
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:i don't see any indication that turing realized a >>>>>>>>>>>>>>>> difference there
On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>> ...
Eh?!an effective enumeration of all turing machines >>>>>>>>>>>>>>>>>>>>>>>> was proven on
turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he >>>>>>>>>>>>>>>>>>>> defines a comptuation
that enumerates out all the numbers, testing each >>>>>>>>>>>>>>>>>>>> one of they represent
a "satisfactory"/"circle-free" machine, and adding >>>>>>>>>>>>>>>>>>>> that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to carefully >>>>>>>>>>>>>>>>>>>> read p247 of turing's
proof and produce the psuedo-code for the machine H, >>>>>>>>>>>>>>>>>>>> assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are you >>>>>>>>>>>>>>>>>>> sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>>> procedures, and understanding the diagonal procedure >>>>>>>>>>>>>>>>>> is critical to understanding the *base* contradiction/ >>>>>>>>>>>>>>>>>> paradox that the rest of his support for godel's >>>>>>>>>>>>>>>>>> result is then built on
And focusing on what is said to be impossible and not >>>>>>>>>>>>>>>>> changing the problem is also important.
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>>>> enumerating the enumeration in the first place. >>>>>>>>>>>>>>>>
Then you zre just showing your stupidity, because YOU >>>>>>>>>>>>>>> can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>>>> enumeration of the computable sequences.
He then points out that he can directly show that the >>>>>>>>>>>>>>> "anti- diagonal" of the (non-effectively computed) >>>>>>>>>>>>>>> enumeration can't be computed but that "This proof, >>>>>>>>>>>>>>> although perfectly sound, has the disadvantage that it >>>>>>>>>>>>>>> may leave the reader with a feeling that 'there must be >>>>>>>>>>>>>>> something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being done. >>>>>>>>>>>>>
I think he is refering he to the standard anti-diagonal >>>>>>>>>>>>> arguement, which shows that since for all n, position n >>>>>>>>>>>>> differs from the value in number n, there can not be any >>>>>>>>>>>>> element that matches the anti- diagonal.
It is just a natural fact of countable infinity, something >>>>>>>>>>>>> it seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with
turing's short diagonal proof, but in writing this post i >>>>>>>>>>>> now find myself in a subtle, yet entirely critical
disagreement:
/let an be the n-th computable sequence, and let φn(m) be >>>>>>>>>>>> the m- th figure in an. Let β be the sequence with 1-φn(m) >>>>>>>>>>>> as its n-th. figure. Since β is computable, there exists a >>>>>>>>>>>> number K [== β] such that 1- φn(n) = φK(n) for all n. >>>>>>>>>>>> Putting n = K, we have 1 = 2φK(K), i.e. 1 is even. This is >>>>>>>>>>>> impossible/
i agree with this proof is far as much as if β was computable >>>>>>>>>> (by TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the
computability of β from the existence of φn(m), THAT'S THE >>>>>>>>>> FALLACy
the fallacy here is assuming that because the direct
diagonal is computable, that one can therefore compute the >>>>>>>>>>>> anti- diagonal using the direct diagonal. the abstract >>>>>>>>>>>> definition makes it look simple, but this ignores the >>>>>>>>>>>> complexities of self- referential analysis (like what turing >>>>>>>>>>>> details on the next page)
But you can, *IF* you have a machine that computes the
diagonal, then just change all the write to the output to >>>>>>>>>>> write the opposite. Note, the "self-reference" that you are >>>>>>>>>>> thinking of stops being a "self- reference" but is a
reference to the original write the diagonal code.
Nope, because your filtering TM (or RTM) will still fail when >>>>>>>>>>> it gets to the number of TURING'S H, as there is no correct >>>>>>>>>>> answer for the machine built by that template.
in both methods i have for rectifying the paradox found in >>>>>>>>>>>> the direct diagonal (either (1) filtering TMs or (2) using >>>>>>>>>>>> RTMs), neither can be used to then compute the anti-diagonal >>>>>>>>>>>
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE >>>>>>>>>> SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered >>>>>>>>>>>> out like turing's paradoxical variation of the direct >>>>>>>>>>>> diagonal would be, and there is no analogous non-paradoxical >>>>>>>>>>>> variation that has a hard coded value that is inverse to >>>>>>>>>>>> what it does return ... such a concept is entirely
nonsensical. a function can only return what it does, it >>>>>>>>>>>> can't also return the inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, and >>>>>>>>>>> thus your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT >>>>>>>>>> *SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs >>>>>>>>>>>> just fails for reasons u'd only understand by working thru >>>>>>>>>>>> the algo urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually >>>>>>>>>>>> computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how to >>>>>>>>>>> do that, as you are thinking the only "reference" can be to >>>>>>>>>>> "self", doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF- >>>>>>>>>> REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H >>>>>>>>>>
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR >>>>>>>>>> INVERSE- H TO BE COMPUTED
He doesn't. You are just showing your stupidity,
one cannot just assume that because the diagonal across >>>>>>>>>>>>>> computable numbers is computable, therefore the anti- >>>>>>>>>>>>>> diagonal across computable numbers is computable... >>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and without >>>>>>>>>>>>> the enumeration, you can't compute either of them.
neither method i have for fixing the diagonal computation >>>>>>>>>>>>>> across the computable numbers can be used to compute the >>>>>>>>>>>>>> inverse diagonal
But your method still doesn't let you compute the
enumeration, and thus you can't actually compute the diagonal. >>>>>>>>>>>>>
Remember, the problem definitions requires that the listing >>>>>>>>>>>>> be a COMPLETE listing of the computable numbers / machine >>>>>>>>>>>>> that compute computable numbers, in some definite order. >>>>>>>>>>>>>
If your enumeration isn't complete, your diagonal isn't >>>>>>>>>>>>> correct.
so while i agree with turing that the anti-diagonal is not >>>>>>>>>>>>>> computable, i don't agree that the normal diagonal is not >>>>>>>>>>>>>> computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all >>>>>>>>>>>>> the other machines, including his original H that doesn't >>>>>>>>>>>>> use your "trick"
But it doesn't.
But instead, he can prove with a more obvious process, >>>>>>>>>>>>>>> that the Decider "D" that could be used to effectively >>>>>>>>>>>>>>> enumerate the sequence of machine that produce computable >>>>>>>>>>>>>>> numbers can not esit.
Thus, he clearly knows the difference, but is pointing >>>>>>>>>>>>>>> out that the attempt to compute the diagonal clearly >>>>>>>>>>>>>>> reveals the issue with effectively enumerating the >>>>>>>>>>>>>>> sequences.
well, he didn't consider that perhaps the proper algo for >>>>>>>>>>>>>> computing the diagonal can avoid the paradox on itself ... >>>>>>>>>>>>>
Your just don't understand that D just can't correctly >>>>>>>>>>>>> decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on >>>>>>>>>>>> turing's H, because my response to this is that D does not >>>>>>>>>>>> need to decide correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT >>>>>>>>>> WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial
recognizer D *can* decide on the fixed H because the fixed H >>>>>>>>> does not try to use any D on itself, so no self-referential >>>>>>>>> paradox is possible in regards to it's own digit on the diagonal >>>>>>>>>
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable >>>>>>>>> AND computable value)
the fixed H is /decidable input/ to partial recognizer D (used >>>>>>>>> in fixed H, which does not call any D itself), and would not be >>>>>>>>> filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and >>>>>>>>> instead returns a hard-coded value for it's own digit on the >>>>>>>>> diagonal, is keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just
anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it >>>>>>>> will fail on that.
bro what part of the non-existence of turing_D or turing_H do you >>>>>>> not understand???
But if your fD exists, then Turing_H that uses your D exists, and >>>>>> will give that D the problem, and fixed_H will call your D on
turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H
gets hung up.
If D errs on Turing_H and calls it not circle_free, then your >>>>>>>> enumeration is missing the circle_free H from its list, and thus >>>>>>>> doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D in >>>>>>> a turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>> satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D
classifier, when actually run actual_turing_H will simply skip
putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the
actual_turing_H isn't an "undecidable input", as it has definite
behavior, just behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of undecidability. >>>>> just because a machine is /undecidable input/ to a particular
classifier does mean we cannot then prove and know ourselves what
the machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs are
not "undecidable", but by the time that they have actually been made
into an input, and thus first had the program created, which
required creating the instance of the decider selctected, they have
definite behavior that other some decider can determine.
It is only that one selected candidate decider chosen to build the
input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a
particular machine and the particular classifier it creates a
structural paradox for,
In other words, you don't understand what "undecidability" actually is. >>>>
Undecidability, means that there does not exist ANY decider that can
get the right answer for all instances of the problem, (or if the
problem has been reduced to just that one input, for that input).
*not* as a general inability to describe a *real* machine's
behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D.
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and prove it's equivalent to fixed_H, so therefore it doesn't fit what u stated was
> But THAT *IS* the meaning of "undecidability".
(you don't have one btw, so please do let me witness ur cope)
IT seesm you just can't read.
see, turing_H's runtime is not describable because it's an /
incomplete specification/ of machine and therefor _does not even
exist_ as a real TM, not because of the hypothesized undecidability >>>>> in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED defined
machine. If the decider "D" actually exists, then H is FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
specification/, as it doesn't handle idiosyncrasies of TM computing,
specifically self- referential set-classification paradoxes
It doesn't need to. It just needs to be built on what you want to
claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as your
equivalent for D, like your partial_recognizer_D, then the template
turing_H can be converted into an actual machine.
actual_turing_H, on the other hand, actually does exist, so we can
know/ prove what it does. it, however, still forms a structural
paradox in regards to partial_recognizer_D which will fail to
classify it as circle-free
But, since you admit to actual_turing_H existing, that means that
since partial_recognizer_D _CAN_ exist, then actual_turing_H _CAN_ exist
And so can the turing_H that uses partial_recognizer_D as its decider D.
The steps the algorithm does is what defines the machine, not the
"names" of the things it "calls", as those names don't actually exist
in the final building of the machine, which is the problem with your
concept, as that is needed to determine your "paradoxical" property.
the names are just shortcuts for the literal values rick, don't be daft
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get
wrong, then it can just return the different answer, and then THAT >>>>>> machine it will get right. The problem is it creates a DIFFERENT
input, that uses this new version of the machine, that it will get >>>>>> wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as
fixed_H, which in turn _does_ have a spot on the diagonal. so
there is no actual need to include a digit from actual_turing_H >>>>>>> on the diagonal, as the number it computes is already included on >>>>>>> the diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of
actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can
possible filter all ALL of that infinite set of machines, but
still accept some machine that computes that particular number
that it would compute with this supposedly correct
partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting
that you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the fairy >>>>>> dust, of assuming that your partial decider that meets ALL your
needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between
fixed_H and actual_turing_H is the way they respectively handle
their own self- references. they function identically when handling >>>>> all other machines
But the problem isn't in it handling the machine fixed_H, but how it
ever handles the results generated by actual_turing_H, which by your
assumption has become a cycle-free machine and ALL equivalent
versions of it have been partially declared to be non-cycle free.
This means that your enumeration of machines you are looking at is,
in fact, incomplete, thus the diagonal you computed is wrong.
fixed_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded
digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE =
satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip
trying to simulate itself and put it's hard coded digit on the
diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip
trying simulate actual_turing_H because partial_recognizer_D fails
to classify it as circle-free
But actual_turing_H is actually circle_free, and thus its results
SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put on the
diagonal, then actual_turing_H _DOES NOT_ need to be in _THAT_ total
enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable numbers,
but of machines that made computable numbers.
same thing moron
Also, your proof is built on the existance of a Russel Teapot, you have
the *specification* /works/, which shows that it's *not* contradiction
for it to be computable
just ASSUMED you cam make this partial_recoginzser_D that can make the
precise "mistakes" you need it to. (They are mistakes because the input
"mistake" is this case is a fallacious non-techincal value judgement on
the return value,
cause they are rather entirely within specification, as FALSE was
specified to be used upon encountering an /undecidable input/, something turing_D did not handle correctly (making it _NOT EXIST_)
that come from Turing_H are, in fact, circle_free because of this
turing_H is built on this partial_recognizdr_D which causes them to
not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to machines, as it
is based on the incorrect definition that assumes a machine can be
something other than itself, and when it changes, it changes other
things that at the meta/template level refered to its old self.
not sure what ur problem is
when actual_turing_H iterates across DN(actual_turing_H) itself, it >>>>> will similarly skip trying to simulate itself, because
partial_recognizer_D fails to classify it as circle-free
Right, so it ALSO is wrong, as it didn't compute the diagonal of the
circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE NUMBERS_,
which is the _ACTUAL PROBLEM_ turing was dealing with,
No, it isn't, as your "set of computable numbers" has duplicates.
it's still a diagonal across all computerate numbers even if it has dpulicates ...
this level of objection is dishonest\
doing so doesn't require /all/ or _even most_ machines to be in the
enumeration, since _most_ are duplicates of some other machine
when actual_turing_H iterates across DN(fixed_H),
partial_recognizer_D *successfully* classifies fixed_H as
satisfactory, so it will simulate fixed_H to it's Kth digit, the
one hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular logic >>>>> in either case), fixed_H also skips simulating itself (also
circular logic), while actual_turing_H actually *does* simulate
fixed_H successfully for it's Kth digit on the diagonal. these
machines compute the *same* number/sequence!
Except that by your partial_recognizer being just partial, it has
dropped required items from the enumeration.
_WHICH IS FINE_
So, you admit it failed???
ACTUAL_TURING_H IS _NOT_ REQUIRED TO BE ON THE DIAGONAL, SO IT'S FINE TO
BE DROPPED
As I said, it dropped REQUIRED items, as this enumeration was of
machines that computed computable numbers.
If you try to add removal of duplicates, then we need to define the
order, and if we do it by number, turing_H will likely be lower in
order, as it doesn't need to generate its own number, and thus is
simpler,
that sentence is nonsense until u actually produce psuedo-code
THis is becides the fact that you still need to show how you plan on
actually detecting all these computationally equivalent machines in
your
fixed_H does not compute a minimum turing-complete set, there are
duplicated sequences, ei machines that compute the same sequence/
number, and _THAT IS FINE_
Why?
BECAUSE AN ENUMERATION THAT INCLUDES ONE OR MORE INSTANCE FOR EACH COMPUTABLE NUMBER STILL ENUMERATES *ALL* COMPUTABLE NUMBERS
And, it was built on the assumption of the existance of a machine,
that actually can not exist (based on Rice's Theorm) and thus is
actually based on the assumption of Unicorns.
partial_recognizer_D is *not* a total decider, so how could rice's
theorum apply???
partial decider to force them to not be classified as circle-free,
even though it turns out that they are (because your decider called
them not).
You don't seem to understand that you don't get to look at the
template used to build the machine, only the resultant machine from
that template.
And, part of the result of that, is that your attempt to try to
define "paradoxical" just fails, as it resolves to just input your
decider is wrong about, in part, because you don't understand what
"decidable" actually means,
see, not only is defining a value in fixed_H that the diagonal
computation can put for itself in diagonal essential in making the
diagonal actually computable, so is partial_recognizer_D skipping
over actual_turing_H to avoid the computability paradox!
In other words, you think it ok to compute the WRONG diagonal.
The needed diagonal was to a COMPLETE enumeration of the circle-free
machine.
NO, IT WAS SPECIFICALLY A COMPLETE ENUMERATION OF _COMPUTABLE
NUMBERS/ SEQUENCES_
Which you don't get.
smh
there is *no* requirement to include a sequence _more than once_, so
there *no* requirement to include all machines in the enumeration!
But there is to show that your partial_recognizer_D can exist, when it
can't.
YOU'VE LOST THE STANDARD PROOF IT DOESN'T EXIST,
WHAT NEW ARGUMENT WOULD THERE BE???
YOU need to prove your "Russel's Teapot" exists.
Your logic just shows that you don't know how to do logic.
That isn't really different then claiming you can prove that every
even number grater than 2 from an infinite set of natural numbers is
the sum of two primes, make you enumeration just the set of prime
numbers, and thus claim that since the only even number is 2, it is
trivially solved.
_FALSE ANALOGY FALLACY_
What is false about it?
THE SUBSET IDENTIFIED BY PARTIAL_RECOGNIZER_D IS NOT "LESS POWERFUL"
THAN THE TOTAL MACHINE SPACE
if the set i'm enumerating includes _AT LEAST ONE INSTANCE_ of _ALL
COMPUTABLE SEQUENCES_
But does it?
YES
then it is a turing-complete subset including all possible TM
computable sequences, and is functionally equivalent to the set of
*all* TMs
But, since your partial_recognizer_D doesn't exist, but was only
assumed to exist, your program doesn't exist,
ultimately, in order to compute a diagonal across computable
numbers, partial_recognizer_D only needs to successfully classify
*one* of the _infinitely_ many machines that compute any given
computable number...
But it isn't the diagonal across some set of computable numbers, but
the diagonal across the ENTIRE set of computable numbers.
Since you dropped some machines, you don't have the needed set.
_*THAT DOES NOT NECESSARILY FOLLOW*_
Then try to define what set you are actually making!!
A SET THAT CONTAINS INFINITE INSTANCES OF ALL COMPUTABLE NUMBERS
The problem is your definition of "paradoxical" doesn't actually work,
as it is based on asking about the results of an input, that changes.
dropping actual_turing_H from the set, for example, is _FINE_ because
it computes _THE SAME SEQUENCE_ as fixed_H, which _IS IN THE SET_ ...
How do you know it includes your fixed_H in the set?
what part of the actual machine makes it different?
THE CODE OF FIXED_H AND ACTUAL_TURING_H IS IN THE THREAD, WHAT ARE YOU TALKING ABOUT???
Remember, in the final machine given as the input, there are not
READ TURING'S PAPER ACTUALLY TO SEE HOW A SUBSET OF TRANSITION FUNCTIONS WITHIN A TURING MACHINE CAN BE LABELED AND REFERRED TO
seperate turing_H and partial_recognizer_D denoted in the machine,
just one big piece of algorithm. Thus nothing to base your criteria on.
goddamn rick smh u could spot a diamond in the rough even if god lead
shit rick, idk how ur still denying what's right under ur nose,
Because what you are putting under my nose is just CRAP based on errors >>>
u by ur hand to one put on a pedestal with a _GIANT_ neon sign
pointing to it...
But the problem is that "your diamond" is actually just a big load of
crap that you misidentify as a diamond, because you don't actually
understand what you are doing.
ur so used to playing contrarian u don't know what truth even looks like anymore, rick
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that >>>>>>>>>>> generates a computable number to build the list of such >>>>>>>>>>> machine to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH >>>>>>>>>> ARE INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH >>>>>>>>>> WILL WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE >>>>>>>>>> INPUT/ TO D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE
COMPUTABLE NUMBERS
It doesn't matter that your new H doesn't get stuck on >>>>>>>>>>>>> itself, it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he >>>>>>>>>>>> doesn't specify what D (or H) needs to do when encountering >>>>>>>>>>>> the / undecidable input/ of H, so therefore both D and H are >>>>>>>>>>>> an incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer it >>>>>>>>>>> must compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE >>>>>>>>>> LIKE THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN
IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS
COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, BECAUSE >>>>>>>>>> IT DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-REFERENTIAL >>>>>>>>>> ANALYSIS, SO IT IS AN /INCOMPLETE SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES >>>>>>>>>> CAN'T BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON??? >>>>>>>>>>
Note, H, when it becomes an actual machine, because we have >>>>>>>>>>> created an actual machine we claim to be the D, has an
answer, and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't >>>>>>>>>>> exist, only machines CORRECTLY meeting the requirements of D >>>>>>>>>>> and H can't exist.
Thus, if your H depends on that D, (or a machine that meets >>>>>>>>>>> its specification) then it also can't exist.
You just make the error of saying the problems don't exist >>>>>>>>>>> because you can't build them, but YOUR machine can, even >>>>>>>>>>> though it has the same problem.
IF D is wrong by deciding it is not circle free, then your >>>>>>>>>>>>> H will compute the wrong diagonal, as the resulting version >>>>>>>>>>>>> of his H WILL be circle free (since it never tries to >>>>>>>>>>>>> simulate itself) and thus DOES produce an computable number >>>>>>>>>>>>> that your computation misses.
Or, if that D is wrong by decing it IS circle free, then >>>>>>>>>>>>> when you H tries to process it, it will get stuck in the >>>>>>>>>>>>> infinite loop.
The problem is that in stepping through the machines in >>>>>>>>>>>>> order, you WILL hit these actual machines built on your >>>>>>>>>>>>> erroneous D (your D must have this flaw, as no D without >>>>>>>>>>>>> exists), and thus you will be wrong on THAT input. IT >>>>>>>>>>>>> doesn't matter if you get a good answer for yourself. >>>>>>>>>>>>>
I doubt it.
idk what he would have said about it, but prolly something >>>>>>>>>>>>>> more substantial than just calling me ignorant repeatedly >>>>>>>>>>>>>
He likely would have gotten frustrated by your idiodic >>>>>>>>>>>>> assertion of bad logic. You would have likely been escorted >>>>>>>>>>>>> out of the meeting as showing you were unqualified and >>>>>>>>>>>>> being a distraction.
Something that seems to be beyond your ignorant >>>>>>>>>>>>>>> understanding.
But the problem is that "paradoxical machines" don't >>>>>>>>>>>>>>> exist in isolation, but only in relationship to a given >>>>>>>>>>>>>>> machine trying to decide them.
H shows that *IF* you can make that enumeration, you >>>>>>>>>>>>>>>>> can make the diagonal, and thus the anti-diagonal. The >>>>>>>>>>>>>>>>> problem is you can't make that enumeration, and >>>>>>>>>>>>>>>>> assuming you can just shows unsoundness.
interestingly: one can only fix the direct diagonal >>>>>>>>>>>>>>>> computation like this
u can't do an analogous fix for the inverse/anti- >>>>>>>>>>>>>>>> diagonal computation. it's not possible hard code a >>>>>>>>>>>>>>>> machine to return an inverted value, a machine can only >>>>>>>>>>>>>>>> return what it does, not the inverse of what it does... >>>>>>>>>>>>>>>>
so if we can filter out paradoxes from the enumeration, >>>>>>>>>>>>>>>> that will leave a direct diagonal computation extant in >>>>>>>>>>>>>>>> that filtered (yet still turing complete list), while >>>>>>>>>>>>>>>> any attempt to compute an inverse diagonal will not be >>>>>>>>>>>>>>>
right. so if ur constructing a diagonal across computable >>>>>>>>>>>>>> numbers then u only need to filter out paradoxes in >>>>>>>>>>>>>> regards to the classifier that classifies them as a >>>>>>>>>>>>>> "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such >>>>>>>>>>>> paradoxes...
In other words, you beleive unquestionably in the existance >>>>>>>>>>> of Russle's teapot until someone can prove it doesn't exist. >>>>>>>>>>
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF
PROVIDING A FUCKING QUOTE TO TURING'S PAPER TO BACK THE
*DIRECT* CLAIM YOU MADE???
For THIS paradox, read the proof. for THIS H, the one given, >>>>>>>>>>> (for which ever version of D you want to try to assume is >>>>>>>>>>> right) D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the
paradox, because there still isn't an answer it can give to H >>>>>>>>>>> that will be correct, as the template for H will always make >>>>>>>>>>> that D wrong.
Note, the problem is you can't actually DEFINE what a
"paradox machine" is, as they aren't actually machines, but >>>>>>>>>>> templates that build machines. And that final machine doesn't >>>>>>>>>>> have actually detectable tell-tales that show it to be from >>>>>>>>>>> that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE >>>>>>>>>> TO WORK WITH, BUT OH WELL
und = () -> halts(und) loop()
They only seem "random" because you don't know what you are >>>>>>>>>>> talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* >>>>>>>>>>>>>> classifiable as satisfactory by said classifier... can >>>>>>>>>>>>>> just be skipped
No, it can only skip those that are not satisfactory, not >>>>>>>>>>>>> those that are but it can not classify as such, or your >>>>>>>>>>>>> enumeration will not be complete, and thus just in error. >>>>>>>>>>>>>
Thus, it needs to be able to correctly classify ALL >>>>>>>>>>>>> machines (as it will be asked about all machines as it >>>>>>>>>>>>> counts through all the descriptions) and thus Turing's H >>>>>>>>>>>>> *WILL* be asked about.
similarly if u want to go a step further an filter out >>>>>>>>>>>>>> computable numbers already included on this diagonal, any >>>>>>>>>>>>>> machine which either *is* computably equivalent OR *is >>>>>>>>>>>>>> not* classifiable in regards to *any* machine already the >>>>>>>>>>>>>> list... can just be skipped
Nope, you can't skip some machines, as you then might lose >>>>>>>>>>>>> some of the computable numbers.
see you can't compute a diagonal across *all* /machines/, >>>>>>>>>>>>>> with said machines, but u can compute a diagonal across >>>>>>>>>>>>>> *all* / computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be >>>>>>>>>>>>> done, since ALL classifiers that attempt it will make an >>>>>>>>>>>>> error, you can't do what you want to do.
nah, (a) computing an enumeration of all /computable
numbers/ is not the same thing as (b) computing the
enumeration of all machines that compute computable numbers. >>>>>>>>>>>> (b) necessarily has duplicates while (a) does not need them. >>>>>>>>>>>> turing's paper wrongly conflates (a) with (b)
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter >>>>>>>>>>> the full list of machibe, which Turing showed can't exist. >>>>>>>>>>>
The problem is such a machine can't handle the classification >>>>>>>>>>> of the H he describes. It doesn't matter that you can make a >>>>>>>>>>> DIFFERENT machine, that you try to deceptively call "H" too >>>>>>>>>>> that it won't have a problem with, when your "H" gets to the >>>>>>>>>>> number of Turing's H, it still has the problem. It can call >>>>>>>>>>> that one not circle-free, and thus omit a circle- free
machine from the list, or call it circle- free, and when even >>>>>>>>>>> YOU try to simulate it the k steps, you get stuck in a loop. >>>>>>>>>>>
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to take >>>>>>>>>>> the code of (a), and invert the symbol put on the tape when >>>>>>>>>>> it is written. Note, it won't be "self-referent", as the "N" >>>>>>>>>>> is uses, is the N of YOUR H, not itself. The code for his (a) >>>>>>>>>>> NEVER reads back what it wrote, so that doesn't affect its >>>>>>>>>>> behavior.
i'm sorry, are you actually saying the machine description >>>>>>>>>>>> does not describe what the machine does???
yes, i still do need to prove my thesis that for any >>>>>>>>>>>>>> paradoxical machine, there exists a functionally
equivalent machine without such paradox
And the problem is that your "paradoxical" isn't actually a >>>>>>>>>>>>> definable property (let alone computable). Part of the >>>>>>>>>>>>> problem is that if you look at just a machine description, >>>>>>>>>>>>> it doesn't (necessarily) tell you about the use of an >>>>>>>>>>>>> "interface" as that use of an interface can be just >>>>>>>>>>>>> inlined, leaving nothing "in the code" to show it exists. >>>>>>>>>>>>
No, that "paradoxical" isn't a definable property of a machine. >>>>>>>>>>
und() includes a structural paradoxical in relation to the >>>>>>>>>> classifier halts(), easily demonstrable thru a short proof >>>>>>>>>> i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A PROPERTY >>>>>>>>>> OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS DEVOTION TO >>>>>>>>>> SOME FUCKING FORM OF INSANITY I CAN'T HONESTLY FATHOM
lol
Sure, because it is just a number. The problem is that you >>>>>>>>>>>>> still have
His specified H, with an actually (incorrect)
implementation of D (which is all that CAN exist) will >>>>>>>>>>>>>>> either be circle- free and thus generate a number (but >>>>>>>>>>>>>>> its D said it isn't, and thus omitted a valid machine >>>>>>>>>>>>>>> from the list) or it isn't circle- free, and fails to >>>>>>>>>>>>>>> computa a number, and thus should have been omitted from >>>>>>>>>>>>>>> the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it is >>>>>>>>>>>>>>> just built on an assuption in error.
so despite turing's worries, the existence of a diagonal >>>>>>>>>>>>>>>> computation does not actually then imply the existence >>>>>>>>>>>>>>>> of an anti- diagonal computation, due the same >>>>>>>>>>>>>>>> particular self- referential weirdness that stumped >>>>>>>>>>>>>>>> turing the first place
But there is no actuall SELF-REFERENCE, so your logic is >>>>>>>>>>>>>>> just based on ERROR.
Your attempt to REDEFINE self-reference to mean >>>>>>>>>>>>>>> processing a number that happens to represent yourself >>>>>>>>>>>>>>> means that you you system "ALL" doesn't actually mean >>>>>>>>>>>>>>> ALL, and thus is just ill- defined.
i'm sorry, you have an issue with me labeling a number >>>>>>>>>>>>>> that directly refers to yourself, as a "self-reference" ??? >>>>>>>>>>>>>
it's a *specific* number that has the currently running >>>>>>>>>>>> machine encoded into it, it's not "just" a number whatever >>>>>>>>>>>> that means
problems with all the "equivalent" machines that have >>>>>>>>>>>>> different numbers.
those are references to functionally equivalent machines, >>>>>>>>>>>> not self- references
Which is why the concept of "self-reference" doesn't work. >>>>>>>>>>>
It may let you filter out the simplest case used in the >>>>>>>>>>>>> proofs, but doesn't solve the actual problem, as the >>>>>>>>>>>>> "Machine Number" doesn't actually fully identify the >>>>>>>>>>>>> problematic cases.
that's not actually true. you can't meaningfully paradox the >>>>>>>>>>>> paradox detector while filter *out* paradoxes to point of >>>>>>>>>>>> creating an actual problem
Then show how you will do it.
This means you need to detect a input that represents a >>>>>>>>>>> machine that uses a computation equivalent to yourself
WHich doesn't exist.
let halts be a halting classifier that takes an input >>>>>>>>>>>> machines and returns true/false whether it halts or not >>>>>>>>>>>>
let paradox_free be a paradox filtering classifier that >>>>>>>>>>>> takes an input classifier, and an input machine to search >>>>>>>>>>>> for paradoxes in regards to: paradox_free(classifier, machine) >>>>>>>>>>>
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox >>>>>>>>>>>> classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case >>>>>>>>>>>> run a halting paradox form making it /undecidable input/ to >>>>>>>>>>>> halts().
And the problem is "paradox_free" can't exist as an always >>>>>>>>>>> correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if >>>>>>>>>>>> case doesn't run and clearly halts(undp) -> TRUE without issues >>>>>>>>>>>>
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a >>>>>>>>>>>> machine is necessary for the particular input->output >>>>>>>>>>>> computation being done, so utilizing the return
paradox_free(halts,undp) -> FALSE to filter out
paradox_free() paradoxes as well as halts() paradoxes will >>>>>>>>>>>> suffice to produce a turing complete subset of machines that >>>>>>>>>>>> can be totally classified by halts()
But we can still build an input that your paradox_free() >>>>>>>>>>> fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE >>>>>>>>>> INCLUDED ON THE DIAGONAL???
No, it is your dependence on being able to assume that
i think a major problem with the consensus perspective is an >>>>>>>>>>>> expectation of a certain cleanliness in the logic as being >>>>>>>>>>>> the only possibility here to compute what we want. TM's >>>>>>>>>>>> can't achieve that, but something slightly messier can. >>>>>>>>>>>
unicorns can solve all your problems.
They are, But you are too stuck in your lies to see it.
lol, i can only 🙏 someone would engage in sound logic with me >>>>>>>>>>>
dear future: u see this shit i'm trying work with???? >>>>>>>>>>>>>> sorry i didn't manage to make progress any faster, but i'm >>>>>>>>>>>>>> literally dragging toddlers kicking and screaming at this >>>>>>>>>>>>>> point jeez...
In other words, in your world "ALL" isn't a word that can >>>>>>>>>>>>> actually be used.
You don't understand that you need to use actual sound logic, >>>>>>>>>>>>
Your "logic" is based on being able to assume that Unicorn's >>>>>>>>>>> exist, and that is a safe assumption until someone can prove >>>>>>>>>>> that Russel's Teapot is not out there. (Even though your >>>>>>>>>>> unicorns HAVE been proved to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't >>>>>>>>>>>>>>> understand what the actual problem is, and your world is >>>>>>>>>>>>>>> just build on things that are lies.
...which u won't find interesting, but i can't fix a >>>>>>>>>>>>>>>> lack of curiosity
🤷
expected was an inexpert audience, supposing that >>>>>>>>>>>>>>>>>>> experts will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>>>>>
IT seems that fundamentally, your world doesn't actually >>>>>>>>>>>>>>> have "computations" as you don't understand the basic >>>>>>>>>>>>>>> requirement that they need to be fully defined in the >>>>>>>>>>>>>>> actions they do.
On 3/6/2026 12:10 PM, Richard Damon wrote:
[...]
You have a lot of patience with Olcott 2.0! It calls you every name in
the book.
On 3/6/26 1:03 PM, Chris M. Thomasson wrote:
On 3/6/2026 12:10 PM, Richard Damon wrote:
[...]
You have a lot of patience with Olcott 2.0! It calls you every name in
the book.
fking pussy can't even read this
--
arising us out of the computing dark ages,
please excuse my pseudo-pyscript,
~ the little crank that could
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:i don't see any indication that turing realized a >>>>>>>>>>>>>>>>> difference there
On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote:
On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> ...
Eh?!an effective enumeration of all turing machines >>>>>>>>>>>>>>>>>>>>>>>>> was proven on
turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he >>>>>>>>>>>>>>>>>>>>> defines a comptuation
that enumerates out all the numbers, testing each >>>>>>>>>>>>>>>>>>>>> one of they represent
a "satisfactory"/"circle-free" machine, and adding >>>>>>>>>>>>>>>>>>>>> that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to carefully >>>>>>>>>>>>>>>>>>>>> read p247 of turing's
proof and produce the psuedo-code for the machine >>>>>>>>>>>>>>>>>>>>> H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are >>>>>>>>>>>>>>>>>>>> you sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>>>> procedures, and understanding the diagonal procedure >>>>>>>>>>>>>>>>>>> is critical to understanding the *base* >>>>>>>>>>>>>>>>>>> contradiction/ paradox that the rest of his support >>>>>>>>>>>>>>>>>>> for godel's result is then built on
And focusing on what is said to be impossible and not >>>>>>>>>>>>>>>>>> changing the problem is also important.
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>>>>> enumerating the enumeration in the first place. >>>>>>>>>>>>>>>>>
Then you zre just showing your stupidity, because YOU >>>>>>>>>>>>>>>> can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>>>>> enumeration of the computable sequences.
He then points out that he can directly show that the >>>>>>>>>>>>>>>> "anti- diagonal" of the (non-effectively computed) >>>>>>>>>>>>>>>> enumeration can't be computed but that "This proof, >>>>>>>>>>>>>>>> although perfectly sound, has the disadvantage that it >>>>>>>>>>>>>>>> may leave the reader with a feeling that 'there must be >>>>>>>>>>>>>>>> something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being >>>>>>>>>>>>>> done.
I think he is refering he to the standard anti-diagonal >>>>>>>>>>>>>> arguement, which shows that since for all n, position n >>>>>>>>>>>>>> differs from the value in number n, there can not be any >>>>>>>>>>>>>> element that matches the anti- diagonal.
It is just a natural fact of countable infinity, something >>>>>>>>>>>>>> it seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with >>>>>>>>>>>>> turing's short diagonal proof, but in writing this post i >>>>>>>>>>>>> now find myself in a subtle, yet entirely critical
disagreement:
/let an be the n-th computable sequence, and let φn(m) be >>>>>>>>>>>>> the m- th figure in an. Let β be the sequence with 1-φn(m) >>>>>>>>>>>>> as its n-th. figure. Since β is computable, there exists a >>>>>>>>>>>>> number K [== β] such that 1- φn(n) = φK(n) for all n. >>>>>>>>>>>>> Putting n = K, we have 1 = 2φK(K), i.e. 1 is even. This is >>>>>>>>>>>>> impossible/
i agree with this proof is far as much as if β was computable >>>>>>>>>>> (by TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the
computability of β from the existence of φn(m), THAT'S THE >>>>>>>>>>> FALLACy
the fallacy here is assuming that because the direct >>>>>>>>>>>>> diagonal is computable, that one can therefore compute the >>>>>>>>>>>>> anti- diagonal using the direct diagonal. the abstract >>>>>>>>>>>>> definition makes it look simple, but this ignores the >>>>>>>>>>>>> complexities of self- referential analysis (like what >>>>>>>>>>>>> turing details on the next page)
But you can, *IF* you have a machine that computes the >>>>>>>>>>>> diagonal, then just change all the write to the output to >>>>>>>>>>>> write the opposite. Note, the "self-reference" that you are >>>>>>>>>>>> thinking of stops being a "self- reference" but is a
reference to the original write the diagonal code.
Nope, because your filtering TM (or RTM) will still fail >>>>>>>>>>>> when it gets to the number of TURING'S H, as there is no >>>>>>>>>>>> correct answer for the machine built by that template.
in both methods i have for rectifying the paradox found in >>>>>>>>>>>>> the direct diagonal (either (1) filtering TMs or (2) using >>>>>>>>>>>>> RTMs), neither can be used to then compute the anti-diagonal >>>>>>>>>>>>
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE >>>>>>>>>>> SPECIFICATION
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered >>>>>>>>>>>>> out like turing's paradoxical variation of the direct >>>>>>>>>>>>> diagonal would be, and there is no analogous non-
paradoxical variation that has a hard coded value that is >>>>>>>>>>>>> inverse to what it does return ... such a concept is >>>>>>>>>>>>> entirely nonsensical. a function can only return what it >>>>>>>>>>>>> does, it can't also return the inverse to what it returns >>>>>>>>>>>>> eh???
But, if D filters it out, then it becomes circle-free, and >>>>>>>>>>>> thus your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT >>>>>>>>>>> *SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with RTMs >>>>>>>>>>>>> just fails for reasons u'd only understand by working thru >>>>>>>>>>>>> the algo urself (p7 of re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/
is just not sufficient evidence that such β is actually >>>>>>>>>>>>> computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how >>>>>>>>>>>> to do that, as you are thinking the only "reference" can be >>>>>>>>>>>> to "self", doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF- >>>>>>>>>>> REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H >>>>>>>>>>>
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS FOR >>>>>>>>>>> INVERSE- H TO BE COMPUTED
He doesn't. You are just showing your stupidity,
one cannot just assume that because the diagonal across >>>>>>>>>>>>>>> computable numbers is computable, therefore the anti- >>>>>>>>>>>>>>> diagonal across computable numbers is computable... >>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and without >>>>>>>>>>>>>> the enumeration, you can't compute either of them. >>>>>>>>>>>>>>
neither method i have for fixing the diagonal computation >>>>>>>>>>>>>>> across the computable numbers can be used to compute the >>>>>>>>>>>>>>> inverse diagonal
But your method still doesn't let you compute the >>>>>>>>>>>>>> enumeration, and thus you can't actually compute the >>>>>>>>>>>>>> diagonal.
Remember, the problem definitions requires that the >>>>>>>>>>>>>> listing be a COMPLETE listing of the computable numbers / >>>>>>>>>>>>>> machine that compute computable numbers, in some definite >>>>>>>>>>>>>> order.
If your enumeration isn't complete, your diagonal isn't >>>>>>>>>>>>>> correct.
so while i agree with turing that the anti-diagonal is >>>>>>>>>>>>>>> not computable, i don't agree that the normal diagonal is >>>>>>>>>>>>>>> not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all >>>>>>>>>>>>>> the other machines, including his original H that doesn't >>>>>>>>>>>>>> use your "trick"
But it doesn't.
But instead, he can prove with a more obvious process, >>>>>>>>>>>>>>>> that the Decider "D" that could be used to effectively >>>>>>>>>>>>>>>> enumerate the sequence of machine that produce >>>>>>>>>>>>>>>> computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing >>>>>>>>>>>>>>>> out that the attempt to compute the diagonal clearly >>>>>>>>>>>>>>>> reveals the issue with effectively enumerating the >>>>>>>>>>>>>>>> sequences.
well, he didn't consider that perhaps the proper algo for >>>>>>>>>>>>>>> computing the diagonal can avoid the paradox on itself ... >>>>>>>>>>>>>>
Your just don't understand that D just can't correctly >>>>>>>>>>>>>> decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on >>>>>>>>>>>>> turing's H, because my response to this is that D does not >>>>>>>>>>>>> need to decide correctly on H to compute a diagonal
Then how do you build YOUR H without that D?
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT >>>>>>>>>>> WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial
recognizer D *can* decide on the fixed H because the fixed H >>>>>>>>>> does not try to use any D on itself, so no self-referential >>>>>>>>>> paradox is possible in regards to it's own digit on the diagonal >>>>>>>>>>
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a dependable >>>>>>>>>> AND computable value)
the fixed H is /decidable input/ to partial recognizer D (used >>>>>>>>>> in fixed H, which does not call any D itself), and would not >>>>>>>>>> be filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and >>>>>>>>>> instead returns a hard-coded value for it's own digit on the >>>>>>>>>> diagonal, is keystone in making it /decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just >>>>>>>>> anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it >>>>>>>>> will fail on that.
bro what part of the non-existence of turing_D or turing_H do >>>>>>>> you not understand???
But if your fD exists, then Turing_H that uses your D exists, and >>>>>>> will give that D the problem, and fixed_H will call your D on
turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H >>>>>>>>> gets hung up.
If D errs on Turing_H and calls it not circle_free, then your >>>>>>>>> enumeration is missing the circle_free H from its list, and >>>>>>>>> thus doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D >>>>>>>> in a turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>> satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D
classifier, when actually run actual_turing_H will simply skip >>>>>>>> putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the
actual_turing_H isn't an "undecidable input", as it has definite >>>>>>> behavior, just behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of
undecidability. just because a machine is /undecidable input/ to a >>>>>> particular classifier does mean we cannot then prove and know
ourselves what the machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs are >>>>> not "undecidable", but by the time that they have actually been
made into an input, and thus first had the program created, which
required creating the instance of the decider selctected, they have >>>>> definite behavior that other some decider can determine.
It is only that one selected candidate decider chosen to build the
input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a
particular machine and the particular classifier it creates a
structural paradox for,
In other words, you don't understand what "undecidability" actually >>>>> is.
Undecidability, means that there does not exist ANY decider that
can get the right answer for all instances of the problem, (or if
the problem has been reduced to just that one input, for that input). >>>>>
*not* as a general inability to describe a *real* machine's
behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D.
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and prove it's
equivalent to fixed_H, so therefore it doesn't fit what u stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a machine, unless you are narrowing the problem to just being about that one machine/input.
And, it is an assertion that NO decider can correctly compute the answer
to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which means you have chosen a particular (broken) implementation of D (since no actually correct D exists), like your partial_recognizer_D, has fulled defined behavior, and a decider can exist that determines it, there is nothing "undecidable" about that input.
It is just that the D that H was built on is wrong. Nothing
"paradoxical" about a machine being wrong.
(you don't have one btw, so please do let me witness ur cope)
IT seesm you just can't read.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
see, turing_H's runtime is not describable because it's an /
incomplete specification/ of machine and therefor _does not even
exist_ as a real TM, not because of the hypothesized
undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED defined
machine. If the decider "D" actually exists, then H is FULLY defined. >>>>
specification/, as it doesn't handle idiosyncrasies of TM computing,
specifically self- referential set-classification paradoxes
It doesn't need to. It just needs to be built on what you want to
claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as your
equivalent for D, like your partial_recognizer_D, then the template
turing_H can be converted into an actual machine.
actual_turing_H, on the other hand, actually does exist, so we can >>>>>> know/ prove what it does. it, however, still forms a structural
paradox in regards to partial_recognizer_D which will fail to
classify it as circle-free
But, since you admit to actual_turing_H existing, that means that
since partial_recognizer_D _CAN_ exist, then actual_turing_H _CAN_
exist
And so can the turing_H that uses partial_recognizer_D as its decider D. >>>
The steps the algorithm does is what defines the machine, not the
"names" of the things it "calls", as those names don't actually exist
in the final building of the machine, which is the problem with your
concept, as that is needed to determine your "paradoxical" property.
the names are just shortcuts for the literal values rick, don't be daft
Right, but the literal values don't have the names on them, and we don't have borders within the machine letting us know we moved from "outer
code of H" into "the code of the D it uses", especially with Turing
Machines were such usages are by necessity "expanded" in-line.
Thus, the "copy" of D that was pulled into H isn't actually the same
thing as the independent machine D that we can talk about, just a
functional equivalent which doesn't even have its own D.N (since that is
a property only of full machines)
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get >>>>>>> wrong, then it can just return the different answer, and then
THAT machine it will get right. The problem is it creates a
DIFFERENT input, that uses this new version of the machine, that >>>>>>> it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as
fixed_H, which in turn _does_ have a spot on the diagonal. so >>>>>>>> there is no actual need to include a digit from actual_turing_H >>>>>>>> on the diagonal, as the number it computes is already included >>>>>>>> on the diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of
actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can
possible filter all ALL of that infinite set of machines, but
still accept some machine that computes that particular number
that it would compute with this supposedly correct
partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting >>>>>>> that you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the fairy >>>>>>> dust, of assuming that your partial decider that meets ALL your >>>>>>> needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between
fixed_H and actual_turing_H is the way they respectively handle
their own self- references. they function identically when
handling all other machines
But the problem isn't in it handling the machine fixed_H, but how
it ever handles the results generated by actual_turing_H, which by
your assumption has become a cycle-free machine and ALL equivalent
versions of it have been partially declared to be non-cycle free.
This means that your enumeration of machines you are looking at is, >>>>> in fact, incomplete, thus the diagonal you computed is wrong.
fixed_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded
digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>> satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip
trying to simulate itself and put it's hard coded digit on the
diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip
trying simulate actual_turing_H because partial_recognizer_D fails >>>>>> to classify it as circle-free
But actual_turing_H is actually circle_free, and thus its results
SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put on
the diagonal, then actual_turing_H _DOES NOT_ need to be in _THAT_
total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable
numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on being sloppy.
Note, Turing SPECIFICALLY pointed out that he was going to the related problem, the problem of:
/but the problem of enumerating computable sequences is equivalent to the problem of finding out whether a given number is the D.N of a circle-
free machine, and we have no general process for doing this in a finite number of steps/
So, HIS enumeration is the complete enumeration of "circle-free
machines", which your proof admittedly doesn't do.
Also, your proof is built on the existance of a Russel Teapot, you have
the *specification* /works/, which shows that it's *not* contradiction
for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion. Your arguement boils down to showing that *IF* you can correctly determine
all machines that fit in a not-definable category (since you concept of [paradoxical can't actually be defined at the machine level, only at how
a machine might have been created at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can make
the precise "mistakes" you need it to. (They are mistakes because the
input
"mistake" is this case is a fallacious non-techincal value judgement
on the return value,
No, since your criteria that you want to use is just an undefinable term
for a machine description.
There is nothing "paradoxical" of the actual machine when created, as it then has fixed and determined behavior that can be decided on. The thing that is paradoxical is the mete-system pattern used to generate the
class of machines as pair with a given decider.
cause they are rather entirely within specification, as FALSE was
specified to be used upon encountering an /undecidable input/,
something turing_D did not handle correctly (making it _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is based on a
actual D that does make a specific decision to the input. And it is just that for any given version of D, we can find a DIFFERENT H that it will
get wrong (the H that was just built on it).
Do you want to claim that ALL these H are "undecidable" even though for
all of them there does exist a decider that gets the right answer?
It seems you just don't know what the word means, and thus you
specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of this
turing_H is built on this partial_recognizdr_D which causes them to
not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to machines, as
it is based on the incorrect definition that assumes a machine can be
something other than itself, and when it changes, it changes other
things that at the meta/template level refered to its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just refuse to
learn.
Try to actually fully DEFINE the word based on the actual properties of
the input machine, and not refering to other machines that are not part
of the input (but would be created as alternate inputs to foil an
alternate decider).
when actual_turing_H iterates across DN(actual_turing_H) itself,
it will similarly skip trying to simulate itself, because
partial_recognizer_D fails to classify it as circle-free
Right, so it ALSO is wrong, as it didn't compute the diagonal of
the circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE NUMBERS_,
which is the _ACTUAL PROBLEM_ turing was dealing with,
No, it isn't, as your "set of computable numbers" has duplicates.
it's still a diagonal across all computerate numbers even if it has
dpulicates ...
No, because an enumeration of a set doesn't HAVE duplicates.
Your probe
this level of objection is dishonest\
No, your using misdefined terms is the real dishonesty.
If you want to show that you can make a diagonal of computable numbers,
you need to build an actual effective enumeration of computable numbers.
And the problem is you can't assume your decider and then use it. That
isn't vallid logic.
It IS valid to assume a decider, and show that such an assumption leads
to a contradiction, to prove that such a decider can not exist. That is
the valid proof by contradiction.
If we can show that if A exist, B must be true, and also that B can not
be true, then we have proven that A can not exist.
Assuming that A exists, and showing that one line of reasoning based on
it doesn't lead to a contradiction, does NOT show that A does (or even
can) exist. That is just fallacious logic.
doing so doesn't require /all/ or _even most_ machines to be in the
enumeration, since _most_ are duplicates of some other machine
when actual_turing_H iterates across DN(fixed_H),
partial_recognizer_D *successfully* classifies fixed_H as
satisfactory, so it will simulate fixed_H to it's Kth digit, the
one hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular
logic in either case), fixed_H also skips simulating itself (also >>>>>> circular logic), while actual_turing_H actually *does* simulate
fixed_H successfully for it's Kth digit on the diagonal. these
machines compute the *same* number/sequence!
Except that by your partial_recognizer being just partial, it has
dropped required items from the enumeration.
_WHICH IS FINE_
So, you admit it failed???
ACTUAL_TURING_H IS _NOT_ REQUIRED TO BE ON THE DIAGONAL, SO IT'S FINE
TO BE DROPPED
If the diagonal is the diagonal of circle-free machines, it is required.
Your problem is you keep on working with strawmen.
If the diagonal is of computable numbers, you need to show how you trim
down the set to not have repeats (in finite time per element) and that
your decider actually exists, you can't just assume it.
As I said, it dropped REQUIRED items, as this enumeration was of
machines that computed computable numbers.
If you try to add removal of duplicates, then we need to define the
order, and if we do it by number, turing_H will likely be lower in
order, as it doesn't need to generate its own number, and thus is
simpler,
that sentence is nonsense until u actually produce psuedo-code
I did, just make my H call your "D" replacement, partial_recognizer_D.
Or, is your problem that you think an enumeration doesn't define an "order"
THis is becides the fact that you still need to show how you plan
on actually detecting all these computationally equivalent machines >>>>> in your
fixed_H does not compute a minimum turing-complete set, there are
duplicated sequences, ei machines that compute the same sequence/
number, and _THAT IS FINE_
Why?
BECAUSE AN ENUMERATION THAT INCLUDES ONE OR MORE INSTANCE FOR EACH
COMPUTABLE NUMBER STILL ENUMERATES *ALL* COMPUTABLE NUMBERS
Nope. Maybe you don't understand what an enumeration.
And enumeration is a one-to-one mapping between the elements and the
Natural Numbers.
By your definition, you can make a countable infinite enumeation of a
finite set, showing that the finite set is infinite.
And, it was built on the assumption of the existance of a machine,
that actually can not exist (based on Rice's Theorm) and thus is
actually based on the assumption of Unicorns.
partial_recognizer_D is *not* a total decider, so how could rice's
theorum apply???
But it *IS* a total decider on your "paradoxical" property.
It can't let ANY of them through, and it must not omit any of the ones
you need.
partial decider to force them to not be classified as circle-free,
even though it turns out that they are (because your decider called >>>>> them not).
You don't seem to understand that you don't get to look at the
template used to build the machine, only the resultant machine from >>>>> that template.
And, part of the result of that, is that your attempt to try to
define "paradoxical" just fails, as it resolves to just input your
decider is wrong about, in part, because you don't understand what
"decidable" actually means,
see, not only is defining a value in fixed_H that the diagonal
computation can put for itself in diagonal essential in making the >>>>>> diagonal actually computable, so is partial_recognizer_D skipping >>>>>> over actual_turing_H to avoid the computability paradox!
In other words, you think it ok to compute the WRONG diagonal.
The needed diagonal was to a COMPLETE enumeration of the circle-
free machine.
NO, IT WAS SPECIFICALLY A COMPLETE ENUMERATION OF _COMPUTABLE
NUMBERS/ SEQUENCES_
Which you don't get.
smh
there is *no* requirement to include a sequence _more than once_, so
there *no* requirement to include all machines in the enumeration!
But there is to show that your partial_recognizer_D can exist, when
it can't.
YOU'VE LOST THE STANDARD PROOF IT DOESN'T EXIST,
But YOUR proof doesn't either, so the standard proof does exist.
WHAT NEW ARGUMENT WOULD THERE BE???
Don't need a NEW argument, as your LIES are just proven to be invalid.
YOU need to prove your "Russel's Teapot" exists.
Your logic just shows that you don't know how to do logic.
That isn't really different then claiming you can prove that every
even number grater than 2 from an infinite set of natural numbers
is the sum of two primes, make you enumeration just the set of
prime numbers, and thus claim that since the only even number is 2, >>>>> it is trivially solved.
_FALSE ANALOGY FALLACY_
What is false about it?
THE SUBSET IDENTIFIED BY PARTIAL_RECOGNIZER_D IS NOT "LESS POWERFUL"
THAN THE TOTAL MACHINE SPACE
Sure it is, as it doesn't make the full space.
if the set i'm enumerating includes _AT LEAST ONE INSTANCE_ of _ALL
COMPUTABLE SEQUENCES_
But does it?
YES
How are you sure.
You have an admittedly PARTIAL recognizer.
How do you know that it accepts at least one instance of a machie for
every computable number.
How does your partial decider distinguish between turing's H and your
fixed H?
then it is a turing-complete subset including all possible TM
computable sequences, and is functionally equivalent to the set of
*all* TMs
But, since your partial_recognizer_D doesn't exist, but was only
assumed to exist, your program doesn't exist,
ultimately, in order to compute a diagonal across computable
numbers, partial_recognizer_D only needs to successfully classify >>>>>> *one* of the _infinitely_ many machines that compute any given
computable number...
But it isn't the diagonal across some set of computable numbers,
but the diagonal across the ENTIRE set of computable numbers.
Since you dropped some machines, you don't have the needed set.
_*THAT DOES NOT NECESSARILY FOLLOW*_
Then try to define what set you are actually making!!
A SET THAT CONTAINS INFINITE INSTANCES OF ALL COMPUTABLE NUMBERS
Which isn't a set.
Boy are you stupid.
Sets contain no duplicates.
The problem is your definition of "paradoxical" doesn't actually
work, as it is based on asking about the results of an input, that
changes.
dropping actual_turing_H from the set, for example, is _FINE_
because it computes _THE SAME SEQUENCE_ as fixed_H, which _IS IN THE
SET_ ...
How do you know it includes your fixed_H in the set?
what part of the actual machine makes it different?
THE CODE OF FIXED_H AND ACTUAL_TURING_H IS IN THE THREAD, WHAT ARE YOU
TALKING ABOUT???
WHich aren't full machines, not until you include the code of partial_recognizer_D,
And, when you define it, how is Actual_turing_H "undecidable", since
decider exist that decide what it will do.
Remember, in the final machine given as the input, there are not
READ TURING'S PAPER ACTUALLY TO SEE HOW A SUBSET OF TRANSITION
FUNCTIONS WITHIN A TURING MACHINE CAN BE LABELED AND REFERRED TO
Yes, and HE Defined what is subset was.
You haven't, as you have a criteria that is just nonsense, as it is
basded on the input changing when you look at a different decider
looking at it.
seperate turing_H and partial_recognizer_D denoted in the machine,
just one big piece of algorithm. Thus nothing to base your criteria on.
shit rick, idk how ur still denying what's right under ur nose,
Because what you are putting under my nose is just CRAP based on
errors
goddamn rick smh u could spot a diamond in the rough even if god
lead u by ur hand to one put on a pedestal with a _GIANT_ neon sign
pointing to it...
But the problem is that "your diamond" is actually just a big load of
crap that you misidentify as a diamond, because you don't actually
understand what you are doing.
ur so used to playing contrarian u don't know what truth even looks
like anymore, rick
Nope. But you live so much in your stupidity, that you think your lies
are valid.
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that >>>>>>>>>>>> generates a computable number to build the list of such >>>>>>>>>>>> machine to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. BOTH >>>>>>>>>>> ARE INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH >>>>>>>>>>> WILL WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE >>>>>>>>>>> INPUT/ TO D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE
COMPUTABLE NUMBERS
It doesn't matter that your new H doesn't get stuck on >>>>>>>>>>>>>> itself, it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he >>>>>>>>>>>>> doesn't specify what D (or H) needs to do when encountering >>>>>>>>>>>>> the / undecidable input/ of H, so therefore both D and H >>>>>>>>>>>>> are an incomplete specifications of a machine
Because "undecidability" doesn't affect the correct answer >>>>>>>>>>>> it must compute.
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL SPACE >>>>>>>>>>> LIKE THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN
IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS >>>>>>>>>>> COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_,
BECAUSE IT DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF-
REFERENTIAL ANALYSIS, SO IT IS AN /INCOMPLETE SPECIFICATION/ >>>>>>>>>>> OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES >>>>>>>>>>> CAN'T BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON??? >>>>>>>>>>>
Note, H, when it becomes an actual machine, because we have >>>>>>>>>>>> created an actual machine we claim to be the D, has an >>>>>>>>>>>> answer, and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't >>>>>>>>>>>> exist, only machines CORRECTLY meeting the requirements of D >>>>>>>>>>>> and H can't exist.
Thus, if your H depends on that D, (or a machine that meets >>>>>>>>>>>> its specification) then it also can't exist.
You just make the error of saying the problems don't exist >>>>>>>>>>>> because you can't build them, but YOUR machine can, even >>>>>>>>>>>> though it has the same problem.
IF D is wrong by deciding it is not circle free, then your >>>>>>>>>>>>>> H will compute the wrong diagonal, as the resulting >>>>>>>>>>>>>> version of his H WILL be circle free (since it never tries >>>>>>>>>>>>>> to simulate itself) and thus DOES produce an computable >>>>>>>>>>>>>> number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then >>>>>>>>>>>>>> when you H tries to process it, it will get stuck in the >>>>>>>>>>>>>> infinite loop.
The problem is that in stepping through the machines in >>>>>>>>>>>>>> order, you WILL hit these actual machines built on your >>>>>>>>>>>>>> erroneous D (your D must have this flaw, as no D without >>>>>>>>>>>>>> exists), and thus you will be wrong on THAT input. IT >>>>>>>>>>>>>> doesn't matter if you get a good answer for yourself. >>>>>>>>>>>>>>
idk what he would have said about it, but prolly >>>>>>>>>>>>>>> something more substantial than just calling me ignorant >>>>>>>>>>>>>>> repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic >>>>>>>>>>>>>> assertion of bad logic. You would have likely been >>>>>>>>>>>>>> escorted out of the meeting as showing you were
unqualified and being a distraction.
Something that seems to be beyond your ignorant >>>>>>>>>>>>>>>> understanding.
But the problem is that "paradoxical machines" don't >>>>>>>>>>>>>>>> exist in isolation, but only in relationship to a given >>>>>>>>>>>>>>>> machine trying to decide them.
H shows that *IF* you can make that enumeration, you >>>>>>>>>>>>>>>>>> can make the diagonal, and thus the anti-diagonal. The >>>>>>>>>>>>>>>>>> problem is you can't make that enumeration, and >>>>>>>>>>>>>>>>>> assuming you can just shows unsoundness.
interestingly: one can only fix the direct diagonal >>>>>>>>>>>>>>>>> computation like this
u can't do an analogous fix for the inverse/anti- >>>>>>>>>>>>>>>>> diagonal computation. it's not possible hard code a >>>>>>>>>>>>>>>>> machine to return an inverted value, a machine can only >>>>>>>>>>>>>>>>> return what it does, not the inverse of what it does... >>>>>>>>>>>>>>>>>
so if we can filter out paradoxes from the enumeration, >>>>>>>>>>>>>>>>> that will leave a direct diagonal computation extant in >>>>>>>>>>>>>>>>> that filtered (yet still turing complete list), while >>>>>>>>>>>>>>>>> any attempt to compute an inverse diagonal will not be >>>>>>>>>>>>>>>>
right. so if ur constructing a diagonal across computable >>>>>>>>>>>>>>> numbers then u only need to filter out paradoxes in >>>>>>>>>>>>>>> regards to the classifier that classifies them as a >>>>>>>>>>>>>>> "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out such >>>>>>>>>>>>> paradoxes...
In other words, you beleive unquestionably in the existance >>>>>>>>>>>> of Russle's teapot until someone can prove it doesn't exist. >>>>>>>>>>>
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF
PROVIDING A FUCKING QUOTE TO TURING'S PAPER TO BACK THE >>>>>>>>>>> *DIRECT* CLAIM YOU MADE???
For THIS paradox, read the proof. for THIS H, the one given, >>>>>>>>>>>> (for which ever version of D you want to try to assume is >>>>>>>>>>>> right) D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the >>>>>>>>>>>> paradox, because there still isn't an answer it can give to >>>>>>>>>>>> H that will be correct, as the template for H will always >>>>>>>>>>>> make that D wrong.
Note, the problem is you can't actually DEFINE what a >>>>>>>>>>>> "paradox machine" is, as they aren't actually machines, but >>>>>>>>>>>> templates that build machines. And that final machine >>>>>>>>>>>> doesn't have actually detectable tell-tales that show it to >>>>>>>>>>>> be from that template.
(also why do always just make random assertions???)
Because I am smart, and know what I am talking about.
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE >>>>>>>>>>> TO WORK WITH, BUT OH WELL
und = () -> halts(und) loop()
They only seem "random" because you don't know what you are >>>>>>>>>>>> talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* >>>>>>>>>>>>>>> classifiable as satisfactory by said classifier... can >>>>>>>>>>>>>>> just be skipped
No, it can only skip those that are not satisfactory, not >>>>>>>>>>>>>> those that are but it can not classify as such, or your >>>>>>>>>>>>>> enumeration will not be complete, and thus just in error. >>>>>>>>>>>>>>
Thus, it needs to be able to correctly classify ALL >>>>>>>>>>>>>> machines (as it will be asked about all machines as it >>>>>>>>>>>>>> counts through all the descriptions) and thus Turing's H >>>>>>>>>>>>>> *WILL* be asked about.
similarly if u want to go a step further an filter out >>>>>>>>>>>>>>> computable numbers already included on this diagonal, any >>>>>>>>>>>>>>> machine which either *is* computably equivalent OR *is >>>>>>>>>>>>>>> not* classifiable in regards to *any* machine already the >>>>>>>>>>>>>>> list... can just be skipped
Nope, you can't skip some machines, as you then might lose >>>>>>>>>>>>>> some of the computable numbers.
see you can't compute a diagonal across *all* /machines/, >>>>>>>>>>>>>>> with said machines, but u can compute a diagonal across >>>>>>>>>>>>>>> *all* / computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be >>>>>>>>>>>>>> done, since ALL classifiers that attempt it will make an >>>>>>>>>>>>>> error, you can't do what you want to do.
nah, (a) computing an enumeration of all /computable >>>>>>>>>>>>> numbers/ is not the same thing as (b) computing the >>>>>>>>>>>>> enumeration of all machines that compute computable >>>>>>>>>>>>> numbers. (b) necessarily has duplicates while (a) does not >>>>>>>>>>>>> need them. turing's paper wrongly conflates (a) with (b) >>>>>>>>>>>>>
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you filter >>>>>>>>>>>> the full list of machibe, which Turing showed can't exist. >>>>>>>>>>>>
The problem is such a machine can't handle the
classification of the H he describes. It doesn't matter that >>>>>>>>>>>> you can make a DIFFERENT machine, that you try to
deceptively call "H" too that it won't have a problem with, >>>>>>>>>>>> when your "H" gets to the number of Turing's H, it still has >>>>>>>>>>>> the problem. It can call that one not circle-free, and thus >>>>>>>>>>>> omit a circle- free machine from the list, or call it >>>>>>>>>>>> circle- free, and when even YOU try to simulate it the k >>>>>>>>>>>> steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to >>>>>>>>>>>> take the code of (a), and invert the symbol put on the tape >>>>>>>>>>>> when it is written. Note, it won't be "self-referent", as >>>>>>>>>>>> the "N" is uses, is the N of YOUR H, not itself. The code >>>>>>>>>>>> for his (a) NEVER reads back what it wrote, so that doesn't >>>>>>>>>>>> affect its behavior.
i'm sorry, are you actually saying the machine description >>>>>>>>>>>>> does not describe what the machine does???
yes, i still do need to prove my thesis that for any >>>>>>>>>>>>>>> paradoxical machine, there exists a functionally >>>>>>>>>>>>>>> equivalent machine without such paradox
And the problem is that your "paradoxical" isn't actually >>>>>>>>>>>>>> a definable property (let alone computable). Part of the >>>>>>>>>>>>>> problem is that if you look at just a machine description, >>>>>>>>>>>>>> it doesn't (necessarily) tell you about the use of an >>>>>>>>>>>>>> "interface" as that use of an interface can be just >>>>>>>>>>>>>> inlined, leaving nothing "in the code" to show it exists. >>>>>>>>>>>>>
No, that "paradoxical" isn't a definable property of a machine. >>>>>>>>>>>
und() includes a structural paradoxical in relation to the >>>>>>>>>>> classifier halts(), easily demonstrable thru a short proof >>>>>>>>>>> i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A
PROPERTY OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS >>>>>>>>>>> DEVOTION TO SOME FUCKING FORM OF INSANITY I CAN'T HONESTLY >>>>>>>>>>> FATHOM
lol
Sure, because it is just a number. The problem is that you >>>>>>>>>>>>>> still have
His specified H, with an actually (incorrect) >>>>>>>>>>>>>>>> implementation of D (which is all that CAN exist) will >>>>>>>>>>>>>>>> either be circle- free and thus generate a number (but >>>>>>>>>>>>>>>> its D said it isn't, and thus omitted a valid machine >>>>>>>>>>>>>>>> from the list) or it isn't circle- free, and fails to >>>>>>>>>>>>>>>> computa a number, and thus should have been omitted from >>>>>>>>>>>>>>>> the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it >>>>>>>>>>>>>>>> is just built on an assuption in error.
so despite turing's worries, the existence of a >>>>>>>>>>>>>>>>> diagonal computation does not actually then imply the >>>>>>>>>>>>>>>>> existence of an anti- diagonal computation, due the >>>>>>>>>>>>>>>>> same particular self- referential weirdness that >>>>>>>>>>>>>>>>> stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic is >>>>>>>>>>>>>>>> just based on ERROR.
Your attempt to REDEFINE self-reference to mean >>>>>>>>>>>>>>>> processing a number that happens to represent yourself >>>>>>>>>>>>>>>> means that you you system "ALL" doesn't actually mean >>>>>>>>>>>>>>>> ALL, and thus is just ill- defined.
i'm sorry, you have an issue with me labeling a number >>>>>>>>>>>>>>> that directly refers to yourself, as a "self-reference" ??? >>>>>>>>>>>>>>
it's a *specific* number that has the currently running >>>>>>>>>>>>> machine encoded into it, it's not "just" a number whatever >>>>>>>>>>>>> that means
problems with all the "equivalent" machines that have >>>>>>>>>>>>>> different numbers.
those are references to functionally equivalent machines, >>>>>>>>>>>>> not self- references
Which is why the concept of "self-reference" doesn't work. >>>>>>>>>>>>
It may let you filter out the simplest case used in the >>>>>>>>>>>>>> proofs, but doesn't solve the actual problem, as the >>>>>>>>>>>>>> "Machine Number" doesn't actually fully identify the >>>>>>>>>>>>>> problematic cases.
that's not actually true. you can't meaningfully paradox >>>>>>>>>>>>> the paradox detector while filter *out* paradoxes to point >>>>>>>>>>>>> of creating an actual problem
Then show how you will do it.
This means you need to detect a input that represents a >>>>>>>>>>>> machine that uses a computation equivalent to yourself >>>>>>>>>>>>
WHich doesn't exist.
let halts be a halting classifier that takes an input >>>>>>>>>>>>> machines and returns true/false whether it halts or not >>>>>>>>>>>>>
let paradox_free be a paradox filtering classifier that >>>>>>>>>>>>> takes an input classifier, and an input machine to search >>>>>>>>>>>>> for paradoxes in regards to: paradox_free(classifier, machine) >>>>>>>>>>>>
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox >>>>>>>>>>>>> classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if case >>>>>>>>>>>>> run a halting paradox form making it /undecidable input/ to >>>>>>>>>>>>> halts().
And the problem is "paradox_free" can't exist as an always >>>>>>>>>>>> correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if >>>>>>>>>>>>> case doesn't run and clearly halts(undp) -> TRUE without >>>>>>>>>>>>> issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within a >>>>>>>>>>>>> machine is necessary for the particular input->output >>>>>>>>>>>>> computation being done, so utilizing the return
paradox_free(halts,undp) -> FALSE to filter out
paradox_free() paradoxes as well as halts() paradoxes will >>>>>>>>>>>>> suffice to produce a turing complete subset of machines >>>>>>>>>>>>> that can be totally classified by halts()
But we can still build an input that your paradox_free() >>>>>>>>>>>> fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE >>>>>>>>>>> INCLUDED ON THE DIAGONAL???
No, it is your dependence on being able to assume that >>>>>>>>>>>> unicorns can solve all your problems.
i think a major problem with the consensus perspective is >>>>>>>>>>>>> an expectation of a certain cleanliness in the logic as >>>>>>>>>>>>> being the only possibility here to compute what we want. >>>>>>>>>>>>> TM's can't achieve that, but something slightly messier can. >>>>>>>>>>>>
They are, But you are too stuck in your lies to see it. >>>>>>>>>>>>
lol, i can only 🙏 someone would engage in sound logic with me >>>>>>>>>>>>
dear future: u see this shit i'm trying work with???? >>>>>>>>>>>>>>> sorry i didn't manage to make progress any faster, but >>>>>>>>>>>>>>> i'm literally dragging toddlers kicking and screaming at >>>>>>>>>>>>>>> this point jeez...
In other words, in your world "ALL" isn't a word that can >>>>>>>>>>>>>> actually be used.
You don't understand that you need to use actual sound logic, >>>>>>>>>>>>>
Your "logic" is based on being able to assume that Unicorn's >>>>>>>>>>>> exist, and that is a safe assumption until someone can prove >>>>>>>>>>>> that Russel's Teapot is not out there. (Even though your >>>>>>>>>>>> unicorns HAVE been proved to not exist).
You are just proving your utter stupidity.
This just shows that you real problem is you don't >>>>>>>>>>>>>>>> understand what the actual problem is, and your world is >>>>>>>>>>>>>>>> just build on things that are lies.
...which u won't find interesting, but i can't fix a >>>>>>>>>>>>>>>>> lack of curiosity
🤷
expected was an inexpert audience, supposing that >>>>>>>>>>>>>>>>>>>> experts will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>>>>>>
IT seems that fundamentally, your world doesn't actually >>>>>>>>>>>>>>>> have "computations" as you don't understand the basic >>>>>>>>>>>>>>>> requirement that they need to be fully defined in the >>>>>>>>>>>>>>>> actions they do.
On 3/6/26 8:09 PM, dart200 wrote:
On 3/6/26 1:03 PM, Chris M. Thomasson wrote:
On 3/6/2026 12:10 PM, Richard Damon wrote:
[...]
You have a lot of patience with Olcott 2.0! It calls you every name
in the book.
fking pussy can't even read this
--
arising us out of the computing dark ages,
please excuse my pseudo-pyscript,
~ the little crank that could
Except what you don't understand, because your mind is too much into the darkness, that you are actually trying to drive computing into the dark
ages of not knowing what it can do because you claim we can just assume
what we want.
You are the just little crank that THINKS it can, but is just lying to itself.
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:i don't see any indication that turing realized a >>>>>>>>>>>>>>>>>> difference there
On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> ...
Eh?!an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on
turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he >>>>>>>>>>>>>>>>>>>>>> defines a comptuation
that enumerates out all the numbers, testing each >>>>>>>>>>>>>>>>>>>>>> one of they represent
a "satisfactory"/"circle-free" machine, and adding >>>>>>>>>>>>>>>>>>>>>> that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to carefully >>>>>>>>>>>>>>>>>>>>>> read p247 of turing's
proof and produce the psuedo-code for the machine >>>>>>>>>>>>>>>>>>>>>> H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are >>>>>>>>>>>>>>>>>>>>> you sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>>>>> procedures, and understanding the diagonal procedure >>>>>>>>>>>>>>>>>>>> is critical to understanding the *base* >>>>>>>>>>>>>>>>>>>> contradiction/ paradox that the rest of his support >>>>>>>>>>>>>>>>>>>> for godel's result is then built on
And focusing on what is said to be impossible and not >>>>>>>>>>>>>>>>>>> changing the problem is also important.
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>>>>>> enumerating the enumeration in the first place. >>>>>>>>>>>>>>>>>>
Then you zre just showing your stupidity, because YOU >>>>>>>>>>>>>>>>> can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>>>>>> enumeration of the computable sequences.
He then points out that he can directly show that the >>>>>>>>>>>>>>>>> "anti- diagonal" of the (non-effectively computed) >>>>>>>>>>>>>>>>> enumeration can't be computed but that "This proof, >>>>>>>>>>>>>>>>> although perfectly sound, has the disadvantage that it >>>>>>>>>>>>>>>>> may leave the reader with a feeling that 'there must be >>>>>>>>>>>>>>>>> something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being >>>>>>>>>>>>>>> done.
I think he is refering he to the standard anti-diagonal >>>>>>>>>>>>>>> arguement, which shows that since for all n, position n >>>>>>>>>>>>>>> differs from the value in number n, there can not be any >>>>>>>>>>>>>>> element that matches the anti- diagonal.
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>> something it seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with >>>>>>>>>>>>>> turing's short diagonal proof, but in writing this post i >>>>>>>>>>>>>> now find myself in a subtle, yet entirely critical >>>>>>>>>>>>>> disagreement:
/let an be the n-th computable sequence, and let φn(m) be >>>>>>>>>>>>>> the m- th figure in an. Let β be the sequence with 1-φn(m) >>>>>>>>>>>>>> as its n-th. figure. Since β is computable, there exists a >>>>>>>>>>>>>> number K [== β] such that 1- φn(n) = φK(n) for all n. >>>>>>>>>>>>>> Putting n = K, we have 1 = 2φK(K), i.e. 1 is even. This is >>>>>>>>>>>>>> impossible/
i agree with this proof is far as much as if β was
computable (by TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the >>>>>>>>>>>> computability of β from the existence of φn(m), THAT'S THE >>>>>>>>>>>> FALLACy
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE >>>>>>>>>>>> SPECIFICATION
the fallacy here is assuming that because the direct >>>>>>>>>>>>>> diagonal is computable, that one can therefore compute the >>>>>>>>>>>>>> anti- diagonal using the direct diagonal. the abstract >>>>>>>>>>>>>> definition makes it look simple, but this ignores the >>>>>>>>>>>>>> complexities of self- referential analysis (like what >>>>>>>>>>>>>> turing details on the next page)
But you can, *IF* you have a machine that computes the >>>>>>>>>>>>> diagonal, then just change all the write to the output to >>>>>>>>>>>>> write the opposite. Note, the "self-reference" that you are >>>>>>>>>>>>> thinking of stops being a "self- reference" but is a >>>>>>>>>>>>> reference to the original write the diagonal code.
Nope, because your filtering TM (or RTM) will still fail >>>>>>>>>>>>> when it gets to the number of TURING'S H, as there is no >>>>>>>>>>>>> correct answer for the machine built by that template. >>>>>>>>>>>>
in both methods i have for rectifying the paradox found in >>>>>>>>>>>>>> the direct diagonal (either (1) filtering TMs or (2) using >>>>>>>>>>>>>> RTMs), neither can be used to then compute the anti-diagonal >>>>>>>>>>>>>
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered >>>>>>>>>>>>>> out like turing's paradoxical variation of the direct >>>>>>>>>>>>>> diagonal would be, and there is no analogous non- >>>>>>>>>>>>>> paradoxical variation that has a hard coded value that is >>>>>>>>>>>>>> inverse to what it does return ... such a concept is >>>>>>>>>>>>>> entirely nonsensical. a function can only return what it >>>>>>>>>>>>>> does, it can't also return the inverse to what it returns >>>>>>>>>>>>>> eh???
But, if D filters it out, then it becomes circle-free, and >>>>>>>>>>>>> thus your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT >>>>>>>>>>>> *SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with >>>>>>>>>>>>>> RTMs just fails for reasons u'd only understand by working >>>>>>>>>>>>>> thru the algo urself (p7 of re: turing's diagonals) >>>>>>>>>>>>>>
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>
is just not sufficient evidence that such β is actually >>>>>>>>>>>>>> computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how >>>>>>>>>>>>> to do that, as you are thinking the only "reference" can be >>>>>>>>>>>>> to "self", doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF- >>>>>>>>>>>> REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H >>>>>>>>>>>>
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS >>>>>>>>>>>> FOR INVERSE- H TO BE COMPUTED
Then how do you build YOUR H without that D?He doesn't. You are just showing your stupidity,
one cannot just assume that because the diagonal across >>>>>>>>>>>>>>>> computable numbers is computable, therefore the anti- >>>>>>>>>>>>>>>> diagonal across computable numbers is computable... >>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and >>>>>>>>>>>>>>> without the enumeration, you can't compute either of them. >>>>>>>>>>>>>>>
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>> computation across the computable numbers can be used to >>>>>>>>>>>>>>>> compute the inverse diagonal
But your method still doesn't let you compute the >>>>>>>>>>>>>>> enumeration, and thus you can't actually compute the >>>>>>>>>>>>>>> diagonal.
Remember, the problem definitions requires that the >>>>>>>>>>>>>>> listing be a COMPLETE listing of the computable numbers / >>>>>>>>>>>>>>> machine that compute computable numbers, in some definite >>>>>>>>>>>>>>> order.
If your enumeration isn't complete, your diagonal isn't >>>>>>>>>>>>>>> correct.
so while i agree with turing that the anti-diagonal is >>>>>>>>>>>>>>>> not computable, i don't agree that the normal diagonal >>>>>>>>>>>>>>>> is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all >>>>>>>>>>>>>>> the other machines, including his original H that doesn't >>>>>>>>>>>>>>> use your "trick"
But instead, he can prove with a more obvious process, >>>>>>>>>>>>>>>>> that the Decider "D" that could be used to effectively >>>>>>>>>>>>>>>>> enumerate the sequence of machine that produce >>>>>>>>>>>>>>>>> computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing >>>>>>>>>>>>>>>>> out that the attempt to compute the diagonal clearly >>>>>>>>>>>>>>>>> reveals the issue with effectively enumerating the >>>>>>>>>>>>>>>>> sequences.
well, he didn't consider that perhaps the proper algo >>>>>>>>>>>>>>>> for computing the diagonal can avoid the paradox on >>>>>>>>>>>>>>>> itself ...
But it doesn't.
Your just don't understand that D just can't correctly >>>>>>>>>>>>>>> decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on >>>>>>>>>>>>>> turing's H, because my response to this is that D does not >>>>>>>>>>>>>> need to decide correctly on H to compute a diagonal >>>>>>>>>>>>>
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT >>>>>>>>>>>> WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial
recognizer D *can* decide on the fixed H because the fixed H >>>>>>>>>>> does not try to use any D on itself, so no self-referential >>>>>>>>>>> paradox is possible in regards to it's own digit on the diagonal >>>>>>>>>>>
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a
dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer D >>>>>>>>>>> (used in fixed H, which does not call any D itself), and >>>>>>>>>>> would not be filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and >>>>>>>>>>> instead returns a hard-coded value for it's own digit on the >>>>>>>>>>> diagonal, is keystone in making it /decidable input/ to D >>>>>>>>>>>
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just >>>>>>>>>> anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it >>>>>>>>>> will fail on that.
bro what part of the non-existence of turing_D or turing_H do >>>>>>>>> you not understand???
But if your fD exists, then Turing_H that uses your D exists, >>>>>>>> and will give that D the problem, and fixed_H will call your D >>>>>>>> on turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H >>>>>>>>>> gets hung up.
If D errs on Turing_H and calls it not circle_free, then your >>>>>>>>>> enumeration is missing the circle_free H from its list, and >>>>>>>>>> thus doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D >>>>>>>>> in a turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>> satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D
classifier, when actually run actual_turing_H will simply skip >>>>>>>>> putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the >>>>>>>> actual_turing_H isn't an "undecidable input", as it has definite >>>>>>>> behavior, just behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of
undecidability. just because a machine is /undecidable input/ to >>>>>>> a particular classifier does mean we cannot then prove and know >>>>>>> ourselves what the machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs
are not "undecidable", but by the time that they have actually
been made into an input, and thus first had the program created,
which required creating the instance of the decider selctected,
they have definite behavior that other some decider can determine. >>>>>>
It is only that one selected candidate decider chosen to build the >>>>>> input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a >>>>>>> particular machine and the particular classifier it creates a
structural paradox for,
In other words, you don't understand what "undecidability"
actually is.
Undecidability, means that there does not exist ANY decider that
can get the right answer for all instances of the problem, (or if >>>>>> the problem has been reduced to just that one input, for that input). >>>>>>
*not* as a general inability to describe a *real* machine's
behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D.
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and prove it's
equivalent to fixed_H, so therefore it doesn't fit what u stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a
machine, unless you are narrowing the problem to just being about that
one machine/input.
because it *only* ever *was* about problematic input to *particular* classifiers...
it's not my problem everyone went for over a century confusing *our*
ability to prove outcomes, with what a /particular/ decider can
output... ya'll got caught up in a rigor of what is fundamentally nonsense
no one has demonstrated a *possible* machine, let alone a *real*
machine, which we can't provably can't prove the outcome for... such a
proof makes the machine _NEITHER POSSIBLE NOR REAL_ so it would be
ludicrous to present such a proof to exist
both turing_H and turing_D are /incomplete specifications/ and therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_
they are just malformed /incomplete specifications/ and _NOTHING MORE_,
and _DESERVE NO FURTHER CONSIDERATION_ beyond the recognition that such
an interface _IS INCOMPLETE_
And, it is an assertion that NO decider can correctly compute the
answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which means
you have chosen a particular (broken) implementation of D (since no
actually correct D exists), like your partial_recognizer_D, has fulled
defined behavior, and a decider can exist that determines it, there is
nothing "undecidable" about that input.
It is just that the D that H was built on is wrong. Nothing
"paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you just _IGNORING_ the specification i specifically created for partial_recognizer_D...
the specification for partial_recognizer_D is when encountering
undecidable input is to utilize FALSE,
this _DOES NOT MEAN_ we cannot know what the /undecidable input/ does,
as a paradox in regards to partial_recognize_D _ONLY AFFECTS_ partial_recognize_D not _OUR ABILITY TO PROVE OUTCOMES_
(you don't have one btw, so please do let me witness ur cope)
IT seesm you just can't read.
It doesn't need to. It just needs to be built on what you want to
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
see, turing_H's runtime is not describable because it's an /
incomplete specification/ of machine and therefor _does not even >>>>>>> exist_ as a real TM, not because of the hypothesized
undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED defined >>>>>> machine. If the decider "D" actually exists, then H is FULLY defined. >>>>>
specification/, as it doesn't handle idiosyncrasies of TM
computing, specifically self- referential set-classification paradoxes >>>>
claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as your
equivalent for D, like your partial_recognizer_D, then the template
turing_H can be converted into an actual machine.
actual_turing_H, on the other hand, actually does exist, so we
can know/ prove what it does. it, however, still forms a
structural paradox in regards to partial_recognizer_D which will >>>>>>> fail to classify it as circle-free
But, since you admit to actual_turing_H existing, that means that
since partial_recognizer_D _CAN_ exist, then actual_turing_H _CAN_
exist
And so can the turing_H that uses partial_recognizer_D as its
decider D.
The steps the algorithm does is what defines the machine, not the
"names" of the things it "calls", as those names don't actually
exist in the final building of the machine, which is the problem
with your concept, as that is needed to determine your "paradoxical"
property.
the names are just shortcuts for the literal values rick, don't be daft
Right, but the literal values don't have the names on them, and we
don't have borders within the machine letting us know we moved from
"outer code of H" into "the code of the D it uses", especially with
Turing Machines were such usages are by necessity "expanded" in-line.
please do actually read turing's paper sometime, he does organize
machine descriptions into functional groups
Thus, the "copy" of D that was pulled into H isn't actually the same
no it can be literally a copy of same description than is the run by a
UTM (universal turing machine) functional group of the machine
thing as the independent machine D that we can talk about, just a
functional equivalent which doesn't even have its own D.N (since that
is a property only of full machines)
ur really pulling at straws here rick
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get >>>>>>>> wrong, then it can just return the different answer, and then >>>>>>>> THAT machine it will get right. The problem is it creates a
DIFFERENT input, that uses this new version of the machine, that >>>>>>>> it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as
fixed_H, which in turn _does_ have a spot on the diagonal. so >>>>>>>>> there is no actual need to include a digit from actual_turing_H >>>>>>>>> on the diagonal, as the number it computes is already included >>>>>>>>> on the diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of >>>>>>>> actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can
possible filter all ALL of that infinite set of machines, but >>>>>>>> still accept some machine that computes that particular number >>>>>>>> that it would compute with this supposedly correct
partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting >>>>>>>> that you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the
fairy dust, of assuming that your partial decider that meets ALL >>>>>>>> your needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between >>>>>>> fixed_H and actual_turing_H is the way they respectively handle >>>>>>> their own self- references. they function identically when
handling all other machines
But the problem isn't in it handling the machine fixed_H, but how >>>>>> it ever handles the results generated by actual_turing_H, which by >>>>>> your assumption has become a cycle-free machine and ALL equivalent >>>>>> versions of it have been partially declared to be non-cycle free.
This means that your enumeration of machines you are looking at
is, in fact, incomplete, thus the diagonal you computed is wrong.
fixed_H = () -> {
N = 1
K = 0
output = [] // written to
F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded
digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>> satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip
trying to simulate itself and put it's hard coded digit on the
diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip
trying simulate actual_turing_H because partial_recognizer_D
fails to classify it as circle-free
But actual_turing_H is actually circle_free, and thus its results >>>>>> SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put on
the diagonal, then actual_turing_H _DOES NOT_ need to be in _THAT_
total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable
numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on being
sloppy.
the only way to compute a diagonal across computable numbers is
simulating each number by the machine that computes it.
Note, Turing SPECIFICALLY pointed out that he was going to the related
problem, the problem of:
/but the problem of enumerating computable sequences is equivalent to
the problem of finding out whether a given number is the D.N of a
circle- free machine, and we have no general process for doing this in
a finite number of steps/
oh wow GREAT CATCH RICK!!!!,
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT ENUMERATING
OUT ALL CIRCLE_FREE MACHINES*,
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER IS A CIRCLE-FREE MACHINE,
we only need to compute that fact for *ONE* machine for any given
computable sequence, _SO A PARTIAL RECOGNIZER IS SUFFICIENT FOR THE JOB_
fuck rick ur so bad at defending this it's actually really useful to me!
(there's *another* specific sentence i have a problem with, where's a
ben when u need him?)
So, HIS enumeration is the complete enumeration of "circle-free
machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration either, why
am i the only one subject to criticism here???
the *specification* /works/, which shows that it's *not*
Also, your proof is built on the existance of a Russel Teapot, you have >>>
contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion. Your
arguement boils down to showing that *IF* you can correctly determine
all machines that fit in a not-definable category (since you concept
of [paradoxical can't actually be defined at the machine level, only
at how a machine might have been created at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can make
the precise "mistakes" you need it to. (They are mistakes because
the input
"mistake" is this case is a fallacious non-techincal value judgement
on the return value,
No, since your criteria that you want to use is just an undefinable
term for a machine description.
There is nothing "paradoxical" of the actual machine when created, as
it then has fixed and determined behavior that can be decided on. The
thing that is paradoxical is the mete-system pattern used to generate
the class of machines as pair with a given decider.
cause they are rather entirely within specification, as FALSE was
specified to be used upon encountering an /undecidable input/,
something turing_D did not handle correctly (making it _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is based on a
actual D that does make a specific decision to the input. And it is
just that for any given version of D, we can find a DIFFERENT H that
it will get wrong (the H that was just built on it).
Do you want to claim that ALL these H are "undecidable" even though for
no... fixed_H contains no paradox, and is therefor not /undecidable
input/ to any classifier
all of them there does exist a decider that gets the right answer?
many machines contain no paradox and are therefore classifiable by all classifiers
It seems you just don't know what the word means, and thus you
specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of this
turing_H is built on this partial_recognizdr_D which causes them to
not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to machines, as
it is based on the incorrect definition that assumes a machine can
be something other than itself, and when it changes, it changes
other things that at the meta/template level refered to its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just refuse to
learn.
Try to actually fully DEFINE the word based on the actual properties
of the input machine, and not refering to other machines that are not
part of the input (but would be created as alternate inputs to foil an
alternate decider).
when actual_turing_H iterates across DN(actual_turing_H) itself, >>>>>>> it will similarly skip trying to simulate itself, because
partial_recognizer_D fails to classify it as circle-free
Right, so it ALSO is wrong, as it didn't compute the diagonal of
the circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE NUMBERS_,
which is the _ACTUAL PROBLEM_ turing was dealing with,
No, it isn't, as your "set of computable numbers" has duplicates.
it's still a diagonal across all computerate numbers even if it has
dpulicates ...
No, because an enumeration of a set doesn't HAVE duplicates.
then the H's as of now are i suppose "generators" instead of enumerators
and turing was wrong about that too, but i don't see u criticizing him
now eh???
Your probe
this level of objection is dishonest\
No, your using misdefined terms is the real dishonesty.
If you want to show that you can make a diagonal of computable
numbers, you need to build an actual effective enumeration of
computable numbers.
And the problem is you can't assume your decider and then use it. That
isn't vallid logic.
but i can show turing was wrong about a series of points, in the most influence math paper of last century,
that's good enough start for me
i can do more once i gain some recognition with that
It IS valid to assume a decider, and show that such an assumption
leads to a contradiction, to prove that such a decider can not exist.
That is the valid proof by contradiction.
If we can show that if A exist, B must be true, and also that B can
not be true, then we have proven that A can not exist.
Assuming that A exists, and showing that one line of reasoning based
on it doesn't lead to a contradiction, does NOT show that A does (or
even can) exist. That is just fallacious logic.
doing so doesn't require /all/ or _even most_ machines to be in the >>>>> enumeration, since _most_ are duplicates of some other machine
when actual_turing_H iterates across DN(fixed_H),
partial_recognizer_D *successfully* classifies fixed_H as
satisfactory, so it will simulate fixed_H to it's Kth digit, the >>>>>>> one hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular
logic in either case), fixed_H also skips simulating itself (also >>>>>>> circular logic), while actual_turing_H actually *does* simulate >>>>>>> fixed_H successfully for it's Kth digit on the diagonal. these
machines compute the *same* number/sequence!
Except that by your partial_recognizer being just partial, it has >>>>>> dropped required items from the enumeration.
_WHICH IS FINE_
So, you admit it failed???
ACTUAL_TURING_H IS _NOT_ REQUIRED TO BE ON THE DIAGONAL, SO IT'S FINE
TO BE DROPPED
If the diagonal is the diagonal of circle-free machines, it is required.
Your problem is you keep on working with strawmen.
If the diagonal is of computable numbers, you need to show how you
trim down the set to not have repeats (in finite time per element) and
that your decider actually exists, you can't just assume it.
As I said, it dropped REQUIRED items, as this enumeration was of
machines that computed computable numbers.
If you try to add removal of duplicates, then we need to define the
order, and if we do it by number, turing_H will likely be lower in
order, as it doesn't need to generate its own number, and thus is
simpler,
that sentence is nonsense until u actually produce psuedo-code
I did, just make my H call your "D" replacement, partial_recognizer_D.
Or, is your problem that you think an enumeration doesn't define an
"order"
THis is becides the fact that you still need to show how you plan >>>>>> on actually detecting all these computationally equivalent
machines in your
fixed_H does not compute a minimum turing-complete set, there are
duplicated sequences, ei machines that compute the same sequence/
number, and _THAT IS FINE_
Why?
BECAUSE AN ENUMERATION THAT INCLUDES ONE OR MORE INSTANCE FOR EACH
COMPUTABLE NUMBER STILL ENUMERATES *ALL* COMPUTABLE NUMBERS
Nope. Maybe you don't understand what an enumeration.
And enumeration is a one-to-one mapping between the elements and the
Natural Numbers.
By your definition, you can make a countable infinite enumeation of a
finite set, showing that the finite set is infinite.
And, it was built on the assumption of the existance of a machine,
that actually can not exist (based on Rice's Theorm) and thus is
actually based on the assumption of Unicorns.
partial_recognizer_D is *not* a total decider, so how could rice's
theorum apply???
But it *IS* a total decider on your "paradoxical" property.
It can't let ANY of them through, and it must not omit any of the ones
you need.
partial decider to force them to not be classified as circle-free, >>>>>> even though it turns out that they are (because your decider
called them not).
You don't seem to understand that you don't get to look at the
template used to build the machine, only the resultant machine
from that template.
And, part of the result of that, is that your attempt to try to
define "paradoxical" just fails, as it resolves to just input your >>>>>> decider is wrong about, in part, because you don't understand what >>>>>> "decidable" actually means,
In other words, you think it ok to compute the WRONG diagonal.
see, not only is defining a value in fixed_H that the diagonal
computation can put for itself in diagonal essential in making
the diagonal actually computable, so is partial_recognizer_D
skipping over actual_turing_H to avoid the computability paradox! >>>>>>
The needed diagonal was to a COMPLETE enumeration of the circle-
free machine.
NO, IT WAS SPECIFICALLY A COMPLETE ENUMERATION OF _COMPUTABLE
NUMBERS/ SEQUENCES_
Which you don't get.
smh
But there is to show that your partial_recognizer_D can exist, when
there is *no* requirement to include a sequence _more than once_,
so there *no* requirement to include all machines in the enumeration! >>>>
it can't.
YOU'VE LOST THE STANDARD PROOF IT DOESN'T EXIST,
But YOUR proof doesn't either, so the standard proof does exist.
WHAT NEW ARGUMENT WOULD THERE BE???
Don't need a NEW argument, as your LIES are just proven to be invalid.
YOU need to prove your "Russel's Teapot" exists.
Your logic just shows that you don't know how to do logic.
That isn't really different then claiming you can prove that every >>>>>> even number grater than 2 from an infinite set of natural numbers >>>>>> is the sum of two primes, make you enumeration just the set of
prime numbers, and thus claim that since the only even number is
2, it is trivially solved.
_FALSE ANALOGY FALLACY_
What is false about it?
THE SUBSET IDENTIFIED BY PARTIAL_RECOGNIZER_D IS NOT "LESS POWERFUL"
THAN THE TOTAL MACHINE SPACE
Sure it is, as it doesn't make the full space.
if the set i'm enumerating includes _AT LEAST ONE INSTANCE_ of _ALL >>>>> COMPUTABLE SEQUENCES_
But does it?
YES
How are you sure.
You have an admittedly PARTIAL recognizer.
How do you know that it accepts at least one instance of a machie for
every computable number.
How does your partial decider distinguish between turing's H and your
fixed H?
then it is a turing-complete subset including all possible TM
computable sequences, and is functionally equivalent to the set of
*all* TMs
But, since your partial_recognizer_D doesn't exist, but was only
assumed to exist, your program doesn't exist,
ultimately, in order to compute a diagonal across computable
numbers, partial_recognizer_D only needs to successfully classify >>>>>>> *one* of the _infinitely_ many machines that compute any given
computable number...
But it isn't the diagonal across some set of computable numbers,
but the diagonal across the ENTIRE set of computable numbers.
Since you dropped some machines, you don't have the needed set.
_*THAT DOES NOT NECESSARILY FOLLOW*_
Then try to define what set you are actually making!!
A SET THAT CONTAINS INFINITE INSTANCES OF ALL COMPUTABLE NUMBERS
Which isn't a set.
Boy are you stupid.
Sets contain no duplicates.
The problem is your definition of "paradoxical" doesn't actually
work, as it is based on asking about the results of an input, that
changes.
dropping actual_turing_H from the set, for example, is _FINE_
because it computes _THE SAME SEQUENCE_ as fixed_H, which _IS IN
THE SET_ ...
How do you know it includes your fixed_H in the set?
what part of the actual machine makes it different?
THE CODE OF FIXED_H AND ACTUAL_TURING_H IS IN THE THREAD, WHAT ARE
YOU TALKING ABOUT???
WHich aren't full machines, not until you include the code of
partial_recognizer_D,
And, when you define it, how is Actual_turing_H "undecidable", since
decider exist that decide what it will do.
Remember, in the final machine given as the input, there are not
READ TURING'S PAPER ACTUALLY TO SEE HOW A SUBSET OF TRANSITION
FUNCTIONS WITHIN A TURING MACHINE CAN BE LABELED AND REFERRED TO
Yes, and HE Defined what is subset was.
You haven't, as you have a criteria that is just nonsense, as it is
basded on the input changing when you look at a different decider
looking at it.
seperate turing_H and partial_recognizer_D denoted in the machine,
just one big piece of algorithm. Thus nothing to base your criteria on. >>>>
shit rick, idk how ur still denying what's right under ur nose,
Because what you are putting under my nose is just CRAP based on
errors
goddamn rick smh u could spot a diamond in the rough even if god
lead u by ur hand to one put on a pedestal with a _GIANT_ neon sign >>>>> pointing to it...
But the problem is that "your diamond" is actually just a big load
of crap that you misidentify as a diamond, because you don't
actually understand what you are doing.
ur so used to playing contrarian u don't know what truth even looks
like anymore, rick
Nope. But you live so much in your stupidity, that you think your lies
are valid.
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that >>>>>>>>>>>>> generates a computable number to build the list of such >>>>>>>>>>>>> machine to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. >>>>>>>>>>>> BOTH ARE INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH >>>>>>>>>>>> WILL WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE >>>>>>>>>>>> INPUT/ TO D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE
COMPUTABLE NUMBERS
Because "undecidability" doesn't affect the correct answer >>>>>>>>>>>>> it must compute.
It doesn't matter that your new H doesn't get stuck on >>>>>>>>>>>>>>> itself, it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he >>>>>>>>>>>>>> doesn't specify what D (or H) needs to do when
encountering the / undecidable input/ of H, so therefore >>>>>>>>>>>>>> both D and H are an incomplete specifications of a machine >>>>>>>>>>>>>
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL >>>>>>>>>>>> SPACE LIKE THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN >>>>>>>>>>>> IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS >>>>>>>>>>>> COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, >>>>>>>>>>>> BECAUSE IT DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF- >>>>>>>>>>>> REFERENTIAL ANALYSIS, SO IT IS AN /INCOMPLETE SPECIFICATION/ >>>>>>>>>>>> OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES >>>>>>>>>>>> CAN'T BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON??? >>>>>>>>>>>>
Note, H, when it becomes an actual machine, because we have >>>>>>>>>>>>> created an actual machine we claim to be the D, has an >>>>>>>>>>>>> answer, and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't >>>>>>>>>>>>> exist, only machines CORRECTLY meeting the requirements of >>>>>>>>>>>>> D and H can't exist.
Thus, if your H depends on that D, (or a machine that meets >>>>>>>>>>>>> its specification) then it also can't exist.
You just make the error of saying the problems don't exist >>>>>>>>>>>>> because you can't build them, but YOUR machine can, even >>>>>>>>>>>>> though it has the same problem.
IF D is wrong by deciding it is not circle free, then >>>>>>>>>>>>>>> your H will compute the wrong diagonal, as the resulting >>>>>>>>>>>>>>> version of his H WILL be circle free (since it never >>>>>>>>>>>>>>> tries to simulate itself) and thus DOES produce an >>>>>>>>>>>>>>> computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then >>>>>>>>>>>>>>> when you H tries to process it, it will get stuck in the >>>>>>>>>>>>>>> infinite loop.
The problem is that in stepping through the machines in >>>>>>>>>>>>>>> order, you WILL hit these actual machines built on your >>>>>>>>>>>>>>> erroneous D (your D must have this flaw, as no D without >>>>>>>>>>>>>>> exists), and thus you will be wrong on THAT input. IT >>>>>>>>>>>>>>> doesn't matter if you get a good answer for yourself. >>>>>>>>>>>>>>>
idk what he would have said about it, but prolly >>>>>>>>>>>>>>>> something more substantial than just calling me ignorant >>>>>>>>>>>>>>>> repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic >>>>>>>>>>>>>>> assertion of bad logic. You would have likely been >>>>>>>>>>>>>>> escorted out of the meeting as showing you were >>>>>>>>>>>>>>> unqualified and being a distraction.
Something that seems to be beyond your ignorant >>>>>>>>>>>>>>>>> understanding.
H shows that *IF* you can make that enumeration, you >>>>>>>>>>>>>>>>>>> can make the diagonal, and thus the anti-diagonal. >>>>>>>>>>>>>>>>>>> The problem is you can't make that enumeration, and >>>>>>>>>>>>>>>>>>> assuming you can just shows unsoundness.
interestingly: one can only fix the direct diagonal >>>>>>>>>>>>>>>>>> computation like this
u can't do an analogous fix for the inverse/anti- >>>>>>>>>>>>>>>>>> diagonal computation. it's not possible hard code a >>>>>>>>>>>>>>>>>> machine to return an inverted value, a machine can >>>>>>>>>>>>>>>>>> only return what it does, not the inverse of what it >>>>>>>>>>>>>>>>>> does...
so if we can filter out paradoxes from the >>>>>>>>>>>>>>>>>> enumeration, that will leave a direct diagonal >>>>>>>>>>>>>>>>>> computation extant in that filtered (yet still turing >>>>>>>>>>>>>>>>>> complete list), while any attempt to compute an >>>>>>>>>>>>>>>>>> inverse diagonal will not be
But the problem is that "paradoxical machines" don't >>>>>>>>>>>>>>>>> exist in isolation, but only in relationship to a given >>>>>>>>>>>>>>>>> machine trying to decide them.
right. so if ur constructing a diagonal across >>>>>>>>>>>>>>>> computable numbers then u only need to filter out >>>>>>>>>>>>>>>> paradoxes in regards to the classifier that classifies >>>>>>>>>>>>>>>> them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out >>>>>>>>>>>>>> such paradoxes...
In other words, you beleive unquestionably in the existance >>>>>>>>>>>>> of Russle's teapot until someone can prove it doesn't exist. >>>>>>>>>>>>
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF >>>>>>>>>>>> PROVIDING A FUCKING QUOTE TO TURING'S PAPER TO BACK THE >>>>>>>>>>>> *DIRECT* CLAIM YOU MADE???
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE >>>>>>>>>>>> TO WORK WITH, BUT OH WELL
For THIS paradox, read the proof. for THIS H, the one >>>>>>>>>>>>> given, (for which ever version of D you want to try to >>>>>>>>>>>>> assume is right) D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the >>>>>>>>>>>>> paradox, because there still isn't an answer it can give to >>>>>>>>>>>>> H that will be correct, as the template for H will always >>>>>>>>>>>>> make that D wrong.
Note, the problem is you can't actually DEFINE what a >>>>>>>>>>>>> "paradox machine" is, as they aren't actually machines, but >>>>>>>>>>>>> templates that build machines. And that final machine >>>>>>>>>>>>> doesn't have actually detectable tell-tales that show it to >>>>>>>>>>>>> be from that template.
Because I am smart, and know what I am talking about. >>>>>>>>>>>>
(also why do always just make random assertions???) >>>>>>>>>>>>>
They only seem "random" because you don't know what you are >>>>>>>>>>>>> talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* >>>>>>>>>>>>>>>> classifiable as satisfactory by said classifier... can >>>>>>>>>>>>>>>> just be skipped
No, it can only skip those that are not satisfactory, not >>>>>>>>>>>>>>> those that are but it can not classify as such, or your >>>>>>>>>>>>>>> enumeration will not be complete, and thus just in error. >>>>>>>>>>>>>>>
Thus, it needs to be able to correctly classify ALL >>>>>>>>>>>>>>> machines (as it will be asked about all machines as it >>>>>>>>>>>>>>> counts through all the descriptions) and thus Turing's H >>>>>>>>>>>>>>> *WILL* be asked about.
similarly if u want to go a step further an filter out >>>>>>>>>>>>>>>> computable numbers already included on this diagonal, >>>>>>>>>>>>>>>> any machine which either *is* computably equivalent OR >>>>>>>>>>>>>>>> *is not* classifiable in regards to *any* machine >>>>>>>>>>>>>>>> already the list... can just be skipped
Nope, you can't skip some machines, as you then might >>>>>>>>>>>>>>> lose some of the computable numbers.
see you can't compute a diagonal across *all* / >>>>>>>>>>>>>>>> machines/, with said machines, but u can compute a >>>>>>>>>>>>>>>> diagonal across *all* / computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be >>>>>>>>>>>>>>> done, since ALL classifiers that attempt it will make an >>>>>>>>>>>>>>> error, you can't do what you want to do.
nah, (a) computing an enumeration of all /computable >>>>>>>>>>>>>> numbers/ is not the same thing as (b) computing the >>>>>>>>>>>>>> enumeration of all machines that compute computable >>>>>>>>>>>>>> numbers. (b) necessarily has duplicates while (a) does not >>>>>>>>>>>>>> need them. turing's paper wrongly conflates (a) with (b) >>>>>>>>>>>>>>
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you >>>>>>>>>>>>> filter the full list of machibe, which Turing showed can't >>>>>>>>>>>>> exist.
The problem is such a machine can't handle the
classification of the H he describes. It doesn't matter >>>>>>>>>>>>> that you can make a DIFFERENT machine, that you try to >>>>>>>>>>>>> deceptively call "H" too that it won't have a problem with, >>>>>>>>>>>>> when your "H" gets to the number of Turing's H, it still >>>>>>>>>>>>> has the problem. It can call that one not circle-free, and >>>>>>>>>>>>> thus omit a circle- free machine from the list, or call it >>>>>>>>>>>>> circle- free, and when even YOU try to simulate it the k >>>>>>>>>>>>> steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to >>>>>>>>>>>>> take the code of (a), and invert the symbol put on the tape >>>>>>>>>>>>> when it is written. Note, it won't be "self-referent", as >>>>>>>>>>>>> the "N" is uses, is the N of YOUR H, not itself. The code >>>>>>>>>>>>> for his (a) NEVER reads back what it wrote, so that doesn't >>>>>>>>>>>>> affect its behavior.
yes, i still do need to prove my thesis that for any >>>>>>>>>>>>>>>> paradoxical machine, there exists a functionally >>>>>>>>>>>>>>>> equivalent machine without such paradox
And the problem is that your "paradoxical" isn't actually >>>>>>>>>>>>>>> a definable property (let alone computable). Part of the >>>>>>>>>>>>>>> problem is that if you look at just a machine
description, it doesn't (necessarily) tell you about the >>>>>>>>>>>>>>> use of an "interface" as that use of an interface can be >>>>>>>>>>>>>>> just inlined, leaving nothing "in the code" to show it >>>>>>>>>>>>>>> exists.
i'm sorry, are you actually saying the machine description >>>>>>>>>>>>>> does not describe what the machine does???
No, that "paradoxical" isn't a definable property of a >>>>>>>>>>>>> machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the >>>>>>>>>>>> classifier halts(), easily demonstrable thru a short proof >>>>>>>>>>>> i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A >>>>>>>>>>>> PROPERTY OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS >>>>>>>>>>>> DEVOTION TO SOME FUCKING FORM OF INSANITY I CAN'T HONESTLY >>>>>>>>>>>> FATHOM
lol
Sure, because it is just a number. The problem is that >>>>>>>>>>>>>>> you still have
His specified H, with an actually (incorrect) >>>>>>>>>>>>>>>>> implementation of D (which is all that CAN exist) will >>>>>>>>>>>>>>>>> either be circle- free and thus generate a number (but >>>>>>>>>>>>>>>>> its D said it isn't, and thus omitted a valid machine >>>>>>>>>>>>>>>>> from the list) or it isn't circle- free, and fails to >>>>>>>>>>>>>>>>> computa a number, and thus should have been omitted >>>>>>>>>>>>>>>>> from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it >>>>>>>>>>>>>>>>> is just built on an assuption in error.
so despite turing's worries, the existence of a >>>>>>>>>>>>>>>>>> diagonal computation does not actually then imply the >>>>>>>>>>>>>>>>>> existence of an anti- diagonal computation, due the >>>>>>>>>>>>>>>>>> same particular self- referential weirdness that >>>>>>>>>>>>>>>>>> stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic >>>>>>>>>>>>>>>>> is just based on ERROR.
Your attempt to REDEFINE self-reference to mean >>>>>>>>>>>>>>>>> processing a number that happens to represent yourself >>>>>>>>>>>>>>>>> means that you you system "ALL" doesn't actually mean >>>>>>>>>>>>>>>>> ALL, and thus is just ill- defined.
i'm sorry, you have an issue with me labeling a number >>>>>>>>>>>>>>>> that directly refers to yourself, as a "self-reference" ??? >>>>>>>>>>>>>>>
it's a *specific* number that has the currently running >>>>>>>>>>>>>> machine encoded into it, it's not "just" a number whatever >>>>>>>>>>>>>> that means
problems with all the "equivalent" machines that have >>>>>>>>>>>>>>> different numbers.
those are references to functionally equivalent machines, >>>>>>>>>>>>>> not self- references
Which is why the concept of "self-reference" doesn't work. >>>>>>>>>>>>>
It may let you filter out the simplest case used in the >>>>>>>>>>>>>>> proofs, but doesn't solve the actual problem, as the >>>>>>>>>>>>>>> "Machine Number" doesn't actually fully identify the >>>>>>>>>>>>>>> problematic cases.
that's not actually true. you can't meaningfully paradox >>>>>>>>>>>>>> the paradox detector while filter *out* paradoxes to point >>>>>>>>>>>>>> of creating an actual problem
Then show how you will do it.
This means you need to detect a input that represents a >>>>>>>>>>>>> machine that uses a computation equivalent to yourself >>>>>>>>>>>>>
let halts be a halting classifier that takes an input >>>>>>>>>>>>>> machines and returns true/false whether it halts or not >>>>>>>>>>>>>>
let paradox_free be a paradox filtering classifier that >>>>>>>>>>>>>> takes an input classifier, and an input machine to search >>>>>>>>>>>>>> for paradoxes in regards to: paradox_free(classifier, >>>>>>>>>>>>>> machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox >>>>>>>>>>>>>> classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if >>>>>>>>>>>>>> case run a halting paradox form making it /undecidable >>>>>>>>>>>>>> input/ to halts().
And the problem is "paradox_free" can't exist as an always >>>>>>>>>>>>> correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if >>>>>>>>>>>>>> case doesn't run and clearly halts(undp) -> TRUE without >>>>>>>>>>>>>> issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within >>>>>>>>>>>>>> a machine is necessary for the particular input->output >>>>>>>>>>>>>> computation being done, so utilizing the return
paradox_free(halts,undp) -> FALSE to filter out
paradox_free() paradoxes as well as halts() paradoxes will >>>>>>>>>>>>>> suffice to produce a turing complete subset of machines >>>>>>>>>>>>>> that can be totally classified by halts()
But we can still build an input that your paradox_free() >>>>>>>>>>>>> fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE >>>>>>>>>>>> INCLUDED ON THE DIAGONAL???
No, it is your dependence on being able to assume that >>>>>>>>>>>>> unicorns can solve all your problems.
i think a major problem with the consensus perspective is >>>>>>>>>>>>>> an expectation of a certain cleanliness in the logic as >>>>>>>>>>>>>> being the only possibility here to compute what we want. >>>>>>>>>>>>>> TM's can't achieve that, but something slightly messier can. >>>>>>>>>>>>>
dear future: u see this shit i'm trying work with???? >>>>>>>>>>>>>>>> sorry i didn't manage to make progress any faster, but >>>>>>>>>>>>>>>> i'm literally dragging toddlers kicking and screaming at >>>>>>>>>>>>>>>> this point jeez...
In other words, in your world "ALL" isn't a word that can >>>>>>>>>>>>>>> actually be used.
You don't understand that you need to use actual sound >>>>>>>>>>>>>>> logic,
lol, i can only 🙏 someone would engage in sound logic >>>>>>>>>>>>>> with me
They are, But you are too stuck in your lies to see it. >>>>>>>>>>>>>
Your "logic" is based on being able to assume that
Unicorn's exist, and that is a safe assumption until >>>>>>>>>>>>> someone can prove that Russel's Teapot is not out there. >>>>>>>>>>>>> (Even though your unicorns HAVE been proved to not exist). >>>>>>>>>>>>>
You are just proving your utter stupidity.
This just shows that you real problem is you don't >>>>>>>>>>>>>>>>> understand what the actual problem is, and your world >>>>>>>>>>>>>>>>> is just build on things that are lies.
...which u won't find interesting, but i can't fix a >>>>>>>>>>>>>>>>>> lack of curiosity
🤷
expected was an inexpert audience, supposing that >>>>>>>>>>>>>>>>>>>>> experts will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>>>>>>>
IT seems that fundamentally, your world doesn't >>>>>>>>>>>>>>>>> actually have "computations" as you don't understand >>>>>>>>>>>>>>>>> the basic requirement that they need to be fully >>>>>>>>>>>>>>>>> defined in the actions they do.
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote:
On 2/27/26 6:09 AM, dart200 wrote:i don't see any indication that turing realized a >>>>>>>>>>>>>>>>>> difference there
On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> ...
Eh?!an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on
turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ of an >>>>>>>>>>>>>>>>>>>>>>> infinitude one by one.
that exactly what turing does in his proof: he >>>>>>>>>>>>>>>>>>>>>> defines a comptuation
that enumerates out all the numbers, testing each >>>>>>>>>>>>>>>>>>>>>> one of they represent
a "satisfactory"/"circle-free" machine, and adding >>>>>>>>>>>>>>>>>>>>>> that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to carefully >>>>>>>>>>>>>>>>>>>>>> read p247 of turing's
proof and produce the psuedo-code for the machine >>>>>>>>>>>>>>>>>>>>>> H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are >>>>>>>>>>>>>>>>>>>>> you sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>>>>> procedures, and understanding the diagonal procedure >>>>>>>>>>>>>>>>>>>> is critical to understanding the *base* >>>>>>>>>>>>>>>>>>>> contradiction/ paradox that the rest of his support >>>>>>>>>>>>>>>>>>>> for godel's result is then built on
And focusing on what is said to be impossible and not >>>>>>>>>>>>>>>>>>> changing the problem is also important.
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>>>>>> enumerating the enumeration in the first place. >>>>>>>>>>>>>>>>>>
Then you zre just showing your stupidity, because YOU >>>>>>>>>>>>>>>>> can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>>>>>> enumeration of the computable sequences.
He then points out that he can directly show that the >>>>>>>>>>>>>>>>> "anti- diagonal" of the (non-effectively computed) >>>>>>>>>>>>>>>>> enumeration can't be computed but that "This proof, >>>>>>>>>>>>>>>>> although perfectly sound, has the disadvantage that it >>>>>>>>>>>>>>>>> may leave the reader with a feeling that 'there must be >>>>>>>>>>>>>>>>> something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being >>>>>>>>>>>>>>> done.
I think he is refering he to the standard anti-diagonal >>>>>>>>>>>>>>> arguement, which shows that since for all n, position n >>>>>>>>>>>>>>> differs from the value in number n, there can not be any >>>>>>>>>>>>>>> element that matches the anti- diagonal.
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>> something it seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with >>>>>>>>>>>>>> turing's short diagonal proof, but in writing this post i >>>>>>>>>>>>>> now find myself in a subtle, yet entirely critical >>>>>>>>>>>>>> disagreement:
/let an be the n-th computable sequence, and let φn(m) be >>>>>>>>>>>>>> the m- th figure in an. Let β be the sequence with 1-φn(m) >>>>>>>>>>>>>> as its n-th. figure. Since β is computable, there exists a >>>>>>>>>>>>>> number K [== β] such that 1- φn(n) = φK(n) for all n. >>>>>>>>>>>>>> Putting n = K, we have 1 = 2φK(K), i.e. 1 is even. This is >>>>>>>>>>>>>> impossible/
i agree with this proof is far as much as if β was
computable (by TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the >>>>>>>>>>>> computability of β from the existence of φn(m), THAT'S THE >>>>>>>>>>>> FALLACy
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE >>>>>>>>>>>> SPECIFICATION
the fallacy here is assuming that because the direct >>>>>>>>>>>>>> diagonal is computable, that one can therefore compute the >>>>>>>>>>>>>> anti- diagonal using the direct diagonal. the abstract >>>>>>>>>>>>>> definition makes it look simple, but this ignores the >>>>>>>>>>>>>> complexities of self- referential analysis (like what >>>>>>>>>>>>>> turing details on the next page)
But you can, *IF* you have a machine that computes the >>>>>>>>>>>>> diagonal, then just change all the write to the output to >>>>>>>>>>>>> write the opposite. Note, the "self-reference" that you are >>>>>>>>>>>>> thinking of stops being a "self- reference" but is a >>>>>>>>>>>>> reference to the original write the diagonal code.
Nope, because your filtering TM (or RTM) will still fail >>>>>>>>>>>>> when it gets to the number of TURING'S H, as there is no >>>>>>>>>>>>> correct answer for the machine built by that template. >>>>>>>>>>>>
in both methods i have for rectifying the paradox found in >>>>>>>>>>>>>> the direct diagonal (either (1) filtering TMs or (2) using >>>>>>>>>>>>>> RTMs), neither can be used to then compute the anti-diagonal >>>>>>>>>>>>>
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is filtered >>>>>>>>>>>>>> out like turing's paradoxical variation of the direct >>>>>>>>>>>>>> diagonal would be, and there is no analogous non- >>>>>>>>>>>>>> paradoxical variation that has a hard coded value that is >>>>>>>>>>>>>> inverse to what it does return ... such a concept is >>>>>>>>>>>>>> entirely nonsensical. a function can only return what it >>>>>>>>>>>>>> does, it can't also return the inverse to what it returns >>>>>>>>>>>>>> eh???
But, if D filters it out, then it becomes circle-free, and >>>>>>>>>>>>> thus your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO IT >>>>>>>>>>>> *SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with >>>>>>>>>>>>>> RTMs just fails for reasons u'd only understand by working >>>>>>>>>>>>>> thru the algo urself (p7 of re: turing's diagonals) >>>>>>>>>>>>>>
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>
is just not sufficient evidence that such β is actually >>>>>>>>>>>>>> computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how >>>>>>>>>>>>> to do that, as you are thinking the only "reference" can be >>>>>>>>>>>>> to "self", doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF- >>>>>>>>>>>> REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H >>>>>>>>>>>>
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS >>>>>>>>>>>> FOR INVERSE- H TO BE COMPUTED
Then how do you build YOUR H without that D?He doesn't. You are just showing your stupidity,
one cannot just assume that because the diagonal across >>>>>>>>>>>>>>>> computable numbers is computable, therefore the anti- >>>>>>>>>>>>>>>> diagonal across computable numbers is computable... >>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and >>>>>>>>>>>>>>> without the enumeration, you can't compute either of them. >>>>>>>>>>>>>>>
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>> computation across the computable numbers can be used to >>>>>>>>>>>>>>>> compute the inverse diagonal
But your method still doesn't let you compute the >>>>>>>>>>>>>>> enumeration, and thus you can't actually compute the >>>>>>>>>>>>>>> diagonal.
Remember, the problem definitions requires that the >>>>>>>>>>>>>>> listing be a COMPLETE listing of the computable numbers / >>>>>>>>>>>>>>> machine that compute computable numbers, in some definite >>>>>>>>>>>>>>> order.
If your enumeration isn't complete, your diagonal isn't >>>>>>>>>>>>>>> correct.
so while i agree with turing that the anti-diagonal is >>>>>>>>>>>>>>>> not computable, i don't agree that the normal diagonal >>>>>>>>>>>>>>>> is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all >>>>>>>>>>>>>>> the other machines, including his original H that doesn't >>>>>>>>>>>>>>> use your "trick"
But instead, he can prove with a more obvious process, >>>>>>>>>>>>>>>>> that the Decider "D" that could be used to effectively >>>>>>>>>>>>>>>>> enumerate the sequence of machine that produce >>>>>>>>>>>>>>>>> computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing >>>>>>>>>>>>>>>>> out that the attempt to compute the diagonal clearly >>>>>>>>>>>>>>>>> reveals the issue with effectively enumerating the >>>>>>>>>>>>>>>>> sequences.
well, he didn't consider that perhaps the proper algo >>>>>>>>>>>>>>>> for computing the diagonal can avoid the paradox on >>>>>>>>>>>>>>>> itself ...
But it doesn't.
Your just don't understand that D just can't correctly >>>>>>>>>>>>>>> decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on >>>>>>>>>>>>>> turing's H, because my response to this is that D does not >>>>>>>>>>>>>> need to decide correctly on H to compute a diagonal >>>>>>>>>>>>>
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT IT >>>>>>>>>>>> WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial
recognizer D *can* decide on the fixed H because the fixed H >>>>>>>>>>> does not try to use any D on itself, so no self-referential >>>>>>>>>>> paradox is possible in regards to it's own digit on the diagonal >>>>>>>>>>>
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a
dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer D >>>>>>>>>>> (used in fixed H, which does not call any D itself), and >>>>>>>>>>> would not be filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, and >>>>>>>>>>> instead returns a hard-coded value for it's own digit on the >>>>>>>>>>> diagonal, is keystone in making it /decidable input/ to D >>>>>>>>>>>
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just >>>>>>>>>> anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it >>>>>>>>>> will fail on that.
bro what part of the non-existence of turing_D or turing_H do >>>>>>>>> you not understand???
But if your fD exists, then Turing_H that uses your D exists, >>>>>>>> and will give that D the problem, and fixed_H will call your D >>>>>>>> on turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H >>>>>>>>>> gets hung up.
If D errs on Turing_H and calls it not circle_free, then your >>>>>>>>>> enumeration is missing the circle_free H from its list, and >>>>>>>>>> thus doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D >>>>>>>>> in a turing_H type machine? that is a fair point
this would give us the machine (that actually does exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to F-
squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>> satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D
classifier, when actually run actual_turing_H will simply skip >>>>>>>>> putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the >>>>>>>> actual_turing_H isn't an "undecidable input", as it has definite >>>>>>>> behavior, just behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of
undecidability. just because a machine is /undecidable input/ to >>>>>>> a particular classifier does mean we cannot then prove and know >>>>>>> ourselves what the machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs
are not "undecidable", but by the time that they have actually
been made into an input, and thus first had the program created,
which required creating the instance of the decider selctected,
they have definite behavior that other some decider can determine. >>>>>>
It is only that one selected candidate decider chosen to build the >>>>>> input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a >>>>>>> particular machine and the particular classifier it creates a
structural paradox for,
In other words, you don't understand what "undecidability"
actually is.
Undecidability, means that there does not exist ANY decider that
can get the right answer for all instances of the problem, (or if >>>>>> the problem has been reduced to just that one input, for that input). >>>>>>
*not* as a general inability to describe a *real* machine's
behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D.
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and prove it's
equivalent to fixed_H, so therefore it doesn't fit what u stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a
machine, unless you are narrowing the problem to just being about that
one machine/input.
because it *only* ever *was* about problematic input to *particular* classifiers...
it's not my problem everyone went for over a century confusing *our*
ability to prove outcomes, with what a /particular/ decider can
output... ya'll got caught up in a rigor of what is fundamentally nonsense
no one has demonstrated a *possible* machine, let alone a *real*
machine, which we can't provably can't prove the outcome for... such a
proof makes the machine _NEITHER POSSIBLE NOR REAL_ so it would be
ludicrous to present such a proof to exist
both turing_H and turing_D are /incomplete specifications/ and therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_
they are just malformed /incomplete specifications/ and _NOTHING MORE_,
and _DESERVE NO FURTHER CONSIDERATION_ beyond the recognition that such
an interface _IS INCOMPLETE_
And, it is an assertion that NO decider can correctly compute the
answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which means
you have chosen a particular (broken) implementation of D (since no
actually correct D exists), like your partial_recognizer_D, has fulled
defined behavior, and a decider can exist that determines it, there is
nothing "undecidable" about that input.
It is just that the D that H was built on is wrong. Nothing
"paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you just _IGNORING_ the specification i specifically created for partial_recognizer_D...
the specification for partial_recognizer_D is when encountering
undecidable input is to utilize FALSE,
this _DOES NOT MEAN_ we cannot know what the /undecidable input/ does,
as a paradox in regards to partial_recognize_D _ONLY AFFECTS_ partial_recognize_D not _OUR ABILITY TO PROVE OUTCOMES_
(you don't have one btw, so please do let me witness ur cope)
IT seesm you just can't read.
It doesn't need to. It just needs to be built on what you want to
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
see, turing_H's runtime is not describable because it's an /
incomplete specification/ of machine and therefor _does not even >>>>>>> exist_ as a real TM, not because of the hypothesized
undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED defined >>>>>> machine. If the decider "D" actually exists, then H is FULLY defined. >>>>>
specification/, as it doesn't handle idiosyncrasies of TM
computing, specifically self- referential set-classification paradoxes >>>>
claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as your
equivalent for D, like your partial_recognizer_D, then the template
turing_H can be converted into an actual machine.
actual_turing_H, on the other hand, actually does exist, so we
can know/ prove what it does. it, however, still forms a
structural paradox in regards to partial_recognizer_D which will >>>>>>> fail to classify it as circle-free
But, since you admit to actual_turing_H existing, that means that
since partial_recognizer_D _CAN_ exist, then actual_turing_H _CAN_
exist
And so can the turing_H that uses partial_recognizer_D as its
decider D.
The steps the algorithm does is what defines the machine, not the
"names" of the things it "calls", as those names don't actually
exist in the final building of the machine, which is the problem
with your concept, as that is needed to determine your "paradoxical"
property.
the names are just shortcuts for the literal values rick, don't be daft
Right, but the literal values don't have the names on them, and we
don't have borders within the machine letting us know we moved from
"outer code of H" into "the code of the D it uses", especially with
Turing Machines were such usages are by necessity "expanded" in-line.
please do actually read turing's paper sometime, he does organize
machine descriptions into functional groups
Thus, the "copy" of D that was pulled into H isn't actually the same
no it can be literally a copy of same description than is the run by a
UTM (universal turing machine) functional group of the machine
thing as the independent machine D that we can talk about, just a
functional equivalent which doesn't even have its own D.N (since that
is a property only of full machines)
ur really pulling at straws here rick
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will get >>>>>>>> wrong, then it can just return the different answer, and then >>>>>>>> THAT machine it will get right. The problem is it creates a
DIFFERENT input, that uses this new version of the machine, that >>>>>>>> it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as
fixed_H, which in turn _does_ have a spot on the diagonal. so >>>>>>>>> there is no actual need to include a digit from actual_turing_H >>>>>>>>> on the diagonal, as the number it computes is already included >>>>>>>>> on the diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of >>>>>>>> actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can
possible filter all ALL of that infinite set of machines, but >>>>>>>> still accept some machine that computes that particular number >>>>>>>> that it would compute with this supposedly correct
partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting >>>>>>>> that you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the
fairy dust, of assuming that your partial decider that meets ALL >>>>>>>> your needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between >>>>>>> fixed_H and actual_turing_H is the way they respectively handle >>>>>>> their own self- references. they function identically when
handling all other machines
But the problem isn't in it handling the machine fixed_H, but how >>>>>> it ever handles the results generated by actual_turing_H, which by >>>>>> your assumption has become a cycle-free machine and ALL equivalent >>>>>> versions of it have been partially declared to be non-cycle free.
This means that your enumeration of machines you are looking at
is, in fact, incomplete, thus the diagonal you computed is wrong.
fixed_H = () -> {
N = 1
K = 0
output = [] // written to
F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded
digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>> satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip
trying to simulate itself and put it's hard coded digit on the
diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip
trying simulate actual_turing_H because partial_recognizer_D
fails to classify it as circle-free
But actual_turing_H is actually circle_free, and thus its results >>>>>> SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put on
the diagonal, then actual_turing_H _DOES NOT_ need to be in _THAT_
total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable
numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on being
sloppy.
the only way to compute a diagonal across computable numbers is
simulating each number by the machine that computes it.
Note, Turing SPECIFICALLY pointed out that he was going to the related
problem, the problem of:
/but the problem of enumerating computable sequences is equivalent to
the problem of finding out whether a given number is the D.N of a
circle- free machine, and we have no general process for doing this in
a finite number of steps/
oh wow GREAT CATCH RICK!!!!,
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT ENUMERATING
OUT ALL CIRCLE_FREE MACHINES*,
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER IS A CIRCLE-FREE MACHINE,
we only need to compute that fact for *ONE* machine for any given
computable sequence, _SO A PARTIAL RECOGNIZER IS SUFFICIENT FOR THE JOB_
fuck rick ur so bad at defending this it's actually really useful to me!
(there's *another* specific sentence i have a problem with, where's a
ben when u need him?)
So, HIS enumeration is the complete enumeration of "circle-free
machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration either, why
am i the only one subject to criticism here???
the *specification* /works/, which shows that it's *not*
Also, your proof is built on the existance of a Russel Teapot, you have >>>
contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion. Your
arguement boils down to showing that *IF* you can correctly determine
all machines that fit in a not-definable category (since you concept
of [paradoxical can't actually be defined at the machine level, only
at how a machine might have been created at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can make
the precise "mistakes" you need it to. (They are mistakes because
the input
"mistake" is this case is a fallacious non-techincal value judgement
on the return value,
No, since your criteria that you want to use is just an undefinable
term for a machine description.
There is nothing "paradoxical" of the actual machine when created, as
it then has fixed and determined behavior that can be decided on. The
thing that is paradoxical is the mete-system pattern used to generate
the class of machines as pair with a given decider.
cause they are rather entirely within specification, as FALSE was
specified to be used upon encountering an /undecidable input/,
something turing_D did not handle correctly (making it _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is based on a
actual D that does make a specific decision to the input. And it is
just that for any given version of D, we can find a DIFFERENT H that
it will get wrong (the H that was just built on it).
Do you want to claim that ALL these H are "undecidable" even though for
no... fixed_H contains no paradox, and is therefor not /undecidable
input/ to any classifier
all of them there does exist a decider that gets the right answer?
many machines contain no paradox and are therefore classifiable by all classifiers
It seems you just don't know what the word means, and thus you
specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of this
turing_H is built on this partial_recognizdr_D which causes them to
not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to machines, as
it is based on the incorrect definition that assumes a machine can
be something other than itself, and when it changes, it changes
other things that at the meta/template level refered to its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just refuse to
learn.
Try to actually fully DEFINE the word based on the actual properties
of the input machine, and not refering to other machines that are not
part of the input (but would be created as alternate inputs to foil an
alternate decider).
when actual_turing_H iterates across DN(actual_turing_H) itself, >>>>>>> it will similarly skip trying to simulate itself, because
partial_recognizer_D fails to classify it as circle-free
Right, so it ALSO is wrong, as it didn't compute the diagonal of
the circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE NUMBERS_,
which is the _ACTUAL PROBLEM_ turing was dealing with,
No, it isn't, as your "set of computable numbers" has duplicates.
it's still a diagonal across all computerate numbers even if it has
dpulicates ...
No, because an enumeration of a set doesn't HAVE duplicates.
then the H's as of now are i suppose "generators" instead of enumerators
and turing was wrong about that too, but i don't see u criticizing him
now eh???
Your probe
this level of objection is dishonest\
No, your using misdefined terms is the real dishonesty.
If you want to show that you can make a diagonal of computable
numbers, you need to build an actual effective enumeration of
computable numbers.
And the problem is you can't assume your decider and then use it. That
isn't vallid logic.
but i can show turing was wrong about a series of points, in the most influence math paper of last century,
that's good enough start for me
i can do more once i gain some recognition with that
It IS valid to assume a decider, and show that such an assumption
leads to a contradiction, to prove that such a decider can not exist.
That is the valid proof by contradiction.
If we can show that if A exist, B must be true, and also that B can
not be true, then we have proven that A can not exist.
Assuming that A exists, and showing that one line of reasoning based
on it doesn't lead to a contradiction, does NOT show that A does (or
even can) exist. That is just fallacious logic.
doing so doesn't require /all/ or _even most_ machines to be in the >>>>> enumeration, since _most_ are duplicates of some other machine
when actual_turing_H iterates across DN(fixed_H),
partial_recognizer_D *successfully* classifies fixed_H as
satisfactory, so it will simulate fixed_H to it's Kth digit, the >>>>>>> one hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular
logic in either case), fixed_H also skips simulating itself (also >>>>>>> circular logic), while actual_turing_H actually *does* simulate >>>>>>> fixed_H successfully for it's Kth digit on the diagonal. these
machines compute the *same* number/sequence!
Except that by your partial_recognizer being just partial, it has >>>>>> dropped required items from the enumeration.
_WHICH IS FINE_
So, you admit it failed???
ACTUAL_TURING_H IS _NOT_ REQUIRED TO BE ON THE DIAGONAL, SO IT'S FINE
TO BE DROPPED
If the diagonal is the diagonal of circle-free machines, it is required.
Your problem is you keep on working with strawmen.
If the diagonal is of computable numbers, you need to show how you
trim down the set to not have repeats (in finite time per element) and
that your decider actually exists, you can't just assume it.
As I said, it dropped REQUIRED items, as this enumeration was of
machines that computed computable numbers.
If you try to add removal of duplicates, then we need to define the
order, and if we do it by number, turing_H will likely be lower in
order, as it doesn't need to generate its own number, and thus is
simpler,
that sentence is nonsense until u actually produce psuedo-code
I did, just make my H call your "D" replacement, partial_recognizer_D.
Or, is your problem that you think an enumeration doesn't define an
"order"
THis is becides the fact that you still need to show how you plan >>>>>> on actually detecting all these computationally equivalent
machines in your
fixed_H does not compute a minimum turing-complete set, there are
duplicated sequences, ei machines that compute the same sequence/
number, and _THAT IS FINE_
Why?
BECAUSE AN ENUMERATION THAT INCLUDES ONE OR MORE INSTANCE FOR EACH
COMPUTABLE NUMBER STILL ENUMERATES *ALL* COMPUTABLE NUMBERS
Nope. Maybe you don't understand what an enumeration.
And enumeration is a one-to-one mapping between the elements and the
Natural Numbers.
By your definition, you can make a countable infinite enumeation of a
finite set, showing that the finite set is infinite.
And, it was built on the assumption of the existance of a machine,
that actually can not exist (based on Rice's Theorm) and thus is
actually based on the assumption of Unicorns.
partial_recognizer_D is *not* a total decider, so how could rice's
theorum apply???
But it *IS* a total decider on your "paradoxical" property.
It can't let ANY of them through, and it must not omit any of the ones
you need.
partial decider to force them to not be classified as circle-free, >>>>>> even though it turns out that they are (because your decider
called them not).
You don't seem to understand that you don't get to look at the
template used to build the machine, only the resultant machine
from that template.
And, part of the result of that, is that your attempt to try to
define "paradoxical" just fails, as it resolves to just input your >>>>>> decider is wrong about, in part, because you don't understand what >>>>>> "decidable" actually means,
In other words, you think it ok to compute the WRONG diagonal.
see, not only is defining a value in fixed_H that the diagonal
computation can put for itself in diagonal essential in making
the diagonal actually computable, so is partial_recognizer_D
skipping over actual_turing_H to avoid the computability paradox! >>>>>>
The needed diagonal was to a COMPLETE enumeration of the circle-
free machine.
NO, IT WAS SPECIFICALLY A COMPLETE ENUMERATION OF _COMPUTABLE
NUMBERS/ SEQUENCES_
Which you don't get.
smh
But there is to show that your partial_recognizer_D can exist, when
there is *no* requirement to include a sequence _more than once_,
so there *no* requirement to include all machines in the enumeration! >>>>
it can't.
YOU'VE LOST THE STANDARD PROOF IT DOESN'T EXIST,
But YOUR proof doesn't either, so the standard proof does exist.
WHAT NEW ARGUMENT WOULD THERE BE???
Don't need a NEW argument, as your LIES are just proven to be invalid.
YOU need to prove your "Russel's Teapot" exists.
Your logic just shows that you don't know how to do logic.
That isn't really different then claiming you can prove that every >>>>>> even number grater than 2 from an infinite set of natural numbers >>>>>> is the sum of two primes, make you enumeration just the set of
prime numbers, and thus claim that since the only even number is
2, it is trivially solved.
_FALSE ANALOGY FALLACY_
What is false about it?
THE SUBSET IDENTIFIED BY PARTIAL_RECOGNIZER_D IS NOT "LESS POWERFUL"
THAN THE TOTAL MACHINE SPACE
Sure it is, as it doesn't make the full space.
if the set i'm enumerating includes _AT LEAST ONE INSTANCE_ of _ALL >>>>> COMPUTABLE SEQUENCES_
But does it?
YES
How are you sure.
You have an admittedly PARTIAL recognizer.
How do you know that it accepts at least one instance of a machie for
every computable number.
How does your partial decider distinguish between turing's H and your
fixed H?
then it is a turing-complete subset including all possible TM
computable sequences, and is functionally equivalent to the set of
*all* TMs
But, since your partial_recognizer_D doesn't exist, but was only
assumed to exist, your program doesn't exist,
ultimately, in order to compute a diagonal across computable
numbers, partial_recognizer_D only needs to successfully classify >>>>>>> *one* of the _infinitely_ many machines that compute any given
computable number...
But it isn't the diagonal across some set of computable numbers,
but the diagonal across the ENTIRE set of computable numbers.
Since you dropped some machines, you don't have the needed set.
_*THAT DOES NOT NECESSARILY FOLLOW*_
Then try to define what set you are actually making!!
A SET THAT CONTAINS INFINITE INSTANCES OF ALL COMPUTABLE NUMBERS
Which isn't a set.
Boy are you stupid.
Sets contain no duplicates.
The problem is your definition of "paradoxical" doesn't actually
work, as it is based on asking about the results of an input, that
changes.
dropping actual_turing_H from the set, for example, is _FINE_
because it computes _THE SAME SEQUENCE_ as fixed_H, which _IS IN
THE SET_ ...
How do you know it includes your fixed_H in the set?
what part of the actual machine makes it different?
THE CODE OF FIXED_H AND ACTUAL_TURING_H IS IN THE THREAD, WHAT ARE
YOU TALKING ABOUT???
WHich aren't full machines, not until you include the code of
partial_recognizer_D,
And, when you define it, how is Actual_turing_H "undecidable", since
decider exist that decide what it will do.
Remember, in the final machine given as the input, there are not
READ TURING'S PAPER ACTUALLY TO SEE HOW A SUBSET OF TRANSITION
FUNCTIONS WITHIN A TURING MACHINE CAN BE LABELED AND REFERRED TO
Yes, and HE Defined what is subset was.
You haven't, as you have a criteria that is just nonsense, as it is
basded on the input changing when you look at a different decider
looking at it.
seperate turing_H and partial_recognizer_D denoted in the machine,
just one big piece of algorithm. Thus nothing to base your criteria on. >>>>
shit rick, idk how ur still denying what's right under ur nose,
Because what you are putting under my nose is just CRAP based on
errors
goddamn rick smh u could spot a diamond in the rough even if god
lead u by ur hand to one put on a pedestal with a _GIANT_ neon sign >>>>> pointing to it...
But the problem is that "your diamond" is actually just a big load
of crap that you misidentify as a diamond, because you don't
actually understand what you are doing.
ur so used to playing contrarian u don't know what truth even looks
like anymore, rick
Nope. But you live so much in your stupidity, that you think your lies
are valid.
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that >>>>>>>>>>>>> generates a computable number to build the list of such >>>>>>>>>>>>> machine to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. >>>>>>>>>>>> BOTH ARE INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, WHICH >>>>>>>>>>>> WILL WORK JUST FINE ON THE ENUMERATION THAT HAS /UNDECIDABLE >>>>>>>>>>>> INPUT/ TO D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE
COMPUTABLE NUMBERS
Because "undecidability" doesn't affect the correct answer >>>>>>>>>>>>> it must compute.
It doesn't matter that your new H doesn't get stuck on >>>>>>>>>>>>>>> itself, it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he >>>>>>>>>>>>>> doesn't specify what D (or H) needs to do when
encountering the / undecidable input/ of H, so therefore >>>>>>>>>>>>>> both D and H are an incomplete specifications of a machine >>>>>>>>>>>>>
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL >>>>>>>>>>>> SPACE LIKE THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN >>>>>>>>>>>> IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS >>>>>>>>>>>> COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, >>>>>>>>>>>> BECAUSE IT DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF- >>>>>>>>>>>> REFERENTIAL ANALYSIS, SO IT IS AN /INCOMPLETE SPECIFICATION/ >>>>>>>>>>>> OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT PARADOXES >>>>>>>>>>>> CAN'T BE DONE, WHERE IN THE FUCK IS THE REFERENCE U MORON??? >>>>>>>>>>>>
Note, H, when it becomes an actual machine, because we have >>>>>>>>>>>>> created an actual machine we claim to be the D, has an >>>>>>>>>>>>> answer, and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't >>>>>>>>>>>>> exist, only machines CORRECTLY meeting the requirements of >>>>>>>>>>>>> D and H can't exist.
Thus, if your H depends on that D, (or a machine that meets >>>>>>>>>>>>> its specification) then it also can't exist.
You just make the error of saying the problems don't exist >>>>>>>>>>>>> because you can't build them, but YOUR machine can, even >>>>>>>>>>>>> though it has the same problem.
IF D is wrong by deciding it is not circle free, then >>>>>>>>>>>>>>> your H will compute the wrong diagonal, as the resulting >>>>>>>>>>>>>>> version of his H WILL be circle free (since it never >>>>>>>>>>>>>>> tries to simulate itself) and thus DOES produce an >>>>>>>>>>>>>>> computable number that your computation misses.
Or, if that D is wrong by decing it IS circle free, then >>>>>>>>>>>>>>> when you H tries to process it, it will get stuck in the >>>>>>>>>>>>>>> infinite loop.
The problem is that in stepping through the machines in >>>>>>>>>>>>>>> order, you WILL hit these actual machines built on your >>>>>>>>>>>>>>> erroneous D (your D must have this flaw, as no D without >>>>>>>>>>>>>>> exists), and thus you will be wrong on THAT input. IT >>>>>>>>>>>>>>> doesn't matter if you get a good answer for yourself. >>>>>>>>>>>>>>>
idk what he would have said about it, but prolly >>>>>>>>>>>>>>>> something more substantial than just calling me ignorant >>>>>>>>>>>>>>>> repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic >>>>>>>>>>>>>>> assertion of bad logic. You would have likely been >>>>>>>>>>>>>>> escorted out of the meeting as showing you were >>>>>>>>>>>>>>> unqualified and being a distraction.
Something that seems to be beyond your ignorant >>>>>>>>>>>>>>>>> understanding.
H shows that *IF* you can make that enumeration, you >>>>>>>>>>>>>>>>>>> can make the diagonal, and thus the anti-diagonal. >>>>>>>>>>>>>>>>>>> The problem is you can't make that enumeration, and >>>>>>>>>>>>>>>>>>> assuming you can just shows unsoundness.
interestingly: one can only fix the direct diagonal >>>>>>>>>>>>>>>>>> computation like this
u can't do an analogous fix for the inverse/anti- >>>>>>>>>>>>>>>>>> diagonal computation. it's not possible hard code a >>>>>>>>>>>>>>>>>> machine to return an inverted value, a machine can >>>>>>>>>>>>>>>>>> only return what it does, not the inverse of what it >>>>>>>>>>>>>>>>>> does...
so if we can filter out paradoxes from the >>>>>>>>>>>>>>>>>> enumeration, that will leave a direct diagonal >>>>>>>>>>>>>>>>>> computation extant in that filtered (yet still turing >>>>>>>>>>>>>>>>>> complete list), while any attempt to compute an >>>>>>>>>>>>>>>>>> inverse diagonal will not be
But the problem is that "paradoxical machines" don't >>>>>>>>>>>>>>>>> exist in isolation, but only in relationship to a given >>>>>>>>>>>>>>>>> machine trying to decide them.
right. so if ur constructing a diagonal across >>>>>>>>>>>>>>>> computable numbers then u only need to filter out >>>>>>>>>>>>>>>> paradoxes in regards to the classifier that classifies >>>>>>>>>>>>>>>> them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out >>>>>>>>>>>>>> such paradoxes...
In other words, you beleive unquestionably in the existance >>>>>>>>>>>>> of Russle's teapot until someone can prove it doesn't exist. >>>>>>>>>>>>
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF >>>>>>>>>>>> PROVIDING A FUCKING QUOTE TO TURING'S PAPER TO BACK THE >>>>>>>>>>>> *DIRECT* CLAIM YOU MADE???
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I HAVE >>>>>>>>>>>> TO WORK WITH, BUT OH WELL
For THIS paradox, read the proof. for THIS H, the one >>>>>>>>>>>>> given, (for which ever version of D you want to try to >>>>>>>>>>>>> assume is right) D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the >>>>>>>>>>>>> paradox, because there still isn't an answer it can give to >>>>>>>>>>>>> H that will be correct, as the template for H will always >>>>>>>>>>>>> make that D wrong.
Note, the problem is you can't actually DEFINE what a >>>>>>>>>>>>> "paradox machine" is, as they aren't actually machines, but >>>>>>>>>>>>> templates that build machines. And that final machine >>>>>>>>>>>>> doesn't have actually detectable tell-tales that show it to >>>>>>>>>>>>> be from that template.
Because I am smart, and know what I am talking about. >>>>>>>>>>>>
(also why do always just make random assertions???) >>>>>>>>>>>>>
They only seem "random" because you don't know what you are >>>>>>>>>>>>> talking about and believe in unicorns (and teapots)
any machine which *is not* "satisfactory" OR *is not* >>>>>>>>>>>>>>>> classifiable as satisfactory by said classifier... can >>>>>>>>>>>>>>>> just be skipped
No, it can only skip those that are not satisfactory, not >>>>>>>>>>>>>>> those that are but it can not classify as such, or your >>>>>>>>>>>>>>> enumeration will not be complete, and thus just in error. >>>>>>>>>>>>>>>
Thus, it needs to be able to correctly classify ALL >>>>>>>>>>>>>>> machines (as it will be asked about all machines as it >>>>>>>>>>>>>>> counts through all the descriptions) and thus Turing's H >>>>>>>>>>>>>>> *WILL* be asked about.
similarly if u want to go a step further an filter out >>>>>>>>>>>>>>>> computable numbers already included on this diagonal, >>>>>>>>>>>>>>>> any machine which either *is* computably equivalent OR >>>>>>>>>>>>>>>> *is not* classifiable in regards to *any* machine >>>>>>>>>>>>>>>> already the list... can just be skipped
Nope, you can't skip some machines, as you then might >>>>>>>>>>>>>>> lose some of the computable numbers.
see you can't compute a diagonal across *all* / >>>>>>>>>>>>>>>> machines/, with said machines, but u can compute a >>>>>>>>>>>>>>>> diagonal across *all* / computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be >>>>>>>>>>>>>>> done, since ALL classifiers that attempt it will make an >>>>>>>>>>>>>>> error, you can't do what you want to do.
nah, (a) computing an enumeration of all /computable >>>>>>>>>>>>>> numbers/ is not the same thing as (b) computing the >>>>>>>>>>>>>> enumeration of all machines that compute computable >>>>>>>>>>>>>> numbers. (b) necessarily has duplicates while (a) does not >>>>>>>>>>>>>> need them. turing's paper wrongly conflates (a) with (b) >>>>>>>>>>>>>>
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you >>>>>>>>>>>>> filter the full list of machibe, which Turing showed can't >>>>>>>>>>>>> exist.
The problem is such a machine can't handle the
classification of the H he describes. It doesn't matter >>>>>>>>>>>>> that you can make a DIFFERENT machine, that you try to >>>>>>>>>>>>> deceptively call "H" too that it won't have a problem with, >>>>>>>>>>>>> when your "H" gets to the number of Turing's H, it still >>>>>>>>>>>>> has the problem. It can call that one not circle-free, and >>>>>>>>>>>>> thus omit a circle- free machine from the list, or call it >>>>>>>>>>>>> circle- free, and when even YOU try to simulate it the k >>>>>>>>>>>>> steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to >>>>>>>>>>>>> take the code of (a), and invert the symbol put on the tape >>>>>>>>>>>>> when it is written. Note, it won't be "self-referent", as >>>>>>>>>>>>> the "N" is uses, is the N of YOUR H, not itself. The code >>>>>>>>>>>>> for his (a) NEVER reads back what it wrote, so that doesn't >>>>>>>>>>>>> affect its behavior.
yes, i still do need to prove my thesis that for any >>>>>>>>>>>>>>>> paradoxical machine, there exists a functionally >>>>>>>>>>>>>>>> equivalent machine without such paradox
And the problem is that your "paradoxical" isn't actually >>>>>>>>>>>>>>> a definable property (let alone computable). Part of the >>>>>>>>>>>>>>> problem is that if you look at just a machine
description, it doesn't (necessarily) tell you about the >>>>>>>>>>>>>>> use of an "interface" as that use of an interface can be >>>>>>>>>>>>>>> just inlined, leaving nothing "in the code" to show it >>>>>>>>>>>>>>> exists.
i'm sorry, are you actually saying the machine description >>>>>>>>>>>>>> does not describe what the machine does???
No, that "paradoxical" isn't a definable property of a >>>>>>>>>>>>> machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the >>>>>>>>>>>> classifier halts(), easily demonstrable thru a short proof >>>>>>>>>>>> i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A >>>>>>>>>>>> PROPERTY OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS >>>>>>>>>>>> DEVOTION TO SOME FUCKING FORM OF INSANITY I CAN'T HONESTLY >>>>>>>>>>>> FATHOM
lol
Sure, because it is just a number. The problem is that >>>>>>>>>>>>>>> you still have
His specified H, with an actually (incorrect) >>>>>>>>>>>>>>>>> implementation of D (which is all that CAN exist) will >>>>>>>>>>>>>>>>> either be circle- free and thus generate a number (but >>>>>>>>>>>>>>>>> its D said it isn't, and thus omitted a valid machine >>>>>>>>>>>>>>>>> from the list) or it isn't circle- free, and fails to >>>>>>>>>>>>>>>>> computa a number, and thus should have been omitted >>>>>>>>>>>>>>>>> from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it >>>>>>>>>>>>>>>>> is just built on an assuption in error.
so despite turing's worries, the existence of a >>>>>>>>>>>>>>>>>> diagonal computation does not actually then imply the >>>>>>>>>>>>>>>>>> existence of an anti- diagonal computation, due the >>>>>>>>>>>>>>>>>> same particular self- referential weirdness that >>>>>>>>>>>>>>>>>> stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic >>>>>>>>>>>>>>>>> is just based on ERROR.
Your attempt to REDEFINE self-reference to mean >>>>>>>>>>>>>>>>> processing a number that happens to represent yourself >>>>>>>>>>>>>>>>> means that you you system "ALL" doesn't actually mean >>>>>>>>>>>>>>>>> ALL, and thus is just ill- defined.
i'm sorry, you have an issue with me labeling a number >>>>>>>>>>>>>>>> that directly refers to yourself, as a "self-reference" ??? >>>>>>>>>>>>>>>
it's a *specific* number that has the currently running >>>>>>>>>>>>>> machine encoded into it, it's not "just" a number whatever >>>>>>>>>>>>>> that means
problems with all the "equivalent" machines that have >>>>>>>>>>>>>>> different numbers.
those are references to functionally equivalent machines, >>>>>>>>>>>>>> not self- references
Which is why the concept of "self-reference" doesn't work. >>>>>>>>>>>>>
It may let you filter out the simplest case used in the >>>>>>>>>>>>>>> proofs, but doesn't solve the actual problem, as the >>>>>>>>>>>>>>> "Machine Number" doesn't actually fully identify the >>>>>>>>>>>>>>> problematic cases.
that's not actually true. you can't meaningfully paradox >>>>>>>>>>>>>> the paradox detector while filter *out* paradoxes to point >>>>>>>>>>>>>> of creating an actual problem
Then show how you will do it.
This means you need to detect a input that represents a >>>>>>>>>>>>> machine that uses a computation equivalent to yourself >>>>>>>>>>>>>
let halts be a halting classifier that takes an input >>>>>>>>>>>>>> machines and returns true/false whether it halts or not >>>>>>>>>>>>>>
let paradox_free be a paradox filtering classifier that >>>>>>>>>>>>>> takes an input classifier, and an input machine to search >>>>>>>>>>>>>> for paradoxes in regards to: paradox_free(classifier, >>>>>>>>>>>>>> machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox >>>>>>>>>>>>>> classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE )
if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if >>>>>>>>>>>>>> case run a halting paradox form making it /undecidable >>>>>>>>>>>>>> input/ to halts().
And the problem is "paradox_free" can't exist as an always >>>>>>>>>>>>> correct decider.
however, if paradox_free(halts,undp) -> TRUE, then that if >>>>>>>>>>>>>> case doesn't run and clearly halts(undp) -> TRUE without >>>>>>>>>>>>>> issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within >>>>>>>>>>>>>> a machine is necessary for the particular input->output >>>>>>>>>>>>>> computation being done, so utilizing the return
paradox_free(halts,undp) -> FALSE to filter out
paradox_free() paradoxes as well as halts() paradoxes will >>>>>>>>>>>>>> suffice to produce a turing complete subset of machines >>>>>>>>>>>>>> that can be totally classified by halts()
But we can still build an input that your paradox_free() >>>>>>>>>>>>> fails on.
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST BE >>>>>>>>>>>> INCLUDED ON THE DIAGONAL???
No, it is your dependence on being able to assume that >>>>>>>>>>>>> unicorns can solve all your problems.
i think a major problem with the consensus perspective is >>>>>>>>>>>>>> an expectation of a certain cleanliness in the logic as >>>>>>>>>>>>>> being the only possibility here to compute what we want. >>>>>>>>>>>>>> TM's can't achieve that, but something slightly messier can. >>>>>>>>>>>>>
dear future: u see this shit i'm trying work with???? >>>>>>>>>>>>>>>> sorry i didn't manage to make progress any faster, but >>>>>>>>>>>>>>>> i'm literally dragging toddlers kicking and screaming at >>>>>>>>>>>>>>>> this point jeez...
In other words, in your world "ALL" isn't a word that can >>>>>>>>>>>>>>> actually be used.
You don't understand that you need to use actual sound >>>>>>>>>>>>>>> logic,
lol, i can only 🙏 someone would engage in sound logic >>>>>>>>>>>>>> with me
They are, But you are too stuck in your lies to see it. >>>>>>>>>>>>>
Your "logic" is based on being able to assume that
Unicorn's exist, and that is a safe assumption until >>>>>>>>>>>>> someone can prove that Russel's Teapot is not out there. >>>>>>>>>>>>> (Even though your unicorns HAVE been proved to not exist). >>>>>>>>>>>>>
You are just proving your utter stupidity.
This just shows that you real problem is you don't >>>>>>>>>>>>>>>>> understand what the actual problem is, and your world >>>>>>>>>>>>>>>>> is just build on things that are lies.
...which u won't find interesting, but i can't fix a >>>>>>>>>>>>>>>>>> lack of curiosity
🤷
expected was an inexpert audience, supposing that >>>>>>>>>>>>>>>>>>>>> experts will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>>>>>>>
IT seems that fundamentally, your world doesn't >>>>>>>>>>>>>>>>> actually have "computations" as you don't understand >>>>>>>>>>>>>>>>> the basic requirement that they need to be fully >>>>>>>>>>>>>>>>> defined in the actions they do.
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote:
i don't see any indication that turing realized a >>>>>>>>>>>>>>>>>>> difference thereOn 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> ...
Eh?!an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on
turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by one? >>>>>>>>>>>>>>>>>>>>>>>>>> Don't tell me you think
yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ of >>>>>>>>>>>>>>>>>>>>>>>> an infinitude one by one.
that exactly what turing does in his proof: he >>>>>>>>>>>>>>>>>>>>>>> defines a comptuation
that enumerates out all the numbers, testing each >>>>>>>>>>>>>>>>>>>>>>> one of they represent
a "satisfactory"/"circle-free" machine, and >>>>>>>>>>>>>>>>>>>>>>> adding that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to carefully >>>>>>>>>>>>>>>>>>>>>>> read p247 of turing's
proof and produce the psuedo-code for the machine >>>>>>>>>>>>>>>>>>>>>>> H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are >>>>>>>>>>>>>>>>>>>>>> you sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>>>>>> procedures, and understanding the diagonal >>>>>>>>>>>>>>>>>>>>> procedure is critical to understanding the *base* >>>>>>>>>>>>>>>>>>>>> contradiction/ paradox that the rest of his support >>>>>>>>>>>>>>>>>>>>> for godel's result is then built on
And focusing on what is said to be impossible and >>>>>>>>>>>>>>>>>>>> not changing the problem is also important. >>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>>>>>>> enumerating the enumeration in the first place. >>>>>>>>>>>>>>>>>>>
Then you zre just showing your stupidity, because YOU >>>>>>>>>>>>>>>>>> can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>>>>>>> enumeration of the computable sequences.
He then points out that he can directly show that the >>>>>>>>>>>>>>>>>> "anti- diagonal" of the (non-effectively computed) >>>>>>>>>>>>>>>>>> enumeration can't be computed but that "This proof, >>>>>>>>>>>>>>>>>> although perfectly sound, has the disadvantage that it >>>>>>>>>>>>>>>>>> may leave the reader with a feeling that 'there must >>>>>>>>>>>>>>>>>> be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is being >>>>>>>>>>>>>>>> done.
I think he is refering he to the standard anti-diagonal >>>>>>>>>>>>>>>> arguement, which shows that since for all n, position n >>>>>>>>>>>>>>>> differs from the value in number n, there can not be any >>>>>>>>>>>>>>>> element that matches the anti- diagonal.
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>>> something it seems you just don't understand.
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with >>>>>>>>>>>>>>> turing's short diagonal proof, but in writing this post i >>>>>>>>>>>>>>> now find myself in a subtle, yet entirely critical >>>>>>>>>>>>>>> disagreement:
/let an be the n-th computable sequence, and let φn(m) be >>>>>>>>>>>>>>> the m- th figure in an. Let β be the sequence with 1- >>>>>>>>>>>>>>> φn(m) as its n-th. figure. Since β is computable, there >>>>>>>>>>>>>>> exists a number K [== β] such that 1- φn(n) = φK(n) for >>>>>>>>>>>>>>> all n. Putting n = K, we have 1 = 2φK(K), i.e. 1 is even. >>>>>>>>>>>>>>> This is impossible/
i agree with this proof is far as much as if β was >>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the >>>>>>>>>>>>> computability of β from the existence of φn(m), THAT'S THE >>>>>>>>>>>>> FALLACy
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE >>>>>>>>>>>>> SPECIFICATION
the fallacy here is assuming that because the direct >>>>>>>>>>>>>>> diagonal is computable, that one can therefore compute >>>>>>>>>>>>>>> the anti- diagonal using the direct diagonal. the >>>>>>>>>>>>>>> abstract definition makes it look simple, but this >>>>>>>>>>>>>>> ignores the complexities of self- referential analysis >>>>>>>>>>>>>>> (like what turing details on the next page)
But you can, *IF* you have a machine that computes the >>>>>>>>>>>>>> diagonal, then just change all the write to the output to >>>>>>>>>>>>>> write the opposite. Note, the "self-reference" that you >>>>>>>>>>>>>> are thinking of stops being a "self- reference" but is a >>>>>>>>>>>>>> reference to the original write the diagonal code. >>>>>>>>>>>>>>
in both methods i have for rectifying the paradox found >>>>>>>>>>>>>>> in the direct diagonal (either (1) filtering TMs or (2) >>>>>>>>>>>>>>> using RTMs), neither can be used to then compute the >>>>>>>>>>>>>>> anti-diagonal
Nope, because your filtering TM (or RTM) will still fail >>>>>>>>>>>>>> when it gets to the number of TURING'S H, as there is no >>>>>>>>>>>>>> correct answer for the machine built by that template. >>>>>>>>>>>>>
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is >>>>>>>>>>>>>>> filtered out like turing's paradoxical variation of the >>>>>>>>>>>>>>> direct diagonal would be, and there is no analogous non- >>>>>>>>>>>>>>> paradoxical variation that has a hard coded value that is >>>>>>>>>>>>>>> inverse to what it does return ... such a concept is >>>>>>>>>>>>>>> entirely nonsensical. a function can only return what it >>>>>>>>>>>>>>> does, it can't also return the inverse to what it returns >>>>>>>>>>>>>>> eh???
But, if D filters it out, then it becomes circle-free, and >>>>>>>>>>>>>> thus your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO >>>>>>>>>>>>> IT *SHOULD* BE FILTERED OUT
in (2) the attempt to compute an inverse diagonal with >>>>>>>>>>>>>>> RTMs just fails for reasons u'd only understand by >>>>>>>>>>>>>>> working thru the algo urself (p7 of re: turing's diagonals) >>>>>>>>>>>>>>>
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>
is just not sufficient evidence that such β is actually >>>>>>>>>>>>>>> computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see how >>>>>>>>>>>>>> to do that, as you are thinking the only "reference" can >>>>>>>>>>>>>> be to "self", doesn't make you argument correct.
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF- >>>>>>>>>>>>> REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H >>>>>>>>>>>>>
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS >>>>>>>>>>>>> FOR INVERSE- H TO BE COMPUTED
Then how do you build YOUR H without that D?He doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>
one cannot just assume that because the diagonal across >>>>>>>>>>>>>>>>> computable numbers is computable, therefore the anti- >>>>>>>>>>>>>>>>> diagonal across computable numbers is computable... >>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and >>>>>>>>>>>>>>>> without the enumeration, you can't compute either of them. >>>>>>>>>>>>>>>>
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>> computation across the computable numbers can be used >>>>>>>>>>>>>>>>> to compute the inverse diagonal
But your method still doesn't let you compute the >>>>>>>>>>>>>>>> enumeration, and thus you can't actually compute the >>>>>>>>>>>>>>>> diagonal.
Remember, the problem definitions requires that the >>>>>>>>>>>>>>>> listing be a COMPLETE listing of the computable >>>>>>>>>>>>>>>> numbers / machine that compute computable numbers, in >>>>>>>>>>>>>>>> some definite order.
If your enumeration isn't complete, your diagonal isn't >>>>>>>>>>>>>>>> correct.
so while i agree with turing that the anti-diagonal is >>>>>>>>>>>>>>>>> not computable, i don't agree that the normal diagonal >>>>>>>>>>>>>>>>> is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on all >>>>>>>>>>>>>>>> the other machines, including his original H that >>>>>>>>>>>>>>>> doesn't use your "trick"
But instead, he can prove with a more obvious process, >>>>>>>>>>>>>>>>>> that the Decider "D" that could be used to effectively >>>>>>>>>>>>>>>>>> enumerate the sequence of machine that produce >>>>>>>>>>>>>>>>>> computable numbers can not esit.
Thus, he clearly knows the difference, but is pointing >>>>>>>>>>>>>>>>>> out that the attempt to compute the diagonal clearly >>>>>>>>>>>>>>>>>> reveals the issue with effectively enumerating the >>>>>>>>>>>>>>>>>> sequences.
well, he didn't consider that perhaps the proper algo >>>>>>>>>>>>>>>>> for computing the diagonal can avoid the paradox on >>>>>>>>>>>>>>>>> itself ...
But it doesn't.
Your just don't understand that D just can't correctly >>>>>>>>>>>>>>>> decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on >>>>>>>>>>>>>>> turing's H, because my response to this is that D does >>>>>>>>>>>>>>> not need to decide correctly on H to compute a diagonal >>>>>>>>>>>>>>
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT >>>>>>>>>>>>> IT WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial >>>>>>>>>>>> recognizer D *can* decide on the fixed H because the fixed H >>>>>>>>>>>> does not try to use any D on itself, so no self-referential >>>>>>>>>>>> paradox is possible in regards to it's own digit on the >>>>>>>>>>>> diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a
dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer D >>>>>>>>>>>> (used in fixed H, which does not call any D itself), and >>>>>>>>>>>> would not be filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, >>>>>>>>>>>> and instead returns a hard-coded value for it's own digit on >>>>>>>>>>>> the diagonal, is keystone in making it /decidable input/ to D >>>>>>>>>>>>
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just >>>>>>>>>>> anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and it >>>>>>>>>>> will fail on that.
bro what part of the non-existence of turing_D or turing_H do >>>>>>>>>> you not understand???
But if your fD exists, then Turing_H that uses your D exists, >>>>>>>>> and will give that D the problem, and fixed_H will call your D >>>>>>>>> on turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H >>>>>>>>>>> gets hung up.
If D errs on Turing_H and calls it not circle_free, then your >>>>>>>>>>> enumeration is missing the circle_free H from its list, and >>>>>>>>>>> thus doesn't make the right diagonal.
now if ur trying to argue: what if we use partial_recognizer_D >>>>>>>>>> in a turing_H type machine? that is a fair point
this would give us the machine (that actually does exist): >>>>>>>>>>
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to
F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>>> satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D >>>>>>>>>> classifier, when actually run actual_turing_H will simply skip >>>>>>>>>> putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the >>>>>>>>> actual_turing_H isn't an "undecidable input", as it has
definite behavior, just behavior that partial_recognizer_D gets >>>>>>>>> wrong.
this is unfortunately a fallacious understanding of
undecidability. just because a machine is /undecidable input/ to >>>>>>>> a particular classifier does mean we cannot then prove and know >>>>>>>> ourselves what the machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs >>>>>>> are not "undecidable", but by the time that they have actually
been made into an input, and thus first had the program created, >>>>>>> which required creating the instance of the decider selctected, >>>>>>> they have definite behavior that other some decider can determine. >>>>>>>
It is only that one selected candidate decider chosen to build
the input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a >>>>>>>> particular machine and the particular classifier it creates a >>>>>>>> structural paradox for,
In other words, you don't understand what "undecidability"
actually is.
Undecidability, means that there does not exist ANY decider that >>>>>>> can get the right answer for all instances of the problem, (or if >>>>>>> the problem has been reduced to just that one input, for that
input).
*not* as a general inability to describe a *real* machine's
behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D.
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and prove
it's equivalent to fixed_H, so therefore it doesn't fit what u
stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a
machine, unless you are narrowing the problem to just being about
that one machine/input.
because it *only* ever *was* about problematic input to *particular*
classifiers...
Except that isn't what "Undecidability" is about. And you are just
proving you don't know what you are talking about.
If we use the correct term, what you are tying to call "undeciability"
is really just not being correct. The problem is a "particualar
classifier" has a fixed set of how it classifies everything. And thus, nothing can be "paradoxical" to it, as we never get to the other side,
The other side you talk about is about if this decider was a different decider, and some how magically the input was also a different input converted by a method not "in system" but only in a meta-system.
it's not my problem everyone went for over a century confusing *our*
ability to prove outcomes, with what a /particular/ decider can
output... ya'll got caught up in a rigor of what is fundamentally
nonsense
Because what a particual decider does, if it isnt' right, is just not interesting,
no one has demonstrated a *possible* machine, let alone a *real*
machine, which we can't provably can't prove the outcome for... such a
proof makes the machine _NEITHER POSSIBLE NOR REAL_ so it would be
ludicrous to present such a proof to exist
Only sort of right. Yes, we can not produce a machine that we know we
can not possibly ever decide if it halts or not, because such a machine
must by definition be non-halting, since ALL halting machines are
provable halting by just running them. Thus, to prove we can't know they halt, we need to prove they don't halt, but then we know their behavior.
The fact we can't demonstrate and example of that class, doesn't mean members of it can't exist.
both turing_H and turing_D are /incomplete specifications/ and
therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_
So? What that proves is that it is IMPOSSIBLE to actually acheive the
goal defined for turing_D, a machine that determines if a given machine
is circle-free.
they are just malformed /incomplete specifications/ and _NOTHING
MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond the recognition
that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such machine
exists.
If you want to be consistant, YOUR partial_decider_D is ALSO incomplete,
as you haven't specified HOW it does what it does, just lie turing
didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such machine
could exist, which he succeeded at, but YOUR claim is that such a thing might be able to exist, which you just have not proven.
And, it is an assertion that NO decider can correctly compute the
answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which means
you have chosen a particular (broken) implementation of D (since no
actually correct D exists), like your partial_recognizer_D, has
fulled defined behavior, and a decider can exist that determines it,
there is nothing "undecidable" about that input.
It is just that the D that H was built on is wrong. Nothing
"paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you just
_IGNORING_ the specification i specifically created for
partial_recognizer_D...
No, your problem is you just are too stupid to understand that your "specification" is just nonsense, as it assumes that you can perform a
test by "changing" the decider and the input to see a results.
But the specification is about a particular input and decider.
the specification for partial_recognizer_D is when encountering
undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your definition is just internally inconsistantly trying to define input that the decider
gets wrong. Since the class the partial decider is
this _DOES NOT MEAN_ we cannot know what the /undecidable input/ does,
as a paradox in regards to partial_recognize_D _ONLY AFFECTS_
partial_recognize_D not _OUR ABILITY TO PROVE OUTCOMES_
In other words, you admit that you words are just lies and don't mean
what they say.
Your "undecidable" isn't about decidability, but correctness of a given decider.
This means that the trivial always false machine meets your
requirements. Any input for which the correct answer for the base
problem would be true, THAT particular decider won't get right, so it is correct in your expanded partial sense to return false.
Right, but the literal values don't have the names on them, and we
(you don't have one btw, so please do let me witness ur cope)
IT seesm you just can't read.
see, turing_H's runtime is not describable because it's an /
incomplete specification/ of machine and therefor _does not even >>>>>>>> exist_ as a real TM, not because of the hypothesized
undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED
defined machine. If the decider "D" actually exists, then H is
FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
specification/, as it doesn't handle idiosyncrasies of TM
computing, specifically self- referential set-classification
paradoxes
It doesn't need to. It just needs to be built on what you want to
claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as your
equivalent for D, like your partial_recognizer_D, then the template >>>>> turing_H can be converted into an actual machine.
since partial_recognizer_D _CAN_ exist, then actual_turing_H _CAN_ >>>>>> exist
actual_turing_H, on the other hand, actually does exist, so we >>>>>>>> can know/ prove what it does. it, however, still forms a
structural paradox in regards to partial_recognizer_D which will >>>>>>>> fail to classify it as circle-free
But, since you admit to actual_turing_H existing, that means that >>>>>>
And so can the turing_H that uses partial_recognizer_D as its
decider D.
The steps the algorithm does is what defines the machine, not the
"names" of the things it "calls", as those names don't actually
exist in the final building of the machine, which is the problem
with your concept, as that is needed to determine your
"paradoxical" property.
the names are just shortcuts for the literal values rick, don't be daft >>>
don't have borders within the machine letting us know we moved from
"outer code of H" into "the code of the D it uses", especially with
Turing Machines were such usages are by necessity "expanded" in-line.
please do actually read turing's paper sometime, he does organize
machine descriptions into functional groups
Yes, at meta level.
Thus, the "copy" of D that was pulled into H isn't actually the same
no it can be literally a copy of same description than is the run by a
UTM (universal turing machine) functional group of the machine
It CAN be, but doesn't NEED to be. And in fact, his H isn't described
that way.
thing as the independent machine D that we can talk about, just a
functional equivalent which doesn't even have its own D.N (since that
is a property only of full machines)
ur really pulling at straws here rick
You really think so?
It seems your problem is you need to work in a tight cage where other
ideas are not allowed that show the holes in your ideas.
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will >>>>>>>>> get wrong, then it can just return the different answer, and >>>>>>>>> then THAT machine it will get right. The problem is it creates >>>>>>>>> a DIFFERENT input, that uses this new version of the machine, >>>>>>>>> that it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as >>>>>>>>>> fixed_H, which in turn _does_ have a spot on the diagonal. so >>>>>>>>>> there is no actual need to include a digit from
actual_turing_H on the diagonal, as the number it computes is >>>>>>>>>> already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix".
And thus if your partial_recoginzer_D accepts ANY verision of >>>>>>>>> actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can >>>>>>>>> possible filter all ALL of that infinite set of machines, but >>>>>>>>> still accept some machine that computes that particular number >>>>>>>>> that it would compute with this supposedly correct
partial_recognizer_D.
You can't ask someone to disprove it, as that is just admitting >>>>>>>>> that you accept that Russel's Teapot must exist.
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the >>>>>>>>> fairy dust, of assuming that your partial decider that meets >>>>>>>>> ALL your needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference between >>>>>>>> fixed_H and actual_turing_H is the way they respectively handle >>>>>>>> their own self- references. they function identically when
handling all other machines
But the problem isn't in it handling the machine fixed_H, but how >>>>>>> it ever handles the results generated by actual_turing_H, which >>>>>>> by your assumption has become a cycle-free machine and ALL
equivalent versions of it have been partially declared to be non- >>>>>>> cycle free.
This means that your enumeration of machines you are looking at >>>>>>> is, in fact, incomplete, thus the diagonal you computed is wrong. >>>>>>>
fixed_H = () -> {
N = 1
K = 0
output = [] // written to
F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded
digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>> satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip >>>>>>>> trying to simulate itself and put it's hard coded digit on the >>>>>>>> diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip >>>>>>>> trying simulate actual_turing_H because partial_recognizer_D
fails to classify it as circle-free
But actual_turing_H is actually circle_free, and thus its results >>>>>>> SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put on >>>>>> the diagonal, then actual_turing_H _DOES NOT_ need to be in _THAT_ >>>>>> total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable
numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on being
sloppy.
the only way to compute a diagonal across computable numbers is
simulating each number by the machine that computes it.
Not the ONLY way, but a good way to try.
But since Turing Shows this can't be done, it can't be done any way.
Note, Turing SPECIFICALLY pointed out that he was going to the
related problem, the problem of:
/but the problem of enumerating computable sequences is equivalent to
the problem of finding out whether a given number is the D.N of a
circle- free machine, and we have no general process for doing this
in a finite number of steps/
oh wow GREAT CATCH RICK!!!!,
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT ENUMERATING
OUT ALL CIRCLE_FREE MACHINES*,
Show it.
THe fact that he quote an equivalence that he doesn't prove he doesn't
make his proof incorrect.
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER IS
A CIRCLE-FREE MACHINE,
Prove it.
we only need to compute that fact for *ONE* machine for any given
computable sequence, _SO A PARTIAL RECOGNIZER IS SUFFICIENT FOR THE JOB_
So do it.
fuck rick ur so bad at defending this it's actually really useful to me!
(there's *another* specific sentence i have a problem with, where's a
ben when u need him?)
All you have shown so far is that you are too stupid to understand the nature of the paper.
Note, it was NOT intended to be the end-all self-contained regerous
proof of the concept. It was a paper, leaning on previous work done, and with the intent of more work to follow to show the basic idea of why something was true.
Thus, not showing how a step works, isn't an error.
Now, to counter his claims, because they are so well accepted, you are
going to need to be rigorous and not based an the fallacy of assuming something and showing that it doesn't lead to a contradiction when you follow a restricted line of reasoining.
So, HIS enumeration is the complete enumeration of "circle-free
machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration either, why
am i the only one subject to criticism here???
Also, your proof is built on the existance of a Russel Teapot, you
have
the *specification* /works/, which shows that it's *not*
contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion. Your
arguement boils down to showing that *IF* you can correctly determine
all machines that fit in a not-definable category (since you concept
of [paradoxical can't actually be defined at the machine level, only
at how a machine might have been created at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can make
the precise "mistakes" you need it to. (They are mistakes because
the input
"mistake" is this case is a fallacious non-techincal value judgement
on the return value,
No, since your criteria that you want to use is just an undefinable
term for a machine description.
There is nothing "paradoxical" of the actual machine when created, as
it then has fixed and determined behavior that can be decided on. The
thing that is paradoxical is the mete-system pattern used to generate
the class of machines as pair with a given decider.
cause they are rather entirely within specification, as FALSE was
specified to be used upon encountering an /undecidable input/,
something turing_D did not handle correctly (making it _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is based on a
actual D that does make a specific decision to the input. And it is
just that for any given version of D, we can find a DIFFERENT H that
it will get wrong (the H that was just built on it).
Do you want to claim that ALL these H are "undecidable" even though for
no... fixed_H contains no paradox, and is therefor not /undecidable
input/ to any classifier
all of them there does exist a decider that gets the right answer?
many machines contain no paradox and are therefore classifiable by all
classifiers
But this proof isn't about the many, but the ALL.
A term it seems you don't understand.
And it isn't the job of others to show you what one you missed, it is
yours to prove that they don't, as YOU are making the claim.
It seems you just don't know what the word means, and thus you
specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of thisnot sure what ur problem is
turing_H is built on this partial_recognizdr_D which causes them to >>>>> not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to machines, as >>>>> it is based on the incorrect definition that assumes a machine can
be something other than itself, and when it changes, it changes
other things that at the meta/template level refered to its old self. >>>>
That you just don't know what your words mean, and you just refuse to
learn.
Try to actually fully DEFINE the word based on the actual properties
of the input machine, and not refering to other machines that are not
part of the input (but would be created as alternate inputs to foil
an alternate decider).
when actual_turing_H iterates across DN(actual_turing_H) itself, >>>>>>>> it will similarly skip trying to simulate itself, because
partial_recognizer_D fails to classify it as circle-free
Right, so it ALSO is wrong, as it didn't compute the diagonal of >>>>>>> the circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE NUMBERS_, >>>>>> which is the _ACTUAL PROBLEM_ turing was dealing with,
No, it isn't, as your "set of computable numbers" has duplicates.
it's still a diagonal across all computerate numbers even if it has
dpulicates ...
No, because an enumeration of a set doesn't HAVE duplicates.
then the H's as of now are i suppose "generators" instead of enumerators >>
and turing was wrong about that too, but i don't see u criticizing him
now eh???
H is a generator, its goal was to generate a particual number.
D is a decider, that allowed H, in its generation, to produce
(internally) the enumeration that its results were defined by,
It seems you don't understand that to reach your goal, sometimes you
have to acheive a sub-goal that is quite different.
Your probe
this level of objection is dishonest\
No, your using misdefined terms is the real dishonesty.
If you want to show that you can make a diagonal of computable
numbers, you need to build an actual effective enumeration of
computable numbers.
And the problem is you can't assume your decider and then use it.
That isn't vallid logic.
but i can show turing was wrong about a series of points, in the most
influence math paper of last century,
No, you haven't.
that's good enough start for me
Then good for you for accepting that you don't know what you are doing,
i can do more once i gain some recognition with that
No, once you get recognition for you stupidity, no one will be
interested in your work./
As I have been pointing out, your whole work is based on BAD DEFINITIONS.
Until you figure out how to actually define (if possible) your concepts,
you have nothing.
It IS valid to assume a decider, and show that such an assumption
leads to a contradiction, to prove that such a decider can not exist.
That is the valid proof by contradiction.
If we can show that if A exist, B must be true, and also that B can
not be true, then we have proven that A can not exist.
Assuming that A exists, and showing that one line of reasoning based
on it doesn't lead to a contradiction, does NOT show that A does (or
even can) exist. That is just fallacious logic.
doing so doesn't require /all/ or _even most_ machines to be in
the enumeration, since _most_ are duplicates of some other machine
when actual_turing_H iterates across DN(fixed_H),
partial_recognizer_D *successfully* classifies fixed_H as
satisfactory, so it will simulate fixed_H to it's Kth digit, the >>>>>>>> one hard-coded into fixed_H
both skip simulating actual_turing_H (which would be circular >>>>>>>> logic in either case), fixed_H also skips simulating itself
(also circular logic), while actual_turing_H actually *does*
simulate fixed_H successfully for it's Kth digit on the
diagonal. these machines compute the *same* number/sequence!
Except that by your partial_recognizer being just partial, it has >>>>>>> dropped required items from the enumeration.
_WHICH IS FINE_
So, you admit it failed???
ACTUAL_TURING_H IS _NOT_ REQUIRED TO BE ON THE DIAGONAL, SO IT'S
FINE TO BE DROPPED
If the diagonal is the diagonal of circle-free machines, it is required. >>>
Your problem is you keep on working with strawmen.
If the diagonal is of computable numbers, you need to show how you
trim down the set to not have repeats (in finite time per element)
and that your decider actually exists, you can't just assume it.
As I said, it dropped REQUIRED items, as this enumeration was of
machines that computed computable numbers.
If you try to add removal of duplicates, then we need to define the >>>>> order, and if we do it by number, turing_H will likely be lower in
order, as it doesn't need to generate its own number, and thus is
simpler,
that sentence is nonsense until u actually produce psuedo-code
I did, just make my H call your "D" replacement, partial_recognizer_D.
Or, is your problem that you think an enumeration doesn't define an
"order"
THis is becides the fact that you still need to show how you plan >>>>>>> on actually detecting all these computationally equivalent
machines in your
fixed_H does not compute a minimum turing-complete set, there are >>>>>> duplicated sequences, ei machines that compute the same sequence/ >>>>>> number, and _THAT IS FINE_
Why?
BECAUSE AN ENUMERATION THAT INCLUDES ONE OR MORE INSTANCE FOR EACH
COMPUTABLE NUMBER STILL ENUMERATES *ALL* COMPUTABLE NUMBERS
Nope. Maybe you don't understand what an enumeration.
And enumeration is a one-to-one mapping between the elements and the
Natural Numbers.
By your definition, you can make a countable infinite enumeation of a
finite set, showing that the finite set is infinite.
And, it was built on the assumption of the existance of a machine,
that actually can not exist (based on Rice's Theorm) and thus is
actually based on the assumption of Unicorns.
partial_recognizer_D is *not* a total decider, so how could rice's
theorum apply???
But it *IS* a total decider on your "paradoxical" property.
It can't let ANY of them through, and it must not omit any of the
ones you need.
partial decider to force them to not be classified as circle-
free, even though it turns out that they are (because your
decider called them not).
You don't seem to understand that you don't get to look at the
template used to build the machine, only the resultant machine
from that template.
And, part of the result of that, is that your attempt to try to >>>>>>> define "paradoxical" just fails, as it resolves to just input
your decider is wrong about, in part, because you don't
understand what "decidable" actually means,
In other words, you think it ok to compute the WRONG diagonal.
see, not only is defining a value in fixed_H that the diagonal >>>>>>>> computation can put for itself in diagonal essential in making >>>>>>>> the diagonal actually computable, so is partial_recognizer_D
skipping over actual_turing_H to avoid the computability paradox! >>>>>>>
The needed diagonal was to a COMPLETE enumeration of the circle- >>>>>>> free machine.
NO, IT WAS SPECIFICALLY A COMPLETE ENUMERATION OF _COMPUTABLE
NUMBERS/ SEQUENCES_
Which you don't get.
smh
But there is to show that your partial_recognizer_D can exist, when >>>>> it can't.
there is *no* requirement to include a sequence _more than once_, >>>>>> so there *no* requirement to include all machines in the enumeration! >>>>>
YOU'VE LOST THE STANDARD PROOF IT DOESN'T EXIST,
But YOUR proof doesn't either, so the standard proof does exist.
WHAT NEW ARGUMENT WOULD THERE BE???
Don't need a NEW argument, as your LIES are just proven to be invalid.
YOU need to prove your "Russel's Teapot" exists.
Your logic just shows that you don't know how to do logic.
That isn't really different then claiming you can prove that
every even number grater than 2 from an infinite set of natural >>>>>>> numbers is the sum of two primes, make you enumeration just the >>>>>>> set of prime numbers, and thus claim that since the only even
number is 2, it is trivially solved.
_FALSE ANALOGY FALLACY_
What is false about it?
THE SUBSET IDENTIFIED BY PARTIAL_RECOGNIZER_D IS NOT "LESS POWERFUL"
THAN THE TOTAL MACHINE SPACE
Sure it is, as it doesn't make the full space.
if the set i'm enumerating includes _AT LEAST ONE INSTANCE_ of
_ALL COMPUTABLE SEQUENCES_
But does it?
YES
How are you sure.
You have an admittedly PARTIAL recognizer.
How do you know that it accepts at least one instance of a machie for
every computable number.
How does your partial decider distinguish between turing's H and your
fixed H?
then it is a turing-complete subset including all possible TM
computable sequences, and is functionally equivalent to the set of >>>>>> *all* TMs
But, since your partial_recognizer_D doesn't exist, but was only
assumed to exist, your program doesn't exist,
ultimately, in order to compute a diagonal across computable
numbers, partial_recognizer_D only needs to successfully
classify *one* of the _infinitely_ many machines that compute >>>>>>>> any given computable number...
But it isn't the diagonal across some set of computable numbers, >>>>>>> but the diagonal across the ENTIRE set of computable numbers.
Since you dropped some machines, you don't have the needed set.
_*THAT DOES NOT NECESSARILY FOLLOW*_
Then try to define what set you are actually making!!
A SET THAT CONTAINS INFINITE INSTANCES OF ALL COMPUTABLE NUMBERS
Which isn't a set.
Boy are you stupid.
Sets contain no duplicates.
The problem is your definition of "paradoxical" doesn't actually
work, as it is based on asking about the results of an input, that
changes.
dropping actual_turing_H from the set, for example, is _FINE_
because it computes _THE SAME SEQUENCE_ as fixed_H, which _IS IN
THE SET_ ...
How do you know it includes your fixed_H in the set?
what part of the actual machine makes it different?
THE CODE OF FIXED_H AND ACTUAL_TURING_H IS IN THE THREAD, WHAT ARE
YOU TALKING ABOUT???
WHich aren't full machines, not until you include the code of
partial_recognizer_D,
And, when you define it, how is Actual_turing_H "undecidable", since
decider exist that decide what it will do.
Remember, in the final machine given as the input, there are not
READ TURING'S PAPER ACTUALLY TO SEE HOW A SUBSET OF TRANSITION
FUNCTIONS WITHIN A TURING MACHINE CAN BE LABELED AND REFERRED TO
Yes, and HE Defined what is subset was.
You haven't, as you have a criteria that is just nonsense, as it is
basded on the input changing when you look at a different decider
looking at it.
seperate turing_H and partial_recognizer_D denoted in the machine,
just one big piece of algorithm. Thus nothing to base your criteria >>>>> on.
Because what you are putting under my nose is just CRAP based on >>>>>>> errors
shit rick, idk how ur still denying what's right under ur nose, >>>>>>>
goddamn rick smh u could spot a diamond in the rough even if god
lead u by ur hand to one put on a pedestal with a _GIANT_ neon
sign pointing to it...
But the problem is that "your diamond" is actually just a big load
of crap that you misidentify as a diamond, because you don't
actually understand what you are doing.
ur so used to playing contrarian u don't know what truth even looks
like anymore, rick
Nope. But you live so much in your stupidity, that you think your
lies are valid.
but this rock is so hard, it's becoming pure diamond
💎💎💎
(someone make a diamond hands emoji eh???)
YOUR H still needs to know if Turing's H is a machine that >>>>>>>>>>>>>> generates a computable number to build the list of such >>>>>>>>>>>>>> machine to define the diagonal to compute.
TURING'S H DOESN'T EXIST. TURING'S D DOESN'T EVEN EXIST. >>>>>>>>>>>>> BOTH ARE INCOMPLETE INCOMPLETE SPECIFICATIONS.>
THE FIXED H DIAGONAL WOULD USE A PARTIAL RECOGNIZER D, >>>>>>>>>>>>> WHICH WILL WORK JUST FINE ON THE ENUMERATION THAT HAS / >>>>>>>>>>>>> UNDECIDABLE INPUT/ TO D FILTERED FROM IT
You seem to not understand the meaning of ALL.
ALL COMPUTABLE NUMBERS =/= ALL MACHINES THAT COMPUTE >>>>>>>>>>>>> COMPUTABLE NUMBERS
Because "undecidability" doesn't affect the correct answer >>>>>>>>>>>>>> it must compute.
It doesn't matter that your new H doesn't get stuck on >>>>>>>>>>>>>>>> itself, it will still error on Turing's H.
turing's H, as it stands, doesn't even exist my dude. he >>>>>>>>>>>>>>> doesn't specify what D (or H) needs to do when
encountering the / undecidable input/ of H, so therefore >>>>>>>>>>>>>>> both D and H are an incomplete specifications of a machine >>>>>>>>>>>>>>
THE SPACE OF TMs MACHINE IS NOT A PERFECT COMPUTATIONAL >>>>>>>>>>>>> SPACE LIKE THEORY KEEPS CONFLATING IT WITH, IT HAS CERTAIN >>>>>>>>>>>>> IDIOSYNCRASIES, AND IT CANNOT COMPUTE EVERYTHING THAT IS >>>>>>>>>>>>> COMPUTABLE
THE DECIDER THAT TURING HYPOTHESIZED _DOES NOT EXIST_, >>>>>>>>>>>>> BECAUSE IT DOES NOT HANDLE THE IDIOSYNCRASIES OF SELF- >>>>>>>>>>>>> REFERENTIAL ANALYSIS, SO IT IS AN /INCOMPLETE
SPECIFICATION/ OF A TM
THAT IS THE *ACTUAL* REASON IT DOES NOT EXIST.
Note, H, when it becomes an actual machine, because we >>>>>>>>>>>>>> have created an actual machine we claim to be the D, has >>>>>>>>>>>>>> an answer, and that D is always wrong.
Thus, it isn't that machines claiming to be D and H can't >>>>>>>>>>>>>> exist, only machines CORRECTLY meeting the requirements of >>>>>>>>>>>>>> D and H can't exist.
Thus, if your H depends on that D, (or a machine that >>>>>>>>>>>>>> meets its specification) then it also can't exist. >>>>>>>>>>>>>>
You just make the error of saying the problems don't exist >>>>>>>>>>>>>> because you can't build them, but YOUR machine can, even >>>>>>>>>>>>>> though it has the same problem.
IF D is wrong by deciding it is not circle free, then >>>>>>>>>>>>>>>> your H will compute the wrong diagonal, as the resulting >>>>>>>>>>>>>>>> version of his H WILL be circle free (since it never >>>>>>>>>>>>>>>> tries to simulate itself) and thus DOES produce an >>>>>>>>>>>>>>>> computable number that your computation misses. >>>>>>>>>>>>>>>>
Or, if that D is wrong by decing it IS circle free, then >>>>>>>>>>>>>>>> when you H tries to process it, it will get stuck in the >>>>>>>>>>>>>>>> infinite loop.
The problem is that in stepping through the machines in >>>>>>>>>>>>>>>> order, you WILL hit these actual machines built on your >>>>>>>>>>>>>>>> erroneous D (your D must have this flaw, as no D without >>>>>>>>>>>>>>>> exists), and thus you will be wrong on THAT input. IT >>>>>>>>>>>>>>>> doesn't matter if you get a good answer for yourself. >>>>>>>>>>>>>>>>
idk what he would have said about it, but prolly >>>>>>>>>>>>>>>>> something more substantial than just calling me >>>>>>>>>>>>>>>>> ignorant repeatedly
I doubt it.
He likely would have gotten frustrated by your idiodic >>>>>>>>>>>>>>>> assertion of bad logic. You would have likely been >>>>>>>>>>>>>>>> escorted out of the meeting as showing you were >>>>>>>>>>>>>>>> unqualified and being a distraction.
Something that seems to be beyond your ignorant >>>>>>>>>>>>>>>>>> understanding.
interestingly: one can only fix the direct diagonal >>>>>>>>>>>>>>>>>>> computation like this
H shows that *IF* you can make that enumeration, you >>>>>>>>>>>>>>>>>>>> can make the diagonal, and thus the anti-diagonal. >>>>>>>>>>>>>>>>>>>> The problem is you can't make that enumeration, and >>>>>>>>>>>>>>>>>>>> assuming you can just shows unsoundness. >>>>>>>>>>>>>>>>>>>
u can't do an analogous fix for the inverse/anti- >>>>>>>>>>>>>>>>>>> diagonal computation. it's not possible hard code a >>>>>>>>>>>>>>>>>>> machine to return an inverted value, a machine can >>>>>>>>>>>>>>>>>>> only return what it does, not the inverse of what it >>>>>>>>>>>>>>>>>>> does...
so if we can filter out paradoxes from the >>>>>>>>>>>>>>>>>>> enumeration, that will leave a direct diagonal >>>>>>>>>>>>>>>>>>> computation extant in that filtered (yet still turing >>>>>>>>>>>>>>>>>>> complete list), while any attempt to compute an >>>>>>>>>>>>>>>>>>> inverse diagonal will not be
But the problem is that "paradoxical machines" don't >>>>>>>>>>>>>>>>>> exist in isolation, but only in relationship to a >>>>>>>>>>>>>>>>>> given machine trying to decide them.
right. so if ur constructing a diagonal across >>>>>>>>>>>>>>>>> computable numbers then u only need to filter out >>>>>>>>>>>>>>>>> paradoxes in regards to the classifier that classifies >>>>>>>>>>>>>>>>> them as a "satisfactory" number
Right, which he shows can not be done.
please do quote where turing shows we can't filter out >>>>>>>>>>>>>>> such paradoxes...
In other words, you beleive unquestionably in the >>>>>>>>>>>>>> existance of Russle's teapot until someone can prove it >>>>>>>>>>>>>> doesn't exist.
BRO U LITERALLY CLAIMED TURING SHOWS FILTERING OUT
PARADOXES CAN'T BE DONE, WHERE IN THE FUCK IS THE REFERENCE >>>>>>>>>>>>> U MORON???
WHY ARE YOU GOING ON ABOUT RUSSEL'S TEAPOT INSTEAD OF >>>>>>>>>>>>> PROVIDING A FUCKING QUOTE TO TURING'S PAPER TO BACK THE >>>>>>>>>>>>> *DIRECT* CLAIM YOU MADE???
I CAN'T BELIEVE THIS IS THE KIND OF DISHONEST GARBAGE I >>>>>>>>>>>>> HAVE TO WORK WITH, BUT OH WELL
For THIS paradox, read the proof. for THIS H, the one >>>>>>>>>>>>>> given, (for which ever version of D you want to try to >>>>>>>>>>>>>> assume is right) D is just wrong.
Thus, it doesn't matter if D can somehow "detect" the >>>>>>>>>>>>>> paradox, because there still isn't an answer it can give >>>>>>>>>>>>>> to H that will be correct, as the template for H will >>>>>>>>>>>>>> always make that D wrong.
Note, the problem is you can't actually DEFINE what a >>>>>>>>>>>>>> "paradox machine" is, as they aren't actually machines, >>>>>>>>>>>>>> but templates that build machines. And that final machine >>>>>>>>>>>>>> doesn't have actually detectable tell-tales that show it >>>>>>>>>>>>>> to be from that template.
Because I am smart, and know what I am talking about. >>>>>>>>>>>>>
(also why do always just make random assertions???) >>>>>>>>>>>>>>
They only seem "random" because you don't know what you >>>>>>>>>>>>>> are talking about and believe in unicorns (and teapots) >>>>>>>>>>>>>>
any machine which *is not* "satisfactory" OR *is not* >>>>>>>>>>>>>>>>> classifiable as satisfactory by said classifier... can >>>>>>>>>>>>>>>>> just be skipped
No, it can only skip those that are not satisfactory, >>>>>>>>>>>>>>>> not those that are but it can not classify as such, or >>>>>>>>>>>>>>>> your enumeration will not be complete, and thus just in >>>>>>>>>>>>>>>> error.
Thus, it needs to be able to correctly classify ALL >>>>>>>>>>>>>>>> machines (as it will be asked about all machines as it >>>>>>>>>>>>>>>> counts through all the descriptions) and thus Turing's H >>>>>>>>>>>>>>>> *WILL* be asked about.
similarly if u want to go a step further an filter out >>>>>>>>>>>>>>>>> computable numbers already included on this diagonal, >>>>>>>>>>>>>>>>> any machine which either *is* computably equivalent OR >>>>>>>>>>>>>>>>> *is not* classifiable in regards to *any* machine >>>>>>>>>>>>>>>>> already the list... can just be skipped
Nope, you can't skip some machines, as you then might >>>>>>>>>>>>>>>> lose some of the computable numbers.
see you can't compute a diagonal across *all* / >>>>>>>>>>>>>>>>> machines/, with said machines, but u can compute a >>>>>>>>>>>>>>>>> diagonal across *all* / computable numbers/
Nope,
Since the enumeration of ALL Computable numbers can't be >>>>>>>>>>>>>>>> done, since ALL classifiers that attempt it will make an >>>>>>>>>>>>>>>> error, you can't do what you want to do.
nah, (a) computing an enumeration of all /computable >>>>>>>>>>>>>>> numbers/ is not the same thing as (b) computing the >>>>>>>>>>>>>>> enumeration of all machines that compute computable >>>>>>>>>>>>>>> numbers. (b) necessarily has duplicates while (a) does >>>>>>>>>>>>>>> not need them. turing's paper wrongly conflates (a) with (b) >>>>>>>>>>>>>>>
i'm pretty sure (a) can be done with TMs
Nope, as your (a) needs the machine D, which lets you >>>>>>>>>>>>>> filter the full list of machibe, which Turing showed can't >>>>>>>>>>>>>> exist.
The problem is such a machine can't handle the
classification of the H he describes. It doesn't matter >>>>>>>>>>>>>> that you can make a DIFFERENT machine, that you try to >>>>>>>>>>>>>> deceptively call "H" too that it won't have a problem >>>>>>>>>>>>>> with, when your "H" gets to the number of Turing's H, it >>>>>>>>>>>>>> still has the problem. It can call that one not circle- >>>>>>>>>>>>>> free, and thus omit a circle- free machine from the list, >>>>>>>>>>>>>> or call it circle- free, and when even YOU try to simulate >>>>>>>>>>>>>> it the k steps, you get stuck in a loop.
(b) probably can't be done with TMs
But if you can do (a), you can do (b). You just need to >>>>>>>>>>>>>> take the code of (a), and invert the symbol put on the >>>>>>>>>>>>>> tape when it is written. Note, it won't be "self- >>>>>>>>>>>>>> referent", as the "N" is uses, is the N of YOUR H, not >>>>>>>>>>>>>> itself. The code for his (a) NEVER reads back what it >>>>>>>>>>>>>> wrote, so that doesn't affect its behavior.
yes, i still do need to prove my thesis that for any >>>>>>>>>>>>>>>>> paradoxical machine, there exists a functionally >>>>>>>>>>>>>>>>> equivalent machine without such paradox
And the problem is that your "paradoxical" isn't >>>>>>>>>>>>>>>> actually a definable property (let alone computable). >>>>>>>>>>>>>>>> Part of the problem is that if you look at just a >>>>>>>>>>>>>>>> machine description, it doesn't (necessarily) tell you >>>>>>>>>>>>>>>> about the use of an "interface" as that use of an >>>>>>>>>>>>>>>> interface can be just inlined, leaving nothing "in the >>>>>>>>>>>>>>>> code" to show it exists.
i'm sorry, are you actually saying the machine
description does not describe what the machine does??? >>>>>>>>>>>>>>
No, that "paradoxical" isn't a definable property of a >>>>>>>>>>>>>> machine.
und = () -> halts(und) loop()
und() includes a structural paradoxical in relation to the >>>>>>>>>>>>> classifier halts(), easily demonstrable thru a short proof >>>>>>>>>>>>> i've done many times
I HAVE NO FUCKING CLUE WHY YOU CAN'T ACCEPT THAT AS A >>>>>>>>>>>>> PROPERTY OF UND(), BESIDES LITERAL BRAIMNDEAD RELIGIOUS >>>>>>>>>>>>> DEVOTION TO SOME FUCKING FORM OF INSANITY I CAN'T HONESTLY >>>>>>>>>>>>> FATHOM
lol
His specified H, with an actually (incorrect) >>>>>>>>>>>>>>>>>> implementation of D (which is all that CAN exist) will >>>>>>>>>>>>>>>>>> either be circle- free and thus generate a number (but >>>>>>>>>>>>>>>>>> its D said it isn't, and thus omitted a valid machine >>>>>>>>>>>>>>>>>> from the list) or it isn't circle- free, and fails to >>>>>>>>>>>>>>>>>> computa a number, and thus should have been omitted >>>>>>>>>>>>>>>>>> from the list but wasn't.
Thus any H that ACTUALLY EXISTS, isn't a "paradox", it >>>>>>>>>>>>>>>>>> is just built on an assuption in error.
so despite turing's worries, the existence of a >>>>>>>>>>>>>>>>>>> diagonal computation does not actually then imply the >>>>>>>>>>>>>>>>>>> existence of an anti- diagonal computation, due the >>>>>>>>>>>>>>>>>>> same particular self- referential weirdness that >>>>>>>>>>>>>>>>>>> stumped turing the first place
But there is no actuall SELF-REFERENCE, so your logic >>>>>>>>>>>>>>>>>> is just based on ERROR.
Your attempt to REDEFINE self-reference to mean >>>>>>>>>>>>>>>>>> processing a number that happens to represent yourself >>>>>>>>>>>>>>>>>> means that you you system "ALL" doesn't actually mean >>>>>>>>>>>>>>>>>> ALL, and thus is just ill- defined.
i'm sorry, you have an issue with me labeling a number >>>>>>>>>>>>>>>>> that directly refers to yourself, as a "self- >>>>>>>>>>>>>>>>> reference" ???
Sure, because it is just a number. The problem is that >>>>>>>>>>>>>>>> you still have
it's a *specific* number that has the currently running >>>>>>>>>>>>>>> machine encoded into it, it's not "just" a number >>>>>>>>>>>>>>> whatever that means
problems with all the "equivalent" machines that have >>>>>>>>>>>>>>>> different numbers.
those are references to functionally equivalent machines, >>>>>>>>>>>>>>> not self- references
Which is why the concept of "self-reference" doesn't work. >>>>>>>>>>>>>>
It may let you filter out the simplest case used in the >>>>>>>>>>>>>>>> proofs, but doesn't solve the actual problem, as the >>>>>>>>>>>>>>>> "Machine Number" doesn't actually fully identify the >>>>>>>>>>>>>>>> problematic cases.
that's not actually true. you can't meaningfully paradox >>>>>>>>>>>>>>> the paradox detector while filter *out* paradoxes to >>>>>>>>>>>>>>> point of creating an actual problem
Then show how you will do it.
This means you need to detect a input that represents a >>>>>>>>>>>>>> machine that uses a computation equivalent to yourself >>>>>>>>>>>>>>
let halts be a halting classifier that takes an input >>>>>>>>>>>>>>> machines and returns true/false whether it halts or not >>>>>>>>>>>>>>>
let paradox_free be a paradox filtering classifier that >>>>>>>>>>>>>>> takes an input classifier, and an input machine to search >>>>>>>>>>>>>>> for paradoxes in regards to: paradox_free(classifier, >>>>>>>>>>>>>>> machine)
WHich doesn't exist.
for example if take our basic halting paradox:
und = () -> halts(und) loop()
then:
paradox_free(halts, und) -> TRUE
but you might think one still can bamboozle the paradox >>>>>>>>>>>>>>> classifier like such:
undp = () -> {
if ( paradox_free(halts, undp) == TRUE ) >>>>>>>>>>>>>>> if ( halts(undp) ) loop()
}
here if paradox_free(halts,undp) -> FALSE, then the if >>>>>>>>>>>>>>> case run a halting paradox form making it /undecidable >>>>>>>>>>>>>>> input/ to halts().
And the problem is "paradox_free" can't exist as an always >>>>>>>>>>>>>> correct decider.
But we can still build an input that your paradox_free() >>>>>>>>>>>>>> fails on.
however, if paradox_free(halts,undp) -> TRUE, then that >>>>>>>>>>>>>>> if case doesn't run and clearly halts(undp) -> TRUE >>>>>>>>>>>>>>> without issues
BUT THAT'S FINE FOR OUR USE CASE, no paradox found within >>>>>>>>>>>>>>> a machine is necessary for the particular input->output >>>>>>>>>>>>>>> computation being done, so utilizing the return >>>>>>>>>>>>>>> paradox_free(halts,undp) -> FALSE to filter out >>>>>>>>>>>>>>> paradox_free() paradoxes as well as halts() paradoxes >>>>>>>>>>>>>>> will suffice to produce a turing complete subset of >>>>>>>>>>>>>>> machines that can be totally classified by halts() >>>>>>>>>>>>>>
SO ACTUALLY DO IT??
WHAT'S THE INPUT THAT CAUSES A FAILURE THAT ACTUALLY MUST >>>>>>>>>>>>> BE INCLUDED ON THE DIAGONAL???
No, it is your dependence on being able to assume that >>>>>>>>>>>>>> unicorns can solve all your problems.
i think a major problem with the consensus perspective is >>>>>>>>>>>>>>> an expectation of a certain cleanliness in the logic as >>>>>>>>>>>>>>> being the only possibility here to compute what we want. >>>>>>>>>>>>>>> TM's can't achieve that, but something slightly messier can. >>>>>>>>>>>>>>
dear future: u see this shit i'm trying work with???? >>>>>>>>>>>>>>>>> sorry i didn't manage to make progress any faster, but >>>>>>>>>>>>>>>>> i'm literally dragging toddlers kicking and screaming >>>>>>>>>>>>>>>>> at this point jeez...
In other words, in your world "ALL" isn't a word that >>>>>>>>>>>>>>>> can actually be used.
You don't understand that you need to use actual sound >>>>>>>>>>>>>>>> logic,
lol, i can only 🙏 someone would engage in sound logic >>>>>>>>>>>>>>> with me
They are, But you are too stuck in your lies to see it. >>>>>>>>>>>>>>
Your "logic" is based on being able to assume that >>>>>>>>>>>>>> Unicorn's exist, and that is a safe assumption until >>>>>>>>>>>>>> someone can prove that Russel's Teapot is not out there. >>>>>>>>>>>>>> (Even though your unicorns HAVE been proved to not exist). >>>>>>>>>>>>>>
You are just proving your utter stupidity.
This just shows that you real problem is you don't >>>>>>>>>>>>>>>>>> understand what the actual problem is, and your world >>>>>>>>>>>>>>>>>> is just build on things that are lies.
...which u won't find interesting, but i can't fix a >>>>>>>>>>>>>>>>>>> lack of curiosity
🤷
expected was an inexpert audience, supposing that >>>>>>>>>>>>>>>>>>>>>> experts will recognise
the relevant mapping to universal quantification? >>>>>>>>>>>>>>>>>>>>>>
IT seems that fundamentally, your world doesn't >>>>>>>>>>>>>>>>>> actually have "computations" as you don't understand >>>>>>>>>>>>>>>>>> the basic requirement that they need to be fully >>>>>>>>>>>>>>>>>> defined in the actions they do.
On 3/7/26 5:52 AM, Richard Damon wrote:
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote:
On 2/28/26 12:38 PM, dart200 wrote:
On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote:
i don't see any indication that turing realized a >>>>>>>>>>>>>>>>>>>> difference thereOn 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> ...And focusing on what is said to be impossible and >>>>>>>>>>>>>>>>>>>>> not changing the problem is also important. >>>>>>>>>>>>>>>>>>>>>
Eh?!yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on >>>>>>>>>>>>>>>>>>>>>>>>>>>> turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by >>>>>>>>>>>>>>>>>>>>>>>>>>> one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ of >>>>>>>>>>>>>>>>>>>>>>>>> an infinitude one by one.
that exactly what turing does in his proof: he >>>>>>>>>>>>>>>>>>>>>>>> defines a comptuation
that enumerates out all the numbers, testing >>>>>>>>>>>>>>>>>>>>>>>> each one of they represent
a "satisfactory"/"circle-free" machine, and >>>>>>>>>>>>>>>>>>>>>>>> adding that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to carefully >>>>>>>>>>>>>>>>>>>>>>>> read p247 of turing's
proof and produce the psuedo-code for the >>>>>>>>>>>>>>>>>>>>>>>> machine H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. Are >>>>>>>>>>>>>>>>>>>>>>> you sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>>>>>>> procedures, and understanding the diagonal >>>>>>>>>>>>>>>>>>>>>> procedure is critical to understanding the *base* >>>>>>>>>>>>>>>>>>>>>> contradiction/ paradox that the rest of his >>>>>>>>>>>>>>>>>>>>>> support for godel's result is then built on >>>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>>>>>>>> enumerating the enumeration in the first place. >>>>>>>>>>>>>>>>>>>>
Then you zre just showing your stupidity, because YOU >>>>>>>>>>>>>>>>>>> can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>>>>>>>> enumeration of the computable sequences. >>>>>>>>>>>>>>>>>>>
He then points out that he can directly show that the >>>>>>>>>>>>>>>>>>> "anti- diagonal" of the (non-effectively computed) >>>>>>>>>>>>>>>>>>> enumeration can't be computed but that "This proof, >>>>>>>>>>>>>>>>>>> although perfectly sound, has the disadvantage that >>>>>>>>>>>>>>>>>>> it may leave the reader with a feeling that 'there >>>>>>>>>>>>>>>>>>> must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is >>>>>>>>>>>>>>>>> being done.
I think he is refering he to the standard anti-diagonal >>>>>>>>>>>>>>>>> arguement, which shows that since for all n, position n >>>>>>>>>>>>>>>>> differs from the value in number n, there can not be >>>>>>>>>>>>>>>>> any element that matches the anti- diagonal. >>>>>>>>>>>>>>>>>
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>>>> something it seems you just don't understand. >>>>>>>>>>>>>>>>>
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with >>>>>>>>>>>>>>>> turing's short diagonal proof, but in writing this post >>>>>>>>>>>>>>>> i now find myself in a subtle, yet entirely critical >>>>>>>>>>>>>>>> disagreement:
/let an be the n-th computable sequence, and let φn(m) >>>>>>>>>>>>>>>> be the m- th figure in an. Let β be the sequence with 1- >>>>>>>>>>>>>>>> φn(m) as its n-th. figure. Since β is computable, there >>>>>>>>>>>>>>>> exists a number K [== β] such that 1- φn(n) = φK(n) for >>>>>>>>>>>>>>>> all n. Putting n = K, we have 1 = 2φK(K), i.e. 1 is >>>>>>>>>>>>>>>> even. This is impossible/
i agree with this proof is far as much as if β was >>>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure,
but what i don't agree is that we can just assume the >>>>>>>>>>>>>> computability of β from the existence of φn(m), THAT'S THE >>>>>>>>>>>>>> FALLACy
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE >>>>>>>>>>>>>> SPECIFICATION
the fallacy here is assuming that because the direct >>>>>>>>>>>>>>>> diagonal is computable, that one can therefore compute >>>>>>>>>>>>>>>> the anti- diagonal using the direct diagonal. the >>>>>>>>>>>>>>>> abstract definition makes it look simple, but this >>>>>>>>>>>>>>>> ignores the complexities of self- referential analysis >>>>>>>>>>>>>>>> (like what turing details on the next page)
But you can, *IF* you have a machine that computes the >>>>>>>>>>>>>>> diagonal, then just change all the write to the output to >>>>>>>>>>>>>>> write the opposite. Note, the "self-reference" that you >>>>>>>>>>>>>>> are thinking of stops being a "self- reference" but is a >>>>>>>>>>>>>>> reference to the original write the diagonal code. >>>>>>>>>>>>>>>
in both methods i have for rectifying the paradox found >>>>>>>>>>>>>>>> in the direct diagonal (either (1) filtering TMs or (2) >>>>>>>>>>>>>>>> using RTMs), neither can be used to then compute the >>>>>>>>>>>>>>>> anti-diagonal
Nope, because your filtering TM (or RTM) will still fail >>>>>>>>>>>>>>> when it gets to the number of TURING'S H, as there is no >>>>>>>>>>>>>>> correct answer for the machine built by that template. >>>>>>>>>>>>>>
(is that where H comes from??? _H_ypothetical???)
in (1) the algo to compute an inverse diagonal is >>>>>>>>>>>>>>>> filtered out like turing's paradoxical variation of the >>>>>>>>>>>>>>>> direct diagonal would be, and there is no analogous non- >>>>>>>>>>>>>>>> paradoxical variation that has a hard coded value that >>>>>>>>>>>>>>>> is inverse to what it does return ... such a concept is >>>>>>>>>>>>>>>> entirely nonsensical. a function can only return what it >>>>>>>>>>>>>>>> does, it can't also return the inverse to what it >>>>>>>>>>>>>>>> returns eh???
But, if D filters it out, then it becomes circle-free, >>>>>>>>>>>>>>> and thus your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO >>>>>>>>>>>>>> IT *SHOULD* BE FILTERED OUT
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH SELF- >>>>>>>>>>>>>> REFERENCE, LIKE THE PARADOX FOUND IN TURING'S HYPOTHETICAL H >>>>>>>>>>>>>>
in (2) the attempt to compute an inverse diagonal with >>>>>>>>>>>>>>>> RTMs just fails for reasons u'd only understand by >>>>>>>>>>>>>>>> working thru the algo urself (p7 of re: turing's diagonals) >>>>>>>>>>>>>>>>
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>>
is just not sufficient evidence that such β is actually >>>>>>>>>>>>>>>> computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see >>>>>>>>>>>>>>> how to do that, as you are thinking the only "reference" >>>>>>>>>>>>>>> can be to "self", doesn't make you argument correct. >>>>>>>>>>>>>>
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS >>>>>>>>>>>>>> FOR INVERSE- H TO BE COMPUTED
Then how do you build YOUR H without that D?He doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>>
one cannot just assume that because the diagonal >>>>>>>>>>>>>>>>>> across computable numbers is computable, therefore the >>>>>>>>>>>>>>>>>> anti- diagonal across computable numbers is computable... >>>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and >>>>>>>>>>>>>>>>> without the enumeration, you can't compute either of them. >>>>>>>>>>>>>>>>>
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>>> computation across the computable numbers can be used >>>>>>>>>>>>>>>>>> to compute the inverse diagonal
But your method still doesn't let you compute the >>>>>>>>>>>>>>>>> enumeration, and thus you can't actually compute the >>>>>>>>>>>>>>>>> diagonal.
Remember, the problem definitions requires that the >>>>>>>>>>>>>>>>> listing be a COMPLETE listing of the computable >>>>>>>>>>>>>>>>> numbers / machine that compute computable numbers, in >>>>>>>>>>>>>>>>> some definite order.
If your enumeration isn't complete, your diagonal isn't >>>>>>>>>>>>>>>>> correct.
so while i agree with turing that the anti-diagonal is >>>>>>>>>>>>>>>>>> not computable, i don't agree that the normal diagonal >>>>>>>>>>>>>>>>>> is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on >>>>>>>>>>>>>>>>> all the other machines, including his original H that >>>>>>>>>>>>>>>>> doesn't use your "trick"
But instead, he can prove with a more obvious >>>>>>>>>>>>>>>>>>> process, that the Decider "D" that could be used to >>>>>>>>>>>>>>>>>>> effectively enumerate the sequence of machine that >>>>>>>>>>>>>>>>>>> produce computable numbers can not esit. >>>>>>>>>>>>>>>>>>>
Thus, he clearly knows the difference, but is >>>>>>>>>>>>>>>>>>> pointing out that the attempt to compute the diagonal >>>>>>>>>>>>>>>>>>> clearly reveals the issue with effectively >>>>>>>>>>>>>>>>>>> enumerating the sequences.
well, he didn't consider that perhaps the proper algo >>>>>>>>>>>>>>>>>> for computing the diagonal can avoid the paradox on >>>>>>>>>>>>>>>>>> itself ...
But it doesn't.
Your just don't understand that D just can't correctly >>>>>>>>>>>>>>>>> decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on >>>>>>>>>>>>>>>> turing's H, because my response to this is that D does >>>>>>>>>>>>>>>> not need to decide correctly on H to compute a diagonal >>>>>>>>>>>>>>>
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT >>>>>>>>>>>>>> IT WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial >>>>>>>>>>>>> recognizer D *can* decide on the fixed H because the fixed >>>>>>>>>>>>> H does not try to use any D on itself, so no self-
referential paradox is possible in regards to it's own >>>>>>>>>>>>> digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a >>>>>>>>>>>>> dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer D >>>>>>>>>>>>> (used in fixed H, which does not call any D itself), and >>>>>>>>>>>>> would not be filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, >>>>>>>>>>>>> and instead returns a hard-coded value for it's own digit >>>>>>>>>>>>> on the diagonal, is keystone in making it /decidable input/ >>>>>>>>>>>>> to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are just >>>>>>>>>>>> anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and >>>>>>>>>>>> it will fail on that.
bro what part of the non-existence of turing_D or turing_H do >>>>>>>>>>> you not understand???
But if your fD exists, then Turing_H that uses your D exists, >>>>>>>>>> and will give that D the problem, and fixed_H will call your D >>>>>>>>>> on turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then fixed_H >>>>>>>>>>>> gets hung up.
If D errs on Turing_H and calls it not circle_free, then >>>>>>>>>>>> your enumeration is missing the circle_free H from its list, >>>>>>>>>>>> and thus doesn't make the right diagonal.
now if ur trying to argue: what if we use
partial_recognizer_D in a turing_H type machine? that is a >>>>>>>>>>> fair point
this would give us the machine (that actually does exist): >>>>>>>>>>>
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to
F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>>>> satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D >>>>>>>>>>> classifier, when actually run actual_turing_H will simply >>>>>>>>>>> skip putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the >>>>>>>>>> actual_turing_H isn't an "undecidable input", as it has
definite behavior, just behavior that partial_recognizer_D >>>>>>>>>> gets wrong.
this is unfortunately a fallacious understanding of
undecidability. just because a machine is /undecidable input/ >>>>>>>>> to a particular classifier does mean we cannot then prove and >>>>>>>>> know ourselves what the machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs >>>>>>>> are not "undecidable", but by the time that they have actually >>>>>>>> been made into an input, and thus first had the program created, >>>>>>>> which required creating the instance of the decider selctected, >>>>>>>> they have definite behavior that other some decider can determine. >>>>>>>>
It is only that one selected candidate decider chosen to build >>>>>>>> the input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between a >>>>>>>>> particular machine and the particular classifier it creates a >>>>>>>>> structural paradox for,
In other words, you don't understand what "undecidability"
actually is.
Undecidability, means that there does not exist ANY decider that >>>>>>>> can get the right answer for all instances of the problem, (or >>>>>>>> if the problem has been reduced to just that one input, for that >>>>>>>> input).
*not* as a general inability to describe a *real* machine's >>>>>>>>> behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D.
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and prove
it's equivalent to fixed_H, so therefore it doesn't fit what u
stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a
machine, unless you are narrowing the problem to just being about
that one machine/input.
because it *only* ever *was* about problematic input to *particular*
classifiers...
Except that isn't what "Undecidability" is about. And you are just
proving you don't know what you are talking about.
If we use the correct term, what you are tying to call "undeciability"
is really just not being correct. The problem is a "particualar
classifier" has a fixed set of how it classifies everything. And thus,
nothing can be "paradoxical" to it, as we never get to the other side,
RICK I'M NEVER GOING TO FIGURE YOU FOR BEING AN UNDYING ASSHOLE, EXCEPT MAYBE ONCE U FINALLY TAKE UR IGNORANCE TO THE GRAVE
The other side you talk about is about if this decider was a different
decider, and some how magically the input was also a different input
converted by a method not "in system" but only in a meta-system.
it's not my problem everyone went for over a century confusing *our*
ability to prove outcomes, with what a /particular/ decider can
output... ya'll got caught up in a rigor of what is fundamentally
nonsense
Because what a particual decider does, if it isnt' right, is just not
interesting,
no one has demonstrated a *possible* machine, let alone a *real*
machine, which we can't provably can't prove the outcome for... such
a proof makes the machine _NEITHER POSSIBLE NOR REAL_ so it would be
ludicrous to present such a proof to exist
Only sort of right. Yes, we can not produce a machine that we know we
can not possibly ever decide if it halts or not, because such a
machine must by definition be non-halting, since ALL halting machines
are provable halting by just running them. Thus, to prove we can't
know they halt, we need to prove they don't halt, but then we know
their behavior.
The fact we can't demonstrate and example of that class, doesn't mean
members of it can't exist.
FUCK UR DAMN TEAPOT RICK
both turing_H and turing_D are /incomplete specifications/ and
therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_
So? What that proves is that it is IMPOSSIBLE to actually acheive the
goal defined for turing_D, a machine that determines if a given
machine is circle-free.
they are just malformed /incomplete specifications/ and _NOTHING
MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond the recognition
that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such machine
exists.
BECAUSE IT'S INCOMPLETE. AN /INCOMPLETE SPECIFICATION/ THAT DOES NOT
HANDLE ALL INPUT SITUATIONS CANNOT EXIST
If you want to be consistant, YOUR partial_decider_D is ALSO incomplete,
IT'S NOT INCOMPLETE AS IT HANDLES ALL INPUT SITUATIONS
as you haven't specified HOW it does what it does, just lie turing
didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such machine
could exist, which he succeeded at, but YOUR claim is that such a
thing might be able to exist, which you just have not proven.
And, it is an assertion that NO decider can correctly compute the
answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which means
you have chosen a particular (broken) implementation of D (since no
actually correct D exists), like your partial_recognizer_D, has
fulled defined behavior, and a decider can exist that determines it,
there is nothing "undecidable" about that input.
It is just that the D that H was built on is wrong. Nothing
"paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you just
_IGNORING_ the specification i specifically created for
partial_recognizer_D...
No, your problem is you just are too stupid to understand that your
"specification" is just nonsense, as it assumes that you can perform a
test by "changing" the decider and the input to see a results.
But the specification is about a particular input and decider.
the specification for partial_recognizer_D is when encountering
undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your definition
is just internally inconsistantly trying to define input that the
decider gets wrong. Since the class the partial decider is
this _DOES NOT MEAN_ we cannot know what the /undecidable input/
does, as a paradox in regards to partial_recognize_D _ONLY AFFECTS_
partial_recognize_D not _OUR ABILITY TO PROVE OUTCOMES_
In other words, you admit that you words are just lies and don't mean
what they say.
Your "undecidable" isn't about decidability, but correctness of a
given decider.
YES, THAT'S ALL UNDECIDABILITY EVER WAS
This means that the trivial always false machine meets your
NO, THE FAILURE TO CLASSIFY MUST COME FROM THE STRUCTURE OF THE INPUT,
NOT AN INCORRECT ALGO OF THE CLASSIFIER, WHY AM I REPEATING ME ON THIS?
requirements. Any input for which the correct answer for the base
problem would be true, THAT particular decider won't get right, so it
is correct in your expanded partial sense to return false.
(you don't have one btw, so please do let me witness ur cope)
IT seesm you just can't read.
see, turing_H's runtime is not describable because it's an / >>>>>>>>> incomplete specification/ of machine and therefor _does not >>>>>>>>> even exist_ as a real TM, not because of the hypothesized
undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED
defined machine. If the decider "D" actually exists, then H is >>>>>>>> FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
specification/, as it doesn't handle idiosyncrasies of TM
computing, specifically self- referential set-classification
paradoxes
It doesn't need to. It just needs to be built on what you want to >>>>>> claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as your
equivalent for D, like your partial_recognizer_D, then the
template turing_H can be converted into an actual machine.
since partial_recognizer_D _CAN_ exist, then actual_turing_H
actual_turing_H, on the other hand, actually does exist, so we >>>>>>>>> can know/ prove what it does. it, however, still forms a
structural paradox in regards to partial_recognizer_D which >>>>>>>>> will fail to classify it as circle-free
But, since you admit to actual_turing_H existing, that means that >>>>>>>
_CAN_ exist
And so can the turing_H that uses partial_recognizer_D as its
decider D.
The steps the algorithm does is what defines the machine, not the >>>>>> "names" of the things it "calls", as those names don't actually
exist in the final building of the machine, which is the problem
with your concept, as that is needed to determine your
"paradoxical" property.
the names are just shortcuts for the literal values rick, don't be
daft
Right, but the literal values don't have the names on them, and we
don't have borders within the machine letting us know we moved from
"outer code of H" into "the code of the D it uses", especially with
Turing Machines were such usages are by necessity "expanded" in-line.
please do actually read turing's paper sometime, he does organize
machine descriptions into functional groups
Yes, at meta level.
Thus, the "copy" of D that was pulled into H isn't actually the same
no it can be literally a copy of same description than is the run by
a UTM (universal turing machine) functional group of the machine
It CAN be, but doesn't NEED to be. And in fact, his H isn't described
that way.
thing as the independent machine D that we can talk about, just a
functional equivalent which doesn't even have its own D.N (since
that is a property only of full machines)
ur really pulling at straws here rick
You really think so?
It seems your problem is you need to work in a tight cage where other
ideas are not allowed that show the holes in your ideas.
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will >>>>>>>>>> get wrong, then it can just return the different answer, and >>>>>>>>>> then THAT machine it will get right. The problem is it creates >>>>>>>>>> a DIFFERENT input, that uses this new version of the machine, >>>>>>>>>> that it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as >>>>>>>>>>> fixed_H, which in turn _does_ have a spot on the diagonal. so >>>>>>>>>>> there is no actual need to include a digit from
actual_turing_H on the diagonal, as the number it computes is >>>>>>>>>>> already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix". >>>>>>>>>>
And thus if your partial_recoginzer_D accepts ANY verision of >>>>>>>>>> actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can >>>>>>>>>> possible filter all ALL of that infinite set of machines, but >>>>>>>>>> still accept some machine that computes that particular number >>>>>>>>>> that it would compute with this supposedly correct
partial_recognizer_D.
You can't ask someone to disprove it, as that is just
admitting that you accept that Russel's Teapot must exist. >>>>>>>>>>
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the >>>>>>>>>> fairy dust, of assuming that your partial decider that meets >>>>>>>>>> ALL your needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference
between fixed_H and actual_turing_H is the way they
respectively handle their own self- references. they function >>>>>>>>> identically when handling all other machines
But the problem isn't in it handling the machine fixed_H, but >>>>>>>> how it ever handles the results generated by actual_turing_H, >>>>>>>> which by your assumption has become a cycle-free machine and ALL >>>>>>>> equivalent versions of it have been partially declared to be
non- cycle free.
This means that your enumeration of machines you are looking at >>>>>>>> is, in fact, incomplete, thus the diagonal you computed is wrong. >>>>>>>>
fixed_H = () -> {
N = 1
K = 0
output = [] // written to
F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard coded
digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>> satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip >>>>>>>>> trying to simulate itself and put it's hard coded digit on the >>>>>>>>> diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip >>>>>>>>> trying simulate actual_turing_H because partial_recognizer_D >>>>>>>>> fails to classify it as circle-free
But actual_turing_H is actually circle_free, and thus its
results SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put on >>>>>>> the diagonal, then actual_turing_H _DOES NOT_ need to be in
_THAT_ total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable
numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on
being sloppy.
the only way to compute a diagonal across computable numbers is
simulating each number by the machine that computes it.
Not the ONLY way, but a good way to try.
But since Turing Shows this can't be done, it can't be done any way.
Note, Turing SPECIFICALLY pointed out that he was going to the
related problem, the problem of:
/but the problem of enumerating computable sequences is equivalent
to the problem of finding out whether a given number is the D.N of a
circle- free machine, and we have no general process for doing this
in a finite number of steps/
oh wow GREAT CATCH RICK!!!!,
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT
ENUMERATING OUT ALL CIRCLE_FREE MACHINES*,
Show it.
what??? bro ur the one who informed me proper enumerations can only have *one* instance of each element enumerated,
and we all know that the total enumeration of machines has infinite
machines producing each computable numbers...
so therefore not only can the enumeration of computable numbers be done without enumerating all circle-free machines, it *MUST* be done without enumerating all circle-free machines, as we can output *ONLY* one
machine for each computable number
that should suffice
THe fact that he quote an equivalence that he doesn't prove he doesn't
make his proof incorrect.
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER
IS A CIRCLE-FREE MACHINE,
Prove it.
there's nothing to prove, the statement is self-evident given the
previous two paragraphs i wrote. we don't need to be able to output it
for *ANY* given machine, we only need to be able to output that fact for *ONE* machine of the infinite which produces *ANY* given compute number
honestly i'm not sure why it's take until year 2026 for someone to point this out
we only need to compute that fact for *ONE* machine for any given
computable sequence, _SO A PARTIAL RECOGNIZER IS SUFFICIENT FOR THE JOB_
So do it.
you ask too much for not recognizing anything i said as true
fuck rick ur so bad at defending this it's actually really useful to me! >>>
(there's *another* specific sentence i have a problem with, where's a
ben when u need him?)
All you have shown so far is that you are too stupid to understand the
nature of the paper.
Note, it was NOT intended to be the end-all self-contained regerous
proof of the concept. It was a paper, leaning on previous work done,
and with the intent of more work to follow to show the basic idea of
why something was true.
Thus, not showing how a step works, isn't an error.
Now, to counter his claims, because they are so well accepted, you are
going to need to be rigorous and not based an the fallacy of assuming
something and showing that it doesn't lead to a contradiction when you
follow a restricted line of reasoining.
So, HIS enumeration is the complete enumeration of "circle-free
machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration either,
why am i the only one subject to criticism here???
no... fixed_H contains no paradox, and is therefor not /undecidable
Also, your proof is built on the existance of a Russel Teapot, you >>>>>> have
the *specification* /works/, which shows that it's *not*
contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion. Your
arguement boils down to showing that *IF* you can correctly
determine all machines that fit in a not-definable category (since
you concept of [paradoxical can't actually be defined at the machine
level, only at how a machine might have been created at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can make >>>>>> the precise "mistakes" you need it to. (They are mistakes because >>>>>> the input
"mistake" is this case is a fallacious non-techincal value
judgement on the return value,
No, since your criteria that you want to use is just an undefinable
term for a machine description.
There is nothing "paradoxical" of the actual machine when created,
as it then has fixed and determined behavior that can be decided on.
The thing that is paradoxical is the mete-system pattern used to
generate the class of machines as pair with a given decider.
cause they are rather entirely within specification, as FALSE was
specified to be used upon encountering an /undecidable input/,
something turing_D did not handle correctly (making it _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is based on a
actual D that does make a specific decision to the input. And it is
just that for any given version of D, we can find a DIFFERENT H that
it will get wrong (the H that was just built on it).
Do you want to claim that ALL these H are "undecidable" even though for >>>
input/ to any classifier
all of them there does exist a decider that gets the right answer?
many machines contain no paradox and are therefore classifiable by
all classifiers
But this proof isn't about the many, but the ALL.
A term it seems you don't understand.
And it isn't the job of others to show you what one you missed, it is
yours to prove that they don't, as YOU are making the claim.
It seems you just don't know what the word means, and thus you
specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of this >>>>>> turing_H is built on this partial_recognizdr_D which causes them
to not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to machines,
as it is based on the incorrect definition that assumes a machine >>>>>> can be something other than itself, and when it changes, it
changes other things that at the meta/template level refered to
its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just refuse
to learn.
Try to actually fully DEFINE the word based on the actual properties
of the input machine, and not refering to other machines that are
not part of the input (but would be created as alternate inputs to
foil an alternate decider).
Right, so it ALSO is wrong, as it didn't compute the diagonal of >>>>>>>> the circle-free machines.
when actual_turing_H iterates across DN(actual_turing_H)
itself, it will similarly skip trying to simulate itself,
because partial_recognizer_D fails to classify it as circle-free >>>>>>>>
fixed_H is computing a diagonal across *all* _COMPUTABLE
NUMBERS_, which is the _ACTUAL PROBLEM_ turing was dealing with,
No, it isn't, as your "set of computable numbers" has duplicates.
it's still a diagonal across all computerate numbers even if it has >>>>> dpulicates ...
No, because an enumeration of a set doesn't HAVE duplicates.
then the H's as of now are i suppose "generators" instead of
enumerators
and turing was wrong about that too, but i don't see u criticizing
him now eh???
H is a generator, its goal was to generate a particual number.
it's goal was a generate a diagonal across all circle-free machines,
which as you said: is *NOT* an enumeration, and an enumeration required duplicating the machines
D is a decider, that allowed H, in its generation, to produce
(internally) the enumeration that its results were defined by,
It seems you don't understand that to reach your goal, sometimes you
have to acheive a sub-goal that is quite different.
On 3/7/26 12:47 PM, dart200 wrote:
On 3/7/26 5:52 AM, Richard Damon wrote:
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 2/28/26 12:38 PM, dart200 wrote:
it is wrong,On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> ...
i don't see any indication that turing realized a >>>>>>>>>>>>>>>>>>>>> difference thereAnd focusing on what is said to be impossible and >>>>>>>>>>>>>>>>>>>>>> not changing the problem is also important. >>>>>>>>>>>>>>>>>>>>>>Eh?!yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>>an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on >>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by >>>>>>>>>>>>>>>>>>>>>>>>>>>> one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ of >>>>>>>>>>>>>>>>>>>>>>>>>> an infinitude one by one.
that exactly what turing does in his proof: he >>>>>>>>>>>>>>>>>>>>>>>>> defines a comptuation
that enumerates out all the numbers, testing >>>>>>>>>>>>>>>>>>>>>>>>> each one of they represent
a "satisfactory"/"circle-free" machine, and >>>>>>>>>>>>>>>>>>>>>>>>> adding that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to >>>>>>>>>>>>>>>>>>>>>>>>> carefully read p247 of turing's >>>>>>>>>>>>>>>>>>>>>>>>> proof and produce the psuedo-code for the >>>>>>>>>>>>>>>>>>>>>>>>> machine H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. >>>>>>>>>>>>>>>>>>>>>>>> Are you sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>>>>>>>> procedures, and understanding the diagonal >>>>>>>>>>>>>>>>>>>>>>> procedure is critical to understanding the *base* >>>>>>>>>>>>>>>>>>>>>>> contradiction/ paradox that the rest of his >>>>>>>>>>>>>>>>>>>>>>> support for godel's result is then built on >>>>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation isn't the >>>>>>>>>>>>>>>>>>>>>> generation of the diagonal itself, but effectively >>>>>>>>>>>>>>>>>>>>>> enumerating the enumeration in the first place. >>>>>>>>>>>>>>>>>>>>>
Then you zre just showing your stupidity, because >>>>>>>>>>>>>>>>>>>> YOU can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>>>>>>>>> enumeration of the computable sequences. >>>>>>>>>>>>>>>>>>>>
He then points out that he can directly show that >>>>>>>>>>>>>>>>>>>> the "anti- diagonal" of the (non-effectively >>>>>>>>>>>>>>>>>>>> computed) enumeration can't be computed but that >>>>>>>>>>>>>>>>>>>> "This proof, although perfectly sound, has the >>>>>>>>>>>>>>>>>>>> disadvantage that it may leave the reader with a >>>>>>>>>>>>>>>>>>>> feeling that 'there must be something wrong'". >>>>>>>>>>>>>>>>>>>
No, YOU are wrong, as you don't understand what is >>>>>>>>>>>>>>>>>> being done.
I think he is refering he to the standard anti- >>>>>>>>>>>>>>>>>> diagonal arguement, which shows that since for all n, >>>>>>>>>>>>>>>>>> position n differs from the value in number n, there >>>>>>>>>>>>>>>>>> can not be any element that matches the anti- diagonal. >>>>>>>>>>>>>>>>>>
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>>>>> something it seems you just don't understand. >>>>>>>>>>>>>>>>>>
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with >>>>>>>>>>>>>>>>> turing's short diagonal proof, but in writing this post >>>>>>>>>>>>>>>>> i now find myself in a subtle, yet entirely critical >>>>>>>>>>>>>>>>> disagreement:
/let an be the n-th computable sequence, and let φn(m) >>>>>>>>>>>>>>>>> be the m- th figure in an. Let β be the sequence with >>>>>>>>>>>>>>>>> 1- φn(m) as its n-th. figure. Since β is computable, >>>>>>>>>>>>>>>>> there exists a number K [== β] such that 1- φn(n) = >>>>>>>>>>>>>>>>> φK(n) for all n. Putting n = K, we have 1 = 2φK(K), >>>>>>>>>>>>>>>>> i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was >>>>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure, >>>>>>>>>>>>>>>
but what i don't agree is that we can just assume the >>>>>>>>>>>>>>> computability of β from the existence of φn(m), THAT'S >>>>>>>>>>>>>>> THE FALLACy
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN INCOMPLETE >>>>>>>>>>>>>>> SPECIFICATION
the fallacy here is assuming that because the direct >>>>>>>>>>>>>>>>> diagonal is computable, that one can therefore compute >>>>>>>>>>>>>>>>> the anti- diagonal using the direct diagonal. the >>>>>>>>>>>>>>>>> abstract definition makes it look simple, but this >>>>>>>>>>>>>>>>> ignores the complexities of self- referential analysis >>>>>>>>>>>>>>>>> (like what turing details on the next page)
But you can, *IF* you have a machine that computes the >>>>>>>>>>>>>>>> diagonal, then just change all the write to the output >>>>>>>>>>>>>>>> to write the opposite. Note, the "self-reference" that >>>>>>>>>>>>>>>> you are thinking of stops being a "self- reference" but >>>>>>>>>>>>>>>> is a reference to the original write the diagonal code. >>>>>>>>>>>>>>>>
in both methods i have for rectifying the paradox found >>>>>>>>>>>>>>>>> in the direct diagonal (either (1) filtering TMs or (2) >>>>>>>>>>>>>>>>> using RTMs), neither can be used to then compute the >>>>>>>>>>>>>>>>> anti-diagonal
Nope, because your filtering TM (or RTM) will still fail >>>>>>>>>>>>>>>> when it gets to the number of TURING'S H, as there is no >>>>>>>>>>>>>>>> correct answer for the machine built by that template. >>>>>>>>>>>>>>>
(is that where H comes from??? _H_ypothetical???) >>>>>>>>>>>>>>>
in (1) the algo to compute an inverse diagonal is >>>>>>>>>>>>>>>>> filtered out like turing's paradoxical variation of the >>>>>>>>>>>>>>>>> direct diagonal would be, and there is no analogous >>>>>>>>>>>>>>>>> non- paradoxical variation that has a hard coded value >>>>>>>>>>>>>>>>> that is inverse to what it does return ... such a >>>>>>>>>>>>>>>>> concept is entirely nonsensical. a function can only >>>>>>>>>>>>>>>>> return what it does, it can't also return the inverse >>>>>>>>>>>>>>>>> to what it returns eh???
But, if D filters it out, then it becomes circle-free, >>>>>>>>>>>>>>>> and thus your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, SO >>>>>>>>>>>>>>> IT *SHOULD* BE FILTERED OUT
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH >>>>>>>>>>>>>>> SELF- REFERENCE, LIKE THE PARADOX FOUND IN TURING'S >>>>>>>>>>>>>>> HYPOTHETICAL H
in (2) the attempt to compute an inverse diagonal with >>>>>>>>>>>>>>>>> RTMs just fails for reasons u'd only understand by >>>>>>>>>>>>>>>>> working thru the algo urself (p7 of re: turing's >>>>>>>>>>>>>>>>> diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>>>
is just not sufficient evidence that such β is actually >>>>>>>>>>>>>>>>> computable given the direct diagonal φn()
And why not. The fact that you are too ignorant to see >>>>>>>>>>>>>>>> how to do that, as you are thinking the only "reference" >>>>>>>>>>>>>>>> can be to "self", doesn't make you argument correct. >>>>>>>>>>>>>>>
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H ALLOWS >>>>>>>>>>>>>>> FOR INVERSE- H TO BE COMPUTED
Then how do you build YOUR H without that D?
one cannot just assume that because the diagonal >>>>>>>>>>>>>>>>>>> across computable numbers is computable, therefore >>>>>>>>>>>>>>>>>>> the anti- diagonal across computable numbers is >>>>>>>>>>>>>>>>>>> computable...
He doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and >>>>>>>>>>>>>>>>>> without the enumeration, you can't compute either of >>>>>>>>>>>>>>>>>> them.
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>>>> computation across the computable numbers can be used >>>>>>>>>>>>>>>>>>> to compute the inverse diagonal
But your method still doesn't let you compute the >>>>>>>>>>>>>>>>>> enumeration, and thus you can't actually compute the >>>>>>>>>>>>>>>>>> diagonal.
Remember, the problem definitions requires that the >>>>>>>>>>>>>>>>>> listing be a COMPLETE listing of the computable >>>>>>>>>>>>>>>>>> numbers / machine that compute computable numbers, in >>>>>>>>>>>>>>>>>> some definite order.
If your enumeration isn't complete, your diagonal >>>>>>>>>>>>>>>>>> isn't correct.
so while i agree with turing that the anti-diagonal >>>>>>>>>>>>>>>>>>> is not computable, i don't agree that the normal >>>>>>>>>>>>>>>>>>> diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on >>>>>>>>>>>>>>>>>> all the other machines, including his original H that >>>>>>>>>>>>>>>>>> doesn't use your "trick"
But instead, he can prove with a more obvious >>>>>>>>>>>>>>>>>>>> process, that the Decider "D" that could be used to >>>>>>>>>>>>>>>>>>>> effectively enumerate the sequence of machine that >>>>>>>>>>>>>>>>>>>> produce computable numbers can not esit. >>>>>>>>>>>>>>>>>>>>
Thus, he clearly knows the difference, but is >>>>>>>>>>>>>>>>>>>> pointing out that the attempt to compute the >>>>>>>>>>>>>>>>>>>> diagonal clearly reveals the issue with effectively >>>>>>>>>>>>>>>>>>>> enumerating the sequences.
well, he didn't consider that perhaps the proper algo >>>>>>>>>>>>>>>>>>> for computing the diagonal can avoid the paradox on >>>>>>>>>>>>>>>>>>> itself ...
But it doesn't.
Your just don't understand that D just can't correctly >>>>>>>>>>>>>>>>>> decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on >>>>>>>>>>>>>>>>> turing's H, because my response to this is that D does >>>>>>>>>>>>>>>>> not need to decide correctly on H to compute a diagonal >>>>>>>>>>>>>>>>
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT THAT >>>>>>>>>>>>>>> IT WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial >>>>>>>>>>>>>> recognizer D *can* decide on the fixed H because the fixed >>>>>>>>>>>>>> H does not try to use any D on itself, so no self- >>>>>>>>>>>>>> referential paradox is possible in regards to it's own >>>>>>>>>>>>>> digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a >>>>>>>>>>>>>> dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer D >>>>>>>>>>>>>> (used in fixed H, which does not call any D itself), and >>>>>>>>>>>>>> would not be filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, >>>>>>>>>>>>>> and instead returns a hard-coded value for it's own digit >>>>>>>>>>>>>> on the diagonal, is keystone in making it /decidable >>>>>>>>>>>>>> input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are >>>>>>>>>>>>> just anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and >>>>>>>>>>>>> it will fail on that.
bro what part of the non-existence of turing_D or turing_H >>>>>>>>>>>> do you not understand???
But if your fD exists, then Turing_H that uses your D exists, >>>>>>>>>>> and will give that D the problem, and fixed_H will call your >>>>>>>>>>> D on turing_H and it still needs to answer about it.
If D errs on Turing_H and calls it circle_free, then >>>>>>>>>>>>> fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, then >>>>>>>>>>>>> your enumeration is missing the circle_free H from its >>>>>>>>>>>>> list, and thus doesn't make the right diagonal.
now if ur trying to argue: what if we use
partial_recognizer_D in a turing_H type machine? that is a >>>>>>>>>>>> fair point
this would give us the machine (that actually does exist): >>>>>>>>>>>>
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written to
F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>>>>> satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D >>>>>>>>>>>> classifier, when actually run actual_turing_H will simply >>>>>>>>>>>> skip putting it's own digit in the computed sequence
But the problem is that for a given partial_recognizer_D, the >>>>>>>>>>> actual_turing_H isn't an "undecidable input", as it has >>>>>>>>>>> definite behavior, just behavior that partial_recognizer_D >>>>>>>>>>> gets wrong.
this is unfortunately a fallacious understanding of
undecidability. just because a machine is /undecidable input/ >>>>>>>>>> to a particular classifier does mean we cannot then prove and >>>>>>>>>> know ourselves what the machine actually does,
But, we aren't looking at "undecidable inputs", as these inputs >>>>>>>>> are not "undecidable", but by the time that they have actually >>>>>>>>> been made into an input, and thus first had the program
created, which required creating the instance of the decider >>>>>>>>> selctected, they have definite behavior that other some decider >>>>>>>>> can determine.
It is only that one selected candidate decider chosen to build >>>>>>>>> the input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between >>>>>>>>>> a particular machine and the particular classifier it creates >>>>>>>>>> a structural paradox for,
In other words, you don't understand what "undecidability"
actually is.
Undecidability, means that there does not exist ANY decider >>>>>>>>> that can get the right answer for all instances of the problem, >>>>>>>>> (or if the problem has been reduced to just that one input, for >>>>>>>>> that input).
*not* as a general inability to describe a *real* machine's >>>>>>>>>> behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D.
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and prove
it's equivalent to fixed_H, so therefore it doesn't fit what u
stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a
machine, unless you are narrowing the problem to just being about
that one machine/input.
because it *only* ever *was* about problematic input to *particular*
classifiers...
Except that isn't what "Undecidability" is about. And you are just
proving you don't know what you are talking about.
If we use the correct term, what you are tying to call
"undeciability" is really just not being correct. The problem is a
"particualar classifier" has a fixed set of how it classifies
everything. And thus, nothing can be "paradoxical" to it, as we never
get to the other side,
RICK I'M NEVER GOING TO FIGURE YOU FOR BEING AN UNDYING ASSHOLE,
EXCEPT MAYBE ONCE U FINALLY TAKE UR IGNORANCE TO THE GRAVE
Right, because logic that is based on error can't unddrstand logic that requires correctness.
The other side you talk about is about if this decider was a
different decider, and some how magically the input was also a
different input converted by a method not "in system" but only in a
meta-system.
it's not my problem everyone went for over a century confusing *our*
ability to prove outcomes, with what a /particular/ decider can
output... ya'll got caught up in a rigor of what is fundamentally
nonsense
Because what a particual decider does, if it isnt' right, is just not
interesting,
no one has demonstrated a *possible* machine, let alone a *real*
machine, which we can't provably can't prove the outcome for... such
a proof makes the machine _NEITHER POSSIBLE NOR REAL_ so it would be
ludicrous to present such a proof to exist
Only sort of right. Yes, we can not produce a machine that we know we
can not possibly ever decide if it halts or not, because such a
machine must by definition be non-halting, since ALL halting machines
are provable halting by just running them. Thus, to prove we can't
know they halt, we need to prove they don't halt, but then we know
their behavior.
The fact we can't demonstrate and example of that class, doesn't mean
members of it can't exist.
FUCK UR DAMN TEAPOT RICK
In other words, you are admitting I am right and need to resort to the fallacy of swearing to distract others from that basic fact.
both turing_H and turing_D are /incomplete specifications/ and
therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_
So? What that proves is that it is IMPOSSIBLE to actually acheive the
goal defined for turing_D, a machine that determines if a given
machine is circle-free.
they are just malformed /incomplete specifications/ and _NOTHING
MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond the recognition
that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such machine
exists.
BECAUSE IT'S INCOMPLETE. AN /INCOMPLETE SPECIFICATION/ THAT DOES NOT
HANDLE ALL INPUT SITUATIONS CANNOT EXIST
The Specification of the INTERFACE is complete, it fully defines what
the given machine must do.
The Specification of the interface is uncomputable, or unrealizable,
because we can't make a machine that fullfills it.
Yes, the specification of the ALGORITHM is incomplete, as we don't
specify how to do all the steps, but then so is yours.
Turing, BECAUSE he used an unspecified algorithm for his D, was able to actually PROVE that the specification for the interface of D was unimplementable.
All your arguement has done is shown that if magic fairy dust powered unicorns exists (which they don't), you could use one of them to built
an (also non-existant) program to compute something that is actually
proven to be uncomputable.
All you have done is proven you don't understand how logic works, and
thus your works SHOULD BE IGNORED.
IT'S NOT INCOMPLETE AS IT HANDLES ALL INPUT SITUATIONS
If you want to be consistant, YOUR partial_decider_D is ALSO incomplete, >>
That isn't what "Incomplete", means, as your partial_decider_D doesn't
exist so it can't handle ANY input situations.
You don't get it, that if you want to put down Turing's machines for not handling cases, you need to also accept that yours doesn't for the same reason, a machine that fully meets your specifaction can't exist.
And part of the reason for yours, is that you have a nonsense term in
your definition, as there is no such thing as a "undecidable input" if
some other decider can get the known right answer.
Part of your problem is you just don't understand what a PROGRAM or ALGORITHM actually is
as you haven't specified HOW it does what it does, just lie turing
didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such machine
could exist, which he succeeded at, but YOUR claim is that such a
thing might be able to exist, which you just have not proven.
And, it is an assertion that NO decider can correctly compute the
answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which
means you have chosen a particular (broken) implementation of D
(since no actually correct D exists), like your
partial_recognizer_D, has fulled defined behavior, and a decider
can exist that determines it, there is nothing "undecidable" about
that input.
It is just that the D that H was built on is wrong. Nothing
"paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you just
_IGNORING_ the specification i specifically created for
partial_recognizer_D...
No, your problem is you just are too stupid to understand that your
"specification" is just nonsense, as it assumes that you can perform
a test by "changing" the decider and the input to see a results.
But the specification is about a particular input and decider.
the specification for partial_recognizer_D is when encountering
undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your definition
is just internally inconsistantly trying to define input that the
decider gets wrong. Since the class the partial decider is
this _DOES NOT MEAN_ we cannot know what the /undecidable input/
does, as a paradox in regards to partial_recognize_D _ONLY AFFECTS_
partial_recognize_D not _OUR ABILITY TO PROVE OUTCOMES_
In other words, you admit that you words are just lies and don't mean
what they say.
Your "undecidable" isn't about decidability, but correctness of a
given decider.
YES, THAT'S ALL UNDECIDABILITY EVER WAS
In other words you are admitting you don't know what you are talking about.
A given decider being incorrect doesn't mean a correct decider doesn't exist. There are many decidable problems, for which incorrect deciders
can exist.
Undecidability is about the being NO correct deciders.
This means that the trivial always false machine meets your
NO, THE FAILURE TO CLASSIFY MUST COME FROM THE STRUCTURE OF THE INPUT,
NOT AN INCORRECT ALGO OF THE CLASSIFIER, WHY AM I REPEATING ME ON THIS?
In other words, you think people/machines can't make mistakes?
Then why are you worried about proofs of correctness.
You don't seem to understand that the whole basis of these proofs isn't about finding *A* input that nothing can decide on, but finding a
pattern that makes an input for a given decider that it won't be able to handle.
That pattern isn't actually something that necessarily remains in the machine after construction, but direction to find a given algorithms
fatal flaw, showing that ALL algorithms for that problem must have a
fatal flaw which can be different for every one.
requirements. Any input for which the correct answer for the base
problem would be true, THAT particular decider won't get right, so it
is correct in your expanded partial sense to return false.
please do actually read turing's paper sometime, he does organize
(you don't have one btw, so please do let me witness ur cope)
IT seesm you just can't read.
see, turing_H's runtime is not describable because it's an / >>>>>>>>>> incomplete specification/ of machine and therefor _does not >>>>>>>>>> even exist_ as a real TM, not because of the hypothesized >>>>>>>>>> undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED
defined machine. If the decider "D" actually exists, then H is >>>>>>>>> FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
specification/, as it doesn't handle idiosyncrasies of TM
computing, specifically self- referential set-classification
paradoxes
It doesn't need to. It just needs to be built on what you want to >>>>>>> claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as your >>>>>>> equivalent for D, like your partial_recognizer_D, then the
template turing_H can be converted into an actual machine.
since partial_recognizer_D _CAN_ exist, then actual_turing_H
actual_turing_H, on the other hand, actually does exist, so we >>>>>>>>>> can know/ prove what it does. it, however, still forms a
structural paradox in regards to partial_recognizer_D which >>>>>>>>>> will fail to classify it as circle-free
But, since you admit to actual_turing_H existing, that means that >>>>>>>>
_CAN_ exist
And so can the turing_H that uses partial_recognizer_D as its
decider D.
The steps the algorithm does is what defines the machine, not the >>>>>>> "names" of the things it "calls", as those names don't actually >>>>>>> exist in the final building of the machine, which is the problem >>>>>>> with your concept, as that is needed to determine your
"paradoxical" property.
the names are just shortcuts for the literal values rick, don't be >>>>>> daft
Right, but the literal values don't have the names on them, and we
don't have borders within the machine letting us know we moved from >>>>> "outer code of H" into "the code of the D it uses", especially with >>>>> Turing Machines were such usages are by necessity "expanded" in-line. >>>>
machine descriptions into functional groups
Yes, at meta level.
no it can be literally a copy of same description than is the run by
Thus, the "copy" of D that was pulled into H isn't actually the same >>>>
a UTM (universal turing machine) functional group of the machine
It CAN be, but doesn't NEED to be. And in fact, his H isn't described
that way.
thing as the independent machine D that we can talk about, just a
functional equivalent which doesn't even have its own D.N (since
that is a property only of full machines)
ur really pulling at straws here rick
You really think so?
It seems your problem is you need to work in a tight cage where other
ideas are not allowed that show the holes in your ideas.
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will >>>>>>>>>>> get wrong, then it can just return the different answer, and >>>>>>>>>>> then THAT machine it will get right. The problem is it
creates a DIFFERENT input, that uses this new version of the >>>>>>>>>>> machine, that it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as >>>>>>>>>>>> fixed_H, which in turn _does_ have a spot on the diagonal. >>>>>>>>>>>> so there is no actual need to include a digit from
actual_turing_H on the diagonal, as the number it computes >>>>>>>>>>>> is already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix". >>>>>>>>>>>
And thus if your partial_recoginzer_D accepts ANY verision of >>>>>>>>>>> actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can >>>>>>>>>>> possible filter all ALL of that infinite set of machines, but >>>>>>>>>>> still accept some machine that computes that particular >>>>>>>>>>> number that it would compute with this supposedly correct >>>>>>>>>>> partial_recognizer_D.
You can't ask someone to disprove it, as that is just
admitting that you accept that Russel's Teapot must exist. >>>>>>>>>>>
goddamn rick, this rock just keeps getting more solid
So you think, because you don't see it is just built on the >>>>>>>>>>> fairy dust, of assuming that your partial decider that meets >>>>>>>>>>> ALL your needs exists,
🤷🤷🤷
to clarify this further: the only algorithmic difference
between fixed_H and actual_turing_H is the way they
respectively handle their own self- references. they function >>>>>>>>>> identically when handling all other machines
But the problem isn't in it handling the machine fixed_H, but >>>>>>>>> how it ever handles the results generated by actual_turing_H, >>>>>>>>> which by your assumption has become a cycle-free machine and >>>>>>>>> ALL equivalent versions of it have been partially declared to >>>>>>>>> be non- cycle free.
This means that your enumeration of machines you are looking at >>>>>>>>> is, in fact, incomplete, thus the diagonal you computed is wrong. >>>>>>>>>
fixed_H = () -> {
N = 1
K = 0
output = [] // written
to F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard
coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>>> satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip >>>>>>>>>> trying to simulate itself and put it's hard coded digit on the >>>>>>>>>> diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will skip >>>>>>>>>> trying simulate actual_turing_H because partial_recognizer_D >>>>>>>>>> fails to classify it as circle-free
But actual_turing_H is actually circle_free, and thus its
results SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put >>>>>>>> on the diagonal, then actual_turing_H _DOES NOT_ need to be in >>>>>>>> _THAT_ total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable
numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on
being sloppy.
the only way to compute a diagonal across computable numbers is
simulating each number by the machine that computes it.
Not the ONLY way, but a good way to try.
But since Turing Shows this can't be done, it can't be done any way.
Note, Turing SPECIFICALLY pointed out that he was going to the
related problem, the problem of:
/but the problem of enumerating computable sequences is equivalent
to the problem of finding out whether a given number is the D.N of
a circle- free machine, and we have no general process for doing
this in a finite number of steps/
oh wow GREAT CATCH RICK!!!!,
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT
ENUMERATING OUT ALL CIRCLE_FREE MACHINES*,
Show it.
what??? bro ur the one who informed me proper enumerations can only
have *one* instance of each element enumerated,
and we all know that the total enumeration of machines has infinite
machines producing each computable numbers...
so therefore not only can the enumeration of computable numbers be
done without enumerating all circle-free machines, it *MUST* be done
without enumerating all circle-free machines, as we can output *ONLY*
one machine for each computable number
that should suffice
In other words, you are admitting you can't do it.
There is no inherent reason you can pull an infinite subset out of a similarly infinite set.
We can enumerate the primes, even though for every prime there is an infinite number of composite numbers that are multiples of it, thus your logic is based on an error.
THe fact that he quote an equivalence that he doesn't prove he
doesn't make his proof incorrect.
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER
IS A CIRCLE-FREE MACHINE,
Prove it.
there's nothing to prove, the statement is self-evident given the
previous two paragraphs i wrote. we don't need to be able to output it
for *ANY* given machine, we only need to be able to output that fact
for *ONE* machine of the infinite which produces *ANY* given compute
number
honestly i'm not sure why it's take until year 2026 for someone to
point this out
In other words, you are just admitting that you understanding of the
problem isn't sufficent to deal with it.
Your problem is you are working on "gut instinct" with a gut that
doesn't understand what it is talking about, because it thinks unicorns exist.
we only need to compute that fact for *ONE* machine for any given
computable sequence, _SO A PARTIAL RECOGNIZER IS SUFFICIENT FOR THE
JOB_
So do it.
you ask too much for not recognizing anything i said as true
Because it isn't, and you have no proofs, because you can't actually
prove a false statement.
Part of your problem is you don't actually know much about what you are talking, so you have nothing to work with,
fuck rick ur so bad at defending this it's actually really useful to
me!
(there's *another* specific sentence i have a problem with, where's
a ben when u need him?)
All you have shown so far is that you are too stupid to understand
the nature of the paper.
Note, it was NOT intended to be the end-all self-contained regerous
proof of the concept. It was a paper, leaning on previous work done,
and with the intent of more work to follow to show the basic idea of
why something was true.
Thus, not showing how a step works, isn't an error.
Now, to counter his claims, because they are so well accepted, you
are going to need to be rigorous and not based an the fallacy of
assuming something and showing that it doesn't lead to a
contradiction when you follow a restricted line of reasoining.
So, HIS enumeration is the complete enumeration of "circle-free
machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration either,
why am i the only one subject to criticism here???
Also, your proof is built on the existance of a Russel Teapot,
you have
the *specification* /works/, which shows that it's *not*
contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion.
Your arguement boils down to showing that *IF* you can correctly
determine all machines that fit in a not-definable category (since
you concept of [paradoxical can't actually be defined at the
machine level, only at how a machine might have been created at a
meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can
make the precise "mistakes" you need it to. (They are mistakes
because the input
"mistake" is this case is a fallacious non-techincal value
judgement on the return value,
No, since your criteria that you want to use is just an undefinable >>>>> term for a machine description.
There is nothing "paradoxical" of the actual machine when created,
as it then has fixed and determined behavior that can be decided
on. The thing that is paradoxical is the mete-system pattern used
to generate the class of machines as pair with a given decider.
But the input is NOT "undecidable" as the actual input is based on
cause they are rather entirely within specification, as FALSE was >>>>>> specified to be used upon encountering an /undecidable input/,
something turing_D did not handle correctly (making it _NOT EXIST_) >>>>>
a actual D that does make a specific decision to the input. And it
is just that for any given version of D, we can find a DIFFERENT H
that it will get wrong (the H that was just built on it).
Do you want to claim that ALL these H are "undecidable" even though >>>>> for
no... fixed_H contains no paradox, and is therefor not /undecidable
input/ to any classifier
all of them there does exist a decider that gets the right answer?
many machines contain no paradox and are therefore classifiable by
all classifiers
But this proof isn't about the many, but the ALL.
A term it seems you don't understand.
And it isn't the job of others to show you what one you missed, it is
yours to prove that they don't, as YOU are making the claim.
It seems you just don't know what the word means, and thus you
specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of this >>>>>>> turing_H is built on this partial_recognizdr_D which causes them >>>>>>> to not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to machines, >>>>>>> as it is based on the incorrect definition that assumes a machine >>>>>>> can be something other than itself, and when it changes, it
changes other things that at the meta/template level refered to >>>>>>> its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just refuse
to learn.
Try to actually fully DEFINE the word based on the actual
properties of the input machine, and not refering to other machines >>>>> that are not part of the input (but would be created as alternate
inputs to foil an alternate decider).
it's still a diagonal across all computerate numbers even if it
No, it isn't, as your "set of computable numbers" has duplicates. >>>>>>
Right, so it ALSO is wrong, as it didn't compute the diagonal >>>>>>>>> of the circle-free machines.
when actual_turing_H iterates across DN(actual_turing_H)
itself, it will similarly skip trying to simulate itself, >>>>>>>>>> because partial_recognizer_D fails to classify it as circle-free >>>>>>>>>
fixed_H is computing a diagonal across *all* _COMPUTABLE
NUMBERS_, which is the _ACTUAL PROBLEM_ turing was dealing with, >>>>>>>
has dpulicates ...
No, because an enumeration of a set doesn't HAVE duplicates.
then the H's as of now are i suppose "generators" instead of
enumerators
and turing was wrong about that too, but i don't see u criticizing
him now eh???
H is a generator, its goal was to generate a particual number.
it's goal was a generate a diagonal across all circle-free machines,
which as you said: is *NOT* an enumeration, and an enumeration
required duplicating the machines
Right, but the result it is generating is BASED (and defined) on a given enumeration which it computes internally).
Your Fixed_H isn't computing anything with a defined specification, as
your "enumeration" that you claim to be based on, isn't actually an enumeration that meets your claimed requirement.
D is a decider, that allowed H, in its generation, to produce
(internally) the enumeration that its results were defined by,
It seems you don't understand that to reach your goal, sometimes you
have to acheive a sub-goal that is quite different.
On 3/8/26 5:20 AM, Richard Damon wrote:
On 3/7/26 12:47 PM, dart200 wrote:
On 3/7/26 5:52 AM, Richard Damon wrote:
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote:
On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 2/28/26 12:38 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> ...
it is wrong,And focusing on what is said to be impossible and >>>>>>>>>>>>>>>>>>>>>>> not changing the problem is also important. >>>>>>>>>>>>>>>>>>>>>>>Eh?!yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>>>an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by >>>>>>>>>>>>>>>>>>>>>>>>>>>>> one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ >>>>>>>>>>>>>>>>>>>>>>>>>>> of an infinitude one by one. >>>>>>>>>>>>>>>>>>>>>>>>>>>
that exactly what turing does in his proof: he >>>>>>>>>>>>>>>>>>>>>>>>>> defines a comptuation
that enumerates out all the numbers, testing >>>>>>>>>>>>>>>>>>>>>>>>>> each one of they represent >>>>>>>>>>>>>>>>>>>>>>>>>> a "satisfactory"/"circle-free" machine, and >>>>>>>>>>>>>>>>>>>>>>>>>> adding that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to >>>>>>>>>>>>>>>>>>>>>>>>>> carefully read p247 of turing's >>>>>>>>>>>>>>>>>>>>>>>>>> proof and produce the psuedo-code for the >>>>>>>>>>>>>>>>>>>>>>>>>> machine H, assuming that
machine D exists
I'll get to it sooner then, because it's mad. >>>>>>>>>>>>>>>>>>>>>>>>> Are you sure he didn't
reason quantified over all but phrase it like a >>>>>>>>>>>>>>>>>>>>>>>>> procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>>>>>>>>> procedures, and understanding the diagonal >>>>>>>>>>>>>>>>>>>>>>>> procedure is critical to understanding the >>>>>>>>>>>>>>>>>>>>>>>> *base* contradiction/ paradox that the rest of >>>>>>>>>>>>>>>>>>>>>>>> his support for godel's result is then built on >>>>>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation isn't >>>>>>>>>>>>>>>>>>>>>>> the generation of the diagonal itself, but >>>>>>>>>>>>>>>>>>>>>>> effectively enumerating the enumeration in the >>>>>>>>>>>>>>>>>>>>>>> first place.
i don't see any indication that turing realized a >>>>>>>>>>>>>>>>>>>>>> difference there
Then you zre just showing your stupidity, because >>>>>>>>>>>>>>>>>>>>> YOU can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not enumerable. >>>>>>>>>>>>>>>>>>>>>
Here is is SPECIFICALLY talking about the effective >>>>>>>>>>>>>>>>>>>>> enumeration of the computable sequences. >>>>>>>>>>>>>>>>>>>>>
He then points out that he can directly show that >>>>>>>>>>>>>>>>>>>>> the "anti- diagonal" of the (non-effectively >>>>>>>>>>>>>>>>>>>>> computed) enumeration can't be computed but that >>>>>>>>>>>>>>>>>>>>> "This proof, although perfectly sound, has the >>>>>>>>>>>>>>>>>>>>> disadvantage that it may leave the reader with a >>>>>>>>>>>>>>>>>>>>> feeling that 'there must be something wrong'". >>>>>>>>>>>>>>>>>>>>
No, YOU are wrong, as you don't understand what is >>>>>>>>>>>>>>>>>>> being done.
I think he is refering he to the standard anti- >>>>>>>>>>>>>>>>>>> diagonal arguement, which shows that since for all n, >>>>>>>>>>>>>>>>>>> position n differs from the value in number n, there >>>>>>>>>>>>>>>>>>> can not be any element that matches the anti- diagonal. >>>>>>>>>>>>>>>>>>>
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>>>>>> something it seems you just don't understand. >>>>>>>>>>>>>>>>>>>
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with >>>>>>>>>>>>>>>>>> turing's short diagonal proof, but in writing this >>>>>>>>>>>>>>>>>> post i now find myself in a subtle, yet entirely >>>>>>>>>>>>>>>>>> critical disagreement:
/let an be the n-th computable sequence, and let φn(m) >>>>>>>>>>>>>>>>>> be the m- th figure in an. Let β be the sequence with >>>>>>>>>>>>>>>>>> 1- φn(m) as its n-th. figure. Since β is computable, >>>>>>>>>>>>>>>>>> there exists a number K [== β] such that 1- φn(n) = >>>>>>>>>>>>>>>>>> φK(n) for all n. Putting n = K, we have 1 = 2φK(K), >>>>>>>>>>>>>>>>>> i.e. 1 is even. This is impossible/
i agree with this proof is far as much as if β was >>>>>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure, >>>>>>>>>>>>>>>>
but what i don't agree is that we can just assume the >>>>>>>>>>>>>>>> computability of β from the existence of φn(m), THAT'S >>>>>>>>>>>>>>>> THE FALLACy
But you can, *IF* you have a machine that computes the >>>>>>>>>>>>>>>>> diagonal, then just change all the write to the output >>>>>>>>>>>>>>>>> to write the opposite. Note, the "self-reference" that >>>>>>>>>>>>>>>>> you are thinking of stops being a "self- reference" but >>>>>>>>>>>>>>>>> is a reference to the original write the diagonal code. >>>>>>>>>>>>>>>>>
the fallacy here is assuming that because the direct >>>>>>>>>>>>>>>>>> diagonal is computable, that one can therefore compute >>>>>>>>>>>>>>>>>> the anti- diagonal using the direct diagonal. the >>>>>>>>>>>>>>>>>> abstract definition makes it look simple, but this >>>>>>>>>>>>>>>>>> ignores the complexities of self- referential analysis >>>>>>>>>>>>>>>>>> (like what turing details on the next page) >>>>>>>>>>>>>>>>>
in both methods i have for rectifying the paradox >>>>>>>>>>>>>>>>>> found in the direct diagonal (either (1) filtering TMs >>>>>>>>>>>>>>>>>> or (2) using RTMs), neither can be used to then >>>>>>>>>>>>>>>>>> compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still >>>>>>>>>>>>>>>>> fail when it gets to the number of TURING'S H, as there >>>>>>>>>>>>>>>>> is no correct answer for the machine built by that >>>>>>>>>>>>>>>>> template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN >>>>>>>>>>>>>>>> INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???) >>>>>>>>>>>>>>>>
in (1) the algo to compute an inverse diagonal is >>>>>>>>>>>>>>>>>> filtered out like turing's paradoxical variation of >>>>>>>>>>>>>>>>>> the direct diagonal would be, and there is no >>>>>>>>>>>>>>>>>> analogous non- paradoxical variation that has a hard >>>>>>>>>>>>>>>>>> coded value that is inverse to what it does return ... >>>>>>>>>>>>>>>>>> such a concept is entirely nonsensical. a function can >>>>>>>>>>>>>>>>>> only return what it does, it can't also return the >>>>>>>>>>>>>>>>>> inverse to what it returns eh???
But, if D filters it out, then it becomes circle-free, >>>>>>>>>>>>>>>>> and thus your enumeration is incomlete.
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, >>>>>>>>>>>>>>>> SO IT *SHOULD* BE FILTERED OUT
And why not. The fact that you are too ignorant to see >>>>>>>>>>>>>>>>> how to do that, as you are thinking the only >>>>>>>>>>>>>>>>> "reference" can be to "self", doesn't make you argument >>>>>>>>>>>>>>>>> correct.
in (2) the attempt to compute an inverse diagonal with >>>>>>>>>>>>>>>>>> RTMs just fails for reasons u'd only understand by >>>>>>>>>>>>>>>>>> working thru the algo urself (p7 of re: turing's >>>>>>>>>>>>>>>>>> diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>>>>
is just not sufficient evidence that such β is >>>>>>>>>>>>>>>>>> actually computable given the direct diagonal φn() >>>>>>>>>>>>>>>>>
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH >>>>>>>>>>>>>>>> SELF- REFERENCE, LIKE THE PARADOX FOUND IN TURING'S >>>>>>>>>>>>>>>> HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H >>>>>>>>>>>>>>>> ALLOWS FOR INVERSE- H TO BE COMPUTED
Then how do you build YOUR H without that D?
one cannot just assume that because the diagonal >>>>>>>>>>>>>>>>>>>> across computable numbers is computable, therefore >>>>>>>>>>>>>>>>>>>> the anti- diagonal across computable numbers is >>>>>>>>>>>>>>>>>>>> computable...
He doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and >>>>>>>>>>>>>>>>>>> without the enumeration, you can't compute either of >>>>>>>>>>>>>>>>>>> them.
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>>>>> computation across the computable numbers can be >>>>>>>>>>>>>>>>>>>> used to compute the inverse diagonal
But your method still doesn't let you compute the >>>>>>>>>>>>>>>>>>> enumeration, and thus you can't actually compute the >>>>>>>>>>>>>>>>>>> diagonal.
Remember, the problem definitions requires that the >>>>>>>>>>>>>>>>>>> listing be a COMPLETE listing of the computable >>>>>>>>>>>>>>>>>>> numbers / machine that compute computable numbers, in >>>>>>>>>>>>>>>>>>> some definite order.
If your enumeration isn't complete, your diagonal >>>>>>>>>>>>>>>>>>> isn't correct.
so while i agree with turing that the anti-diagonal >>>>>>>>>>>>>>>>>>>> is not computable, i don't agree that the normal >>>>>>>>>>>>>>>>>>>> diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on >>>>>>>>>>>>>>>>>>> all the other machines, including his original H that >>>>>>>>>>>>>>>>>>> doesn't use your "trick"
But instead, he can prove with a more obvious >>>>>>>>>>>>>>>>>>>>> process, that the Decider "D" that could be used to >>>>>>>>>>>>>>>>>>>>> effectively enumerate the sequence of machine that >>>>>>>>>>>>>>>>>>>>> produce computable numbers can not esit. >>>>>>>>>>>>>>>>>>>>>
Thus, he clearly knows the difference, but is >>>>>>>>>>>>>>>>>>>>> pointing out that the attempt to compute the >>>>>>>>>>>>>>>>>>>>> diagonal clearly reveals the issue with effectively >>>>>>>>>>>>>>>>>>>>> enumerating the sequences.
well, he didn't consider that perhaps the proper >>>>>>>>>>>>>>>>>>>> algo for computing the diagonal can avoid the >>>>>>>>>>>>>>>>>>>> paradox on itself ...
But it doesn't.
Your just don't understand that D just can't >>>>>>>>>>>>>>>>>>> correctly decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly on >>>>>>>>>>>>>>>>>> turing's H, because my response to this is that D does >>>>>>>>>>>>>>>>>> not need to decide correctly on H to compute a diagonal >>>>>>>>>>>>>>>>>
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT >>>>>>>>>>>>>>>> THAT IT WOULDN'T BE USED
incredibly, ironically, AND not co-incidently: partial >>>>>>>>>>>>>>> recognizer D *can* decide on the fixed H because the >>>>>>>>>>>>>>> fixed H does not try to use any D on itself, so no self- >>>>>>>>>>>>>>> referential paradox is possible in regards to it's own >>>>>>>>>>>>>>> digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a >>>>>>>>>>>>>>> dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer D >>>>>>>>>>>>>>> (used in fixed H, which does not call any D itself), and >>>>>>>>>>>>>>> would not be filtered out by paradox detectors
the fact fixed H does not use any classifier D on itself, >>>>>>>>>>>>>>> and instead returns a hard-coded value for it's own digit >>>>>>>>>>>>>>> on the diagonal, is keystone in making it /decidable >>>>>>>>>>>>>>> input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are >>>>>>>>>>>>>> just anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H and >>>>>>>>>>>>>> it will fail on that.
bro what part of the non-existence of turing_D or turing_H >>>>>>>>>>>>> do you not understand???
But if your fD exists, then Turing_H that uses your D >>>>>>>>>>>> exists, and will give that D the problem, and fixed_H will >>>>>>>>>>>> call your D on turing_H and it still needs to answer about it. >>>>>>>>>>>>
If D errs on Turing_H and calls it circle_free, then >>>>>>>>>>>>>> fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, then >>>>>>>>>>>>>> your enumeration is missing the circle_free H from its >>>>>>>>>>>>>> list, and thus doesn't make the right diagonal.
now if ur trying to argue: what if we use
partial_recognizer_D in a turing_H type machine? that is a >>>>>>>>>>>>> fair point
this would give us the machine (that actually does exist): >>>>>>>>>>>>>
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written
to F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>>>>>> satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE
because actual_turing_H is an /undecidable input/ to any D >>>>>>>>>>>>> classifier, when actually run actual_turing_H will simply >>>>>>>>>>>>> skip putting it's own digit in the computed sequence >>>>>>>>>>>>>
But the problem is that for a given partial_recognizer_D, >>>>>>>>>>>> the actual_turing_H isn't an "undecidable input", as it has >>>>>>>>>>>> definite behavior, just behavior that partial_recognizer_D >>>>>>>>>>>> gets wrong.
this is unfortunately a fallacious understanding of
undecidability. just because a machine is /undecidable input/ >>>>>>>>>>> to a particular classifier does mean we cannot then prove and >>>>>>>>>>> know ourselves what the machine actually does,
But, we aren't looking at "undecidable inputs", as these
inputs are not "undecidable", but by the time that they have >>>>>>>>>> actually been made into an input, and thus first had the
program created, which required creating the instance of the >>>>>>>>>> decider selctected, they have definite behavior that other >>>>>>>>>> some decider can determine.
It is only that one selected candidate decider chosen to build >>>>>>>>>> the input that is shown to get the wrong answer.
undecidability only actually exists as a relationship between >>>>>>>>>>> a particular machine and the particular classifier it creates >>>>>>>>>>> a structural paradox for,
In other words, you don't understand what "undecidability" >>>>>>>>>> actually is.
Undecidability, means that there does not exist ANY decider >>>>>>>>>> that can get the right answer for all instances of the
problem, (or if the problem has been reduced to just that one >>>>>>>>>> input, for that input).
*not* as a general inability to describe a *real* machine's >>>>>>>>>>> behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D.
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and prove >>>>>>> it's equivalent to fixed_H, so therefore it doesn't fit what u
stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a
machine, unless you are narrowing the problem to just being about >>>>>> that one machine/input.
because it *only* ever *was* about problematic input to
*particular* classifiers...
Except that isn't what "Undecidability" is about. And you are just
proving you don't know what you are talking about.
If we use the correct term, what you are tying to call
"undeciability" is really just not being correct. The problem is a
"particualar classifier" has a fixed set of how it classifies
everything. And thus, nothing can be "paradoxical" to it, as we
never get to the other side,
RICK I'M NEVER GOING TO FIGURE YOU FOR BEING AN UNDYING ASSHOLE,
EXCEPT MAYBE ONCE U FINALLY TAKE UR IGNORANCE TO THE GRAVE
Right, because logic that is based on error can't unddrstand logic
that requires correctness.
The other side you talk about is about if this decider was a
different decider, and some how magically the input was also a
different input converted by a method not "in system" but only in a
meta-system.
it's not my problem everyone went for over a century confusing
*our* ability to prove outcomes, with what a /particular/ decider
can output... ya'll got caught up in a rigor of what is
fundamentally nonsense
Because what a particual decider does, if it isnt' right, is just
not interesting,
no one has demonstrated a *possible* machine, let alone a *real*
machine, which we can't provably can't prove the outcome for...
such a proof makes the machine _NEITHER POSSIBLE NOR REAL_ so it
would be ludicrous to present such a proof to exist
Only sort of right. Yes, we can not produce a machine that we know
we can not possibly ever decide if it halts or not, because such a
machine must by definition be non-halting, since ALL halting
machines are provable halting by just running them. Thus, to prove
we can't know they halt, we need to prove they don't halt, but then
we know their behavior.
The fact we can't demonstrate and example of that class, doesn't
mean members of it can't exist.
FUCK UR DAMN TEAPOT RICK
In other words, you are admitting I am right and need to resort to the
fallacy of swearing to distract others from that basic fact.
i don't have a response anymore to the level of stupidity required to
keep bleating on and on about being so certain of machines it would be a contradiction to even produce an example of...
both turing_H and turing_D are /incomplete specifications/ and
therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_
So? What that proves is that it is IMPOSSIBLE to actually acheive
the goal defined for turing_D, a machine that determines if a given
machine is circle-free.
they are just malformed /incomplete specifications/ and _NOTHING
MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond the
recognition that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such machine
exists.
BECAUSE IT'S INCOMPLETE. AN /INCOMPLETE SPECIFICATION/ THAT DOES NOT
HANDLE ALL INPUT SITUATIONS CANNOT EXIST
The Specification of the INTERFACE is complete, it fully defines what
the given machine must do.
imagine trying to argue a machine that can't exist, because it didn't
handle all input situations, as "complete"
The Specification of the interface is uncomputable, or unrealizable,
because we can't make a machine that fullfills it.
Yes, the specification of the ALGORITHM is incomplete, as we don't
specify how to do all the steps, but then so is yours.
Turing, BECAUSE he used an unspecified algorithm for his D, was able
to actually PROVE that the specification for the interface of D was
unimplementable.
All your arguement has done is shown that if magic fairy dust powered
unicorns exists (which they don't), you could use one of them to built
an (also non-existant) program to compute something that is actually
proven to be uncomputable.
All you have done is proven you don't understand how logic works, and
thus your works SHOULD BE IGNORED.
If you want to be consistant, YOUR partial_decider_D is ALSO
incomplete,
IT'S NOT INCOMPLETE AS IT HANDLES ALL INPUT SITUATIONS
That isn't what "Incomplete", means, as your partial_decider_D doesn't
exist so it can't handle ANY input situations.
YOU HAVEN'T DEMONSTRATED A CONTRADICTION WITH PARTIAL_RECOGNIZER_D, SO
YOU HAVEN'T SHOWN IT TO NOT EXIST...
You don't get it, that if you want to put down Turing's machines for
not handling cases, you need to also accept that yours doesn't for the
same reason, a machine that fully meets your specifaction can't exist.
And part of the reason for yours, is that you have a nonsense term in
your definition, as there is no such thing as a "undecidable input" if
some other decider can get the known right answer.
Part of your problem is you just don't understand what a PROGRAM or
ALGORITHM actually is
as you haven't specified HOW it does what it does, just lie turing
didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such machine
could exist, which he succeeded at, but YOUR claim is that such a
thing might be able to exist, which you just have not proven.
And, it is an assertion that NO decider can correctly compute the >>>>>> answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which
means you have chosen a particular (broken) implementation of D
(since no actually correct D exists), like your
partial_recognizer_D, has fulled defined behavior, and a decider
can exist that determines it, there is nothing "undecidable" about >>>>>> that input.
It is just that the D that H was built on is wrong. Nothing
"paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you
just _IGNORING_ the specification i specifically created for
partial_recognizer_D...
No, your problem is you just are too stupid to understand that your
"specification" is just nonsense, as it assumes that you can perform
a test by "changing" the decider and the input to see a results.
But the specification is about a particular input and decider.
the specification for partial_recognizer_D is when encountering
undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your
definition is just internally inconsistantly trying to define input
that the decider gets wrong. Since the class the partial decider is
this _DOES NOT MEAN_ we cannot know what the /undecidable input/
does, as a paradox in regards to partial_recognize_D _ONLY AFFECTS_ >>>>> partial_recognize_D not _OUR ABILITY TO PROVE OUTCOMES_
In other words, you admit that you words are just lies and don't
mean what they say.
Your "undecidable" isn't about decidability, but correctness of a
given decider.
YES, THAT'S ALL UNDECIDABILITY EVER WAS
In other words you are admitting you don't know what you are talking
about.
A given decider being incorrect doesn't mean a correct decider doesn't
exist. There are many decidable problems, for which incorrect deciders
can exist.
Undecidability is about the being NO correct deciders.
YOU HAVEN'T EVEN SHOW A MACHINE WHICH HAS NO CORRECT CLASSIFIERS FOR...
This means that the trivial always false machine meets your
NO, THE FAILURE TO CLASSIFY MUST COME FROM THE STRUCTURE OF THE
INPUT, NOT AN INCORRECT ALGO OF THE CLASSIFIER, WHY AM I REPEATING ME
ON THIS?
In other words, you think people/machines can't make mistakes?
BEING UNABLE TO CLASSIFY /UNDECIDABLE INPUT/ IS _NOT_ A MISTAKE ON
BEHALF OF THE CLASSIFIER
Then why are you worried about proofs of correctness.
You don't seem to understand that the whole basis of these proofs
isn't about finding *A* input that nothing can decide on, but finding
a pattern that makes an input for a given decider that it won't be
able to handle.
That pattern isn't actually something that necessarily remains in the
machine after construction, but direction to find a given algorithms
fatal flaw, showing that ALL algorithms for that problem must have a
fatal flaw which can be different for every one.
requirements. Any input for which the correct answer for the base
problem would be true, THAT particular decider won't get right, so
it is correct in your expanded partial sense to return false.
IT seesm you just can't read.
(you don't have one btw, so please do let me witness ur cope) >>>>>>>>
see, turing_H's runtime is not describable because it's an / >>>>>>>>>>> incomplete specification/ of machine and therefor _does not >>>>>>>>>>> even exist_ as a real TM, not because of the hypothesized >>>>>>>>>>> undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED >>>>>>>>>> defined machine. If the decider "D" actually exists, then H is >>>>>>>>>> FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
specification/, as it doesn't handle idiosyncrasies of TM
computing, specifically self- referential set-classification >>>>>>>>> paradoxes
It doesn't need to. It just needs to be built on what you want >>>>>>>> to claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as your >>>>>>>> equivalent for D, like your partial_recognizer_D, then the
template turing_H can be converted into an actual machine.
since partial_recognizer_D _CAN_ exist, then actual_turing_H >>>>>>>>> _CAN_ exist
actual_turing_H, on the other hand, actually does exist, so >>>>>>>>>>> we can know/ prove what it does. it, however, still forms a >>>>>>>>>>> structural paradox in regards to partial_recognizer_D which >>>>>>>>>>> will fail to classify it as circle-free
But, since you admit to actual_turing_H existing, that means that >>>>>>>>>
And so can the turing_H that uses partial_recognizer_D as its >>>>>>>> decider D.
The steps the algorithm does is what defines the machine, not >>>>>>>> the "names" of the things it "calls", as those names don't
actually exist in the final building of the machine, which is >>>>>>>> the problem with your concept, as that is needed to determine >>>>>>>> your "paradoxical" property.
the names are just shortcuts for the literal values rick, don't >>>>>>> be daft
Right, but the literal values don't have the names on them, and we >>>>>> don't have borders within the machine letting us know we moved
from "outer code of H" into "the code of the D it uses",
especially with Turing Machines were such usages are by necessity >>>>>> "expanded" in-line.
please do actually read turing's paper sometime, he does organize
machine descriptions into functional groups
Yes, at meta level.
no it can be literally a copy of same description than is the run
Thus, the "copy" of D that was pulled into H isn't actually the same >>>>>
by a UTM (universal turing machine) functional group of the machine
It CAN be, but doesn't NEED to be. And in fact, his H isn't
described that way.
thing as the independent machine D that we can talk about, just a >>>>>> functional equivalent which doesn't even have its own D.N (since
that is a property only of full machines)
ur really pulling at straws here rick
You really think so?
It seems your problem is you need to work in a tight cage where
other ideas are not allowed that show the holes in your ideas.
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm will >>>>>>>>>>>> get wrong, then it can just return the different answer, and >>>>>>>>>>>> then THAT machine it will get right. The problem is it >>>>>>>>>>>> creates a DIFFERENT input, that uses this new version of the >>>>>>>>>>>> machine, that it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as >>>>>>>>>>>>> fixed_H, which in turn _does_ have a spot on the diagonal. >>>>>>>>>>>>> so there is no actual need to include a digit from
actual_turing_H on the diagonal, as the number it computes >>>>>>>>>>>>> is already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix". >>>>>>>>>>>>
And thus if your partial_recoginzer_D accepts ANY verision >>>>>>>>>>>> of actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can >>>>>>>>>>>> possible filter all ALL of that infinite set of machines, >>>>>>>>>>>> but still accept some machine that computes that particular >>>>>>>>>>>> number that it would compute with this supposedly correct >>>>>>>>>>>> partial_recognizer_D.
You can't ask someone to disprove it, as that is just >>>>>>>>>>>> admitting that you accept that Russel's Teapot must exist. >>>>>>>>>>>>
So you think, because you don't see it is just built on the >>>>>>>>>>>> fairy dust, of assuming that your partial decider that meets >>>>>>>>>>>> ALL your needs exists,
goddamn rick, this rock just keeps getting more solid >>>>>>>>>>>>
🤷🤷🤷
to clarify this further: the only algorithmic difference >>>>>>>>>>> between fixed_H and actual_turing_H is the way they
respectively handle their own self- references. they function >>>>>>>>>>> identically when handling all other machines
But the problem isn't in it handling the machine fixed_H, but >>>>>>>>>> how it ever handles the results generated by actual_turing_H, >>>>>>>>>> which by your assumption has become a cycle-free machine and >>>>>>>>>> ALL equivalent versions of it have been partially declared to >>>>>>>>>> be non- cycle free.
This means that your enumeration of machines you are looking >>>>>>>>>> at is, in fact, incomplete, thus the diagonal you computed is >>>>>>>>>> wrong.
fixed_H = () -> {
N = 1
K = 0
output = [] // written
to F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard
coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>>>> satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will skip >>>>>>>>>>> trying to simulate itself and put it's hard coded digit on >>>>>>>>>>> the diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will >>>>>>>>>>> skip trying simulate actual_turing_H because
partial_recognizer_D fails to classify it as circle-free
But actual_turing_H is actually circle_free, and thus its >>>>>>>>>> results SHOULD have been on the diagonal.
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put >>>>>>>>> on the diagonal, then actual_turing_H _DOES NOT_ need to be in >>>>>>>>> _THAT_ total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable >>>>>>>> numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on
being sloppy.
the only way to compute a diagonal across computable numbers is
simulating each number by the machine that computes it.
Not the ONLY way, but a good way to try.
But since Turing Shows this can't be done, it can't be done any way.
Note, Turing SPECIFICALLY pointed out that he was going to the
related problem, the problem of:
/but the problem of enumerating computable sequences is equivalent >>>>>> to the problem of finding out whether a given number is the D.N of >>>>>> a circle- free machine, and we have no general process for doing
this in a finite number of steps/
oh wow GREAT CATCH RICK!!!!,
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT
ENUMERATING OUT ALL CIRCLE_FREE MACHINES*,
Show it.
what??? bro ur the one who informed me proper enumerations can only
have *one* instance of each element enumerated,
and we all know that the total enumeration of machines has infinite
machines producing each computable numbers...
so therefore not only can the enumeration of computable numbers be
done without enumerating all circle-free machines, it *MUST* be done
without enumerating all circle-free machines, as we can output *ONLY*
one machine for each computable number
that should suffice
In other words, you are admitting you can't do it.
no richard, we've reached a point where u are unable to correctly and honestly identify a self-evident concept:
FOR _ANY_ GIVEN COMPUTABLE NUMBER, THERE ARE _INFINITE_ MACHINES WHICH COMPUTE IT ...
WE ONLY NEED TO CLASSIFY _ONE_ OF THOSE MACHINES TO ADD THAT GIVEN COMPUTABLE NUMBER THE LIST, THE REST NOT ONLY CAN, BUT MUST, BE IGNORED
THEREFOR, WE DO _NOT_ NEED A CLASSIFIER WITH THE POWER TO COMPUTE
WHETHER _ANY_ GIVEN MACHINE IS A COMPUTABLE NUMBER, WE ONLY NEED A CLASSIFIER WITH ENOUGH POWER TO CLASSIFIER _ONE OF THE INFINITE_
MACHINES FOR _ANY_ GIVEN COMPUTABLE NUMBER
There is no inherent reason you can pull an infinite subset out of a
similarly infinite set.
We can enumerate the primes, even though for every prime there is an
infinite number of composite numbers that are multiples of it, thus
your logic is based on an error.
THe fact that he quote an equivalence that he doesn't prove he
doesn't make his proof incorrect.
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER >>>>> IS A CIRCLE-FREE MACHINE,
Prove it.
there's nothing to prove, the statement is self-evident given the
previous two paragraphs i wrote. we don't need to be able to output
it for *ANY* given machine, we only need to be able to output that
fact for *ONE* machine of the infinite which produces *ANY* given
compute number
honestly i'm not sure why it's take until year 2026 for someone to
point this out
In other words, you are just admitting that you understanding of the
problem isn't sufficent to deal with it.
no richard, we've reached a point where u are unable to correctly and honestly identify a self-evident concept:
THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER IS A CIRCLE-FREE MACHINE, AS WE DO NOT NEED TO IDENTIFY _ALL_ CIRCLE-FREE MACHINES, JUST _ONE FOR EACH_ COMPUTABLE NUMBER
Your problem is you are working on "gut instinct" with a gut that
doesn't understand what it is talking about, because it thinks
unicorns exist.
we only need to compute that fact for *ONE* machine for any given
computable sequence, _SO A PARTIAL RECOGNIZER IS SUFFICIENT FOR THE >>>>> JOB_
So do it.
you ask too much for not recognizing anything i said as true
no richard, we've reached a point where u are unable to correctly and honestly identify a self-evident concept
Because it isn't, and you have no proofs, because you can't actually
prove a false statement.
Part of your problem is you don't actually know much about what you
are talking, so you have nothing to work with,
fuck rick ur so bad at defending this it's actually really useful
to me!
(there's *another* specific sentence i have a problem with, where's >>>>> a ben when u need him?)
All you have shown so far is that you are too stupid to understand
the nature of the paper.
Note, it was NOT intended to be the end-all self-contained regerous
proof of the concept. It was a paper, leaning on previous work done,
and with the intent of more work to follow to show the basic idea of
why something was true.
Thus, not showing how a step works, isn't an error.
Now, to counter his claims, because they are so well accepted, you
are going to need to be rigorous and not based an the fallacy of
assuming something and showing that it doesn't lead to a
contradiction when you follow a restricted line of reasoining.
So, HIS enumeration is the complete enumeration of "circle-free
machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration either,
why am i the only one subject to criticism here???
Also, your proof is built on the existance of a Russel Teapot, >>>>>>>> you have
the *specification* /works/, which shows that it's *not*
contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion.
Your arguement boils down to showing that *IF* you can correctly
determine all machines that fit in a not-definable category (since >>>>>> you concept of [paradoxical can't actually be defined at the
machine level, only at how a machine might have been created at a >>>>>> meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can >>>>>>>> make the precise "mistakes" you need it to. (They are mistakes >>>>>>>> because the input
"mistake" is this case is a fallacious non-techincal value
judgement on the return value,
No, since your criteria that you want to use is just an
undefinable term for a machine description.
There is nothing "paradoxical" of the actual machine when created, >>>>>> as it then has fixed and determined behavior that can be decided
on. The thing that is paradoxical is the mete-system pattern used >>>>>> to generate the class of machines as pair with a given decider.
But the input is NOT "undecidable" as the actual input is based on >>>>>> a actual D that does make a specific decision to the input. And it >>>>>> is just that for any given version of D, we can find a DIFFERENT H >>>>>> that it will get wrong (the H that was just built on it).
cause they are rather entirely within specification, as FALSE was >>>>>>> specified to be used upon encountering an /undecidable input/,
something turing_D did not handle correctly (making it _NOT EXIST_) >>>>>>
Do you want to claim that ALL these H are "undecidable" even
though for
no... fixed_H contains no paradox, and is therefor not /undecidable >>>>> input/ to any classifier
all of them there does exist a decider that gets the right answer?
many machines contain no paradox and are therefore classifiable by
all classifiers
But this proof isn't about the many, but the ALL.
A term it seems you don't understand.
And it isn't the job of others to show you what one you missed, it
is yours to prove that they don't, as YOU are making the claim.
It seems you just don't know what the word means, and thus you
specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of
this turing_H is built on this partial_recognizdr_D which causes >>>>>>>> them to not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to machines, >>>>>>>> as it is based on the incorrect definition that assumes a
machine can be something other than itself, and when it changes, >>>>>>>> it changes other things that at the meta/template level refered >>>>>>>> to its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just refuse >>>>>> to learn.
Try to actually fully DEFINE the word based on the actual
properties of the input machine, and not refering to other
machines that are not part of the input (but would be created as
alternate inputs to foil an alternate decider).
it's still a diagonal across all computerate numbers even if it >>>>>>> has dpulicates ...
No, it isn't, as your "set of computable numbers" has duplicates. >>>>>>>
Right, so it ALSO is wrong, as it didn't compute the diagonal >>>>>>>>>> of the circle-free machines.
when actual_turing_H iterates across DN(actual_turing_H) >>>>>>>>>>> itself, it will similarly skip trying to simulate itself, >>>>>>>>>>> because partial_recognizer_D fails to classify it as circle-free >>>>>>>>>>
fixed_H is computing a diagonal across *all* _COMPUTABLE
NUMBERS_, which is the _ACTUAL PROBLEM_ turing was dealing with, >>>>>>>>
No, because an enumeration of a set doesn't HAVE duplicates.
then the H's as of now are i suppose "generators" instead of
enumerators
and turing was wrong about that too, but i don't see u criticizing
him now eh???
H is a generator, its goal was to generate a particual number.
it's goal was a generate a diagonal across all circle-free machines,
which as you said: is *NOT* an enumeration, and an enumeration
required duplicating the machines
Right, but the result it is generating is BASED (and defined) on a
given enumeration which it computes internally).
Your Fixed_H isn't computing anything with a defined specification, as
your "enumeration" that you claim to be based on, isn't actually an
enumeration that meets your claimed requirement.
D is a decider, that allowed H, in its generation, to produce
(internally) the enumeration that its results were defined by,
It seems you don't understand that to reach your goal, sometimes you
have to acheive a sub-goal that is quite different.
On 3/8/26 11:27 AM, dart200 wrote:
On 3/8/26 5:20 AM, Richard Damon wrote:
On 3/7/26 12:47 PM, dart200 wrote:
On 3/7/26 5:52 AM, Richard Damon wrote:
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:i agree with this proof is far as much as if β was >>>>>>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure, >>>>>>>>>>>>>>>>>
On 2/28/26 6:24 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/28/26 12:38 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> ...
it is wrong,And focusing on what is said to be impossible >>>>>>>>>>>>>>>>>>>>>>>> and not changing the problem is also important. >>>>>>>>>>>>>>>>>>>>>>>>Eh?!yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>>>>an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing's original paper and can be reused >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> anywhere...
You think you can test all of them one by >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ >>>>>>>>>>>>>>>>>>>>>>>>>>>> of an infinitude one by one. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
that exactly what turing does in his proof: >>>>>>>>>>>>>>>>>>>>>>>>>>> he defines a comptuation >>>>>>>>>>>>>>>>>>>>>>>>>>> that enumerates out all the numbers, testing >>>>>>>>>>>>>>>>>>>>>>>>>>> each one of they represent >>>>>>>>>>>>>>>>>>>>>>>>>>> a "satisfactory"/"circle-free" machine, and >>>>>>>>>>>>>>>>>>>>>>>>>>> adding that to diagonal
across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to >>>>>>>>>>>>>>>>>>>>>>>>>>> carefully read p247 of turing's >>>>>>>>>>>>>>>>>>>>>>>>>>> proof and produce the psuedo-code for the >>>>>>>>>>>>>>>>>>>>>>>>>>> machine H, assuming that >>>>>>>>>>>>>>>>>>>>>>>>>>> machine D exists
I'll get to it sooner then, because it's mad. >>>>>>>>>>>>>>>>>>>>>>>>>> Are you sure he didn't
reason quantified over all but phrase it like >>>>>>>>>>>>>>>>>>>>>>>>>> a procedure for what he
the theory of computation is the theory of such >>>>>>>>>>>>>>>>>>>>>>>>> procedures, and understanding the diagonal >>>>>>>>>>>>>>>>>>>>>>>>> procedure is critical to understanding the >>>>>>>>>>>>>>>>>>>>>>>>> *base* contradiction/ paradox that the rest of >>>>>>>>>>>>>>>>>>>>>>>>> his support for godel's result is then built on >>>>>>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation isn't >>>>>>>>>>>>>>>>>>>>>>>> the generation of the diagonal itself, but >>>>>>>>>>>>>>>>>>>>>>>> effectively enumerating the enumeration in the >>>>>>>>>>>>>>>>>>>>>>>> first place.
i don't see any indication that turing realized a >>>>>>>>>>>>>>>>>>>>>>> difference there
Then you zre just showing your stupidity, because >>>>>>>>>>>>>>>>>>>>>> YOU can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not >>>>>>>>>>>>>>>>>>>>>> enumerable.
Here is is SPECIFICALLY talking about the >>>>>>>>>>>>>>>>>>>>>> effective enumeration of the computable sequences. >>>>>>>>>>>>>>>>>>>>>>
He then points out that he can directly show that >>>>>>>>>>>>>>>>>>>>>> the "anti- diagonal" of the (non-effectively >>>>>>>>>>>>>>>>>>>>>> computed) enumeration can't be computed but that >>>>>>>>>>>>>>>>>>>>>> "This proof, although perfectly sound, has the >>>>>>>>>>>>>>>>>>>>>> disadvantage that it may leave the reader with a >>>>>>>>>>>>>>>>>>>>>> feeling that 'there must be something wrong'". >>>>>>>>>>>>>>>>>>>>>
No, YOU are wrong, as you don't understand what is >>>>>>>>>>>>>>>>>>>> being done.
I think he is refering he to the standard anti- >>>>>>>>>>>>>>>>>>>> diagonal arguement, which shows that since for all >>>>>>>>>>>>>>>>>>>> n, position n differs from the value in number n, >>>>>>>>>>>>>>>>>>>> there can not be any element that matches the anti- >>>>>>>>>>>>>>>>>>>> diagonal.
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>>>>>>> something it seems you just don't understand. >>>>>>>>>>>>>>>>>>>>
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with >>>>>>>>>>>>>>>>>>> turing's short diagonal proof, but in writing this >>>>>>>>>>>>>>>>>>> post i now find myself in a subtle, yet entirely >>>>>>>>>>>>>>>>>>> critical disagreement:
/let an be the n-th computable sequence, and let >>>>>>>>>>>>>>>>>>> φn(m) be the m- th figure in an. Let β be the >>>>>>>>>>>>>>>>>>> sequence with 1- φn(m) as its n-th. figure. Since β >>>>>>>>>>>>>>>>>>> is computable, there exists a number K [== β] such >>>>>>>>>>>>>>>>>>> that 1- φn(n) = φK(n) for all n. Putting n = K, we >>>>>>>>>>>>>>>>>>> have 1 = 2φK(K), i.e. 1 is even. This is impossible/ >>>>>>>>>>>>>>>>>
but what i don't agree is that we can just assume the >>>>>>>>>>>>>>>>> computability of β from the existence of φn(m), THAT'S >>>>>>>>>>>>>>>>> THE FALLACy
But you can, *IF* you have a machine that computes the >>>>>>>>>>>>>>>>>> diagonal, then just change all the write to the output >>>>>>>>>>>>>>>>>> to write the opposite. Note, the "self-reference" that >>>>>>>>>>>>>>>>>> you are thinking of stops being a "self- reference" >>>>>>>>>>>>>>>>>> but is a reference to the original write the diagonal >>>>>>>>>>>>>>>>>> code.
the fallacy here is assuming that because the direct >>>>>>>>>>>>>>>>>>> diagonal is computable, that one can therefore >>>>>>>>>>>>>>>>>>> compute the anti- diagonal using the direct diagonal. >>>>>>>>>>>>>>>>>>> the abstract definition makes it look simple, but >>>>>>>>>>>>>>>>>>> this ignores the complexities of self- referential >>>>>>>>>>>>>>>>>>> analysis (like what turing details on the next page) >>>>>>>>>>>>>>>>>>
in both methods i have for rectifying the paradox >>>>>>>>>>>>>>>>>>> found in the direct diagonal (either (1) filtering >>>>>>>>>>>>>>>>>>> TMs or (2) using RTMs), neither can be used to then >>>>>>>>>>>>>>>>>>> compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still >>>>>>>>>>>>>>>>>> fail when it gets to the number of TURING'S H, as >>>>>>>>>>>>>>>>>> there is no correct answer for the machine built by >>>>>>>>>>>>>>>>>> that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN >>>>>>>>>>>>>>>>> INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???) >>>>>>>>>>>>>>>>>
But, if D filters it out, then it becomes circle-free, >>>>>>>>>>>>>>>>>> and thus your enumeration is incomlete.
in (1) the algo to compute an inverse diagonal is >>>>>>>>>>>>>>>>>>> filtered out like turing's paradoxical variation of >>>>>>>>>>>>>>>>>>> the direct diagonal would be, and there is no >>>>>>>>>>>>>>>>>>> analogous non- paradoxical variation that has a hard >>>>>>>>>>>>>>>>>>> coded value that is inverse to what it does >>>>>>>>>>>>>>>>>>> return ... such a concept is entirely nonsensical. a >>>>>>>>>>>>>>>>>>> function can only return what it does, it can't also >>>>>>>>>>>>>>>>>>> return the inverse to what it returns eh??? >>>>>>>>>>>>>>>>>>
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, >>>>>>>>>>>>>>>>> SO IT *SHOULD* BE FILTERED OUT
And why not. The fact that you are too ignorant to see >>>>>>>>>>>>>>>>>> how to do that, as you are thinking the only >>>>>>>>>>>>>>>>>> "reference" can be to "self", doesn't make you >>>>>>>>>>>>>>>>>> argument correct.
in (2) the attempt to compute an inverse diagonal >>>>>>>>>>>>>>>>>>> with RTMs just fails for reasons u'd only understand >>>>>>>>>>>>>>>>>>> by working thru the algo urself (p7 of re: turing's >>>>>>>>>>>>>>>>>>> diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>>>>>
is just not sufficient evidence that such β is >>>>>>>>>>>>>>>>>>> actually computable given the direct diagonal φn() >>>>>>>>>>>>>>>>>>
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH >>>>>>>>>>>>>>>>> SELF- REFERENCE, LIKE THE PARADOX FOUND IN TURING'S >>>>>>>>>>>>>>>>> HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H >>>>>>>>>>>>>>>>> ALLOWS FOR INVERSE- H TO BE COMPUTED
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT >>>>>>>>>>>>>>>>> THAT IT WOULDN'T BE USED
one cannot just assume that because the diagonal >>>>>>>>>>>>>>>>>>>>> across computable numbers is computable, therefore >>>>>>>>>>>>>>>>>>>>> the anti- diagonal across computable numbers is >>>>>>>>>>>>>>>>>>>>> computable...
He doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and >>>>>>>>>>>>>>>>>>>> without the enumeration, you can't compute either of >>>>>>>>>>>>>>>>>>>> them.
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>>>>>> computation across the computable numbers can be >>>>>>>>>>>>>>>>>>>>> used to compute the inverse diagonal
But your method still doesn't let you compute the >>>>>>>>>>>>>>>>>>>> enumeration, and thus you can't actually compute the >>>>>>>>>>>>>>>>>>>> diagonal.
Remember, the problem definitions requires that the >>>>>>>>>>>>>>>>>>>> listing be a COMPLETE listing of the computable >>>>>>>>>>>>>>>>>>>> numbers / machine that compute computable numbers, >>>>>>>>>>>>>>>>>>>> in some definite order.
If your enumeration isn't complete, your diagonal >>>>>>>>>>>>>>>>>>>> isn't correct.
so while i agree with turing that the anti-diagonal >>>>>>>>>>>>>>>>>>>>> is not computable, i don't agree that the normal >>>>>>>>>>>>>>>>>>>>> diagonal is not computable
Why?
How does D decide on the original H?
Your modified H still needs a correct D to decide on >>>>>>>>>>>>>>>>>>>> all the other machines, including his original H >>>>>>>>>>>>>>>>>>>> that doesn't use your "trick"
well, he didn't consider that perhaps the proper >>>>>>>>>>>>>>>>>>>>> algo for computing the diagonal can avoid the >>>>>>>>>>>>>>>>>>>>> paradox on itself ...
But instead, he can prove with a more obvious >>>>>>>>>>>>>>>>>>>>>> process, that the Decider "D" that could be used >>>>>>>>>>>>>>>>>>>>>> to effectively enumerate the sequence of machine >>>>>>>>>>>>>>>>>>>>>> that produce computable numbers can not esit. >>>>>>>>>>>>>>>>>>>>>>
Thus, he clearly knows the difference, but is >>>>>>>>>>>>>>>>>>>>>> pointing out that the attempt to compute the >>>>>>>>>>>>>>>>>>>>>> diagonal clearly reveals the issue with >>>>>>>>>>>>>>>>>>>>>> effectively enumerating the sequences. >>>>>>>>>>>>>>>>>>>>>
But it doesn't.
Your just don't understand that D just can't >>>>>>>>>>>>>>>>>>>> correctly decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly >>>>>>>>>>>>>>>>>>> on turing's H, because my response to this is that D >>>>>>>>>>>>>>>>>>> does not need to decide correctly on H to compute a >>>>>>>>>>>>>>>>>>> diagonal
Then how do you build YOUR H without that D? >>>>>>>>>>>>>>>>>
incredibly, ironically, AND not co-incidently: partial >>>>>>>>>>>>>>>> recognizer D *can* decide on the fixed H because the >>>>>>>>>>>>>>>> fixed H does not try to use any D on itself, so no self- >>>>>>>>>>>>>>>> referential paradox is possible in regards to it's own >>>>>>>>>>>>>>>> digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a >>>>>>>>>>>>>>>> dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer D >>>>>>>>>>>>>>>> (used in fixed H, which does not call any D itself), and >>>>>>>>>>>>>>>> would not be filtered out by paradox detectors >>>>>>>>>>>>>>>>
the fact fixed H does not use any classifier D on >>>>>>>>>>>>>>>> itself, and instead returns a hard-coded value for it's >>>>>>>>>>>>>>>> own digit on the diagonal, is keystone in making it / >>>>>>>>>>>>>>>> decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are >>>>>>>>>>>>>>> just anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H >>>>>>>>>>>>>>> and it will fail on that.
bro what part of the non-existence of turing_D or turing_H >>>>>>>>>>>>>> do you not understand???
But if your fD exists, then Turing_H that uses your D >>>>>>>>>>>>> exists, and will give that D the problem, and fixed_H will >>>>>>>>>>>>> call your D on turing_H and it still needs to answer about it. >>>>>>>>>>>>>
If D errs on Turing_H and calls it circle_free, then >>>>>>>>>>>>>>> fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, then >>>>>>>>>>>>>>> your enumeration is missing the circle_free H from its >>>>>>>>>>>>>>> list, and thus doesn't make the right diagonal.
now if ur trying to argue: what if we use
partial_recognizer_D in a turing_H type machine? that is a >>>>>>>>>>>>>> fair point
this would give us the machine (that actually does exist): >>>>>>>>>>>>>>
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written
to F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>>>>>>> satisfactory
K += 1
digit = simulate_Kth_digit(N,K)
output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE >>>>>>>>>>>>>>
because actual_turing_H is an /undecidable input/ to any D >>>>>>>>>>>>>> classifier, when actually run actual_turing_H will simply >>>>>>>>>>>>>> skip putting it's own digit in the computed sequence >>>>>>>>>>>>>>
But the problem is that for a given partial_recognizer_D, >>>>>>>>>>>>> the actual_turing_H isn't an "undecidable input", as it has >>>>>>>>>>>>> definite behavior, just behavior that partial_recognizer_D >>>>>>>>>>>>> gets wrong.
this is unfortunately a fallacious understanding of
undecidability. just because a machine is /undecidable >>>>>>>>>>>> input/ to a particular classifier does mean we cannot then >>>>>>>>>>>> prove and know ourselves what the machine actually does, >>>>>>>>>>>>
But, we aren't looking at "undecidable inputs", as these >>>>>>>>>>> inputs are not "undecidable", but by the time that they have >>>>>>>>>>> actually been made into an input, and thus first had the >>>>>>>>>>> program created, which required creating the instance of the >>>>>>>>>>> decider selctected, they have definite behavior that other >>>>>>>>>>> some decider can determine.
It is only that one selected candidate decider chosen to >>>>>>>>>>> build the input that is shown to get the wrong answer.
undecidability only actually exists as a relationship >>>>>>>>>>>> between a particular machine and the particular classifier >>>>>>>>>>>> it creates a structural paradox for,
In other words, you don't understand what "undecidability" >>>>>>>>>>> actually is.
Undecidability, means that there does not exist ANY decider >>>>>>>>>>> that can get the right answer for all instances of the
problem, (or if the problem has been reduced to just that one >>>>>>>>>>> input, for that input).
*not* as a general inability to describe a *real* machine's >>>>>>>>>>>> behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D. >>>>>>>>>
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and prove >>>>>>>> it's equivalent to fixed_H, so therefore it doesn't fit what u >>>>>>>> stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a
machine, unless you are narrowing the problem to just being about >>>>>>> that one machine/input.
because it *only* ever *was* about problematic input to
*particular* classifiers...
Except that isn't what "Undecidability" is about. And you are just
proving you don't know what you are talking about.
If we use the correct term, what you are tying to call
"undeciability" is really just not being correct. The problem is a
"particualar classifier" has a fixed set of how it classifies
everything. And thus, nothing can be "paradoxical" to it, as we
never get to the other side,
RICK I'M NEVER GOING TO FIGURE YOU FOR BEING AN UNDYING ASSHOLE,
EXCEPT MAYBE ONCE U FINALLY TAKE UR IGNORANCE TO THE GRAVE
Right, because logic that is based on error can't unddrstand logic
that requires correctness.
The other side you talk about is about if this decider was a
different decider, and some how magically the input was also a
different input converted by a method not "in system" but only in a >>>>> meta-system.
it's not my problem everyone went for over a century confusing
*our* ability to prove outcomes, with what a /particular/ decider >>>>>> can output... ya'll got caught up in a rigor of what is
fundamentally nonsense
Because what a particual decider does, if it isnt' right, is just
not interesting,
no one has demonstrated a *possible* machine, let alone a *real*
machine, which we can't provably can't prove the outcome for...
such a proof makes the machine _NEITHER POSSIBLE NOR REAL_ so it
would be ludicrous to present such a proof to exist
Only sort of right. Yes, we can not produce a machine that we know
we can not possibly ever decide if it halts or not, because such a
machine must by definition be non-halting, since ALL halting
machines are provable halting by just running them. Thus, to prove
we can't know they halt, we need to prove they don't halt, but then >>>>> we know their behavior.
The fact we can't demonstrate and example of that class, doesn't
mean members of it can't exist.
FUCK UR DAMN TEAPOT RICK
In other words, you are admitting I am right and need to resort to
the fallacy of swearing to distract others from that basic fact.
i don't have a response anymore to the level of stupidity required to
keep bleating on and on about being so certain of machines it would be
a contradiction to even produce an example of...
In other words, you are just addmitting that you are too stupid to understand the possiblity of something being unknowable.
It seems you are incapable of thinking abstractly, of even being able to conceive of something you didn't invent without an example.
But you are also incapable of thinking concreately, because you don't understand the need to show that there is a reason you need to be able
to show that the machines you imagine are possible.
I will point out AGAIN, that these unknowable machines are not needed to
be understood to understand the uncomputable nature of the problem (They just come out as an effect of that).
both turing_H and turing_D are /incomplete specifications/ and
therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_
So? What that proves is that it is IMPOSSIBLE to actually acheive
the goal defined for turing_D, a machine that determines if a given >>>>> machine is circle-free.
they are just malformed /incomplete specifications/ and _NOTHING
MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond the
recognition that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such
machine exists.
BECAUSE IT'S INCOMPLETE. AN /INCOMPLETE SPECIFICATION/ THAT DOES NOT
HANDLE ALL INPUT SITUATIONS CANNOT EXIST
The Specification of the INTERFACE is complete, it fully defines what
the given machine must do.
imagine trying to argue a machine that can't exist, because it didn't
handle all input situations, as "complete"
In other words, you can't understand the difference between a
specification of an interface, and the specification of a machine.
By your own logic, YOUR machine is INCOMPLETE, as you can't give the
actual specific algorithm it uses. Your "logic" is based on ASSUMING
that something is computable, with absolutely ZERO evidence that it is.
In other words, you live in a world of fantasy
The Specification of the interface is uncomputable, or unrealizable,
because we can't make a machine that fullfills it.
Yes, the specification of the ALGORITHM is incomplete, as we don't
specify how to do all the steps, but then so is yours.
Turing, BECAUSE he used an unspecified algorithm for his D, was able
to actually PROVE that the specification for the interface of D was
unimplementable.
All your arguement has done is shown that if magic fairy dust powered
unicorns exists (which they don't), you could use one of them to
built an (also non-existant) program to compute something that is
actually proven to be uncomputable.
All you have done is proven you don't understand how logic works, and
thus your works SHOULD BE IGNORED.
If you want to be consistant, YOUR partial_decider_D is ALSO
incomplete,
IT'S NOT INCOMPLETE AS IT HANDLES ALL INPUT SITUATIONS
That isn't what "Incomplete", means, as your partial_decider_D
doesn't exist so it can't handle ANY input situations.
YOU HAVEN'T DEMONSTRATED A CONTRADICTION WITH PARTIAL_RECOGNIZER_D, SO
YOU HAVEN'T SHOWN IT TO NOT EXIST...
In other words, you assert the existance of Russel's teapot.
The "contradiction" is in the fact that you concept of "undecidable
input" just is nonsense.
How is it different then inputs that the specific decider is just wrong able.
Describe how you define that property based on JUST the specific input itself and the specific decider.
Your world is based on variable constants and similar nonsense.
You don't get it, that if you want to put down Turing's machines for
not handling cases, you need to also accept that yours doesn't for
the same reason, a machine that fully meets your specifaction can't
exist.
And part of the reason for yours, is that you have a nonsense term in
your definition, as there is no such thing as a "undecidable input"
if some other decider can get the known right answer.
Part of your problem is you just don't understand what a PROGRAM or
ALGORITHM actually is
as you haven't specified HOW it does what it does, just lie turing
didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such
machine could exist, which he succeeded at, but YOUR claim is that
such a thing might be able to exist, which you just have not proven. >>>>>
And, it is an assertion that NO decider can correctly compute the >>>>>>> answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which
means you have chosen a particular (broken) implementation of D >>>>>>> (since no actually correct D exists), like your
partial_recognizer_D, has fulled defined behavior, and a decider >>>>>>> can exist that determines it, there is nothing "undecidable"
about that input.
It is just that the D that H was built on is wrong. Nothing
"paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you
just _IGNORING_ the specification i specifically created for
partial_recognizer_D...
No, your problem is you just are too stupid to understand that your >>>>> "specification" is just nonsense, as it assumes that you can
perform a test by "changing" the decider and the input to see a
results.
But the specification is about a particular input and decider.
the specification for partial_recognizer_D is when encountering
undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your
definition is just internally inconsistantly trying to define input >>>>> that the decider gets wrong. Since the class the partial decider is
this _DOES NOT MEAN_ we cannot know what the /undecidable input/
does, as a paradox in regards to partial_recognize_D _ONLY
AFFECTS_ partial_recognize_D not _OUR ABILITY TO PROVE OUTCOMES_
In other words, you admit that you words are just lies and don't
mean what they say.
Your "undecidable" isn't about decidability, but correctness of a
given decider.
YES, THAT'S ALL UNDECIDABILITY EVER WAS
In other words you are admitting you don't know what you are talking
about.
A given decider being incorrect doesn't mean a correct decider
doesn't exist. There are many decidable problems, for which incorrect
deciders can exist.
Undecidability is about the being NO correct deciders.
YOU HAVEN'T EVEN SHOW A MACHINE WHICH HAS NO CORRECT CLASSIFIERS FOR...
Because I don't need to.
All that is needed to prove a problem is uncomputable is to show that
you can make an input for any given decider that it is wrong about.
You don't seem to understand the problem.
This means that the trivial always false machine meets your
NO, THE FAILURE TO CLASSIFY MUST COME FROM THE STRUCTURE OF THE
INPUT, NOT AN INCORRECT ALGO OF THE CLASSIFIER, WHY AM I REPEATING
ME ON THIS?
In other words, you think people/machines can't make mistakes?
BEING UNABLE TO CLASSIFY /UNDECIDABLE INPUT/ IS _NOT_ A MISTAKE ON
BEHALF OF THE CLASSIFIER
But the input isn't "undecidable", as that isn't an actual property of
an input.
Your logic is just based on category errors, based on your own stupidity.
Then why are you worried about proofs of correctness.
You don't seem to understand that the whole basis of these proofs
isn't about finding *A* input that nothing can decide on, but finding
a pattern that makes an input for a given decider that it won't be
able to handle.
That pattern isn't actually something that necessarily remains in the
machine after construction, but direction to find a given algorithms
fatal flaw, showing that ALL algorithms for that problem must have a
fatal flaw which can be different for every one.
requirements. Any input for which the correct answer for the base
problem would be true, THAT particular decider won't get right, so
it is correct in your expanded partial sense to return false.
IT seesm you just can't read.
(you don't have one btw, so please do let me witness ur cope) >>>>>>>>>
see, turing_H's runtime is not describable because it's an / >>>>>>>>>>>> incomplete specification/ of machine and therefor _does not >>>>>>>>>>>> even exist_ as a real TM, not because of the hypothesized >>>>>>>>>>>> undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED >>>>>>>>>>> defined machine. If the decider "D" actually exists, then H >>>>>>>>>>> is FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete
specification/, as it doesn't handle idiosyncrasies of TM >>>>>>>>>> computing, specifically self- referential set-classification >>>>>>>>>> paradoxes
It doesn't need to. It just needs to be built on what you want >>>>>>>>> to claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as your >>>>>>>>> equivalent for D, like your partial_recognizer_D, then the
template turing_H can be converted into an actual machine.
actual_turing_H, on the other hand, actually does exist, so >>>>>>>>>>>> we can know/ prove what it does. it, however, still forms a >>>>>>>>>>>> structural paradox in regards to partial_recognizer_D which >>>>>>>>>>>> will fail to classify it as circle-free
But, since you admit to actual_turing_H existing, that means >>>>>>>>>>> that
since partial_recognizer_D _CAN_ exist, then actual_turing_H >>>>>>>>>> _CAN_ exist
And so can the turing_H that uses partial_recognizer_D as its >>>>>>>>> decider D.
The steps the algorithm does is what defines the machine, not >>>>>>>>> the "names" of the things it "calls", as those names don't
actually exist in the final building of the machine, which is >>>>>>>>> the problem with your concept, as that is needed to determine >>>>>>>>> your "paradoxical" property.
the names are just shortcuts for the literal values rick, don't >>>>>>>> be daft
Right, but the literal values don't have the names on them, and >>>>>>> we don't have borders within the machine letting us know we moved >>>>>>> from "outer code of H" into "the code of the D it uses",
especially with Turing Machines were such usages are by necessity >>>>>>> "expanded" in-line.
please do actually read turing's paper sometime, he does organize >>>>>> machine descriptions into functional groups
Yes, at meta level.
It CAN be, but doesn't NEED to be. And in fact, his H isn't
no it can be literally a copy of same description than is the run >>>>>> by a UTM (universal turing machine) functional group of the machine >>>>>
Thus, the "copy" of D that was pulled into H isn't actually the same >>>>>>
described that way.
thing as the independent machine D that we can talk about, just a >>>>>>> functional equivalent which doesn't even have its own D.N (since >>>>>>> that is a property only of full machines)
ur really pulling at straws here rick
You really think so?
It seems your problem is you need to work in a tight cage where
other ideas are not allowed that show the holes in your ideas.
pratial_recognizer_D has an actual and fixed algorithm.
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm >>>>>>>>>>>>> will get wrong, then it can just return the different >>>>>>>>>>>>> answer, and then THAT machine it will get right. The >>>>>>>>>>>>> problem is it creates a DIFFERENT input, that uses this new >>>>>>>>>>>>> version of the machine, that it will get wrong.
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence as >>>>>>>>>>>>>> fixed_H, which in turn _does_ have a spot on the diagonal. >>>>>>>>>>>>>> so there is no actual need to include a digit from >>>>>>>>>>>>>> actual_turing_H on the diagonal, as the number it computes >>>>>>>>>>>>>> is already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix". >>>>>>>>>>>>>
And thus if your partial_recoginzer_D accepts ANY verision >>>>>>>>>>>>> of actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D can >>>>>>>>>>>>> possible filter all ALL of that infinite set of machines, >>>>>>>>>>>>> but still accept some machine that computes that particular >>>>>>>>>>>>> number that it would compute with this supposedly correct >>>>>>>>>>>>> partial_recognizer_D.
You can't ask someone to disprove it, as that is just >>>>>>>>>>>>> admitting that you accept that Russel's Teapot must exist. >>>>>>>>>>>>>
So you think, because you don't see it is just built on the >>>>>>>>>>>>> fairy dust, of assuming that your partial decider that >>>>>>>>>>>>> meets ALL your needs exists,
goddamn rick, this rock just keeps getting more solid >>>>>>>>>>>>>
🤷🤷🤷
to clarify this further: the only algorithmic difference >>>>>>>>>>>> between fixed_H and actual_turing_H is the way they
respectively handle their own self- references. they
function identically when handling all other machines
But the problem isn't in it handling the machine fixed_H, but >>>>>>>>>>> how it ever handles the results generated by actual_turing_H, >>>>>>>>>>> which by your assumption has become a cycle-free machine and >>>>>>>>>>> ALL equivalent versions of it have been partially declared to >>>>>>>>>>> be non- cycle free.
This means that your enumeration of machines you are looking >>>>>>>>>>> at is, in fact, incomplete, thus the diagonal you computed is >>>>>>>>>>> wrong.
But actual_turing_H is actually circle_free, and thus its >>>>>>>>>>> results SHOULD have been on the diagonal.
fixed_H = () -> {
N = 1
K = 0
output = [] // written
to F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard
coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE = >>>>>>>>>>>> satisfactory
K += 1
output.push(simulate_Kth_digit(N,K))
}
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will >>>>>>>>>>>> skip trying to simulate itself and put it's hard coded digit >>>>>>>>>>>> on the diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will >>>>>>>>>>>> skip trying simulate actual_turing_H because
partial_recognizer_D fails to classify it as circle-free >>>>>>>>>>>
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is put >>>>>>>>>> on the diagonal, then actual_turing_H _DOES NOT_ need to be in >>>>>>>>>> _THAT_ total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable >>>>>>>>> numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on >>>>>>> being sloppy.
the only way to compute a diagonal across computable numbers is
simulating each number by the machine that computes it.
Not the ONLY way, but a good way to try.
But since Turing Shows this can't be done, it can't be done any way. >>>>>
Note, Turing SPECIFICALLY pointed out that he was going to the
related problem, the problem of:
/but the problem of enumerating computable sequences is
equivalent to the problem of finding out whether a given number >>>>>>> is the D.N of a circle- free machine, and we have no general
process for doing this in a finite number of steps/
oh wow GREAT CATCH RICK!!!!,
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT
ENUMERATING OUT ALL CIRCLE_FREE MACHINES*,
Show it.
what??? bro ur the one who informed me proper enumerations can only
have *one* instance of each element enumerated,
and we all know that the total enumeration of machines has infinite
machines producing each computable numbers...
so therefore not only can the enumeration of computable numbers be
done without enumerating all circle-free machines, it *MUST* be done
without enumerating all circle-free machines, as we can output
*ONLY* one machine for each computable number
that should suffice
In other words, you are admitting you can't do it.
no richard, we've reached a point where u are unable to correctly and
honestly identify a self-evident concept:
FOR _ANY_ GIVEN COMPUTABLE NUMBER, THERE ARE _INFINITE_ MACHINES WHICH
COMPUTE IT ...
I fully understand that.
WE ONLY NEED TO CLASSIFY _ONE_ OF THOSE MACHINES TO ADD THAT GIVEN
COMPUTABLE NUMBER THE LIST, THE REST NOT ONLY CAN, BUT MUST, BE IGNORED
Right, and you need to PROVE that you can.
THEREFOR, WE DO _NOT_ NEED A CLASSIFIER WITH THE POWER TO COMPUTE
WHETHER _ANY_ GIVEN MACHINE IS A COMPUTABLE NUMBER, WE ONLY NEED A
CLASSIFIER WITH ENOUGH POWER TO CLASSIFIER _ONE OF THE INFINITE_
MACHINES FOR _ANY_ GIVEN COMPUTABLE NUMBER
Right, you need to PROVE that you can build a classifier that will
select at least one machine for every computable number.
You can't just assume your decider exists, you need to show HOW it works.
Since you concept is based on identifing the "undeciable inputs", you
need to show how you actually will do it, which means you need to define
it in a way that is detectable.
There is no inherent reason you can pull an infinite subset out of a
similarly infinite set.
We can enumerate the primes, even though for every prime there is an
infinite number of composite numbers that are multiples of it, thus
your logic is based on an error.
THe fact that he quote an equivalence that he doesn't prove he
doesn't make his proof incorrect.
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN
NUMBER IS A CIRCLE-FREE MACHINE,
Prove it.
there's nothing to prove, the statement is self-evident given the
previous two paragraphs i wrote. we don't need to be able to output
it for *ANY* given machine, we only need to be able to output that
fact for *ONE* machine of the infinite which produces *ANY* given
compute number
honestly i'm not sure why it's take until year 2026 for someone to
point this out
In other words, you are just admitting that you understanding of the
problem isn't sufficent to deal with it.
no richard, we've reached a point where u are unable to correctly and
honestly identify a self-evident concept:
THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER IS
A CIRCLE-FREE MACHINE, AS WE DO NOT NEED TO IDENTIFY _ALL_ CIRCLE-FREE
MACHINES, JUST _ONE FOR EACH_ COMPUTABLE NUMBER
So, you need to show that you CAN do that.
Turing points out that they are computationally related, and I will
beleive him more that you how claims you can do one without showing how.
You have ADMITTED your ignorance of the field, having looked at only a
very few of the papers, I suspect there was something discussed
somewhere about this, even if it wasn't as famous of a paper (as it
wasn't so eathshaking by itself).
Your problem is you are working on "gut instinct" with a gut that
doesn't understand what it is talking about, because it thinks
unicorns exist.
we only need to compute that fact for *ONE* machine for any given >>>>>> computable sequence, _SO A PARTIAL RECOGNIZER IS SUFFICIENT FOR
THE JOB_
So do it.
you ask too much for not recognizing anything i said as true
no richard, we've reached a point where u are unable to correctly and
honestly identify a self-evident concept
Because is is just wrong, being based on category errors.
As long as you just ASSUME you can do what you claim, you are just
making a ASS out of yourself, proving that your "self-evident" is really
a case o "self-delusion"
Because it isn't, and you have no proofs, because you can't actually
prove a false statement.
Part of your problem is you don't actually know much about what you
are talking, so you have nothing to work with,
fuck rick ur so bad at defending this it's actually really useful >>>>>> to me!
(there's *another* specific sentence i have a problem with,
where's a ben when u need him?)
All you have shown so far is that you are too stupid to understand
the nature of the paper.
Note, it was NOT intended to be the end-all self-contained regerous >>>>> proof of the concept. It was a paper, leaning on previous work
done, and with the intent of more work to follow to show the basic
idea of why something was true.
Thus, not showing how a step works, isn't an error.
Now, to counter his claims, because they are so well accepted, you
are going to need to be rigorous and not based an the fallacy of
assuming something and showing that it doesn't lead to a
contradiction when you follow a restricted line of reasoining.
So, HIS enumeration is the complete enumeration of "circle-free >>>>>>> machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration either, >>>>>> why am i the only one subject to criticism here???
Also, your proof is built on the existance of a Russel Teapot, >>>>>>>>> you have
the *specification* /works/, which shows that it's *not*
contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion. >>>>>>> Your arguement boils down to showing that *IF* you can correctly >>>>>>> determine all machines that fit in a not-definable category
(since you concept of [paradoxical can't actually be defined at >>>>>>> the machine level, only at how a machine might have been created >>>>>>> at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can >>>>>>>>> make the precise "mistakes" you need it to. (They are mistakes >>>>>>>>> because the input
"mistake" is this case is a fallacious non-techincal value
judgement on the return value,
No, since your criteria that you want to use is just an
undefinable term for a machine description.
There is nothing "paradoxical" of the actual machine when
created, as it then has fixed and determined behavior that can be >>>>>>> decided on. The thing that is paradoxical is the mete-system
pattern used to generate the class of machines as pair with a
given decider.
cause they are rather entirely within specification, as FALSE >>>>>>>> was specified to be used upon encountering an /undecidable
input/, something turing_D did not handle correctly (making it >>>>>>>> _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is based >>>>>>> on a actual D that does make a specific decision to the input.
And it is just that for any given version of D, we can find a
DIFFERENT H that it will get wrong (the H that was just built on >>>>>>> it).
Do you want to claim that ALL these H are "undecidable" even
though for
no... fixed_H contains no paradox, and is therefor not /
undecidable input/ to any classifier
all of them there does exist a decider that gets the right answer? >>>>>>many machines contain no paradox and are therefore classifiable by >>>>>> all classifiers
But this proof isn't about the many, but the ALL.
A term it seems you don't understand.
And it isn't the job of others to show you what one you missed, it
is yours to prove that they don't, as YOU are making the claim.
It seems you just don't know what the word means, and thus you
specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of >>>>>>>>> this turing_H is built on this partial_recognizdr_D which
causes them to not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to
machines, as it is based on the incorrect definition that
assumes a machine can be something other than itself, and when >>>>>>>>> it changes, it changes other things that at the meta/template >>>>>>>>> level refered to its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just
refuse to learn.
Try to actually fully DEFINE the word based on the actual
properties of the input machine, and not refering to other
machines that are not part of the input (but would be created as >>>>>>> alternate inputs to foil an alternate decider).
it's still a diagonal across all computerate numbers even if it >>>>>>>> has dpulicates ...
No, it isn't, as your "set of computable numbers" has duplicates. >>>>>>>>
when actual_turing_H iterates across DN(actual_turing_H) >>>>>>>>>>>> itself, it will similarly skip trying to simulate itself, >>>>>>>>>>>> because partial_recognizer_D fails to classify it as circle- >>>>>>>>>>>> free
Right, so it ALSO is wrong, as it didn't compute the diagonal >>>>>>>>>>> of the circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE
NUMBERS_, which is the _ACTUAL PROBLEM_ turing was dealing with, >>>>>>>>>
No, because an enumeration of a set doesn't HAVE duplicates.
then the H's as of now are i suppose "generators" instead of
enumerators
and turing was wrong about that too, but i don't see u criticizing >>>>>> him now eh???
H is a generator, its goal was to generate a particual number.
it's goal was a generate a diagonal across all circle-free machines,
which as you said: is *NOT* an enumeration, and an enumeration
required duplicating the machines
Right, but the result it is generating is BASED (and defined) on a
given enumeration which it computes internally).
Your Fixed_H isn't computing anything with a defined specification,
as your "enumeration" that you claim to be based on, isn't actually
an enumeration that meets your claimed requirement.
D is a decider, that allowed H, in its generation, to produce
(internally) the enumeration that its results were defined by,
It seems you don't understand that to reach your goal, sometimes
you have to acheive a sub-goal that is quite different.
On 3/8/26 5:14 PM, Richard Damon wrote:
On 3/8/26 11:27 AM, dart200 wrote:
On 3/8/26 5:20 AM, Richard Damon wrote:
On 3/7/26 12:47 PM, dart200 wrote:
On 3/7/26 5:52 AM, Richard Damon wrote:
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote:
On 3/3/26 3:55 AM, dart200 wrote:i agree with this proof is far as much as if β was >>>>>>>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure, >>>>>>>>>>>>>>>>>>
On 2/28/26 6:24 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 2/28/26 8:24 PM, dart200 wrote:
On 2/28/26 2:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 12:38 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
it is wrong,And focusing on what is said to be impossible >>>>>>>>>>>>>>>>>>>>>>>>> and not changing the problem is also important. >>>>>>>>>>>>>>>>>>>>>>>>>Eh?!On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> ...yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing's original paper and can be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> reused anywhere... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>You think you can test all of them one by >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test /all/ >>>>>>>>>>>>>>>>>>>>>>>>>>>>> of an infinitude one by one. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
that exactly what turing does in his proof: >>>>>>>>>>>>>>>>>>>>>>>>>>>> he defines a comptuation >>>>>>>>>>>>>>>>>>>>>>>>>>>> that enumerates out all the numbers, testing >>>>>>>>>>>>>>>>>>>>>>>>>>>> each one of they represent >>>>>>>>>>>>>>>>>>>>>>>>>>>> a "satisfactory"/"circle-free" machine, and >>>>>>>>>>>>>>>>>>>>>>>>>>>> adding that to diagonal >>>>>>>>>>>>>>>>>>>>>>>>>>>> across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to >>>>>>>>>>>>>>>>>>>>>>>>>>>> carefully read p247 of turing's >>>>>>>>>>>>>>>>>>>>>>>>>>>> proof and produce the psuedo-code for the >>>>>>>>>>>>>>>>>>>>>>>>>>>> machine H, assuming that >>>>>>>>>>>>>>>>>>>>>>>>>>>> machine D exists
I'll get to it sooner then, because it's mad. >>>>>>>>>>>>>>>>>>>>>>>>>>> Are you sure he didn't
reason quantified over all but phrase it like >>>>>>>>>>>>>>>>>>>>>>>>>>> a procedure for what he
the theory of computation is the theory of >>>>>>>>>>>>>>>>>>>>>>>>>> such procedures, and understanding the >>>>>>>>>>>>>>>>>>>>>>>>>> diagonal procedure is critical to >>>>>>>>>>>>>>>>>>>>>>>>>> understanding the *base* contradiction/ >>>>>>>>>>>>>>>>>>>>>>>>>> paradox that the rest of his support for >>>>>>>>>>>>>>>>>>>>>>>>>> godel's result is then built on >>>>>>>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation isn't >>>>>>>>>>>>>>>>>>>>>>>>> the generation of the diagonal itself, but >>>>>>>>>>>>>>>>>>>>>>>>> effectively enumerating the enumeration in the >>>>>>>>>>>>>>>>>>>>>>>>> first place.
i don't see any indication that turing realized >>>>>>>>>>>>>>>>>>>>>>>> a difference there
Then you zre just showing your stupidity, because >>>>>>>>>>>>>>>>>>>>>>> YOU can't tell the difference.
After all, on page 246 he says:
The computable sequences are therefore not >>>>>>>>>>>>>>>>>>>>>>> enumerable.
Here is is SPECIFICALLY talking about the >>>>>>>>>>>>>>>>>>>>>>> effective enumeration of the computable sequences. >>>>>>>>>>>>>>>>>>>>>>>
He then points out that he can directly show that >>>>>>>>>>>>>>>>>>>>>>> the "anti- diagonal" of the (non-effectively >>>>>>>>>>>>>>>>>>>>>>> computed) enumeration can't be computed but that >>>>>>>>>>>>>>>>>>>>>>> "This proof, although perfectly sound, has the >>>>>>>>>>>>>>>>>>>>>>> disadvantage that it may leave the reader with a >>>>>>>>>>>>>>>>>>>>>>> feeling that 'there must be something wrong'". >>>>>>>>>>>>>>>>>>>>>>
No, YOU are wrong, as you don't understand what is >>>>>>>>>>>>>>>>>>>>> being done.
I think he is refering he to the standard anti- >>>>>>>>>>>>>>>>>>>>> diagonal arguement, which shows that since for all >>>>>>>>>>>>>>>>>>>>> n, position n differs from the value in number n, >>>>>>>>>>>>>>>>>>>>> there can not be any element that matches the anti- >>>>>>>>>>>>>>>>>>>>> diagonal.
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>>>>>>>> something it seems you just don't understand. >>>>>>>>>>>>>>>>>>>>>
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed with >>>>>>>>>>>>>>>>>>>> turing's short diagonal proof, but in writing this >>>>>>>>>>>>>>>>>>>> post i now find myself in a subtle, yet entirely >>>>>>>>>>>>>>>>>>>> critical disagreement:
/let an be the n-th computable sequence, and let >>>>>>>>>>>>>>>>>>>> φn(m) be the m- th figure in an. Let β be the >>>>>>>>>>>>>>>>>>>> sequence with 1- φn(m) as its n-th. figure. Since β >>>>>>>>>>>>>>>>>>>> is computable, there exists a number K [== β] such >>>>>>>>>>>>>>>>>>>> that 1- φn(n) = φK(n) for all n. Putting n = K, we >>>>>>>>>>>>>>>>>>>> have 1 = 2φK(K), i.e. 1 is even. This is impossible/ >>>>>>>>>>>>>>>>>>
but what i don't agree is that we can just assume the >>>>>>>>>>>>>>>>>> computability of β from the existence of φn(m), THAT'S >>>>>>>>>>>>>>>>>> THE FALLACy
the fallacy here is assuming that because the direct >>>>>>>>>>>>>>>>>>>> diagonal is computable, that one can therefore >>>>>>>>>>>>>>>>>>>> compute the anti- diagonal using the direct >>>>>>>>>>>>>>>>>>>> diagonal. the abstract definition makes it look >>>>>>>>>>>>>>>>>>>> simple, but this ignores the complexities of self- >>>>>>>>>>>>>>>>>>>> referential analysis (like what turing details on >>>>>>>>>>>>>>>>>>>> the next page)
But you can, *IF* you have a machine that computes >>>>>>>>>>>>>>>>>>> the diagonal, then just change all the write to the >>>>>>>>>>>>>>>>>>> output to write the opposite. Note, the "self- >>>>>>>>>>>>>>>>>>> reference" that you are thinking of stops being a >>>>>>>>>>>>>>>>>>> "self- reference" but is a reference to the original >>>>>>>>>>>>>>>>>>> write the diagonal code.
in both methods i have for rectifying the paradox >>>>>>>>>>>>>>>>>>>> found in the direct diagonal (either (1) filtering >>>>>>>>>>>>>>>>>>>> TMs or (2) using RTMs), neither can be used to then >>>>>>>>>>>>>>>>>>>> compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still >>>>>>>>>>>>>>>>>>> fail when it gets to the number of TURING'S H, as >>>>>>>>>>>>>>>>>>> there is no correct answer for the machine built by >>>>>>>>>>>>>>>>>>> that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN >>>>>>>>>>>>>>>>>> INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???) >>>>>>>>>>>>>>>>>>
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY TMs, >>>>>>>>>>>>>>>>>> SO IT *SHOULD* BE FILTERED OUT
But, if D filters it out, then it becomes circle- >>>>>>>>>>>>>>>>>>> free, and thus your enumeration is incomlete. >>>>>>>>>>>>>>>>>>
in (1) the algo to compute an inverse diagonal is >>>>>>>>>>>>>>>>>>>> filtered out like turing's paradoxical variation of >>>>>>>>>>>>>>>>>>>> the direct diagonal would be, and there is no >>>>>>>>>>>>>>>>>>>> analogous non- paradoxical variation that has a hard >>>>>>>>>>>>>>>>>>>> coded value that is inverse to what it does >>>>>>>>>>>>>>>>>>>> return ... such a concept is entirely nonsensical. a >>>>>>>>>>>>>>>>>>>> function can only return what it does, it can't also >>>>>>>>>>>>>>>>>>>> return the inverse to what it returns eh??? >>>>>>>>>>>>>>>>>>>
And why not. The fact that you are too ignorant to >>>>>>>>>>>>>>>>>>> see how to do that, as you are thinking the only >>>>>>>>>>>>>>>>>>> "reference" can be to "self", doesn't make you >>>>>>>>>>>>>>>>>>> argument correct.
in (2) the attempt to compute an inverse diagonal >>>>>>>>>>>>>>>>>>>> with RTMs just fails for reasons u'd only understand >>>>>>>>>>>>>>>>>>>> by working thru the algo urself (p7 of re: turing's >>>>>>>>>>>>>>>>>>>> diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>>>>>>
is just not sufficient evidence that such β is >>>>>>>>>>>>>>>>>>>> actually computable given the direct diagonal φn() >>>>>>>>>>>>>>>>>>>
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH >>>>>>>>>>>>>>>>>> SELF- REFERENCE, LIKE THE PARADOX FOUND IN TURING'S >>>>>>>>>>>>>>>>>> HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H >>>>>>>>>>>>>>>>>> ALLOWS FOR INVERSE- H TO BE COMPUTED
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT >>>>>>>>>>>>>>>>>> THAT IT WOULDN'T BE USED
one cannot just assume that because the diagonal >>>>>>>>>>>>>>>>>>>>>> across computable numbers is computable, therefore >>>>>>>>>>>>>>>>>>>>>> the anti- diagonal across computable numbers is >>>>>>>>>>>>>>>>>>>>>> computable...
He doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and >>>>>>>>>>>>>>>>>>>>> without the enumeration, you can't compute either >>>>>>>>>>>>>>>>>>>>> of them.
But your method still doesn't let you compute the >>>>>>>>>>>>>>>>>>>>> enumeration, and thus you can't actually compute >>>>>>>>>>>>>>>>>>>>> the diagonal.
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>>>>>>> computation across the computable numbers can be >>>>>>>>>>>>>>>>>>>>>> used to compute the inverse diagonal >>>>>>>>>>>>>>>>>>>>>
Remember, the problem definitions requires that the >>>>>>>>>>>>>>>>>>>>> listing be a COMPLETE listing of the computable >>>>>>>>>>>>>>>>>>>>> numbers / machine that compute computable numbers, >>>>>>>>>>>>>>>>>>>>> in some definite order.
If your enumeration isn't complete, your diagonal >>>>>>>>>>>>>>>>>>>>> isn't correct.
so while i agree with turing that the anti- >>>>>>>>>>>>>>>>>>>>>> diagonal is not computable, i don't agree that the >>>>>>>>>>>>>>>>>>>>>> normal diagonal is not computable
Why?
How does D decide on the original H? >>>>>>>>>>>>>>>>>>>>>
Your modified H still needs a correct D to decide >>>>>>>>>>>>>>>>>>>>> on all the other machines, including his original H >>>>>>>>>>>>>>>>>>>>> that doesn't use your "trick"
well, he didn't consider that perhaps the proper >>>>>>>>>>>>>>>>>>>>>> algo for computing the diagonal can avoid the >>>>>>>>>>>>>>>>>>>>>> paradox on itself ...
But instead, he can prove with a more obvious >>>>>>>>>>>>>>>>>>>>>>> process, that the Decider "D" that could be used >>>>>>>>>>>>>>>>>>>>>>> to effectively enumerate the sequence of machine >>>>>>>>>>>>>>>>>>>>>>> that produce computable numbers can not esit. >>>>>>>>>>>>>>>>>>>>>>>
Thus, he clearly knows the difference, but is >>>>>>>>>>>>>>>>>>>>>>> pointing out that the attempt to compute the >>>>>>>>>>>>>>>>>>>>>>> diagonal clearly reveals the issue with >>>>>>>>>>>>>>>>>>>>>>> effectively enumerating the sequences. >>>>>>>>>>>>>>>>>>>>>>
But it doesn't.
Your just don't understand that D just can't >>>>>>>>>>>>>>>>>>>>> correctly decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly >>>>>>>>>>>>>>>>>>>> on turing's H, because my response to this is that D >>>>>>>>>>>>>>>>>>>> does not need to decide correctly on H to compute a >>>>>>>>>>>>>>>>>>>> diagonal
Then how do you build YOUR H without that D? >>>>>>>>>>>>>>>>>>
incredibly, ironically, AND not co-incidently: partial >>>>>>>>>>>>>>>>> recognizer D *can* decide on the fixed H because the >>>>>>>>>>>>>>>>> fixed H does not try to use any D on itself, so no >>>>>>>>>>>>>>>>> self- referential paradox is possible in regards to >>>>>>>>>>>>>>>>> it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a >>>>>>>>>>>>>>>>> dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer >>>>>>>>>>>>>>>>> D (used in fixed H, which does not call any D itself), >>>>>>>>>>>>>>>>> and would not be filtered out by paradox detectors >>>>>>>>>>>>>>>>>
the fact fixed H does not use any classifier D on >>>>>>>>>>>>>>>>> itself, and instead returns a hard-coded value for it's >>>>>>>>>>>>>>>>> own digit on the diagonal, is keystone in making it / >>>>>>>>>>>>>>>>> decidable input/ to D
holy fuck rick idk what to tell u anymore,
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are >>>>>>>>>>>>>>>> just anouncing that you thing errors are ok.
The problem is fixed_H still need to run D on turing_H >>>>>>>>>>>>>>>> and it will fail on that.
bro what part of the non-existence of turing_D or >>>>>>>>>>>>>>> turing_H do you not understand???
But if your fD exists, then Turing_H that uses your D >>>>>>>>>>>>>> exists, and will give that D the problem, and fixed_H will >>>>>>>>>>>>>> call your D on turing_H and it still needs to answer about >>>>>>>>>>>>>> it.
now if ur trying to argue: what if we use
If D errs on Turing_H and calls it circle_free, then >>>>>>>>>>>>>>>> fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, then >>>>>>>>>>>>>>>> your enumeration is missing the circle_free H from its >>>>>>>>>>>>>>>> list, and thus doesn't make the right diagonal. >>>>>>>>>>>>>>>
partial_recognizer_D in a turing_H type machine? that is >>>>>>>>>>>>>>> a fair point
this would give us the machine (that actually does exist): >>>>>>>>>>>>>>>
actual_turing_H = () -> {
N = 1
K = 0
output = [] // written
to F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE =
satisfactory
K += 1
digit = simulate_Kth_digit(N,K) >>>>>>>>>>>>>>> output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE >>>>>>>>>>>>>>>
because actual_turing_H is an /undecidable input/ to any >>>>>>>>>>>>>>> D classifier, when actually run actual_turing_H will >>>>>>>>>>>>>>> simply skip putting it's own digit in the computed sequence >>>>>>>>>>>>>>>
But the problem is that for a given partial_recognizer_D, >>>>>>>>>>>>>> the actual_turing_H isn't an "undecidable input", as it >>>>>>>>>>>>>> has definite behavior, just behavior that
partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of >>>>>>>>>>>>> undecidability. just because a machine is /undecidable >>>>>>>>>>>>> input/ to a particular classifier does mean we cannot then >>>>>>>>>>>>> prove and know ourselves what the machine actually does, >>>>>>>>>>>>>
But, we aren't looking at "undecidable inputs", as these >>>>>>>>>>>> inputs are not "undecidable", but by the time that they have >>>>>>>>>>>> actually been made into an input, and thus first had the >>>>>>>>>>>> program created, which required creating the instance of the >>>>>>>>>>>> decider selctected, they have definite behavior that other >>>>>>>>>>>> some decider can determine.
It is only that one selected candidate decider chosen to >>>>>>>>>>>> build the input that is shown to get the wrong answer. >>>>>>>>>>>>
undecidability only actually exists as a relationship >>>>>>>>>>>>> between a particular machine and the particular classifier >>>>>>>>>>>>> it creates a structural paradox for,
In other words, you don't understand what "undecidability" >>>>>>>>>>>> actually is.
Undecidability, means that there does not exist ANY decider >>>>>>>>>>>> that can get the right answer for all instances of the >>>>>>>>>>>> problem, (or if the problem has been reduced to just that >>>>>>>>>>>> one input, for that input).
*not* as a general inability to describe a *real* machine's >>>>>>>>>>>>> behavior that u are trying to treat it as,
But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D. >>>>>>>>>>
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and
prove it's equivalent to fixed_H, so therefore it doesn't fit >>>>>>>>> what u stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a >>>>>>>> machine, unless you are narrowing the problem to just being
about that one machine/input.
because it *only* ever *was* about problematic input to
*particular* classifiers...
Except that isn't what "Undecidability" is about. And you are just >>>>>> proving you don't know what you are talking about.
If we use the correct term, what you are tying to call
"undeciability" is really just not being correct. The problem is a >>>>>> "particualar classifier" has a fixed set of how it classifies
everything. And thus, nothing can be "paradoxical" to it, as we
never get to the other side,
RICK I'M NEVER GOING TO FIGURE YOU FOR BEING AN UNDYING ASSHOLE,
EXCEPT MAYBE ONCE U FINALLY TAKE UR IGNORANCE TO THE GRAVE
Right, because logic that is based on error can't unddrstand logic
that requires correctness.
The other side you talk about is about if this decider was a
different decider, and some how magically the input was also a
different input converted by a method not "in system" but only in >>>>>> a meta-system.
it's not my problem everyone went for over a century confusing
*our* ability to prove outcomes, with what a /particular/ decider >>>>>>> can output... ya'll got caught up in a rigor of what is
fundamentally nonsense
Because what a particual decider does, if it isnt' right, is just >>>>>> not interesting,
no one has demonstrated a *possible* machine, let alone a *real* >>>>>>> machine, which we can't provably can't prove the outcome for... >>>>>>> such a proof makes the machine _NEITHER POSSIBLE NOR REAL_ so it >>>>>>> would be ludicrous to present such a proof to exist
Only sort of right. Yes, we can not produce a machine that we know >>>>>> we can not possibly ever decide if it halts or not, because such a >>>>>> machine must by definition be non-halting, since ALL halting
machines are provable halting by just running them. Thus, to prove >>>>>> we can't know they halt, we need to prove they don't halt, but
then we know their behavior.
The fact we can't demonstrate and example of that class, doesn't
mean members of it can't exist.
FUCK UR DAMN TEAPOT RICK
In other words, you are admitting I am right and need to resort to
the fallacy of swearing to distract others from that basic fact.
i don't have a response anymore to the level of stupidity required to
keep bleating on and on about being so certain of machines it would
be a contradiction to even produce an example of...
In other words, you are just addmitting that you are too stupid to
understand the possiblity of something being unknowable.
all u have is an ill conceived notion based on the fallacy that _one_ classifier's failure to classify, can be overgeneralized to a _total_ inability to generally prove outcomes, ultimately founded on the
unproven thesis that all computability is encompassed by TM computing
i don't need to respond to that further
It seems you are incapable of thinking abstractly, of even being able
to conceive of something you didn't invent without an example.
But you are also incapable of thinking concreately, because you don't
understand the need to show that there is a reason you need to be able
to show that the machines you imagine are possible.
I will point out AGAIN, that these unknowable machines are not needed
to be understood to understand the uncomputable nature of the problem
(They just come out as an effect of that).
both turing_H and turing_D are /incomplete specifications/ and
therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_
So? What that proves is that it is IMPOSSIBLE to actually acheive >>>>>> the goal defined for turing_D, a machine that determines if a
given machine is circle-free.
they are just malformed /incomplete specifications/ and _NOTHING >>>>>>> MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond the
recognition that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such
machine exists.
BECAUSE IT'S INCOMPLETE. AN /INCOMPLETE SPECIFICATION/ THAT DOES
NOT HANDLE ALL INPUT SITUATIONS CANNOT EXIST
The Specification of the INTERFACE is complete, it fully defines
what the given machine must do.
imagine trying to argue a machine that can't exist, because it didn't
handle all input situations, as "complete"
In other words, you can't understand the difference between a
specification of an interface, and the specification of a machine.
IT DOESN'T "SPECIFY A MACHINE" BECAUSE IT'S _NOT_ A MACHINE DESCRIPTION,
IT _SPECIFIES AN INTERFACE_, ONE THAT IS TOO INCOMPLETE TO BE COMPUTED
BY A MACHINE
By your own logic, YOUR machine is INCOMPLETE, as you can't give the
actual specific algorithm it uses. Your "logic" is based on ASSUMING
that something is computable, with absolutely ZERO evidence that it is.
In other words, you live in a world of fantasy
The Specification of the interface is uncomputable, or unrealizable,
because we can't make a machine that fullfills it.
Yes, the specification of the ALGORITHM is incomplete, as we don't
specify how to do all the steps, but then so is yours.
Turing, BECAUSE he used an unspecified algorithm for his D, was able
to actually PROVE that the specification for the interface of D was
unimplementable.
All your arguement has done is shown that if magic fairy dust
powered unicorns exists (which they don't), you could use one of
them to built an (also non-existant) program to compute something
that is actually proven to be uncomputable.
All you have done is proven you don't understand how logic works,
and thus your works SHOULD BE IGNORED.
If you want to be consistant, YOUR partial_decider_D is ALSO
incomplete,
IT'S NOT INCOMPLETE AS IT HANDLES ALL INPUT SITUATIONS
That isn't what "Incomplete", means, as your partial_decider_D
doesn't exist so it can't handle ANY input situations.
YOU HAVEN'T DEMONSTRATED A CONTRADICTION WITH PARTIAL_RECOGNIZER_D,
SO YOU HAVEN'T SHOWN IT TO NOT EXIST...
In other words, you assert the existance of Russel's teapot.
The "contradiction" is in the fact that you concept of "undecidable
input" just is nonsense.
u know exactly what i'm talking about when i say /undecidable input/, u
just refuse to acknowledge anything i say as correct. because as soon as that damn breaks, ur whole position gets flooded with truth
How is it different then inputs that the specific decider is just
wrong able.
Describe how you define that property based on JUST the specific input
itself and the specific decider.
Your world is based on variable constants and similar nonsense.
You don't get it, that if you want to put down Turing's machines for
not handling cases, you need to also accept that yours doesn't for
the same reason, a machine that fully meets your specifaction can't
exist.
And part of the reason for yours, is that you have a nonsense term
in your definition, as there is no such thing as a "undecidable
input" if some other decider can get the known right answer.
Part of your problem is you just don't understand what a PROGRAM or
ALGORITHM actually is
as you haven't specified HOW it does what it does, just lie turing >>>>>> didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such
machine could exist, which he succeeded at, but YOUR claim is that >>>>>> such a thing might be able to exist, which you just have not proven. >>>>>>
And, it is an assertion that NO decider can correctly compute >>>>>>>> the answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which >>>>>>>> means you have chosen a particular (broken) implementation of D >>>>>>>> (since no actually correct D exists), like your
partial_recognizer_D, has fulled defined behavior, and a decider >>>>>>>> can exist that determines it, there is nothing "undecidable"
about that input.
It is just that the D that H was built on is wrong. Nothing
"paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you >>>>>>> just _IGNORING_ the specification i specifically created for
partial_recognizer_D...
No, your problem is you just are too stupid to understand that
your "specification" is just nonsense, as it assumes that you can >>>>>> perform a test by "changing" the decider and the input to see a
results.
But the specification is about a particular input and decider.
the specification for partial_recognizer_D is when encountering >>>>>>> undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your
definition is just internally inconsistantly trying to define
input that the decider gets wrong. Since the class the partial
decider is
this _DOES NOT MEAN_ we cannot know what the /undecidable input/ >>>>>>> does, as a paradox in regards to partial_recognize_D _ONLY
AFFECTS_ partial_recognize_D not _OUR ABILITY TO PROVE OUTCOMES_
In other words, you admit that you words are just lies and don't
mean what they say.
Your "undecidable" isn't about decidability, but correctness of a >>>>>> given decider.
YES, THAT'S ALL UNDECIDABILITY EVER WAS
In other words you are admitting you don't know what you are talking
about.
A given decider being incorrect doesn't mean a correct decider
doesn't exist. There are many decidable problems, for which
incorrect deciders can exist.
Undecidability is about the being NO correct deciders.
YOU HAVEN'T EVEN SHOW A MACHINE WHICH HAS NO CORRECT CLASSIFIERS FOR...
Because I don't need to.
YOU'VE ONLY SHOWN THAT FOR ANY CLASSIFIER THERE EXISTS A MACHINE IT
CAN'T CLASSIFY...
NOT THAT THERE EXISTS A MACHINE THAT CAN'T BE CLASSIFIED...
All that is needed to prove a problem is uncomputable is to show that
you can make an input for any given decider that it is wrong about.
THAT ONLY SHOWS THAT THE PROBLEM CAN'T BE COMPUTED BY A _SINGLE_ MACHINE
You don't seem to understand the problem.
YOU DON'T UNDERSTAND WHAT'S BEEN PROVEN
This means that the trivial always false machine meets your
NO, THE FAILURE TO CLASSIFY MUST COME FROM THE STRUCTURE OF THE
INPUT, NOT AN INCORRECT ALGO OF THE CLASSIFIER, WHY AM I REPEATING
ME ON THIS?
In other words, you think people/machines can't make mistakes?
BEING UNABLE TO CLASSIFY /UNDECIDABLE INPUT/ IS _NOT_ A MISTAKE ON
BEHALF OF THE CLASSIFIER
But the input isn't "undecidable", as that isn't an actual property of
an input.
IT IS _CAUSED_ *BY THE STRUCTURE OF THE INPUT*
Your logic is just based on category errors, based on your own stupidity.
Then why are you worried about proofs of correctness.
You don't seem to understand that the whole basis of these proofs
isn't about finding *A* input that nothing can decide on, but
finding a pattern that makes an input for a given decider that it
won't be able to handle.
That pattern isn't actually something that necessarily remains in
the machine after construction, but direction to find a given
algorithms fatal flaw, showing that ALL algorithms for that problem
must have a fatal flaw which can be different for every one.
requirements. Any input for which the correct answer for the base >>>>>> problem would be true, THAT particular decider won't get right, so >>>>>> it is correct in your expanded partial sense to return false.
IT seesm you just can't read.
(you don't have one btw, so please do let me witness ur cope) >>>>>>>>>>
see, turing_H's runtime is not describable because it's >>>>>>>>>>>>> an / incomplete specification/ of machine and therefor >>>>>>>>>>>>> _does not even exist_ as a real TM, not because of the >>>>>>>>>>>>> hypothesized undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED >>>>>>>>>>>> defined machine. If the decider "D" actually exists, then H >>>>>>>>>>>> is FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete >>>>>>>>>>> specification/, as it doesn't handle idiosyncrasies of TM >>>>>>>>>>> computing, specifically self- referential set-classification >>>>>>>>>>> paradoxes
It doesn't need to. It just needs to be built on what you want >>>>>>>>>> to claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as >>>>>>>>>> your equivalent for D, like your partial_recognizer_D, then >>>>>>>>>> the template turing_H can be converted into an actual machine. >>>>>>>>>>
actual_turing_H, on the other hand, actually does exist, so >>>>>>>>>>>>> we can know/ prove what it does. it, however, still forms a >>>>>>>>>>>>> structural paradox in regards to partial_recognizer_D which >>>>>>>>>>>>> will fail to classify it as circle-free
But, since you admit to actual_turing_H existing, that means >>>>>>>>>>>> that
since partial_recognizer_D _CAN_ exist, then actual_turing_H >>>>>>>>>>> _CAN_ exist
And so can the turing_H that uses partial_recognizer_D as its >>>>>>>>>> decider D.
The steps the algorithm does is what defines the machine, not >>>>>>>>>> the "names" of the things it "calls", as those names don't >>>>>>>>>> actually exist in the final building of the machine, which is >>>>>>>>>> the problem with your concept, as that is needed to determine >>>>>>>>>> your "paradoxical" property.
the names are just shortcuts for the literal values rick, don't >>>>>>>>> be daft
Right, but the literal values don't have the names on them, and >>>>>>>> we don't have borders within the machine letting us know we
moved from "outer code of H" into "the code of the D it uses", >>>>>>>> especially with Turing Machines were such usages are by
necessity "expanded" in-line.
please do actually read turing's paper sometime, he does organize >>>>>>> machine descriptions into functional groups
Yes, at meta level.
It CAN be, but doesn't NEED to be. And in fact, his H isn't
Thus, the "copy" of D that was pulled into H isn't actually the >>>>>>>> same
no it can be literally a copy of same description than is the run >>>>>>> by a UTM (universal turing machine) functional group of the machine >>>>>>
described that way.
thing as the independent machine D that we can talk about, just >>>>>>>> a functional equivalent which doesn't even have its own D.N
(since that is a property only of full machines)
ur really pulling at straws here rick
You really think so?
It seems your problem is you need to work in a tight cage where
other ideas are not allowed that show the holes in your ideas.
pratial_recognizer_D has an actual and fixed algorithm. >>>>>>>>>>>>
But the problem isn't in it handling the machine fixed_H, >>>>>>>>>>>> but how it ever handles the results generated by
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm >>>>>>>>>>>>>> will get wrong, then it can just return the different >>>>>>>>>>>>>> answer, and then THAT machine it will get right. The >>>>>>>>>>>>>> problem is it creates a DIFFERENT input, that uses this >>>>>>>>>>>>>> new version of the machine, that it will get wrong. >>>>>>>>>>>>>>
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence >>>>>>>>>>>>>>> as fixed_H, which in turn _does_ have a spot on the >>>>>>>>>>>>>>> diagonal. so there is no actual need to include a digit >>>>>>>>>>>>>>> from actual_turing_H on the diagonal, as the number it >>>>>>>>>>>>>>> computes is already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix". >>>>>>>>>>>>>>
And thus if your partial_recoginzer_D accepts ANY verision >>>>>>>>>>>>>> of actual_turing_H, your fixed_H will get stuck.
So, now, you need to show how your partial_recognizer_D >>>>>>>>>>>>>> can possible filter all ALL of that infinite set of >>>>>>>>>>>>>> machines, but still accept some machine that computes that >>>>>>>>>>>>>> particular number that it would compute with this >>>>>>>>>>>>>> supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just >>>>>>>>>>>>>> admitting that you accept that Russel's Teapot must exist. >>>>>>>>>>>>>>
So you think, because you don't see it is just built on >>>>>>>>>>>>>> the fairy dust, of assuming that your partial decider that >>>>>>>>>>>>>> meets ALL your needs exists,
goddamn rick, this rock just keeps getting more solid >>>>>>>>>>>>>>
🤷🤷🤷
to clarify this further: the only algorithmic difference >>>>>>>>>>>>> between fixed_H and actual_turing_H is the way they >>>>>>>>>>>>> respectively handle their own self- references. they >>>>>>>>>>>>> function identically when handling all other machines >>>>>>>>>>>>
actual_turing_H, which by your assumption has become a >>>>>>>>>>>> cycle-free machine and ALL equivalent versions of it have >>>>>>>>>>>> been partially declared to be non- cycle free.
This means that your enumeration of machines you are looking >>>>>>>>>>>> at is, in fact, incomplete, thus the diagonal you computed >>>>>>>>>>>> is wrong.
But actual_turing_H is actually circle_free, and thus its >>>>>>>>>>>> results SHOULD have been on the diagonal.
fixed_H = () -> {
N = 1
K = 0
output = [] //
written to F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard
coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE =
satisfactory
K += 1
output.push(simulate_Kth_digit(N,K)) >>>>>>>>>>>>> }
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will >>>>>>>>>>>>> skip trying to simulate itself and put it's hard coded >>>>>>>>>>>>> digit on the diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will >>>>>>>>>>>>> skip trying simulate actual_turing_H because
partial_recognizer_D fails to classify it as circle-free >>>>>>>>>>>>
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is >>>>>>>>>>> put on the diagonal, then actual_turing_H _DOES NOT_ need to >>>>>>>>>>> be in _THAT_ total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable >>>>>>>>>> numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on >>>>>>>> being sloppy.
the only way to compute a diagonal across computable numbers is >>>>>>> simulating each number by the machine that computes it.
Not the ONLY way, but a good way to try.
But since Turing Shows this can't be done, it can't be done any way. >>>>>>
Note, Turing SPECIFICALLY pointed out that he was going to the >>>>>>>> related problem, the problem of:
/but the problem of enumerating computable sequences is
equivalent to the problem of finding out whether a given number >>>>>>>> is the D.N of a circle- free machine, and we have no general
process for doing this in a finite number of steps/
oh wow GREAT CATCH RICK!!!!,
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT
ENUMERATING OUT ALL CIRCLE_FREE MACHINES*,
Show it.
what??? bro ur the one who informed me proper enumerations can only >>>>> have *one* instance of each element enumerated,
and we all know that the total enumeration of machines has infinite >>>>> machines producing each computable numbers...
so therefore not only can the enumeration of computable numbers be
done without enumerating all circle-free machines, it *MUST* be
done without enumerating all circle-free machines, as we can output >>>>> *ONLY* one machine for each computable number
that should suffice
In other words, you are admitting you can't do it.
no richard, we've reached a point where u are unable to correctly and
honestly identify a self-evident concept:
FOR _ANY_ GIVEN COMPUTABLE NUMBER, THERE ARE _INFINITE_ MACHINES
WHICH COMPUTE IT ...
I fully understand that.
WE ONLY NEED TO CLASSIFY _ONE_ OF THOSE MACHINES TO ADD THAT GIVEN
COMPUTABLE NUMBER THE LIST, THE REST NOT ONLY CAN, BUT MUST, BE IGNORED
Right, and you need to PROVE that you can.
IT'S ENOUGH TO REFUTE A *KEYSTONE* POINT IN TURING'S PAPER
THE REST OF THE PAPER IS PRETTY MUCH VOID DUE TO TURING CONFLATING THE ENUMERATION OF COMPUTABLE NUMBERS WITH THAT OF MACHINES THAT COMPUTE COMPUTATIONAL NUMBERS
THEREFOR, WE DO _NOT_ NEED A CLASSIFIER WITH THE POWER TO COMPUTE
WHETHER _ANY_ GIVEN MACHINE IS A COMPUTABLE NUMBER, WE ONLY NEED A
CLASSIFIER WITH ENOUGH POWER TO CLASSIFIER _ONE OF THE INFINITE_
MACHINES FOR _ANY_ GIVEN COMPUTABLE NUMBER
Right, you need to PROVE that you can build a classifier that will
select at least one machine for every computable number.
You can't just assume your decider exists, you need to show HOW it works.
Since you concept is based on identifing the "undeciable inputs", you
need to show how you actually will do it, which means you need to
define it in a way that is detectable.
i already demonstrated the basic idea several times
actually building a partial_recognizer for all possible computable
numbers is a research question far outside the scope of this discussion since it involves open questions in number theory we don't even have
answers to
There is no inherent reason you can pull an infinite subset out of a
similarly infinite set.
We can enumerate the primes, even though for every prime there is an
infinite number of composite numbers that are multiples of it, thus
your logic is based on an error.
THe fact that he quote an equivalence that he doesn't prove he
doesn't make his proof incorrect.
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN
NUMBER IS A CIRCLE-FREE MACHINE,
Prove it.
there's nothing to prove, the statement is self-evident given the
previous two paragraphs i wrote. we don't need to be able to output >>>>> it for *ANY* given machine, we only need to be able to output that
fact for *ONE* machine of the infinite which produces *ANY* given
compute number
honestly i'm not sure why it's take until year 2026 for someone to
point this out
In other words, you are just admitting that you understanding of the
problem isn't sufficent to deal with it.
no richard, we've reached a point where u are unable to correctly and
honestly identify a self-evident concept:
THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER
IS A CIRCLE-FREE MACHINE, AS WE DO NOT NEED TO IDENTIFY _ALL_ CIRCLE-
FREE MACHINES, JUST _ONE FOR EACH_ COMPUTABLE NUMBER
So, you need to show that you CAN do that.
Turing points out that they are computationally related, and I will
being related does _not_ make them the _same problem_,
and turing's proof *depends* on them being the _same problem_
beleive him more that you how claims you can do one without showing how.
You have ADMITTED your ignorance of the field, having looked at only a
very few of the papers, I suspect there was something discussed
this is suspicion that is easily dealt with using search tools. but
instead of backing up that suspicion ur just continuing to be suspect,
and i'm never gunna buy it
somewhere about this, even if it wasn't as famous of a paper (as it
wasn't so eathshaking by itself).
Your problem is you are working on "gut instinct" with a gut that
doesn't understand what it is talking about, because it thinks
unicorns exist.
we only need to compute that fact for *ONE* machine for any given >>>>>>> computable sequence, _SO A PARTIAL RECOGNIZER IS SUFFICIENT FOR >>>>>>> THE JOB_
So do it.
you ask too much for not recognizing anything i said as true
no richard, we've reached a point where u are unable to correctly and
honestly identify a self-evident concept
Because is is just wrong, being based on category errors.
As long as you just ASSUME you can do what you claim, you are just
making a ASS out of yourself, proving that your "self-evident" is
really a case o "self-delusion"
Because it isn't, and you have no proofs, because you can't actually
prove a false statement.
Part of your problem is you don't actually know much about what you
are talking, so you have nothing to work with,
fuck rick ur so bad at defending this it's actually really useful >>>>>>> to me!
(there's *another* specific sentence i have a problem with,
where's a ben when u need him?)
All you have shown so far is that you are too stupid to understand >>>>>> the nature of the paper.
Note, it was NOT intended to be the end-all self-contained
regerous proof of the concept. It was a paper, leaning on previous >>>>>> work done, and with the intent of more work to follow to show the >>>>>> basic idea of why something was true.
Thus, not showing how a step works, isn't an error.
Now, to counter his claims, because they are so well accepted, you >>>>>> are going to need to be rigorous and not based an the fallacy of
assuming something and showing that it doesn't lead to a
contradiction when you follow a restricted line of reasoining.
So, HIS enumeration is the complete enumeration of "circle-free >>>>>>>> machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration
either, why am i the only one subject to criticism here???
Also, your proof is built on the existance of a Russel Teapot, >>>>>>>>>> you have
the *specification* /works/, which shows that it's *not*
contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion. >>>>>>>> Your arguement boils down to showing that *IF* you can correctly >>>>>>>> determine all machines that fit in a not-definable category
(since you concept of [paradoxical can't actually be defined at >>>>>>>> the machine level, only at how a machine might have been created >>>>>>>> at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can >>>>>>>>>> make the precise "mistakes" you need it to. (They are mistakes >>>>>>>>>> because the input
"mistake" is this case is a fallacious non-techincal value
judgement on the return value,
No, since your criteria that you want to use is just an
undefinable term for a machine description.
There is nothing "paradoxical" of the actual machine when
created, as it then has fixed and determined behavior that can >>>>>>>> be decided on. The thing that is paradoxical is the mete-system >>>>>>>> pattern used to generate the class of machines as pair with a >>>>>>>> given decider.
cause they are rather entirely within specification, as FALSE >>>>>>>>> was specified to be used upon encountering an /undecidable
input/, something turing_D did not handle correctly (making it >>>>>>>>> _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is based >>>>>>>> on a actual D that does make a specific decision to the input. >>>>>>>> And it is just that for any given version of D, we can find a >>>>>>>> DIFFERENT H that it will get wrong (the H that was just built on >>>>>>>> it).
Do you want to claim that ALL these H are "undecidable" even
though for
no... fixed_H contains no paradox, and is therefor not /
undecidable input/ to any classifier
all of them there does exist a decider that gets the right answer? >>>>>>>many machines contain no paradox and are therefore classifiable >>>>>>> by all classifiers
But this proof isn't about the many, but the ALL.
A term it seems you don't understand.
And it isn't the job of others to show you what one you missed, it >>>>>> is yours to prove that they don't, as YOU are making the claim.
It seems you just don't know what the word means, and thus you >>>>>>>> specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of >>>>>>>>>> this turing_H is built on this partial_recognizdr_D which >>>>>>>>>> causes them to not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to
machines, as it is based on the incorrect definition that >>>>>>>>>> assumes a machine can be something other than itself, and when >>>>>>>>>> it changes, it changes other things that at the meta/template >>>>>>>>>> level refered to its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just
refuse to learn.
Try to actually fully DEFINE the word based on the actual
properties of the input machine, and not refering to other
machines that are not part of the input (but would be created as >>>>>>>> alternate inputs to foil an alternate decider).
it's still a diagonal across all computerate numbers even if it >>>>>>>>> has dpulicates ...
No, it isn't, as your "set of computable numbers" has duplicates. >>>>>>>>>
when actual_turing_H iterates across DN(actual_turing_H) >>>>>>>>>>>>> itself, it will similarly skip trying to simulate itself, >>>>>>>>>>>>> because partial_recognizer_D fails to classify it as >>>>>>>>>>>>> circle- free
Right, so it ALSO is wrong, as it didn't compute the
diagonal of the circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE >>>>>>>>>>> NUMBERS_, which is the _ACTUAL PROBLEM_ turing was dealing with, >>>>>>>>>>
No, because an enumeration of a set doesn't HAVE duplicates.
then the H's as of now are i suppose "generators" instead of
enumerators
and turing was wrong about that too, but i don't see u
criticizing him now eh???
H is a generator, its goal was to generate a particual number.
it's goal was a generate a diagonal across all circle-free machines, >>>>>
which as you said: is *NOT* an enumeration, and an enumeration
required duplicating the machines
Right, but the result it is generating is BASED (and defined) on a
given enumeration which it computes internally).
Your Fixed_H isn't computing anything with a defined specification,
as your "enumeration" that you claim to be based on, isn't actually
an enumeration that meets your claimed requirement.
D is a decider, that allowed H, in its generation, to produce
(internally) the enumeration that its results were defined by,
It seems you don't understand that to reach your goal, sometimes
you have to acheive a sub-goal that is quite different.
On 3/9/26 12:32 AM, dart200 wrote:
On 3/8/26 5:14 PM, Richard Damon wrote:
On 3/8/26 11:27 AM, dart200 wrote:
On 3/8/26 5:20 AM, Richard Damon wrote:
On 3/7/26 12:47 PM, dart200 wrote:
On 3/7/26 5:52 AM, Richard Damon wrote:
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/3/26 3:55 AM, dart200 wrote:
i agree with this proof is far as much as if β was >>>>>>>>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure, >>>>>>>>>>>>>>>>>>>On 2/28/26 6:24 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 2/28/26 8:24 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 2:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 12:38 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
And focusing on what is said to be impossible >>>>>>>>>>>>>>>>>>>>>>>>>> and not changing the problem is also important. >>>>>>>>>>>>>>>>>>>>>>>>>>the theory of computation is the theory of >>>>>>>>>>>>>>>>>>>>>>>>>>> such procedures, and understanding the >>>>>>>>>>>>>>>>>>>>>>>>>>> diagonal procedure is critical to >>>>>>>>>>>>>>>>>>>>>>>>>>> understanding the *base* contradiction/ >>>>>>>>>>>>>>>>>>>>>>>>>>> paradox that the rest of his support for >>>>>>>>>>>>>>>>>>>>>>>>>>> godel's result is then built on >>>>>>>>>>>>>>>>>>>>>>>>>>Eh?!On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ...yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing's original paper and can be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> reused anywhere... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>You think you can test all of them one >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> by one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test / >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> all/ of an infinitude one by one. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
that exactly what turing does in his proof: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> he defines a comptuation >>>>>>>>>>>>>>>>>>>>>>>>>>>>> that enumerates out all the numbers, >>>>>>>>>>>>>>>>>>>>>>>>>>>>> testing each one of they represent >>>>>>>>>>>>>>>>>>>>>>>>>>>>> a "satisfactory"/"circle-free" machine, and >>>>>>>>>>>>>>>>>>>>>>>>>>>>> adding that to diagonal >>>>>>>>>>>>>>>>>>>>>>>>>>>>> across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> carefully read p247 of turing's >>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof and produce the psuedo-code for the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine H, assuming that >>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine D exists
I'll get to it sooner then, because it's >>>>>>>>>>>>>>>>>>>>>>>>>>>> mad. Are you sure he didn't >>>>>>>>>>>>>>>>>>>>>>>>>>>> reason quantified over all but phrase it >>>>>>>>>>>>>>>>>>>>>>>>>>>> like a procedure for what he >>>>>>>>>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation isn't >>>>>>>>>>>>>>>>>>>>>>>>>> the generation of the diagonal itself, but >>>>>>>>>>>>>>>>>>>>>>>>>> effectively enumerating the enumeration in the >>>>>>>>>>>>>>>>>>>>>>>>>> first place.
i don't see any indication that turing realized >>>>>>>>>>>>>>>>>>>>>>>>> a difference there
Then you zre just showing your stupidity, >>>>>>>>>>>>>>>>>>>>>>>> because YOU can't tell the difference. >>>>>>>>>>>>>>>>>>>>>>>>
After all, on page 246 he says: >>>>>>>>>>>>>>>>>>>>>>>>
The computable sequences are therefore not >>>>>>>>>>>>>>>>>>>>>>>> enumerable.
Here is is SPECIFICALLY talking about the >>>>>>>>>>>>>>>>>>>>>>>> effective enumeration of the computable sequences. >>>>>>>>>>>>>>>>>>>>>>>>
He then points out that he can directly show >>>>>>>>>>>>>>>>>>>>>>>> that the "anti- diagonal" of the (non- >>>>>>>>>>>>>>>>>>>>>>>> effectively computed) enumeration can't be >>>>>>>>>>>>>>>>>>>>>>>> computed but that "This proof, although >>>>>>>>>>>>>>>>>>>>>>>> perfectly sound, has the disadvantage that it >>>>>>>>>>>>>>>>>>>>>>>> may leave the reader with a feeling that 'there >>>>>>>>>>>>>>>>>>>>>>>> must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what is >>>>>>>>>>>>>>>>>>>>>> being done.
I think he is refering he to the standard anti- >>>>>>>>>>>>>>>>>>>>>> diagonal arguement, which shows that since for all >>>>>>>>>>>>>>>>>>>>>> n, position n differs from the value in number n, >>>>>>>>>>>>>>>>>>>>>> there can not be any element that matches the >>>>>>>>>>>>>>>>>>>>>> anti- diagonal.
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>>>>>>>>> something it seems you just don't understand. >>>>>>>>>>>>>>>>>>>>>>
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed >>>>>>>>>>>>>>>>>>>>> with turing's short diagonal proof, but in writing >>>>>>>>>>>>>>>>>>>>> this post i now find myself in a subtle, yet >>>>>>>>>>>>>>>>>>>>> entirely critical disagreement:
/let an be the n-th computable sequence, and let >>>>>>>>>>>>>>>>>>>>> φn(m) be the m- th figure in an. Let β be the >>>>>>>>>>>>>>>>>>>>> sequence with 1- φn(m) as its n-th. figure. Since β >>>>>>>>>>>>>>>>>>>>> is computable, there exists a number K [== β] such >>>>>>>>>>>>>>>>>>>>> that 1- φn(n) = φK(n) for all n. Putting n = K, we >>>>>>>>>>>>>>>>>>>>> have 1 = 2φK(K), i.e. 1 is even. This is impossible/ >>>>>>>>>>>>>>>>>>>
but what i don't agree is that we can just assume the >>>>>>>>>>>>>>>>>>> computability of β from the existence of φn(m), >>>>>>>>>>>>>>>>>>> THAT'S THE FALLACy
the fallacy here is assuming that because the >>>>>>>>>>>>>>>>>>>>> direct diagonal is computable, that one can >>>>>>>>>>>>>>>>>>>>> therefore compute the anti- diagonal using the >>>>>>>>>>>>>>>>>>>>> direct diagonal. the abstract definition makes it >>>>>>>>>>>>>>>>>>>>> look simple, but this ignores the complexities of >>>>>>>>>>>>>>>>>>>>> self- referential analysis (like what turing >>>>>>>>>>>>>>>>>>>>> details on the next page)
But you can, *IF* you have a machine that computes >>>>>>>>>>>>>>>>>>>> the diagonal, then just change all the write to the >>>>>>>>>>>>>>>>>>>> output to write the opposite. Note, the "self- >>>>>>>>>>>>>>>>>>>> reference" that you are thinking of stops being a >>>>>>>>>>>>>>>>>>>> "self- reference" but is a reference to the original >>>>>>>>>>>>>>>>>>>> write the diagonal code.
in both methods i have for rectifying the paradox >>>>>>>>>>>>>>>>>>>>> found in the direct diagonal (either (1) filtering >>>>>>>>>>>>>>>>>>>>> TMs or (2) using RTMs), neither can be used to then >>>>>>>>>>>>>>>>>>>>> compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still >>>>>>>>>>>>>>>>>>>> fail when it gets to the number of TURING'S H, as >>>>>>>>>>>>>>>>>>>> there is no correct answer for the machine built by >>>>>>>>>>>>>>>>>>>> that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN >>>>>>>>>>>>>>>>>>> INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???) >>>>>>>>>>>>>>>>>>>
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY >>>>>>>>>>>>>>>>>>> TMs, SO IT *SHOULD* BE FILTERED OUT
But, if D filters it out, then it becomes circle- >>>>>>>>>>>>>>>>>>>> free, and thus your enumeration is incomlete. >>>>>>>>>>>>>>>>>>>
in (1) the algo to compute an inverse diagonal is >>>>>>>>>>>>>>>>>>>>> filtered out like turing's paradoxical variation of >>>>>>>>>>>>>>>>>>>>> the direct diagonal would be, and there is no >>>>>>>>>>>>>>>>>>>>> analogous non- paradoxical variation that has a >>>>>>>>>>>>>>>>>>>>> hard coded value that is inverse to what it does >>>>>>>>>>>>>>>>>>>>> return ... such a concept is entirely nonsensical. >>>>>>>>>>>>>>>>>>>>> a function can only return what it does, it can't >>>>>>>>>>>>>>>>>>>>> also return the inverse to what it returns eh??? >>>>>>>>>>>>>>>>>>>>
And why not. The fact that you are too ignorant to >>>>>>>>>>>>>>>>>>>> see how to do that, as you are thinking the only >>>>>>>>>>>>>>>>>>>> "reference" can be to "self", doesn't make you >>>>>>>>>>>>>>>>>>>> argument correct.
in (2) the attempt to compute an inverse diagonal >>>>>>>>>>>>>>>>>>>>> with RTMs just fails for reasons u'd only >>>>>>>>>>>>>>>>>>>>> understand by working thru the algo urself (p7 of >>>>>>>>>>>>>>>>>>>>> re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>>>>>>>
is just not sufficient evidence that such β is >>>>>>>>>>>>>>>>>>>>> actually computable given the direct diagonal φn() >>>>>>>>>>>>>>>>>>>>
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH >>>>>>>>>>>>>>>>>>> SELF- REFERENCE, LIKE THE PARADOX FOUND IN TURING'S >>>>>>>>>>>>>>>>>>> HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H >>>>>>>>>>>>>>>>>>> ALLOWS FOR INVERSE- H TO BE COMPUTED
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT >>>>>>>>>>>>>>>>>>> THAT IT WOULDN'T BE USED
one cannot just assume that because the diagonal >>>>>>>>>>>>>>>>>>>>>>> across computable numbers is computable, >>>>>>>>>>>>>>>>>>>>>>> therefore the anti- diagonal across computable >>>>>>>>>>>>>>>>>>>>>>> numbers is computable...
He doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, and >>>>>>>>>>>>>>>>>>>>>> without the enumeration, you can't compute either >>>>>>>>>>>>>>>>>>>>>> of them.
But your method still doesn't let you compute the >>>>>>>>>>>>>>>>>>>>>> enumeration, and thus you can't actually compute >>>>>>>>>>>>>>>>>>>>>> the diagonal.
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>>>>>>>> computation across the computable numbers can be >>>>>>>>>>>>>>>>>>>>>>> used to compute the inverse diagonal >>>>>>>>>>>>>>>>>>>>>>
Remember, the problem definitions requires that >>>>>>>>>>>>>>>>>>>>>> the listing be a COMPLETE listing of the >>>>>>>>>>>>>>>>>>>>>> computable numbers / machine that compute >>>>>>>>>>>>>>>>>>>>>> computable numbers, in some definite order. >>>>>>>>>>>>>>>>>>>>>>
If your enumeration isn't complete, your diagonal >>>>>>>>>>>>>>>>>>>>>> isn't correct.
Why?
so while i agree with turing that the anti- >>>>>>>>>>>>>>>>>>>>>>> diagonal is not computable, i don't agree that >>>>>>>>>>>>>>>>>>>>>>> the normal diagonal is not computable >>>>>>>>>>>>>>>>>>>>>>
How does D decide on the original H? >>>>>>>>>>>>>>>>>>>>>>
Your modified H still needs a correct D to decide >>>>>>>>>>>>>>>>>>>>>> on all the other machines, including his original >>>>>>>>>>>>>>>>>>>>>> H that doesn't use your "trick"
well, he didn't consider that perhaps the proper >>>>>>>>>>>>>>>>>>>>>>> algo for computing the diagonal can avoid the >>>>>>>>>>>>>>>>>>>>>>> paradox on itself ...
But instead, he can prove with a more obvious >>>>>>>>>>>>>>>>>>>>>>>> process, that the Decider "D" that could be used >>>>>>>>>>>>>>>>>>>>>>>> to effectively enumerate the sequence of machine >>>>>>>>>>>>>>>>>>>>>>>> that produce computable numbers can not esit. >>>>>>>>>>>>>>>>>>>>>>>>
Thus, he clearly knows the difference, but is >>>>>>>>>>>>>>>>>>>>>>>> pointing out that the attempt to compute the >>>>>>>>>>>>>>>>>>>>>>>> diagonal clearly reveals the issue with >>>>>>>>>>>>>>>>>>>>>>>> effectively enumerating the sequences. >>>>>>>>>>>>>>>>>>>>>>>
But it doesn't.
Your just don't understand that D just can't >>>>>>>>>>>>>>>>>>>>>> correctly decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide correctly >>>>>>>>>>>>>>>>>>>>> on turing's H, because my response to this is that >>>>>>>>>>>>>>>>>>>>> D does not need to decide correctly on H to compute >>>>>>>>>>>>>>>>>>>>> a diagonal
Then how do you build YOUR H without that D? >>>>>>>>>>>>>>>>>>>
incredibly, ironically, AND not co-incidently: partial >>>>>>>>>>>>>>>>>> recognizer D *can* decide on the fixed H because the >>>>>>>>>>>>>>>>>> fixed H does not try to use any D on itself, so no >>>>>>>>>>>>>>>>>> self- referential paradox is possible in regards to >>>>>>>>>>>>>>>>>> it's own digit on the diagonal
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a >>>>>>>>>>>>>>>>>> dependable AND computable value)
the fixed H is /decidable input/ to partial recognizer >>>>>>>>>>>>>>>>>> D (used in fixed H, which does not call any D itself), >>>>>>>>>>>>>>>>>> and would not be filtered out by paradox detectors >>>>>>>>>>>>>>>>>>
the fact fixed H does not use any classifier D on >>>>>>>>>>>>>>>>>> itself, and instead returns a hard-coded value for >>>>>>>>>>>>>>>>>> it's own digit on the diagonal, is keystone in making >>>>>>>>>>>>>>>>>> it / decidable input/ to D
holy fuck rick idk what to tell u anymore, >>>>>>>>>>>>>>>>>>
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you are >>>>>>>>>>>>>>>>> just anouncing that you thing errors are ok. >>>>>>>>>>>>>>>>>
The problem is fixed_H still need to run D on turing_H >>>>>>>>>>>>>>>>> and it will fail on that.
bro what part of the non-existence of turing_D or >>>>>>>>>>>>>>>> turing_H do you not understand???
But if your fD exists, then Turing_H that uses your D >>>>>>>>>>>>>>> exists, and will give that D the problem, and fixed_H >>>>>>>>>>>>>>> will call your D on turing_H and it still needs to answer >>>>>>>>>>>>>>> about it.
now if ur trying to argue: what if we use
If D errs on Turing_H and calls it circle_free, then >>>>>>>>>>>>>>>>> fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, >>>>>>>>>>>>>>>>> then your enumeration is missing the circle_free H from >>>>>>>>>>>>>>>>> its list, and thus doesn't make the right diagonal. >>>>>>>>>>>>>>>>
partial_recognizer_D in a turing_H type machine? that is >>>>>>>>>>>>>>>> a fair point
this would give us the machine (that actually does exist): >>>>>>>>>>>>>>>>
actual_turing_H = () -> {
N = 1
K = 0
output = [] //
written to F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE =
satisfactory
K += 1
digit = simulate_Kth_digit(N,K) >>>>>>>>>>>>>>>> output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE >>>>>>>>>>>>>>>>
because actual_turing_H is an /undecidable input/ to any >>>>>>>>>>>>>>>> D classifier, when actually run actual_turing_H will >>>>>>>>>>>>>>>> simply skip putting it's own digit in the computed sequence >>>>>>>>>>>>>>>>
But the problem is that for a given partial_recognizer_D, >>>>>>>>>>>>>>> the actual_turing_H isn't an "undecidable input", as it >>>>>>>>>>>>>>> has definite behavior, just behavior that
partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of >>>>>>>>>>>>>> undecidability. just because a machine is /undecidable >>>>>>>>>>>>>> input/ to a particular classifier does mean we cannot then >>>>>>>>>>>>>> prove and know ourselves what the machine actually does, >>>>>>>>>>>>>>
But, we aren't looking at "undecidable inputs", as these >>>>>>>>>>>>> inputs are not "undecidable", but by the time that they >>>>>>>>>>>>> have actually been made into an input, and thus first had >>>>>>>>>>>>> the program created, which required creating the instance >>>>>>>>>>>>> of the decider selctected, they have definite behavior that >>>>>>>>>>>>> other some decider can determine.
It is only that one selected candidate decider chosen to >>>>>>>>>>>>> build the input that is shown to get the wrong answer. >>>>>>>>>>>>>
undecidability only actually exists as a relationship >>>>>>>>>>>>>> between a particular machine and the particular classifier >>>>>>>>>>>>>> it creates a structural paradox for,
In other words, you don't understand what "undecidability" >>>>>>>>>>>>> actually is.
Undecidability, means that there does not exist ANY decider >>>>>>>>>>>>> that can get the right answer for all instances of the >>>>>>>>>>>>> problem, (or if the problem has been reduced to just that >>>>>>>>>>>>> one input, for that input).
*not* as a general inability to describe a *real* >>>>>>>>>>>>>> machine's behavior that u are trying to treat it as, >>>>>>>>>>>>>But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D. >>>>>>>>>>>
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and >>>>>>>>>> prove it's equivalent to fixed_H, so therefore it doesn't fit >>>>>>>>>> what u stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a >>>>>>>>> machine, unless you are narrowing the problem to just being >>>>>>>>> about that one machine/input.
because it *only* ever *was* about problematic input to
*particular* classifiers...
Except that isn't what "Undecidability" is about. And you are
just proving you don't know what you are talking about.
If we use the correct term, what you are tying to call
"undeciability" is really just not being correct. The problem is >>>>>>> a "particualar classifier" has a fixed set of how it classifies >>>>>>> everything. And thus, nothing can be "paradoxical" to it, as we >>>>>>> never get to the other side,
RICK I'M NEVER GOING TO FIGURE YOU FOR BEING AN UNDYING ASSHOLE,
EXCEPT MAYBE ONCE U FINALLY TAKE UR IGNORANCE TO THE GRAVE
Right, because logic that is based on error can't unddrstand logic
that requires correctness.
The other side you talk about is about if this decider was a
different decider, and some how magically the input was also a
different input converted by a method not "in system" but only in >>>>>>> a meta-system.
it's not my problem everyone went for over a century confusing >>>>>>>> *our* ability to prove outcomes, with what a /particular/
decider can output... ya'll got caught up in a rigor of what is >>>>>>>> fundamentally nonsense
Because what a particual decider does, if it isnt' right, is just >>>>>>> not interesting,
no one has demonstrated a *possible* machine, let alone a *real* >>>>>>>> machine, which we can't provably can't prove the outcome for... >>>>>>>> such a proof makes the machine _NEITHER POSSIBLE NOR REAL_ so it >>>>>>>> would be ludicrous to present such a proof to exist
Only sort of right. Yes, we can not produce a machine that we
know we can not possibly ever decide if it halts or not, because >>>>>>> such a machine must by definition be non-halting, since ALL
halting machines are provable halting by just running them. Thus, >>>>>>> to prove we can't know they halt, we need to prove they don't
halt, but then we know their behavior.
The fact we can't demonstrate and example of that class, doesn't >>>>>>> mean members of it can't exist.
FUCK UR DAMN TEAPOT RICK
In other words, you are admitting I am right and need to resort to
the fallacy of swearing to distract others from that basic fact.
i don't have a response anymore to the level of stupidity required
to keep bleating on and on about being so certain of machines it
would be a contradiction to even produce an example of...
In other words, you are just addmitting that you are too stupid to
understand the possiblity of something being unknowable.
all u have is an ill conceived notion based on the fallacy that _one_
classifier's failure to classify, can be overgeneralized to a _total_
inability to generally prove outcomes, ultimately founded on the
unproven thesis that all computability is encompassed by TM computing
But it isn't just ONE classifier. He shows how to make an input that
defeats ANY GIVEN classifier.
ANY classifier tht tries to determine this classification, will be
defeated by the input built by that template on it.
Thus, ALL such classifiers will be wrong.
Note, the arguement is not actually dependent on these being Turing Machines, but can be generalized to ANY method based on the general
concept of "computing".
i don't need to respond to that further
Sure you do, or you are just admitting that you don't have an answer.
It seems you are incapable of thinking abstractly, of even being able
to conceive of something you didn't invent without an example.
But you are also incapable of thinking concreately, because you don't
understand the need to show that there is a reason you need to be
able to show that the machines you imagine are possible.
I will point out AGAIN, that these unknowable machines are not needed
to be understood to understand the uncomputable nature of the problem
(They just come out as an effect of that).
both turing_H and turing_D are /incomplete specifications/ and >>>>>>>> therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_
So? What that proves is that it is IMPOSSIBLE to actually acheive >>>>>>> the goal defined for turing_D, a machine that determines if a
given machine is circle-free.
they are just malformed /incomplete specifications/ and _NOTHING >>>>>>>> MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond the
recognition that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such
machine exists.
BECAUSE IT'S INCOMPLETE. AN /INCOMPLETE SPECIFICATION/ THAT DOES
NOT HANDLE ALL INPUT SITUATIONS CANNOT EXIST
The Specification of the INTERFACE is complete, it fully defines
what the given machine must do.
imagine trying to argue a machine that can't exist, because it
didn't handle all input situations, as "complete"
In other words, you can't understand the difference between a
specification of an interface, and the specification of a machine.
IT DOESN'T "SPECIFY A MACHINE" BECAUSE IT'S _NOT_ A MACHINE DESCRIPTION,
IT _SPECIFIES AN INTERFACE_, ONE THAT IS TOO INCOMPLETE TO BE COMPUTED
BY A MACHINE
So, neither is your "machine" an actual machine, as all YOU have is an "interface" that determines your partial classification that isn't
actually implemented.
Turing's H, CAN be built from ANY actual machine that we want to try to claim might be a possible implementation of Turing's D. And all of these WILL BE actual machines, and ALL of them are shown to not produce the correct answer for Turing's H. Thus, there can not be any machine that
does what D is supposed to do.
On the other hand, YOU try to claim your fixed_H is an atual machine,
but it isn't until you actually define your partial_decider_D, which you can't do, so your's isn't a machine either.
And, it turns out that any machine that you want to try to claim is an implementation of your partial decider will either accept a machine that will hang your fixed_H, or will not accept any machine for some
computable number.
We may not be able to directly prove this like Turing
Did, but Rice's proof shows that it can't be done, and your arguement
that we need to find it is just you exercising the fallacy of asserting
that Russel's Teapot must exist if we can't prove it doesn't.
By your own logic, YOUR machine is INCOMPLETE, as you can't give the
actual specific algorithm it uses. Your "logic" is based on ASSUMING
that something is computable, with absolutely ZERO evidence that it is.
In other words, you live in a world of fantasy
The Specification of the interface is uncomputable, or
unrealizable, because we can't make a machine that fullfills it.
Yes, the specification of the ALGORITHM is incomplete, as we don't
specify how to do all the steps, but then so is yours.
Turing, BECAUSE he used an unspecified algorithm for his D, was
able to actually PROVE that the specification for the interface of
D was unimplementable.
All your arguement has done is shown that if magic fairy dust
powered unicorns exists (which they don't), you could use one of
them to built an (also non-existant) program to compute something
that is actually proven to be uncomputable.
All you have done is proven you don't understand how logic works,
and thus your works SHOULD BE IGNORED.
If you want to be consistant, YOUR partial_decider_D is ALSO
incomplete,
IT'S NOT INCOMPLETE AS IT HANDLES ALL INPUT SITUATIONS
That isn't what "Incomplete", means, as your partial_decider_D
doesn't exist so it can't handle ANY input situations.
YOU HAVEN'T DEMONSTRATED A CONTRADICTION WITH PARTIAL_RECOGNIZER_D,
SO YOU HAVEN'T SHOWN IT TO NOT EXIST...
In other words, you assert the existance of Russel's teapot.
The "contradiction" is in the fact that you concept of "undecidable
input" just is nonsense.
u know exactly what i'm talking about when i say /undecidable input/,
u just refuse to acknowledge anything i say as correct. because as
soon as that damn breaks, ur whole position gets flooded with truth
Right, I know you mean an input that the given decider is just wrong about.
Any other definition is just a lie.
How is it different then inputs that the specific decider is just
wrong able.
Describe how you define that property based on JUST the specific
input itself and the specific decider.
Your world is based on variable constants and similar nonsense.
Because I don't need to.
You don't get it, that if you want to put down Turing's machines
for not handling cases, you need to also accept that yours doesn't
for the same reason, a machine that fully meets your specifaction
can't exist.
And part of the reason for yours, is that you have a nonsense term
in your definition, as there is no such thing as a "undecidable
input" if some other decider can get the known right answer.
Part of your problem is you just don't understand what a PROGRAM or >>>>> ALGORITHM actually is
as you haven't specified HOW it does what it does, just lie
turing didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such
machine could exist, which he succeeded at, but YOUR claim is
that such a thing might be able to exist, which you just have not >>>>>>> proven.
And, it is an assertion that NO decider can correctly compute >>>>>>>>> the answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which >>>>>>>>> means you have chosen a particular (broken) implementation of D >>>>>>>>> (since no actually correct D exists), like your
partial_recognizer_D, has fulled defined behavior, and a
decider can exist that determines it, there is nothing
"undecidable" about that input.
It is just that the D that H was built on is wrong. Nothing >>>>>>>>> "paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you >>>>>>>> just _IGNORING_ the specification i specifically created for
partial_recognizer_D...
No, your problem is you just are too stupid to understand that
your "specification" is just nonsense, as it assumes that you can >>>>>>> perform a test by "changing" the decider and the input to see a >>>>>>> results.
But the specification is about a particular input and decider.
the specification for partial_recognizer_D is when encountering >>>>>>>> undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your
definition is just internally inconsistantly trying to define
input that the decider gets wrong. Since the class the partial
decider is
In other words, you admit that you words are just lies and don't >>>>>>> mean what they say.
this _DOES NOT MEAN_ we cannot know what the /undecidable input/ >>>>>>>> does, as a paradox in regards to partial_recognize_D _ONLY
AFFECTS_ partial_recognize_D not _OUR ABILITY TO PROVE OUTCOMES_ >>>>>>>
Your "undecidable" isn't about decidability, but correctness of a >>>>>>> given decider.
YES, THAT'S ALL UNDECIDABILITY EVER WAS
In other words you are admitting you don't know what you are
talking about.
A given decider being incorrect doesn't mean a correct decider
doesn't exist. There are many decidable problems, for which
incorrect deciders can exist.
Undecidability is about the being NO correct deciders.
YOU HAVEN'T EVEN SHOW A MACHINE WHICH HAS NO CORRECT CLASSIFIERS FOR... >>>
YOU'VE ONLY SHOWN THAT FOR ANY CLASSIFIER THERE EXISTS A MACHINE IT
CAN'T CLASSIFY...
NOT THAT THERE EXISTS A MACHINE THAT CAN'T BE CLASSIFIED...
But I don't need to.
The uncomputability of that classification is proven by the first.
The existance of the second class of machine is just something more,
that is clearly beyond your understanding.
All that is needed to prove a problem is uncomputable is to show that
you can make an input for any given decider that it is wrong about.
THAT ONLY SHOWS THAT THE PROBLEM CAN'T BE COMPUTED BY A _SINGLE_ MACHINE
Right, or any effectively enumerable set of machine, since we can then
build the single machine that tests all of them.
And that *IS* the definition of computable,
You don't seem to understand the problem.
YOU DON'T UNDERSTAND WHAT'S BEEN PROVEN
Sure I do, you just admitted what I have said is proven, that the
problem is not computable, which means there does not exist a single
machine that correctly computes all the answers.
This means that the trivial always false machine meets your
NO, THE FAILURE TO CLASSIFY MUST COME FROM THE STRUCTURE OF THE
INPUT, NOT AN INCORRECT ALGO OF THE CLASSIFIER, WHY AM I REPEATING >>>>>> ME ON THIS?
In other words, you think people/machines can't make mistakes?
BEING UNABLE TO CLASSIFY /UNDECIDABLE INPUT/ IS _NOT_ A MISTAKE ON
BEHALF OF THE CLASSIFIER
But the input isn't "undecidable", as that isn't an actual property
of an input.
IT IS _CAUSED_ *BY THE STRUCTURE OF THE INPUT*
No, as the input is fully decidable by some other decider, and thus
there is nothing "undecidable" about it.
It is just that the method will always find a flaw in the algorithm of
that specific decider, making it just wrong.
Your logic is just based on category errors, based on your own
stupidity.
Then why are you worried about proofs of correctness.
You don't seem to understand that the whole basis of these proofs
isn't about finding *A* input that nothing can decide on, but
finding a pattern that makes an input for a given decider that it
won't be able to handle.
That pattern isn't actually something that necessarily remains in
the machine after construction, but direction to find a given
algorithms fatal flaw, showing that ALL algorithms for that problem >>>>> must have a fatal flaw which can be different for every one.
requirements. Any input for which the correct answer for the base >>>>>>> problem would be true, THAT particular decider won't get right, >>>>>>> so it is correct in your expanded partial sense to return false. >>>>>>>
IT seesm you just can't read.
(you don't have one btw, so please do let me witness ur cope) >>>>>>>>>>>
see, turing_H's runtime is not describable because it's >>>>>>>>>>>>>> an / incomplete specification/ of machine and therefor >>>>>>>>>>>>>> _does not even exist_ as a real TM, not because of the >>>>>>>>>>>>>> hypothesized undecidability in relation to some D
It is ONLY "incomplete", because it is built on a PRESUMED >>>>>>>>>>>>> defined machine. If the decider "D" actually exists, then H >>>>>>>>>>>>> is FULLY defined.
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete >>>>>>>>>>>> specification/, as it doesn't handle idiosyncrasies of TM >>>>>>>>>>>> computing, specifically self- referential set-classification >>>>>>>>>>>> paradoxes
It doesn't need to. It just needs to be built on what you >>>>>>>>>>> want to claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as >>>>>>>>>>> your equivalent for D, like your partial_recognizer_D, then >>>>>>>>>>> the template turing_H can be converted into an actual machine. >>>>>>>>>>>
actual_turing_H, on the other hand, actually does exist, >>>>>>>>>>>>>> so we can know/ prove what it does. it, however, still >>>>>>>>>>>>>> forms a structural paradox in regards to
partial_recognizer_D which will fail to classify it as >>>>>>>>>>>>>> circle-free
But, since you admit to actual_turing_H existing, that >>>>>>>>>>>>> means that
since partial_recognizer_D _CAN_ exist, then actual_turing_H >>>>>>>>>>>> _CAN_ exist
And so can the turing_H that uses partial_recognizer_D as its >>>>>>>>>>> decider D.
The steps the algorithm does is what defines the machine, not >>>>>>>>>>> the "names" of the things it "calls", as those names don't >>>>>>>>>>> actually exist in the final building of the machine, which is >>>>>>>>>>> the problem with your concept, as that is needed to determine >>>>>>>>>>> your "paradoxical" property.
the names are just shortcuts for the literal values rick, >>>>>>>>>> don't be daft
Right, but the literal values don't have the names on them, and >>>>>>>>> we don't have borders within the machine letting us know we >>>>>>>>> moved from "outer code of H" into "the code of the D it uses", >>>>>>>>> especially with Turing Machines were such usages are by
necessity "expanded" in-line.
please do actually read turing's paper sometime, he does
organize machine descriptions into functional groups
Yes, at meta level.
Thus, the "copy" of D that was pulled into H isn't actually the >>>>>>>>> same
no it can be literally a copy of same description than is the >>>>>>>> run by a UTM (universal turing machine) functional group of the >>>>>>>> machine
It CAN be, but doesn't NEED to be. And in fact, his H isn't
described that way.
thing as the independent machine D that we can talk about, just >>>>>>>>> a functional equivalent which doesn't even have its own D.N >>>>>>>>> (since that is a property only of full machines)
ur really pulling at straws here rick
You really think so?
It seems your problem is you need to work in a tight cage where >>>>>>> other ideas are not allowed that show the holes in your ideas.
pratial_recognizer_D has an actual and fixed algorithm. >>>>>>>>>>>>>
But the problem isn't in it handling the machine fixed_H, >>>>>>>>>>>>> but how it ever handles the results generated by
Thus, your basic criteria is a category error.
If a D can determine which machines its basic algorithm >>>>>>>>>>>>>>> will get wrong, then it can just return the different >>>>>>>>>>>>>>> answer, and then THAT machine it will get right. The >>>>>>>>>>>>>>> problem is it creates a DIFFERENT input, that uses this >>>>>>>>>>>>>>> new version of the machine, that it will get wrong. >>>>>>>>>>>>>>>
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence >>>>>>>>>>>>>>>> as fixed_H, which in turn _does_ have a spot on the >>>>>>>>>>>>>>>> diagonal. so there is no actual need to include a digit >>>>>>>>>>>>>>>> from actual_turing_H on the diagonal, as the number it >>>>>>>>>>>>>>>> computes is already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix". >>>>>>>>>>>>>>>
And thus if your partial_recoginzer_D accepts ANY >>>>>>>>>>>>>>> verision of actual_turing_H, your fixed_H will get stuck. >>>>>>>>>>>>>>>
So, now, you need to show how your partial_recognizer_D >>>>>>>>>>>>>>> can possible filter all ALL of that infinite set of >>>>>>>>>>>>>>> machines, but still accept some machine that computes >>>>>>>>>>>>>>> that particular number that it would compute with this >>>>>>>>>>>>>>> supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just >>>>>>>>>>>>>>> admitting that you accept that Russel's Teapot must exist. >>>>>>>>>>>>>>>
So you think, because you don't see it is just built on >>>>>>>>>>>>>>> the fairy dust, of assuming that your partial decider >>>>>>>>>>>>>>> that meets ALL your needs exists,
goddamn rick, this rock just keeps getting more solid >>>>>>>>>>>>>>>
🤷🤷🤷
to clarify this further: the only algorithmic difference >>>>>>>>>>>>>> between fixed_H and actual_turing_H is the way they >>>>>>>>>>>>>> respectively handle their own self- references. they >>>>>>>>>>>>>> function identically when handling all other machines >>>>>>>>>>>>>
actual_turing_H, which by your assumption has become a >>>>>>>>>>>>> cycle-free machine and ALL equivalent versions of it have >>>>>>>>>>>>> been partially declared to be non- cycle free.
This means that your enumeration of machines you are >>>>>>>>>>>>> looking at is, in fact, incomplete, thus the diagonal you >>>>>>>>>>>>> computed is wrong.
But actual_turing_H is actually circle_free, and thus its >>>>>>>>>>>>> results SHOULD have been on the diagonal.
fixed_H = () -> {
N = 1
K = 0
output = [] //
written to F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard
coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE >>>>>>>>>>>>>> = satisfactory
K += 1
output.push(simulate_Kth_digit(N,K)) >>>>>>>>>>>>>> }
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will >>>>>>>>>>>>>> skip trying to simulate itself and put it's hard coded >>>>>>>>>>>>>> digit on the diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will >>>>>>>>>>>>>> skip trying simulate actual_turing_H because
partial_recognizer_D fails to classify it as circle-free >>>>>>>>>>>>>
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is >>>>>>>>>>>> put on the diagonal, then actual_turing_H _DOES NOT_ need to >>>>>>>>>>>> be in _THAT_ total enumeration of _COMPUTABLE NUMBERS_
But your computation wasn't actually a diagonal of computable >>>>>>>>>>> numbers, but of machines that made computable numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends on >>>>>>>>> being sloppy.
the only way to compute a diagonal across computable numbers is >>>>>>>> simulating each number by the machine that computes it.
Not the ONLY way, but a good way to try.
But since Turing Shows this can't be done, it can't be done any way. >>>>>>>
Note, Turing SPECIFICALLY pointed out that he was going to the >>>>>>>>> related problem, the problem of:
/but the problem of enumerating computable sequences is
equivalent to the problem of finding out whether a given number >>>>>>>>> is the D.N of a circle- free machine, and we have no general >>>>>>>>> process for doing this in a finite number of steps/
oh wow GREAT CATCH RICK!!!!,
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT
ENUMERATING OUT ALL CIRCLE_FREE MACHINES*,
Show it.
what??? bro ur the one who informed me proper enumerations can
only have *one* instance of each element enumerated,
and we all know that the total enumeration of machines has
infinite machines producing each computable numbers...
so therefore not only can the enumeration of computable numbers be >>>>>> done without enumerating all circle-free machines, it *MUST* be
done without enumerating all circle-free machines, as we can
output *ONLY* one machine for each computable number
that should suffice
In other words, you are admitting you can't do it.
no richard, we've reached a point where u are unable to correctly
and honestly identify a self-evident concept:
FOR _ANY_ GIVEN COMPUTABLE NUMBER, THERE ARE _INFINITE_ MACHINES
WHICH COMPUTE IT ...
I fully understand that.
Right, and you need to PROVE that you can.
WE ONLY NEED TO CLASSIFY _ONE_ OF THOSE MACHINES TO ADD THAT GIVEN
COMPUTABLE NUMBER THE LIST, THE REST NOT ONLY CAN, BUT MUST, BE IGNORED >>>
IT'S ENOUGH TO REFUTE A *KEYSTONE* POINT IN TURING'S PAPER
Nope, claiming that there might exist a machine, that doesn't actually exist, doesn't prove anything.
THE REST OF THE PAPER IS PRETTY MUCH VOID DUE TO TURING CONFLATING THE
ENUMERATION OF COMPUTABLE NUMBERS WITH THAT OF MACHINES THAT COMPUTE
COMPUTATIONAL NUMBERS
No, he shows that there does exist a problem that a decider can not
decide on. That shows that the Hilbert Entscheidungsproblem can not have
a solution, which is his real goal.
Note, on page 246 he sketches out the proof, noting that *IF* the
computable numbers WERE effective enumerable (what he calls just enumerable), then we could compute that anti-diagonal by thus doing that enumeration and selecting the appropriate digits.
He also points out that this gets to the feeling that "there must be something wrong" (even though the logic is perfectly correct).>>
THEREFOR, WE DO _NOT_ NEED A CLASSIFIER WITH THE POWER TO COMPUTE
WHETHER _ANY_ GIVEN MACHINE IS A COMPUTABLE NUMBER, WE ONLY NEED A
CLASSIFIER WITH ENOUGH POWER TO CLASSIFIER _ONE OF THE INFINITE_
MACHINES FOR _ANY_ GIVEN COMPUTABLE NUMBER
Right, you need to PROVE that you can build a classifier that will
select at least one machine for every computable number.
You can't just assume your decider exists, you need to show HOW it
works.
Since you concept is based on identifing the "undeciable inputs", you
need to show how you actually will do it, which means you need to
define it in a way that is detectable.
i already demonstrated the basic idea several times
No, you haven't, as your "demonstartion" always violates the defintion
of a machine and an input.
How is your definition anything different that an input that this
machine gets wrong.
After all, when you talk about changing the decider, that doesn't
actualy change the input, since that changed decider is NOT the original decider, and thus the actual input doesn't change.
actually building a partial_recognizer for all possible computable
numbers is a research question far outside the scope of this
discussion since it involves open questions in number theory we don't
even have answers to
No, it is a fundamental assumption of your claim, that is in direct opposition to the proofs.
Your arguement is just that you must be allowed to just assume Russel's teapot exist without proof.
There is no inherent reason you can pull an infinite subset out of
a similarly infinite set.
We can enumerate the primes, even though for every prime there is
an infinite number of composite numbers that are multiples of it,
thus your logic is based on an error.
THe fact that he quote an equivalence that he doesn't prove he
doesn't make his proof incorrect.
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_ >>>>>>>> EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN
NUMBER IS A CIRCLE-FREE MACHINE,
Prove it.
there's nothing to prove, the statement is self-evident given the >>>>>> previous two paragraphs i wrote. we don't need to be able to
output it for *ANY* given machine, we only need to be able to
output that fact for *ONE* machine of the infinite which produces >>>>>> *ANY* given compute number
honestly i'm not sure why it's take until year 2026 for someone to >>>>>> point this out
In other words, you are just admitting that you understanding of
the problem isn't sufficent to deal with it.
no richard, we've reached a point where u are unable to correctly
and honestly identify a self-evident concept:
THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER
IS A CIRCLE-FREE MACHINE, AS WE DO NOT NEED TO IDENTIFY _ALL_
CIRCLE- FREE MACHINES, JUST _ONE FOR EACH_ COMPUTABLE NUMBER
So, you need to show that you CAN do that.
Turing points out that they are computationally related, and I will
being related does _not_ make them the _same problem_,
Right, and he doesn't say they are. he says they are an "equivalent" problem, i.e there is a strong relation to each other (so they either
both have or don't have an answer), that he has presumably proven
somewhere else.
Calling it an error without proof just shows your stupidity.
and turing's proof *depends* on them being the _same problem_
No, since his ultimate goal is the proof is about the Hilbert Entscheidungsproblem, which was the big question of the day.
To show is proof is wrong, *YOU* need to show that they problems are NOT equivalent.
beleive him more that you how claims you can do one without showing how. >>>
You have ADMITTED your ignorance of the field, having looked at only
a very few of the papers, I suspect there was something discussed
this is suspicion that is easily dealt with using search tools. but
instead of backing up that suspicion ur just continuing to be suspect,
and i'm never gunna buy it
Really? and how much of the early papers have actually been preserved
and put into something searchable?
How much of the actual presentations of the various conferences have
been transcribed and put into search engines?
The fact that he makes the passing reference to it, and no one in that
error jumped on that is a good sign that it was accepted.
As I said, prove him wrong or that the conclusion is wrong by ACTUALLY showing how to do what he claims can't be done. And them needs more than assuming you can make a decider that does the real work.
That is you just living off unicorn milk.
somewhere about this, even if it wasn't as famous of a paper (as it
wasn't so eathshaking by itself).
Your problem is you are working on "gut instinct" with a gut that
doesn't understand what it is talking about, because it thinks
unicorns exist.
we only need to compute that fact for *ONE* machine for any
given computable sequence, _SO A PARTIAL RECOGNIZER IS
SUFFICIENT FOR THE JOB_
So do it.
you ask too much for not recognizing anything i said as true
no richard, we've reached a point where u are unable to correctly
and honestly identify a self-evident concept
Because is is just wrong, being based on category errors.
As long as you just ASSUME you can do what you claim, you are just
making a ASS out of yourself, proving that your "self-evident" is
really a case o "self-delusion"
Because it isn't, and you have no proofs, because you can't
actually prove a false statement.
Part of your problem is you don't actually know much about what you >>>>> are talking, so you have nothing to work with,
fuck rick ur so bad at defending this it's actually really
useful to me!
(there's *another* specific sentence i have a problem with,
where's a ben when u need him?)
All you have shown so far is that you are too stupid to
understand the nature of the paper.
Note, it was NOT intended to be the end-all self-contained
regerous proof of the concept. It was a paper, leaning on
previous work done, and with the intent of more work to follow to >>>>>>> show the basic idea of why something was true.
Thus, not showing how a step works, isn't an error.
Now, to counter his claims, because they are so well accepted,
you are going to need to be rigorous and not based an the fallacy >>>>>>> of assuming something and showing that it doesn't lead to a
contradiction when you follow a restricted line of reasoining.
So, HIS enumeration is the complete enumeration of "circle-free >>>>>>>>> machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration
either, why am i the only one subject to criticism here???
Also, your proof is built on the existance of a Russel
Teapot, you have
the *specification* /works/, which shows that it's *not*
contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the conclusion. >>>>>>>>> Your arguement boils down to showing that *IF* you can
correctly determine all machines that fit in a not-definable >>>>>>>>> category (since you concept of [paradoxical can't actually be >>>>>>>>> defined at the machine level, only at how a machine might have >>>>>>>>> been created at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that can >>>>>>>>>>> make the precise "mistakes" you need it to. (They are
mistakes because the input
"mistake" is this case is a fallacious non-techincal value >>>>>>>>>> judgement on the return value,
No, since your criteria that you want to use is just an
undefinable term for a machine description.
There is nothing "paradoxical" of the actual machine when
created, as it then has fixed and determined behavior that can >>>>>>>>> be decided on. The thing that is paradoxical is the mete-system >>>>>>>>> pattern used to generate the class of machines as pair with a >>>>>>>>> given decider.
cause they are rather entirely within specification, as FALSE >>>>>>>>>> was specified to be used upon encountering an /undecidable >>>>>>>>>> input/, something turing_D did not handle correctly (making it >>>>>>>>>> _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is based >>>>>>>>> on a actual D that does make a specific decision to the input. >>>>>>>>> And it is just that for any given version of D, we can find a >>>>>>>>> DIFFERENT H that it will get wrong (the H that was just built >>>>>>>>> on it).
Do you want to claim that ALL these H are "undecidable" even >>>>>>>>> though for
no... fixed_H contains no paradox, and is therefor not /
undecidable input/ to any classifier
all of them there does exist a decider that gets the right answer? >>>>>>>>many machines contain no paradox and are therefore classifiable >>>>>>>> by all classifiers
But this proof isn't about the many, but the ALL.
A term it seems you don't understand.
And it isn't the job of others to show you what one you missed, >>>>>>> it is yours to prove that they don't, as YOU are making the claim. >>>>>>>
It seems you just don't know what the word means, and thus you >>>>>>>>> specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of >>>>>>>>>>> this turing_H is built on this partial_recognizdr_D which >>>>>>>>>>> causes them to not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to
machines, as it is based on the incorrect definition that >>>>>>>>>>> assumes a machine can be something other than itself, and >>>>>>>>>>> when it changes, it changes other things that at the meta/ >>>>>>>>>>> template level refered to its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just >>>>>>>>> refuse to learn.
Try to actually fully DEFINE the word based on the actual
properties of the input machine, and not refering to other
machines that are not part of the input (but would be created >>>>>>>>> as alternate inputs to foil an alternate decider).
when actual_turing_H iterates across DN(actual_turing_H) >>>>>>>>>>>>>> itself, it will similarly skip trying to simulate itself, >>>>>>>>>>>>>> because partial_recognizer_D fails to classify it as >>>>>>>>>>>>>> circle- free
Right, so it ALSO is wrong, as it didn't compute the >>>>>>>>>>>>> diagonal of the circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE >>>>>>>>>>>> NUMBERS_, which is the _ACTUAL PROBLEM_ turing was dealing >>>>>>>>>>>> with,
No, it isn't, as your "set of computable numbers" has
duplicates.
it's still a diagonal across all computerate numbers even if >>>>>>>>>> it has dpulicates ...
No, because an enumeration of a set doesn't HAVE duplicates.
then the H's as of now are i suppose "generators" instead of
enumerators
and turing was wrong about that too, but i don't see u
criticizing him now eh???
H is a generator, its goal was to generate a particual number.
it's goal was a generate a diagonal across all circle-free machines, >>>>>>
which as you said: is *NOT* an enumeration, and an enumeration
required duplicating the machines
Right, but the result it is generating is BASED (and defined) on a
given enumeration which it computes internally).
Your Fixed_H isn't computing anything with a defined specification, >>>>> as your "enumeration" that you claim to be based on, isn't actually >>>>> an enumeration that meets your claimed requirement.
D is a decider, that allowed H, in its generation, to produce
(internally) the enumeration that its results were defined by,
It seems you don't understand that to reach your goal, sometimes >>>>>>> you have to acheive a sub-goal that is quite different.
On 3/9/26 5:03 PM, Richard Damon wrote:
On 3/9/26 12:32 AM, dart200 wrote:
On 3/8/26 5:14 PM, Richard Damon wrote:
On 3/8/26 11:27 AM, dart200 wrote:
On 3/8/26 5:20 AM, Richard Damon wrote:
On 3/7/26 12:47 PM, dart200 wrote:
On 3/7/26 5:52 AM, Richard Damon wrote:
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 3/3/26 3:55 AM, dart200 wrote:
On 2/28/26 6:24 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 8:24 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 2:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 12:38 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
Then you zre just showing your stupidity, >>>>>>>>>>>>>>>>>>>>>>>>> because YOU can't tell the difference. >>>>>>>>>>>>>>>>>>>>>>>>>And focusing on what is said to be impossible >>>>>>>>>>>>>>>>>>>>>>>>>>> and not changing the problem is also important. >>>>>>>>>>>>>>>>>>>>>>>>>>>the theory of computation is the theory of >>>>>>>>>>>>>>>>>>>>>>>>>>>> such procedures, and understanding the >>>>>>>>>>>>>>>>>>>>>>>>>>>> diagonal procedure is critical to >>>>>>>>>>>>>>>>>>>>>>>>>>>> understanding the *base* contradiction/ >>>>>>>>>>>>>>>>>>>>>>>>>>>> paradox that the rest of his support for >>>>>>>>>>>>>>>>>>>>>>>>>>>> godel's result is then built on >>>>>>>>>>>>>>>>>>>>>>>>>>>Eh?!On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ...yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
an effective enumeration of all turing >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machines was proven on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing's original paper and can be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> reused anywhere... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>You think you can test all of them one >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> by one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test / >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> all/ of an infinitude one by one. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
that exactly what turing does in his >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof: he defines a comptuation >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that enumerates out all the numbers, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> testing each one of they represent >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a "satisfactory"/"circle-free" machine, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and adding that to diagonal >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> carefully read p247 of turing's >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof and produce the psuedo-code for the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine H, assuming that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine D exists
I'll get to it sooner then, because it's >>>>>>>>>>>>>>>>>>>>>>>>>>>>> mad. Are you sure he didn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>> reason quantified over all but phrase it >>>>>>>>>>>>>>>>>>>>>>>>>>>>> like a procedure for what he >>>>>>>>>>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation >>>>>>>>>>>>>>>>>>>>>>>>>>> isn't the generation of the diagonal itself, >>>>>>>>>>>>>>>>>>>>>>>>>>> but effectively enumerating the enumeration >>>>>>>>>>>>>>>>>>>>>>>>>>> in the first place.
i don't see any indication that turing >>>>>>>>>>>>>>>>>>>>>>>>>> realized a difference there >>>>>>>>>>>>>>>>>>>>>>>>>
After all, on page 246 he says: >>>>>>>>>>>>>>>>>>>>>>>>>
The computable sequences are therefore not >>>>>>>>>>>>>>>>>>>>>>>>> enumerable.
Here is is SPECIFICALLY talking about the >>>>>>>>>>>>>>>>>>>>>>>>> effective enumeration of the computable sequences. >>>>>>>>>>>>>>>>>>>>>>>>>
He then points out that he can directly show >>>>>>>>>>>>>>>>>>>>>>>>> that the "anti- diagonal" of the (non- >>>>>>>>>>>>>>>>>>>>>>>>> effectively computed) enumeration can't be >>>>>>>>>>>>>>>>>>>>>>>>> computed but that "This proof, although >>>>>>>>>>>>>>>>>>>>>>>>> perfectly sound, has the disadvantage that it >>>>>>>>>>>>>>>>>>>>>>>>> may leave the reader with a feeling that 'there >>>>>>>>>>>>>>>>>>>>>>>>> must be something wrong'".
it is wrong,
No, YOU are wrong, as you don't understand what >>>>>>>>>>>>>>>>>>>>>>> is being done.
I think he is refering he to the standard anti- >>>>>>>>>>>>>>>>>>>>>>> diagonal arguement, which shows that since for >>>>>>>>>>>>>>>>>>>>>>> all n, position n differs from the value in >>>>>>>>>>>>>>>>>>>>>>> number n, there can not be any element that >>>>>>>>>>>>>>>>>>>>>>> matches the anti- diagonal.
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>>>>>>>>>> something it seems you just don't understand. >>>>>>>>>>>>>>>>>>>>>>>
Show how that is actually wrong.
wow, u know up until now, i thot i fully agreed >>>>>>>>>>>>>>>>>>>>>> with turing's short diagonal proof, but in writing >>>>>>>>>>>>>>>>>>>>>> this post i now find myself in a subtle, yet >>>>>>>>>>>>>>>>>>>>>> entirely critical disagreement:
/let an be the n-th computable sequence, and let >>>>>>>>>>>>>>>>>>>>>> φn(m) be the m- th figure in an. Let β be the >>>>>>>>>>>>>>>>>>>>>> sequence with 1- φn(m) as its n-th. figure. Since >>>>>>>>>>>>>>>>>>>>>> β is computable, there exists a number K [== β] >>>>>>>>>>>>>>>>>>>>>> such that 1- φn(n) = φK(n) for all n. Putting n = >>>>>>>>>>>>>>>>>>>>>> K, we have 1 = 2φK(K), i.e. 1 is even. This is >>>>>>>>>>>>>>>>>>>>>> impossible/
i agree with this proof is far as much as if β was >>>>>>>>>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure, >>>>>>>>>>>>>>>>>>>>
but what i don't agree is that we can just assume >>>>>>>>>>>>>>>>>>>> the computability of β from the existence of φn(m), >>>>>>>>>>>>>>>>>>>> THAT'S THE FALLACy
the fallacy here is assuming that because the >>>>>>>>>>>>>>>>>>>>>> direct diagonal is computable, that one can >>>>>>>>>>>>>>>>>>>>>> therefore compute the anti- diagonal using the >>>>>>>>>>>>>>>>>>>>>> direct diagonal. the abstract definition makes it >>>>>>>>>>>>>>>>>>>>>> look simple, but this ignores the complexities of >>>>>>>>>>>>>>>>>>>>>> self- referential analysis (like what turing >>>>>>>>>>>>>>>>>>>>>> details on the next page)
But you can, *IF* you have a machine that computes >>>>>>>>>>>>>>>>>>>>> the diagonal, then just change all the write to the >>>>>>>>>>>>>>>>>>>>> output to write the opposite. Note, the "self- >>>>>>>>>>>>>>>>>>>>> reference" that you are thinking of stops being a >>>>>>>>>>>>>>>>>>>>> "self- reference" but is a reference to the >>>>>>>>>>>>>>>>>>>>> original write the diagonal code.
in both methods i have for rectifying the paradox >>>>>>>>>>>>>>>>>>>>>> found in the direct diagonal (either (1) filtering >>>>>>>>>>>>>>>>>>>>>> TMs or (2) using RTMs), neither can be used to >>>>>>>>>>>>>>>>>>>>>> then compute the anti-diagonal
Nope, because your filtering TM (or RTM) will still >>>>>>>>>>>>>>>>>>>>> fail when it gets to the number of TURING'S H, as >>>>>>>>>>>>>>>>>>>>> there is no correct answer for the machine built by >>>>>>>>>>>>>>>>>>>>> that template.
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN >>>>>>>>>>>>>>>>>>>> INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???) >>>>>>>>>>>>>>>>>>>>
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY >>>>>>>>>>>>>>>>>>>> TMs, SO IT *SHOULD* BE FILTERED OUT
But, if D filters it out, then it becomes circle- >>>>>>>>>>>>>>>>>>>>> free, and thus your enumeration is incomlete. >>>>>>>>>>>>>>>>>>>>
in (1) the algo to compute an inverse diagonal is >>>>>>>>>>>>>>>>>>>>>> filtered out like turing's paradoxical variation >>>>>>>>>>>>>>>>>>>>>> of the direct diagonal would be, and there is no >>>>>>>>>>>>>>>>>>>>>> analogous non- paradoxical variation that has a >>>>>>>>>>>>>>>>>>>>>> hard coded value that is inverse to what it does >>>>>>>>>>>>>>>>>>>>>> return ... such a concept is entirely nonsensical. >>>>>>>>>>>>>>>>>>>>>> a function can only return what it does, it can't >>>>>>>>>>>>>>>>>>>>>> also return the inverse to what it returns eh??? >>>>>>>>>>>>>>>>>>>>>
And why not. The fact that you are too ignorant to >>>>>>>>>>>>>>>>>>>>> see how to do that, as you are thinking the only >>>>>>>>>>>>>>>>>>>>> "reference" can be to "self", doesn't make you >>>>>>>>>>>>>>>>>>>>> argument correct.
in (2) the attempt to compute an inverse diagonal >>>>>>>>>>>>>>>>>>>>>> with RTMs just fails for reasons u'd only >>>>>>>>>>>>>>>>>>>>>> understand by working thru the algo urself (p7 of >>>>>>>>>>>>>>>>>>>>>> re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>>>>>>>>
is just not sufficient evidence that such β is >>>>>>>>>>>>>>>>>>>>>> actually computable given the direct diagonal φn() >>>>>>>>>>>>>>>>>>>>>
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED WITH >>>>>>>>>>>>>>>>>>>> SELF- REFERENCE, LIKE THE PARADOX FOUND IN TURING'S >>>>>>>>>>>>>>>>>>>> HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H >>>>>>>>>>>>>>>>>>>> ALLOWS FOR INVERSE- H TO BE COMPUTED
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, NOT >>>>>>>>>>>>>>>>>>>> THAT IT WOULDN'T BE USED
one cannot just assume that because the diagonal >>>>>>>>>>>>>>>>>>>>>>>> across computable numbers is computable, >>>>>>>>>>>>>>>>>>>>>>>> therefore the anti- diagonal across computable >>>>>>>>>>>>>>>>>>>>>>>> numbers is computable...
He doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, >>>>>>>>>>>>>>>>>>>>>>> and without the enumeration, you can't compute >>>>>>>>>>>>>>>>>>>>>>> either of them.
But your method still doesn't let you compute the >>>>>>>>>>>>>>>>>>>>>>> enumeration, and thus you can't actually compute >>>>>>>>>>>>>>>>>>>>>>> the diagonal.
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>>>>>>>>> computation across the computable numbers can be >>>>>>>>>>>>>>>>>>>>>>>> used to compute the inverse diagonal >>>>>>>>>>>>>>>>>>>>>>>
Remember, the problem definitions requires that >>>>>>>>>>>>>>>>>>>>>>> the listing be a COMPLETE listing of the >>>>>>>>>>>>>>>>>>>>>>> computable numbers / machine that compute >>>>>>>>>>>>>>>>>>>>>>> computable numbers, in some definite order. >>>>>>>>>>>>>>>>>>>>>>>
If your enumeration isn't complete, your diagonal >>>>>>>>>>>>>>>>>>>>>>> isn't correct.
Why?
so while i agree with turing that the anti- >>>>>>>>>>>>>>>>>>>>>>>> diagonal is not computable, i don't agree that >>>>>>>>>>>>>>>>>>>>>>>> the normal diagonal is not computable >>>>>>>>>>>>>>>>>>>>>>>
How does D decide on the original H? >>>>>>>>>>>>>>>>>>>>>>>
Your modified H still needs a correct D to decide >>>>>>>>>>>>>>>>>>>>>>> on all the other machines, including his original >>>>>>>>>>>>>>>>>>>>>>> H that doesn't use your "trick"
well, he didn't consider that perhaps the proper >>>>>>>>>>>>>>>>>>>>>>>> algo for computing the diagonal can avoid the >>>>>>>>>>>>>>>>>>>>>>>> paradox on itself ...
But instead, he can prove with a more obvious >>>>>>>>>>>>>>>>>>>>>>>>> process, that the Decider "D" that could be >>>>>>>>>>>>>>>>>>>>>>>>> used to effectively enumerate the sequence of >>>>>>>>>>>>>>>>>>>>>>>>> machine that produce computable numbers can not >>>>>>>>>>>>>>>>>>>>>>>>> esit.
Thus, he clearly knows the difference, but is >>>>>>>>>>>>>>>>>>>>>>>>> pointing out that the attempt to compute the >>>>>>>>>>>>>>>>>>>>>>>>> diagonal clearly reveals the issue with >>>>>>>>>>>>>>>>>>>>>>>>> effectively enumerating the sequences. >>>>>>>>>>>>>>>>>>>>>>>>
But it doesn't.
Your just don't understand that D just can't >>>>>>>>>>>>>>>>>>>>>>> correctly decide on his given H.
no idea why ur claiming that
i clearly understand that D cannot decide >>>>>>>>>>>>>>>>>>>>>> correctly on turing's H, because my response to >>>>>>>>>>>>>>>>>>>>>> this is that D does not need to decide correctly >>>>>>>>>>>>>>>>>>>>>> on H to compute a diagonal
Then how do you build YOUR H without that D? >>>>>>>>>>>>>>>>>>>>
incredibly, ironically, AND not co-incidently: >>>>>>>>>>>>>>>>>>> partial recognizer D *can* decide on the fixed H >>>>>>>>>>>>>>>>>>> because the fixed H does not try to use any D on >>>>>>>>>>>>>>>>>>> itself, so no self- referential paradox is possible >>>>>>>>>>>>>>>>>>> in regards to it's own digit on the diagonal >>>>>>>>>>>>>>>>>>>
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a >>>>>>>>>>>>>>>>>>> dependable AND computable value)
the fixed H is /decidable input/ to partial >>>>>>>>>>>>>>>>>>> recognizer D (used in fixed H, which does not call >>>>>>>>>>>>>>>>>>> any D itself), and would not be filtered out by >>>>>>>>>>>>>>>>>>> paradox detectors
the fact fixed H does not use any classifier D on >>>>>>>>>>>>>>>>>>> itself, and instead returns a hard-coded value for >>>>>>>>>>>>>>>>>>> it's own digit on the diagonal, is keystone in making >>>>>>>>>>>>>>>>>>> it / decidable input/ to D
holy fuck rick idk what to tell u anymore, >>>>>>>>>>>>>>>>>>>
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you >>>>>>>>>>>>>>>>>> are just anouncing that you thing errors are ok. >>>>>>>>>>>>>>>>>>
The problem is fixed_H still need to run D on turing_H >>>>>>>>>>>>>>>>>> and it will fail on that.
bro what part of the non-existence of turing_D or >>>>>>>>>>>>>>>>> turing_H do you not understand???
But if your fD exists, then Turing_H that uses your D >>>>>>>>>>>>>>>> exists, and will give that D the problem, and fixed_H >>>>>>>>>>>>>>>> will call your D on turing_H and it still needs to >>>>>>>>>>>>>>>> answer about it.
now if ur trying to argue: what if we use
If D errs on Turing_H and calls it circle_free, then >>>>>>>>>>>>>>>>>> fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, >>>>>>>>>>>>>>>>>> then your enumeration is missing the circle_free H >>>>>>>>>>>>>>>>>> from its list, and thus doesn't make the right diagonal. >>>>>>>>>>>>>>>>>
partial_recognizer_D in a turing_H type machine? that >>>>>>>>>>>>>>>>> is a fair point
this would give us the machine (that actually does exist): >>>>>>>>>>>>>>>>>
actual_turing_H = () -> {
N = 1
K = 0
output = [] //
written to F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE
= satisfactory
K += 1
digit = simulate_Kth_digit(N,K) >>>>>>>>>>>>>>>>> output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE >>>>>>>>>>>>>>>>>
because actual_turing_H is an /undecidable input/ to >>>>>>>>>>>>>>>>> any D classifier, when actually run actual_turing_H >>>>>>>>>>>>>>>>> will simply skip putting it's own digit in the computed >>>>>>>>>>>>>>>>> sequence
But the problem is that for a given
partial_recognizer_D, the actual_turing_H isn't an >>>>>>>>>>>>>>>> "undecidable input", as it has definite behavior, just >>>>>>>>>>>>>>>> behavior that partial_recognizer_D gets wrong.
this is unfortunately a fallacious understanding of >>>>>>>>>>>>>>> undecidability. just because a machine is /undecidable >>>>>>>>>>>>>>> input/ to a particular classifier does mean we cannot >>>>>>>>>>>>>>> then prove and know ourselves what the machine actually >>>>>>>>>>>>>>> does,
But, we aren't looking at "undecidable inputs", as these >>>>>>>>>>>>>> inputs are not "undecidable", but by the time that they >>>>>>>>>>>>>> have actually been made into an input, and thus first had >>>>>>>>>>>>>> the program created, which required creating the instance >>>>>>>>>>>>>> of the decider selctected, they have definite behavior >>>>>>>>>>>>>> that other some decider can determine.
It is only that one selected candidate decider chosen to >>>>>>>>>>>>>> build the input that is shown to get the wrong answer. >>>>>>>>>>>>>>
undecidability only actually exists as a relationship >>>>>>>>>>>>>>> between a particular machine and the particular >>>>>>>>>>>>>>> classifier it creates a structural paradox for,
In other words, you don't understand what "undecidability" >>>>>>>>>>>>>> actually is.
Undecidability, means that there does not exist ANY >>>>>>>>>>>>>> decider that can get the right answer for all instances of >>>>>>>>>>>>>> the problem, (or if the problem has been reduced to just >>>>>>>>>>>>>> that one input, for that input).
*not* as a general inability to describe a *real* >>>>>>>>>>>>>>> machine's behavior that u are trying to treat it as, >>>>>>>>>>>>>>But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D. >>>>>>>>>>>>
what *actual* example _THAT CAN EXIST_ do u have???
The one that calls your partial_recoginzer_D as its D.
except we can describe the behavior of actual_turing_H and >>>>>>>>>>> prove it's equivalent to fixed_H, so therefore it doesn't fit >>>>>>>>>>> what u stated was
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or a >>>>>>>>>> machine, unless you are narrowing the problem to just being >>>>>>>>>> about that one machine/input.
because it *only* ever *was* about problematic input to
*particular* classifiers...
Except that isn't what "Undecidability" is about. And you are >>>>>>>> just proving you don't know what you are talking about.
If we use the correct term, what you are tying to call
"undeciability" is really just not being correct. The problem is >>>>>>>> a "particualar classifier" has a fixed set of how it classifies >>>>>>>> everything. And thus, nothing can be "paradoxical" to it, as we >>>>>>>> never get to the other side,
RICK I'M NEVER GOING TO FIGURE YOU FOR BEING AN UNDYING ASSHOLE, >>>>>>> EXCEPT MAYBE ONCE U FINALLY TAKE UR IGNORANCE TO THE GRAVE
Right, because logic that is based on error can't unddrstand logic >>>>>> that requires correctness.
The other side you talk about is about if this decider was a
different decider, and some how magically the input was also a >>>>>>>> different input converted by a method not "in system" but only >>>>>>>> in a meta-system.
it's not my problem everyone went for over a century confusing >>>>>>>>> *our* ability to prove outcomes, with what a /particular/
decider can output... ya'll got caught up in a rigor of what is >>>>>>>>> fundamentally nonsense
Because what a particual decider does, if it isnt' right, is
just not interesting,
Only sort of right. Yes, we can not produce a machine that we >>>>>>>> know we can not possibly ever decide if it halts or not, because >>>>>>>> such a machine must by definition be non-halting, since ALL
no one has demonstrated a *possible* machine, let alone a
*real* machine, which we can't provably can't prove the outcome >>>>>>>>> for... such a proof makes the machine _NEITHER POSSIBLE NOR >>>>>>>>> REAL_ so it would be ludicrous to present such a proof to exist >>>>>>>>
halting machines are provable halting by just running them.
Thus, to prove we can't know they halt, we need to prove they >>>>>>>> don't halt, but then we know their behavior.
The fact we can't demonstrate and example of that class, doesn't >>>>>>>> mean members of it can't exist.
FUCK UR DAMN TEAPOT RICK
In other words, you are admitting I am right and need to resort to >>>>>> the fallacy of swearing to distract others from that basic fact.
i don't have a response anymore to the level of stupidity required
to keep bleating on and on about being so certain of machines it
would be a contradiction to even produce an example of...
In other words, you are just addmitting that you are too stupid to
understand the possiblity of something being unknowable.
all u have is an ill conceived notion based on the fallacy that _one_
classifier's failure to classify, can be overgeneralized to a _total_
inability to generally prove outcomes, ultimately founded on the
unproven thesis that all computability is encompassed by TM computing
But it isn't just ONE classifier. He shows how to make an input that
defeats ANY GIVEN classifier.
ANY classifier tht tries to determine this classification, will be
defeated by the input built by that template on it.
Thus, ALL such classifiers will be wrong.
Note, the arguement is not actually dependent on these being Turing
Machines, but can be generalized to ANY method based on the general
concept of "computing".
i don't need to respond to that further
Sure you do, or you are just admitting that you don't have an answer.
nope 🤷
It seems you are incapable of thinking abstractly, of even being
able to conceive of something you didn't invent without an example.
But you are also incapable of thinking concreately, because you
don't understand the need to show that there is a reason you need to
be able to show that the machines you imagine are possible.
I will point out AGAIN, that these unknowable machines are not
needed to be understood to understand the uncomputable nature of the
problem (They just come out as an effect of that).
So? What that proves is that it is IMPOSSIBLE to actually
both turing_H and turing_D are /incomplete specifications/ and >>>>>>>>> therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_ >>>>>>>>
acheive the goal defined for turing_D, a machine that determines >>>>>>>> if a given machine is circle-free.
they are just malformed /incomplete specifications/ and
_NOTHING MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond >>>>>>>>> the recognition that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such
machine exists.
BECAUSE IT'S INCOMPLETE. AN /INCOMPLETE SPECIFICATION/ THAT DOES >>>>>>> NOT HANDLE ALL INPUT SITUATIONS CANNOT EXIST
The Specification of the INTERFACE is complete, it fully defines
what the given machine must do.
imagine trying to argue a machine that can't exist, because it
didn't handle all input situations, as "complete"
In other words, you can't understand the difference between a
specification of an interface, and the specification of a machine.
IT DOESN'T "SPECIFY A MACHINE" BECAUSE IT'S _NOT_ A MACHINE DESCRIPTION, >>>
IT _SPECIFIES AN INTERFACE_, ONE THAT IS TOO INCOMPLETE TO BE
COMPUTED BY A MACHINE
So, neither is your "machine" an actual machine, as all YOU have is an
"interface" that determines your partial classification that isn't
actually implemented.
AN INTERFACE THAT CAN EXIST IS A FUCKTON MORE THAN WHAT YOU'VE GOT
Turing's H, CAN be built from ANY actual machine that we want to try
to claim might be a possible implementation of Turing's D. And all of
these WILL BE actual machines, and ALL of them are shown to not
produce the correct answer for Turing's H. Thus, there can not be any
machine that does what D is supposed to do.
On the other hand, YOU try to claim your fixed_H is an atual machine,
but it isn't until you actually define your partial_decider_D, which
you can't do, so your's isn't a machine either.
IT'S NOT A CONTRADICTION FOR THOSE MACHINES TO EXIST, EI: THE
SPECIFICATION IS NOT INCOMPLETE,
UNLIKE THE INCOMPLETE SPECIFICATION YOU PARADE AROUND AS MEANINGFUL
And, it turns out that any machine that you want to try to claim is an
implementation of your partial decider will either accept a machine
that will hang your fixed_H, or will not accept any machine for some
computable number.
_TOTALLY UNPROVEN STATEMENT_
We may not be able to directly prove this like Turing
WTF??? IF U CAN'T PROVE IT THEN STOP ASSERTING IT AS TRUE
Did, but Rice's proof shows that it can't be done, and your arguement
RICE'S PROOF IS _BASED_ ON TURING'S PROOF, YOU CAN'T USE IT INSTEAD OF TURING'S ... LOL
fking tard
that we need to find it is just you exercising the fallacy of
asserting that Russel's Teapot must exist if we can't prove it doesn't.
By your own logic, YOUR machine is INCOMPLETE, as you can't give the
actual specific algorithm it uses. Your "logic" is based on ASSUMING
that something is computable, with absolutely ZERO evidence that it is. >>>>
In other words, you live in a world of fantasy
The Specification of the interface is uncomputable, or
unrealizable, because we can't make a machine that fullfills it.
Yes, the specification of the ALGORITHM is incomplete, as we don't >>>>>> specify how to do all the steps, but then so is yours.
Turing, BECAUSE he used an unspecified algorithm for his D, was
able to actually PROVE that the specification for the interface of >>>>>> D was unimplementable.
All your arguement has done is shown that if magic fairy dust
powered unicorns exists (which they don't), you could use one of
them to built an (also non-existant) program to compute something >>>>>> that is actually proven to be uncomputable.
All you have done is proven you don't understand how logic works, >>>>>> and thus your works SHOULD BE IGNORED.
If you want to be consistant, YOUR partial_decider_D is ALSO
incomplete,
IT'S NOT INCOMPLETE AS IT HANDLES ALL INPUT SITUATIONS
That isn't what "Incomplete", means, as your partial_decider_D
doesn't exist so it can't handle ANY input situations.
YOU HAVEN'T DEMONSTRATED A CONTRADICTION WITH PARTIAL_RECOGNIZER_D, >>>>> SO YOU HAVEN'T SHOWN IT TO NOT EXIST...
In other words, you assert the existance of Russel's teapot.
The "contradiction" is in the fact that you concept of "undecidable
input" just is nonsense.
u know exactly what i'm talking about when i say /undecidable input/,
u just refuse to acknowledge anything i say as correct. because as
soon as that damn breaks, ur whole position gets flooded with truth
Right, I know you mean an input that the given decider is just wrong
about.
LOL, BEING IN SPEC IT NOT _WRONG_
Any other definition is just a lie.
LOL, BEING IN SPEC IS NOT A _LIE_
How is it different then inputs that the specific decider is just
wrong able.
Describe how you define that property based on JUST the specific
input itself and the specific decider.
Your world is based on variable constants and similar nonsense.
You don't get it, that if you want to put down Turing's machines
for not handling cases, you need to also accept that yours doesn't >>>>>> for the same reason, a machine that fully meets your specifaction >>>>>> can't exist.
And part of the reason for yours, is that you have a nonsense term >>>>>> in your definition, as there is no such thing as a "undecidable
input" if some other decider can get the known right answer.
Part of your problem is you just don't understand what a PROGRAM
or ALGORITHM actually is
as you haven't specified HOW it does what it does, just lie
turing didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such
machine could exist, which he succeeded at, but YOUR claim is >>>>>>>> that such a thing might be able to exist, which you just have >>>>>>>> not proven.
And, it is an assertion that NO decider can correctly compute >>>>>>>>>> the answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, which >>>>>>>>>> means you have chosen a particular (broken) implementation of >>>>>>>>>> D (since no actually correct D exists), like your
partial_recognizer_D, has fulled defined behavior, and a
decider can exist that determines it, there is nothing
"undecidable" about that input.
It is just that the D that H was built on is wrong. Nothing >>>>>>>>>> "paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on you >>>>>>>>> just _IGNORING_ the specification i specifically created for >>>>>>>>> partial_recognizer_D...
No, your problem is you just are too stupid to understand that >>>>>>>> your "specification" is just nonsense, as it assumes that you >>>>>>>> can perform a test by "changing" the decider and the input to >>>>>>>> see a results.
But the specification is about a particular input and decider. >>>>>>>>
the specification for partial_recognizer_D is when encountering >>>>>>>>> undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your
definition is just internally inconsistantly trying to define >>>>>>>> input that the decider gets wrong. Since the class the partial >>>>>>>> decider is
this _DOES NOT MEAN_ we cannot know what the /undecidable
input/ does, as a paradox in regards to partial_recognize_D >>>>>>>>> _ONLY AFFECTS_ partial_recognize_D not _OUR ABILITY TO PROVE >>>>>>>>> OUTCOMES_
In other words, you admit that you words are just lies and don't >>>>>>>> mean what they say.
Your "undecidable" isn't about decidability, but correctness of >>>>>>>> a given decider.
YES, THAT'S ALL UNDECIDABILITY EVER WAS
In other words you are admitting you don't know what you are
talking about.
A given decider being incorrect doesn't mean a correct decider
doesn't exist. There are many decidable problems, for which
incorrect deciders can exist.
Undecidability is about the being NO correct deciders.
YOU HAVEN'T EVEN SHOW A MACHINE WHICH HAS NO CORRECT CLASSIFIERS
FOR...
Because I don't need to.
YOU'VE ONLY SHOWN THAT FOR ANY CLASSIFIER THERE EXISTS A MACHINE IT
CAN'T CLASSIFY...
NOT THAT THERE EXISTS A MACHINE THAT CAN'T BE CLASSIFIED...
But I don't need to.
The uncomputability of that classification is proven by the first.
ALL THAT'S BEEN SHOWN IS _ONE_ MACHINE CAN'T CLASSIFY ALL MACHINEs.
IT DOES NOT THE SHOW THERE IS MACHINE THAT *CANNOT* BE CLASSIFIED BY
*ANY* MACHINE
THOSE ARE _DIFFERENT_ CLAIMS, REGARDLESS OF U JUST CONFLATING THEM
BECAUSE UR GETTING DESPERATE
The existance of the second class of machine is just something more,
that is clearly beyond your understanding.
All that is needed to prove a problem is uncomputable is to show
that you can make an input for any given decider that it is wrong
about.
THAT ONLY SHOWS THAT THE PROBLEM CAN'T BE COMPUTED BY A _SINGLE_ MACHINE
Right, or any effectively enumerable set of machine, since we can then
build the single machine that tests all of them.
And that *IS* the definition of computable,
BECAUSE UR JUST ASSUMING THE CT-THESIS AS TRUE, WHICH IS NOT PROVEN
You don't seem to understand the problem.
YOU DON'T UNDERSTAND WHAT'S BEEN PROVEN
Sure I do, you just admitted what I have said is proven, that the
LOL, UR CLAIMS ARE GETTING MORE GARBAGE BY THE POST
problem is not computable, which means there does not exist a single
machine that correctly computes all the answers.
This means that the trivial always false machine meets your
NO, THE FAILURE TO CLASSIFY MUST COME FROM THE STRUCTURE OF THE >>>>>>> INPUT, NOT AN INCORRECT ALGO OF THE CLASSIFIER, WHY AM I
REPEATING ME ON THIS?
In other words, you think people/machines can't make mistakes?
BEING UNABLE TO CLASSIFY /UNDECIDABLE INPUT/ IS _NOT_ A MISTAKE ON
BEHALF OF THE CLASSIFIER
But the input isn't "undecidable", as that isn't an actual property
of an input.
IT IS _CAUSED_ *BY THE STRUCTURE OF THE INPUT*
No, as the input is fully decidable by some other decider, and thus
there is nothing "undecidable" about it.
RIGHT, THAT'S BECAUSE UNDECIDABILITY IS ONLY DEFINED AS A RELATIONSHIP BETWEEN A MACHINE AND A PARTICULAR CLASSIFIER PARADOXES BY THAT MACHINE
THERE _NEVER_ HAS BEEN _ANY_ OTHER FORM OF PROVEN UNDECIDABILITY WITHIN COMPUTING,
AND UR DOING FUCK ALL TO PROVE ANOTHER FORM EXISTS
It is just that the method will always find a flaw in the algorithm of
that specific decider, making it just wrong.
Your logic is just based on category errors, based on your own
stupidity.
Then why are you worried about proofs of correctness.
You don't seem to understand that the whole basis of these proofs >>>>>> isn't about finding *A* input that nothing can decide on, but
finding a pattern that makes an input for a given decider that it >>>>>> won't be able to handle.
That pattern isn't actually something that necessarily remains in >>>>>> the machine after construction, but direction to find a given
algorithms fatal flaw, showing that ALL algorithms for that
problem must have a fatal flaw which can be different for every one. >>>>>>
requirements. Any input for which the correct answer for the
base problem would be true, THAT particular decider won't get >>>>>>>> right, so it is correct in your expanded partial sense to return >>>>>>>> false.
IT seesm you just can't read.
(you don't have one btw, so please do let me witness ur cope) >>>>>>>>>>>>
It is ONLY "incomplete", because it is built on a PRESUMED >>>>>>>>>>>>>> defined machine. If the decider "D" actually exists, then >>>>>>>>>>>>>> H is FULLY defined.
see, turing_H's runtime is not describable because it's >>>>>>>>>>>>>>> an / incomplete specification/ of machine and therefor >>>>>>>>>>>>>>> _does not even exist_ as a real TM, not because of the >>>>>>>>>>>>>>> hypothesized undecidability in relation to some D >>>>>>>>>>>>>>
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete >>>>>>>>>>>>> specification/, as it doesn't handle idiosyncrasies of TM >>>>>>>>>>>>> computing, specifically self- referential set-
classification paradoxes
It doesn't need to. It just needs to be built on what you >>>>>>>>>>>> want to claim to be your answer for the concept of D.
As soon as you define what machine you are going to use as >>>>>>>>>>>> your equivalent for D, like your partial_recognizer_D, then >>>>>>>>>>>> the template turing_H can be converted into an actual machine. >>>>>>>>>>>>
actual_turing_H, on the other hand, actually does exist, >>>>>>>>>>>>>>> so we can know/ prove what it does. it, however, still >>>>>>>>>>>>>>> forms a structural paradox in regards to
partial_recognizer_D which will fail to classify it as >>>>>>>>>>>>>>> circle-free
But, since you admit to actual_turing_H existing, that >>>>>>>>>>>>>> means that
since partial_recognizer_D _CAN_ exist, then
actual_turing_H _CAN_ exist
And so can the turing_H that uses partial_recognizer_D as >>>>>>>>>>>> its decider D.
The steps the algorithm does is what defines the machine, >>>>>>>>>>>> not the "names" of the things it "calls", as those names >>>>>>>>>>>> don't actually exist in the final building of the machine, >>>>>>>>>>>> which is the problem with your concept, as that is needed to >>>>>>>>>>>> determine your "paradoxical" property.
the names are just shortcuts for the literal values rick, >>>>>>>>>>> don't be daft
Right, but the literal values don't have the names on them, >>>>>>>>>> and we don't have borders within the machine letting us know >>>>>>>>>> we moved from "outer code of H" into "the code of the D it >>>>>>>>>> uses", especially with Turing Machines were such usages are by >>>>>>>>>> necessity "expanded" in-line.
please do actually read turing's paper sometime, he does
organize machine descriptions into functional groups
Yes, at meta level.
Thus, the "copy" of D that was pulled into H isn't actually >>>>>>>>>> the same
no it can be literally a copy of same description than is the >>>>>>>>> run by a UTM (universal turing machine) functional group of the >>>>>>>>> machine
It CAN be, but doesn't NEED to be. And in fact, his H isn't
described that way.
thing as the independent machine D that we can talk about, >>>>>>>>>> just a functional equivalent which doesn't even have its own >>>>>>>>>> D.N (since that is a property only of full machines)
ur really pulling at straws here rick
You really think so?
It seems your problem is you need to work in a tight cage where >>>>>>>> other ideas are not allowed that show the holes in your ideas. >>>>>>>>
But your computation wasn't actually a diagonal of
pratial_recognizer_D has an actual and fixed algorithm. >>>>>>>>>>>>>>
But the problem isn't in it handling the machine fixed_H, >>>>>>>>>>>>>> but how it ever handles the results generated by
Thus, your basic criteria is a category error. >>>>>>>>>>>>>>>>
If a D can determine which machines its basic algorithm >>>>>>>>>>>>>>>> will get wrong, then it can just return the different >>>>>>>>>>>>>>>> answer, and then THAT machine it will get right. The >>>>>>>>>>>>>>>> problem is it creates a DIFFERENT input, that uses this >>>>>>>>>>>>>>>> new version of the machine, that it will get wrong. >>>>>>>>>>>>>>>>
WHICH IS FINE, actually
actual_turing_H ultimately computes the *same* sequence >>>>>>>>>>>>>>>>> as fixed_H, which in turn _does_ have a spot on the >>>>>>>>>>>>>>>>> diagonal. so there is no actual need to include a digit >>>>>>>>>>>>>>>>> from actual_turing_H on the diagonal, as the number it >>>>>>>>>>>>>>>>> computes is already included on the diagonal
But not as actual_turing_H, which doesn't have your "fix". >>>>>>>>>>>>>>>>
And thus if your partial_recoginzer_D accepts ANY >>>>>>>>>>>>>>>> verision of actual_turing_H, your fixed_H will get stuck. >>>>>>>>>>>>>>>>
So, now, you need to show how your partial_recognizer_D >>>>>>>>>>>>>>>> can possible filter all ALL of that infinite set of >>>>>>>>>>>>>>>> machines, but still accept some machine that computes >>>>>>>>>>>>>>>> that particular number that it would compute with this >>>>>>>>>>>>>>>> supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just >>>>>>>>>>>>>>>> admitting that you accept that Russel's Teapot must exist. >>>>>>>>>>>>>>>>
So you think, because you don't see it is just built on >>>>>>>>>>>>>>>> the fairy dust, of assuming that your partial decider >>>>>>>>>>>>>>>> that meets ALL your needs exists,
goddamn rick, this rock just keeps getting more solid >>>>>>>>>>>>>>>>
🤷🤷🤷
to clarify this further: the only algorithmic difference >>>>>>>>>>>>>>> between fixed_H and actual_turing_H is the way they >>>>>>>>>>>>>>> respectively handle their own self- references. they >>>>>>>>>>>>>>> function identically when handling all other machines >>>>>>>>>>>>>>
actual_turing_H, which by your assumption has become a >>>>>>>>>>>>>> cycle-free machine and ALL equivalent versions of it have >>>>>>>>>>>>>> been partially declared to be non- cycle free.
This means that your enumeration of machines you are >>>>>>>>>>>>>> looking at is, in fact, incomplete, thus the diagonal you >>>>>>>>>>>>>> computed is wrong.
But actual_turing_H is actually circle_free, and thus its >>>>>>>>>>>>>> results SHOULD have been on the diagonal.
fixed_H = () -> {
N = 1
K = 0
output = [] //
written to F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) // hard
coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // TRUE
= satisfactory
K += 1
output.push(simulate_Kth_digit(N,K)) >>>>>>>>>>>>>>> }
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will >>>>>>>>>>>>>>> skip trying to simulate itself and put it's hard coded >>>>>>>>>>>>>>> digit on the diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it will >>>>>>>>>>>>>>> skip trying simulate actual_turing_H because
partial_recognizer_D fails to classify it as circle-free >>>>>>>>>>>>>>
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is >>>>>>>>>>>>> put on the diagonal, then actual_turing_H _DOES NOT_ need >>>>>>>>>>>>> to be in _THAT_ total enumeration of _COMPUTABLE NUMBERS_ >>>>>>>>>>>>
computable numbers, but of machines that made computable >>>>>>>>>>>> numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends >>>>>>>>>> on being sloppy.
the only way to compute a diagonal across computable numbers is >>>>>>>>> simulating each number by the machine that computes it.
Not the ONLY way, but a good way to try.
But since Turing Shows this can't be done, it can't be done any >>>>>>>> way.
oh wow GREAT CATCH RICK!!!!,
Note, Turing SPECIFICALLY pointed out that he was going to the >>>>>>>>>> related problem, the problem of:
/but the problem of enumerating computable sequences is
equivalent to the problem of finding out whether a given
number is the D.N of a circle- free machine, and we have no >>>>>>>>>> general process for doing this in a finite number of steps/ >>>>>>>>>
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT
ENUMERATING OUT ALL CIRCLE_FREE MACHINES*,
Show it.
what??? bro ur the one who informed me proper enumerations can
only have *one* instance of each element enumerated,
and we all know that the total enumeration of machines has
infinite machines producing each computable numbers...
so therefore not only can the enumeration of computable numbers >>>>>>> be done without enumerating all circle-free machines, it *MUST* >>>>>>> be done without enumerating all circle-free machines, as we can >>>>>>> output *ONLY* one machine for each computable number
that should suffice
In other words, you are admitting you can't do it.
no richard, we've reached a point where u are unable to correctly
and honestly identify a self-evident concept:
FOR _ANY_ GIVEN COMPUTABLE NUMBER, THERE ARE _INFINITE_ MACHINES
WHICH COMPUTE IT ...
I fully understand that.
WE ONLY NEED TO CLASSIFY _ONE_ OF THOSE MACHINES TO ADD THAT GIVEN
COMPUTABLE NUMBER THE LIST, THE REST NOT ONLY CAN, BUT MUST, BE
IGNORED
Right, and you need to PROVE that you can.
IT'S ENOUGH TO REFUTE A *KEYSTONE* POINT IN TURING'S PAPER
Nope, claiming that there might exist a machine, that doesn't actually
exist, doesn't prove anything.
WHAT??? TURING CONFLATING THE ENUMERATION OF ALL CIRCLE-FREE MACHINES
WITH THAT OF JUST COMPUTABLE NUMBERS IS JUST WRONG, REGARDLESS OF WHAT
IS FOUND TO BE POSSIBLE AFTER
fking tard LOL
THE REST OF THE PAPER IS PRETTY MUCH VOID DUE TO TURING CONFLATING
THE ENUMERATION OF COMPUTABLE NUMBERS WITH THAT OF MACHINES THAT
COMPUTE COMPUTATIONAL NUMBERS
No, he shows that there does exist a problem that a decider can not
decide on. That shows that the Hilbert Entscheidungsproblem can not
have a solution, which is his real goal.
Note, on page 246 he sketches out the proof, noting that *IF* the
computable numbers WERE effective enumerable (what he calls just
enumerable), then we could compute that anti-diagonal by thus doing
that enumeration and selecting the appropriate digits.
_THAT IS ALSO A FALLACY_
THE IDIOSYNCRASIES OF TM COMPUTING DO NOT ALLOW THE USE OF A DIAGONAL TO THEN COMPUTE AN ANTI-DIAGONAL
YOU'RE NOT GOING TO UNDERSTAND WHY UNTIL YOU WORK THRU THE FIX I MADE TO MAKE THE DIAGONAL COMPUTABLE, AND WORK THRU WHY THAT FIX DOES NOT WORK
FOR THE ANTI-DIAGONAL COMPUTATION _EVEN HAVE A DIAGONAL AT HAND_
He also points out that this gets to the feeling that "there must be
something wrong" (even though the logic is perfectly correct).>>
THEREFOR, WE DO _NOT_ NEED A CLASSIFIER WITH THE POWER TO COMPUTE
WHETHER _ANY_ GIVEN MACHINE IS A COMPUTABLE NUMBER, WE ONLY NEED A
CLASSIFIER WITH ENOUGH POWER TO CLASSIFIER _ONE OF THE INFINITE_
MACHINES FOR _ANY_ GIVEN COMPUTABLE NUMBER
Right, you need to PROVE that you can build a classifier that will
select at least one machine for every computable number.
You can't just assume your decider exists, you need to show HOW it
works.
Since you concept is based on identifing the "undeciable inputs",
you need to show how you actually will do it, which means you need
to define it in a way that is detectable.
i already demonstrated the basic idea several times
No, you haven't, as your "demonstartion" always violates the defintion
of a machine and an input.
How is your definition anything different that an input that this
machine gets wrong.
After all, when you talk about changing the decider, that doesn't
actualy change the input, since that changed decider is NOT the
original decider, and thus the actual input doesn't change.
actually building a partial_recognizer for all possible computable
numbers is a research question far outside the scope of this
discussion since it involves open questions in number theory we don't
even have answers to
No, it is a fundamental assumption of your claim, that is in direct
opposition to the proofs.
Your arguement is just that you must be allowed to just assume
Russel's teapot exist without proof.
There is no inherent reason you can pull an infinite subset out of >>>>>> a similarly infinite set.
We can enumerate the primes, even though for every prime there is >>>>>> an infinite number of composite numbers that are multiples of it, >>>>>> thus your logic is based on an error.
THe fact that he quote an equivalence that he doesn't prove he >>>>>>>> doesn't make his proof incorrect.
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_ >>>>>>>>> EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN >>>>>>>>> NUMBER IS A CIRCLE-FREE MACHINE,
Prove it.
there's nothing to prove, the statement is self-evident given the >>>>>>> previous two paragraphs i wrote. we don't need to be able to
output it for *ANY* given machine, we only need to be able to
output that fact for *ONE* machine of the infinite which produces >>>>>>> *ANY* given compute number
honestly i'm not sure why it's take until year 2026 for someone >>>>>>> to point this out
In other words, you are just admitting that you understanding of
the problem isn't sufficent to deal with it.
no richard, we've reached a point where u are unable to correctly
and honestly identify a self-evident concept:
THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN NUMBER >>>>> IS A CIRCLE-FREE MACHINE, AS WE DO NOT NEED TO IDENTIFY _ALL_
CIRCLE- FREE MACHINES, JUST _ONE FOR EACH_ COMPUTABLE NUMBER
So, you need to show that you CAN do that.
Turing points out that they are computationally related, and I will
being related does _not_ make them the _same problem_,
Right, and he doesn't say they are. he says they are an "equivalent"
problem, i.e there is a strong relation to each other (so they either
both have or don't have an answer), that he has presumably proven
somewhere else.
Calling it an error without proof just shows your stupidity.
FOR SOMEONE WHO A) UNDERSTANDS THE BASIC CONCEPTS INVOLVED, AND B) CAN
THINK CRITICALLY,
THE PROOF IS LITERALLY THE TWO PARAGRAPHS I ALREADY WROTE
and turing's proof *depends* on them being the _same problem_
No, since his ultimate goal is the proof is about the Hilbert
Entscheidungsproblem, which was the big question of the day.
IF HE'S ACTUALLY FAILED TO PROVE THE COMPUTABLE SEQUENCES AS NOT
ENUMERABLE THEN THE PROOF FALLS APART, I'M SORRY
To show is proof is wrong, *YOU* need to show that they problems are
NOT equivalent.
I ALREADY DID, SO YET AGAIN:
ENUMERATING THE COMPUTABLE SEQUENCES ONLY REQUIRES THE LIMITED POWER TO IDENTIFY _ONE_ CIRCLE-FREE MACHINE FOR EACH COMPUTABLE SEQUENCE, _NOT_
ALL CIRCLE-FREE MACHINES
> THAT'S LITERALLY THE PROOF RIGHT THERE
>
> #god
beleive him more that you how claims you can do one without showing
how.
You have ADMITTED your ignorance of the field, having looked at only
a very few of the papers, I suspect there was something discussed
this is suspicion that is easily dealt with using search tools. but
instead of backing up that suspicion ur just continuing to be
suspect, and i'm never gunna buy it
Really? and how much of the early papers have actually been preserved
and put into something searchable?
How much of the actual presentations of the various conferences have
been transcribed and put into search engines?
The fact that he makes the passing reference to it, and no one in that
error jumped on that is a good sign that it was accepted.
As I said, prove him wrong or that the conclusion is wrong by ACTUALLY
showing how to do what he claims can't be done. And them needs more
than assuming you can make a decider that does the real work.
That is you just living off unicorn milk.
somewhere about this, even if it wasn't as famous of a paper (as it
wasn't so eathshaking by itself).
Your problem is you are working on "gut instinct" with a gut that >>>>>> doesn't understand what it is talking about, because it thinks
unicorns exist.
we only need to compute that fact for *ONE* machine for any >>>>>>>>> given computable sequence, _SO A PARTIAL RECOGNIZER IS
SUFFICIENT FOR THE JOB_
So do it.
you ask too much for not recognizing anything i said as true
no richard, we've reached a point where u are unable to correctly
and honestly identify a self-evident concept
Because is is just wrong, being based on category errors.
As long as you just ASSUME you can do what you claim, you are just
making a ASS out of yourself, proving that your "self-evident" is
really a case o "self-delusion"
Because it isn't, and you have no proofs, because you can't
actually prove a false statement.
Part of your problem is you don't actually know much about what
you are talking, so you have nothing to work with,
fuck rick ur so bad at defending this it's actually really
useful to me!
(there's *another* specific sentence i have a problem with, >>>>>>>>> where's a ben when u need him?)
All you have shown so far is that you are too stupid to
understand the nature of the paper.
Note, it was NOT intended to be the end-all self-contained
regerous proof of the concept. It was a paper, leaning on
previous work done, and with the intent of more work to follow >>>>>>>> to show the basic idea of why something was true.
Thus, not showing how a step works, isn't an error.
Now, to counter his claims, because they are so well accepted, >>>>>>>> you are going to need to be rigorous and not based an the
fallacy of assuming something and showing that it doesn't lead >>>>>>>> to a contradiction when you follow a restricted line of reasoining. >>>>>>>>
So, HIS enumeration is the complete enumeration of "circle- >>>>>>>>>> free machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration
either, why am i the only one subject to criticism here???
Also, your proof is built on the existance of a Russel >>>>>>>>>>>> Teapot, you have
the *specification* /works/, which shows that it's *not* >>>>>>>>>>> contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the
conclusion. Your arguement boils down to showing that *IF* you >>>>>>>>>> can correctly determine all machines that fit in a not-
definable category (since you concept of [paradoxical can't >>>>>>>>>> actually be defined at the machine level, only at how a
machine might have been created at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that >>>>>>>>>>>> can make the precise "mistakes" you need it to. (They are >>>>>>>>>>>> mistakes because the input
"mistake" is this case is a fallacious non-techincal value >>>>>>>>>>> judgement on the return value,
No, since your criteria that you want to use is just an
undefinable term for a machine description.
There is nothing "paradoxical" of the actual machine when >>>>>>>>>> created, as it then has fixed and determined behavior that can >>>>>>>>>> be decided on. The thing that is paradoxical is the mete- >>>>>>>>>> system pattern used to generate the class of machines as pair >>>>>>>>>> with a given decider.
cause they are rather entirely within specification, as FALSE >>>>>>>>>>> was specified to be used upon encountering an /undecidable >>>>>>>>>>> input/, something turing_D did not handle correctly (making >>>>>>>>>>> it _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is >>>>>>>>>> based on a actual D that does make a specific decision to the >>>>>>>>>> input. And it is just that for any given version of D, we can >>>>>>>>>> find a DIFFERENT H that it will get wrong (the H that was just >>>>>>>>>> built on it).
Do you want to claim that ALL these H are "undecidable" even >>>>>>>>>> though for
no... fixed_H contains no paradox, and is therefor not /
undecidable input/ to any classifier
all of them there does exist a decider that gets the right >>>>>>>>>> answer?
many machines contain no paradox and are therefore classifiable >>>>>>>>> by all classifiers
But this proof isn't about the many, but the ALL.
A term it seems you don't understand.
And it isn't the job of others to show you what one you missed, >>>>>>>> it is yours to prove that they don't, as YOU are making the claim. >>>>>>>>
then the H's as of now are i suppose "generators" instead of >>>>>>>>> enumerators
It seems you just don't know what the word means, and thus you >>>>>>>>>> specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because of >>>>>>>>>>>> this turing_H is built on this partial_recognizdr_D which >>>>>>>>>>>> causes them to not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to >>>>>>>>>>>> machines, as it is based on the incorrect definition that >>>>>>>>>>>> assumes a machine can be something other than itself, and >>>>>>>>>>>> when it changes, it changes other things that at the meta/ >>>>>>>>>>>> template level refered to its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just >>>>>>>>>> refuse to learn.
Try to actually fully DEFINE the word based on the actual >>>>>>>>>> properties of the input machine, and not refering to other >>>>>>>>>> machines that are not part of the input (but would be created >>>>>>>>>> as alternate inputs to foil an alternate decider).
when actual_turing_H iterates across DN(actual_turing_H) >>>>>>>>>>>>>>> itself, it will similarly skip trying to simulate itself, >>>>>>>>>>>>>>> because partial_recognizer_D fails to classify it as >>>>>>>>>>>>>>> circle- free
Right, so it ALSO is wrong, as it didn't compute the >>>>>>>>>>>>>> diagonal of the circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE >>>>>>>>>>>>> NUMBERS_, which is the _ACTUAL PROBLEM_ turing was dealing >>>>>>>>>>>>> with,
No, it isn't, as your "set of computable numbers" has >>>>>>>>>>>> duplicates.
it's still a diagonal across all computerate numbers even if >>>>>>>>>>> it has dpulicates ...
No, because an enumeration of a set doesn't HAVE duplicates. >>>>>>>>>
and turing was wrong about that too, but i don't see u
criticizing him now eh???
H is a generator, its goal was to generate a particual number.
it's goal was a generate a diagonal across all circle-free machines, >>>>>>>
which as you said: is *NOT* an enumeration, and an enumeration
required duplicating the machines
Right, but the result it is generating is BASED (and defined) on a >>>>>> given enumeration which it computes internally).
Your Fixed_H isn't computing anything with a defined
specification, as your "enumeration" that you claim to be based
on, isn't actually an enumeration that meets your claimed
requirement.
D is a decider, that allowed H, in its generation, to produce >>>>>>>> (internally) the enumeration that its results were defined by, >>>>>>>>
It seems you don't understand that to reach your goal, sometimes >>>>>>>> you have to acheive a sub-goal that is quite different.
On 3/9/26 9:02 PM, dart200 wrote:
On 3/9/26 5:03 PM, Richard Damon wrote:
On 3/9/26 12:32 AM, dart200 wrote:
On 3/8/26 5:14 PM, Richard Damon wrote:
On 3/8/26 11:27 AM, dart200 wrote:
On 3/8/26 5:20 AM, Richard Damon wrote:
On 3/7/26 12:47 PM, dart200 wrote:i don't have a response anymore to the level of stupidity required >>>>>> to keep bleating on and on about being so certain of machines it
On 3/7/26 5:52 AM, Richard Damon wrote:
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:
On 3/5/26 11:06 PM, dart200 wrote:except we can describe the behavior of actual_turing_H and >>>>>>>>>>>> prove it's equivalent to fixed_H, so therefore it doesn't >>>>>>>>>>>> fit what u stated was
On 3/5/26 6:27 PM, Richard Damon wrote:
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:this is unfortunately a fallacious understanding of >>>>>>>>>>>>>>>> undecidability. just because a machine is /undecidable >>>>>>>>>>>>>>>> input/ to a particular classifier does mean we cannot >>>>>>>>>>>>>>>> then prove and know ourselves what the machine actually >>>>>>>>>>>>>>>> does,
On 3/4/26 4:47 AM, Richard Damon wrote:
On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/3/26 3:55 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 6:24 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 8:24 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 2:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 12:38 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
wow, u know up until now, i thot i fully agreed >>>>>>>>>>>>>>>>>>>>>>> with turing's short diagonal proof, but in >>>>>>>>>>>>>>>>>>>>>>> writing this post i now find myself in a subtle, >>>>>>>>>>>>>>>>>>>>>>> yet entirely critical disagreement: >>>>>>>>>>>>>>>>>>>>>>>it is wrong,Then you zre just showing your stupidity, >>>>>>>>>>>>>>>>>>>>>>>>>> because YOU can't tell the difference. >>>>>>>>>>>>>>>>>>>>>>>>>>And focusing on what is said to be >>>>>>>>>>>>>>>>>>>>>>>>>>>> impossible and not changing the problem is >>>>>>>>>>>>>>>>>>>>>>>>>>>> also important.the theory of computation is the theory of >>>>>>>>>>>>>>>>>>>>>>>>>>>>> such procedures, and understanding the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> diagonal procedure is critical to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> understanding the *base* contradiction/ >>>>>>>>>>>>>>>>>>>>>>>>>>>>> paradox that the rest of his support for >>>>>>>>>>>>>>>>>>>>>>>>>>>>> godel's result is then built on >>>>>>>>>>>>>>>>>>>>>>>>>>>>On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. Thomasson >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
Eh?!On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ...yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
an effective enumeration of all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing machines was proven on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing's original paper and can be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> reused anywhere... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>You think you can test all of them one >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> by one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test / >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> all/ of an infinitude one by one. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
that exactly what turing does in his >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof: he defines a comptuation >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that enumerates out all the numbers, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> testing each one of they represent >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a "satisfactory"/"circle-free" machine, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and adding that to diagonal >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> carefully read p247 of turing's >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof and produce the psuedo-code for the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine H, assuming that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine D exists
I'll get to it sooner then, because it's >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mad. Are you sure he didn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> reason quantified over all but phrase it >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> like a procedure for what he >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation >>>>>>>>>>>>>>>>>>>>>>>>>>>> isn't the generation of the diagonal itself, >>>>>>>>>>>>>>>>>>>>>>>>>>>> but effectively enumerating the enumeration >>>>>>>>>>>>>>>>>>>>>>>>>>>> in the first place.
i don't see any indication that turing >>>>>>>>>>>>>>>>>>>>>>>>>>> realized a difference there >>>>>>>>>>>>>>>>>>>>>>>>>>
After all, on page 246 he says: >>>>>>>>>>>>>>>>>>>>>>>>>>
The computable sequences are therefore not >>>>>>>>>>>>>>>>>>>>>>>>>> enumerable.
Here is is SPECIFICALLY talking about the >>>>>>>>>>>>>>>>>>>>>>>>>> effective enumeration of the computable >>>>>>>>>>>>>>>>>>>>>>>>>> sequences.
He then points out that he can directly show >>>>>>>>>>>>>>>>>>>>>>>>>> that the "anti- diagonal" of the (non- >>>>>>>>>>>>>>>>>>>>>>>>>> effectively computed) enumeration can't be >>>>>>>>>>>>>>>>>>>>>>>>>> computed but that "This proof, although >>>>>>>>>>>>>>>>>>>>>>>>>> perfectly sound, has the disadvantage that it >>>>>>>>>>>>>>>>>>>>>>>>>> may leave the reader with a feeling that >>>>>>>>>>>>>>>>>>>>>>>>>> 'there must be something wrong'". >>>>>>>>>>>>>>>>>>>>>>>>>
No, YOU are wrong, as you don't understand what >>>>>>>>>>>>>>>>>>>>>>>> is being done.
I think he is refering he to the standard anti- >>>>>>>>>>>>>>>>>>>>>>>> diagonal arguement, which shows that since for >>>>>>>>>>>>>>>>>>>>>>>> all n, position n differs from the value in >>>>>>>>>>>>>>>>>>>>>>>> number n, there can not be any element that >>>>>>>>>>>>>>>>>>>>>>>> matches the anti- diagonal.
It is just a natural fact of countable infinity, >>>>>>>>>>>>>>>>>>>>>>>> something it seems you just don't understand. >>>>>>>>>>>>>>>>>>>>>>>>
Show how that is actually wrong. >>>>>>>>>>>>>>>>>>>>>>>
/let an be the n-th computable sequence, and let >>>>>>>>>>>>>>>>>>>>>>> φn(m) be the m- th figure in an. Let β be the >>>>>>>>>>>>>>>>>>>>>>> sequence with 1- φn(m) as its n-th. figure. Since >>>>>>>>>>>>>>>>>>>>>>> β is computable, there exists a number K [== β] >>>>>>>>>>>>>>>>>>>>>>> such that 1- φn(n) = φK(n) for all n. Putting n = >>>>>>>>>>>>>>>>>>>>>>> K, we have 1 = 2φK(K), i.e. 1 is even. This is >>>>>>>>>>>>>>>>>>>>>>> impossible/
i agree with this proof is far as much as if β was >>>>>>>>>>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure, >>>>>>>>>>>>>>>>>>>>>
but what i don't agree is that we can just assume >>>>>>>>>>>>>>>>>>>>> the computability of β from the existence of φn(m), >>>>>>>>>>>>>>>>>>>>> THAT'S THE FALLACy
the fallacy here is assuming that because the >>>>>>>>>>>>>>>>>>>>>>> direct diagonal is computable, that one can >>>>>>>>>>>>>>>>>>>>>>> therefore compute the anti- diagonal using the >>>>>>>>>>>>>>>>>>>>>>> direct diagonal. the abstract definition makes it >>>>>>>>>>>>>>>>>>>>>>> look simple, but this ignores the complexities of >>>>>>>>>>>>>>>>>>>>>>> self- referential analysis (like what turing >>>>>>>>>>>>>>>>>>>>>>> details on the next page)
But you can, *IF* you have a machine that computes >>>>>>>>>>>>>>>>>>>>>> the diagonal, then just change all the write to >>>>>>>>>>>>>>>>>>>>>> the output to write the opposite. Note, the "self- >>>>>>>>>>>>>>>>>>>>>> reference" that you are thinking of stops being a >>>>>>>>>>>>>>>>>>>>>> "self- reference" but is a reference to the >>>>>>>>>>>>>>>>>>>>>> original write the diagonal code.
Nope, because your filtering TM (or RTM) will >>>>>>>>>>>>>>>>>>>>>> still fail when it gets to the number of TURING'S >>>>>>>>>>>>>>>>>>>>>> H, as there is no correct answer for the machine >>>>>>>>>>>>>>>>>>>>>> built by that template.
in both methods i have for rectifying the paradox >>>>>>>>>>>>>>>>>>>>>>> found in the direct diagonal (either (1) >>>>>>>>>>>>>>>>>>>>>>> filtering TMs or (2) using RTMs), neither can be >>>>>>>>>>>>>>>>>>>>>>> used to then compute the anti-diagonal >>>>>>>>>>>>>>>>>>>>>>
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN >>>>>>>>>>>>>>>>>>>>> INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???) >>>>>>>>>>>>>>>>>>>>>
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY >>>>>>>>>>>>>>>>>>>>> TMs, SO IT *SHOULD* BE FILTERED OUT
in (1) the algo to compute an inverse diagonal is >>>>>>>>>>>>>>>>>>>>>>> filtered out like turing's paradoxical variation >>>>>>>>>>>>>>>>>>>>>>> of the direct diagonal would be, and there is no >>>>>>>>>>>>>>>>>>>>>>> analogous non- paradoxical variation that has a >>>>>>>>>>>>>>>>>>>>>>> hard coded value that is inverse to what it does >>>>>>>>>>>>>>>>>>>>>>> return ... such a concept is entirely >>>>>>>>>>>>>>>>>>>>>>> nonsensical. a function can only return what it >>>>>>>>>>>>>>>>>>>>>>> does, it can't also return the inverse to what it >>>>>>>>>>>>>>>>>>>>>>> returns eh???
But, if D filters it out, then it becomes circle- >>>>>>>>>>>>>>>>>>>>>> free, and thus your enumeration is incomlete. >>>>>>>>>>>>>>>>>>>>>
And why not. The fact that you are too ignorant to >>>>>>>>>>>>>>>>>>>>>> see how to do that, as you are thinking the only >>>>>>>>>>>>>>>>>>>>>> "reference" can be to "self", doesn't make you >>>>>>>>>>>>>>>>>>>>>> argument correct.
in (2) the attempt to compute an inverse diagonal >>>>>>>>>>>>>>>>>>>>>>> with RTMs just fails for reasons u'd only >>>>>>>>>>>>>>>>>>>>>>> understand by working thru the algo urself (p7 of >>>>>>>>>>>>>>>>>>>>>>> re: turing's diagonals)
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>>>>>>>>>
is just not sufficient evidence that such β is >>>>>>>>>>>>>>>>>>>>>>> actually computable given the direct diagonal φn() >>>>>>>>>>>>>>>>>>>>>>
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED >>>>>>>>>>>>>>>>>>>>> WITH SELF- REFERENCE, LIKE THE PARADOX FOUND IN >>>>>>>>>>>>>>>>>>>>> TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H >>>>>>>>>>>>>>>>>>>>> ALLOWS FOR INVERSE- H TO BE COMPUTED >>>>>>>>>>>>>>>>>>>>>
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, >>>>>>>>>>>>>>>>>>>>> NOT THAT IT WOULDN'T BE USED
no idea why ur claiming thatHe doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>>>>>>>>>
one cannot just assume that because the >>>>>>>>>>>>>>>>>>>>>>>>> diagonal across computable numbers is >>>>>>>>>>>>>>>>>>>>>>>>> computable, therefore the anti- diagonal across >>>>>>>>>>>>>>>>>>>>>>>>> computable numbers is computable... >>>>>>>>>>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, >>>>>>>>>>>>>>>>>>>>>>>> and without the enumeration, you can't compute >>>>>>>>>>>>>>>>>>>>>>>> either of them.
But your method still doesn't let you compute >>>>>>>>>>>>>>>>>>>>>>>> the enumeration, and thus you can't actually >>>>>>>>>>>>>>>>>>>>>>>> compute the diagonal.
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>>>>>>>>>> computation across the computable numbers can >>>>>>>>>>>>>>>>>>>>>>>>> be used to compute the inverse diagonal >>>>>>>>>>>>>>>>>>>>>>>>
Remember, the problem definitions requires that >>>>>>>>>>>>>>>>>>>>>>>> the listing be a COMPLETE listing of the >>>>>>>>>>>>>>>>>>>>>>>> computable numbers / machine that compute >>>>>>>>>>>>>>>>>>>>>>>> computable numbers, in some definite order. >>>>>>>>>>>>>>>>>>>>>>>>
If your enumeration isn't complete, your >>>>>>>>>>>>>>>>>>>>>>>> diagonal isn't correct.
Why?
so while i agree with turing that the anti- >>>>>>>>>>>>>>>>>>>>>>>>> diagonal is not computable, i don't agree that >>>>>>>>>>>>>>>>>>>>>>>>> the normal diagonal is not computable >>>>>>>>>>>>>>>>>>>>>>>>
How does D decide on the original H? >>>>>>>>>>>>>>>>>>>>>>>>
Your modified H still needs a correct D to >>>>>>>>>>>>>>>>>>>>>>>> decide on all the other machines, including his >>>>>>>>>>>>>>>>>>>>>>>> original H that doesn't use your "trick" >>>>>>>>>>>>>>>>>>>>>>>>
But it doesn't.
well, he didn't consider that perhaps the >>>>>>>>>>>>>>>>>>>>>>>>> proper algo for computing the diagonal can >>>>>>>>>>>>>>>>>>>>>>>>> avoid the paradox on itself ... >>>>>>>>>>>>>>>>>>>>>>>>
But instead, he can prove with a more obvious >>>>>>>>>>>>>>>>>>>>>>>>>> process, that the Decider "D" that could be >>>>>>>>>>>>>>>>>>>>>>>>>> used to effectively enumerate the sequence of >>>>>>>>>>>>>>>>>>>>>>>>>> machine that produce computable numbers can >>>>>>>>>>>>>>>>>>>>>>>>>> not esit.
Thus, he clearly knows the difference, but is >>>>>>>>>>>>>>>>>>>>>>>>>> pointing out that the attempt to compute the >>>>>>>>>>>>>>>>>>>>>>>>>> diagonal clearly reveals the issue with >>>>>>>>>>>>>>>>>>>>>>>>>> effectively enumerating the sequences. >>>>>>>>>>>>>>>>>>>>>>>>>
Your just don't understand that D just can't >>>>>>>>>>>>>>>>>>>>>>>> correctly decide on his given H. >>>>>>>>>>>>>>>>>>>>>>>
i clearly understand that D cannot decide >>>>>>>>>>>>>>>>>>>>>>> correctly on turing's H, because my response to >>>>>>>>>>>>>>>>>>>>>>> this is that D does not need to decide correctly >>>>>>>>>>>>>>>>>>>>>>> on H to compute a diagonal
Then how do you build YOUR H without that D? >>>>>>>>>>>>>>>>>>>>>
incredibly, ironically, AND not co-incidently: >>>>>>>>>>>>>>>>>>>> partial recognizer D *can* decide on the fixed H >>>>>>>>>>>>>>>>>>>> because the fixed H does not try to use any D on >>>>>>>>>>>>>>>>>>>> itself, so no self- referential paradox is possible >>>>>>>>>>>>>>>>>>>> in regards to it's own digit on the diagonal >>>>>>>>>>>>>>>>>>>>
partial_recognizer_D(fixed_H) -> TRUE
which is a decidably TRUE return value (therefore a >>>>>>>>>>>>>>>>>>>> dependable AND computable value)
the fixed H is /decidable input/ to partial >>>>>>>>>>>>>>>>>>>> recognizer D (used in fixed H, which does not call >>>>>>>>>>>>>>>>>>>> any D itself), and would not be filtered out by >>>>>>>>>>>>>>>>>>>> paradox detectors
the fact fixed H does not use any classifier D on >>>>>>>>>>>>>>>>>>>> itself, and instead returns a hard-coded value for >>>>>>>>>>>>>>>>>>>> it's own digit on the diagonal, is keystone in >>>>>>>>>>>>>>>>>>>> making it / decidable input/ to D
holy fuck rick idk what to tell u anymore, >>>>>>>>>>>>>>>>>>>>
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you >>>>>>>>>>>>>>>>>>> are just anouncing that you thing errors are ok. >>>>>>>>>>>>>>>>>>>
The problem is fixed_H still need to run D on >>>>>>>>>>>>>>>>>>> turing_H and it will fail on that.
bro what part of the non-existence of turing_D or >>>>>>>>>>>>>>>>>> turing_H do you not understand???
But if your fD exists, then Turing_H that uses your D >>>>>>>>>>>>>>>>> exists, and will give that D the problem, and fixed_H >>>>>>>>>>>>>>>>> will call your D on turing_H and it still needs to >>>>>>>>>>>>>>>>> answer about it.
now if ur trying to argue: what if we use >>>>>>>>>>>>>>>>>> partial_recognizer_D in a turing_H type machine? that >>>>>>>>>>>>>>>>>> is a fair point
If D errs on Turing_H and calls it circle_free, then >>>>>>>>>>>>>>>>>>> fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, >>>>>>>>>>>>>>>>>>> then your enumeration is missing the circle_free H >>>>>>>>>>>>>>>>>>> from its list, and thus doesn't make the right diagonal. >>>>>>>>>>>>>>>>>>
this would give us the machine (that actually does >>>>>>>>>>>>>>>>>> exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] //
written to F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // TRUE
= satisfactory
K += 1
digit = simulate_Kth_digit(N,K) >>>>>>>>>>>>>>>>>> output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE >>>>>>>>>>>>>>>>>>
because actual_turing_H is an /undecidable input/ to >>>>>>>>>>>>>>>>>> any D classifier, when actually run actual_turing_H >>>>>>>>>>>>>>>>>> will simply skip putting it's own digit in the >>>>>>>>>>>>>>>>>> computed sequence
But the problem is that for a given
partial_recognizer_D, the actual_turing_H isn't an >>>>>>>>>>>>>>>>> "undecidable input", as it has definite behavior, just >>>>>>>>>>>>>>>>> behavior that partial_recognizer_D gets wrong. >>>>>>>>>>>>>>>>
But, we aren't looking at "undecidable inputs", as these >>>>>>>>>>>>>>> inputs are not "undecidable", but by the time that they >>>>>>>>>>>>>>> have actually been made into an input, and thus first had >>>>>>>>>>>>>>> the program created, which required creating the instance >>>>>>>>>>>>>>> of the decider selctected, they have definite behavior >>>>>>>>>>>>>>> that other some decider can determine.
It is only that one selected candidate decider chosen to >>>>>>>>>>>>>>> build the input that is shown to get the wrong answer. >>>>>>>>>>>>>>>
undecidability only actually exists as a relationship >>>>>>>>>>>>>>>> between a particular machine and the particular >>>>>>>>>>>>>>>> classifier it creates a structural paradox for, >>>>>>>>>>>>>>>>
In other words, you don't understand what
"undecidability" actually is.
Undecidability, means that there does not exist ANY >>>>>>>>>>>>>>> decider that can get the right answer for all instances >>>>>>>>>>>>>>> of the problem, (or if the problem has been reduced to >>>>>>>>>>>>>>> just that one input, for that input).
*not* as a general inability to describe a *real* >>>>>>>>>>>>>>>> machine's behavior that u are trying to treat it as, >>>>>>>>>>>>>>>But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example,
But it DOES exist if you have a machine you claim to be a D. >>>>>>>>>>>>>
The one that calls your partial_recoginzer_D as its D. >>>>>>>>>>>>
what *actual* example _THAT CAN EXIST_ do u have??? >>>>>>>>>>>>>
> But THAT *IS* the meaning of "undecidability".
Undecidability as a property of a PROBLEM, not an "input" or >>>>>>>>>>> a machine, unless you are narrowing the problem to just being >>>>>>>>>>> about that one machine/input.
because it *only* ever *was* about problematic input to
*particular* classifiers...
Except that isn't what "Undecidability" is about. And you are >>>>>>>>> just proving you don't know what you are talking about.
If we use the correct term, what you are tying to call
"undeciability" is really just not being correct. The problem >>>>>>>>> is a "particualar classifier" has a fixed set of how it
classifies everything. And thus, nothing can be "paradoxical" >>>>>>>>> to it, as we never get to the other side,
RICK I'M NEVER GOING TO FIGURE YOU FOR BEING AN UNDYING ASSHOLE, >>>>>>>> EXCEPT MAYBE ONCE U FINALLY TAKE UR IGNORANCE TO THE GRAVE
Right, because logic that is based on error can't unddrstand
logic that requires correctness.
The other side you talk about is about if this decider was a >>>>>>>>> different decider, and some how magically the input was also a >>>>>>>>> different input converted by a method not "in system" but only >>>>>>>>> in a meta-system.
it's not my problem everyone went for over a century confusing >>>>>>>>>> *our* ability to prove outcomes, with what a /particular/ >>>>>>>>>> decider can output... ya'll got caught up in a rigor of what >>>>>>>>>> is fundamentally nonsense
Because what a particual decider does, if it isnt' right, is >>>>>>>>> just not interesting,
no one has demonstrated a *possible* machine, let alone a >>>>>>>>>> *real* machine, which we can't provably can't prove the
outcome for... such a proof makes the machine _NEITHER
POSSIBLE NOR REAL_ so it would be ludicrous to present such a >>>>>>>>>> proof to exist
Only sort of right. Yes, we can not produce a machine that we >>>>>>>>> know we can not possibly ever decide if it halts or not,
because such a machine must by definition be non-halting, since >>>>>>>>> ALL halting machines are provable halting by just running them. >>>>>>>>> Thus, to prove we can't know they halt, we need to prove they >>>>>>>>> don't halt, but then we know their behavior.
The fact we can't demonstrate and example of that class,
doesn't mean members of it can't exist.
FUCK UR DAMN TEAPOT RICK
In other words, you are admitting I am right and need to resort >>>>>>> to the fallacy of swearing to distract others from that basic fact. >>>>>>
would be a contradiction to even produce an example of...
In other words, you are just addmitting that you are too stupid to
understand the possiblity of something being unknowable.
all u have is an ill conceived notion based on the fallacy that
_one_ classifier's failure to classify, can be overgeneralized to a
_total_ inability to generally prove outcomes, ultimately founded on
the unproven thesis that all computability is encompassed by TM
computing
But it isn't just ONE classifier. He shows how to make an input that
defeats ANY GIVEN classifier.
ANY classifier tht tries to determine this classification, will be
defeated by the input built by that template on it.
Thus, ALL such classifiers will be wrong.
Note, the arguement is not actually dependent on these being Turing
Machines, but can be generalized to ANY method based on the general
concept of "computing".
i don't need to respond to that further
Sure you do, or you are just admitting that you don't have an answer.
nope 🤷
In other words, you are just admitting you don't know how logic works.
You are just admitting you are giving up on logic.
It seems you are incapable of thinking abstractly, of even being
able to conceive of something you didn't invent without an example.
But you are also incapable of thinking concreately, because you
don't understand the need to show that there is a reason you need
to be able to show that the machines you imagine are possible.
I will point out AGAIN, that these unknowable machines are not
needed to be understood to understand the uncomputable nature of
the problem (They just come out as an effect of that).
So? What that proves is that it is IMPOSSIBLE to actually
both turing_H and turing_D are /incomplete specifications/ and >>>>>>>>>> therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL MACHINES_ >>>>>>>>>
acheive the goal defined for turing_D, a machine that
determines if a given machine is circle-free.
they are just malformed /incomplete specifications/ and
_NOTHING MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond >>>>>>>>>> the recognition that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such >>>>>>>>> machine exists.
BECAUSE IT'S INCOMPLETE. AN /INCOMPLETE SPECIFICATION/ THAT DOES >>>>>>>> NOT HANDLE ALL INPUT SITUATIONS CANNOT EXIST
The Specification of the INTERFACE is complete, it fully defines >>>>>>> what the given machine must do.
imagine trying to argue a machine that can't exist, because it
didn't handle all input situations, as "complete"
In other words, you can't understand the difference between a
specification of an interface, and the specification of a machine.
IT DOESN'T "SPECIFY A MACHINE" BECAUSE IT'S _NOT_ A MACHINE
DESCRIPTION,
IT _SPECIFIES AN INTERFACE_, ONE THAT IS TOO INCOMPLETE TO BE
COMPUTED BY A MACHINE
So, neither is your "machine" an actual machine, as all YOU have is
an "interface" that determines your partial classification that isn't
actually implemented.
AN INTERFACE THAT CAN EXIST IS A FUCKTON MORE THAN WHAT YOU'VE GOT
But it CAN'T work, as there is no actual machine to implement it.
Just because you haven't seen the reason it doesn't exist, doesn't mean
that it can exist.
Prove me wrong, and show how you ARE going to make a machine satisfy the interface.
Your problem is you don't understand that you need to use FAcTS and not
just imaginary ideas.
Turing's H, CAN be built from ANY actual machine that we want to try
to claim might be a possible implementation of Turing's D. And all of
these WILL BE actual machines, and ALL of them are shown to not
produce the correct answer for Turing's H. Thus, there can not be any
machine that does what D is supposed to do.
On the other hand, YOU try to claim your fixed_H is an atual machine,
but it isn't until you actually define your partial_decider_D, which
you can't do, so your's isn't a machine either.
IT'S NOT A CONTRADICTION FOR THOSE MACHINES TO EXIST, EI: THE
SPECIFICATION IS NOT INCOMPLETE,
Yes it is, you just haven't found it.
UNLIKE THE INCOMPLETE SPECIFICATION YOU PARADE AROUND AS MEANINGFUL
But yours still don't exist.
To ask me to prove otherwise is to assert that Russel's teapot does exist.
And, it turns out that any machine that you want to try to claim is
an implementation of your partial decider will either accept a
machine that will hang your fixed_H, or will not accept any machine
for some computable number.
_TOTALLY UNPROVEN STATEMENT_
Prove by Rice.
Of cource, your problem is you assume that you can disprove a theorm by assuming the existance of a machine that doesn't actually exist.
We may not be able to directly prove this like Turing
WTF??? IF U CAN'T PROVE IT THEN STOP ASSERTING IT AS TRUE
But it *IS* proven by Rice's theorem.
Just like Turing's proof is correct, even if you can't understand the
full nature of the proof.
Did, but Rice's proof shows that it can't be done, and your arguement
RICE'S PROOF IS _BASED_ ON TURING'S PROOF, YOU CAN'T USE IT INSTEAD OF
TURING'S ... LOL
Wrong. There are NUMEROUS variations of the proof, and you can't
actually prove any of them actually incorrect, just that you don't understand some of the steps.
Of course, YOUR "proof" is based on groundless assertions, so is even
weaker than the proof you are trying to refute.
If you can assume you can make your decider, Turing can assert that two problems are the same.
The difference is that Turings claim can be proven, while yours can't be.
fking tard
It seeems you are just screwing yourself, as your only basis for
claiming them wrong is that you are smoking the magic fairy dust from
the Unicorn.
that we need to find it is just you exercising the fallacy of
asserting that Russel's Teapot must exist if we can't prove it doesn't.
By your own logic, YOUR machine is INCOMPLETE, as you can't give
the actual specific algorithm it uses. Your "logic" is based on
ASSUMING that something is computable, with absolutely ZERO
evidence that it is.
In other words, you live in a world of fantasy
The Specification of the interface is uncomputable, or
unrealizable, because we can't make a machine that fullfills it. >>>>>>>
Yes, the specification of the ALGORITHM is incomplete, as we
don't specify how to do all the steps, but then so is yours.
Turing, BECAUSE he used an unspecified algorithm for his D, was >>>>>>> able to actually PROVE that the specification for the interface >>>>>>> of D was unimplementable.
All your arguement has done is shown that if magic fairy dust
powered unicorns exists (which they don't), you could use one of >>>>>>> them to built an (also non-existant) program to compute something >>>>>>> that is actually proven to be uncomputable.
All you have done is proven you don't understand how logic works, >>>>>>> and thus your works SHOULD BE IGNORED.
If you want to be consistant, YOUR partial_decider_D is ALSO >>>>>>>>> incomplete,
IT'S NOT INCOMPLETE AS IT HANDLES ALL INPUT SITUATIONS
That isn't what "Incomplete", means, as your partial_decider_D
doesn't exist so it can't handle ANY input situations.
YOU HAVEN'T DEMONSTRATED A CONTRADICTION WITH
PARTIAL_RECOGNIZER_D, SO YOU HAVEN'T SHOWN IT TO NOT EXIST...
In other words, you assert the existance of Russel's teapot.
The "contradiction" is in the fact that you concept of "undecidable >>>>> input" just is nonsense.
u know exactly what i'm talking about when i say /undecidable
input/, u just refuse to acknowledge anything i say as correct.
because as soon as that damn breaks, ur whole position gets flooded
with truth
Right, I know you mean an input that the given decider is just wrong
about.
LOL, BEING IN SPEC IT NOT _WRONG_
NOt when the spec itself is a nonsense lie based on the misuse of words.
Since a machine exists that can decide on that input, it is NOT "undecidable".
The word that means that answer you machine gave doesn't satisy the BASE property you are talking about is that it was just wrong.
"Can't" is not proper grammer for the behavior of a given machine, it
just doesn't, as the only thing that a given machine CAN do is what it
does, so there is no possibility for it to do otherwise.
Any other definition is just a lie.
LOL, BEING IN SPEC IS NOT A _LIE_
If the spec is a lie it is.
There is no such thing as an "undecidable input"
Just wrong machines.
And stupid dart's
How is it different then inputs that the specific decider is just
wrong able.
Describe how you define that property based on JUST the specific
input itself and the specific decider.
Your world is based on variable constants and similar nonsense.
You don't get it, that if you want to put down Turing's machines >>>>>>> for not handling cases, you need to also accept that yours
doesn't for the same reason, a machine that fully meets your
specifaction can't exist.
And part of the reason for yours, is that you have a nonsense
term in your definition, as there is no such thing as a
"undecidable input" if some other decider can get the known right >>>>>>> answer.
Part of your problem is you just don't understand what a PROGRAM >>>>>>> or ALGORITHM actually is
as you haven't specified HOW it does what it does, just lie >>>>>>>>> turing didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such >>>>>>>>> machine could exist, which he succeeded at, but YOUR claim is >>>>>>>>> that such a thing might be able to exist, which you just have >>>>>>>>> not proven.
And, it is an assertion that NO decider can correctly compute >>>>>>>>>>> the answer to that problem for all the possible inputs.
Since a given ACTUAL machine described as your turing_H, >>>>>>>>>>> which means you have chosen a particular (broken)
implementation of D (since no actually correct D exists), >>>>>>>>>>> like your partial_recognizer_D, has fulled defined behavior, >>>>>>>>>>> and a decider can exist that determines it, there is nothing >>>>>>>>>>> "undecidable" about that input.
It is just that the D that H was built on is wrong. Nothing >>>>>>>>>>> "paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on >>>>>>>>>> you just _IGNORING_ the specification i specifically created >>>>>>>>>> for partial_recognizer_D...
No, your problem is you just are too stupid to understand that >>>>>>>>> your "specification" is just nonsense, as it assumes that you >>>>>>>>> can perform a test by "changing" the decider and the input to >>>>>>>>> see a results.
But the specification is about a particular input and decider. >>>>>>>>>
the specification for partial_recognizer_D is when
encountering undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your
definition is just internally inconsistantly trying to define >>>>>>>>> input that the decider gets wrong. Since the class the partial >>>>>>>>> decider is
this _DOES NOT MEAN_ we cannot know what the /undecidable >>>>>>>>>> input/ does, as a paradox in regards to partial_recognize_D >>>>>>>>>> _ONLY AFFECTS_ partial_recognize_D not _OUR ABILITY TO PROVE >>>>>>>>>> OUTCOMES_
In other words, you admit that you words are just lies and
don't mean what they say.
Your "undecidable" isn't about decidability, but correctness of >>>>>>>>> a given decider.
YES, THAT'S ALL UNDECIDABILITY EVER WAS
In other words you are admitting you don't know what you are
talking about.
A given decider being incorrect doesn't mean a correct decider
doesn't exist. There are many decidable problems, for which
incorrect deciders can exist.
Undecidability is about the being NO correct deciders.
YOU HAVEN'T EVEN SHOW A MACHINE WHICH HAS NO CORRECT CLASSIFIERS
FOR...
Because I don't need to.
YOU'VE ONLY SHOWN THAT FOR ANY CLASSIFIER THERE EXISTS A MACHINE IT
CAN'T CLASSIFY...
NOT THAT THERE EXISTS A MACHINE THAT CAN'T BE CLASSIFIED...
But I don't need to.
The uncomputability of that classification is proven by the first.
ALL THAT'S BEEN SHOWN IS _ONE_ MACHINE CAN'T CLASSIFY ALL MACHINEs.
And that the arguement can apply to ANY machine, since there was no restriction on what could be viewed as D
IT DOES NOT THE SHOW THERE IS MACHINE THAT *CANNOT* BE CLASSIFIED BY
*ANY* MACHINE
Which it doesn't need to do.
All you are doiing is proving you really don't know what you are talking about.
THOSE ARE _DIFFERENT_ CLAIMS, REGARDLESS OF U JUST CONFLATING THEM
BECAUSE UR GETTING DESPERATE
And who is conflaiting them?
Yes, I pointed out as an aside, that there exists machines that no know
to be always correct partial decider can decide on, but that is NOT what Turing is proving here, and thus claiming he doesn't succeed at that
just shows you don't know what you are talking about.
The existance of the second class of machine is just something more,
that is clearly beyond your understanding.
All that is needed to prove a problem is uncomputable is to show
that you can make an input for any given decider that it is wrong
about.
THAT ONLY SHOWS THAT THE PROBLEM CAN'T BE COMPUTED BY A _SINGLE_
MACHINE
Right, or any effectively enumerable set of machine, since we can
then build the single machine that tests all of them.
And that *IS* the definition of computable,
BECAUSE UR JUST ASSUMING THE CT-THESIS AS TRUE, WHICH IS NOT PROVEN
Nope. It seems you are just to stupid to understand the logic.
Part of the problem is you just don't understand what the words mean.
You don't seem to understand the problem.
YOU DON'T UNDERSTAND WHAT'S BEEN PROVEN
Sure I do, you just admitted what I have said is proven, that the
LOL, UR CLAIMS ARE GETTING MORE GARBAGE BY THE POST
Nope, you are just showing how stupid you are, because you don't know
what the words you are trying to use mean.
problem is not computable, which means there does not exist a single
machine that correctly computes all the answers.
This means that the trivial always false machine meets your
NO, THE FAILURE TO CLASSIFY MUST COME FROM THE STRUCTURE OF THE >>>>>>>> INPUT, NOT AN INCORRECT ALGO OF THE CLASSIFIER, WHY AM I
REPEATING ME ON THIS?
In other words, you think people/machines can't make mistakes?
BEING UNABLE TO CLASSIFY /UNDECIDABLE INPUT/ IS _NOT_ A MISTAKE ON >>>>>> BEHALF OF THE CLASSIFIER
But the input isn't "undecidable", as that isn't an actual property >>>>> of an input.
IT IS _CAUSED_ *BY THE STRUCTURE OF THE INPUT*
No, as the input is fully decidable by some other decider, and thus
there is nothing "undecidable" about it.
RIGHT, THAT'S BECAUSE UNDECIDABILITY IS ONLY DEFINED AS A RELATIONSHIP
BETWEEN A MACHINE AND A PARTICULAR CLASSIFIER PARADOXES BY THAT MACHINE
Nope.
You are sure stupid about the meaning of the words.
"Undecidability" is the property of a Problem / Classification rule to
be properl carried out by ANY possible algorithm.
THERE _NEVER_ HAS BEEN _ANY_ OTHER FORM OF PROVEN UNDECIDABILITY
WITHIN COMPUTING,
And where did you get your definition?
Maybe you should read some of the elementary statements about it, like
at https://en.wikipedia.org/wiki/Undecidable_problem
"An undecidable problem is a decision problem for which it is proved to
be impossible to construct an algorithm that always leads to a correct yes-or-no answer. "
Note, no mention about the relationship to a machine (except that such a machine doesn't exist).
AND UR DOING FUCK ALL TO PROVE ANOTHER FORM EXISTS
Since your definition is just your own lies, YOU are the one making
stuff up.
It is just that the method will always find a flaw in the algorithm
of that specific decider, making it just wrong.
Your logic is just based on category errors, based on your own
stupidity.
Then why are you worried about proofs of correctness.
You don't seem to understand that the whole basis of these proofs >>>>>>> isn't about finding *A* input that nothing can decide on, but
finding a pattern that makes an input for a given decider that it >>>>>>> won't be able to handle.
That pattern isn't actually something that necessarily remains in >>>>>>> the machine after construction, but direction to find a given
algorithms fatal flaw, showing that ALL algorithms for that
problem must have a fatal flaw which can be different for every one. >>>>>>>
requirements. Any input for which the correct answer for the >>>>>>>>> base problem would be true, THAT particular decider won't get >>>>>>>>> right, so it is correct in your expanded partial sense to
return false.
IT seesm you just can't read.
(you don't have one btw, so please do let me witness ur cope) >>>>>>>>>>>>>
It is ONLY "incomplete", because it is built on a >>>>>>>>>>>>>>> PRESUMED defined machine. If the decider "D" actually >>>>>>>>>>>>>>> exists, then H is FULLY defined.
see, turing_H's runtime is not describable because it's >>>>>>>>>>>>>>>> an / incomplete specification/ of machine and therefor >>>>>>>>>>>>>>>> _does not even exist_ as a real TM, not because of the >>>>>>>>>>>>>>>> hypothesized undecidability in relation to some D >>>>>>>>>>>>>>>
turing_D _DOES NOT EXIST_ as it is *also* an /incomplete >>>>>>>>>>>>>> specification/, as it doesn't handle idiosyncrasies of TM >>>>>>>>>>>>>> computing, specifically self- referential set-
classification paradoxes
It doesn't need to. It just needs to be built on what you >>>>>>>>>>>>> want to claim to be your answer for the concept of D. >>>>>>>>>>>>>
As soon as you define what machine you are going to use as >>>>>>>>>>>>> your equivalent for D, like your partial_recognizer_D, then >>>>>>>>>>>>> the template turing_H can be converted into an actual machine. >>>>>>>>>>>>>
actual_turing_H, on the other hand, actually does exist, >>>>>>>>>>>>>>>> so we can know/ prove what it does. it, however, still >>>>>>>>>>>>>>>> forms a structural paradox in regards to
partial_recognizer_D which will fail to classify it as >>>>>>>>>>>>>>>> circle-free
But, since you admit to actual_turing_H existing, that >>>>>>>>>>>>>>> means that
since partial_recognizer_D _CAN_ exist, then
actual_turing_H _CAN_ exist
And so can the turing_H that uses partial_recognizer_D as >>>>>>>>>>>>> its decider D.
The steps the algorithm does is what defines the machine, >>>>>>>>>>>>> not the "names" of the things it "calls", as those names >>>>>>>>>>>>> don't actually exist in the final building of the machine, >>>>>>>>>>>>> which is the problem with your concept, as that is needed >>>>>>>>>>>>> to determine your "paradoxical" property.
the names are just shortcuts for the literal values rick, >>>>>>>>>>>> don't be daft
Right, but the literal values don't have the names on them, >>>>>>>>>>> and we don't have borders within the machine letting us know >>>>>>>>>>> we moved from "outer code of H" into "the code of the D it >>>>>>>>>>> uses", especially with Turing Machines were such usages are >>>>>>>>>>> by necessity "expanded" in-line.
please do actually read turing's paper sometime, he does
organize machine descriptions into functional groups
Yes, at meta level.
Thus, the "copy" of D that was pulled into H isn't actually >>>>>>>>>>> the same
no it can be literally a copy of same description than is the >>>>>>>>>> run by a UTM (universal turing machine) functional group of >>>>>>>>>> the machine
It CAN be, but doesn't NEED to be. And in fact, his H isn't >>>>>>>>> described that way.
thing as the independent machine D that we can talk about, >>>>>>>>>>> just a functional equivalent which doesn't even have its own >>>>>>>>>>> D.N (since that is a property only of full machines)
ur really pulling at straws here rick
You really think so?
It seems your problem is you need to work in a tight cage where >>>>>>>>> other ideas are not allowed that show the holes in your ideas. >>>>>>>>>
But your computation wasn't actually a diagonal of
pratial_recognizer_D has an actual and fixed algorithm. >>>>>>>>>>>>>>>
But the problem isn't in it handling the machine fixed_H, >>>>>>>>>>>>>>> but how it ever handles the results generated by >>>>>>>>>>>>>>> actual_turing_H, which by your assumption has become a >>>>>>>>>>>>>>> cycle-free machine and ALL equivalent versions of it have >>>>>>>>>>>>>>> been partially declared to be non- cycle free.
Thus, your basic criteria is a category error. >>>>>>>>>>>>>>>>>
If a D can determine which machines its basic algorithm >>>>>>>>>>>>>>>>> will get wrong, then it can just return the different >>>>>>>>>>>>>>>>> answer, and then THAT machine it will get right. The >>>>>>>>>>>>>>>>> problem is it creates a DIFFERENT input, that uses this >>>>>>>>>>>>>>>>> new version of the machine, that it will get wrong. >>>>>>>>>>>>>>>>>
WHICH IS FINE, actuallyBut not as actual_turing_H, which doesn't have your "fix". >>>>>>>>>>>>>>>>>
actual_turing_H ultimately computes the *same* >>>>>>>>>>>>>>>>>> sequence as fixed_H, which in turn _does_ have a spot >>>>>>>>>>>>>>>>>> on the diagonal. so there is no actual need to include >>>>>>>>>>>>>>>>>> a digit from actual_turing_H on the diagonal, as the >>>>>>>>>>>>>>>>>> number it computes is already included on the diagonal >>>>>>>>>>>>>>>>>
And thus if your partial_recoginzer_D accepts ANY >>>>>>>>>>>>>>>>> verision of actual_turing_H, your fixed_H will get stuck. >>>>>>>>>>>>>>>>>
So, now, you need to show how your partial_recognizer_D >>>>>>>>>>>>>>>>> can possible filter all ALL of that infinite set of >>>>>>>>>>>>>>>>> machines, but still accept some machine that computes >>>>>>>>>>>>>>>>> that particular number that it would compute with this >>>>>>>>>>>>>>>>> supposedly correct partial_recognizer_D.
You can't ask someone to disprove it, as that is just >>>>>>>>>>>>>>>>> admitting that you accept that Russel's Teapot must exist. >>>>>>>>>>>>>>>>>
So you think, because you don't see it is just built on >>>>>>>>>>>>>>>>> the fairy dust, of assuming that your partial decider >>>>>>>>>>>>>>>>> that meets ALL your needs exists,
goddamn rick, this rock just keeps getting more solid >>>>>>>>>>>>>>>>>
🤷🤷🤷
to clarify this further: the only algorithmic difference >>>>>>>>>>>>>>>> between fixed_H and actual_turing_H is the way they >>>>>>>>>>>>>>>> respectively handle their own self- references. they >>>>>>>>>>>>>>>> function identically when handling all other machines >>>>>>>>>>>>>>>
This means that your enumeration of machines you are >>>>>>>>>>>>>>> looking at is, in fact, incomplete, thus the diagonal you >>>>>>>>>>>>>>> computed is wrong.
But actual_turing_H is actually circle_free, and thus its >>>>>>>>>>>>>>> results SHOULD have been on the diagonal.
fixed_H = () -> {
N = 1
K = 0
output = [] //
written to F- squares
do {
if (N == DN(fixed_H) {
K += 1
output.push(0) //
hard coded digit 0
} elif (partial_recognizer_D(N) == TRUE) { // >>>>>>>>>>>>>>>> TRUE = satisfactory
K += 1
output.push(simulate_Kth_digit(N,K)) >>>>>>>>>>>>>>>> }
N += 1
}
}
when fixed_H iterates across DN(fixed_H) itself, it will >>>>>>>>>>>>>>>> skip trying to simulate itself and put it's hard coded >>>>>>>>>>>>>>>> digit on the diagonal, which for fixed_H is 0
when fixed_H iterates across DN(actual_turing_H), it >>>>>>>>>>>>>>>> will skip trying simulate actual_turing_H because >>>>>>>>>>>>>>>> partial_recognizer_D fails to classify it as circle-free >>>>>>>>>>>>>>>
it computes _THE SAME NUMBER_ as fixed_H, so if fixed_H is >>>>>>>>>>>>>> put on the diagonal, then actual_turing_H _DOES NOT_ need >>>>>>>>>>>>>> to be in _THAT_ total enumeration of _COMPUTABLE NUMBERS_ >>>>>>>>>>>>>
computable numbers, but of machines that made computable >>>>>>>>>>>>> numbers.
same thing moron
Nope, I guess you are just admitting that your logic depends >>>>>>>>>>> on being sloppy.
the only way to compute a diagonal across computable numbers >>>>>>>>>> is simulating each number by the machine that computes it.
Not the ONLY way, but a good way to try.
But since Turing Shows this can't be done, it can't be done any >>>>>>>>> way.
oh wow GREAT CATCH RICK!!!!,
Note, Turing SPECIFICALLY pointed out that he was going to >>>>>>>>>>> the related problem, the problem of:
/but the problem of enumerating computable sequences is >>>>>>>>>>> equivalent to the problem of finding out whether a given >>>>>>>>>>> number is the D.N of a circle- free machine, and we have no >>>>>>>>>>> general process for doing this in a finite number of steps/ >>>>>>>>>>
_THAT'S ANOTHER ERROR IN HIS PROOF_
*WE CAN ENUMERATE OUT ALL COMPUTATIONAL SEQUENCES, WITHOUT >>>>>>>>>> ENUMERATING OUT ALL CIRCLE_FREE MACHINES*,
Show it.
what??? bro ur the one who informed me proper enumerations can >>>>>>>> only have *one* instance of each element enumerated,
and we all know that the total enumeration of machines has
infinite machines producing each computable numbers...
so therefore not only can the enumeration of computable numbers >>>>>>>> be done without enumerating all circle-free machines, it *MUST* >>>>>>>> be done without enumerating all circle-free machines, as we can >>>>>>>> output *ONLY* one machine for each computable number
that should suffice
In other words, you are admitting you can't do it.
no richard, we've reached a point where u are unable to correctly >>>>>> and honestly identify a self-evident concept:
FOR _ANY_ GIVEN COMPUTABLE NUMBER, THERE ARE _INFINITE_ MACHINES
WHICH COMPUTE IT ...
I fully understand that.
WE ONLY NEED TO CLASSIFY _ONE_ OF THOSE MACHINES TO ADD THAT GIVEN >>>>>> COMPUTABLE NUMBER THE LIST, THE REST NOT ONLY CAN, BUT MUST, BE
IGNORED
Right, and you need to PROVE that you can.
IT'S ENOUGH TO REFUTE A *KEYSTONE* POINT IN TURING'S PAPER
Nope, claiming that there might exist a machine, that doesn't
actually exist, doesn't prove anything.
WHAT??? TURING CONFLATING THE ENUMERATION OF ALL CIRCLE-FREE MACHINES
WITH THAT OF JUST COMPUTABLE NUMBERS IS JUST WRONG, REGARDLESS OF WHAT
IS FOUND TO BE POSSIBLE AFTER
Can you show that they are not related?
Or, is this just another "fact" out of your imagination.
fking tard LOL
THE REST OF THE PAPER IS PRETTY MUCH VOID DUE TO TURING CONFLATING
THE ENUMERATION OF COMPUTABLE NUMBERS WITH THAT OF MACHINES THAT
COMPUTE COMPUTATIONAL NUMBERS
No, he shows that there does exist a problem that a decider can not
decide on. That shows that the Hilbert Entscheidungsproblem can not
have a solution, which is his real goal.
Note, on page 246 he sketches out the proof, noting that *IF* the
computable numbers WERE effective enumerable (what he calls just
enumerable), then we could compute that anti-diagonal by thus doing
that enumeration and selecting the appropriate digits.
_THAT IS ALSO A FALLACY_
THE IDIOSYNCRASIES OF TM COMPUTING DO NOT ALLOW THE USE OF A DIAGONAL
TO THEN COMPUTE AN ANTI-DIAGONAL
Sure it does.
Say you have a machine that prints out the diagonal of the enumeration.
Take that machine, and just invert each write to one of the unchangeable cells, and reverse the operation of any read from an unchangable cell.
Such a machine will trace the EXACT same path of states as the original,
and output the anti-diagonal instead of the diagonal.
If not, where does it go wrong?
YOU'RE NOT GOING TO UNDERSTAND WHY UNTIL YOU WORK THRU THE FIX I MADE
TO MAKE THE DIAGONAL COMPUTABLE, AND WORK THRU WHY THAT FIX DOES NOT
WORK FOR THE ANTI-DIAGONAL COMPUTATION _EVEN HAVE A DIAGONAL AT HAND_
So, what is wrong with what I said?
The "anti-diagonal" program checks for the number of the diagonal
program, but if you can make the self-referential program, you can make
the other-referential program just like it.
He also points out that this gets to the feeling that "there must be
something wrong" (even though the logic is perfectly correct).>>
THEREFOR, WE DO _NOT_ NEED A CLASSIFIER WITH THE POWER TO COMPUTE >>>>>> WHETHER _ANY_ GIVEN MACHINE IS A COMPUTABLE NUMBER, WE ONLY NEED A >>>>>> CLASSIFIER WITH ENOUGH POWER TO CLASSIFIER _ONE OF THE INFINITE_
MACHINES FOR _ANY_ GIVEN COMPUTABLE NUMBER
Right, you need to PROVE that you can build a classifier that will
select at least one machine for every computable number.
You can't just assume your decider exists, you need to show HOW it
works.
Since you concept is based on identifing the "undeciable inputs",
you need to show how you actually will do it, which means you need
to define it in a way that is detectable.
i already demonstrated the basic idea several times
No, you haven't, as your "demonstartion" always violates the
defintion of a machine and an input.
How is your definition anything different that an input that this
machine gets wrong.
After all, when you talk about changing the decider, that doesn't
actualy change the input, since that changed decider is NOT the
original decider, and thus the actual input doesn't change.
actually building a partial_recognizer for all possible computable
numbers is a research question far outside the scope of this
discussion since it involves open questions in number theory we
don't even have answers to
No, it is a fundamental assumption of your claim, that is in direct
opposition to the proofs.
Your arguement is just that you must be allowed to just assume
Russel's teapot exist without proof.
There is no inherent reason you can pull an infinite subset out >>>>>>> of a similarly infinite set.
We can enumerate the primes, even though for every prime there is >>>>>>> an infinite number of composite numbers that are multiples of it, >>>>>>> thus your logic is based on an error.
THe fact that he quote an equivalence that he doesn't prove he >>>>>>>>> doesn't make his proof incorrect.
SO THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_ >>>>>>>>>> EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN >>>>>>>>>> NUMBER IS A CIRCLE-FREE MACHINE,
Prove it.
there's nothing to prove, the statement is self-evident given >>>>>>>> the previous two paragraphs i wrote. we don't need to be able to >>>>>>>> output it for *ANY* given machine, we only need to be able to >>>>>>>> output that fact for *ONE* machine of the infinite which
produces *ANY* given compute number
honestly i'm not sure why it's take until year 2026 for someone >>>>>>>> to point this out
In other words, you are just admitting that you understanding of >>>>>>> the problem isn't sufficent to deal with it.
no richard, we've reached a point where u are unable to correctly >>>>>> and honestly identify a self-evident concept:
THE PROBLEM OF ENUMERATING COMPUTATIONAL SEQUENCES IS _NOT_
EQUIVALENT TO THE PROBLEM OF FINDING OUT WHETHER _ANY_ GIVEN
NUMBER IS A CIRCLE-FREE MACHINE, AS WE DO NOT NEED TO IDENTIFY
_ALL_ CIRCLE- FREE MACHINES, JUST _ONE FOR EACH_ COMPUTABLE NUMBER
So, you need to show that you CAN do that.
Turing points out that they are computationally related, and I will
being related does _not_ make them the _same problem_,
Right, and he doesn't say they are. he says they are an "equivalent"
problem, i.e there is a strong relation to each other (so they either
both have or don't have an answer), that he has presumably proven
somewhere else.
Calling it an error without proof just shows your stupidity.
FOR SOMEONE WHO A) UNDERSTANDS THE BASIC CONCEPTS INVOLVED, AND B) CAN
THINK CRITICALLY,
THE PROOF IS LITERALLY THE TWO PARAGRAPHS I ALREADY WROTE
Nope, it assume the existance of something that isn't actualy proven to exist, and turns out can't be.
You seem to fail at basic knowledge of how logic works, which explains
why you logic is just so stupid.
and turing's proof *depends* on them being the _same problem_
No, since his ultimate goal is the proof is about the Hilbert
Entscheidungsproblem, which was the big question of the day.
IF HE'S ACTUALLY FAILED TO PROVE THE COMPUTABLE SEQUENCES AS NOT
ENUMERABLE THEN THE PROOF FALLS APART, I'M SORRY
He proved the circle-free machies are not enumerable, which is at least
a similar problem, which also proves his point.
To show is proof is wrong, *YOU* need to show that they problems are
NOT equivalent.
I ALREADY DID, SO YET AGAIN:
ENUMERATING THE COMPUTABLE SEQUENCES ONLY REQUIRES THE LIMITED POWER
TO IDENTIFY _ONE_ CIRCLE-FREE MACHINE FOR EACH COMPUTABLE SEQUENCE,
_NOT_ ALL CIRCLE-FREE MACHINES
Which you can't prove you can do.
> THAT'S LITERALLY THE PROOF RIGHT THERE
>
> #god
So, how do you prove that your partial decider does that?
Your "proof" is based on basic fallacies, showing you fail to understand basic logic.
beleive him more that you how claims you can do one without showing >>>>> how.this is suspicion that is easily dealt with using search tools. but
You have ADMITTED your ignorance of the field, having looked at
only a very few of the papers, I suspect there was something discussed >>>>
instead of backing up that suspicion ur just continuing to be
suspect, and i'm never gunna buy it
Really? and how much of the early papers have actually been preserved
and put into something searchable?
How much of the actual presentations of the various conferences have
been transcribed and put into search engines?
The fact that he makes the passing reference to it, and no one in
that error jumped on that is a good sign that it was accepted.
As I said, prove him wrong or that the conclusion is wrong by
ACTUALLY showing how to do what he claims can't be done. And them
needs more than assuming you can make a decider that does the real work. >>>
That is you just living off unicorn milk.
somewhere about this, even if it wasn't as famous of a paper (as it >>>>> wasn't so eathshaking by itself).
Your problem is you are working on "gut instinct" with a gut that >>>>>>> doesn't understand what it is talking about, because it thinks
unicorns exist.
we only need to compute that fact for *ONE* machine for any >>>>>>>>>> given computable sequence, _SO A PARTIAL RECOGNIZER IS
SUFFICIENT FOR THE JOB_
So do it.
you ask too much for not recognizing anything i said as true
no richard, we've reached a point where u are unable to correctly >>>>>> and honestly identify a self-evident concept
Because is is just wrong, being based on category errors.
As long as you just ASSUME you can do what you claim, you are just
making a ASS out of yourself, proving that your "self-evident" is
really a case o "self-delusion"
Because it isn't, and you have no proofs, because you can't
actually prove a false statement.
Part of your problem is you don't actually know much about what >>>>>>> you are talking, so you have nothing to work with,
it's goal was a generate a diagonal across all circle-free
fuck rick ur so bad at defending this it's actually really >>>>>>>>>> useful to me!
(there's *another* specific sentence i have a problem with, >>>>>>>>>> where's a ben when u need him?)
All you have shown so far is that you are too stupid to
understand the nature of the paper.
Note, it was NOT intended to be the end-all self-contained
regerous proof of the concept. It was a paper, leaning on
previous work done, and with the intent of more work to follow >>>>>>>>> to show the basic idea of why something was true.
Thus, not showing how a step works, isn't an error.
Now, to counter his claims, because they are so well accepted, >>>>>>>>> you are going to need to be rigorous and not based an the
fallacy of assuming something and showing that it doesn't lead >>>>>>>>> to a contradiction when you follow a restricted line of
reasoining.
So, HIS enumeration is the complete enumeration of "circle- >>>>>>>>>>> free machines", which your proof admittedly doesn't do.
lol, turing_H does nothing to de-duplicate the enumeration >>>>>>>>>> either, why am i the only one subject to criticism here??? >>>>>>>>>>>
Also, your proof is built on the existance of a Russel >>>>>>>>>>>>> Teapot, you have
the *specification* /works/, which shows that it's *not* >>>>>>>>>>>> contradiction for it to be computable
Nope, as it is based on the fallacy of assuming the
conclusion. Your arguement boils down to showing that *IF* >>>>>>>>>>> you can correctly determine all machines that fit in a not- >>>>>>>>>>> definable category (since you concept of [paradoxical can't >>>>>>>>>>> actually be defined at the machine level, only at how a >>>>>>>>>>> machine might have been created at a meta-level.
just ASSUMED you cam make this partial_recoginzser_D that >>>>>>>>>>>>> can make the precise "mistakes" you need it to. (They are >>>>>>>>>>>>> mistakes because the input
"mistake" is this case is a fallacious non-techincal value >>>>>>>>>>>> judgement on the return value,
No, since your criteria that you want to use is just an >>>>>>>>>>> undefinable term for a machine description.
There is nothing "paradoxical" of the actual machine when >>>>>>>>>>> created, as it then has fixed and determined behavior that >>>>>>>>>>> can be decided on. The thing that is paradoxical is the mete- >>>>>>>>>>> system pattern used to generate the class of machines as pair >>>>>>>>>>> with a given decider.
cause they are rather entirely within specification, as >>>>>>>>>>>> FALSE was specified to be used upon encountering an / >>>>>>>>>>>> undecidable input/, something turing_D did not handle >>>>>>>>>>>> correctly (making it _NOT EXIST_)
But the input is NOT "undecidable" as the actual input is >>>>>>>>>>> based on a actual D that does make a specific decision to the >>>>>>>>>>> input. And it is just that for any given version of D, we can >>>>>>>>>>> find a DIFFERENT H that it will get wrong (the H that was >>>>>>>>>>> just built on it).
Do you want to claim that ALL these H are "undecidable" even >>>>>>>>>>> though for
no... fixed_H contains no paradox, and is therefor not /
undecidable input/ to any classifier
all of them there does exist a decider that gets the right >>>>>>>>>>> answer?
many machines contain no paradox and are therefore
classifiable by all classifiers
But this proof isn't about the many, but the ALL.
A term it seems you don't understand.
And it isn't the job of others to show you what one you missed, >>>>>>>>> it is yours to prove that they don't, as YOU are making the claim. >>>>>>>>>
then the H's as of now are i suppose "generators" instead of >>>>>>>>>> enumerators
It seems you just don't know what the word means, and thus >>>>>>>>>>> you specification is actually just non-sense.
that come from Turing_H are, in fact, circle_free because >>>>>>>>>>>>> of this turing_H is built on this partial_recognizdr_D >>>>>>>>>>>>> which causes them to not hang up on themselves.
Your "definition" of paradoxical just doesn't apply to >>>>>>>>>>>>> machines, as it is based on the incorrect definition that >>>>>>>>>>>>> assumes a machine can be something other than itself, and >>>>>>>>>>>>> when it changes, it changes other things that at the meta/ >>>>>>>>>>>>> template level refered to its old self.
not sure what ur problem is
That you just don't know what your words mean, and you just >>>>>>>>>>> refuse to learn.
Try to actually fully DEFINE the word based on the actual >>>>>>>>>>> properties of the input machine, and not refering to other >>>>>>>>>>> machines that are not part of the input (but would be created >>>>>>>>>>> as alternate inputs to foil an alternate decider).
when actual_turing_H iterates across DN(actual_turing_H) >>>>>>>>>>>>>>>> itself, it will similarly skip trying to simulate >>>>>>>>>>>>>>>> itself, because partial_recognizer_D fails to classify >>>>>>>>>>>>>>>> it as circle- free
Right, so it ALSO is wrong, as it didn't compute the >>>>>>>>>>>>>>> diagonal of the circle-free machines.
fixed_H is computing a diagonal across *all* _COMPUTABLE >>>>>>>>>>>>>> NUMBERS_, which is the _ACTUAL PROBLEM_ turing was dealing >>>>>>>>>>>>>> with,
No, it isn't, as your "set of computable numbers" has >>>>>>>>>>>>> duplicates.
it's still a diagonal across all computerate numbers even if >>>>>>>>>>>> it has dpulicates ...
No, because an enumeration of a set doesn't HAVE duplicates. >>>>>>>>>>
and turing was wrong about that too, but i don't see u
criticizing him now eh???
H is a generator, its goal was to generate a particual number. >>>>>>>>
machines,
which as you said: is *NOT* an enumeration, and an enumeration >>>>>>>> required duplicating the machines
Right, but the result it is generating is BASED (and defined) on >>>>>>> a given enumeration which it computes internally).
Your Fixed_H isn't computing anything with a defined
specification, as your "enumeration" that you claim to be based >>>>>>> on, isn't actually an enumeration that meets your claimed
requirement.
D is a decider, that allowed H, in its generation, to produce >>>>>>>>> (internally) the enumeration that its results were defined by, >>>>>>>>>
It seems you don't understand that to reach your goal,
sometimes you have to acheive a sub-goal that is quite different. >>>>>>>>>
On 3/9/26 7:15 PM, Richard Damon wrote:
On 3/9/26 9:02 PM, dart200 wrote:
On 3/9/26 5:03 PM, Richard Damon wrote:
On 3/9/26 12:32 AM, dart200 wrote:
On 3/8/26 5:14 PM, Richard Damon wrote:
On 3/8/26 11:27 AM, dart200 wrote:
On 3/8/26 5:20 AM, Richard Damon wrote:
On 3/7/26 12:47 PM, dart200 wrote:i don't have a response anymore to the level of stupidity
On 3/7/26 5:52 AM, Richard Damon wrote:
On 3/7/26 2:16 AM, dart200 wrote:
On 3/6/26 12:10 PM, Richard Damon wrote:
On 3/6/26 12:26 PM, dart200 wrote:
On 3/6/26 3:52 AM, Richard Damon wrote:Undecidability as a property of a PROBLEM, not an "input" or >>>>>>>>>>>> a machine, unless you are narrowing the problem to just >>>>>>>>>>>> being about that one machine/input.
On 3/5/26 11:06 PM, dart200 wrote:except we can describe the behavior of actual_turing_H and >>>>>>>>>>>>> prove it's equivalent to fixed_H, so therefore it doesn't >>>>>>>>>>>>> fit what u stated was
On 3/5/26 6:27 PM, Richard Damon wrote:But it DOES exist if you have a machine you claim to be a D. >>>>>>>>>>>>>>
On 3/5/26 1:57 PM, dart200 wrote:
On 3/5/26 4:28 AM, Richard Damon wrote:
On 3/5/26 12:17 AM, dart200 wrote:this is unfortunately a fallacious understanding of >>>>>>>>>>>>>>>>> undecidability. just because a machine is /undecidable >>>>>>>>>>>>>>>>> input/ to a particular classifier does mean we cannot >>>>>>>>>>>>>>>>> then prove and know ourselves what the machine actually >>>>>>>>>>>>>>>>> does,
On 3/4/26 4:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/4/26 2:40 AM, dart200 wrote:
On 3/3/26 8:42 PM, dart200 wrote:
On 3/3/26 4:49 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/3/26 3:55 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 6:24 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 8:24 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 2:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 12:38 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/28/26 5:21 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 6:09 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/27/26 2:51 AM, Tristan Wibberley wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 24/02/2026 21:30, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/24/26 11:38 AM, Tristan Wibberley >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
wow, u know up until now, i thot i fully agreed >>>>>>>>>>>>>>>>>>>>>>>> with turing's short diagonal proof, but in >>>>>>>>>>>>>>>>>>>>>>>> writing this post i now find myself in a subtle, >>>>>>>>>>>>>>>>>>>>>>>> yet entirely critical disagreement: >>>>>>>>>>>>>>>>>>>>>>>>it is wrong,Then you zre just showing your stupidity, >>>>>>>>>>>>>>>>>>>>>>>>>>> because YOU can't tell the difference. >>>>>>>>>>>>>>>>>>>>>>>>>>>i don't see any indication that turing >>>>>>>>>>>>>>>>>>>>>>>>>>>> realized a difference there >>>>>>>>>>>>>>>>>>>>>>>>>>>And focusing on what is said to be >>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossible and not changing the problem is >>>>>>>>>>>>>>>>>>>>>>>>>>>>> also important.the theory of computation is the theory of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> such procedures, and understanding the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> diagonal procedure is critical to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understanding the *base* contradiction/ >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> paradox that the rest of his support for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> godel's result is then built on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>On 22/02/2026 21:08, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/26 12:49 PM, Chris M. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Thomasson wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 2/22/2026 9:04 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ...
Eh?!yes that's what diagonal proofs do... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>an effective enumeration of all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing machines was proven on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> turing's original paper and can be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> reused anywhere... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>You think you can test all of them >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> one by one? Don't tell me you think >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A test is a procedure! You can't test / >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> all/ of an infinitude one by one. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
that exactly what turing does in his >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof: he defines a comptuation >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that enumerates out all the numbers, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> testing each one of they represent >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a "satisfactory"/"circle-free" machine, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and adding that to diagonal >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> across defined across computable numbers >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
it really would be a great exercise to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> carefully read p247 of turing's >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof and produce the psuedo-code for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the machine H, assuming that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine D exists >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I'll get to it sooner then, because it's >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mad. Are you sure he didn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> reason quantified over all but phrase it >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> like a procedure for what he >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The problem with the diagonal generation >>>>>>>>>>>>>>>>>>>>>>>>>>>>> isn't the generation of the diagonal >>>>>>>>>>>>>>>>>>>>>>>>>>>>> itself, but effectively enumerating the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumeration in the first place. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
After all, on page 246 he says: >>>>>>>>>>>>>>>>>>>>>>>>>>>
The computable sequences are therefore not >>>>>>>>>>>>>>>>>>>>>>>>>>> enumerable.
Here is is SPECIFICALLY talking about the >>>>>>>>>>>>>>>>>>>>>>>>>>> effective enumeration of the computable >>>>>>>>>>>>>>>>>>>>>>>>>>> sequences.
He then points out that he can directly show >>>>>>>>>>>>>>>>>>>>>>>>>>> that the "anti- diagonal" of the (non- >>>>>>>>>>>>>>>>>>>>>>>>>>> effectively computed) enumeration can't be >>>>>>>>>>>>>>>>>>>>>>>>>>> computed but that "This proof, although >>>>>>>>>>>>>>>>>>>>>>>>>>> perfectly sound, has the disadvantage that it >>>>>>>>>>>>>>>>>>>>>>>>>>> may leave the reader with a feeling that >>>>>>>>>>>>>>>>>>>>>>>>>>> 'there must be something wrong'". >>>>>>>>>>>>>>>>>>>>>>>>>>
No, YOU are wrong, as you don't understand what >>>>>>>>>>>>>>>>>>>>>>>>> is being done.
I think he is refering he to the standard anti- >>>>>>>>>>>>>>>>>>>>>>>>> diagonal arguement, which shows that since for >>>>>>>>>>>>>>>>>>>>>>>>> all n, position n differs from the value in >>>>>>>>>>>>>>>>>>>>>>>>> number n, there can not be any element that >>>>>>>>>>>>>>>>>>>>>>>>> matches the anti- diagonal.
It is just a natural fact of countable >>>>>>>>>>>>>>>>>>>>>>>>> infinity, something it seems you just don't >>>>>>>>>>>>>>>>>>>>>>>>> understand.
Show how that is actually wrong. >>>>>>>>>>>>>>>>>>>>>>>>
/let an be the n-th computable sequence, and let >>>>>>>>>>>>>>>>>>>>>>>> φn(m) be the m- th figure in an. Let β be the >>>>>>>>>>>>>>>>>>>>>>>> sequence with 1- φn(m) as its n-th. figure. >>>>>>>>>>>>>>>>>>>>>>>> Since β is computable, there exists a number K >>>>>>>>>>>>>>>>>>>>>>>> [== β] such that 1- φn(n) = φK(n) for all n. >>>>>>>>>>>>>>>>>>>>>>>> Putting n = K, we have 1 = 2φK(K), i.e. 1 is >>>>>>>>>>>>>>>>>>>>>>>> even. This is impossible/
i agree with this proof is far as much as if β was >>>>>>>>>>>>>>>>>>>>>> computable (by TMs) we'd have a problem for sure, >>>>>>>>>>>>>>>>>>>>>>
but what i don't agree is that we can just assume >>>>>>>>>>>>>>>>>>>>>> the computability of β from the existence of >>>>>>>>>>>>>>>>>>>>>> φn(m), THAT'S THE FALLACy
But you can, *IF* you have a machine that >>>>>>>>>>>>>>>>>>>>>>> computes the diagonal, then just change all the >>>>>>>>>>>>>>>>>>>>>>> write to the output to write the opposite. Note, >>>>>>>>>>>>>>>>>>>>>>> the "self- reference" that you are thinking of >>>>>>>>>>>>>>>>>>>>>>> stops being a "self- reference" but is a >>>>>>>>>>>>>>>>>>>>>>> reference to the original write the diagonal code. >>>>>>>>>>>>>>>>>>>>>>>
the fallacy here is assuming that because the >>>>>>>>>>>>>>>>>>>>>>>> direct diagonal is computable, that one can >>>>>>>>>>>>>>>>>>>>>>>> therefore compute the anti- diagonal using the >>>>>>>>>>>>>>>>>>>>>>>> direct diagonal. the abstract definition makes >>>>>>>>>>>>>>>>>>>>>>>> it look simple, but this ignores the >>>>>>>>>>>>>>>>>>>>>>>> complexities of self- referential analysis (like >>>>>>>>>>>>>>>>>>>>>>>> what turing details on the next page) >>>>>>>>>>>>>>>>>>>>>>>
Nope, because your filtering TM (or RTM) will >>>>>>>>>>>>>>>>>>>>>>> still fail when it gets to the number of TURING'S >>>>>>>>>>>>>>>>>>>>>>> H, as there is no correct answer for the machine >>>>>>>>>>>>>>>>>>>>>>> built by that template.
in both methods i have for rectifying the >>>>>>>>>>>>>>>>>>>>>>>> paradox found in the direct diagonal (either (1) >>>>>>>>>>>>>>>>>>>>>>>> filtering TMs or (2) using RTMs), neither can be >>>>>>>>>>>>>>>>>>>>>>>> used to then compute the anti-diagonal >>>>>>>>>>>>>>>>>>>>>>>
TURING'S HYPOTHETICAL H DOESN'T EXIST, IT'S AN >>>>>>>>>>>>>>>>>>>>>> INCOMPLETE SPECIFICATION
(is that where H comes from??? _H_ypothetical???) >>>>>>>>>>>>>>>>>>>>>>
THE ANTI-DIAGONAL IS *NOT* A COMPUTABLE NUMBER BY >>>>>>>>>>>>>>>>>>>>>> TMs, SO IT *SHOULD* BE FILTERED OUT >>>>>>>>>>>>>>>>>>>>>>
in (1) the algo to compute an inverse diagonal >>>>>>>>>>>>>>>>>>>>>>>> is filtered out like turing's paradoxical >>>>>>>>>>>>>>>>>>>>>>>> variation of the direct diagonal would be, and >>>>>>>>>>>>>>>>>>>>>>>> there is no analogous non- paradoxical variation >>>>>>>>>>>>>>>>>>>>>>>> that has a hard coded value that is inverse to >>>>>>>>>>>>>>>>>>>>>>>> what it does return ... such a concept is >>>>>>>>>>>>>>>>>>>>>>>> entirely nonsensical. a function can only return >>>>>>>>>>>>>>>>>>>>>>>> what it does, it can't also return the inverse >>>>>>>>>>>>>>>>>>>>>>>> to what it returns eh???
But, if D filters it out, then it becomes circle- >>>>>>>>>>>>>>>>>>>>>>> free, and thus your enumeration is incomlete. >>>>>>>>>>>>>>>>>>>>>>
And why not. The fact that you are too ignorant >>>>>>>>>>>>>>>>>>>>>>> to see how to do that, as you are thinking the >>>>>>>>>>>>>>>>>>>>>>> only "reference" can be to "self", doesn't make >>>>>>>>>>>>>>>>>>>>>>> you argument correct.
in (2) the attempt to compute an inverse >>>>>>>>>>>>>>>>>>>>>>>> diagonal with RTMs just fails for reasons u'd >>>>>>>>>>>>>>>>>>>>>>>> only understand by working thru the algo urself >>>>>>>>>>>>>>>>>>>>>>>> (p7 of re: turing's diagonals) >>>>>>>>>>>>>>>>>>>>>>>>
the premise:
/Let β be the sequence with 1-φn(m) as its n-th/ >>>>>>>>>>>>>>>>>>>>>>>>
is just not sufficient evidence that such β is >>>>>>>>>>>>>>>>>>>>>>>> actually computable given the direct diagonal φn() >>>>>>>>>>>>>>>>>>>>>>>
BECAUSE IT DOESN'T CAPTURE THE NUANCES INVOLVED >>>>>>>>>>>>>>>>>>>>>> WITH SELF- REFERENCE, LIKE THE PARADOX FOUND IN >>>>>>>>>>>>>>>>>>>>>> TURING'S HYPOTHETICAL H
NEITHER METHOD I'VE HYPOTHESIZED ABOUT AN EXTANT H >>>>>>>>>>>>>>>>>>>>>> ALLOWS FOR INVERSE- H TO BE COMPUTED >>>>>>>>>>>>>>>>>>>>>>
I SAID I D DIDN'T NEED TO DECIDE ON THE FIXED H, >>>>>>>>>>>>>>>>>>>>>> NOT THAT IT WOULDN'T BE USED
no idea why ur claiming thatHe doesn't. You are just showing your stupidity, >>>>>>>>>>>>>>>>>>>>>>>>
one cannot just assume that because the >>>>>>>>>>>>>>>>>>>>>>>>>> diagonal across computable numbers is >>>>>>>>>>>>>>>>>>>>>>>>>> computable, therefore the anti- diagonal >>>>>>>>>>>>>>>>>>>>>>>>>> across computable numbers is computable... >>>>>>>>>>>>>>>>>>>>>>>>>
He is proving the enumeration is uncomputable, >>>>>>>>>>>>>>>>>>>>>>>>> and without the enumeration, you can't compute >>>>>>>>>>>>>>>>>>>>>>>>> either of them.
But your method still doesn't let you compute >>>>>>>>>>>>>>>>>>>>>>>>> the enumeration, and thus you can't actually >>>>>>>>>>>>>>>>>>>>>>>>> compute the diagonal.
neither method i have for fixing the diagonal >>>>>>>>>>>>>>>>>>>>>>>>>> computation across the computable numbers can >>>>>>>>>>>>>>>>>>>>>>>>>> be used to compute the inverse diagonal >>>>>>>>>>>>>>>>>>>>>>>>>
Remember, the problem definitions requires that >>>>>>>>>>>>>>>>>>>>>>>>> the listing be a COMPLETE listing of the >>>>>>>>>>>>>>>>>>>>>>>>> computable numbers / machine that compute >>>>>>>>>>>>>>>>>>>>>>>>> computable numbers, in some definite order. >>>>>>>>>>>>>>>>>>>>>>>>>
If your enumeration isn't complete, your >>>>>>>>>>>>>>>>>>>>>>>>> diagonal isn't correct.
Why?
so while i agree with turing that the anti- >>>>>>>>>>>>>>>>>>>>>>>>>> diagonal is not computable, i don't agree that >>>>>>>>>>>>>>>>>>>>>>>>>> the normal diagonal is not computable >>>>>>>>>>>>>>>>>>>>>>>>>
How does D decide on the original H? >>>>>>>>>>>>>>>>>>>>>>>>>
Your modified H still needs a correct D to >>>>>>>>>>>>>>>>>>>>>>>>> decide on all the other machines, including his >>>>>>>>>>>>>>>>>>>>>>>>> original H that doesn't use your "trick" >>>>>>>>>>>>>>>>>>>>>>>>>
But it doesn't.
well, he didn't consider that perhaps the >>>>>>>>>>>>>>>>>>>>>>>>>> proper algo for computing the diagonal can >>>>>>>>>>>>>>>>>>>>>>>>>> avoid the paradox on itself ... >>>>>>>>>>>>>>>>>>>>>>>>>
But instead, he can prove with a more obvious >>>>>>>>>>>>>>>>>>>>>>>>>>> process, that the Decider "D" that could be >>>>>>>>>>>>>>>>>>>>>>>>>>> used to effectively enumerate the sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>> machine that produce computable numbers can >>>>>>>>>>>>>>>>>>>>>>>>>>> not esit.
Thus, he clearly knows the difference, but is >>>>>>>>>>>>>>>>>>>>>>>>>>> pointing out that the attempt to compute the >>>>>>>>>>>>>>>>>>>>>>>>>>> diagonal clearly reveals the issue with >>>>>>>>>>>>>>>>>>>>>>>>>>> effectively enumerating the sequences. >>>>>>>>>>>>>>>>>>>>>>>>>>
Your just don't understand that D just can't >>>>>>>>>>>>>>>>>>>>>>>>> correctly decide on his given H. >>>>>>>>>>>>>>>>>>>>>>>>
i clearly understand that D cannot decide >>>>>>>>>>>>>>>>>>>>>>>> correctly on turing's H, because my response to >>>>>>>>>>>>>>>>>>>>>>>> this is that D does not need to decide correctly >>>>>>>>>>>>>>>>>>>>>>>> on H to compute a diagonal
Then how do you build YOUR H without that D? >>>>>>>>>>>>>>>>>>>>>>
incredibly, ironically, AND not co-incidently: >>>>>>>>>>>>>>>>>>>>> partial recognizer D *can* decide on the fixed H >>>>>>>>>>>>>>>>>>>>> because the fixed H does not try to use any D on >>>>>>>>>>>>>>>>>>>>> itself, so no self- referential paradox is possible >>>>>>>>>>>>>>>>>>>>> in regards to it's own digit on the diagonal >>>>>>>>>>>>>>>>>>>>>
partial_recognizer_D(fixed_H) -> TRUE >>>>>>>>>>>>>>>>>>>>>
which is a decidably TRUE return value (therefore a >>>>>>>>>>>>>>>>>>>>> dependable AND computable value)
the fixed H is /decidable input/ to partial >>>>>>>>>>>>>>>>>>>>> recognizer D (used in fixed H, which does not call >>>>>>>>>>>>>>>>>>>>> any D itself), and would not be filtered out by >>>>>>>>>>>>>>>>>>>>> paradox detectors
the fact fixed H does not use any classifier D on >>>>>>>>>>>>>>>>>>>>> itself, and instead returns a hard-coded value for >>>>>>>>>>>>>>>>>>>>> it's own digit on the diagonal, is keystone in >>>>>>>>>>>>>>>>>>>>> making it / decidable input/ to D
holy fuck rick idk what to tell u anymore, >>>>>>>>>>>>>>>>>>>>>
that shit is rock solid theory.
🤷🤷🤷
And your "enumeration" isn't complete, and thus you >>>>>>>>>>>>>>>>>>>> are just anouncing that you thing errors are ok. >>>>>>>>>>>>>>>>>>>>
The problem is fixed_H still need to run D on >>>>>>>>>>>>>>>>>>>> turing_H and it will fail on that.
bro what part of the non-existence of turing_D or >>>>>>>>>>>>>>>>>>> turing_H do you not understand???
But if your fD exists, then Turing_H that uses your D >>>>>>>>>>>>>>>>>> exists, and will give that D the problem, and fixed_H >>>>>>>>>>>>>>>>>> will call your D on turing_H and it still needs to >>>>>>>>>>>>>>>>>> answer about it.
If D errs on Turing_H and calls it circle_free, then >>>>>>>>>>>>>>>>>>>> fixed_H gets hung up.
If D errs on Turing_H and calls it not circle_free, >>>>>>>>>>>>>>>>>>>> then your enumeration is missing the circle_free H >>>>>>>>>>>>>>>>>>>> from its list, and thus doesn't make the right >>>>>>>>>>>>>>>>>>>> diagonal.
now if ur trying to argue: what if we use >>>>>>>>>>>>>>>>>>> partial_recognizer_D in a turing_H type machine? that >>>>>>>>>>>>>>>>>>> is a fair point
this would give us the machine (that actually does >>>>>>>>>>>>>>>>>>> exist):
actual_turing_H = () -> {
N = 1
K = 0
output = [] //
written to F- squares
do {
if (partial_recognizer_D(N) == TRUE) { // >>>>>>>>>>>>>>>>>>> TRUE = satisfactory
K += 1
digit = simulate_Kth_digit(N,K) >>>>>>>>>>>>>>>>>>> output.push(digit)
}
N += 1
}
}
well since
partial_recognizer_D(actual_turing_H) -> FALSE >>>>>>>>>>>>>>>>>>>
because actual_turing_H is an /undecidable input/ to >>>>>>>>>>>>>>>>>>> any D classifier, when actually run actual_turing_H >>>>>>>>>>>>>>>>>>> will simply skip putting it's own digit in the >>>>>>>>>>>>>>>>>>> computed sequence
But the problem is that for a given
partial_recognizer_D, the actual_turing_H isn't an >>>>>>>>>>>>>>>>>> "undecidable input", as it has definite behavior, just >>>>>>>>>>>>>>>>>> behavior that partial_recognizer_D gets wrong. >>>>>>>>>>>>>>>>>
But, we aren't looking at "undecidable inputs", as these >>>>>>>>>>>>>>>> inputs are not "undecidable", but by the time that they >>>>>>>>>>>>>>>> have actually been made into an input, and thus first >>>>>>>>>>>>>>>> had the program created, which required creating the >>>>>>>>>>>>>>>> instance of the decider selctected, they have definite >>>>>>>>>>>>>>>> behavior that other some decider can determine. >>>>>>>>>>>>>>>>
It is only that one selected candidate decider chosen to >>>>>>>>>>>>>>>> build the input that is shown to get the wrong answer. >>>>>>>>>>>>>>>>
undecidability only actually exists as a relationship >>>>>>>>>>>>>>>>> between a particular machine and the particular >>>>>>>>>>>>>>>>> classifier it creates a structural paradox for, >>>>>>>>>>>>>>>>>
In other words, you don't understand what
"undecidability" actually is.
Undecidability, means that there does not exist ANY >>>>>>>>>>>>>>>> decider that can get the right answer for all instances >>>>>>>>>>>>>>>> of the problem, (or if the problem has been reduced to >>>>>>>>>>>>>>>> just that one input, for that input).
*not* as a general inability to describe a *real* >>>>>>>>>>>>>>>>> machine's behavior that u are trying to treat it as, >>>>>>>>>>>>>>>>But THAT *IS* the meaning of "undecidability".
well u haven't demonstrated such a machine
turing_H _DOES NOT EXIST_ so it's not an example, >>>>>>>>>>>>>>
The one that calls your partial_recoginzer_D as its D. >>>>>>>>>>>>>
what *actual* example _THAT CAN EXIST_ do u have??? >>>>>>>>>>>>>>
> But THAT *IS* the meaning of "undecidability". >>>>>>>>>>>>
because it *only* ever *was* about problematic input to >>>>>>>>>>> *particular* classifiers...
Except that isn't what "Undecidability" is about. And you are >>>>>>>>>> just proving you don't know what you are talking about.
If we use the correct term, what you are tying to call
"undeciability" is really just not being correct. The problem >>>>>>>>>> is a "particualar classifier" has a fixed set of how it
classifies everything. And thus, nothing can be "paradoxical" >>>>>>>>>> to it, as we never get to the other side,
RICK I'M NEVER GOING TO FIGURE YOU FOR BEING AN UNDYING
ASSHOLE, EXCEPT MAYBE ONCE U FINALLY TAKE UR IGNORANCE TO THE >>>>>>>>> GRAVE
Right, because logic that is based on error can't unddrstand
logic that requires correctness.
The other side you talk about is about if this decider was a >>>>>>>>>> different decider, and some how magically the input was also a >>>>>>>>>> different input converted by a method not "in system" but only >>>>>>>>>> in a meta-system.
it's not my problem everyone went for over a century
confusing *our* ability to prove outcomes, with what a / >>>>>>>>>>> particular/ decider can output... ya'll got caught up in a >>>>>>>>>>> rigor of what is fundamentally nonsense
Because what a particual decider does, if it isnt' right, is >>>>>>>>>> just not interesting,
no one has demonstrated a *possible* machine, let alone a >>>>>>>>>>> *real* machine, which we can't provably can't prove the >>>>>>>>>>> outcome for... such a proof makes the machine _NEITHER
POSSIBLE NOR REAL_ so it would be ludicrous to present such a >>>>>>>>>>> proof to exist
Only sort of right. Yes, we can not produce a machine that we >>>>>>>>>> know we can not possibly ever decide if it halts or not,
because such a machine must by definition be non-halting, >>>>>>>>>> since ALL halting machines are provable halting by just
running them. Thus, to prove we can't know they halt, we need >>>>>>>>>> to prove they don't halt, but then we know their behavior. >>>>>>>>>>
The fact we can't demonstrate and example of that class,
doesn't mean members of it can't exist.
FUCK UR DAMN TEAPOT RICK
In other words, you are admitting I am right and need to resort >>>>>>>> to the fallacy of swearing to distract others from that basic fact. >>>>>>>
required to keep bleating on and on about being so certain of
machines it would be a contradiction to even produce an example >>>>>>> of...
In other words, you are just addmitting that you are too stupid to >>>>>> understand the possiblity of something being unknowable.
all u have is an ill conceived notion based on the fallacy that
_one_ classifier's failure to classify, can be overgeneralized to a >>>>> _total_ inability to generally prove outcomes, ultimately founded
on the unproven thesis that all computability is encompassed by TM
computing
But it isn't just ONE classifier. He shows how to make an input that
defeats ANY GIVEN classifier.
ANY classifier tht tries to determine this classification, will be
defeated by the input built by that template on it.
Thus, ALL such classifiers will be wrong.
Note, the arguement is not actually dependent on these being Turing
Machines, but can be generalized to ANY method based on the general
concept of "computing".
i don't need to respond to that further
Sure you do, or you are just admitting that you don't have an answer.
nope 🤷
In other words, you are just admitting you don't know how logic works.
you are just admitting you don't know how gaslighting works...
You are just admitting you are giving up on logic.
It seems you are incapable of thinking abstractly, of even being
able to conceive of something you didn't invent without an example. >>>>>>
But you are also incapable of thinking concreately, because you
don't understand the need to show that there is a reason you need >>>>>> to be able to show that the machines you imagine are possible.
I will point out AGAIN, that these unknowable machines are not
needed to be understood to understand the uncomputable nature of
the problem (They just come out as an effect of that).
both turing_H and turing_D are /incomplete specifications/ >>>>>>>>>>> and therefore _AREN'T POSSIBLE MACHINES_ let alone _REAL >>>>>>>>>>> MACHINES_
So? What that proves is that it is IMPOSSIBLE to actually >>>>>>>>>> acheive the goal defined for turing_D, a machine that
determines if a given machine is circle-free.
they are just malformed /incomplete specifications/ and >>>>>>>>>>> _NOTHING MORE_, and _DESERVE NO FURTHER CONSIDERATION_ beyond >>>>>>>>>>> the recognition that such an interface _IS INCOMPLETE_
No, it prove that the interface is uncomputable, and no such >>>>>>>>>> machine exists.
BECAUSE IT'S INCOMPLETE. AN /INCOMPLETE SPECIFICATION/ THAT >>>>>>>>> DOES NOT HANDLE ALL INPUT SITUATIONS CANNOT EXIST
The Specification of the INTERFACE is complete, it fully defines >>>>>>>> what the given machine must do.
imagine trying to argue a machine that can't exist, because it
didn't handle all input situations, as "complete"
In other words, you can't understand the difference between a
specification of an interface, and the specification of a machine.
IT DOESN'T "SPECIFY A MACHINE" BECAUSE IT'S _NOT_ A MACHINE
DESCRIPTION,
IT _SPECIFIES AN INTERFACE_, ONE THAT IS TOO INCOMPLETE TO BE
COMPUTED BY A MACHINE
So, neither is your "machine" an actual machine, as all YOU have is
an "interface" that determines your partial classification that
isn't actually implemented.
AN INTERFACE THAT CAN EXIST IS A FUCKTON MORE THAN WHAT YOU'VE GOT
But it CAN'T work, as there is no actual machine to implement it.
WHAT EVIDENCE DO YOU HAVE THAT IT _CAN'T_ WORK???
Just because you haven't seen the reason it doesn't exist, doesn't
mean that it can exist.
Prove me wrong, and show how you ARE going to make a machine satisfy
the interface.
LOL WAT? FIRST YOU OUTRIGHT CLAIM IT _CAN'T_ WORK, AND NEXT YOUR ASKING
ME TO SHOW _HOW_ IT WORKS?
THAT'S NOT A SINCERE POSITION BRO
Your problem is you don't understand that you need to use FAcTS and
not just imaginary ideas.
Turing's H, CAN be built from ANY actual machine that we want to try
to claim might be a possible implementation of Turing's D. And all
of these WILL BE actual machines, and ALL of them are shown to not
produce the correct answer for Turing's H. Thus, there can not be
any machine that does what D is supposed to do.
On the other hand, YOU try to claim your fixed_H is an atual
machine, but it isn't until you actually define your
partial_decider_D, which you can't do, so your's isn't a machine
either.
IT'S NOT A CONTRADICTION FOR THOSE MACHINES TO EXIST, EI: THE
SPECIFICATION IS NOT INCOMPLETE,
Yes it is, you just haven't found it.
WHAT????
HOW COULD YOU KNOW I HAVEN'T FOUND IT YET, IF YOU HAVEN'T FOUND IT YET????
UNLIKE THE INCOMPLETE SPECIFICATION YOU PARADE AROUND AS MEANINGFUL
But yours still don't exist.
To ask me to prove otherwise is to assert that Russel's teapot does
exist.
DIDN'T REALIZED ASKING FOR PROOF OF YOUR CLAIMS IS NOW RUSSEL'S TEAPOT
THOT WE WERE DOING MATH HERE WERE ONLY PROOFS MATTER
And, it turns out that any machine that you want to try to claim is
an implementation of your partial decider will either accept a
machine that will hang your fixed_H, or will not accept any machine
for some computable number.
_TOTALLY UNPROVEN STATEMENT_
Prove by Rice.
RICE DID NOT EVEN REMOTELY ADDRESS THE CLASSIFICATION POWER OF PARTIAL RECOGNIZERS...
U THINK U CAN JUST SPOUT RANDOM REFERENCE???
Of cource, your problem is you assume that you can disprove a theorm
by assuming the existance of a machine that doesn't actually exist.
We may not be able to directly prove this like Turing
WTF??? IF U CAN'T PROVE IT THEN STOP ASSERTING IT AS TRUE
But it *IS* proven by Rice's theorem.
Just like Turing's proof is correct, even if you can't understand the
full nature of the proof.
LOL BRO OK, NOW UR JUST MAKING RANDOM CLAIMS AT ME
Did, but Rice's proof shows that it can't be done, and your arguement
RICE'S PROOF IS _BASED_ ON TURING'S PROOF, YOU CAN'T USE IT INSTEAD
OF TURING'S ... LOL
Wrong. There are NUMEROUS variations of the proof, and you can't
actually prove any of them actually incorrect, just that you don't
understand some of the steps.
Of course, YOUR "proof" is based on groundless assertions, so is even
weaker than the proof you are trying to refute.
If you can assume you can make your decider, Turing can assert that
two problems are the same.
... except they still aren't ...
you can cope about it all you want, but it ain't never gunna be true.
enumeration computable numbers does not require enumerating all circle-
free machines, anyone who can't accept that is corrupt
The difference is that Turings claim can be proven, while yours can't be.
fking tard
It seeems you are just screwing yourself, as your only basis for
claiming them wrong is that you are smoking the magic fairy dust from
the Unicorn.
that we need to find it is just you exercising the fallacy of
asserting that Russel's Teapot must exist if we can't prove it doesn't. >>>>
By your own logic, YOUR machine is INCOMPLETE, as you can't give
the actual specific algorithm it uses. Your "logic" is based on
ASSUMING that something is computable, with absolutely ZERO
evidence that it is.
In other words, you live in a world of fantasy
The Specification of the interface is uncomputable, or
unrealizable, because we can't make a machine that fullfills it. >>>>>>>>
Yes, the specification of the ALGORITHM is incomplete, as we
don't specify how to do all the steps, but then so is yours.
Turing, BECAUSE he used an unspecified algorithm for his D, was >>>>>>>> able to actually PROVE that the specification for the interface >>>>>>>> of D was unimplementable.
All your arguement has done is shown that if magic fairy dust >>>>>>>> powered unicorns exists (which they don't), you could use one of >>>>>>>> them to built an (also non-existant) program to compute
something that is actually proven to be uncomputable.
All you have done is proven you don't understand how logic
works, and thus your works SHOULD BE IGNORED.
If you want to be consistant, YOUR partial_decider_D is ALSO >>>>>>>>>> incomplete,
IT'S NOT INCOMPLETE AS IT HANDLES ALL INPUT SITUATIONS
That isn't what "Incomplete", means, as your partial_decider_D >>>>>>>> doesn't exist so it can't handle ANY input situations.
YOU HAVEN'T DEMONSTRATED A CONTRADICTION WITH
PARTIAL_RECOGNIZER_D, SO YOU HAVEN'T SHOWN IT TO NOT EXIST...
In other words, you assert the existance of Russel's teapot.
The "contradiction" is in the fact that you concept of
"undecidable input" just is nonsense.
u know exactly what i'm talking about when i say /undecidable
input/, u just refuse to acknowledge anything i say as correct.
because as soon as that damn breaks, ur whole position gets flooded >>>>> with truth
Right, I know you mean an input that the given decider is just wrong
about.
LOL, BEING IN SPEC IT NOT _WRONG_
NOt when the spec itself is a nonsense lie based on the misuse of words.
ok bro
Since a machine exists that can decide on that input, it is NOT
"undecidable".
The word that means that answer you machine gave doesn't satisy the
BASE property you are talking about is that it was just wrong.
"Can't" is not proper grammer for the behavior of a given machine, it
just doesn't, as the only thing that a given machine CAN do is what it
does, so there is no possibility for it to do otherwise.
Any other definition is just a lie.
LOL, BEING IN SPEC IS NOT A _LIE_
If the spec is a lie it is.
There is no such thing as an "undecidable input"
ok bro
Just wrong machines.
And stupid dart's
How is it different then inputs that the specific decider is just >>>>>> wrong able.
Describe how you define that property based on JUST the specific
input itself and the specific decider.
Your world is based on variable constants and similar nonsense.
You don't get it, that if you want to put down Turing's machines >>>>>>>> for not handling cases, you need to also accept that yours
doesn't for the same reason, a machine that fully meets your
specifaction can't exist.
And part of the reason for yours, is that you have a nonsense >>>>>>>> term in your definition, as there is no such thing as a
"undecidable input" if some other decider can get the known
right answer.
Part of your problem is you just don't understand what a PROGRAM >>>>>>>> or ALGORITHM actually is
as you haven't specified HOW it does what it does, just lie >>>>>>>>>> turing didn't define how his D was to do its work.
Your problem is that Turing's goal was to prove that no such >>>>>>>>>> machine could exist, which he succeeded at, but YOUR claim is >>>>>>>>>> that such a thing might be able to exist, which you just have >>>>>>>>>> not proven.
And, it is an assertion that NO decider can correctly >>>>>>>>>>>> compute the answer to that problem for all the possible inputs. >>>>>>>>>>>>
Since a given ACTUAL machine described as your turing_H, >>>>>>>>>>>> which means you have chosen a particular (broken)
implementation of D (since no actually correct D exists), >>>>>>>>>>>> like your partial_recognizer_D, has fulled defined behavior, >>>>>>>>>>>> and a decider can exist that determines it, there is nothing >>>>>>>>>>>> "undecidable" about that input.
It is just that the D that H was built on is wrong. Nothing >>>>>>>>>>>> "paradoxical" about a machine being wrong.
again, wrong is a _NON-TECHNICAL VALUE ASSESSMENT_ based on >>>>>>>>>>> you just _IGNORING_ the specification i specifically created >>>>>>>>>>> for partial_recognizer_D...
No, your problem is you just are too stupid to understand that >>>>>>>>>> your "specification" is just nonsense, as it assumes that you >>>>>>>>>> can perform a test by "changing" the decider and the input to >>>>>>>>>> see a results.
But the specification is about a particular input and decider. >>>>>>>>>>
the specification for partial_recognizer_D is when
encountering undecidable input is to utilize FALSE,
But there is no such thing as an "undecidable input", your >>>>>>>>>> definition is just internally inconsistantly trying to define >>>>>>>>>> input that the decider gets wrong. Since the class the partial >>>>>>>>>> decider is
this _DOES NOT MEAN_ we cannot know what the /undecidable >>>>>>>>>>> input/ does, as a paradox in regards to partial_recognize_D >>>>>>>>>>> _ONLY AFFECTS_ partial_recognize_D not _OUR ABILITY TO PROVE >>>>>>>>>>> OUTCOMES_
In other words, you admit that you words are just lies and >>>>>>>>>> don't mean what they say.
Your "undecidable" isn't about decidability, but correctness >>>>>>>>>> of a given decider.
YES, THAT'S ALL UNDECIDABILITY EVER WAS
In other words you are admitting you don't know what you are
talking about.
A given decider being incorrect doesn't mean a correct decider >>>>>>>> doesn't exist. There are many decidable problems, for which
incorrect deciders can exist.
Undecidability is about the being NO correct deciders.
YOU HAVEN'T EVEN SHOW A MACHINE WHICH HAS NO CORRECT CLASSIFIERS >>>>>>> FOR...
Because I don't need to.
YOU'VE ONLY SHOWN THAT FOR ANY CLASSIFIER THERE EXISTS A MACHINE IT >>>>> CAN'T CLASSIFY...
NOT THAT THERE EXISTS A MACHINE THAT CAN'T BE CLASSIFIED...
But I don't need to.
The uncomputability of that classification is proven by the first.
ALL THAT'S BEEN SHOWN IS _ONE_ MACHINE CAN'T CLASSIFY ALL MACHINEs.
And that the arguement can apply to ANY machine, since there was no
restriction on what could be viewed as D
IT DOES NOT THE SHOW THERE IS MACHINE THAT *CANNOT* BE CLASSIFIED BY
*ANY* MACHINE
Which it doesn't need to do.
ok bro
All you are doiing is proving you really don't know what you are
talking about.
THOSE ARE _DIFFERENT_ CLAIMS, REGARDLESS OF U JUST CONFLATING THEM
BECAUSE UR GETTING DESPERATE
And who is conflaiting them?
my god u can't even keep track of the last sentence i read?
lol ok bro
Yes, I pointed out as an aside, that there exists machines that no
know to be always correct partial decider can decide on, but that is
NOT what Turing is proving here, and thus claiming he doesn't succeed
at that just shows you don't know what you are talking about.
The existance of the second class of machine is just something more,
that is clearly beyond your understanding.
All that is needed to prove a problem is uncomputable is to show
that you can make an input for any given decider that it is wrong >>>>>> about.
THAT ONLY SHOWS THAT THE PROBLEM CAN'T BE COMPUTED BY A _SINGLE_
MACHINE
Right, or any effectively enumerable set of machine, since we can
then build the single machine that tests all of them.
And that *IS* the definition of computable,
BECAUSE UR JUST ASSUMING THE CT-THESIS AS TRUE, WHICH IS NOT PROVEN
Nope. It seems you are just to stupid to understand the logic.
Part of the problem is you just don't understand what the words mean.
You don't seem to understand the problem.
YOU DON'T UNDERSTAND WHAT'S BEEN PROVEN
Sure I do, you just admitted what I have said is proven, that the
LOL, UR CLAIMS ARE GETTING MORE GARBAGE BY THE POST
Nope, you are just showing how stupid you are, because you don't know
what the words you are trying to use mean.
lol, ur claims really are getting more garbage by the post
problem is not computable, which means there does not exist a single
machine that correctly computes all the answers.
NO, THE FAILURE TO CLASSIFY MUST COME FROM THE STRUCTURE OF THE >>>>>>>>> INPUT, NOT AN INCORRECT ALGO OF THE CLASSIFIER, WHY AM I
This means that the trivial always false machine meets your >>>>>>>>>
REPEATING ME ON THIS?
In other words, you think people/machines can't make mistakes?
BEING UNABLE TO CLASSIFY /UNDECIDABLE INPUT/ IS _NOT_ A MISTAKE >>>>>>> ON BEHALF OF THE CLASSIFIER
But the input isn't "undecidable", as that isn't an actual
property of an input.
IT IS _CAUSED_ *BY THE STRUCTURE OF THE INPUT*
No, as the input is fully decidable by some other decider, and thus
there is nothing "undecidable" about it.
RIGHT, THAT'S BECAUSE UNDECIDABILITY IS ONLY DEFINED AS A
RELATIONSHIP BETWEEN A MACHINE AND A PARTICULAR CLASSIFIER PARADOXES
BY THAT MACHINE
Nope.
You are sure stupid about the meaning of the words.
"Undecidability" is the property of a Problem / Classification rule to
be properl carried out by ANY possible algorithm.
THERE _NEVER_ HAS BEEN _ANY_ OTHER FORM OF PROVEN UNDECIDABILITY
WITHIN COMPUTING,
And where did you get your definition?
origin fallacy
Maybe you should read some of the elementary statements about it, like
at https://en.wikipedia.org/wiki/Undecidable_problem
also origin fallacy
"An undecidable problem is a decision problem for which it is proved
to be impossible to construct an algorithm that always leads to a
correct yes-or-no answer. "
that's a sentence, doesn't mean it's correct
Note, no mention about the relationship to a machine (except that such
a machine doesn't exist).
because it conflates algo with machine, due to assuming the ct-thesis as true
--- Synchronet 3.21d-Linux NewsLink 1.2
AND UR DOING FUCK ALL TO PROVE ANOTHER FORM EXISTS
Since your definition is just your own lies, YOU are the one making
stuff up.
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