The following claim from p246 of Turing’s seminal paper On Computable Numbers is a fallacy:
/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/
Annoyingly, the linguists markup of a leading asterisk is also taken in >usenet formatting markup for a bullet mark:
The following claim from p246 of Turing’s seminal paper On Computable Numbers is a fallacy:You've missed something out. That something is something like "If I
/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-What's an "infinite circle-free machine"? All the theory we are talking
free machine, and we have no general process for doing this in a finite number of steps/
For any given computable sequence, there are _infinite_ circle-free
machines ....
.... which compute that particular sequence. Not only can variousI think you'll need to prove that you can identify a machine that
machines differ significantly in the specific steps to produce the
same output, machines can be changed in superficial ways that do not meaningfully affect the steps of computation, akin to modern no-op
statements or unreachable code
The problem of enumerating computable sequences, however, only depends
on successfully identifying _one_ circle-free machine that computes any given computable sequences. While identifying more than one can
certainly be done, it is _not_ a requirement for enumerating computable sequences, as _one_ machine computing a sequence /suffices to output any
and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_Your given reason fails to rule out the equivalence. For a start, under
actually equivalent to a _general process_ of enumerating circle-free machines, as there is no need to identify all circle-free machines which compute any given computable sequence
Said problem is only equivalent to a _limited process_ of enumerating circle-free machines. The machine which identifies circle-free machinesAgain, can this machine exist? It seems to me that the "limited power"
only needs the limited power of determining _at least one_ circle-free machine for any given computable sequence, _not all_ machines for any
given computable sequence
Because of this fallacy, the proof found on the following p247, where an ill-defined machine 𝓗 (which attempts and fails to compute the direct diagonal β’) is found to be undecidable in respect to circle-freeIf this were a fallacy, you should write it up properly, get it
decider 𝓓; does not then prove an impossibility for enumerating computable sequences. As the problem of enumerating /all circle-free machines/ is _not_ equivalent to that of enumerating /just computable sequences/
----
arising us out of the computing dark ages,
please excuse my pseudo-pyscript,
~ the little crank that could
[ Followup-To: set ]
In comp.theory dart200 <user7160@newsgrouper.org.invalid> wrote:
The following claim from p246 of Turing’s seminal paper On Computable
Numbers is a fallacy:
You've missed something out. That something is something like "If I understand correctly" or "As far as I can see". Without such a qualification, your statement just looks like extreme hubris.
It is vanishingly unlikely that you have understood correctly or you
have seen far enough. If there were a flaw in Turing's 1936 paper, and
it were subtle enough to escape detection by the millions of specialists
who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
/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/
For any given computable sequence, there are _infinite_ circle-free
machines ....
What's an "infinite circle-free machine"?
All the theory we are talking about deals only with finite machines.
.... which compute that particular sequence. Not only can various
machines differ significantly in the specific steps to produce the
same output, machines can be changed in superficial ways that do not
meaningfully affect the steps of computation, akin to modern no-op
statements or unreachable code
The problem of enumerating computable sequences, however, only depends
on successfully identifying _one_ circle-free machine that computes any
given computable sequences. While identifying more than one can
certainly be done, it is _not_ a requirement for enumerating computable
sequences, as _one_ machine computing a sequence /suffices to output any
and all digits of that sequence/
I think you'll need to prove that you can identify a machine that
computes a given computable sequence. By definition of "computable"
there is such a machine, but how do you "identify" it? Suppose you had
a computable number and a machine, how would you test whether or not
that machine generates the number?
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-free
machines, as there is no need to identify all circle-free machines which
compute any given computable sequence
Your given reason fails to rule out the equivalence. For a start, under
what relationship does/doesn't this equivalence hold?
Said problem is only equivalent to a _limited process_ of enumerating
circle-free machines. The machine which identifies circle-free machines
only needs the limited power of determining _at least one_ circle-free
machine for any given computable sequence, _not all_ machines for any
given computable sequence
Again, can this machine exist? It seems to me that the "limited power"
you describe is not at all limited. It is known that there is no
machine which can ascertain whether or not two turing machines produce
the same output. So the ability to discard machines redundant in this
sense will not exist either.
Because of this fallacy, the proof found on the following p247, where an
ill-defined machine 𝓗 (which attempts and fails to compute the direct
diagonal β’) is found to be undecidable in respect to circle-free
decider 𝓓; does not then prove an impossibility for enumerating
computable sequences. As the problem of enumerating /all circle-free
machines/ is _not_ equivalent to that of enumerating /just computable
sequences/
If this were a fallacy, you should write it up properly, get it
published and become famous. Far more likely, however, is that you have
just failed to understand what Turing's paper means.
--
arising us out of the computing dark ages,
please excuse my pseudo-pyscript,
~ the little crank that could
If there were a flaw in Turing's 1936 paper, and^^^^
it were subtle enough to escape detection by the millions of specialists^^^^^
who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
I think you'll need to prove that you can identify a machine that
computes a given computable sequence. By definition of "computable"
there is such a machine, but how do you "identify" it? Suppose you had
a computable number and a machine, how would you test whether or not
that machine generates the number?
On 10/03/2026 16:51, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On Computable Numbers is a fallacy:
/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/
I am delighted to see that your usage of forward-slashes is valid as an example of usenet text formatting markup but frustrated that it's so
close to being an example of the linguist's forward-slash markup to
indicate a wrong example.
Alas, the meaning of the sentence is what is wrong, although perhaps the syntax is technically wrong somehow too, and perhaps some intrinsic
semantic problem is present too so we can say they are the linguists
error markers too after all.
Annoyingly, the linguists markup of a leading asterisk is also taken in usenet formatting markup for a bullet mark:
* this nonsense sentence is
/too crap this is sentence a/
It is known that there is no
machine which can ascertain whether or not two turing machines produce
the same output.
The following claim from p246 of Turing’s seminal paper On Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-free
machines which compute that particular sequence. Not only can various machines differ significantly in the specific steps to produce the same output, machines can be changed in superficial ways that do not
meaningfully affect the steps of computation, akin to modern no-op statements or unreachable code
The problem of enumerating computable sequences, however, only depends
on successfully identifying _one_ circle-free machine that computes any given computable sequences. While identifying more than one can
certainly be done, it is _not_ a requirement for enumerating computable sequences, as _one_ machine computing a sequence /suffices to output any
and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-free machines, as there is no need to identify all circle-free machines which compute any given computable sequence
Said problem is only equivalent to a _limited process_ of enumerating circle-free machines. The machine which identifies circle-free machines
only needs the limited power of determining _at least one_ circle-free machine for any given computable sequence, _not all_ machines for any
given computable sequence
Because of this fallacy, the proof found on the following p247, where an ill-defined machine 𝓗 (which attempts and fails to compute the direct diagonal β’) is found to be undecidable in respect to circle-free
decider 𝓓; does not then prove an impossibility for enumerating computable sequences. As the problem of enumerating /all circle-free machines/ is _not_ equivalent to that of enumerating /just computable sequences/
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On Computable
Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-free
machines which compute that particular sequence. Not only can various
machines differ significantly in the specific steps to produce the
same output, machines can be changed in superficial ways that do not
meaningfully affect the steps of computation, akin to modern no-op
statements or unreachable code
The problem of enumerating computable sequences, however, only depends
on successfully identifying _one_ circle-free machine that computes
any given computable sequences. While identifying more than one can
certainly be done, it is _not_ a requirement for enumerating
computable sequences, as _one_ machine computing a sequence /suffices
to output any and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-free
machines, as there is no need to identify all circle-free machines
which compute any given computable sequence
Which just shows that you don't understand what the word "Equivalent"
means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can have two "Equivalent classification problems" that by using different classes,
come to the same result, that there exist uncomputable problems.
Said problem is only equivalent to a _limited process_ of enumerating
circle-free machines. The machine which identifies circle-free
machines only needs the limited power of determining _at least one_
circle-free machine for any given computable sequence, _not all_
machines for any given computable sequence
Because of this fallacy, the proof found on the following p247, where
an ill-defined machine 𝓗 (which attempts and fails to compute the
direct diagonal β’) is found to be undecidable in respect to circle-
free decider 𝓓; does not then prove an impossibility for enumerating
computable sequences. As the problem of enumerating /all circle-free
machines/ is _not_ equivalent to that of enumerating /just computable
sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a PRD
to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of the
number given to it.
2) No machine that it accepts will fail to produce a computable number,
and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least one number that represents a machine that computes it.
So, if PRD exists, we can build a machine that computes an anti-diagonal
by testing each number in sequence with PRD, and for each number that it accepts, it will simulate that machine until that machine generates k
digits of output, k being the number of values accepted to this point,
and then it outputs the opposite digit of the kth digit generated by the
nth machine.
This machine must produce a computable number, as it only simulates and
uses the output of machines that PRD accepted, so by 1, PRD answered,
and by 2 that machine can be simulated for as long as we want, and we
WILL get to the desired digit.
Now, since this machine generates a computable number, by condition 3,
there must exist a finite number n that represents a machine that
generates it that PRD will accept, and thus our machine will simulate
for k digits (which will be less than n) and output the opposite value.
Thus, whatever n you want to claim is the machine that generates the
same computable number doesn't, and thus there can not exist a PRD that
does what you claim.
In fact, this method works for ANY method you may want to claim allows
you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic requires you
to just be able to assume things exist that don't.
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to compute
the direct diagonal β’) is found to be undecidable in respect to circle-free decider 𝓓; does not then prove an impossibility for enumerating computable sequences.
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to compute
the direct diagonal β’) is found to be undecidable in respect to
circle-free decider 𝓓; does not then prove an impossibility for
enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably undecidable,
that must mean any “better-defined” machine that also satisfies those “ill-defined” criteria must be provably undecidable.
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On Computable >>> Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-free
machines which compute that particular sequence. Not only can various
machines differ significantly in the specific steps to produce the
same output, machines can be changed in superficial ways that do not
meaningfully affect the steps of computation, akin to modern no-op
statements or unreachable code
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine that
computes any given computable sequences. While identifying more than
one can certainly be done, it is _not_ a requirement for enumerating
computable sequences, as _one_ machine computing a sequence /suffices
to output any and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-free
machines, as there is no need to identify all circle-free machines
which compute any given computable sequence
Which just shows that you don't understand what the word "Equivalent"
means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can have
two "Equivalent classification problems" that by using different
classes, come to the same result, that there exist uncomputable problems.
put more clearly: enumerating computable sequences requires enumerating
only _and not more than_ a *subset* of circle-free machines that does
_not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle-free machine
Said problem is only equivalent to a _limited process_ of enumerating
circle-free machines. The machine which identifies circle-free
machines only needs the limited power of determining _at least one_
circle-free machine for any given computable sequence, _not all_
machines for any given computable sequence
Because of this fallacy, the proof found on the following p247, where
an ill-defined machine 𝓗 (which attempts and fails to compute the
direct diagonal β’) is found to be undecidable in respect to circle- >>> free decider 𝓓; does not then prove an impossibility for enumerating >>> computable sequences. As the problem of enumerating /all circle-free
machines/ is _not_ equivalent to that of enumerating /just computable
sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a
PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of the
number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least
one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each
number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values
accepted to this point, and then it outputs the opposite digit of the
kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates
and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we
want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition 3,
there must exist a finite number n that represents a machine that
generates it that PRD will accept, and thus our machine will simulate
for k digits (which will be less than n) and output the opposite value.
Thus, whatever n you want to claim is the machine that generates the
same computable number doesn't, and thus there can not exist a PRD
that does what you claim.
In fact, this method works for ANY method you may want to claim allows
you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic requires
you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made.
*having a machine that computes a diagonal, does not actually then imply it's usable to compute an anti-diagonal*
and the reason that's true is *same* self-referential weirdness that
stumped turing on the original paper
consider fixed_H again, since that is the form of diagonal computation
that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each output
anti_digit = 1-digit // of the machine fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
what would happen if PRD(anti_fixed_H) => TRUE?
what would happen if PRD(anti_fixed_H) => FALSE?
does either case actually output a true total anti-diagonal???
(hint: no)
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-free
machines which compute that particular sequence. Not only can
various machines differ significantly in the specific steps to
produce the same output, machines can be changed in superficial ways
that do not meaningfully affect the steps of computation, akin to
modern no-op statements or unreachable code
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine that
computes any given computable sequences. While identifying more than
one can certainly be done, it is _not_ a requirement for enumerating
computable sequences, as _one_ machine computing a sequence /
suffices to output any and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-
free machines, as there is no need to identify all circle-free
machines which compute any given computable sequence
Which just shows that you don't understand what the word "Equivalent"
means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can have
two "Equivalent classification problems" that by using different
classes, come to the same result, that there exist uncomputable
problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle-free
machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining _at
least one_ circle-free machine for any given computable sequence,
_not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to compute >>>> the direct diagonal β’) is found to be undecidable in respect to
circle- free decider 𝓓; does not then prove an impossibility for
enumerating computable sequences. As the problem of enumerating /all
circle-free machines/ is _not_ equivalent to that of enumerating /
just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a
PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of the
number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least
one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have no
proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each
number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values
accepted to this point, and then it outputs the opposite digit of the
kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates
and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we
want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition
3, there must exist a finite number n that represents a machine that
generates it that PRD will accept, and thus our machine will simulate
for k digits (which will be less than n) and output the opposite value.
Thus, whatever n you want to claim is the machine that generates the
same computable number doesn't, and thus there can not exist a PRD
that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic requires
you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made.
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
and the reason that's true is *same* self-referential weirdness that
stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change it to
an equivalent program that prints the anti-diagonal.
consider fixed_H again, since that is the form of diagonal computation
that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each output
anti_digit = 1-digit // of the machine fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
what would happen if PRD(anti_fixed_H) => TRUE?
what would happen if PRD(anti_fixed_H) => FALSE?
But we don't care what PRD(anti_fixed_H) is.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D can
not possible acheive its requirements to accept at least one machine
that generates a computable number, and thus, the "enumeration" you
generate from it is not complete, but only partial, and thus your whole claim falls apart.
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-free
machines which compute that particular sequence. Not only can
various machines differ significantly in the specific steps to
produce the same output, machines can be changed in superficial
ways that do not meaningfully affect the steps of computation, akin
to modern no-op statements or unreachable code
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine that
computes any given computable sequences. While identifying more
than one can certainly be done, it is _not_ a requirement for
enumerating computable sequences, as _one_ machine computing a
sequence / suffices to output any and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-
free machines, as there is no need to identify all circle-free
machines which compute any given computable sequence
Which just shows that you don't understand what the word
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can have
two "Equivalent classification problems" that by using different
classes, come to the same result, that there exist uncomputable
problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle-free
machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining
_at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in
respect to circle- free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences. As the problem
of enumerating /all circle-free machines/ is _not_ equivalent to
that of enumerating / just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a
PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of
the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least
one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have
no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each
number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values
accepted to this point, and then it outputs the opposite digit of
the kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates
and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we
want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition
3, there must exist a finite number n that represents a machine that
generates it that PRD will accept, and thus our machine will
simulate for k digits (which will be less than n) and output the
opposite value.
Thus, whatever n you want to claim is the machine that generates the
same computable number doesn't, and thus there can not exist a PRD
that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic requires
you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made.
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding eh???)
and the reason that's true is *same* self-referential weirdness that
stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change it
to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
consider fixed_H again, since that is the form of diagonal computation
that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each output
anti_digit = 1-digit // of the machine fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay attention:
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H), simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H), fixed_H
will skip simulate anti_fixed_H. this means that anti_fixed_H will also
skip producing an anti-digit to it's own output ... so it still doesn't compute a total anti-diagonal
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the
machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D can
not possible acheive its requirements to accept at least one machine
that generates a computable number, and thus, the "enumeration" you
generate from it is not complete, but only partial, and thus your
whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox, let's
try "fixing" anti_H by handling it's self-ref akin to fixed_H. for this
i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here???
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ... which
is *not* a true anti-diagonal
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it outputs
...
therefore, you can't fix the anti-diagonal computation like you can for
the diagonal computation. the self-referential weirdness that stumped
turing is fixable _only_ for the diagonal, _not_ for the anti-diagonal,
and therefore the anti-diagonal is still _not computable_
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-free >>>>> machines which compute that particular sequence. Not only can
various machines differ significantly in the specific steps to
produce the same output, machines can be changed in superficial
ways that do not meaningfully affect the steps of computation, akin >>>>> to modern no-op statements or unreachable code
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine that
computes any given computable sequences. While identifying more
than one can certainly be done, it is _not_ a requirement for
enumerating computable sequences, as _one_ machine computing a
sequence / suffices to output any and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-
free machines, as there is no need to identify all circle-free
machines which compute any given computable sequence
Which just shows that you don't understand what the word
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can have
two "Equivalent classification problems" that by using different
classes, come to the same result, that there exist uncomputable
problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle-free
machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining
_at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to compute >>>>> the direct diagonal β’) is found to be undecidable in respect to >>>>> circle- free decider 𝓓; does not then prove an impossibility for >>>>> enumerating computable sequences. As the problem of enumerating /
all circle-free machines/ is _not_ equivalent to that of
enumerating / just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a
PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of
the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least
one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have
no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each
number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values
accepted to this point, and then it outputs the opposite digit of
the kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates
and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we
want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition
3, there must exist a finite number n that represents a machine that
generates it that PRD will accept, and thus our machine will
simulate for k digits (which will be less than n) and output the
opposite value.
Thus, whatever n you want to claim is the machine that generates the
same computable number doesn't, and thus there can not exist a PRD
that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic requires
you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made.
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding eh???)
and the reason that's true is *same* self-referential weirdness that
stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change it
to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
consider fixed_H again, since that is the form of diagonal computation
that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each output
anti_digit = 1-digit // of the machine fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay attention:
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H), simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H), fixed_H
will skip simulate anti_fixed_H. this means that anti_fixed_H will also
skip producing an anti-digit to it's own output ... so it still doesn't compute a total anti-diagonal
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the
machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D can
not possible acheive its requirements to accept at least one machine
that generates a computable number, and thus, the "enumeration" you
generate from it is not complete, but only partial, and thus your
whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox, let's
try "fixing" anti_H by handling it's self-ref akin to fixed_H. for this
i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here???
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ... which
is *not* a true anti-diagonal
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it
outputs ...
therefore, you can't fix the anti-diagonal computation like you can for
the diagonal computation. the self-referential weirdness that stumped
turing is fixable _only_ for the diagonal, _not_ for the anti-diagonal,
and therefore the anti-diagonal is still _not computable_
[ Followup-To: set ]
In comp.theory dart200 <user7160@newsgrouper.org.invalid> wrote:
The following claim from p246 of Turing’s seminal paper On Computable
Numbers is a fallacy:
You've missed something out. That something is something like "If I understand correctly" or "As far as I can see". Without such a qualification, your statement just looks like extreme hubris.
It is vanishingly unlikely that you have understood correctly or you
have seen far enough. If there were a flaw in Turing's 1936 paper, and
it were subtle enough to escape detection by the millions of specialists
who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
Alan Mackenzie <acm@muc.de> writes:
[ Followup-To: set ]
In comp.theory dart200 <user7160@newsgrouper.org.invalid> wrote:
The following claim from p246 of Turings seminal paper On Computable
Numbers is a fallacy:
You've missed something out. That something is something like "If I
understand correctly" or "As far as I can see". Without such a
qualification, your statement just looks like extreme hubris.
It is vanishingly unlikely that you have understood correctly or you
have seen far enough. If there were a flaw in Turing's 1936 paper, and
it were subtle enough to escape detection by the millions of specialists
who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
And then there's the context. He has chosen to present this startling discovery in a place similar (and sometimes grander) false claims are
made every single day. It's as if he had dug a huge diamond and has
decided to flog it on a market stall in Dalston.
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_
circle-free machines which compute that particular sequence. Not
only can various machines differ significantly in the specific
steps to produce the same output, machines can be changed in
superficial ways that do not meaningfully affect the steps of
computation, akin to modern no-op statements or unreachable code
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine that >>>>>> computes any given computable sequences. While identifying more
than one can certainly be done, it is _not_ a requirement for
enumerating computable sequences, as _one_ machine computing a
sequence / suffices to output any and all digits of that sequence/ >>>>>>
The problem of enumerating computable sequences is therefore _not_ >>>>>> actually equivalent to a _general process_ of enumerating circle-
free machines, as there is no need to identify all circle-free
machines which compute any given computable sequence
Which just shows that you don't understand what the word
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can
have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_
circle-free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining
_at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in
respect to circle- free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences. As the problem >>>>>> of enumerating / all circle-free machines/ is _not_ equivalent to
that of enumerating / just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a
PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of
the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least
one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have
no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each
number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values
accepted to this point, and then it outputs the opposite digit of
the kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates
and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we
want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition
3, there must exist a finite number n that represents a machine
that generates it that PRD will accept, and thus our machine will
simulate for k digits (which will be less than n) and output the
opposite value.
Thus, whatever n you want to claim is the machine that generates
the same computable number doesn't, and thus there can not exist a
PRD that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made. >>>>
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness that
stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change it
to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal computation >>>> that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each output >>>> anti_digit = 1-digit // of the machine fixed_H >>>> output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a non-circle free input.
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H
will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the requirement
you gave it, so fixed_H did not actually compute a digonal of an
enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the
machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D
can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only partial,
and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H.
for this i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here???
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program DID,
at least it would if PRD exists.
Thus, your PRD disappear in the same contradiction that Turing_D and
Turing_H does, and thus so does your fixed_H
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you don't
get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the
anti-diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts is
what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
On 13/03/2026 00:20, Ben Bacarisse wrote:
Alan Mackenzie <acm@muc.de> writes:
[ Followup-To: set ]
In comp.theory dart200 <user7160@newsgrouper.org.invalid> wrote:
The following claim from p246 of Turing’s seminal paper On Computable >>>> Numbers is a fallacy:
You've missed something out. That something is something like "If I
understand correctly" or "As far as I can see". Without such a
qualification, your statement just looks like extreme hubris.
It is vanishingly unlikely that you have understood correctly or you
have seen far enough. If there were a flaw in Turing's 1936 paper, and >>> it were subtle enough to escape detection by the millions of specialists >>> who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
And then there's the context. He has chosen to present this startling
discovery in a place similar (and sometimes grander) false claims are
made every single day. It's as if he had dug a huge diamond and has
decided to flog it on a market stall in Dalston.
One problem here is that dart200 chooses to criticise Turing's
historical paper, and most responders are not familiar with the (by
modern standards) clunky terminology, and have possibly not even read
the paper(?). And dart200 seems not to be good at following the thread
of reasoning, and cannot "fill in gaps in reasoning" himself, as he
lacks the mathematical maturity (no shame in this - Turing was writing
for his peers). So while he may have identified a short-falling in Turing's "presentation" of the material, he misunderstands the impact on
the bigger picture of the paper.
Anyhow, what's going on on those couple of pages? Assuming I'm
following correctly:
Turing:
| 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
Problem A: (computing an) enumeration of (all) computable sequences
Problem B: computing whether any number is the D.N of a circle-free machine
Claim: There is no machine that solves B.
Turing suggests that Problems A and B are equivalent. Indeed they are "logically" equivalent, in the sense that both can be proven unsolvable
(by TMs), but normally I'd take Turing's phrasing as suggest that there
are clear arguments that relate a given solution to Problem A to a / consequential/ solution to Problem B, and vice-versa. Is this the case?
Clearly if we have a solution to Problem B, we can construct /from it/ a solution to Problem A. Turing effectively does this on page 247. So we have one direction covered, no problem.
But if we have a TM solving Problem A, how would we construct /from it/
a TM solving Problem B? I don't see an obvious argument working along those lines. dart200 seems to be mostly making this point. I doubt
Turing had such an argument, unless I'm missing something obvious.
(Hey, happens...)
OTOH, we can show /regardless of Problem A/ that Problem B has no
solution, which is the specific claim Turing is discussing on the next page. He is not presenting a proof that Problem A has no solution!
So on the next page (247), Turing presents a proof that there is no
machine which solves Problem B. He naturally starts with assuming such a machind D exists, and from that constructs a new machine H, and
ultimately reaches a contradiction in its behaviour, finally concluding "...Thus both verdicts are impossible, and we conclude that there can be
no such machine D".
That's clear enough - the proof is showing there's no solution to
Problem B. dart200 makes a big point (his "diamond"?) that the proof
fails to show there is no solution to Problem A, but Turing does not
present it as such, so no big deal!
Regarding Problem A. the obvious proof for this would follow the lines
of the usual diagonal argument. If we had a machine enumerating a
sequence of D.N.s of machines computing computable numbers, then we
could apply the diagonal argument to construct a new machine that
computes a new computable number which is not represented in the
original D.N enumeration.
Worded another way, we can show that any /computable/ enumeration of the computable numbers must be incomplete. Or more succintly, Problem A has
no solution.
Turing does talk a bit about all this on page 246, without really
spelling it out in a more formal proof.
So even if we grant that Turing's presentation could have been a little clearer when he talks about the two problems being "equivalent", there
is no further impact of this on his paper. dart200 seems to have just
lost the thread a bit re where Turing is going, and thinks his
"discovery" is more significant than it really is.
Mike.
Alan Mackenzie <acm@muc.de> writes:
[ Followup-To: set ]
In comp.theory dart200 <user7160@newsgrouper.org.invalid> wrote:
The following claim from p246 of Turing’s seminal paper On Computable
Numbers is a fallacy:
You've missed something out. That something is something like "If I
understand correctly" or "As far as I can see". Without such a
qualification, your statement just looks like extreme hubris.
It is vanishingly unlikely that you have understood correctly or you
have seen far enough. If there were a flaw in Turing's 1936 paper, and
it were subtle enough to escape detection by the millions of specialists
who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
And then there's the context. He has chosen to present this startling discovery in a place similar (and sometimes grander) false claims are
made every single day. It's as if he had dug a huge diamond and has
decided to flog it on a market stall in Dalston.
On 03/12/2026 03:53 PM, Richard Damon wrote:
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_
circle-free machines which compute that particular sequence. Not >>>>>>> only can various machines differ significantly in the specific
steps to produce the same output, machines can be changed in
superficial ways that do not meaningfully affect the steps of
computation, akin to modern no-op statements or unreachable code >>>>>>>
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine that >>>>>>> computes any given computable sequences. While identifying more
than one can certainly be done, it is _not_ a requirement for
enumerating computable sequences, as _one_ machine computing a
sequence / suffices to output any and all digits of that sequence/ >>>>>>>
The problem of enumerating computable sequences is therefore _not_ >>>>>>> actually equivalent to a _general process_ of enumerating circle- >>>>>>> free machines, as there is no need to identify all circle-free
machines which compute any given computable sequence
Which just shows that you don't understand what the word
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can
have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_
circle-free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining >>>>>>> _at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in >>>>>>> respect to circle- free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences. As the problem >>>>>>> of enumerating / all circle-free machines/ is _not_ equivalent to >>>>>>> that of enumerating / just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a >>>>>> PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of
the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least >>>>>> one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have
no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each >>>>>> number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values >>>>>> accepted to this point, and then it outputs the opposite digit of
the kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates >>>>>> and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we >>>>>> want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition >>>>>> 3, there must exist a finite number n that represents a machine
that generates it that PRD will accept, and thus our machine will
simulate for k digits (which will be less than n) and output the
opposite value.
Thus, whatever n you want to claim is the machine that generates
the same computable number doesn't, and thus there can not exist a >>>>>> PRD that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made. >>>>>
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding
eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness that >>>>> stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change it
to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal computation >>>>> that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit >>>>> K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each
output
anti_digit = 1-digit // of the machine
fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay
attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a
non-circle free input.
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H
will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the requirement
you gave it, so fixed_H did not actually compute a digonal of an
enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the
machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D
can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only partial,
and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H.
for this i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here???
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program DID,
at least it would if PRD exists.
Thus, your PRD disappear in the same contradiction that Turing_D and
Turing_H does, and thus so does your fixed_H
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you don't
get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the
anti-diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts is
what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
The only proper counterexample to uncountability of continuous domains
is "sweep".
There are basically two sides to it, line-drawing between 0 and 1,
and the rationals being HUGE, "line-reals" and "signal-reals".
Then, getting into the laws of large numbers, and models of arithmetic,
the infinite has plenty much, much larger numbers than given bounds.
On 03/12/2026 03:53 PM, Richard Damon wrote:
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_
circle-free machines which compute that particular sequence. Not >>>>>>> only can various machines differ significantly in the specific
steps to produce the same output, machines can be changed in
superficial ways that do not meaningfully affect the steps of
computation, akin to modern no-op statements or unreachable code >>>>>>>
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine that >>>>>>> computes any given computable sequences. While identifying more
than one can certainly be done, it is _not_ a requirement for
enumerating computable sequences, as _one_ machine computing a
sequence / suffices to output any and all digits of that sequence/ >>>>>>>
The problem of enumerating computable sequences is therefore _not_ >>>>>>> actually equivalent to a _general process_ of enumerating circle- >>>>>>> free machines, as there is no need to identify all circle-free
machines which compute any given computable sequence
Which just shows that you don't understand what the word
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can
have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_
circle-free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining >>>>>>> _at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in >>>>>>> respect to circle- free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences. As the problem >>>>>>> of enumerating / all circle-free machines/ is _not_ equivalent to >>>>>>> that of enumerating / just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a >>>>>> PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of
the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least >>>>>> one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have
no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each >>>>>> number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values >>>>>> accepted to this point, and then it outputs the opposite digit of
the kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates >>>>>> and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we >>>>>> want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition >>>>>> 3, there must exist a finite number n that represents a machine
that generates it that PRD will accept, and thus our machine will
simulate for k digits (which will be less than n) and output the
opposite value.
Thus, whatever n you want to claim is the machine that generates
the same computable number doesn't, and thus there can not exist a >>>>>> PRD that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made. >>>>>
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding
eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness that >>>>> stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change it
to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal computation >>>>> that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit >>>>> K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each
output
anti_digit = 1-digit // of the machine
fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay
attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a
non-circle free input.
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H
will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the requirement
you gave it, so fixed_H did not actually compute a digonal of an
enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the
machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D
can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only partial,
and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H.
for this i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here???
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program DID,
at least it would if PRD exists.
Thus, your PRD disappear in the same contradiction that Turing_D and
Turing_H does, and thus so does your fixed_H
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you don't
get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the
anti-diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts is
what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
The only proper counterexample to uncountability of continuous domains
is "sweep".
There are basically two sides to it, line-drawing between 0 and 1,
and the rationals being HUGE, "line-reals" and "signal-reals".
Then, getting into the laws of large numbers, and models of arithmetic,
the infinite has plenty much, much larger numbers than given bounds.
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in
respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably undecidable,
that must mean any “better-defined” machine that also satisfies
those “ill-defined” criteria must be provably undecidable.
the "better-defined" machine don't satisfy the criteria to be undecidable
On 3/13/26 12:23 AM, Ross Finlayson wrote:
On 03/12/2026 03:53 PM, Richard Damon wrote:
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:i'm sorry richard, it does not actually do that
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_
circle-free machines which compute that particular sequence. Not >>>>>>>> only can various machines differ significantly in the specific >>>>>>>> steps to produce the same output, machines can be changed in
superficial ways that do not meaningfully affect the steps of
computation, akin to modern no-op statements or unreachable code >>>>>>>>
The problem of enumerating computable sequences, however, only >>>>>>>> depends on successfully identifying _one_ circle-free machine that >>>>>>>> computes any given computable sequences. While identifying more >>>>>>>> than one can certainly be done, it is _not_ a requirement for
enumerating computable sequences, as _one_ machine computing a >>>>>>>> sequence / suffices to output any and all digits of that sequence/ >>>>>>>>
The problem of enumerating computable sequences is therefore _not_ >>>>>>>> actually equivalent to a _general process_ of enumerating circle- >>>>>>>> free machines, as there is no need to identify all circle-free >>>>>>>> machines which compute any given computable sequence
Which just shows that you don't understand what the word
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very >>>>>>> different methods and path create the exact same output, we can
have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_
circle-free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies >>>>>>>> circle-free machines only needs the limited power of determining >>>>>>>> _at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence >>>>>>>>
Because of this fallacy, the proof found on the following p247, >>>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>> respect to circle- free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences. As the problem >>>>>>>> of enumerating / all circle-free machines/ is _not_ equivalent to >>>>>>>> that of enumerating / just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a >>>>>>> PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of >>>>>>> the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever. >>>>>>>
3) For EVERY Computable Number that exist, PRD will accept at least >>>>>>> one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have >>>>>> no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti- >>>>>>> diagonal by testing each number in sequence with PRD, and for each >>>>>>> number that it accepts, it will simulate that machine until that >>>>>>> machine generates k digits of output, k being the number of values >>>>>>> accepted to this point, and then it outputs the opposite digit of >>>>>>> the kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates >>>>>>> and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we >>>>>>> want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition >>>>>>> 3, there must exist a finite number n that represents a machine
that generates it that PRD will accept, and thus our machine will >>>>>>> simulate for k digits (which will be less than n) and output the >>>>>>> opposite value.
Thus, whatever n you want to claim is the machine that generates >>>>>>> the same computable number doesn't, and thus there can not exist a >>>>>>> PRD that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers. >>>>>>>
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping >>>>>> someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly
made.
*having a machine that computes a diagonal, does not actually then >>>>>> imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists. >>>>
(please do actually read the whole post in depth before responding
eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness that >>>>>> stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change it >>>>> to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal
computation
that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory >>>>>> output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each
output
anti_digit = 1-digit // of the machine
fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay
attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a
non-circle free input.
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H
will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the requirement
you gave it, so fixed_H did not actually compute a digonal of an
enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the >>>>> machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus >>>>> what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D
can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only partial, >>>>> and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H.
for this i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here??? >>>> K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program DID,
at least it would if PRD exists.
Thus, your PRD disappear in the same contradiction that Turing_D and
Turing_H does, and thus so does your fixed_H
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you don't
get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the
anti-diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts is >>> what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
The only proper counterexample to uncountability of continuous domains
is "sweep".
Who was talking about unCOUNTability.
It was unCOMPUTability.
There are basically two sides to it, line-drawing between 0 and 1,
and the rationals being HUGE, "line-reals" and "signal-reals".
Then, getting into the laws of large numbers, and models of arithmetic,
the infinite has plenty much, much larger numbers than given bounds.
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in
respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably undecidable,
that must mean any “better-defined” machine that also satisfies
those “ill-defined” criteria must be provably undecidable.
the "better-defined" machine don't satisfy the criteria to be undecidable
But they’re a subset of the “ill-defined” set that Turing was considering, are they not?
Unless you’re considering an entirely different set, in which case
your argument has nothing to do with Turing.
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in
respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably undecidable, >>>> that must mean any “better-defined” machine that also satisfies
those “ill-defined” criteria must be provably undecidable.
the "better-defined" machine don't satisfy the criteria to be
undecidable
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of them as
the same is a _fallacy_
specifically, there are *infinitely* many circle-free machines for each *computable sequence*. circle-free machines forms a /surjection/ onto computational sequences, not bijection. it's a many to one relationship,
not equatable.
enumerating circle-free machines requires enumerating *all* possible circular machines generally,
but enumerating only computable numbers only *requires* a categorical
subset of circle-free machines, and strictly so for any given
enumeration. only _one_ circle-machine per computable sequence is
required to enumerate the sequence, and you _cannot_ "enumerate" _more
than one_ per every computable sequence. enumerating _only_ computable sequences is a _lesser_ problem than enumerating circle-free machines
the contradiction turing demonstrated is only guaranteed to exist when totally enumerating out circle-machines, with a turing machine,
he did not prove the problem exists while enumerating only *one* circle- free machine per computable sequence.
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in
respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably undecidable, >>>>> that must mean any “better-defined” machine that also satisfies
those “ill-defined” criteria must be provably undecidable.
the "better-defined" machine don't satisfy the criteria to be
undecidable
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of them as
the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he must have been talking about.
He doesn't say the two machines generated by the two problems are in any
way equivalent, he says that the PROBLEMS are equivalent,
specifically, there are *infinitely* many circle-free machines for
each *computable sequence*. circle-free machines forms a /surjection/
onto computational sequences, not bijection. it's a many to one
relationship, not equatable.
enumerating circle-free machines requires enumerating *all* possible
circular machines generally,
but enumerating only computable numbers only *requires* a categorical
subset of circle-free machines, and strictly so for any given
enumeration. only _one_ circle-machine per computable sequence is
required to enumerate the sequence, and you _cannot_ "enumerate" _more
than one_ per every computable sequence. enumerating _only_ computable
sequences is a _lesser_ problem than enumerating circle-free machines
But only fractionally so, but since both problems are infinitely
impossible, that fraction doesn't matter.
the contradiction turing demonstrated is only guaranteed to exist when
totally enumerating out circle-machines, with a turing machine,
he did not prove the problem exists while enumerating only *one*
circle- free machine per computable sequence.
And he didn't actually claim to.
But he does point out that that is provable, but such a proof will seem "wrong" to those like you, even though it is actually correct.
After all, if we have a method to compute the enumeration of the
computable numbers, that means we have a method to compute the kth
digits of the nth number.
And thus we can, from that method, build a machine to compute the
diagonal or the anti-diagonal by using that method, finding the kth
digit of the kth number and output it or its opposite.
We can then ask the question, which number in this sequence is that anti-diagonal, which since we just built the computation that computes
it, must be in the list. But it can't be in the list, as if it was the
kth number, its kth digit differs from that anti-diagonal which it was supposed to be.
Your problem is you keep on living in a world where you don't need to
prove things but can just assume them true, and proof is for some time later.
Your enumeration MUST be incomplete, and thus isn't what you want it to
be, no matter how much you pray to the purple magic fairy dust powered unicorn for it to be.
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in >>>>>>> respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably undecidable, >>>>>> that must mean any “better-defined” machine that also satisfies >>>>>> those “ill-defined” criteria must be provably undecidable.
the "better-defined" machine don't satisfy the criteria to be
undecidable
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of them as
the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he must
have been talking about.
| 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
He doesn't say the two machines generated by the two problems are in
any way equivalent, he says that the PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable sequences, then
u could use that solution to determine whether any given machine is circle-free ...
and if so could be used to enumerate the circle-free machines,
making the problem of enumerating the sets equivalent,
which implies the sets are equivalent, or contain the same number of elements.
this is a fallacy, as circle-free machines forms a surjection onto computable numbers
specifically, there are *infinitely* many circle-free machines for
each *computable sequence*. circle-free machines forms a /surjection/
onto computational sequences, not bijection. it's a many to one
relationship, not equatable.
enumerating circle-free machines requires enumerating *all* possible
circular machines generally,
but enumerating only computable numbers only *requires* a categorical
subset of circle-free machines, and strictly so for any given
enumeration. only _one_ circle-machine per computable sequence is
required to enumerate the sequence, and you _cannot_ "enumerate"
_more than one_ per every computable sequence. enumerating _only_
computable sequences is a _lesser_ problem than enumerating circle-
free machines
But only fractionally so, but since both problems are infinitely
impossible, that fraction doesn't matter.
the contradiction turing demonstrated is only guaranteed to exist
when totally enumerating out circle-machines, with a turing machine,
he did not prove the problem exists while enumerating only *one*
circle- free machine per computable sequence.
And he didn't actually claim to.
But he does point out that that is provable, but such a proof will
seem "wrong" to those like you, even though it is actually correct.
After all, if we have a method to compute the enumeration of the
computable numbers, that means we have a method to compute the kth
digits of the nth number.
And thus we can, from that method, build a machine to compute the
diagonal or the anti-diagonal by using that method, finding the kth
digit of the kth number and output it or its opposite.
We can then ask the question, which number in this sequence is that
anti-diagonal, which since we just built the computation that computes
it, must be in the list. But it can't be in the list, as if it was the
kth number, its kth digit differs from that anti-diagonal which it was
supposed to be.
Your problem is you keep on living in a world where you don't need to
prove things but can just assume them true, and proof is for some time
later.
Your enumeration MUST be incomplete, and thus isn't what you want it
to be, no matter how much you pray to the purple magic fairy dust
powered unicorn for it to be.
the diagonalization argument is only _one_ of the proofs against the effective enumerability of computable numbers,
because turing equates enumerating computable numbers with that of circle-free machines, the paradox forms the _second_
realizing the fallacy of equating the problems refutes the _second_
proof not the first. the first is addressed in my other response to you
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-
free machines which compute that particular sequence. Not only can >>>>>> various machines differ significantly in the specific steps to
produce the same output, machines can be changed in superficial
ways that do not meaningfully affect the steps of computation,
akin to modern no-op statements or unreachable code
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine that >>>>>> computes any given computable sequences. While identifying more
than one can certainly be done, it is _not_ a requirement for
enumerating computable sequences, as _one_ machine computing a
sequence / suffices to output any and all digits of that sequence/ >>>>>>
The problem of enumerating computable sequences is therefore _not_ >>>>>> actually equivalent to a _general process_ of enumerating circle- >>>>>> free machines, as there is no need to identify all circle-free
machines which compute any given computable sequence
Which just shows that you don't understand what the word
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can
have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle-
free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining
_at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in
respect to circle- free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences. As the problem >>>>>> of enumerating / all circle-free machines/ is _not_ equivalent to >>>>>> that of enumerating / just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a >>>>> PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of
the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least >>>>> one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have
no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each
number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values
accepted to this point, and then it outputs the opposite digit of
the kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates >>>>> and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we
want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition >>>>> 3, there must exist a finite number n that represents a machine
that generates it that PRD will accept, and thus our machine will
simulate for k digits (which will be less than n) and output the
opposite value.
Thus, whatever n you want to claim is the machine that generates
the same computable number doesn't, and thus there can not exist a
PRD that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made. >>>>
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness that
stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change it
to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal computation >>>> that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each output
anti_digit = 1-digit // of the machine fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a non-circle free input.
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H
will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the requirement
you gave it, so fixed_H did not actually compute a digonal of an
enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the
machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D
can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only partial,
and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H.
for this i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here???
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program DID,
at least it would if PRD exists.
Thus, your PRD disappear in the same contradiction that Turing_D and Turing_H does, and thus so does your fixed_H--
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you don't
get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the anti-
diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts is what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247, >>>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>> respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably undecidable, >>>>>>> that must mean any “better-defined” machine that also satisfies >>>>>>> those “ill-defined” criteria must be provably undecidable.
the "better-defined" machine don't satisfy the criteria to be
undecidable
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in which case >>>>> your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of them
as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he must
have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine or even
using the same basic algorithm.
He doesn't say the two machines generated by the two problems are in
any way equivalent, he says that the PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable sequences,
then u could use that solution to determine whether any given machine
is circle-free ...
No, he his saying the problems are equivalent as to the nature
and if so could be used to enumerate the circle-free machines,
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
which implies the sets are equivalent, or contain the same number of
elements.
How do you get that? They aren't even counting the same sort of thing.
And the set ARE the same size, Countably infinite.
Just like the Natural Numbers, and many subsets of it like the evens,
the odds, the primes, the perfect numbers and such. ALL the sets have
"the same number of elements" even though some a proper subsets of others.
this is a fallacy, as circle-free machines forms a surjection onto
computable numbers
As do the even numbers to the whole set of Natural Numbers.
You don't seem to understand the nature of infinite sets.
specifically, there are *infinitely* many circle-free machines for
each *computable sequence*. circle-free machines forms a /
surjection/ onto computational sequences, not bijection. it's a many
to one relationship, not equatable.
enumerating circle-free machines requires enumerating *all* possible
circular machines generally,
but enumerating only computable numbers only *requires* a
categorical subset of circle-free machines, and strictly so for any
given enumeration. only _one_ circle-machine per computable sequence
is required to enumerate the sequence, and you _cannot_ "enumerate"
_more than one_ per every computable sequence. enumerating _only_
computable sequences is a _lesser_ problem than enumerating circle-
free machines
But only fractionally so, but since both problems are infinitely
impossible, that fraction doesn't matter.
the contradiction turing demonstrated is only guaranteed to exist
when totally enumerating out circle-machines, with a turing machine,
he did not prove the problem exists while enumerating only *one*
circle- free machine per computable sequence.
And he didn't actually claim to.
But he does point out that that is provable, but such a proof will
seem "wrong" to those like you, even though it is actually correct.
After all, if we have a method to compute the enumeration of the
computable numbers, that means we have a method to compute the kth
digits of the nth number.
And thus we can, from that method, build a machine to compute the
diagonal or the anti-diagonal by using that method, finding the kth
digit of the kth number and output it or its opposite.
We can then ask the question, which number in this sequence is that
anti-diagonal, which since we just built the computation that
computes it, must be in the list. But it can't be in the list, as if
it was the kth number, its kth digit differs from that anti-diagonal
which it was supposed to be.
Your problem is you keep on living in a world where you don't need to
prove things but can just assume them true, and proof is for some
time later.
Your enumeration MUST be incomplete, and thus isn't what you want it
to be, no matter how much you pray to the purple magic fairy dust
powered unicorn for it to be.
the diagonalization argument is only _one_ of the proofs against the
effective enumerability of computable numbers,
And it works, so they are not enumerable (in the sense used there).
This means that your claims that they can be is based on error, and your refusal to accept that shows your stupidity.
because turing equates enumerating computable numbers with that of
circle-free machines, the paradox forms the _second_
realizing the fallacy of equating the problems refutes the _second_
proof not the first. the first is addressed in my other response to you
That you don't understand what Turing said, doesn't make it a fallacy.
It seems your arguement is based on the strawman fallacy.
THe people he was writing to, understood the nature of the equivalence
he was writing about, and new how to take the step between the two
problems.
The fat you don't, doesn't make his claim wrong, it puts it over your head.
The fact that you still claim that the machine PROVEN to exist if your
PRD exist, that computes what you admit is uncomputable, but still you
claim your PRD can exist, shows that you are not thinking logically,
because you are fixated on something you don't understand.
On 3/12/26 3:53 PM, Richard Damon wrote:
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-
free machines which compute that particular sequence. Not only
can various machines differ significantly in the specific steps
to produce the same output, machines can be changed in
superficial ways that do not meaningfully affect the steps of
computation, akin to modern no-op statements or unreachable code >>>>>>>
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine
that computes any given computable sequences. While identifying
more than one can certainly be done, it is _not_ a requirement
for enumerating computable sequences, as _one_ machine computing >>>>>>> a sequence / suffices to output any and all digits of that sequence/ >>>>>>>
The problem of enumerating computable sequences is therefore
_not_ actually equivalent to a _general process_ of enumerating
circle- free machines, as there is no need to identify all
circle-free machines which compute any given computable sequence
Which just shows that you don't understand what the word
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can
have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle-
free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining >>>>>>> _at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in >>>>>>> respect to circle- free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences. As the
problem of enumerating / all circle-free machines/ is _not_
equivalent to that of enumerating / just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for
a PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of
the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at
least one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have
no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each >>>>>> number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values >>>>>> accepted to this point, and then it outputs the opposite digit of
the kth digit generated by the nth machine.
This machine must produce a computable number, as it only
simulates and uses the output of machines that PRD accepted, so by >>>>>> 1, PRD answered, and by 2 that machine can be simulated for as
long as we want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by
condition 3, there must exist a finite number n that represents a
machine that generates it that PRD will accept, and thus our
machine will simulate for k digits (which will be less than n) and >>>>>> output the opposite value.
Thus, whatever n you want to claim is the machine that generates
the same computable number doesn't, and thus there can not exist a >>>>>> PRD that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made. >>>>>
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding
eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness
that stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change
it to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal computation >>>>> that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each
output
anti_digit = 1-digit // of the machine
fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay
attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a
non-circle free input.
obviously PRD(anti_fixed_H) cannot return TRUE, so therefore it returns
FALSE
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H
will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
there is:
fixed_anti_fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_fixed_H) { // handle self-ref
N += 1
continue // skip including itself
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
this computes the same thing as anti_fixed_H(), but is decidable by PRD. PRD(fixed_anti_fixed_H) returns TRUE
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
... err yes, the total anti-diagonal is _not_ computable. the _closest_
we can get is a sequence that includes the inverse for all computable sequences _except_ to the anti-diagonal computation itself
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the
requirement you gave it, so fixed_H did not actually compute a digonal
of an enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the
machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D
can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only
partial, and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H.
for this i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here???
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program
DID, at least it would if PRD exists.
no it didn't. it skips computing an inverse to it's own computation
Thus, your PRD disappear in the same contradiction that Turing_D and
Turing_H does, and thus so does your fixed_H
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you
don't get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the
anti- diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts
is what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247, >>>>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>>> respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably undecidable, >>>>>>>> that must mean any “better-defined” machine that also satisfies >>>>>>>> those “ill-defined” criteria must be provably undecidable.
the "better-defined" machine don't satisfy the criteria to be
undecidable
But they’re a subset of the “ill-defined” set that Turing was >>>>>> considering, are they not?
Unless you’re considering an entirely different set, in which case >>>>>> your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of them
as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he must
have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine or even
using the same basic algorithm.
He doesn't say the two machines generated by the two problems are in
any way equivalent, he says that the PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable sequences,
then u could use that solution to determine whether any given machine
is circle-free ...
No, he his saying the problems are equivalent as to the nature
and if so could be used to enumerate the circle-free machines,
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B
AND
a solution to B can be used to produce a solution to A
turing is wrong about this. a solution to enumerating circle-free
machines can be used to produce a solution to enumerating computable numbers, but the reverse is *NOT* true
which implies the sets are equivalent, or contain the same number of
elements.
How do you get that? They aren't even counting the same sort of thing.
And the set ARE the same size, Countably infinite.
yet *cirle-free* machines forms a /surjection/ onto *computable sequences*
it's a many to one relationship. the problem of identifying ONE machine
for each computable sequence is not the same as identifying ALL machines
for each computable sequence
Just like the Natural Numbers, and many subsets of it like the evens,
the odds, the primes, the perfect numbers and such. ALL the sets have
"the same number of elements" even though some a proper subsets of
others.
this is a fallacy, as circle-free machines forms a surjection onto
computable numbers
As do the even numbers to the whole set of Natural Numbers.
You don't seem to understand the nature of infinite sets.
specifically, there are *infinitely* many circle-free machines for
each *computable sequence*. circle-free machines forms a /
surjection/ onto computational sequences, not bijection. it's a
many to one relationship, not equatable.
enumerating circle-free machines requires enumerating *all*
possible circular machines generally,
but enumerating only computable numbers only *requires* a
categorical subset of circle-free machines, and strictly so for any >>>>> given enumeration. only _one_ circle-machine per computable
sequence is required to enumerate the sequence, and you _cannot_
"enumerate" _more than one_ per every computable sequence.
enumerating _only_ computable sequences is a _lesser_ problem than
enumerating circle- free machines
But only fractionally so, but since both problems are infinitely
impossible, that fraction doesn't matter.
the contradiction turing demonstrated is only guaranteed to exist
when totally enumerating out circle-machines, with a turing machine, >>>>>
he did not prove the problem exists while enumerating only *one*
circle- free machine per computable sequence.
And he didn't actually claim to.
But he does point out that that is provable, but such a proof will
seem "wrong" to those like you, even though it is actually correct.
After all, if we have a method to compute the enumeration of the
computable numbers, that means we have a method to compute the kth
digits of the nth number.
And thus we can, from that method, build a machine to compute the
diagonal or the anti-diagonal by using that method, finding the kth
digit of the kth number and output it or its opposite.
We can then ask the question, which number in this sequence is that
anti-diagonal, which since we just built the computation that
computes it, must be in the list. But it can't be in the list, as if
it was the kth number, its kth digit differs from that anti-diagonal
which it was supposed to be.
Your problem is you keep on living in a world where you don't need
to prove things but can just assume them true, and proof is for some
time later.
Your enumeration MUST be incomplete, and thus isn't what you want it
to be, no matter how much you pray to the purple magic fairy dust
powered unicorn for it to be.
the diagonalization argument is only _one_ of the proofs against the
effective enumerability of computable numbers,
And it works, so they are not enumerable (in the sense used there).
u did not show that PRD can be used to compute an anti-diagonal because
u failed to recognize what happens when the anti-diagonal computation
tries to enumerate itself to produce a digit opposite to what it does
return ...
This means that your claims that they can be is based on error, and
your refusal to accept that shows your stupidity.
because turing equates enumerating computable numbers with that of
circle-free machines, the paradox forms the _second_
realizing the fallacy of equating the problems refutes the _second_
proof not the first. the first is addressed in my other response to you
That you don't understand what Turing said, doesn't make it a fallacy.
It seems your arguement is based on the strawman fallacy.
THe people he was writing to, understood the nature of the equivalence
he was writing about, and new how to take the step between the two
problems.
The fat you don't, doesn't make his claim wrong, it puts it over your
head.
The fact that you still claim that the machine PROVEN to exist if your
PRD exist, that computes what you admit is uncomputable, but still you
claim your PRD can exist, shows that you are not thinking logically,
because you are fixated on something you don't understand.
On 3/12/26 3:53 PM, Richard Damon wrote:
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle- >>>>>>> free machines which compute that particular sequence. Not only
can various machines differ significantly in the specific steps >>>>>>> to produce the same output, machines can be changed in
superficial ways that do not meaningfully affect the steps of
computation, akin to modern no-op statements or unreachable code >>>>>>>
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine
that computes any given computable sequences. While identifying >>>>>>> more than one can certainly be done, it is _not_ a requirement
for enumerating computable sequences, as _one_ machine computing >>>>>>> a sequence / suffices to output any and all digits of that sequence/ >>>>>>>
The problem of enumerating computable sequences is therefore
_not_ actually equivalent to a _general process_ of enumerating >>>>>>> circle- free machines, as there is no need to identify all
circle-free machines which compute any given computable sequence
Which just shows that you don't understand what the word
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can
have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle-
free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining >>>>>>> _at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247, >>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in >>>>>>> respect to circle- free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences. As the
problem of enumerating / all circle-free machines/ is _not_
equivalent to that of enumerating / just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for >>>>>> a PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of >>>>>> the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at
least one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have >>>>> no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each >>>>>> number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values >>>>>> accepted to this point, and then it outputs the opposite digit of >>>>>> the kth digit generated by the nth machine.
This machine must produce a computable number, as it only
simulates and uses the output of machines that PRD accepted, so by >>>>>> 1, PRD answered, and by 2 that machine can be simulated for as
long as we want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by
condition 3, there must exist a finite number n that represents a >>>>>> machine that generates it that PRD will accept, and thus our
machine will simulate for k digits (which will be less than n) and >>>>>> output the opposite value.
Thus, whatever n you want to claim is the machine that generates
the same computable number doesn't, and thus there can not exist a >>>>>> PRD that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping >>>>> someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made. >>>>>
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding
eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness
that stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change
it to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal computation >>>>> that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit >>>>> K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each
output
anti_digit = 1-digit // of the machine
fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay
attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a
non-circle free input.
obviously PRD(anti_fixed_H) cannot return TRUE, so therefore it returns FALSE
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H
will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
there is:
fixed_anti_fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_fixed_H) { // handle self-ref
N += 1
continue // skip including itself
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
this computes the same thing as anti_fixed_H(), but is decidable by PRD. PRD(fixed_anti_fixed_H) returns TRUE
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
... err yes, the total anti-diagonal is _not_ computable. the _closest_
we can get is a sequence that includes the inverse for all computable sequences _except_ to the anti-diagonal computation itself
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the
requirement you gave it, so fixed_H did not actually compute a digonal
of an enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the
machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D
can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only
partial, and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H.
for this i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here???
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program
DID, at least it would if PRD exists.
no it didn't. it skips computing an inverse to it's own computation
Thus, your PRD disappear in the same contradiction that Turing_D and
Turing_H does, and thus so does your fixed_H
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you
don't get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the
anti- diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts
is what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
the "better-defined" machine don't satisfy the criteria to be >>>>>>>> undecidable
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247, >>>>>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>>>> respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>> impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably undecidable,
that must mean any “better-defined” machine that also satisfies >>>>>>>>> those “ill-defined” criteria must be provably undecidable. >>>>>>>>
But they’re a subset of the “ill-defined” set that Turing was >>>>>>> considering, are they not?
Unless you’re considering an entirely different set, in which case >>>>>>> your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of them >>>>>> as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he must >>>>> have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine or
even using the same basic algorithm.
He doesn't say the two machines generated by the two problems are
in any way equivalent, he says that the PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable sequences,
then u could use that solution to determine whether any given
machine is circle-free ...
No, he his saying the problems are equivalent as to the nature
and if so could be used to enumerate the circle-free machines,
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they produce the
same "answer".
Since the problem is the question of "Can" you do something,
turing is wrong about this. a solution to enumerating circle-free
machines can be used to produce a solution to enumerating computable
numbers, but the reverse is *NOT* true
But it doesn't need to.
You are just falling into the definist fallacy, as you are not using the right definition, but the one YOU want to use.
which implies the sets are equivalent, or contain the same number of
elements.
How do you get that? They aren't even counting the same sort of thing.
And the set ARE the same size, Countably infinite.
yet *cirle-free* machines forms a /surjection/ onto *computable
sequences*
it's a many to one relationship. the problem of identifying ONE
machine for each computable sequence is not the same as identifying
ALL machines for each computable sequence
But both sets are the same size.
And he didn't say it was the SAME problem, just an equivalent one.
Just like the Natural Numbers, and many subsets of it like the evens,
the odds, the primes, the perfect numbers and such. ALL the sets have
"the same number of elements" even though some a proper subsets of
others.
this is a fallacy, as circle-free machines forms a surjection onto
computable numbers
As do the even numbers to the whole set of Natural Numbers.
You don't seem to understand the nature of infinite sets.
specifically, there are *infinitely* many circle-free machines for >>>>>> each *computable sequence*. circle-free machines forms a /
surjection/ onto computational sequences, not bijection. it's a
many to one relationship, not equatable.
enumerating circle-free machines requires enumerating *all*
possible circular machines generally,
but enumerating only computable numbers only *requires* a
categorical subset of circle-free machines, and strictly so for
any given enumeration. only _one_ circle-machine per computable
sequence is required to enumerate the sequence, and you _cannot_
"enumerate" _more than one_ per every computable sequence.
enumerating _only_ computable sequences is a _lesser_ problem than >>>>>> enumerating circle- free machines
But only fractionally so, but since both problems are infinitely
impossible, that fraction doesn't matter.
the contradiction turing demonstrated is only guaranteed to exist >>>>>> when totally enumerating out circle-machines, with a turing machine, >>>>>>
he did not prove the problem exists while enumerating only *one*
circle- free machine per computable sequence.
And he didn't actually claim to.
But he does point out that that is provable, but such a proof will
seem "wrong" to those like you, even though it is actually correct.
After all, if we have a method to compute the enumeration of the
computable numbers, that means we have a method to compute the kth
digits of the nth number.
And thus we can, from that method, build a machine to compute the
diagonal or the anti-diagonal by using that method, finding the kth >>>>> digit of the kth number and output it or its opposite.
We can then ask the question, which number in this sequence is that >>>>> anti-diagonal, which since we just built the computation that
computes it, must be in the list. But it can't be in the list, as
if it was the kth number, its kth digit differs from that anti-
diagonal which it was supposed to be.
Your problem is you keep on living in a world where you don't need
to prove things but can just assume them true, and proof is for
some time later.
Your enumeration MUST be incomplete, and thus isn't what you want
it to be, no matter how much you pray to the purple magic fairy
dust powered unicorn for it to be.
the diagonalization argument is only _one_ of the proofs against the
effective enumerability of computable numbers,
And it works, so they are not enumerable (in the sense used there).
u did not show that PRD can be used to compute an anti-diagonal
because u failed to recognize what happens when the anti-diagonal
computation tries to enumerate itself to produce a digit opposite to
what it does return ...
Well, if PRD selects the anti-diagonal program as an acceptable program
and it fails, then PRD just failed, as it can only accept programs that
are cycle-free.
The problem is by your specification, it needs to select some machine
that produces the exact same computation, but that machine WILL be the
kth one selected, and its kth digit MUST be different then the anti- diagonal machine, so there can't be such a machine selected.
Thus your PRD fails at the task of selecting AT LEAST ONE machine that computes every computable number.
You aren't paying attention to what the problem is.
This just shows your fundamental error in how you reason.
This means that your claims that they can be is based on error, and
your refusal to accept that shows your stupidity.
That you don't understand what Turing said, doesn't make it a fallacy.
because turing equates enumerating computable numbers with that of
circle-free machines, the paradox forms the _second_
realizing the fallacy of equating the problems refutes the _second_
proof not the first. the first is addressed in my other response to you >>>
It seems your arguement is based on the strawman fallacy.
THe people he was writing to, understood the nature of the
equivalence he was writing about, and new how to take the step
between the two problems.
The fat you don't, doesn't make his claim wrong, it puts it over your
head.
The fact that you still claim that the machine PROVEN to exist if
your PRD exist, that computes what you admit is uncomputable, but
still you claim your PRD can exist, shows that you are not thinking
logically, because you are fixated on something you don't understand.
On 3/12/26 7:18 PM, Mike Terry wrote:
On 13/03/2026 00:20, Ben Bacarisse wrote:
Alan Mackenzie <acm@muc.de> writes:
[ Followup-To: set ]
In comp.theory dart200 <user7160@newsgrouper.org.invalid> wrote:
The following claim from p246 of Turings seminal paper On Computable >>>>> Numbers is a fallacy:
You've missed something out. That something is something like "If I
understand correctly" or "As far as I can see". Without such a
qualification, your statement just looks like extreme hubris.
It is vanishingly unlikely that you have understood correctly or you
have seen far enough. If there were a flaw in Turing's 1936 paper, and >>>> it were subtle enough to escape detection by the millions of specialists >>>> who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
And then there's the context. He has chosen to present this startling
discovery in a place similar (and sometimes grander) false claims are
made every single day. It's as if he had dug a huge diamond and has
decided to flog it on a market stall in Dalston.
One problem here is that dart200 chooses to criticise Turing's historical paper, and most
responders are not familiar with the (by modern standards) clunky terminology, and have possibly
not even read the paper(?). And dart200 seems not to be good at following the thread of
reasoning, and cannot "fill in gaps in reasoning" himself, as he lacks the mathematical maturity
(no shame in this - Turing was writing for his peers). So while he may have identified a
short-falling in Turing's "presentation" of the material, he misunderstands the impact on the
bigger picture of the paper.
i mean, the any fallacy like that exists should be a big deal no???
Anyhow, what's going on on those couple of pages? Assuming I'm following correctly:
Turing:
| 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
Problem A: (computing an) enumeration of (all) computable sequences
Problem B: computing whether any number is the D.N of a circle-free machine >> Claim: There is no machine that solves B.
Turing suggests that Problems A and B are equivalent. Indeed they are "logically" equivalent, in
the sense that both can be proven unsolvable
turing only proved B unsolvable
(by TMs), but normally I'd take Turing's phrasing as suggest that there are clear arguments that
relate a given solution to Problem A to a / consequential/ solution to Problem B, and vice-versa.
Is this the case?
Clearly if we have a solution to Problem B, we can construct /from it/ a solution to Problem A.
Turing effectively does this on page 247. So we have one direction covered, no problem.
we actually he doesn't do that really.
he doesn't produce an enumeration of computable sequence which would require deduplicating
equivalent machines such that each sequence only appears once...
he stopped at the fact that Problem B is uncomputable
But if we have a TM solving Problem A, how would we construct /from it/ a TM solving Problem B? I
don't see an obvious argument working along those lines. dart200 seems to be mostly making this
point. I doubt
correct
Turing had such an argument, unless I'm missing something obvious. (Hey, happens...)
he doesn't
OTOH, we can show /regardless of Problem A/ that Problem B has no solution, which is the specific
claim Turing is discussing on the next page. He is not presenting a proof that Problem A has no
solution!
So on the next page (247), Turing presents a proof that there is no machine which solves Problem
B. He naturally starts with assuming such a machind D exists, and from that constructs a new
machine H, and ultimately reaches a contradiction in its behaviour, finally concluding "...Thus
both verdicts are impossible, and we conclude that there can be no such machine D".
That's clear enough - the proof is showing there's no solution to Problem B. dart200 makes a big
point (his "diamond"?) that the proof fails to show there is no solution to Problem A, but Turing
does not present it as such, so no big deal!
actually turing makes a big deal about problem A not being solvable on p246, and presents 2 proofs
in support of this:
the diagonal type argument on p246,
and the paradox type argument on p247,
both are fallacies, neither present sufficient evidence that Problem A is unsolvable
Regarding Problem A. the obvious proof for this would follow the lines of the usual diagonal
argument. If we had a machine enumerating a sequence of D.N.s of machines computing computable
numbers, then we could apply the diagonal argument to construct a new machine that computes a new
computable number which is not represented in the original D.N enumeration.
that's the 2nd fallacy turing makes (one even more surprising tbh), that i detailed in a recent
reply to rick if ur curious
i'll post that as another thread too, but essentially the self-referential weirdness that stumped
turing _can_ be fixed for the direct diagonal. one can use a self-reference to avoid infinite
recursion by just returning a hard-coded answer when the diagonal iterates upon itself,
but cannot this _cannot_ be done for the anti-diagonal because one cannot hard code a digit opposite
to what machine does return ... such a concept in nonsense
Worded another way, we can show that any /computable/ enumeration of the computable numbers must
be incomplete. Or more succintly, Problem A has no solution.
Turing does talk a bit about all this on page 246, without really spelling it out in a more formal
proof.
So even if we grant that Turing's presentation could have been a little clearer when he talks
about the two problems being "equivalent", there is no further impact of this on his paper.
dart200 seems to have just lost the thread a bit re where Turing is going, and thinks his
"discovery" is more significant than it really is.
i believe if computable sequences haven't been actually been proven non-enumerable, then the rest of
turing's paper falls apart
On 13/03/2026 04:26, dart200 wrote:
On 3/12/26 7:18 PM, Mike Terry wrote:
On 13/03/2026 00:20, Ben Bacarisse wrote:
Alan Mackenzie <acm@muc.de> writes:
[ Followup-To: set ]
In comp.theory dart200 <user7160@newsgrouper.org.invalid> wrote:
The following claim from p246 of Turing’s seminal paper On Computable >>>>>> Numbers is a fallacy:
You've missed something out. That something is something like "If I >>>>> understand correctly" or "As far as I can see". Without such a
qualification, your statement just looks like extreme hubris.
It is vanishingly unlikely that you have understood correctly or you >>>>> have seen far enough. If there were a flaw in Turing's 1936 paper, >>>>> and
it were subtle enough to escape detection by the millions of
specialists
who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
And then there's the context. He has chosen to present this startling >>>> discovery in a place similar (and sometimes grander) false claims are
made every single day. It's as if he had dug a huge diamond and has
decided to flog it on a market stall in Dalston.
One problem here is that dart200 chooses to criticise Turing's
historical paper, and most responders are not familiar with the (by
modern standards) clunky terminology, and have possibly not even read
the paper(?). And dart200 seems not to be good at following the
thread of reasoning, and cannot "fill in gaps in reasoning" himself,
as he lacks the mathematical maturity (no shame in this - Turing was
writing for his peers). So while he may have identified a short-
falling in Turing's "presentation" of the material, he misunderstands
the impact on the bigger picture of the paper.
i mean, the any fallacy like that exists should be a big deal no???
No, not really. Turing claims that both problem A and B below have no solution. He spells out a proof of B, and points out that the diagonal argument establishes A. The proof for B is fine, and a diagonal
argument does indeed establish that no TM can satisfy A. So both these claims are in fact correct.
You seem to disagree with this, but you simply don't understand what's
going on...
Anyhow, what's going on on those couple of pages? Assuming I'm
following correctly:
Turing:
| 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
Problem A: (computing an) enumeration of (all) computable sequences
Problem B: computing whether any number is the D.N of a circle-free
machine
Claim: There is no machine that solves B.
Turing suggests that Problems A and B are equivalent. Indeed they
are "logically" equivalent, in the sense that both can be proven
unsolvable
turing only proved B unsolvable
(by TMs), but normally I'd take Turing's phrasing as suggest that
there are clear arguments that relate a given solution to Problem A
to a / consequential/ solution to Problem B, and vice-versa. Is this
the case?
Clearly if we have a solution to Problem B, we can construct /from
it/ a solution to Problem A. Turing effectively does this on page
247. So we have one direction covered, no problem.
we actually he doesn't do that really.
he doesn't produce an enumeration of computable sequence which would
require deduplicating equivalent machines such that each sequence only
appears once...
he stopped at the fact that Problem B is uncomputable
But if we have a TM solving Problem A, how would we construct /from
it/ a TM solving Problem B? I don't see an obvious argument working
along those lines. dart200 seems to be mostly making this point. I
doubt
correct
Turing had such an argument, unless I'm missing something obvious.
(Hey, happens...)
he doesn't
OTOH, we can show /regardless of Problem A/ that Problem B has no
solution, which is the specific claim Turing is discussing on the
next page. He is not presenting a proof that Problem A has no solution! >>>
So on the next page (247), Turing presents a proof that there is no
machine which solves Problem B. He naturally starts with assuming
such a machind D exists, and from that constructs a new machine H,
and ultimately reaches a contradiction in its behaviour, finally
concluding "...Thus both verdicts are impossible, and we conclude
that there can be no such machine D".
That's clear enough - the proof is showing there's no solution to
Problem B. dart200 makes a big point (his "diamond"?) that the proof
fails to show there is no solution to Problem A, but Turing does not
present it as such, so no big deal!
actually turing makes a big deal about problem A not being solvable on
p246, and presents 2 proofs in support of this:
the diagonal type argument on p246,
This establishes that Problem A has no solution. He does not really
spell this out in detail, but his readers would understand it.
Actually, TM starts section 8 (p246) with
Turing:
| It may be thought that arguments which prove that the real numbers
| are not enumerable would also prove that the computable numbers and
| sequences cannot be enumerable.
So the "big deal" is pointing out the flaw in that kind of reasoning, because he sees some subtlety there that merits discussion. (Clearly
the computable numbers /are/ enumerable in the sense that there exists a
1-1 correspondence with the natural numbers.) Still, what people
consider the "big deal" is subjective, so it's pointless to argue on
that point...
and the paradox type argument on p247,
He does not present that as a proof that Problem A has no solution - it
is a proof that Problem B (identifying circle-free programs) has no solution.
both are fallacies, neither present sufficient evidence that Problem A
is unsolvable
That is just your misunderstanding. You lack the maths background to
make any judgement on what Turing was actually saying and thinking.
Regarding Problem A. the obvious proof for this would follow the
lines of the usual diagonal argument. If we had a machine
enumerating a sequence of D.N.s of machines computing computable
numbers, then we could apply the diagonal argument to construct a new
machine that computes a new computable number which is not
represented in the original D.N enumeration.
that's the 2nd fallacy turing makes (one even more surprising tbh),
that i detailed in a recent reply to rick if ur curious
No, that argument is fine.
i'll post that as another thread too, but essentially the self-
referential weirdness that stumped turing _can_ be fixed for the
direct diagonal. one can use a self-reference to avoid infinite
recursion by just returning a hard-coded answer when the diagonal
iterates upon itself,
but cannot this _cannot_ be done for the anti-diagonal because one
cannot hard code a digit opposite to what machine does return ... such
a concept in nonsense
Sorry, but that's just gibberish.
Sounds like the sort of thing PO would say!
Worded another way, we can show that any /computable/ enumeration of
the computable numbers must be incomplete. Or more succintly,
Problem A has no solution.
Turing does talk a bit about all this on page 246, without really
spelling it out in a more formal proof.
So even if we grant that Turing's presentation could have been a
little clearer when he talks about the two problems being
"equivalent", there is no further impact of this on his paper.
dart200 seems to have just lost the thread a bit re where Turing is
going, and thinks his "discovery" is more significant than it really is. >>>
i believe if computable sequences haven't been actually been proven
non-enumerable, then the rest of turing's paper falls apart
But computable sequences /can't/ be (programatically) enumerated as a diagonal argument shows. They are enumerable in the set-theoretic
sense, just not the "computable" sense.
Perhaps you should identify further parts of his paper that you think
have "fallen apart"? Otherwise this just reads as you being confused,
and certainly no big deal. (I probably won't respond any further...)
Mike.
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
the "better-defined" machine don't satisfy the criteria to be >>>>>>>>> undecidable
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247, >>>>>>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>>>>> respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>>> impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably >>>>>>>>>> undecidable,
that must mean any “better-defined” machine that also satisfies >>>>>>>>>> those “ill-defined” criteria must be provably undecidable. >>>>>>>>>
But they’re a subset of the “ill-defined” set that Turing was >>>>>>>> considering, are they not?
Unless you’re considering an entirely different set, in which case >>>>>>>> your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of
them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he
must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine or
even using the same basic algorithm.
He doesn't say the two machines generated by the two problems are >>>>>> in any way equivalent, he says that the PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable sequences, >>>>> then u could use that solution to determine whether any given
machine is circle-free ...
No, he his saying the problems are equivalent as to the nature
and if so could be used to enumerate the circle-free machines,
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they produce the
same "answer".
Since the problem is the question of "Can" you do something,
turing is wrong about this. a solution to enumerating circle-free
machines can be used to produce a solution to enumerating computable
numbers, but the reverse is *NOT* true
But it doesn't need to.
yes it does, rick
You are just falling into the definist fallacy, as you are not using
the right definition, but the one YOU want to use.
which implies the sets are equivalent, or contain the same number
of elements.
How do you get that? They aren't even counting the same sort of thing. >>>>
And the set ARE the same size, Countably infinite.
yet *cirle-free* machines forms a /surjection/ onto *computable
sequences*
it's a many to one relationship. the problem of identifying ONE
machine for each computable sequence is not the same as identifying
ALL machines for each computable sequence
But both sets are the same size.
they are the same cardinality, rick, but that does _not_ imply they are
the same problem. in fact _all_ computable sequences are the same cardinality, but clearly they do not all represent the _same_ problem...
a solution to the problem of computable numbers only needs to recognize *ONE* circle-free machine per computable number, not *ALL* circle-free machines per comparable number
recognizing a *subset* is _NOT_ the same problem as recognizing a
*whole* set
that's just a fallacy whether accept it or not rick
And he didn't say it was the SAME problem, just an equivalent one.
Just like the Natural Numbers, and many subsets of it like the
evens, the odds, the primes, the perfect numbers and such. ALL the
sets have "the same number of elements" even though some a proper
subsets of others.
this is a fallacy, as circle-free machines forms a surjection onto
computable numbers
As do the even numbers to the whole set of Natural Numbers.
You don't seem to understand the nature of infinite sets.
specifically, there are *infinitely* many circle-free machines
for each *computable sequence*. circle-free machines forms a /
surjection/ onto computational sequences, not bijection. it's a >>>>>>> many to one relationship, not equatable.
enumerating circle-free machines requires enumerating *all*
possible circular machines generally,
but enumerating only computable numbers only *requires* a
categorical subset of circle-free machines, and strictly so for >>>>>>> any given enumeration. only _one_ circle-machine per computable >>>>>>> sequence is required to enumerate the sequence, and you _cannot_ >>>>>>> "enumerate" _more than one_ per every computable sequence.
enumerating _only_ computable sequences is a _lesser_ problem
than enumerating circle- free machines
But only fractionally so, but since both problems are infinitely
impossible, that fraction doesn't matter.
the contradiction turing demonstrated is only guaranteed to exist >>>>>>> when totally enumerating out circle-machines, with a turing machine, >>>>>>>
he did not prove the problem exists while enumerating only *one* >>>>>>> circle- free machine per computable sequence.
And he didn't actually claim to.
But he does point out that that is provable, but such a proof will >>>>>> seem "wrong" to those like you, even though it is actually correct. >>>>>>
After all, if we have a method to compute the enumeration of the
computable numbers, that means we have a method to compute the kth >>>>>> digits of the nth number.
And thus we can, from that method, build a machine to compute the >>>>>> diagonal or the anti-diagonal by using that method, finding the
kth digit of the kth number and output it or its opposite.
We can then ask the question, which number in this sequence is
that anti-diagonal, which since we just built the computation that >>>>>> computes it, must be in the list. But it can't be in the list, as >>>>>> if it was the kth number, its kth digit differs from that anti-
diagonal which it was supposed to be.
Your problem is you keep on living in a world where you don't need >>>>>> to prove things but can just assume them true, and proof is for
some time later.
Your enumeration MUST be incomplete, and thus isn't what you want >>>>>> it to be, no matter how much you pray to the purple magic fairy
dust powered unicorn for it to be.
the diagonalization argument is only _one_ of the proofs against
the effective enumerability of computable numbers,
And it works, so they are not enumerable (in the sense used there).
u did not show that PRD can be used to compute an anti-diagonal
because u failed to recognize what happens when the anti-diagonal
computation tries to enumerate itself to produce a digit opposite to
what it does return ...
Well, if PRD selects the anti-diagonal program as an acceptable
program and it fails, then PRD just failed, as it can only accept
programs that are cycle-free.
u have the incredible ability to only acknowledged things that suit ur stance: which is that turing could do no wrong,
cause regardless, *the anti-diagonal was not made computable by assuming
the computability of the diagonal*
that assumption was a fallacy, and similarly subverts turing's papers
moving the goal post in how to deal with these almost anti-diagonals is
a question for further research that i have not yet dealt with yet.
i've come this far, it's hard for me to believe this will be the thing
that finally stumbles me into what ... giving up???
is that what u want rick??? for me to come this far in pointing out
flaws and give up now???
lol
keep up that copium that ur apparently beloved theory doesn't have
massive shitpile of fallacy at it's core...
The problem is by your specification, it needs to select some machine
that produces the exact same computation, but that machine WILL be the
kth one selected, and its kth digit MUST be different then the anti-
diagonal machine, so there can't be such a machine selected.
Thus your PRD fails at the task of selecting AT LEAST ONE machine that
computes every computable number.
You aren't paying attention to what the problem is.
This just shows your fundamental error in how you reason.
This means that your claims that they can be is based on error, and
your refusal to accept that shows your stupidity.
because turing equates enumerating computable numbers with that of
circle-free machines, the paradox forms the _second_
realizing the fallacy of equating the problems refutes the _second_ >>>>> proof not the first. the first is addressed in my other response to >>>>> you
That you don't understand what Turing said, doesn't make it a fallacy. >>>>
It seems your arguement is based on the strawman fallacy.
THe people he was writing to, understood the nature of the
equivalence he was writing about, and new how to take the step
between the two problems.
The fat you don't, doesn't make his claim wrong, it puts it over
your head.
The fact that you still claim that the machine PROVEN to exist if
your PRD exist, that computes what you admit is uncomputable, but
still you claim your PRD can exist, shows that you are not thinking
logically, because you are fixated on something you don't understand.
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in
respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably
undecidable, that must mean any “better-defined” machine that
also satisfies those “ill-defined” criteria must be provably
undecidable.
the "better-defined" machine don't satisfy the criteria to be
undecidable
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
On Fri, 13 Mar 2026 00:30:24 -0700, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in
respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably
undecidable, that must mean any “better-defined” machine that
also satisfies those “ill-defined” criteria must be provably
undecidable.
the "better-defined" machine don't satisfy the criteria to be
undecidable
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
So where does the “ill-defined machine” come in? That was the one you
said “is found to be undecidable”. If your machine is “better-defined”, that that means it must satisfy the same criteria,
and then some.
Those original same criteria would render your machine to be just as undecidable as Turing’s. QED.
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
the "better-defined" machine don't satisfy the criteria to be >>>>>>>>>> undecidable
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247, >>>>>>>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>>>>>> respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>>>> impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably >>>>>>>>>>> undecidable,
that must mean any “better-defined” machine that also satisfies >>>>>>>>>>> those “ill-defined” criteria must be provably undecidable. >>>>>>>>>>
But they’re a subset of the “ill-defined” set that Turing was >>>>>>>>> considering, are they not?
Unless you’re considering an entirely different set, in which case >>>>>>>>> your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of >>>>>>>> them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he
must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine or
even using the same basic algorithm.
He doesn't say the two machines generated by the two problems are >>>>>>> in any way equivalent, he says that the PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable
sequences, then u could use that solution to determine whether any >>>>>> given machine is circle-free ...
No, he his saying the problems are equivalent as to the nature
and if so could be used to enumerate the circle-free machines,
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they produce
the same "answer".
Since the problem is the question of "Can" you do something,
turing is wrong about this. a solution to enumerating circle-free
machines can be used to produce a solution to enumerating computable
numbers, but the reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and thus are
trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates circle-ftee machihes could be used to enumerate computable numbers.
In fact, you are just contradicting yourself because you forget what you mean, as having an enumberation of circle-free machines has the
potential to make a list of computable numbers by eliminating the duplicates.
So, eat your own dog food.
You are just falling into the definist fallacy, as you are not using
the right definition, but the one YOU want to use.
which implies the sets are equivalent, or contain the same number >>>>>> of elements.
How do you get that? They aren't even counting the same sort of thing. >>>>>
And the set ARE the same size, Countably infinite.
yet *cirle-free* machines forms a /surjection/ onto *computable
sequences*
it's a many to one relationship. the problem of identifying ONE
machine for each computable sequence is not the same as identifying
ALL machines for each computable sequence
But both sets are the same size.
they are the same cardinality, rick, but that does _not_ imply they
are the same problem. in fact _all_ computable sequences are the same
cardinality, but clearly they do not all represent the _same_ problem...
And who said they were?
SAME is NOT "Equivalent"
a solution to the problem of computable numbers only needs to
recognize *ONE* circle-free machine per computable number, not *ALL*
circle-free machines per comparable number
Right, so is a different machine, but faces the same problem.
recognizing a *subset* is _NOT_ the same problem as recognizing a
*whole* set
Right.
that's just a fallacy whether accept it or not rick
But the fact that you machine doesn't produce the entire set of
computable numbers jsut proves you are wrong.
Your PRD can't produce in its selection of machines that are circle-
free, a machine that produces the number value that anti-fixed-H produces.
And anti-fixed-H, since it only simulates machine confirmed to be circle-free by PRD, does produce a computable number.
Just one that isn't in the claim complete enumeration done by PRD.
And he didn't say it was the SAME problem, just an equivalent one.
Just like the Natural Numbers, and many subsets of it like the
evens, the odds, the primes, the perfect numbers and such. ALL the
sets have "the same number of elements" even though some a proper
subsets of others.
this is a fallacy, as circle-free machines forms a surjection onto >>>>>> computable numbers
As do the even numbers to the whole set of Natural Numbers.
You don't seem to understand the nature of infinite sets.
specifically, there are *infinitely* many circle-free machines >>>>>>>> for each *computable sequence*. circle-free machines forms a / >>>>>>>> surjection/ onto computational sequences, not bijection. it's a >>>>>>>> many to one relationship, not equatable.
enumerating circle-free machines requires enumerating *all*
possible circular machines generally,
but enumerating only computable numbers only *requires* a
categorical subset of circle-free machines, and strictly so for >>>>>>>> any given enumeration. only _one_ circle-machine per computable >>>>>>>> sequence is required to enumerate the sequence, and you _cannot_ >>>>>>>> "enumerate" _more than one_ per every computable sequence.
enumerating _only_ computable sequences is a _lesser_ problem >>>>>>>> than enumerating circle- free machines
But only fractionally so, but since both problems are infinitely >>>>>>> impossible, that fraction doesn't matter.
the contradiction turing demonstrated is only guaranteed to
exist when totally enumerating out circle-machines, with a
turing machine,
he did not prove the problem exists while enumerating only *one* >>>>>>>> circle- free machine per computable sequence.
And he didn't actually claim to.
But he does point out that that is provable, but such a proof
will seem "wrong" to those like you, even though it is actually >>>>>>> correct.
After all, if we have a method to compute the enumeration of the >>>>>>> computable numbers, that means we have a method to compute the
kth digits of the nth number.
And thus we can, from that method, build a machine to compute the >>>>>>> diagonal or the anti-diagonal by using that method, finding the >>>>>>> kth digit of the kth number and output it or its opposite.
We can then ask the question, which number in this sequence is
that anti-diagonal, which since we just built the computation
that computes it, must be in the list. But it can't be in the
list, as if it was the kth number, its kth digit differs from
that anti- diagonal which it was supposed to be.
Your problem is you keep on living in a world where you don't
need to prove things but can just assume them true, and proof is >>>>>>> for some time later.
Your enumeration MUST be incomplete, and thus isn't what you want >>>>>>> it to be, no matter how much you pray to the purple magic fairy >>>>>>> dust powered unicorn for it to be.
the diagonalization argument is only _one_ of the proofs against
the effective enumerability of computable numbers,
And it works, so they are not enumerable (in the sense used there).
u did not show that PRD can be used to compute an anti-diagonal
because u failed to recognize what happens when the anti-diagonal
computation tries to enumerate itself to produce a digit opposite to
what it does return ...
Well, if PRD selects the anti-diagonal program as an acceptable
program and it fails, then PRD just failed, as it can only accept
programs that are cycle-free.
u have the incredible ability to only acknowledged things that suit ur
stance: which is that turing could do no wrong,
cause regardless, *the anti-diagonal was not made computable by
assuming the computability of the diagonal*
Sure it is. Prove that it isn't.
that assumption was a fallacy, and similarly subverts turing's papers
What is the actual ERROR in what I said to do?
I( seems you don't have an answer to that.
moving the goal post in how to deal with these almost anti-diagonals
is a question for further research that i have not yet dealt with yet.
But we don't care about the almost diagonal or anti-diagonal, only the ACTUAL one, as defined by the enumeration your decider generates.
i've come this far, it's hard for me to believe this will be the thing
that finally stumbles me into what ... giving up???
If being wrong doesn't stop you, that just proves that you are just a pathological liar.
is that what u want rick??? for me to come this far in pointing out
flaws and give up now???
better to give up now, and have a chance to find a right path then to
spend. you life like Olcott gaslit by your own lies.
lol
keep up that copium that ur apparently beloved theory doesn't have
massive shitpile of fallacy at it's core...
The massive shitpile is in your work, not the theory. But of course to
shit, logic seems unbearable.
That fact that you don't actually understand what the basic words mean
just show how bad your base is.
The problem is by your specification, it needs to select some machine
that produces the exact same computation, but that machine WILL be
the kth one selected, and its kth digit MUST be different then the
anti- diagonal machine, so there can't be such a machine selected.
Thus your PRD fails at the task of selecting AT LEAST ONE machine
that computes every computable number.
You aren't paying attention to what the problem is.
This just shows your fundamental error in how you reason.
This means that your claims that they can be is based on error, and >>>>> your refusal to accept that shows your stupidity.
because turing equates enumerating computable numbers with that of >>>>>> circle-free machines, the paradox forms the _second_
realizing the fallacy of equating the problems refutes the
_second_ proof not the first. the first is addressed in my other
response to you
That you don't understand what Turing said, doesn't make it a fallacy. >>>>>
It seems your arguement is based on the strawman fallacy.
THe people he was writing to, understood the nature of the
equivalence he was writing about, and new how to take the step
between the two problems.
The fat you don't, doesn't make his claim wrong, it puts it over
your head.
The fact that you still claim that the machine PROVEN to exist if
your PRD exist, that computes what you admit is uncomputable, but
still you claim your PRD can exist, shows that you are not thinking >>>>> logically, because you are fixated on something you don't understand. >>>>>
On 03/13/2026 10:25 AM, dart200 wrote:
On 3/12/26 3:53 PM, Richard Damon wrote:
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:i'm sorry richard, it does not actually do that
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper OnWhich just shows that you don't understand what the word
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle- >>>>>>>> free machines which compute that particular sequence. Not only >>>>>>>> can various machines differ significantly in the specific steps >>>>>>>> to produce the same output, machines can be changed in
superficial ways that do not meaningfully affect the steps of
computation, akin to modern no-op statements or unreachable code >>>>>>>>
The problem of enumerating computable sequences, however, only >>>>>>>> depends on successfully identifying _one_ circle-free machine
that computes any given computable sequences. While identifying >>>>>>>> more than one can certainly be done, it is _not_ a requirement >>>>>>>> for enumerating computable sequences, as _one_ machine computing >>>>>>>> a sequence / suffices to output any and all digits of that
sequence/
The problem of enumerating computable sequences is therefore
_not_ actually equivalent to a _general process_ of enumerating >>>>>>>> circle- free machines, as there is no need to identify all
circle-free machines which compute any given computable sequence >>>>>>>
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very >>>>>>> different methods and path create the exact same output, we can
have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle-
free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies >>>>>>>> circle-free machines only needs the limited power of determining >>>>>>>> _at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence >>>>>>>>
Because of this fallacy, the proof found on the following p247, >>>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>> respect to circle- free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences. As the
problem of enumerating / all circle-free machines/ is _not_
equivalent to that of enumerating / just computable sequences/ >>>>>>>>
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for >>>>>>> a PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of >>>>>>> the number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever. >>>>>>>
3) For EVERY Computable Number that exist, PRD will accept at
least one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have >>>>>> no proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti- >>>>>>> diagonal by testing each number in sequence with PRD, and for each >>>>>>> number that it accepts, it will simulate that machine until that >>>>>>> machine generates k digits of output, k being the number of values >>>>>>> accepted to this point, and then it outputs the opposite digit of >>>>>>> the kth digit generated by the nth machine.
This machine must produce a computable number, as it only
simulates and uses the output of machines that PRD accepted, so by >>>>>>> 1, PRD answered, and by 2 that machine can be simulated for as
long as we want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by
condition 3, there must exist a finite number n that represents a >>>>>>> machine that generates it that PRD will accept, and thus our
machine will simulate for k digits (which will be less than n) and >>>>>>> output the opposite value.
Thus, whatever n you want to claim is the machine that generates >>>>>>> the same computable number doesn't, and thus there can not exist a >>>>>>> PRD that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers. >>>>>>>
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping >>>>>> someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly
made.
*having a machine that computes a diagonal, does not actually then >>>>>> imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists. >>>>
(please do actually read the whole post in depth before responding
eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness
that stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change
it to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal
computation
that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory >>>>>> output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each
output
anti_digit = 1-digit // of the machine
fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay
attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a
non-circle free input.
obviously PRD(anti_fixed_H) cannot return TRUE, so therefore it returns
FALSE
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H
will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
there is:
fixed_anti_fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_fixed_H) { // handle self-ref
N += 1
continue // skip including itself
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
this computes the same thing as anti_fixed_H(), but is decidable by PRD.
PRD(fixed_anti_fixed_H) returns TRUE
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
... err yes, the total anti-diagonal is _not_ computable. the _closest_
we can get is a sequence that includes the inverse for all computable
sequences _except_ to the anti-diagonal computation itself
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the
requirement you gave it, so fixed_H did not actually compute a digonal
of an enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the >>>>> machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus >>>>> what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D
can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only
partial, and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H.
for this i will combine the two machines for simplicity, i will leave
implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here??? >>>> K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we
need to define what it's digit on the diagonal will be. but if that's
its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program
DID, at least it would if PRD exists.
no it didn't. it skips computing an inverse to it's own computation
Thus, your PRD disappear in the same contradiction that Turing_D and
Turing_H does, and thus so does your fixed_H
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output,
you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you
don't get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the
anti- diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts
is what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
The usual accounts that the infinite is not computable
doesn't contradict that each finite is computable.
Then it gets directly into accounts as modeled by
ordinary set theory.
These are then among matters of "quantifier disambiguation"
against "impredicativity".
There are various models of the Halting Problem which
is also what is being discussed, or the Branching Problem,
that vary, about usually enough "supertasks".
There are many well-known "approximation algorithms
to NP-hard problems" that reduce them to polynomial.
On 3/13/26 6:24 PM, Lawrence D’Oliveiro wrote:
^^^^^^^^^^So where does the “ill-defined machine” come in? That was the one you
he equates the two sets on p246
on p247 he constructs machine H that attempts to construct a diagonal
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
the "better-defined" machine don't satisfy the criteria to be >>>>>>>>>>> undecidable
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following >>>>>>>>>>>>> p247,
where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>>>>>>> respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>>>>> impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably >>>>>>>>>>>> undecidable,
that must mean any “better-defined” machine that also satisfies
those “ill-defined” criteria must be provably undecidable. >>>>>>>>>>>
But they’re a subset of the “ill-defined” set that Turing was >>>>>>>>>> considering, are they not?
Unless you’re considering an entirely different set, in which >>>>>>>>>> case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of >>>>>>>>> them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he >>>>>>>> must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine or >>>>>> even using the same basic algorithm.
He doesn't say the two machines generated by the two problems >>>>>>>> are in any way equivalent, he says that the PROBLEMS are
equivalent,
he's literally saying that if u can enumerate computable
sequences, then u could use that solution to determine whether
any given machine is circle-free ...
No, he his saying the problems are equivalent as to the nature
and if so could be used to enumerate the circle-free machines,
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they produce
the same "answer".
Since the problem is the question of "Can" you do something,
turing is wrong about this. a solution to enumerating circle-free
machines can be used to produce a solution to enumerating
computable numbers, but the reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and thus are
trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates circle-ftee
machihes could be used to enumerate computable numbers.
my god rick, please fucking read the not even whole paper, but at least
the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING PAGES
WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that,
| if this general process exists [for circle-free machines]
| then there is a machine which computes β
do i need to spell out why with even more detail???
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, produce a general process to output whether a machine is circle-free or not
the _first fallacy_ is that because that isn't actually equivalent to enumerating computable sequences (which is a lesser problem that only
needs to recognize a subset of circle-free machines), ruling out a
general process for deciding circle-free machine does _not_ actually
rule out a general process for enumerating computable numbers
the _second fallacy_ is assuming that just because a diagonal is
computable, the anti-diagonal becomes computable
that one seems obvious on the surface, but when we actually dig into the details, one _cannot_ use a diagonal machine to produce an anti-diagonal machine
(it either gets stuck looping trying to find it's own anti-digit, or it skips over itself not producing a total anti-diagonal. there's no way to hard-code a digit for itself on the anti-diagonal, like you can with the diagonal)
In fact, you are just contradicting yourself because you forget what
you mean, as having an enumberation of circle-free machines has the
potential to make a list of computable numbers by eliminating the
duplicates.
So, eat your own dog food.
You are just falling into the definist fallacy, as you are not using
the right definition, but the one YOU want to use.
which implies the sets are equivalent, or contain the same number >>>>>>> of elements.
How do you get that? They aren't even counting the same sort of
thing.
And the set ARE the same size, Countably infinite.
yet *cirle-free* machines forms a /surjection/ onto *computable
sequences*
it's a many to one relationship. the problem of identifying ONE
machine for each computable sequence is not the same as identifying >>>>> ALL machines for each computable sequence
But both sets are the same size.
they are the same cardinality, rick, but that does _not_ imply they
are the same problem. in fact _all_ computable sequences are the same
cardinality, but clearly they do not all represent the _same_ problem...
And who said they were?
SAME is NOT "Equivalent"
a solution to the problem of computable numbers only needs to
recognize *ONE* circle-free machine per computable number, not *ALL*
circle-free machines per comparable number
Right, so is a different machine, but faces the same problem.
turing did not prove that, and neither did you
recognizing a *subset* is _NOT_ the same problem as recognizing a
*whole* set
Right.
that's just a fallacy whether accept it or not rick
But the fact that you machine doesn't produce the entire set of
computable numbers jsut proves you are wrong.
Your PRD can't produce in its selection of machines that are circle-
free, a machine that produces the number value that anti-fixed-H
produces.
goal posts rick. i'm currently just demonstrating fallacies in turing's proof
i haven't proven that a single machine can enumerate all computable sequences, that's still just a working thesis
right now the almost anti-diagonals seem troubling, but i haven't
actually put time into reckoning about it
And anti-fixed-H, since it only simulates machine confirmed to be
circle-free by PRD, does produce a computable number.
Just one that isn't in the claim complete enumeration done by PRD.
And he didn't say it was the SAME problem, just an equivalent one.
u did not show that PRD can be used to compute an anti-diagonal
Just like the Natural Numbers, and many subsets of it like the
evens, the odds, the primes, the perfect numbers and such. ALL the >>>>>> sets have "the same number of elements" even though some a proper >>>>>> subsets of others.
this is a fallacy, as circle-free machines forms a surjection
onto computable numbers
As do the even numbers to the whole set of Natural Numbers.
You don't seem to understand the nature of infinite sets.
specifically, there are *infinitely* many circle-free machines >>>>>>>>> for each *computable sequence*. circle-free machines forms a / >>>>>>>>> surjection/ onto computational sequences, not bijection. it's a >>>>>>>>> many to one relationship, not equatable.
enumerating circle-free machines requires enumerating *all* >>>>>>>>> possible circular machines generally,
but enumerating only computable numbers only *requires* a
categorical subset of circle-free machines, and strictly so for >>>>>>>>> any given enumeration. only _one_ circle-machine per computable >>>>>>>>> sequence is required to enumerate the sequence, and you
_cannot_ "enumerate" _more than one_ per every computable
sequence. enumerating _only_ computable sequences is a _lesser_ >>>>>>>>> problem than enumerating circle- free machines
But only fractionally so, but since both problems are infinitely >>>>>>>> impossible, that fraction doesn't matter.
the contradiction turing demonstrated is only guaranteed to >>>>>>>>> exist when totally enumerating out circle-machines, with a
turing machine,
he did not prove the problem exists while enumerating only
*one* circle- free machine per computable sequence.
And he didn't actually claim to.
But he does point out that that is provable, but such a proof >>>>>>>> will seem "wrong" to those like you, even though it is actually >>>>>>>> correct.
After all, if we have a method to compute the enumeration of the >>>>>>>> computable numbers, that means we have a method to compute the >>>>>>>> kth digits of the nth number.
And thus we can, from that method, build a machine to compute >>>>>>>> the diagonal or the anti-diagonal by using that method, finding >>>>>>>> the kth digit of the kth number and output it or its opposite. >>>>>>>>
We can then ask the question, which number in this sequence is >>>>>>>> that anti-diagonal, which since we just built the computation >>>>>>>> that computes it, must be in the list. But it can't be in the >>>>>>>> list, as if it was the kth number, its kth digit differs from >>>>>>>> that anti- diagonal which it was supposed to be.
Your problem is you keep on living in a world where you don't >>>>>>>> need to prove things but can just assume them true, and proof is >>>>>>>> for some time later.
Your enumeration MUST be incomplete, and thus isn't what you
want it to be, no matter how much you pray to the purple magic >>>>>>>> fairy dust powered unicorn for it to be.
the diagonalization argument is only _one_ of the proofs against >>>>>>> the effective enumerability of computable numbers,
And it works, so they are not enumerable (in the sense used there). >>>>>
because u failed to recognize what happens when the anti-diagonal
computation tries to enumerate itself to produce a digit opposite
to what it does return ...
Well, if PRD selects the anti-diagonal program as an acceptable
program and it fails, then PRD just failed, as it can only accept
programs that are cycle-free.
u have the incredible ability to only acknowledged things that suit
ur stance: which is that turing could do no wrong,
cause regardless, *the anti-diagonal was not made computable by
assuming the computability of the diagonal*
Sure it is. Prove that it isn't.
please reread the psuedo-code again
that assumption was a fallacy, and similarly subverts turing's papers
What is the actual ERROR in what I said to do?
I( seems you don't have an answer to that.
moving the goal post in how to deal with these almost anti-diagonals
is a question for further research that i have not yet dealt with yet.
But we don't care about the almost diagonal or anti-diagonal, only the
ACTUAL one, as defined by the enumeration your decider generates.
i've come this far, it's hard for me to believe this will be the
thing that finally stumbles me into what ... giving up???
If being wrong doesn't stop you, that just proves that you are just a
pathological liar.
i have gone through many various phases of different wrongs my convos
with you, it's about redirecting effort and being flexible obviously,
ur being black and white about world that is in actuality incredibly
full of color
try not to be so dull eh???
is that what u want rick??? for me to come this far in pointing out
flaws and give up now???
better to give up now, and have a chance to find a right path then to
spend. you life like Olcott gaslit by your own lies.
not until i have the proof richard ¯\_(ツ)_/¯
lol
keep up that copium that ur apparently beloved theory doesn't have
massive shitpile of fallacy at it's core...
The massive shitpile is in your work, not the theory. But of course to
shit, logic seems unbearable.
That fact that you don't actually understand what the basic words mean
just show how bad your base is.
The problem is by your specification, it needs to select some
machine that produces the exact same computation, but that machine
WILL be the kth one selected, and its kth digit MUST be different
then the anti- diagonal machine, so there can't be such a machine
selected.
Thus your PRD fails at the task of selecting AT LEAST ONE machine
that computes every computable number.
You aren't paying attention to what the problem is.
This just shows your fundamental error in how you reason.
This means that your claims that they can be is based on error,
and your refusal to accept that shows your stupidity.
because turing equates enumerating computable numbers with that >>>>>>> of circle-free machines, the paradox forms the _second_
realizing the fallacy of equating the problems refutes the
_second_ proof not the first. the first is addressed in my other >>>>>>> response to you
That you don't understand what Turing said, doesn't make it a
fallacy.
It seems your arguement is based on the strawman fallacy.
THe people he was writing to, understood the nature of the
equivalence he was writing about, and new how to take the step
between the two problems.
The fat you don't, doesn't make his claim wrong, it puts it over
your head.
The fact that you still claim that the machine PROVEN to exist if >>>>>> your PRD exist, that computes what you admit is uncomputable, but >>>>>> still you claim your PRD can exist, shows that you are not
thinking logically, because you are fixated on something you don't >>>>>> understand.
On 3/13/26 6:24 PM, Lawrence D’Oliveiro wrote:
On Fri, 13 Mar 2026 00:30:24 -0700, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to
compute the direct diagonal β’) is found to be undecidable in >>>>>>> respect to circle-free decider 𝓓; does not then prove an
impossibility for enumerating computable sequences.
But if the machine can be “ill-defined”, yet provably
undecidable, that must mean any “better-defined” machine that
also satisfies those “ill-defined” criteria must be provably
undecidable.
the "better-defined" machine don't satisfy the criteria to be
undecidable
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
So where does the “ill-defined machine” come in? That was the one you
he equates the two sets on p246
on p247 he constructs machine H that attempts to construct a diagonal
across all circle-free machines, called H, using the circle-free decider
D to disprove the effective enumerability of circle-free machines (and
in his view therefore the effective enumerability of computable sequences)
the pseudo-code for said machine is:
H = () -> {
N = 0
K = 0
do {
if (D(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
it's found undecidable because:
if D(H) => TRUE, then H will try to simulate it's own Kth digit, which
is not defined anywhere, and get caught in an infinite loop (a
"circular" machine)
if D(H) => FALSE, then H will skip trying to simulate it's own kth
digit, and proceed in a circle-free fashion, albeit without actually computing a full diagonal
the reason i said H is ill-defined is because as it stands it has not
been specified well enough to implement, as D itself has not been
specified well enough to implement. the idiosyncrasy of self-referential computing is held up as a contradiction and no further work is done to refine the specification... so it's not actually handled by the
interface, and D as it stands cannot be implemented. and therefore
neither can H.
neither represent actual machines in the total machine enumeration
said “is found to be undecidable”. If your machine is
“better-defined”, that that means it must satisfy the same criteria,
for a corrected D i propose one where /undecidable input/ (which is
input that cannot be assigned a TRUE or FALSE value based on the
semantic property), is merged in with FALSE to create a *partial_recognizer_D* (PRD)
PRD can positivity recognize only a fixed subset of circle-free machines using TRUE, and does not recognize any circular machines using FALSE, a trade off made so it doesn't get stuck on machines that are not
recognizable by it one way or the other, _and can actually exist_
a major operational thesis i'm working with (but admittedly have not
proven yet) is that the fixed subset of circle-free machines which can
be recognized by a single partial recognizer (like PRD) is turning-
complete in that it still encompasses all possible circle-free sequences (remember there are infinite machines which compute any given sequence)
all the corrected H does is use a self-reference to avoid simulating
itself to find that digit which was never defined for itself... and just output a digit for itself on the diagonal (which in turn does define
it's digit on the diagonal)
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // self-ref
output 0 // hard coded digit
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
and then some.
Those original same criteria would render your machine to be just as
undecidable as Turing’s. QED.
no, because the undecidability is independent and superfluous to what
the machine actually computes
On 14/03/2026 04:25, dart200 wrote:
On 3/13/26 6:24 PM, Lawrence D’Oliveiro wrote:
...
^^^^^^^^^^So where does the “ill-defined machine” come in? That was the one you >>he equates the two sets on p246
on p247 he constructs machine H that attempts to construct a diagonal
Perhaps I didn't pay close enough attention but I thought he defined constraints on the variable H rather than constructing a machine.
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
the "better-defined" machine don't satisfy the criteria to >>>>>>>>>>>> be undecidable
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote:
Because of this fallacy, the proof found on the following >>>>>>>>>>>>>> p247,But if the machine can be “ill-defined”, yet provably >>>>>>>>>>>>> undecidable,
where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>>>>>>>> respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>>>>>> impossibility for enumerating computable sequences. >>>>>>>>>>>>>
that must mean any “better-defined” machine that also >>>>>>>>>>>>> satisfies
those “ill-defined” criteria must be provably undecidable. >>>>>>>>>>>>
But they’re a subset of the “ill-defined” set that Turing was >>>>>>>>>>> considering, are they not?
Unless you’re considering an entirely different set, in which >>>>>>>>>>> case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of >>>>>>>>>> them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he >>>>>>>>> must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine or >>>>>>> even using the same basic algorithm.
He doesn't say the two machines generated by the two problems >>>>>>>>> are in any way equivalent, he says that the PROBLEMS are
equivalent,
he's literally saying that if u can enumerate computable
sequences, then u could use that solution to determine whether >>>>>>>> any given machine is circle-free ...
No, he his saying the problems are equivalent as to the nature
and if so could be used to enumerate the circle-free machines, >>>>>>>>
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they produce
the same "answer".
Since the problem is the question of "Can" you do something,
turing is wrong about this. a solution to enumerating circle-free >>>>>> machines can be used to produce a solution to enumerating
computable numbers, but the reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and thus are
trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates circle-
ftee machihes could be used to enumerate computable numbers.
my god rick, please fucking read the not even whole paper, but at
least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING PAGES
WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that,
| if this general process exists [for circle-free machines]
| then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the results of the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, produce a
general process to output whether a machine is circle-free or not
the _first fallacy_ is that because that isn't actually equivalent to
enumerating computable sequences (which is a lesser problem that only
needs to recognize a subset of circle-free machines), ruling out a
general process for deciding circle-free machine does _not_ actually
rule out a general process for enumerating computable numbers
A fallacy in your mind, because you don't understand what he means by equivalent.
You are just falling for your own definist fallacy,.
the _second fallacy_ is assuming that just because a diagonal is
computable, the anti-diagonal becomes computable
But it is.
that one seems obvious on the surface, but when we actually dig into
the details, one _cannot_ use a diagonal machine to produce an anti-
diagonal machine
Sure you can. The fact that you think not is part of your problem.
As I said, if computing the diagonal is possible, then to compute the anti-diagonal, just reverse the value written on the permanent output,
and if you ever read from the permanent output, reverse your decisions
on it.
WHY can't you do that?. Note, in your self-reference hack, you don't
change the number references (so it is no longer actually its own
number, but the number of the machine it was based on).
(it either gets stuck looping trying to find it's own anti-digit, or
it skips over itself not producing a total anti-diagonal. there's no
way to hard-code a digit for itself on the anti-diagonal, like you can
with the diagonal)
Which is what proves that the enumeration can't include it, and thus it can't actually be a computable number.
The decider creating the enumeration is what has the problem. If it
accepts the anti-program itself, then that program becomes non-circle
free, and it accepted a machine that is not in the enumeration.
If it skips the anti-program, then that program IS circle-free and it
fails to enumerate the whole list.
Just like in your "fix", you let the decider skip some machines, as
Turings H can be decided on, your decider must be allowed to also skip
some machines to avoid misclassifying anti-H. But, you still need to
accept SOME machine that computes that value, but *ANY* machine that it accepts will have at least one digit different than what anti-H
computes, as by the structure of antu-H, the machine will be the k'th machine processed by anti-H, and anti-H will differ by it in at least
the k'th digit.
It doesn't matter that you can make a machine that computes an almost diagonal, that is in the list, the problem is that anti-H still exists,
and its output never appears in your enumeration, and thus it is just
not complete, and you proof fails.
In fact, you are just contradicting yourself because you forget what
you mean, as having an enumberation of circle-free machines has the
potential to make a list of computable numbers by eliminating the
duplicates.
So, eat your own dog food.
You are just falling into the definist fallacy, as you are not
using the right definition, but the one YOU want to use.
which implies the sets are equivalent, or contain the same
number of elements.
How do you get that? They aren't even counting the same sort of >>>>>>> thing.
And the set ARE the same size, Countably infinite.
yet *cirle-free* machines forms a /surjection/ onto *computable
sequences*
it's a many to one relationship. the problem of identifying ONE
machine for each computable sequence is not the same as
identifying ALL machines for each computable sequence
But both sets are the same size.
they are the same cardinality, rick, but that does _not_ imply they
are the same problem. in fact _all_ computable sequences are the
same cardinality, but clearly they do not all represent the _same_
problem...
And who said they were?
SAME is NOT "Equivalent"
a solution to the problem of computable numbers only needs to
recognize *ONE* circle-free machine per computable number, not *ALL*
circle-free machines per comparable number
Right, so is a different machine, but faces the same problem.
turing did not prove that, and neither did you
I showed (as Turing does with his implication) that the computable
number problem also doesn't have a machine that computes its anti-diagonal.
As that output isn't in the enumeration.
This is just the result that he refered to that "seems wrong, but is actually right". You are just the sort of person he knew couldn't accept that actually rigerous proof.
recognizing a *subset* is _NOT_ the same problem as recognizing a
*whole* set
Right.
that's just a fallacy whether accept it or not rick
But the fact that you machine doesn't produce the entire set of
computable numbers jsut proves you are wrong.
Your PRD can't produce in its selection of machines that are circle-
free, a machine that produces the number value that anti-fixed-H
produces.
goal posts rick. i'm currently just demonstrating fallacies in
turing's proof
Then show ACTUAL errors where his conclusion isn't actually following
his claim.
Not where you put words in his mouth by misinterpreting what he says.
Your own stupidity and ignorance is NOT a fallacy for Turing but creates them in your work.
i haven't proven that a single machine can enumerate all computable
sequences, that's still just a working thesis
Which has been proven to be impossible.
right now the almost anti-diagonals seem troubling, but i haven't
actually put time into reckoning about it
We don't care abotu "almost". And your problem is time won't solve the issue, as you have gone down a one-way road of error.
And anti-fixed-H, since it only simulates machine confirmed to be
circle-free by PRD, does produce a computable number.
Just one that isn't in the claim complete enumeration done by PRD.
And he didn't say it was the SAME problem, just an equivalent one.
u did not show that PRD can be used to compute an anti-diagonal
Just like the Natural Numbers, and many subsets of it like the
evens, the odds, the primes, the perfect numbers and such. ALL
the sets have "the same number of elements" even though some a
proper subsets of others.
this is a fallacy, as circle-free machines forms a surjection >>>>>>>> onto computable numbers
As do the even numbers to the whole set of Natural Numbers.
You don't seem to understand the nature of infinite sets.
specifically, there are *infinitely* many circle-free machines >>>>>>>>>> for each *computable sequence*. circle-free machines forms a / >>>>>>>>>> surjection/ onto computational sequences, not bijection. it's >>>>>>>>>> a many to one relationship, not equatable.
enumerating circle-free machines requires enumerating *all* >>>>>>>>>> possible circular machines generally,
but enumerating only computable numbers only *requires* a >>>>>>>>>> categorical subset of circle-free machines, and strictly so >>>>>>>>>> for any given enumeration. only _one_ circle-machine per
computable sequence is required to enumerate the sequence, and >>>>>>>>>> you _cannot_ "enumerate" _more than one_ per every computable >>>>>>>>>> sequence. enumerating _only_ computable sequences is a
_lesser_ problem than enumerating circle- free machines
But only fractionally so, but since both problems are
infinitely impossible, that fraction doesn't matter.
the contradiction turing demonstrated is only guaranteed to >>>>>>>>>> exist when totally enumerating out circle-machines, with a >>>>>>>>>> turing machine,
he did not prove the problem exists while enumerating only >>>>>>>>>> *one* circle- free machine per computable sequence.
And he didn't actually claim to.
But he does point out that that is provable, but such a proof >>>>>>>>> will seem "wrong" to those like you, even though it is actually >>>>>>>>> correct.
After all, if we have a method to compute the enumeration of >>>>>>>>> the computable numbers, that means we have a method to compute >>>>>>>>> the kth digits of the nth number.
And thus we can, from that method, build a machine to compute >>>>>>>>> the diagonal or the anti-diagonal by using that method, finding >>>>>>>>> the kth digit of the kth number and output it or its opposite. >>>>>>>>>
We can then ask the question, which number in this sequence is >>>>>>>>> that anti-diagonal, which since we just built the computation >>>>>>>>> that computes it, must be in the list. But it can't be in the >>>>>>>>> list, as if it was the kth number, its kth digit differs from >>>>>>>>> that anti- diagonal which it was supposed to be.
Your problem is you keep on living in a world where you don't >>>>>>>>> need to prove things but can just assume them true, and proof >>>>>>>>> is for some time later.
Your enumeration MUST be incomplete, and thus isn't what you >>>>>>>>> want it to be, no matter how much you pray to the purple magic >>>>>>>>> fairy dust powered unicorn for it to be.
the diagonalization argument is only _one_ of the proofs against >>>>>>>> the effective enumerability of computable numbers,
And it works, so they are not enumerable (in the sense used there). >>>>>>
because u failed to recognize what happens when the anti-diagonal >>>>>> computation tries to enumerate itself to produce a digit opposite >>>>>> to what it does return ...
Well, if PRD selects the anti-diagonal program as an acceptable
program and it fails, then PRD just failed, as it can only accept
programs that are cycle-free.
u have the incredible ability to only acknowledged things that suit
ur stance: which is that turing could do no wrong,
cause regardless, *the anti-diagonal was not made computable by
assuming the computability of the diagonal*
Sure it is. Prove that it isn't.
please reread the psuedo-code again
that assumption was a fallacy, and similarly subverts turing's papers
What is the actual ERROR in what I said to do?
I( seems you don't have an answer to that.
moving the goal post in how to deal with these almost anti-diagonals
is a question for further research that i have not yet dealt with yet. >>>>
But we don't care about the almost diagonal or anti-diagonal, only
the ACTUAL one, as defined by the enumeration your decider generates.
i've come this far, it's hard for me to believe this will be the
thing that finally stumbles me into what ... giving up???
If being wrong doesn't stop you, that just proves that you are just a
pathological liar.
i have gone through many various phases of different wrongs my convos
with you, it's about redirecting effort and being flexible obviously,
ur being black and white about world that is in actuality incredibly
full of color
try not to be so dull eh???
is that what u want rick??? for me to come this far in pointing out
flaws and give up now???
better to give up now, and have a chance to find a right path then to
spend. you life like Olcott gaslit by your own lies.
not until i have the proof richard ¯\_(ツ)_/¯
lol
keep up that copium that ur apparently beloved theory doesn't have
massive shitpile of fallacy at it's core...
The massive shitpile is in your work, not the theory. But of course
to shit, logic seems unbearable.
That fact that you don't actually understand what the basic words
mean just show how bad your base is.
The problem is by your specification, it needs to select some
machine that produces the exact same computation, but that machine
WILL be the kth one selected, and its kth digit MUST be different
then the anti- diagonal machine, so there can't be such a machine
selected.
Thus your PRD fails at the task of selecting AT LEAST ONE machine
that computes every computable number.
You aren't paying attention to what the problem is.
This just shows your fundamental error in how you reason.
This means that your claims that they can be is based on error, >>>>>>> and your refusal to accept that shows your stupidity.
because turing equates enumerating computable numbers with that >>>>>>>> of circle-free machines, the paradox forms the _second_
realizing the fallacy of equating the problems refutes the
_second_ proof not the first. the first is addressed in my other >>>>>>>> response to you
That you don't understand what Turing said, doesn't make it a
fallacy.
It seems your arguement is based on the strawman fallacy.
THe people he was writing to, understood the nature of the
equivalence he was writing about, and new how to take the step
between the two problems.
The fat you don't, doesn't make his claim wrong, it puts it over >>>>>>> your head.
The fact that you still claim that the machine PROVEN to exist if >>>>>>> your PRD exist, that computes what you admit is uncomputable, but >>>>>>> still you claim your PRD can exist, shows that you are not
thinking logically, because you are fixated on something you
don't understand.
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote:
the "better-defined" machine don't satisfy the criteria to >>>>>>>>>>>>> be undecidable
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>
Because of this fallacy, the proof found on the following >>>>>>>>>>>>>>> p247,But if the machine can be “ill-defined”, yet provably >>>>>>>>>>>>>> undecidable,
where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>>>>>>>> compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>>>>>>> impossibility for enumerating computable sequences. >>>>>>>>>>>>>>
that must mean any “better-defined” machine that also >>>>>>>>>>>>>> satisfies
those “ill-defined” criteria must be provably undecidable. >>>>>>>>>>>>>
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in >>>>>>>>>>>> which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution of >>>>>>>>>>> them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks he >>>>>>>>>> must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine >>>>>>>> or even using the same basic algorithm.
He doesn't say the two machines generated by the two problems >>>>>>>>>> are in any way equivalent, he says that the PROBLEMS are
equivalent,
he's literally saying that if u can enumerate computable
sequences, then u could use that solution to determine whether >>>>>>>>> any given machine is circle-free ...
No, he his saying the problems are equivalent as to the nature >>>>>>>>
and if so could be used to enumerate the circle-free machines, >>>>>>>>>
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they produce >>>>>> the same "answer".
Since the problem is the question of "Can" you do something,
turing is wrong about this. a solution to enumerating circle-free >>>>>>> machines can be used to produce a solution to enumerating
computable numbers, but the reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and thus
are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates circle-
ftee machihes could be used to enumerate computable numbers.
my god rick, please fucking read the not even whole paper, but at
least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING PAGES
WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that,
| if this general process exists [for circle-free machines]
| then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the results of the
enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, produce a
general process to output whether a machine is circle-free or not
the _first fallacy_ is that because that isn't actually equivalent to
enumerating computable sequences (which is a lesser problem that only
needs to recognize a subset of circle-free machines), ruling out a
general process for deciding circle-free machine does _not_ actually
rule out a general process for enumerating computable numbers
A fallacy in your mind, because you don't understand what he means by
equivalent.
how can computing a _subset_ of circle-free machines be equivalent to compute a _total_ set of circle-free machines...???
You are just falling for your own definist fallacy,.
the _second fallacy_ is assuming that just because a diagonal is
computable, the anti-diagonal becomes computable
But it is.
that one seems obvious on the surface, but when we actually dig into
the details, one _cannot_ use a diagonal machine to produce an anti-
diagonal machine
Sure you can. The fact that you think not is part of your problem.
As I said, if computing the diagonal is possible, then to compute the
anti-diagonal, just reverse the value written on the permanent output,
and if you ever read from the permanent output, reverse your decisions
on it.
WHY can't you do that?. Note, in your self-reference hack, you don't
because it doesn't compute an anti-digit for itself and therefore does
not form a total anti-diagonal. the anti-diagonal is still uncomputable
even if a general process to enumerate computable numbers exists
the other problems ur bringing up do not refute this point
change the number references (so it is no longer actually its own
number, but the number of the machine it was based on).
(it either gets stuck looping trying to find it's own anti-digit, or
it skips over itself not producing a total anti-diagonal. there's no
way to hard-code a digit for itself on the anti-diagonal, like you
can with the diagonal)
Which is what proves that the enumeration can't include it, and thus
it can't actually be a computable number.
The decider creating the enumeration is what has the problem. If it
accepts the anti-program itself, then that program becomes non-circle
free, and it accepted a machine that is not in the enumeration.
If it skips the anti-program, then that program IS circle-free and it
fails to enumerate the whole list.
Just like in your "fix", you let the decider skip some machines, as
Turings H can be decided on, your decider must be allowed to also skip
some machines to avoid misclassifying anti-H. But, you still need to
accept SOME machine that computes that value, but *ANY* machine that
it accepts will have at least one digit different than what anti-H
computes, as by the structure of antu-H, the machine will be the k'th
machine processed by anti-H, and anti-H will differ by it in at least
the k'th digit.
It doesn't matter that you can make a machine that computes an almost
diagonal, that is in the list, the problem is that anti-H still
exists, and its output never appears in your enumeration, and thus it
is just not complete, and you proof fails.
The following claim from p246 of Turing’s seminal paper On Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-free
machines which compute that particular sequence. Not only can various machines differ significantly in the specific steps to produce the same output, machines can be changed in superficial ways that do not
meaningfully affect the steps of computation, akin to modern no-op statements or unreachable code
The problem of enumerating computable sequences, however, only depends
on successfully identifying _one_ circle-free machine that computes any given computable sequences. While identifying more than one can
certainly be done, it is _not_ a requirement for enumerating computable sequences, as _one_ machine computing a sequence /suffices to output any
and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-free machines, as there is no need to identify all circle-free machines which compute any given computable sequence
Said problem is only equivalent to a _limited process_ of enumerating circle-free machines. The machine which identifies circle-free machines
only needs the limited power of determining _at least one_ circle-free machine for any given computable sequence, _not all_ machines for any
given computable sequence
Because of this fallacy, the proof found on the following p247, where an ill-defined machine 𝓗 (which attempts and fails to compute the direct diagonal β’) is found to be undecidable in respect to circle-free
decider 𝓓; does not then prove an impossibility for enumerating computable sequences. As the problem of enumerating /all circle-free machines/ is _not_ equivalent to that of enumerating /just computable sequences/
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote:
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>the "better-defined" machine don't satisfy the criteria to >>>>>>>>>>>>>> be undecidable
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>> following p247,But if the machine can be “ill-defined”, yet provably >>>>>>>>>>>>>>> undecidable,
where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>>>>>>>>> compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>>>>>>>> impossibility for enumerating computable sequences. >>>>>>>>>>>>>>>
that must mean any “better-defined” machine that also >>>>>>>>>>>>>>> satisfies
those “ill-defined” criteria must be provably undecidable. >>>>>>>>>>>>>>
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in >>>>>>>>>>>>> which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution >>>>>>>>>>>> of them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks >>>>>>>>>>> he must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine >>>>>>>>> or even using the same basic algorithm.
He doesn't say the two machines generated by the two problems >>>>>>>>>>> are in any way equivalent, he says that the PROBLEMS are >>>>>>>>>>> equivalent,
he's literally saying that if u can enumerate computable
sequences, then u could use that solution to determine whether >>>>>>>>>> any given machine is circle-free ...
No, he his saying the problems are equivalent as to the nature >>>>>>>>>
and if so could be used to enumerate the circle-free machines, >>>>>>>>>>
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they
produce the same "answer".
Since the problem is the question of "Can" you do something,
turing is wrong about this. a solution to enumerating circle- >>>>>>>> free machines can be used to produce a solution to enumerating >>>>>>>> computable numbers, but the reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and thus
are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates circle-
ftee machihes could be used to enumerate computable numbers.
my god rick, please fucking read the not even whole paper, but at
least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING PAGES
WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that,
| if this general process exists [for circle-free machines]
| then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the results of
the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, produce
a general process to output whether a machine is circle-free or not
the _first fallacy_ is that because that isn't actually equivalent
to enumerating computable sequences (which is a lesser problem that
only needs to recognize a subset of circle-free machines), ruling
out a general process for deciding circle-free machine does _not_
actually rule out a general process for enumerating computable numbers
A fallacy in your mind, because you don't understand what he means by
equivalent.
how can computing a _subset_ of circle-free machines be equivalent to
compute a _total_ set of circle-free machines...???
Who said they were equivalent COMPUTATIONS.
The problem of creating the computations are equivalent PROBLEMS.
Your problem is in you ignorance, you don't know what he is talking about.
You are just falling for your own definist fallacy,.
the _second fallacy_ is assuming that just because a diagonal is
computable, the anti-diagonal becomes computable
But it is.
that one seems obvious on the surface, but when we actually dig into
the details, one _cannot_ use a diagonal machine to produce an anti-
diagonal machine
Sure you can. The fact that you think not is part of your problem.
As I said, if computing the diagonal is possible, then to compute the
anti-diagonal, just reverse the value written on the permanent
output, and if you ever read from the permanent output, reverse your
decisions on it.
WHY can't you do that?. Note, in your self-reference hack, you don't
because it doesn't compute an anti-digit for itself and therefore does
not form a total anti-diagonal. the anti-diagonal is still
uncomputable even if a general process to enumerate computable numbers
exists
But it only doesn't because your PRD fails to provide the needed enumeration.
PRD CAN'T tell it to enumerate itself, or it actively fails at accepting only circle-free machines. (The OTHER PRD that does creates a DIFFERENT anti-fixed-H that isn't circle free).
anti-fixed-H does compute the anti-diagonal for the (incomplete)
enumeration that your actual PRD generates, and shows that it missed an computable number.
the other problems ur bringing up do not refute this point
Sure it does. Your "enumeration" as generated by any actual PRD is just incomplete. You CAN'T make a PRD that meets your specification, as it
turns out that anti-fixed-H proves that it is an uncomputable
specification.
change the number references (so it is no longer actually its own
number, but the number of the machine it was based on).
(it either gets stuck looping trying to find it's own anti-digit, or
it skips over itself not producing a total anti-diagonal. there's no
way to hard-code a digit for itself on the anti-diagonal, like you
can with the diagonal)
Which is what proves that the enumeration can't include it, and thus
it can't actually be a computable number.
The decider creating the enumeration is what has the problem. If it
accepts the anti-program itself, then that program becomes non-circle
free, and it accepted a machine that is not in the enumeration.
If it skips the anti-program, then that program IS circle-free and it
fails to enumerate the whole list.
Just like in your "fix", you let the decider skip some machines, as
Turings H can be decided on, your decider must be allowed to also
skip some machines to avoid misclassifying anti-H. But, you still
need to accept SOME machine that computes that value, but *ANY*
machine that it accepts will have at least one digit different than
what anti-H computes, as by the structure of antu-H, the machine will
be the k'th machine processed by anti-H, and anti-H will differ by it
in at least the k'th digit.
It doesn't matter that you can make a machine that computes an almost
diagonal, that is in the list, the problem is that anti-H still
exists, and its output never appears in your enumeration, and thus it
is just not complete, and you proof fails.
All you are doing is demonstrating that you are fundamentally not
understand the nature of the field you are trying to talk about.
You don't seem to understand what an actual computation is, and that machines can't be built on "interfaces", only actual implementations.--
This things like anti-fixed-h, once you talk about them, imply that you
have chosen your implementation for PRD, and thus your enumeration is
FIXED and can't be changed without reseting the problem. Templates like
H, or fixed-H, or anti-fixed-H don't actualy become "machines" until you establish your claimed decider they are going to be based on.
If your claimed PRD doesn't put (its) anti-fixed-H in the enumeration,
then anti-fixed-H isn't wrong to not put that anti-digit in its output.
And thus it generates a computable number that no machine that PRD
accepts generates, and thus that PRD is just failing to meet its specification
If your claimed PRD tries to accept its anti-fixed-H in the enumeration, then because of the problem you point out, it just fails to accept only circle-free machines. It runs into the problem that ITS version of anti- fixed-H is a different machine then the anti-fixed-H based on the other
PRD that skipped it, and these machines will have different values of N.
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On
Computable Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-free
machines which compute that particular sequence. Not only can
various machines differ significantly in the specific steps to
produce the same output, machines can be changed in superficial ways
that do not meaningfully affect the steps of computation, akin to
modern no-op statements or unreachable code
The problem of enumerating computable sequences, however, only
depends on successfully identifying _one_ circle-free machine that
computes any given computable sequences. While identifying more than
one can certainly be done, it is _not_ a requirement for enumerating
computable sequences, as _one_ machine computing a sequence /
suffices to output any and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-
free machines, as there is no need to identify all circle-free
machines which compute any given computable sequence
Which just shows that you don't understand what the word "Equivalent"
means here.
Just like we can have "Equivalent" Turing Machines, that by very
different methods and path create the exact same output, we can have
two "Equivalent classification problems" that by using different
classes, come to the same result, that there exist uncomputable
problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle-free
machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies
circle-free machines only needs the limited power of determining _at
least one_ circle-free machine for any given computable sequence,
_not all_ machines for any given computable sequence
Because of this fallacy, the proof found on the following p247,
where an ill-defined machine 𝓗 (which attempts and fails to compute >>>> the direct diagonal β’) is found to be undecidable in respect to
circle- free decider 𝓓; does not then prove an impossibility for
enumerating computable sequences. As the problem of enumerating /all
circle-free machines/ is _not_ equivalent to that of enumerating /
just computable sequences/
And, your "partial_recognizer_D" also can be proved to have an
uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for a
PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of the
number given to it.
2) No machine that it accepts will fail to produce a computable
number, and thus will continue to run and produce output forever.
3) For EVERY Computable Number that exist, PRD will accept at least
one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have no
proven yet) can be computable by some PRD
So, if PRD exists, we can build a machine that computes an anti-
diagonal by testing each number in sequence with PRD, and for each
number that it accepts, it will simulate that machine until that
machine generates k digits of output, k being the number of values
accepted to this point, and then it outputs the opposite digit of the
kth digit generated by the nth machine.
This machine must produce a computable number, as it only simulates
and uses the output of machines that PRD accepted, so by 1, PRD
answered, and by 2 that machine can be simulated for as long as we
want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by condition
3, there must exist a finite number n that represents a machine that
generates it that PRD will accept, and thus our machine will simulate
for k digits (which will be less than n) and output the opposite value.
Thus, whatever n you want to claim is the machine that generates the
same computable number doesn't, and thus there can not exist a PRD
that does what you claim.
In fact, this method works for ANY method you may want to claim
allows you to compute the enumeration of the computable numbers.
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic requires
you to just be able to assume things exist that don't.
that's a great, actually coherent objection rick, that i was hoping
someone would mention it! NO INSULTS REQUIRED!
it's unfortunately also a fallacy, and one that turing similarly made.
*having a machine that computes a diagonal, does not actually then
imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD exists.
and the reason that's true is *same* self-referential weirdness that
stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change it to
an equivalent program that prints the anti-diagonal.
consider fixed_H again, since that is the form of diagonal computation
that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each output
anti_digit = 1-digit // of the machine fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
what would happen if PRD(anti_fixed_H) => TRUE?
what would happen if PRD(anti_fixed_H) => FALSE?
But we don't care what PRD(anti_fixed_H) is.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus
what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D can
not possible acheive its requirements to accept at least one machine
that generates a computable number, and thus, the "enumeration" you
generate from it is not complete, but only partial, and thus your whole claim falls apart.
On 03/13/2026 10:53 AM, Ross Finlayson wrote:
On 03/13/2026 10:25 AM, dart200 wrote:
On 3/12/26 3:53 PM, Richard Damon wrote:
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On >>>>>>>>> Computable Numbers is a fallacy:Which just shows that you don't understand what the word
/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/
For any given computable sequence, there are _infinite_ circle- >>>>>>>>> free machines which compute that particular sequence. Not only >>>>>>>>> can various machines differ significantly in the specific steps >>>>>>>>> to produce the same output, machines can be changed in
superficial ways that do not meaningfully affect the steps of >>>>>>>>> computation, akin to modern no-op statements or unreachable code >>>>>>>>>
The problem of enumerating computable sequences, however, only >>>>>>>>> depends on successfully identifying _one_ circle-free machine >>>>>>>>> that computes any given computable sequences. While identifying >>>>>>>>> more than one can certainly be done, it is _not_ a requirement >>>>>>>>> for enumerating computable sequences, as _one_ machine computing >>>>>>>>> a sequence / suffices to output any and all digits of that
sequence/
The problem of enumerating computable sequences is therefore >>>>>>>>> _not_ actually equivalent to a _general process_ of enumerating >>>>>>>>> circle- free machines, as there is no need to identify all
circle-free machines which compute any given computable sequence >>>>>>>>
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very >>>>>>>> different methods and path create the exact same output, we can >>>>>>>> have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle- >>>>>>> free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies >>>>>>>>> circle-free machines only needs the limited power of determining >>>>>>>>> _at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence >>>>>>>>>
Because of this fallacy, the proof found on the following p247, >>>>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>>> respect to circle- free decider 𝓓; does not then prove an >>>>>>>>> impossibility for enumerating computable sequences. As the
problem of enumerating / all circle-free machines/ is _not_
equivalent to that of enumerating / just computable sequences/ >>>>>>>>>
And, your "partial_recognizer_D" also can be proved to have an >>>>>>>> uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for >>>>>>>> a PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of >>>>>>>> the number given to it.
2) No machine that it accepts will fail to produce a computable >>>>>>>> number, and thus will continue to run and produce output forever. >>>>>>>>
3) For EVERY Computable Number that exist, PRD will accept at
least one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have >>>>>>> no proven yet) can be computable by some PRD
that's a great, actually coherent objection rick, that i was hoping >>>>>>> someone would mention it! NO INSULTS REQUIRED!
So, if PRD exists, we can build a machine that computes an anti- >>>>>>>> diagonal by testing each number in sequence with PRD, and for each >>>>>>>> number that it accepts, it will simulate that machine until that >>>>>>>> machine generates k digits of output, k being the number of values >>>>>>>> accepted to this point, and then it outputs the opposite digit of >>>>>>>> the kth digit generated by the nth machine.
This machine must produce a computable number, as it only
simulates and uses the output of machines that PRD accepted, so by >>>>>>>> 1, PRD answered, and by 2 that machine can be simulated for as >>>>>>>> long as we want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by
condition 3, there must exist a finite number n that represents a >>>>>>>> machine that generates it that PRD will accept, and thus our
machine will simulate for k digits (which will be less than n) and >>>>>>>> output the opposite value.
Thus, whatever n you want to claim is the machine that generates >>>>>>>> the same computable number doesn't, and thus there can not exist a >>>>>>>> PRD that does what you claim.
In fact, this method works for ANY method you may want to claim >>>>>>>> allows you to compute the enumeration of the computable numbers. >>>>>>>>
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't. >>>>>>>
it's unfortunately also a fallacy, and one that turing similarly >>>>>>> made.
*having a machine that computes a diagonal, does not actually then >>>>>>> imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD
exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding
eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness
that stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change >>>>>> it to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal
computation
that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output simulate_kth_digit(N,K) // kth digit >>>>>>> K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each
output
anti_digit = 1-digit // of the machine
fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay
attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a
non-circle free input.
obviously PRD(anti_fixed_H) cannot return TRUE, so therefore it returns
FALSE
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H >>>>> will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
there is:
fixed_anti_fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_fixed_H) { // handle self-ref >>> N += 1
continue // skip including itself
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
this computes the same thing as anti_fixed_H(), but is decidable by PRD. >>> PRD(fixed_anti_fixed_H) returns TRUE
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
... err yes, the total anti-diagonal is _not_ computable. the _closest_
we can get is a sequence that includes the inverse for all computable
sequences _except_ to the anti-diagonal computation itself
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the
requirement you gave it, so fixed_H did not actually compute a digonal >>>> of an enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the >>>>>> machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus >>>>>> what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D >>>>>> can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only
partial, and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H. >>>>> for this i will combine the two machines for simplicity, i will leave >>>>> implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here???
K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we >>>>> need to define what it's digit on the diagonal will be. but if that's >>>>> its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program
DID, at least it would if PRD exists.
no it didn't. it skips computing an inverse to it's own computation
Thus, your PRD disappear in the same contradiction that Turing_D and
Turing_H does, and thus so does your fixed_H
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output, >>>>> you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you
don't get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the
anti- diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts
is what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
The usual accounts that the infinite is not computable
doesn't contradict that each finite is computable.
Then it gets directly into accounts as modeled by
ordinary set theory.
These are then among matters of "quantifier disambiguation"
against "impredicativity".
There are various models of the Halting Problem which
is also what is being discussed, or the Branching Problem,
that vary, about usually enough "supertasks".
There are many well-known "approximation algorithms
to NP-hard problems" that reduce them to polynomial.
So, are you going to acknowledge these sorts relations
to the usual accounts of "computability" and "countability"?
The ac-countability?
Problem B: computing whether any number is the D.N of a circle-free
machine
On 3/14/26 2:05 AM, Tristan Wibberley wrote:
On 14/03/2026 04:25, dart200 wrote:
On 3/13/26 6:24 PM, Lawrence D’Oliveiro wrote:
...
^^^^^^^^^^So where does the “ill-defined machine” come in? That was the one you >>>he equates the two sets on p246
on p247 he constructs machine H that attempts to construct a diagonal
Perhaps I didn't pay close enough attention but I thought he defined
constraints on the variable H rather than constructing a machine.
the entire page 247 is discussing machine H
it can be hard to the H symbols cause the scanned pages aren't great
On 13/03/2026 02:18, Mike Terry wrote:
Problem B: computing whether any number is the D.N of a circle-free
machine
ah, sweet ambiguity.
Is there a number that's the D.N of a circle-free machine?
or
Is a given number the D.N of a circle-free machine?
On 13/03/2026 02:18, Mike Terry wrote:
Problem B: computing whether any number is the D.N of a circle-free
machine
ah, sweet ambiguity.
Is there a number that's the D.N of a circle-free machine?
or
Is a given number the D.N of a circle-free machine?
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote:
On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>the "better-defined" machine don't satisfy the criteria >>>>>>>>>>>>>>> to be undecidable
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>>> following p247,But if the machine can be “ill-defined”, yet provably >>>>>>>>>>>>>>>> undecidable,
where an ill-defined machine 𝓗 (which attempts and >>>>>>>>>>>>>>>>> fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>>>>>>>>> impossibility for enumerating computable sequences. >>>>>>>>>>>>>>>>
that must mean any “better-defined” machine that also >>>>>>>>>>>>>>>> satisfies
those “ill-defined” criteria must be provably undecidable. >>>>>>>>>>>>>>>
But they’re a subset of the “ill-defined” set that Turing was
considering, are they not?
Unless you’re considering an entirely different set, in >>>>>>>>>>>>>> which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution >>>>>>>>>>>>> of them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks >>>>>>>>>>>> he must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here.
After all Functional Equivalence doesn't mean the same machine >>>>>>>>>> or even using the same basic algorithm.
He doesn't say the two machines generated by the two
problems are in any way equivalent, he says that the
PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable >>>>>>>>>>> sequences, then u could use that solution to determine
whether any given machine is circle-free ...
No, he his saying the problems are equivalent as to the nature >>>>>>>>>>
and if so could be used to enumerate the circle-free machines, >>>>>>>>>>>
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they
produce the same "answer".
Since the problem is the question of "Can" you do something,
turing is wrong about this. a solution to enumerating circle- >>>>>>>>> free machines can be used to produce a solution to enumerating >>>>>>>>> computable numbers, but the reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and thus >>>>>> are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates circle- >>>>>> ftee machihes could be used to enumerate computable numbers.
my god rick, please fucking read the not even whole paper, but at
least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING PAGES >>>>> WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that,
| if this general process exists [for circle-free machines]
| then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the results of
the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, produce >>>>> a general process to output whether a machine is circle-free or not
A fallacy in your mind, because you don't understand what he means
the _first fallacy_ is that because that isn't actually equivalent
to enumerating computable sequences (which is a lesser problem that >>>>> only needs to recognize a subset of circle-free machines), ruling
out a general process for deciding circle-free machine does _not_
actually rule out a general process for enumerating computable numbers >>>>
by equivalent.
how can computing a _subset_ of circle-free machines be equivalent to
compute a _total_ set of circle-free machines...???
Who said they were equivalent COMPUTATIONS.
| 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
The problem of creating the computations are equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's pretty ridiculous richard
if problems are equivalent then a solution to A can be used to solve B
and vise versa ...
if u don't agree with this then u can move right the fuck along with ur willful ignorance and gaslighting dick
Your problem is in you ignorance, you don't know what he is talking
about.
You are just falling for your own definist fallacy,.
the _second fallacy_ is assuming that just because a diagonal is
computable, the anti-diagonal becomes computable
But it is.
that one seems obvious on the surface, but when we actually dig
into the details, one _cannot_ use a diagonal machine to produce an >>>>> anti- diagonal machine
Sure you can. The fact that you think not is part of your problem.
As I said, if computing the diagonal is possible, then to compute
the anti-diagonal, just reverse the value written on the permanent
output, and if you ever read from the permanent output, reverse your
decisions on it.
WHY can't you do that?. Note, in your self-reference hack, you don't
because it doesn't compute an anti-digit for itself and therefore
does not form a total anti-diagonal. the anti-diagonal is still
uncomputable even if a general process to enumerate computable
numbers exists
But it only doesn't because your PRD fails to provide the needed
enumeration.
if it did it would produce a cyclical machine, so obviously it can't,
u haven't proven no machine can compute the same sequence
PRD CAN'T tell it to enumerate itself, or it actively fails at
accepting only circle-free machines. (The OTHER PRD that does creates
a DIFFERENT anti-fixed-H that isn't circle free).
anti-fixed-H does compute the anti-diagonal for the (incomplete)
enumeration that your actual PRD generates, and shows that it missed
an computable number.
the other problems ur bringing up do not refute this point
Sure it does. Your "enumeration" as generated by any actual PRD is
just incomplete. You CAN'T make a PRD that meets your specification,
as it turns out that anti-fixed-H proves that it is an uncomputable
specification.
change the number references (so it is no longer actually its own
number, but the number of the machine it was based on).
(it either gets stuck looping trying to find it's own anti-digit,
or it skips over itself not producing a total anti-diagonal.
there's no way to hard-code a digit for itself on the anti-
diagonal, like you can with the diagonal)
Which is what proves that the enumeration can't include it, and thus
it can't actually be a computable number.
The decider creating the enumeration is what has the problem. If it
accepts the anti-program itself, then that program becomes non-
circle free, and it accepted a machine that is not in the enumeration. >>>>
If it skips the anti-program, then that program IS circle-free and
it fails to enumerate the whole list.
Just like in your "fix", you let the decider skip some machines, as
Turings H can be decided on, your decider must be allowed to also
skip some machines to avoid misclassifying anti-H. But, you still
need to accept SOME machine that computes that value, but *ANY*
machine that it accepts will have at least one digit different than
what anti-H computes, as by the structure of antu-H, the machine
will be the k'th machine processed by anti-H, and anti-H will differ
by it in at least the k'th digit.
It doesn't matter that you can make a machine that computes an
almost diagonal, that is in the list, the problem is that anti-H
still exists, and its output never appears in your enumeration, and
thus it is just not complete, and you proof fails.
All you are doing is demonstrating that you are fundamentally not
understand the nature of the field you are trying to talk about.
take ur ignorance to grave dick, i've come a long way to showing fallacy
and the fact u can't recognize it is a you problem rick, not a me problem
people like you are why mortality is still a necessary facet of this
species
how about u try being constructive instead of destructive for once?
You don't seem to understand what an actual computation is, and that
machines can't be built on "interfaces", only actual implementations.
This things like anti-fixed-h, once you talk about them, imply that
you have chosen your implementation for PRD, and thus your enumeration
is FIXED and can't be changed without reseting the problem. Templates
like H, or fixed-H, or anti-fixed-H don't actualy become "machines"
until you establish your claimed decider they are going to be based on.
If your claimed PRD doesn't put (its) anti-fixed-H in the enumeration,
then anti-fixed-H isn't wrong to not put that anti-digit in its
output. And thus it generates a computable number that no machine that
PRD accepts generates, and thus that PRD is just failing to meet its
specification
If your claimed PRD tries to accept its anti-fixed-H in the
enumeration, then because of the problem you point out, it just fails
to accept only circle-free machines. It runs into the problem that ITS
version of anti- fixed-H is a different machine then the anti-fixed-H
based on the other PRD that skipped it, and these machines will have
different values of N.
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:my god rick, please fucking read the not even whole paper, but at >>>>>> least the _section_ rick please...
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>>>> following p247,But if the machine can be “ill-defined”, yet provably >>>>>>>>>>>>>>>>> undecidable,
where an ill-defined machine 𝓗 (which attempts and >>>>>>>>>>>>>>>>>> fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>>>>>>>>>> impossibility for enumerating computable sequences. >>>>>>>>>>>>>>>>>
that must mean any “better-defined” machine that also >>>>>>>>>>>>>>>>> satisfies
those “ill-defined” criteria must be provably undecidable.
the "better-defined" machine don't satisfy the criteria >>>>>>>>>>>>>>>> to be undecidable
But they’re a subset of the “ill-defined” set that Turing
was
considering, are they not?
Unless you’re considering an entirely different set, in >>>>>>>>>>>>>>> which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the solution >>>>>>>>>>>>>> of them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain thinks >>>>>>>>>>>>> he must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here. >>>>>>>>>>>
After all Functional Equivalence doesn't mean the same
machine or even using the same basic algorithm.
He doesn't say the two machines generated by the two >>>>>>>>>>>>> problems are in any way equivalent, he says that the >>>>>>>>>>>>> PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable >>>>>>>>>>>> sequences, then u could use that solution to determine >>>>>>>>>>>> whether any given machine is circle-free ...
No, he his saying the problems are equivalent as to the nature >>>>>>>>>>>
and if so could be used to enumerate the circle-free machines, >>>>>>>>>>>>
making the problem of enumerating the sets equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B >>>>>>>>>>
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they
produce the same "answer".
Since the problem is the question of "Can" you do something, >>>>>>>>>
turing is wrong about this. a solution to enumerating circle- >>>>>>>>>> free machines can be used to produce a solution to enumerating >>>>>>>>>> computable numbers, but the reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and thus >>>>>>> are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates
circle- ftee machihes could be used to enumerate computable numbers. >>>>>>
i'm tired of answering questions that ARE ON THE SAME FUCKING
PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that,
| if this general process exists [for circle-free machines]
| then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the results of
the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine,
produce a general process to output whether a machine is circle-
free or not
the _first fallacy_ is that because that isn't actually equivalent >>>>>> to enumerating computable sequences (which is a lesser problem
that only needs to recognize a subset of circle-free machines),
ruling out a general process for deciding circle-free machine does >>>>>> _not_ actually rule out a general process for enumerating
computable numbers
A fallacy in your mind, because you don't understand what he means
by equivalent.
how can computing a _subset_ of circle-free machines be equivalent
to compute a _total_ set of circle-free machines...???
Who said they were equivalent COMPUTATIONS.
| 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
Right, equivLWNR PROBLEM, which means both problems are either solvable
or not (under all applicable models).
IT seems you are just showing you don't know what the word means,
because you are just ignornat.
The problem of creating the computations are equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's pretty ridiculous richard
Because I am not, you are gaslighting yourself with your false
definitions that you try to insist on.
if problems are equivalent then a solution to A can be used to solve B
and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck along with
ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the field, a
field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better than people
who actually know something.
In a word, Dunning-Kruger
Your problem is in you ignorance, you don't know what he is talking
about.
because it doesn't compute an anti-digit for itself and therefore
You are just falling for your own definist fallacy,.
the _second fallacy_ is assuming that just because a diagonal is
computable, the anti-diagonal becomes computable
But it is.
that one seems obvious on the surface, but when we actually dig
into the details, one _cannot_ use a diagonal machine to produce
an anti- diagonal machine
Sure you can. The fact that you think not is part of your problem.
As I said, if computing the diagonal is possible, then to compute
the anti-diagonal, just reverse the value written on the permanent
output, and if you ever read from the permanent output, reverse
your decisions on it.
WHY can't you do that?. Note, in your self-reference hack, you don't >>>>
does not form a total anti-diagonal. the anti-diagonal is still
uncomputable even if a general process to enumerate computable
numbers exists
But it only doesn't because your PRD fails to provide the needed
enumeration.
if it did it would produce a cyclical machine, so obviously it can't,
And thus you admit that it can't meet the needed requirements, but still insist that it does.
It seems you don't understand what you are talking about.
u haven't proven no machine can compute the same sequence
But that isn't the question.
No machine that PRD accepts does, which is all that is needed to show
that no variation of PRD can compute what you claim it does.
It seems you still don't understand how basic logic works.
PRD CAN'T tell it to enumerate itself, or it actively fails at
accepting only circle-free machines. (The OTHER PRD that does creates
a DIFFERENT anti-fixed-H that isn't circle free).
anti-fixed-H does compute the anti-diagonal for the (incomplete)
enumeration that your actual PRD generates, and shows that it missed
an computable number.
the other problems ur bringing up do not refute this point
Sure it does. Your "enumeration" as generated by any actual PRD is
just incomplete. You CAN'T make a PRD that meets your specification,
as it turns out that anti-fixed-H proves that it is an uncomputable
specification.
change the number references (so it is no longer actually its own
number, but the number of the machine it was based on).
(it either gets stuck looping trying to find it's own anti-digit, >>>>>> or it skips over itself not producing a total anti-diagonal.
there's no way to hard-code a digit for itself on the anti-
diagonal, like you can with the diagonal)
Which is what proves that the enumeration can't include it, and
thus it can't actually be a computable number.
The decider creating the enumeration is what has the problem. If it >>>>> accepts the anti-program itself, then that program becomes non-
circle free, and it accepted a machine that is not in the enumeration. >>>>>
If it skips the anti-program, then that program IS circle-free and
it fails to enumerate the whole list.
Just like in your "fix", you let the decider skip some machines, as >>>>> Turings H can be decided on, your decider must be allowed to also
skip some machines to avoid misclassifying anti-H. But, you still
need to accept SOME machine that computes that value, but *ANY*
machine that it accepts will have at least one digit different than >>>>> what anti-H computes, as by the structure of antu-H, the machine
will be the k'th machine processed by anti-H, and anti-H will
differ by it in at least the k'th digit.
It doesn't matter that you can make a machine that computes an
almost diagonal, that is in the list, the problem is that anti-H
still exists, and its output never appears in your enumeration, and >>>>> thus it is just not complete, and you proof fails.
All you are doing is demonstrating that you are fundamentally not
understand the nature of the field you are trying to talk about.
take ur ignorance to grave dick, i've come a long way to showing
fallacy and the fact u can't recognize it is a you problem rick, not a
me problem
THe problem is YOUR ignorance, and beleiving in the impossible.
You have ADMITTED to your ignorance, but still insist that you are
smarted than the experts, whose work you don't understand.
That show just how STUPID you really are.
people like you are why mortality is still a necessary facet of this
species
how about u try being constructive instead of destructive for once?
I am, I am showing that reality is more important than claiming you can
do what has been proved impossible.
It seems your mind can't handle the fact that you are wrong, because you feel the need to believe in the impossible.
That has DOOMED your to a life of failure.
Maybe YOU are the one that needs to change, from a life based on lying
to yourself and the world, to one that accepts reality, and seeks to do
the best you can.
You don't seem to understand what an actual computation is, and that
machines can't be built on "interfaces", only actual implementations.
This things like anti-fixed-h, once you talk about them, imply that
you have chosen your implementation for PRD, and thus your
enumeration is FIXED and can't be changed without reseting the
problem. Templates like H, or fixed-H, or anti-fixed-H don't actualy
become "machines" until you establish your claimed decider they are
going to be based on.
If your claimed PRD doesn't put (its) anti-fixed-H in the
enumeration, then anti-fixed-H isn't wrong to not put that anti-digit
in its output. And thus it generates a computable number that no
machine that PRD accepts generates, and thus that PRD is just failing
to meet its specification
If your claimed PRD tries to accept its anti-fixed-H in the
enumeration, then because of the problem you point out, it just fails
to accept only circle-free machines. It runs into the problem that
ITS version of anti- fixed-H is a different machine then the anti-
fixed-H based on the other PRD that skipped it, and these machines
will have different values of N.
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>>>>> following p247,But if the machine can be “ill-defined”, yet provably >>>>>>>>>>>>>>>>>> undecidable,
where an ill-defined machine 𝓗 (which attempts and >>>>>>>>>>>>>>>>>>> fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then prove an >>>>>>>>>>>>>>>>>>> impossibility for enumerating computable sequences. >>>>>>>>>>>>>>>>>>
that must mean any “better-defined” machine that also >>>>>>>>>>>>>>>>>> satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the criteria >>>>>>>>>>>>>>>>> to be undecidable
But they’re a subset of the “ill-defined” set that >>>>>>>>>>>>>>>> Turing was
considering, are they not?
Unless you’re considering an entirely different set, in >>>>>>>>>>>>>>>> which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the >>>>>>>>>>>>>>> solution of them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain >>>>>>>>>>>>>> thinks he must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here. >>>>>>>>>>>>
After all Functional Equivalence doesn't mean the same >>>>>>>>>>>> machine or even using the same basic algorithm.
He doesn't say the two machines generated by the two >>>>>>>>>>>>>> problems are in any way equivalent, he says that the >>>>>>>>>>>>>> PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable >>>>>>>>>>>>> sequences, then u could use that solution to determine >>>>>>>>>>>>> whether any given machine is circle-free ...
No, he his saying the problems are equivalent as to the nature >>>>>>>>>>>>
That isn't what "equivalent" means.
and if so could be used to enumerate the circle-free machines, >>>>>>>>>>>>>
making the problem of enumerating the sets equivalent, >>>>>>>>>>>>
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B >>>>>>>>>>>
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they >>>>>>>>>> produce the same "answer".
Since the problem is the question of "Can" you do something, >>>>>>>>>>
But it doesn't need to.
turing is wrong about this. a solution to enumerating circle- >>>>>>>>>>> free machines can be used to produce a solution to
enumerating computable numbers, but the reverse is *NOT* true >>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and
thus are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates
circle- ftee machihes could be used to enumerate computable
numbers.
my god rick, please fucking read the not even whole paper, but at >>>>>>> least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING
PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that, >>>>>>> | if this general process exists [for circle-free machines]
| then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the results of >>>>>> the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine,
produce a general process to output whether a machine is circle- >>>>>>> free or not
the _first fallacy_ is that because that isn't actually
equivalent to enumerating computable sequences (which is a lesser >>>>>>> problem that only needs to recognize a subset of circle-free
machines), ruling out a general process for deciding circle-free >>>>>>> machine does _not_ actually rule out a general process for
enumerating computable numbers
A fallacy in your mind, because you don't understand what he means >>>>>> by equivalent.
how can computing a _subset_ of circle-free machines be equivalent
to compute a _total_ set of circle-free machines...???
Who said they were equivalent COMPUTATIONS.
| 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
Right, equivLWNR PROBLEM, which means both problems are either
solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other...
which is a fallacy in this case, they are not equivalent problems
IT seems you are just showing you don't know what the word means,
because you are just ignornat.
ur an ass dick
The problem of creating the computations are equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's pretty ridiculous richard
Because I am not, you are gaslighting yourself with your false
definitions that you try to insist on.
if problems are equivalent then a solution to A can be used to solve
B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck along with
ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the field, a
field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better than
people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Your problem is in you ignorance, you don't know what he is talking
about.
because it doesn't compute an anti-digit for itself and therefore
You are just falling for your own definist fallacy,.
the _second fallacy_ is assuming that just because a diagonal is >>>>>>> computable, the anti-diagonal becomes computable
But it is.
that one seems obvious on the surface, but when we actually dig >>>>>>> into the details, one _cannot_ use a diagonal machine to produce >>>>>>> an anti- diagonal machine
Sure you can. The fact that you think not is part of your problem. >>>>>>
As I said, if computing the diagonal is possible, then to compute >>>>>> the anti-diagonal, just reverse the value written on the permanent >>>>>> output, and if you ever read from the permanent output, reverse
your decisions on it.
WHY can't you do that?. Note, in your self-reference hack, you don't >>>>>
does not form a total anti-diagonal. the anti-diagonal is still
uncomputable even if a general process to enumerate computable
numbers exists
But it only doesn't because your PRD fails to provide the needed
enumeration.
if it did it would produce a cyclical machine, so obviously it can't,
And thus you admit that it can't meet the needed requirements, but
still insist that it does.
It seems you don't understand what you are talking about.
u haven't proven no machine can compute the same sequence
But that isn't the question.
that is a question, that u haven't proven
No machine that PRD accepts does, which is all that is needed to show
that no variation of PRD can compute what you claim it does.
It seems you still don't understand how basic logic works.
PRD CAN'T tell it to enumerate itself, or it actively fails at
accepting only circle-free machines. (The OTHER PRD that does
creates a DIFFERENT anti-fixed-H that isn't circle free).
anti-fixed-H does compute the anti-diagonal for the (incomplete)
enumeration that your actual PRD generates, and shows that it missed
an computable number.
the other problems ur bringing up do not refute this point
Sure it does. Your "enumeration" as generated by any actual PRD is
just incomplete. You CAN'T make a PRD that meets your specification,
as it turns out that anti-fixed-H proves that it is an uncomputable
specification.
change the number references (so it is no longer actually its own >>>>>> number, but the number of the machine it was based on).
(it either gets stuck looping trying to find it's own anti-digit, >>>>>>> or it skips over itself not producing a total anti-diagonal.
there's no way to hard-code a digit for itself on the anti-
diagonal, like you can with the diagonal)
Which is what proves that the enumeration can't include it, and
thus it can't actually be a computable number.
The decider creating the enumeration is what has the problem. If
it accepts the anti-program itself, then that program becomes non- >>>>>> circle free, and it accepted a machine that is not in the
enumeration.
If it skips the anti-program, then that program IS circle-free and >>>>>> it fails to enumerate the whole list.
Just like in your "fix", you let the decider skip some machines,
as Turings H can be decided on, your decider must be allowed to
also skip some machines to avoid misclassifying anti-H. But, you
still need to accept SOME machine that computes that value, but
*ANY* machine that it accepts will have at least one digit
different than what anti-H computes, as by the structure of antu- >>>>>> H, the machine will be the k'th machine processed by anti-H, and
anti-H will differ by it in at least the k'th digit.
It doesn't matter that you can make a machine that computes an
almost diagonal, that is in the list, the problem is that anti-H
still exists, and its output never appears in your enumeration,
and thus it is just not complete, and you proof fails.
All you are doing is demonstrating that you are fundamentally not
understand the nature of the field you are trying to talk about.
take ur ignorance to grave dick, i've come a long way to showing
fallacy and the fact u can't recognize it is a you problem rick, not
a me problem
THe problem is YOUR ignorance, and beleiving in the impossible.
why in the fuck are you people so resistant to working on things???
the fact u can't even admit the fallacies i pointed out is fucking rich,
u _have_ to paint _everything_ i said as wrong,
and i know that is just not fair dick
it's just u being a ungodly sinner
You have ADMITTED to your ignorance, but still insist that you are
smarted than the experts, whose work you don't understand.
That show just how STUPID you really are.
people like you are why mortality is still a necessary facet of this
species
how about u try being constructive instead of destructive for once?
I am, I am showing that reality is more important than claiming you
can do what has been proved impossible.
It seems your mind can't handle the fact that you are wrong, because
you feel the need to believe in the impossible.
That has DOOMED your to a life of failure.
my god rick u are fucked indeed
> do u even deserve to know truth???
>
> #god
Maybe YOU are the one that needs to change, from a life based on lying
to yourself and the world, to one that accepts reality, and seeks to
do the best you can.
You don't seem to understand what an actual computation is, and that
machines can't be built on "interfaces", only actual implementations.
This things like anti-fixed-h, once you talk about them, imply that
you have chosen your implementation for PRD, and thus your
enumeration is FIXED and can't be changed without reseting the
problem. Templates like H, or fixed-H, or anti-fixed-H don't actualy
become "machines" until you establish your claimed decider they are
going to be based on.
If your claimed PRD doesn't put (its) anti-fixed-H in the
enumeration, then anti-fixed-H isn't wrong to not put that anti-
digit in its output. And thus it generates a computable number that
no machine that PRD accepts generates, and thus that PRD is just
failing to meet its specification
If your claimed PRD tries to accept its anti-fixed-H in the
enumeration, then because of the problem you point out, it just
fails to accept only circle-free machines. It runs into the problem
that ITS version of anti- fixed-H is a different machine then the
anti- fixed-H based on the other PRD that skipped it, and these
machines will have different values of N.
On 03/13/2026 10:53 AM, Ross Finlayson wrote:
On 03/13/2026 10:25 AM, dart200 wrote:
On 3/12/26 3:53 PM, Richard Damon wrote:
On 3/12/26 2:00 PM, dart200 wrote:
On 3/12/26 3:50 AM, Richard Damon wrote:
On 3/12/26 12:15 AM, dart200 wrote:
On 3/11/26 3:53 PM, Richard Damon wrote:
On 3/10/26 12:51 PM, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On >>>>>>>>> Computable Numbers is a fallacy:Which just shows that you don't understand what the word
/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/
For any given computable sequence, there are _infinite_ circle- >>>>>>>>> free machines which compute that particular sequence. Not only >>>>>>>>> can various machines differ significantly in the specific steps >>>>>>>>> to produce the same output, machines can be changed in
superficial ways that do not meaningfully affect the steps of >>>>>>>>> computation, akin to modern no-op statements or unreachable code >>>>>>>>>
The problem of enumerating computable sequences, however, only >>>>>>>>> depends on successfully identifying _one_ circle-free machine >>>>>>>>> that computes any given computable sequences. While identifying >>>>>>>>> more than one can certainly be done, it is _not_ a requirement >>>>>>>>> for enumerating computable sequences, as _one_ machine computing >>>>>>>>> a sequence / suffices to output any and all digits of that
sequence/
The problem of enumerating computable sequences is therefore >>>>>>>>> _not_ actually equivalent to a _general process_ of enumerating >>>>>>>>> circle- free machines, as there is no need to identify all
circle-free machines which compute any given computable sequence >>>>>>>>
"Equivalent" means here.
Just like we can have "Equivalent" Turing Machines, that by very >>>>>>>> different methods and path create the exact same output, we can >>>>>>>> have two "Equivalent classification problems" that by using
different classes, come to the same result, that there exist
uncomputable problems.
put more clearly: enumerating computable sequences requires
enumerating only _and not more than_ a *subset* of circle-free
machines that does _not_ include _all_ circle-free machines,
which is just _not_ the same problem as enumerating _all_ circle- >>>>>>> free machine
Said problem is only equivalent to a _limited process_ of
enumerating circle-free machines. The machine which identifies >>>>>>>>> circle-free machines only needs the limited power of determining >>>>>>>>> _at least one_ circle-free machine for any given computable
sequence, _not all_ machines for any given computable sequence >>>>>>>>>
Because of this fallacy, the proof found on the following p247, >>>>>>>>> where an ill-defined machine 𝓗 (which attempts and fails to >>>>>>>>> compute the direct diagonal β’) is found to be undecidable in >>>>>>>>> respect to circle- free decider 𝓓; does not then prove an >>>>>>>>> impossibility for enumerating computable sequences. As the
problem of enumerating / all circle-free machines/ is _not_
equivalent to that of enumerating / just computable sequences/ >>>>>>>>>
And, your "partial_recognizer_D" also can be proved to have an >>>>>>>> uncomputable interface, and thus doesn't exist either.
Just to be clear, your claim is that you think it is possible for >>>>>>>> a PRD to exist that meets the following specifications:
1) It will ALWAYS generate an answer in finite time regardless of >>>>>>>> the number given to it.
2) No machine that it accepts will fail to produce a computable >>>>>>>> number, and thus will continue to run and produce output forever. >>>>>>>>
3) For EVERY Computable Number that exist, PRD will accept at
least one number that represents a machine that computes it.
that is a well-defined set of requirements that i believe (but have >>>>>>> no proven yet) can be computable by some PRD
that's a great, actually coherent objection rick, that i was hoping >>>>>>> someone would mention it! NO INSULTS REQUIRED!
So, if PRD exists, we can build a machine that computes an anti- >>>>>>>> diagonal by testing each number in sequence with PRD, and for each >>>>>>>> number that it accepts, it will simulate that machine until that >>>>>>>> machine generates k digits of output, k being the number of values >>>>>>>> accepted to this point, and then it outputs the opposite digit of >>>>>>>> the kth digit generated by the nth machine.
This machine must produce a computable number, as it only
simulates and uses the output of machines that PRD accepted, so by >>>>>>>> 1, PRD answered, and by 2 that machine can be simulated for as >>>>>>>> long as we want, and we WILL get to the desired digit.
Now, since this machine generates a computable number, by
condition 3, there must exist a finite number n that represents a >>>>>>>> machine that generates it that PRD will accept, and thus our
machine will simulate for k digits (which will be less than n) and >>>>>>>> output the opposite value.
Thus, whatever n you want to claim is the machine that generates >>>>>>>> the same computable number doesn't, and thus there can not exist a >>>>>>>> PRD that does what you claim.
In fact, this method works for ANY method you may want to claim >>>>>>>> allows you to compute the enumeration of the computable numbers. >>>>>>>>
This just shows that there likely IS some sort of equivalence
relationship between these two problems.
Of course, you are going to ignore this, because your logic
requires you to just be able to assume things exist that don't. >>>>>>>
it's unfortunately also a fallacy, and one that turing similarly >>>>>>> made.
*having a machine that computes a diagonal, does not actually then >>>>>>> imply it's usable to compute an anti-diagonal*
But My machine DOES compute the anti-diagonal, at least if PRD
exists.
i'm sorry richard, it does not actually do that
(please do actually read the whole post in depth before responding
eh???)
How?
It printd EXACTLY the opposite of your fixed_HJ
and the reason that's true is *same* self-referential weirdness
that stumped turing on the original paper
And HOWEVER you define your "self-referential" program, can change >>>>>> it to an equivalent program that prints the anti-diagonal.
that's not actually possible rick
No, it appears what isn't possible is for you to understand how these
things work, bcause you seem to be inherently STUPID.
consider fixed_H again, since that is the form of diagonal
computation
that PRD can accept
fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_H) { // handle self-ref
output 0 // hard coded digit 0 >>>>>>> K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory >>>>>>> output simulate_kth_digit(N,K) // kth digit
K += 1
}
N += 1
}
}
let us try using this to produce an anti-diagonal
anti_fixed_H = () -> {
for (digit in fixed_H()) { // run loop for each >>>>>>> output
anti_digit = 1-digit // of the machine
fixed_H
output anti-digit
}
}
does this actually compute an anti-diagonal, rick?
Why not?
i didn't ask those questions without reason bro, please do pay
attention:
Since the answer to does it print the anti-diagonal,
what would happen if PRD(anti_fixed_H) => TRUE?
if PRD(anti_fixed_H) => TRUE, then when N == DN(anti_fixed_H),
simulate(DN(anti_fixed_H), K) will run, which will try to simulate
fixed_H() to that K, and get caught in an infinite loop making it
"circular" ... so it doesn't compute an anti-diagonal
And thus PRD fails to meet its specification, by returning true for a
non-circle free input.
obviously PRD(anti_fixed_H) cannot return TRUE, so therefore it returns
FALSE
Sorry, but you don't seem to understand what a REQURIEMENT is.
what would happen if PRD(anti_fixed_H) => FALSE?
if PRD(anti_fixed_H) => FALSE, then when N == DN(anti_fixed_H),
fixed_H will skip simulate anti_fixed_H. this means that anti_fixed_H >>>>> will also skip producing an anti-digit to it's own output ... so it
still doesn't compute a total anti-diagonal
But if PRD(anti_fixed_H) => false, then the output of that machine at
THAT place isn't part of the output.
But there must be SOME machine that is supposed to produce the same
there is:
fixed_anti_fixed_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_fixed_H) { // handle self-ref
N += 1
continue // skip including itself
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
this computes the same thing as anti_fixed_H(), but is decidable by PRD. >>> PRD(fixed_anti_fixed_H) returns TRUE
output as anti_fixed_H, but what ever machine that is, its kth digit
will be wrong.
... err yes, the total anti-diagonal is _not_ computable. the _closest_
we can get is a sequence that includes the inverse for all computable
sequences _except_ to the anti-diagonal computation itself
But we don't care what PRD(anti_fixed_H) is.
clearly PRD(anti_fixed_H) => FALSE
Right, and NO machine accepted by PRD generetes the same computable
number as anti_fixed_H, and thus PRD just fails to meet the
requirement you gave it, so fixed_H did not actually compute a digonal >>>> of an enumeration that includes ALL computable numbers.
Thus, your claim that it does is just a pure LIE.
does either case actually output a true total anti-diagonal???
Sure, as it is exactly the opposite of the diagonal.
(hint: no)
Why not?
Which digit on the diagonal is NOT the opposite of what fixed_H
determined to be the diagonal based on PRD's enumeation order of the >>>>>> machines?
And where in that list of machines is one that computes the anti-
diagonal number that anti_fixed_D computed?
You seem to be forgetting that your PRD is what determines the
enumeration order of the machines and the digits generated, and thus >>>>>> what would be called the diagonal or the anti-diagonal.
Your problem seems to be that you aren't willing to think.
I have PROVEN that your claimed interface for partial_recognizer_D >>>>>> can not possible acheive its requirements to accept at least one
machine that generates a computable number, and thus, the
"enumeration" you generate from it is not complete, but only
partial, and thus your whole claim falls apart.
but ok bud, i still hear u coping about not "fixing" the paradox,
let's try "fixing" anti_H by handling it's self-ref akin to fixed_H. >>>>> for this i will combine the two machines for simplicity, i will leave >>>>> implementing this in the aforementioned examples for the reader:
You don't get to change it.
That is just saying you are going to ignore that you are wrong, which
just shows you don't value truth.
fixed_anti_H = () -> {
N = 0
K = 0
do {
if (N == DN(fixed_anti_H) { // handle self-ref
output ??? // what do we put here??? >>>>> K += 1
} elif (PRD(N) == TRUE) { // TRUE = satisfactory
output 1-sim_kth_digit(N,K) // kth anti-digit
K += 1
}
N += 1
}
}
see rick... in order to avoid the paradox by handling the self-ref we >>>>> need to define what it's digit on the diagonal will be. but if that's >>>>> its the output digit for it's Kth spot on the diagonal, then that's
_not_ the anti-digit to the Kth spot on the diagonal, meaning it
computes an anti-diagonal to every machine but it's own output ...
which is *not* a true anti-diagonal
Right, you didn't compute the proper anti-diagonal, but my program
DID, at least it would if PRD exists.
no it didn't. it skips computing an inverse to it's own computation
Thus, your PRD disappear in the same contradiction that Turing_D and
Turing_H does, and thus so does your fixed_H
in order for fixed_anti_H to be a true anti-diagonal we'd need to
somehow statically define an output that is inverse to what is
statically defined as the output ... and that's just a totally
nonsensical concept rick. you can only statically define it's output, >>>>> you can't statically define an output that is inverse to what it
outputs ...
No, my does, at least it would if PRD exists.
Since its sole purpose is to prove that your PRD can't exist, you
don't get to "fix" it.
therefore, you can't fix the anti-diagonal computation like you can
for the diagonal computation. the self-referential weirdness that
stumped turing is fixable _only_ for the diagonal, _not_ for the
anti- diagonal,
and therefore the anti-diagonal is still _not computable_
No, the existance of the anti-diagonal program existing if PRD exixts
is what proves that you PRD can't exist.
Just as Turing shows that his D can't exist.
Thus your system is just as non-existant as Turings, so either you
accept his proof or you need to discard yours.
The usual accounts that the infinite is not computable
doesn't contradict that each finite is computable.
Then it gets directly into accounts as modeled by
ordinary set theory.
These are then among matters of "quantifier disambiguation"
against "impredicativity".
There are various models of the Halting Problem which
is also what is being discussed, or the Branching Problem,
that vary, about usually enough "supertasks".
There are many well-known "approximation algorithms
to NP-hard problems" that reduce them to polynomial.
So, are you going to acknowledge these sorts relations
to the usual accounts of "computability" and "countability"?
The ac-countability?
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>>>>>> following p247,But if the machine can be “ill-defined”, yet provably >>>>>>>>>>>>>>>>>>> undecidable,
where an ill-defined machine 𝓗 (which attempts and >>>>>>>>>>>>>>>>>>>> fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then >>>>>>>>>>>>>>>>>>>> prove an
impossibility for enumerating computable sequences. >>>>>>>>>>>>>>>>>>>
that must mean any “better-defined” machine that also >>>>>>>>>>>>>>>>>>> satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the >>>>>>>>>>>>>>>>>> criteria to be undecidable
But they’re a subset of the “ill-defined” set that >>>>>>>>>>>>>>>>> Turing was
considering, are they not?
Unless you’re considering an entirely different set, in >>>>>>>>>>>>>>>>> which case
your argument has nothing to do with Turing.
there are two sets being conflated here:
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the >>>>>>>>>>>>>>>> solution of them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain >>>>>>>>>>>>>>> thinks he must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here. >>>>>>>>>>>>>
After all Functional Equivalence doesn't mean the same >>>>>>>>>>>>> machine or even using the same basic algorithm.
He doesn't say the two machines generated by the two >>>>>>>>>>>>>>> problems are in any way equivalent, he says that the >>>>>>>>>>>>>>> PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable >>>>>>>>>>>>>> sequences, then u could use that solution to determine >>>>>>>>>>>>>> whether any given machine is circle-free ...
No, he his saying the problems are equivalent as to the nature >>>>>>>>>>>>>
That isn't what "equivalent" means.
and if so could be used to enumerate the circle-free >>>>>>>>>>>>>> machines,
making the problem of enumerating the sets equivalent, >>>>>>>>>>>>>
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B >>>>>>>>>>>>
AND
a solution to B can be used to produce a solution to A
Where do you get that definition?
Two problems are logically eqivalent if in all models they >>>>>>>>>>> produce the same "answer".
Since the problem is the question of "Can" you do something, >>>>>>>>>>>
But it doesn't need to.
turing is wrong about this. a solution to enumerating >>>>>>>>>>>> circle- free machines can be used to produce a solution to >>>>>>>>>>>> enumerating computable numbers, but the reverse is *NOT* true >>>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and >>>>>>>>> thus are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates
circle- ftee machihes could be used to enumerate computable >>>>>>>>> numbers.
my god rick, please fucking read the not even whole paper, but >>>>>>>> at least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING >>>>>>>> PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that, >>>>>>>> | if this general process exists [for circle-free machines]
| then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the results >>>>>>> of the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine,
produce a general process to output whether a machine is circle- >>>>>>>> free or not
the _first fallacy_ is that because that isn't actually
equivalent to enumerating computable sequences (which is a
lesser problem that only needs to recognize a subset of circle- >>>>>>>> free machines), ruling out a general process for deciding
circle-free machine does _not_ actually rule out a general
process for enumerating computable numbers
A fallacy in your mind, because you don't understand what he
means by equivalent.
how can computing a _subset_ of circle-free machines be equivalent >>>>>> to compute a _total_ set of circle-free machines...???
Who said they were equivalent COMPUTATIONS.
| 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
Right, equivLWNR PROBLEM, which means both problems are either
solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other...
Nope.
Where are you getting your definitions? Because you are using the wrong ones.
All you are doing is proving your stubborn refusal to learn what you are talking about, and that you don't care you are ignorant.
which is a fallacy in this case, they are not equivalent problems
Sure they are, you just don't know what that means as you continue to
hang on to your errors because you don't understand the language you are reading.
IT seems you are just showing you don't know what the word means,
because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed out, as it
shows how much of an ass you are.
The problem of creating the computations are equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's pretty ridiculous
richard
Because I am not, you are gaslighting yourself with your false
definitions that you try to insist on.
if problems are equivalent then a solution to A can be used to solve
B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck along with
ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the field, a
field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better than
people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect.
Your problem is in you ignorance, you don't know what he is talking >>>>> about.
because it doesn't compute an anti-digit for itself and therefore >>>>>> does not form a total anti-diagonal. the anti-diagonal is still
You are just falling for your own definist fallacy,.
the _second fallacy_ is assuming that just because a diagonal is >>>>>>>> computable, the anti-diagonal becomes computable
But it is.
that one seems obvious on the surface, but when we actually dig >>>>>>>> into the details, one _cannot_ use a diagonal machine to produce >>>>>>>> an anti- diagonal machine
Sure you can. The fact that you think not is part of your problem. >>>>>>>
As I said, if computing the diagonal is possible, then to compute >>>>>>> the anti-diagonal, just reverse the value written on the
permanent output, and if you ever read from the permanent output, >>>>>>> reverse your decisions on it.
WHY can't you do that?. Note, in your self-reference hack, you don't >>>>>>
uncomputable even if a general process to enumerate computable
numbers exists
But it only doesn't because your PRD fails to provide the needed
enumeration.
if it did it would produce a cyclical machine, so obviously it can't,
And thus you admit that it can't meet the needed requirements, but
still insist that it does.
It seems you don't understand what you are talking about.
u haven't proven no machine can compute the same sequence
But that isn't the question.
that is a question, that u haven't proven
And a question that is just a strawman.
Of course other machines will compute that string, just none of the one
that PRD accepts.
Your "logic" is just based on fallacies.
It seems you don't understand how logic works.
But then, you think the world is run by fairy magic powered unicorns, so anything can be true if you wish hard enough.
No machine that PRD accepts does, which is all that is needed to show
that no variation of PRD can compute what you claim it does.
It seems you still don't understand how basic logic works.
PRD CAN'T tell it to enumerate itself, or it actively fails at
accepting only circle-free machines. (The OTHER PRD that does
creates a DIFFERENT anti-fixed-H that isn't circle free).
anti-fixed-H does compute the anti-diagonal for the (incomplete)
enumeration that your actual PRD generates, and shows that it
missed an computable number.
the other problems ur bringing up do not refute this point
Sure it does. Your "enumeration" as generated by any actual PRD is
just incomplete. You CAN'T make a PRD that meets your
specification, as it turns out that anti-fixed-H proves that it is
an uncomputable specification.
change the number references (so it is no longer actually its own >>>>>>> number, but the number of the machine it was based on).
(it either gets stuck looping trying to find it's own anti-
digit, or it skips over itself not producing a total anti-
diagonal. there's no way to hard-code a digit for itself on the >>>>>>>> anti- diagonal, like you can with the diagonal)
Which is what proves that the enumeration can't include it, and >>>>>>> thus it can't actually be a computable number.
The decider creating the enumeration is what has the problem. If >>>>>>> it accepts the anti-program itself, then that program becomes
non- circle free, and it accepted a machine that is not in the
enumeration.
If it skips the anti-program, then that program IS circle-free
and it fails to enumerate the whole list.
Just like in your "fix", you let the decider skip some machines, >>>>>>> as Turings H can be decided on, your decider must be allowed to >>>>>>> also skip some machines to avoid misclassifying anti-H. But, you >>>>>>> still need to accept SOME machine that computes that value, but >>>>>>> *ANY* machine that it accepts will have at least one digit
different than what anti-H computes, as by the structure of antu- >>>>>>> H, the machine will be the k'th machine processed by anti-H, and >>>>>>> anti-H will differ by it in at least the k'th digit.
It doesn't matter that you can make a machine that computes an
almost diagonal, that is in the list, the problem is that anti-H >>>>>>> still exists, and its output never appears in your enumeration, >>>>>>> and thus it is just not complete, and you proof fails.
All you are doing is demonstrating that you are fundamentally not
understand the nature of the field you are trying to talk about.
take ur ignorance to grave dick, i've come a long way to showing
fallacy and the fact u can't recognize it is a you problem rick, not
a me problem
THe problem is YOUR ignorance, and beleiving in the impossible.
why in the fuck are you people so resistant to working on things???
We resist working on lies
the fact u can't even admit the fallacies i pointed out is fucking
rich, u _have_ to paint _everything_ i said as wrong,
Because they aren't actually fallacies, just your own misunderstanding.
Why do you not accept that your ideas are based on impossible things?
and i know that is just not fair dick
it's just u being a ungodly sinner
I'm not the one claiming things that can't be proven.
You have ADMITTED to your ignorance, but still insist that you are
smarted than the experts, whose work you don't understand.
That show just how STUPID you really are.
people like you are why mortality is still a necessary facet of this
species
how about u try being constructive instead of destructive for once?
I am, I am showing that reality is more important than claiming you
can do what has been proved impossible.
It seems your mind can't handle the fact that you are wrong, because
you feel the need to believe in the impossible.
That has DOOMED your to a life of failure.
my god rick u are fucked indeed
> do u even deserve to know truth???
>
> #god
That is a fair question to YOU.
I showed how your PRD can't accept at least one copy of every computable number, but you refuse to look at that, by going to strawman.
anti-fixed-H *DOES* compute your anti-diagonal of the enumeration that
your PRD generates.
It also can't be generated by any of the machines accepted, because it
can't be any of the rows of that enumeration.
Thus, PRG doesn't enumerate machines for ALL of the computable numbers.
Thus, your idea blew up in a puff of purple smoke.
But, you are so stupid, you refuse to see that, but just curse the light that is showing you your errors.
Maybe YOU are the one that needs to change, from a life based on
lying to yourself and the world, to one that accepts reality, and
seeks to do the best you can.
You don't seem to understand what an actual computation is, and
that machines can't be built on "interfaces", only actual
implementations.
This things like anti-fixed-h, once you talk about them, imply that >>>>> you have chosen your implementation for PRD, and thus your
enumeration is FIXED and can't be changed without reseting the
problem. Templates like H, or fixed-H, or anti-fixed-H don't
actualy become "machines" until you establish your claimed decider
they are going to be based on.
If your claimed PRD doesn't put (its) anti-fixed-H in the
enumeration, then anti-fixed-H isn't wrong to not put that anti-
digit in its output. And thus it generates a computable number that >>>>> no machine that PRD accepts generates, and thus that PRD is just
failing to meet its specification
If your claimed PRD tries to accept its anti-fixed-H in the
enumeration, then because of the problem you point out, it just
fails to accept only circle-free machines. It runs into the problem >>>>> that ITS version of anti- fixed-H is a different machine then the
anti- fixed-H based on the other PRD that skipped it, and these
machines will have different values of N.
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:Who said they were equivalent COMPUTATIONS.
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:Where do you get that definition?
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>
there are two sets being conflated here:On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>>>>>>> following p247,But if the machine can be “ill-defined”, yet >>>>>>>>>>>>>>>>>>>> provably undecidable,
where an ill-defined machine 𝓗 (which attempts and >>>>>>>>>>>>>>>>>>>>> fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then >>>>>>>>>>>>>>>>>>>>> prove an
impossibility for enumerating computable sequences. >>>>>>>>>>>>>>>>>>>>
that must mean any “better-defined” machine that >>>>>>>>>>>>>>>>>>>> also satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the >>>>>>>>>>>>>>>>>>> criteria to be undecidable
But they’re a subset of the “ill-defined” set that >>>>>>>>>>>>>>>>>> Turing was
considering, are they not?
Unless you’re considering an entirely different set, >>>>>>>>>>>>>>>>>> in which case
your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the >>>>>>>>>>>>>>>>> solution of them as the same is a _fallacy_
But he didn't, that is just what your ignorant brain >>>>>>>>>>>>>>>> thinks he must have been talking about.
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here. >>>>>>>>>>>>>>
After all Functional Equivalence doesn't mean the same >>>>>>>>>>>>>> machine or even using the same basic algorithm.
He doesn't say the two machines generated by the two >>>>>>>>>>>>>>>> problems are in any way equivalent, he says that the >>>>>>>>>>>>>>>> PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable >>>>>>>>>>>>>>> sequences, then u could use that solution to determine >>>>>>>>>>>>>>> whether any given machine is circle-free ...
No, he his saying the problems are equivalent as to the >>>>>>>>>>>>>> nature
That isn't what "equivalent" means.
and if so could be used to enumerate the circle-free >>>>>>>>>>>>>>> machines,
making the problem of enumerating the sets equivalent, >>>>>>>>>>>>>>
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B >>>>>>>>>>>>>
AND
a solution to B can be used to produce a solution to A >>>>>>>>>>>>
Two problems are logically eqivalent if in all models they >>>>>>>>>>>> produce the same "answer".
Since the problem is the question of "Can" you do something, >>>>>>>>>>>>
But it doesn't need to.
turing is wrong about this. a solution to enumerating >>>>>>>>>>>>> circle- free machines can be used to produce a solution to >>>>>>>>>>>>> enumerating computable numbers, but the reverse is *NOT* true >>>>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and >>>>>>>>>> thus are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates >>>>>>>>>> circle- ftee machihes could be used to enumerate computable >>>>>>>>>> numbers.
my god rick, please fucking read the not even whole paper, but >>>>>>>>> at least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING >>>>>>>>> PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that, >>>>>>>>> | if this general process exists [for circle-free machines]
| then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the results >>>>>>>> of the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, >>>>>>>>> produce a general process to output whether a machine is
circle- free or not
the _first fallacy_ is that because that isn't actually
equivalent to enumerating computable sequences (which is a
lesser problem that only needs to recognize a subset of circle- >>>>>>>>> free machines), ruling out a general process for deciding
circle-free machine does _not_ actually rule out a general
process for enumerating computable numbers
A fallacy in your mind, because you don't understand what he
means by equivalent.
how can computing a _subset_ of circle-free machines be
equivalent to compute a _total_ set of circle-free machines...??? >>>>>>
| 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
Right, equivLWNR PROBLEM, which means both problems are either
solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other...
Nope.
Where are you getting your definitions? Because you are using the
wrong ones.
All you are doing is proving your stubborn refusal to learn what you
are talking about, and that you don't care you are ignorant.
which is a fallacy in this case, they are not equivalent problems
Sure they are, you just don't know what that means as you continue to
hang on to your errors because you don't understand the language you
are reading.
IT seems you are just showing you don't know what the word means,
because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed out, as it
shows how much of an ass you are.
The problem of creating the computations are equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's pretty ridiculous
richard
Because I am not, you are gaslighting yourself with your false
definitions that you try to insist on.
if problems are equivalent then a solution to A can be used to
solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck along
with ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the field,
a field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better than
people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect.
ur whole response is just a shitpile of insults and fallacies
can't wait to see u take ur ignorance to the grave dick
And thus you admit that it can't meet the needed requirements, but
Your problem is in you ignorance, you don't know what he is
talking about.
You are just falling for your own definist fallacy,.
the _second fallacy_ is assuming that just because a diagonal >>>>>>>>> is computable, the anti-diagonal becomes computable
But it is.
that one seems obvious on the surface, but when we actually dig >>>>>>>>> into the details, one _cannot_ use a diagonal machine to
produce an anti- diagonal machine
Sure you can. The fact that you think not is part of your problem. >>>>>>>>
As I said, if computing the diagonal is possible, then to
compute the anti-diagonal, just reverse the value written on the >>>>>>>> permanent output, and if you ever read from the permanent
output, reverse your decisions on it.
WHY can't you do that?. Note, in your self-reference hack, you >>>>>>>> don't
because it doesn't compute an anti-digit for itself and therefore >>>>>>> does not form a total anti-diagonal. the anti-diagonal is still >>>>>>> uncomputable even if a general process to enumerate computable
numbers exists
But it only doesn't because your PRD fails to provide the needed
enumeration.
if it did it would produce a cyclical machine, so obviously it can't, >>>>
still insist that it does.
It seems you don't understand what you are talking about.
u haven't proven no machine can compute the same sequence
But that isn't the question.
that is a question, that u haven't proven
And a question that is just a strawman.
Of course other machines will compute that string, just none of the
one that PRD accepts.
Your "logic" is just based on fallacies.
It seems you don't understand how logic works.
But then, you think the world is run by fairy magic powered unicorns,
so anything can be true if you wish hard enough.
No machine that PRD accepts does, which is all that is needed to
show that no variation of PRD can compute what you claim it does.
It seems you still don't understand how basic logic works.
PRD CAN'T tell it to enumerate itself, or it actively fails at
accepting only circle-free machines. (The OTHER PRD that does
creates a DIFFERENT anti-fixed-H that isn't circle free).
anti-fixed-H does compute the anti-diagonal for the (incomplete)
enumeration that your actual PRD generates, and shows that it
missed an computable number.
the other problems ur bringing up do not refute this point
Sure it does. Your "enumeration" as generated by any actual PRD is >>>>>> just incomplete. You CAN'T make a PRD that meets your
specification, as it turns out that anti-fixed-H proves that it is >>>>>> an uncomputable specification.
change the number references (so it is no longer actually its >>>>>>>> own number, but the number of the machine it was based on).
(it either gets stuck looping trying to find it's own anti- >>>>>>>>> digit, or it skips over itself not producing a total anti-
diagonal. there's no way to hard-code a digit for itself on the >>>>>>>>> anti- diagonal, like you can with the diagonal)
Which is what proves that the enumeration can't include it, and >>>>>>>> thus it can't actually be a computable number.
The decider creating the enumeration is what has the problem. If >>>>>>>> it accepts the anti-program itself, then that program becomes >>>>>>>> non- circle free, and it accepted a machine that is not in the >>>>>>>> enumeration.
If it skips the anti-program, then that program IS circle-free >>>>>>>> and it fails to enumerate the whole list.
Just like in your "fix", you let the decider skip some machines, >>>>>>>> as Turings H can be decided on, your decider must be allowed to >>>>>>>> also skip some machines to avoid misclassifying anti-H. But, you >>>>>>>> still need to accept SOME machine that computes that value, but >>>>>>>> *ANY* machine that it accepts will have at least one digit
different than what anti-H computes, as by the structure of
antu- H, the machine will be the k'th machine processed by anti- >>>>>>>> H, and anti-H will differ by it in at least the k'th digit.
It doesn't matter that you can make a machine that computes an >>>>>>>> almost diagonal, that is in the list, the problem is that anti-H >>>>>>>> still exists, and its output never appears in your enumeration, >>>>>>>> and thus it is just not complete, and you proof fails.
All you are doing is demonstrating that you are fundamentally not >>>>>> understand the nature of the field you are trying to talk about.
take ur ignorance to grave dick, i've come a long way to showing
fallacy and the fact u can't recognize it is a you problem rick,
not a me problem
THe problem is YOUR ignorance, and beleiving in the impossible.
why in the fuck are you people so resistant to working on things???
We resist working on lies
the fact u can't even admit the fallacies i pointed out is fucking
rich, u _have_ to paint _everything_ i said as wrong,
Because they aren't actually fallacies, just your own misunderstanding.
Why do you not accept that your ideas are based on impossible things?
and i know that is just not fair dick
it's just u being a ungodly sinner
I'm not the one claiming things that can't be proven.
You have ADMITTED to your ignorance, but still insist that you are
smarted than the experts, whose work you don't understand.
That show just how STUPID you really are.
people like you are why mortality is still a necessary facet of
this species
how about u try being constructive instead of destructive for once?
I am, I am showing that reality is more important than claiming you
can do what has been proved impossible.
It seems your mind can't handle the fact that you are wrong, because
you feel the need to believe in the impossible.
That has DOOMED your to a life of failure.
my god rick u are fucked indeed
> do u even deserve to know truth???
>
> #god
That is a fair question to YOU.
I showed how your PRD can't accept at least one copy of every
computable number, but you refuse to look at that, by going to strawman.
anti-fixed-H *DOES* compute your anti-diagonal of the enumeration that
your PRD generates.
It also can't be generated by any of the machines accepted, because it
can't be any of the rows of that enumeration.
Thus, PRG doesn't enumerate machines for ALL of the computable numbers.
Thus, your idea blew up in a puff of purple smoke.
But, you are so stupid, you refuse to see that, but just curse the
light that is showing you your errors.
Maybe YOU are the one that needs to change, from a life based on
lying to yourself and the world, to one that accepts reality, and
seeks to do the best you can.
You don't seem to understand what an actual computation is, and
that machines can't be built on "interfaces", only actual
implementations.
This things like anti-fixed-h, once you talk about them, imply
that you have chosen your implementation for PRD, and thus your
enumeration is FIXED and can't be changed without reseting the
problem. Templates like H, or fixed-H, or anti-fixed-H don't
actualy become "machines" until you establish your claimed decider >>>>>> they are going to be based on.
If your claimed PRD doesn't put (its) anti-fixed-H in the
enumeration, then anti-fixed-H isn't wrong to not put that anti-
digit in its output. And thus it generates a computable number
that no machine that PRD accepts generates, and thus that PRD is
just failing to meet its specification
If your claimed PRD tries to accept its anti-fixed-H in the
enumeration, then because of the problem you point out, it just
fails to accept only circle-free machines. It runs into the
problem that ITS version of anti- fixed-H is a different machine
then the anti- fixed-H based on the other PRD that skipped it, and >>>>>> these machines will have different values of N.
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:Who said they were equivalent COMPUTATIONS.
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:Where do you get that definition?
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>But he didn't, that is just what your ignorant brain >>>>>>>>>>>>>>>>> thinks he must have been talking about.
there are two sets being conflated here:On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>>>>>>>> following p247,But if the machine can be “ill-defined”, yet >>>>>>>>>>>>>>>>>>>>> provably undecidable,
where an ill-defined machine 𝓗 (which attempts and >>>>>>>>>>>>>>>>>>>>>> fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then >>>>>>>>>>>>>>>>>>>>>> prove an
impossibility for enumerating computable sequences. >>>>>>>>>>>>>>>>>>>>>
that must mean any “better-defined” machine that >>>>>>>>>>>>>>>>>>>>> also satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the >>>>>>>>>>>>>>>>>>>> criteria to be undecidable
But they’re a subset of the “ill-defined” set that >>>>>>>>>>>>>>>>>>> Turing was
considering, are they not?
Unless you’re considering an entirely different set, >>>>>>>>>>>>>>>>>>> in which case
your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the >>>>>>>>>>>>>>>>>> solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here. >>>>>>>>>>>>>>>
After all Functional Equivalence doesn't mean the same >>>>>>>>>>>>>>> machine or even using the same basic algorithm.
He doesn't say the two machines generated by the two >>>>>>>>>>>>>>>>> problems are in any way equivalent, he says that the >>>>>>>>>>>>>>>>> PROBLEMS are equivalent,
he's literally saying that if u can enumerate computable >>>>>>>>>>>>>>>> sequences, then u could use that solution to determine >>>>>>>>>>>>>>>> whether any given machine is circle-free ...
No, he his saying the problems are equivalent as to the >>>>>>>>>>>>>>> nature
That isn't what "equivalent" means.
and if so could be used to enumerate the circle-free >>>>>>>>>>>>>>>> machines,
making the problem of enumerating the sets equivalent, >>>>>>>>>>>>>>>
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B >>>>>>>>>>>>>>
AND
a solution to B can be used to produce a solution to A >>>>>>>>>>>>>
Two problems are logically eqivalent if in all models they >>>>>>>>>>>>> produce the same "answer".
Since the problem is the question of "Can" you do something, >>>>>>>>>>>>>
But it doesn't need to.
turing is wrong about this. a solution to enumerating >>>>>>>>>>>>>> circle- free machines can be used to produce a solution to >>>>>>>>>>>>>> enumerating computable numbers, but the reverse is *NOT* true >>>>>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and >>>>>>>>>>> thus are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates >>>>>>>>>>> circle- ftee machihes could be used to enumerate computable >>>>>>>>>>> numbers.
my god rick, please fucking read the not even whole paper, but >>>>>>>>>> at least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING >>>>>>>>>> PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that, >>>>>>>>>> | if this general process exists [for circle-free machines] >>>>>>>>>> | then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the results >>>>>>>>> of the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, >>>>>>>>>> produce a general process to output whether a machine is
circle- free or not
the _first fallacy_ is that because that isn't actually
equivalent to enumerating computable sequences (which is a >>>>>>>>>> lesser problem that only needs to recognize a subset of
circle- free machines), ruling out a general process for
deciding circle-free machine does _not_ actually rule out a >>>>>>>>>> general process for enumerating computable numbers
A fallacy in your mind, because you don't understand what he >>>>>>>>> means by equivalent.
how can computing a _subset_ of circle-free machines be
equivalent to compute a _total_ set of circle-free machines...??? >>>>>>>
| 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
Right, equivLWNR PROBLEM, which means both problems are either
solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other...
Nope.
Where are you getting your definitions? Because you are using the
wrong ones.
All you are doing is proving your stubborn refusal to learn what you
are talking about, and that you don't care you are ignorant.
which is a fallacy in this case, they are not equivalent problems
Sure they are, you just don't know what that means as you continue to
hang on to your errors because you don't understand the language you
are reading.
IT seems you are just showing you don't know what the word means,
because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed out, as it
shows how much of an ass you are.
idk why ur gaslighting me about this, but it's pretty ridiculous
The problem of creating the computations are equivalent PROBLEMS. >>>>>>
richard
Because I am not, you are gaslighting yourself with your false
definitions that you try to insist on.
if problems are equivalent then a solution to A can be used to
solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck along
with ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the field, >>>>> a field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better than
people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect.
ur whole response is just a shitpile of insults and fallacies
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that claims I am wrong?
All you have so far is a wrong definition of EQUIVALENT.
And you claim fallacy when I ask you for your source.
I guess you whole arguement is just based on the fallacy of appeal to "authority" where that "authority" is just ignorant you.
So, who is going to the grave in ignorance.
The one that actually KNOWS something, and can point to sources.
Or the idiot that claims asking for sources is a fallacy?
And thus you admit that it can't meet the needed requirements, but
Your problem is in you ignorance, you don't know what he is
talking about.
You are just falling for your own definist fallacy,.
the _second fallacy_ is assuming that just because a diagonal >>>>>>>>>> is computable, the anti-diagonal becomes computable
But it is.
that one seems obvious on the surface, but when we actually >>>>>>>>>> dig into the details, one _cannot_ use a diagonal machine to >>>>>>>>>> produce an anti- diagonal machine
Sure you can. The fact that you think not is part of your problem. >>>>>>>>>
As I said, if computing the diagonal is possible, then to
compute the anti-diagonal, just reverse the value written on >>>>>>>>> the permanent output, and if you ever read from the permanent >>>>>>>>> output, reverse your decisions on it.
WHY can't you do that?. Note, in your self-reference hack, you >>>>>>>>> don't
because it doesn't compute an anti-digit for itself and
therefore does not form a total anti-diagonal. the anti-diagonal >>>>>>>> is still uncomputable even if a general process to enumerate
computable numbers exists
But it only doesn't because your PRD fails to provide the needed >>>>>>> enumeration.
if it did it would produce a cyclical machine, so obviously it can't, >>>>>
still insist that it does.
It seems you don't understand what you are talking about.
u haven't proven no machine can compute the same sequence
But that isn't the question.
that is a question, that u haven't proven
And a question that is just a strawman.
Of course other machines will compute that string, just none of the
one that PRD accepts.
Your "logic" is just based on fallacies.
It seems you don't understand how logic works.
But then, you think the world is run by fairy magic powered unicorns,
so anything can be true if you wish hard enough.
No machine that PRD accepts does, which is all that is needed to
show that no variation of PRD can compute what you claim it does.
It seems you still don't understand how basic logic works.
PRD CAN'T tell it to enumerate itself, or it actively fails at
accepting only circle-free machines. (The OTHER PRD that does
creates a DIFFERENT anti-fixed-H that isn't circle free).
anti-fixed-H does compute the anti-diagonal for the (incomplete) >>>>>>> enumeration that your actual PRD generates, and shows that it
missed an computable number.
the other problems ur bringing up do not refute this point
Sure it does. Your "enumeration" as generated by any actual PRD >>>>>>> is just incomplete. You CAN'T make a PRD that meets your
specification, as it turns out that anti-fixed-H proves that it >>>>>>> is an uncomputable specification.
change the number references (so it is no longer actually its >>>>>>>>> own number, but the number of the machine it was based on).
(it either gets stuck looping trying to find it's own anti- >>>>>>>>>> digit, or it skips over itself not producing a total anti- >>>>>>>>>> diagonal. there's no way to hard-code a digit for itself on >>>>>>>>>> the anti- diagonal, like you can with the diagonal)
Which is what proves that the enumeration can't include it, and >>>>>>>>> thus it can't actually be a computable number.
The decider creating the enumeration is what has the problem. >>>>>>>>> If it accepts the anti-program itself, then that program
becomes non- circle free, and it accepted a machine that is not >>>>>>>>> in the enumeration.
If it skips the anti-program, then that program IS circle-free >>>>>>>>> and it fails to enumerate the whole list.
Just like in your "fix", you let the decider skip some
machines, as Turings H can be decided on, your decider must be >>>>>>>>> allowed to also skip some machines to avoid misclassifying
anti-H. But, you still need to accept SOME machine that
computes that value, but *ANY* machine that it accepts will >>>>>>>>> have at least one digit different than what anti-H computes, as >>>>>>>>> by the structure of antu- H, the machine will be the k'th
machine processed by anti- H, and anti-H will differ by it in >>>>>>>>> at least the k'th digit.
It doesn't matter that you can make a machine that computes an >>>>>>>>> almost diagonal, that is in the list, the problem is that anti- >>>>>>>>> H still exists, and its output never appears in your
enumeration, and thus it is just not complete, and you proof >>>>>>>>> fails.
All you are doing is demonstrating that you are fundamentally not >>>>>>> understand the nature of the field you are trying to talk about.
take ur ignorance to grave dick, i've come a long way to showing
fallacy and the fact u can't recognize it is a you problem rick,
not a me problem
THe problem is YOUR ignorance, and beleiving in the impossible.
why in the fuck are you people so resistant to working on things???
We resist working on lies
the fact u can't even admit the fallacies i pointed out is fucking
rich, u _have_ to paint _everything_ i said as wrong,
Because they aren't actually fallacies, just your own misunderstanding.
Why do you not accept that your ideas are based on impossible things?
and i know that is just not fair dick
it's just u being a ungodly sinner
I'm not the one claiming things that can't be proven.
You have ADMITTED to your ignorance, but still insist that you are
smarted than the experts, whose work you don't understand.
That show just how STUPID you really are.
I am, I am showing that reality is more important than claiming you >>>>> can do what has been proved impossible.
people like you are why mortality is still a necessary facet of
this species
how about u try being constructive instead of destructive for once? >>>>>
It seems your mind can't handle the fact that you are wrong,
because you feel the need to believe in the impossible.
That has DOOMED your to a life of failure.
my god rick u are fucked indeed
> do u even deserve to know truth???
>
> #god
That is a fair question to YOU.
I showed how your PRD can't accept at least one copy of every
computable number, but you refuse to look at that, by going to strawman. >>>
anti-fixed-H *DOES* compute your anti-diagonal of the enumeration
that your PRD generates.
It also can't be generated by any of the machines accepted, because
it can't be any of the rows of that enumeration.
Thus, PRG doesn't enumerate machines for ALL of the computable numbers.
Thus, your idea blew up in a puff of purple smoke.
But, you are so stupid, you refuse to see that, but just curse the
light that is showing you your errors.
Maybe YOU are the one that needs to change, from a life based on
lying to yourself and the world, to one that accepts reality, and
seeks to do the best you can.
You don't seem to understand what an actual computation is, and >>>>>>> that machines can't be built on "interfaces", only actual
implementations.
This things like anti-fixed-h, once you talk about them, imply
that you have chosen your implementation for PRD, and thus your >>>>>>> enumeration is FIXED and can't be changed without reseting the
problem. Templates like H, or fixed-H, or anti-fixed-H don't
actualy become "machines" until you establish your claimed
decider they are going to be based on.
If your claimed PRD doesn't put (its) anti-fixed-H in the
enumeration, then anti-fixed-H isn't wrong to not put that anti- >>>>>>> digit in its output. And thus it generates a computable number
that no machine that PRD accepts generates, and thus that PRD is >>>>>>> just failing to meet its specification
If your claimed PRD tries to accept its anti-fixed-H in the
enumeration, then because of the problem you point out, it just >>>>>>> fails to accept only circle-free machines. It runs into the
problem that ITS version of anti- fixed-H is a different machine >>>>>>> then the anti- fixed-H based on the other PRD that skipped it,
and these machines will have different values of N.
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:Who said they were equivalent COMPUTATIONS.
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:Where do you get that definition?
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>But he didn't, that is just what your ignorant brain >>>>>>>>>>>>>>>>>> thinks he must have been talking about.
there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>>>>>>>>> following p247,But if the machine can be “ill-defined”, yet >>>>>>>>>>>>>>>>>>>>>> provably undecidable,
where an ill-defined machine 𝓗 (which attempts >>>>>>>>>>>>>>>>>>>>>>> and fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then >>>>>>>>>>>>>>>>>>>>>>> prove an
impossibility for enumerating computable sequences. >>>>>>>>>>>>>>>>>>>>>>
that must mean any “better-defined” machine that >>>>>>>>>>>>>>>>>>>>>> also satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the >>>>>>>>>>>>>>>>>>>>> criteria to be undecidable
But they’re a subset of the “ill-defined” set that >>>>>>>>>>>>>>>>>>>> Turing was
considering, are they not?
Unless you’re considering an entirely different set, >>>>>>>>>>>>>>>>>>>> in which case
your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the >>>>>>>>>>>>>>>>>>> solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>>
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here. >>>>>>>>>>>>>>>>
After all Functional Equivalence doesn't mean the same >>>>>>>>>>>>>>>> machine or even using the same basic algorithm. >>>>>>>>>>>>>>>>
No, he his saying the problems are equivalent as to the >>>>>>>>>>>>>>>> nature
He doesn't say the two machines generated by the two >>>>>>>>>>>>>>>>>> problems are in any way equivalent, he says that the >>>>>>>>>>>>>>>>>> PROBLEMS are equivalent,
he's literally saying that if u can enumerate >>>>>>>>>>>>>>>>> computable sequences, then u could use that solution to >>>>>>>>>>>>>>>>> determine whether any given machine is circle-free ... >>>>>>>>>>>>>>>>
That isn't what "equivalent" means.
and if so could be used to enumerate the circle-free >>>>>>>>>>>>>>>>> machines,
making the problem of enumerating the sets equivalent, >>>>>>>>>>>>>>>>
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B >>>>>>>>>>>>>>>
AND
a solution to B can be used to produce a solution to A >>>>>>>>>>>>>>
Two problems are logically eqivalent if in all models they >>>>>>>>>>>>>> produce the same "answer".
Since the problem is the question of "Can" you do something, >>>>>>>>>>>>>>
turing is wrong about this. a solution to enumerating >>>>>>>>>>>>>>> circle- free machines can be used to produce a solution >>>>>>>>>>>>>>> to enumerating computable numbers, but the reverse is >>>>>>>>>>>>>>> *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, and >>>>>>>>>>>> thus are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates >>>>>>>>>>>> circle- ftee machihes could be used to enumerate computable >>>>>>>>>>>> numbers.
my god rick, please fucking read the not even whole paper, >>>>>>>>>>> but at least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME FUCKING >>>>>>>>>>> PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing that, >>>>>>>>>>> | if this general process exists [for circle-free machines] >>>>>>>>>>> | then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the
results of the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, >>>>>>>>>>> produce a general process to output whether a machine is >>>>>>>>>>> circle- free or not
the _first fallacy_ is that because that isn't actually >>>>>>>>>>> equivalent to enumerating computable sequences (which is a >>>>>>>>>>> lesser problem that only needs to recognize a subset of >>>>>>>>>>> circle- free machines), ruling out a general process for >>>>>>>>>>> deciding circle-free machine does _not_ actually rule out a >>>>>>>>>>> general process for enumerating computable numbers
A fallacy in your mind, because you don't understand what he >>>>>>>>>> means by equivalent.
how can computing a _subset_ of circle-free machines be
equivalent to compute a _total_ set of circle-free machines...??? >>>>>>>>
| 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
Right, equivLWNR PROBLEM, which means both problems are either
solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other...
Nope.
Where are you getting your definitions? Because you are using the
wrong ones.
All you are doing is proving your stubborn refusal to learn what you
are talking about, and that you don't care you are ignorant.
which is a fallacy in this case, they are not equivalent problems
Sure they are, you just don't know what that means as you continue
to hang on to your errors because you don't understand the language
you are reading.
IT seems you are just showing you don't know what the word means, >>>>>> because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed out, as
it shows how much of an ass you are.
idk why ur gaslighting me about this, but it's pretty ridiculous >>>>>>> richard
The problem of creating the computations are equivalent PROBLEMS. >>>>>>>
Because I am not, you are gaslighting yourself with your false
definitions that you try to insist on.
if problems are equivalent then a solution to A can be used to
solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck along >>>>>>> with ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the
field, a field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better than >>>>>> people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect.
ur whole response is just a shitpile of insults and fallacies
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that claims I
am wrong?
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always without agreeing more than a spattering of times over literally months of engagement,
on top of heaps of unjustified insults completely unbecoming of anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:Who said they were equivalent COMPUTATIONS.
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:Where do you get that definition?
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote:
On 3/13/26 3:30 AM, dart200 wrote:
On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>But he didn't, that is just what your ignorant brain >>>>>>>>>>>>>>>>>>> thinks he must have been talking about.
there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>>>>>>>>>> following p247,But if the machine can be “ill-defined”, yet >>>>>>>>>>>>>>>>>>>>>>> provably undecidable,
where an ill-defined machine 𝓗 (which attempts >>>>>>>>>>>>>>>>>>>>>>>> and fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then >>>>>>>>>>>>>>>>>>>>>>>> prove an
impossibility for enumerating computable sequences. >>>>>>>>>>>>>>>>>>>>>>>
that must mean any “better-defined” machine that >>>>>>>>>>>>>>>>>>>>>>> also satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the >>>>>>>>>>>>>>>>>>>>>> criteria to be undecidable
But they’re a subset of the “ill-defined” set that >>>>>>>>>>>>>>>>>>>>> Turing was
considering, are they not?
Unless you’re considering an entirely different >>>>>>>>>>>>>>>>>>>>> set, in which case
your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the >>>>>>>>>>>>>>>>>>>> solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>>>
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here. >>>>>>>>>>>>>>>>>
After all Functional Equivalence doesn't mean the same >>>>>>>>>>>>>>>>> machine or even using the same basic algorithm. >>>>>>>>>>>>>>>>>
No, he his saying the problems are equivalent as to the >>>>>>>>>>>>>>>>> nature
He doesn't say the two machines generated by the two >>>>>>>>>>>>>>>>>>> problems are in any way equivalent, he says that the >>>>>>>>>>>>>>>>>>> PROBLEMS are equivalent,
he's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>> computable sequences, then u could use that solution >>>>>>>>>>>>>>>>>> to determine whether any given machine is circle-free ... >>>>>>>>>>>>>>>>>
That isn't what "equivalent" means.
and if so could be used to enumerate the circle-free >>>>>>>>>>>>>>>>>> machines,
making the problem of enumerating the sets equivalent, >>>>>>>>>>>>>>>>>
if problem A and B are equivalent:
then a solution to A can be used to produce a solution to B >>>>>>>>>>>>>>>>
AND
a solution to B can be used to produce a solution to A >>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all models >>>>>>>>>>>>>>> they produce the same "answer".
Since the problem is the question of "Can" you do something, >>>>>>>>>>>>>>>
turing is wrong about this. a solution to enumerating >>>>>>>>>>>>>>>> circle- free machines can be used to produce a solution >>>>>>>>>>>>>>>> to enumerating computable numbers, but the reverse is >>>>>>>>>>>>>>>> *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, >>>>>>>>>>>>> and thus are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates >>>>>>>>>>>>> circle- ftee machihes could be used to enumerate computable >>>>>>>>>>>>> numbers.
my god rick, please fucking read the not even whole paper, >>>>>>>>>>>> but at least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME >>>>>>>>>>>> FUCKING PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing >>>>>>>>>>>> that,
| if this general process exists [for circle-free machines] >>>>>>>>>>>> | then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the >>>>>>>>>>> results of the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree...
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, >>>>>>>>>>>> produce a general process to output whether a machine is >>>>>>>>>>>> circle- free or not
the _first fallacy_ is that because that isn't actually >>>>>>>>>>>> equivalent to enumerating computable sequences (which is a >>>>>>>>>>>> lesser problem that only needs to recognize a subset of >>>>>>>>>>>> circle- free machines), ruling out a general process for >>>>>>>>>>>> deciding circle-free machine does _not_ actually rule out a >>>>>>>>>>>> general process for enumerating computable numbers
A fallacy in your mind, because you don't understand what he >>>>>>>>>>> means by equivalent.
how can computing a _subset_ of circle-free machines be
equivalent to compute a _total_ set of circle-free machines...??? >>>>>>>>>
| 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
Right, equivLWNR PROBLEM, which means both problems are either
solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other...
Nope.
Where are you getting your definitions? Because you are using the
wrong ones.
All you are doing is proving your stubborn refusal to learn what
you are talking about, and that you don't care you are ignorant.
which is a fallacy in this case, they are not equivalent problems
Sure they are, you just don't know what that means as you continue
to hang on to your errors because you don't understand the language >>>>> you are reading.
IT seems you are just showing you don't know what the word means, >>>>>>> because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed out, as
it shows how much of an ass you are.
idk why ur gaslighting me about this, but it's pretty ridiculous >>>>>>>> richard
The problem of creating the computations are equivalent PROBLEMS. >>>>>>>>
Because I am not, you are gaslighting yourself with your false
definitions that you try to insist on.
if problems are equivalent then a solution to A can be used to >>>>>>>> solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck along >>>>>>>> with ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the
field, a field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better than >>>>>>> people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect.
ur whole response is just a shitpile of insults and fallacies
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that claims I
am wrong?
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always without
agreeing more than a spattering of times over literally months of
engagement,
on top of heaps of unjustified insults completely unbecoming of anyone
engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
If you can show an actual error I am making, with sources to back up--
your claims, present them.
The problem is you KNOW that you don't know what you are talking about because you have ADMITTED to not actually studing more that a few
papers, but you think you are smarter than the people who wrote them.
YOU are the one flying to the grave in a crashing plane of ignorance.
I will note, that just like with Peter Olcott, YOU are the one that
started the insults, showing whose mind is in the gutter.
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote:Where do you get that definition?
On 3/13/26 12:41 PM, dart200 wrote:
On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote:
| the problem of enumerating computable sequences is >>>>>>>>>>>>>>>>>>> equivalentOn 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>But he didn't, that is just what your ignorant brain >>>>>>>>>>>>>>>>>>>> thinks he must have been talking about. >>>>>>>>>>>>>>>>>>>
there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found on the >>>>>>>>>>>>>>>>>>>>>>>>> following p247,
where an ill-defined machine 𝓗 (which attempts >>>>>>>>>>>>>>>>>>>>>>>>> and fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not then >>>>>>>>>>>>>>>>>>>>>>>>> prove an
impossibility for enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>> sequences.
But if the machine can be “ill-defined”, yet >>>>>>>>>>>>>>>>>>>>>>>> provably undecidable,
that must mean any “better-defined” machine that >>>>>>>>>>>>>>>>>>>>>>>> also satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the >>>>>>>>>>>>>>>>>>>>>>> criteria to be undecidable
But they’re a subset of the “ill-defined” set that
Turing was
considering, are they not?
Unless you’re considering an entirely different >>>>>>>>>>>>>>>>>>>>>> set, in which case
your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the >>>>>>>>>>>>>>>>>>>>> solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here. >>>>>>>>>>>>>>>>>>
After all Functional Equivalence doesn't mean the same >>>>>>>>>>>>>>>>>> machine or even using the same basic algorithm. >>>>>>>>>>>>>>>>>>
He doesn't say the two machines generated by the two >>>>>>>>>>>>>>>>>>>> problems are in any way equivalent, he says that the >>>>>>>>>>>>>>>>>>>> PROBLEMS are equivalent,
he's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>>> computable sequences, then u could use that solution >>>>>>>>>>>>>>>>>>> to determine whether any given machine is circle- >>>>>>>>>>>>>>>>>>> free ...
No, he his saying the problems are equivalent as to >>>>>>>>>>>>>>>>>> the nature
That isn't what "equivalent" means.
and if so could be used to enumerate the circle-free >>>>>>>>>>>>>>>>>>> machines,
making the problem of enumerating the sets equivalent, >>>>>>>>>>>>>>>>>>
if problem A and B are equivalent:
then a solution to A can be used to produce a solution >>>>>>>>>>>>>>>>> to B
AND
a solution to B can be used to produce a solution to A >>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all models >>>>>>>>>>>>>>>> they produce the same "answer".
Since the problem is the question of "Can" you do >>>>>>>>>>>>>>>> something,
turing is wrong about this. a solution to enumerating >>>>>>>>>>>>>>>>> circle- free machines can be used to produce a solution >>>>>>>>>>>>>>>>> to enumerating computable numbers, but the reverse is >>>>>>>>>>>>>>>>> *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, >>>>>>>>>>>>>> and thus are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that generates >>>>>>>>>>>>>> circle- ftee machihes could be used to enumerate
computable numbers.
my god rick, please fucking read the not even whole paper, >>>>>>>>>>>>> but at least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME >>>>>>>>>>>>> FUCKING PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing >>>>>>>>>>>>> that,
| if this general process exists [for circle-free machines] >>>>>>>>>>>>> | then there is a machine which computes β
do i need to spell out why with even more detail???
And B is the machine that computes the diagonals of the >>>>>>>>>>>> results of the enumeration of circle-free machines.
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing machine, >>>>>>>>>>>>> produce a general process to output whether a machine is >>>>>>>>>>>>> circle- free or not
the _first fallacy_ is that because that isn't actually >>>>>>>>>>>>> equivalent to enumerating computable sequences (which is a >>>>>>>>>>>>> lesser problem that only needs to recognize a subset of >>>>>>>>>>>>> circle- free machines), ruling out a general process for >>>>>>>>>>>>> deciding circle-free machine does _not_ actually rule out a >>>>>>>>>>>>> general process for enumerating computable numbers
A fallacy in your mind, because you don't understand what he >>>>>>>>>>>> means by equivalent.
how can computing a _subset_ of circle-free machines be >>>>>>>>>>> equivalent to compute a _total_ set of circle-free
machines...???
Who said they were equivalent COMPUTATIONS.
| 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
Right, equivLWNR PROBLEM, which means both problems are either >>>>>>>> solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other...
Nope.
Where are you getting your definitions? Because you are using the >>>>>> wrong ones.
All you are doing is proving your stubborn refusal to learn what
you are talking about, and that you don't care you are ignorant.
Sure they are, you just don't know what that means as you continue >>>>>> to hang on to your errors because you don't understand the
which is a fallacy in this case, they are not equivalent problems >>>>>>
language you are reading.
IT seems you are just showing you don't know what the word
means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed out, as >>>>>> it shows how much of an ass you are.
idk why ur gaslighting me about this, but it's pretty
The problem of creating the computations are equivalent PROBLEMS. >>>>>>>>>
ridiculous richard
Because I am not, you are gaslighting yourself with your false >>>>>>>> definitions that you try to insist on.
if problems are equivalent then a solution to A can be used to >>>>>>>>> solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck along >>>>>>>>> with ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the
field, a field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better
than people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect.
ur whole response is just a shitpile of insults and fallacies
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that claims
I am wrong?
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always
without agreeing more than a spattering of times over literally
months of engagement,
on top of heaps of unjustified insults completely unbecoming of
anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject stupidity where u "win" when the opponent gives on u being and endless fucking troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_
u've never "won" an argument here in the decades u wasted ur life here
get back to helping america bomb muzzies for their joo overlords, that's
all ur good for
If you can show an actual error I am making, with sources to back up
your claims, present them.
The problem is you KNOW that you don't know what you are talking about
because you have ADMITTED to not actually studing more that a few
papers, but you think you are smarter than the people who wrote them.
YOU are the one flying to the grave in a crashing plane of ignorance.
I will note, that just like with Peter Olcott, YOU are the one that
started the insults, showing whose mind is in the gutter.
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:Nope.
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:And B is the machine that computes the diagonals of the >>>>>>>>>>>>> results of the enumeration of circle-free machines.
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote:Where do you get that definition?
On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote:
| the problem of enumerating computable sequences is >>>>>>>>>>>>>>>>>>>> equivalentOn 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>But he didn't, that is just what your ignorant >>>>>>>>>>>>>>>>>>>>> brain thinks he must have been talking about. >>>>>>>>>>>>>>>>>>>>
there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found on >>>>>>>>>>>>>>>>>>>>>>>>>> the following p247,
where an ill-defined machine 𝓗 (which attempts >>>>>>>>>>>>>>>>>>>>>>>>>> and fails to
compute the direct diagonal β’) is found to be >>>>>>>>>>>>>>>>>>>>>>>>>> undecidable in
respect to circle-free decider 𝓓; does not >>>>>>>>>>>>>>>>>>>>>>>>>> then prove an
impossibility for enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>>> sequences.
But if the machine can be “ill-defined”, yet >>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable,
that must mean any “better-defined” machine >>>>>>>>>>>>>>>>>>>>>>>>> that also satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the >>>>>>>>>>>>>>>>>>>>>>>> criteria to be undecidable
But they’re a subset of the “ill-defined” set >>>>>>>>>>>>>>>>>>>>>>> that Turing was
considering, are they not?
Unless you’re considering an entirely different >>>>>>>>>>>>>>>>>>>>>>> set, in which case
your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the >>>>>>>>>>>>>>>>>>>>>> solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means here. >>>>>>>>>>>>>>>>>>>
After all Functional Equivalence doesn't mean the >>>>>>>>>>>>>>>>>>> same machine or even using the same basic algorithm. >>>>>>>>>>>>>>>>>>>
He doesn't say the two machines generated by the >>>>>>>>>>>>>>>>>>>>> two problems are in any way equivalent, he says >>>>>>>>>>>>>>>>>>>>> that the PROBLEMS are equivalent,
he's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>>>> computable sequences, then u could use that solution >>>>>>>>>>>>>>>>>>>> to determine whether any given machine is circle- >>>>>>>>>>>>>>>>>>>> free ...
No, he his saying the problems are equivalent as to >>>>>>>>>>>>>>>>>>> the nature
That isn't what "equivalent" means.
and if so could be used to enumerate the circle-free >>>>>>>>>>>>>>>>>>>> machines,
making the problem of enumerating the sets equivalent, >>>>>>>>>>>>>>>>>>>
if problem A and B are equivalent:
then a solution to A can be used to produce a solution >>>>>>>>>>>>>>>>>> to B
AND
a solution to B can be used to produce a solution to A >>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all models >>>>>>>>>>>>>>>>> they produce the same "answer".
Since the problem is the question of "Can" you do >>>>>>>>>>>>>>>>> something,
turing is wrong about this. a solution to enumerating >>>>>>>>>>>>>>>>>> circle- free machines can be used to produce a >>>>>>>>>>>>>>>>>> solution to enumerating computable numbers, but the >>>>>>>>>>>>>>>>>> reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, >>>>>>>>>>>>>>> and thus are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even whole paper, >>>>>>>>>>>>>> but at least the _section_ rick please...
i'm tired of answering questions that ARE ON THE SAME >>>>>>>>>>>>>> FUCKING PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by showing >>>>>>>>>>>>>> that,
| if this general process exists [for circle-free machines] >>>>>>>>>>>>>> | then there is a machine which computes β
do i need to spell out why with even more detail??? >>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines
<=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing >>>>>>>>>>>>>> machine, produce a general process to output whether a >>>>>>>>>>>>>> machine is circle- free or not
A fallacy in your mind, because you don't understand what >>>>>>>>>>>>> he means by equivalent.
the _first fallacy_ is that because that isn't actually >>>>>>>>>>>>>> equivalent to enumerating computable sequences (which is a >>>>>>>>>>>>>> lesser problem that only needs to recognize a subset of >>>>>>>>>>>>>> circle- free machines), ruling out a general process for >>>>>>>>>>>>>> deciding circle-free machine does _not_ actually rule out >>>>>>>>>>>>>> a general process for enumerating computable numbers >>>>>>>>>>>>>
how can computing a _subset_ of circle-free machines be >>>>>>>>>>>> equivalent to compute a _total_ set of circle-free
machines...???
Who said they were equivalent COMPUTATIONS.
| 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
Right, equivLWNR PROBLEM, which means both problems are either >>>>>>>>> solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other... >>>>>>>
Where are you getting your definitions? Because you are using the >>>>>>> wrong ones.
All you are doing is proving your stubborn refusal to learn what >>>>>>> you are talking about, and that you don't care you are ignorant. >>>>>>>
Sure they are, you just don't know what that means as you
which is a fallacy in this case, they are not equivalent problems >>>>>>>
continue to hang on to your errors because you don't understand >>>>>>> the language you are reading.
IT seems you are just showing you don't know what the word
means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed out, >>>>>>> as it shows how much of an ass you are.
The problem of creating the computations are equivalent >>>>>>>>>>> PROBLEMS.
idk why ur gaslighting me about this, but it's pretty
ridiculous richard
Because I am not, you are gaslighting yourself with your false >>>>>>>>> definitions that you try to insist on.
if problems are equivalent then a solution to A can be used to >>>>>>>>>> solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck >>>>>>>>>> along with ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the >>>>>>>>> field, a field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better >>>>>>>>> than people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect.
ur whole response is just a shitpile of insults and fallacies
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that claims >>>>> I am wrong?
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always
without agreeing more than a spattering of times over literally
months of engagement,
on top of heaps of unjustified insults completely unbecoming of
anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject stupidity where
u "win" when the opponent gives on u being and endless fucking troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_
Nope, if the opponent is as brain dead as you show yourself, it isn't a
matter of convincing you, but protecting the naive from you lies.
It seems you are just admitting that you are stuck in your lies and just can't think because, like Olcott, you have successfully gatlit yourself
into being convinced of your lies.
u've never "won" an argument here in the decades u wasted ur life here
get back to helping america bomb muzzies for their joo overlords,
that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but your bluster.
Sorry, that won't cut it for begging people to give you money to carry
your ignorant ideas further, as you are just showing there isn't
anything to base going farther on.
If you can show an actual error I am making, with sources to back up
your claims, present them.
The problem is you KNOW that you don't know what you are talking
about because you have ADMITTED to not actually studing more that a
few papers, but you think you are smarter than the people who wrote
them.
YOU are the one flying to the grave in a crashing plane of ignorance.
I will note, that just like with Peter Olcott, YOU are the one that
started the insults, showing whose mind is in the gutter.
Hi,
Somebody just changed the Vanilla Prolog
meta interpreter from:
solve(true) :- !.
solve((A,B)) :- !, solve(A), solve(B).
solve(H) :- clause(H, B), solve(B).
Into a cycle checking interpreter. It makes
certain Datalog programs and queries complete,
but it doesn't make Horn clauses complete:
solve(A) :- solve(A, []).
solve(true, _) :- !.
solve((A,B), L) :- !, solve(A, L), solve(B, L).
solve(A, L) :- member(B, L), A =@= B, !, fail.
solve(H, L) :- clause(H, B), solve(B, [H|L]).
Bye
P.S.: Here is a proof for Datalog:
Since Datalog has only constants and variables,
no function symbols at all, there are only finitely
many literals at runtime modulo (=@=)/2.
Q.E.D.
dart200 schrieb:
The following claim from p246 of Turing’s seminal paper On Computable
Numbers is a fallacy:
/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/
For any given computable sequence, there are _infinite_ circle-free
machines which compute that particular sequence. Not only can various
machines differ significantly in the specific steps to produce the
same output, machines can be changed in superficial ways that do not
meaningfully affect the steps of computation, akin to modern no-op
statements or unreachable code
The problem of enumerating computable sequences, however, only depends
on successfully identifying _one_ circle-free machine that computes
any given computable sequences. While identifying more than one can
certainly be done, it is _not_ a requirement for enumerating
computable sequences, as _one_ machine computing a sequence /suffices
to output any and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-free
machines, as there is no need to identify all circle-free machines
which compute any given computable sequence
Said problem is only equivalent to a _limited process_ of enumerating
circle-free machines. The machine which identifies circle-free
machines only needs the limited power of determining _at least one_
circle-free machine for any given computable sequence, _not all_
machines for any given computable sequence
Because of this fallacy, the proof found on the following p247, where
an ill-defined machine 𝓗 (which attempts and fails to compute the
direct diagonal β’) is found to be undecidable in respect to
circle-free decider 𝓓; does not then prove an impossibility for
enumerating computable sequences. As the problem of enumerating /all
circle-free machines/ is _not_ equivalent to that of enumerating /just
computable sequences/
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:ur whole response is just a shitpile of insults and fallacies
On 3/14/26 8:08 PM, Richard Damon wrote:Nope.
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:And B is the machine that computes the diagonals of the >>>>>>>>>>>>>> results of the enumeration of circle-free machines. >>>>>>>>>>>>>>
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote:
On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote:Where do you get that definition?
On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>
| the problem of enumerating computable sequences >>>>>>>>>>>>>>>>>>>>> is equivalentBut he didn't, that is just what your ignorant >>>>>>>>>>>>>>>>>>>>>> brain thinks he must have been talking about. >>>>>>>>>>>>>>>>>>>>>there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 >>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
Because of this fallacy, the proof found on >>>>>>>>>>>>>>>>>>>>>>>>>>> the following p247,
where an ill-defined machine 𝓗 (which >>>>>>>>>>>>>>>>>>>>>>>>>>> attempts and fails to
compute the direct diagonal β’) is found to >>>>>>>>>>>>>>>>>>>>>>>>>>> be undecidable in
respect to circle-free decider 𝓓; does not >>>>>>>>>>>>>>>>>>>>>>>>>>> then prove an
impossibility for enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>>>> sequences.
But if the machine can be “ill-defined”, yet >>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable,
that must mean any “better-defined” machine >>>>>>>>>>>>>>>>>>>>>>>>>> that also satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the >>>>>>>>>>>>>>>>>>>>>>>>> criteria to be undecidable
But they’re a subset of the “ill-defined” set >>>>>>>>>>>>>>>>>>>>>>>> that Turing was
considering, are they not?
Unless you’re considering an entirely different >>>>>>>>>>>>>>>>>>>>>>>> set, in which case
your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating the >>>>>>>>>>>>>>>>>>>>>>> solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>>
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means >>>>>>>>>>>>>>>>>>>> here.
After all Functional Equivalence doesn't mean the >>>>>>>>>>>>>>>>>>>> same machine or even using the same basic algorithm. >>>>>>>>>>>>>>>>>>>>
He doesn't say the two machines generated by the >>>>>>>>>>>>>>>>>>>>>> two problems are in any way equivalent, he says >>>>>>>>>>>>>>>>>>>>>> that the PROBLEMS are equivalent,
he's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>>>>> computable sequences, then u could use that >>>>>>>>>>>>>>>>>>>>> solution to determine whether any given machine is >>>>>>>>>>>>>>>>>>>>> circle- free ...
No, he his saying the problems are equivalent as to >>>>>>>>>>>>>>>>>>>> the nature
That isn't what "equivalent" means.
and if so could be used to enumerate the circle- >>>>>>>>>>>>>>>>>>>>> free machines,
making the problem of enumerating the sets equivalent, >>>>>>>>>>>>>>>>>>>>
if problem A and B are equivalent:
then a solution to A can be used to produce a >>>>>>>>>>>>>>>>>>> solution to B
AND
a solution to B can be used to produce a solution to A >>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all models >>>>>>>>>>>>>>>>>> they produce the same "answer".
Since the problem is the question of "Can" you do >>>>>>>>>>>>>>>>>> something,
turing is wrong about this. a solution to enumerating >>>>>>>>>>>>>>>>>>> circle- free machines can be used to produce a >>>>>>>>>>>>>>>>>>> solution to enumerating computable numbers, but the >>>>>>>>>>>>>>>>>>> reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was saying, >>>>>>>>>>>>>>>> and thus are trying to kill a strawman.
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even whole >>>>>>>>>>>>>>> paper, but at least the _section_ rick please... >>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON THE SAME >>>>>>>>>>>>>>> FUCKING PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by >>>>>>>>>>>>>>> showing that,
| if this general process exists [for circle-free machines] >>>>>>>>>>>>>>> | then there is a machine which computes β
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>> <=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_
sure he demonstrated that we cannot, with a turing >>>>>>>>>>>>>>> machine, produce a general process to output whether a >>>>>>>>>>>>>>> machine is circle- free or not
A fallacy in your mind, because you don't understand what >>>>>>>>>>>>>> he means by equivalent.
the _first fallacy_ is that because that isn't actually >>>>>>>>>>>>>>> equivalent to enumerating computable sequences (which is >>>>>>>>>>>>>>> a lesser problem that only needs to recognize a subset of >>>>>>>>>>>>>>> circle- free machines), ruling out a general process for >>>>>>>>>>>>>>> deciding circle-free machine does _not_ actually rule out >>>>>>>>>>>>>>> a general process for enumerating computable numbers >>>>>>>>>>>>>>
how can computing a _subset_ of circle-free machines be >>>>>>>>>>>>> equivalent to compute a _total_ set of circle-free
machines...???
Who said they were equivalent COMPUTATIONS.
| 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
Right, equivLWNR PROBLEM, which means both problems are either >>>>>>>>>> solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other... >>>>>>>>
Where are you getting your definitions? Because you are using >>>>>>>> the wrong ones.
All you are doing is proving your stubborn refusal to learn what >>>>>>>> you are talking about, and that you don't care you are ignorant. >>>>>>>>
Sure they are, you just don't know what that means as you
which is a fallacy in this case, they are not equivalent problems >>>>>>>>
continue to hang on to your errors because you don't understand >>>>>>>> the language you are reading.
IT seems you are just showing you don't know what the word >>>>>>>>>> means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed out, >>>>>>>> as it shows how much of an ass you are.
The problem of creating the computations are equivalent >>>>>>>>>>>> PROBLEMS.
idk why ur gaslighting me about this, but it's pretty
ridiculous richard
Because I am not, you are gaslighting yourself with your false >>>>>>>>>> definitions that you try to insist on.
if problems are equivalent then a solution to A can be used >>>>>>>>>>> to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck >>>>>>>>>>> along with ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the >>>>>>>>>> field, a field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better >>>>>>>>>> than people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect. >>>>>>>
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that
claims I am wrong?
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always
without agreeing more than a spattering of times over literally
months of engagement,
on top of heaps of unjustified insults completely unbecoming of
anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject stupidity where
u "win" when the opponent gives on u being and endless fucking troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_
Nope, if the opponent is as brain dead as you show yourself, it isn't a
calling me brain dead is incredibly toxic,
matter of convincing you, but protecting the naive from you lies.
this group is bunch of boomers who spent decades losing arguments
amongst themselves...
who in the fuck here is "naive"???
It seems you are just admitting that you are stuck in your lies and
just can't think because, like Olcott, you have successfully gatlit
yourself into being convinced of your lies.
i demonstrated two distinct fallacies in turing's paper, that really
aren't the hard to understand,
regardless of whether turing's conclusions are ultimately correct or not:
the fallacies are still fallacies
u've never "won" an argument here in the decades u wasted ur life here
get back to helping america bomb muzzies for their joo overlords,
that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but your bluster.
Sorry, that won't cut it for begging people to give you money to carry
who am i begging??? who around here even has money to give??? 😂😂😂
dicks and making a random ass accusations:
name a more iconic duo
your ignorant ideas further, as you are just showing there isn't
anything to base going farther on.
If you can show an actual error I am making, with sources to back up
your claims, present them.
The problem is you KNOW that you don't know what you are talking
about because you have ADMITTED to not actually studing more that a
few papers, but you think you are smarter than the people who wrote
them.
YOU are the one flying to the grave in a crashing plane of ignorance.
I will note, that just like with Peter Olcott, YOU are the one that
started the insults, showing whose mind is in the gutter.
that's a total L bro. if u fail to convince ur opponent, u _lose_
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:ur whole response is just a shitpile of insults and fallacies
On 3/14/26 8:08 PM, Richard Damon wrote:Nope.
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:And B is the machine that computes the diagonals of the >>>>>>>>>>>>>>> results of the enumeration of circle-free machines. >>>>>>>>>>>>>>>
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>Where do you get that definition?
| the problem of enumerating computable sequences >>>>>>>>>>>>>>>>>>>>>> is equivalentBut he didn't, that is just what your ignorant >>>>>>>>>>>>>>>>>>>>>>> brain thinks he must have been talking about. >>>>>>>>>>>>>>>>>>>>>>there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 >>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
Because of this fallacy, the proof found on >>>>>>>>>>>>>>>>>>>>>>>>>>>> the following p247,
where an ill-defined machine 𝓗 (which >>>>>>>>>>>>>>>>>>>>>>>>>>>> attempts and fails to
compute the direct diagonal β’) is found to >>>>>>>>>>>>>>>>>>>>>>>>>>>> be undecidable in
respect to circle-free decider 𝓓; does not >>>>>>>>>>>>>>>>>>>>>>>>>>>> then prove an
impossibility for enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>>>>> sequences.
But if the machine can be “ill-defined”, yet >>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable,
that must mean any “better-defined” machine >>>>>>>>>>>>>>>>>>>>>>>>>>> that also satisfies
those “ill-defined” criteria must be provably >>>>>>>>>>>>>>>>>>>>>>>>>>> undecidable.
the "better-defined" machine don't satisfy the >>>>>>>>>>>>>>>>>>>>>>>>>> criteria to be undecidable
But they’re a subset of the “ill-defined” set >>>>>>>>>>>>>>>>>>>>>>>>> that Turing was
considering, are they not?
Unless you’re considering an entirely different >>>>>>>>>>>>>>>>>>>>>>>>> set, in which case
your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating >>>>>>>>>>>>>>>>>>>>>>>> the solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>>>
| 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
EQUIVALENT. Not the SAME.
I guess you don't understand what EQUIVALENT means >>>>>>>>>>>>>>>>>>>>> here.
After all Functional Equivalence doesn't mean the >>>>>>>>>>>>>>>>>>>>> same machine or even using the same basic algorithm. >>>>>>>>>>>>>>>>>>>>>
he's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>>>>>> computable sequences, then u could use that >>>>>>>>>>>>>>>>>>>>>> solution to determine whether any given machine is >>>>>>>>>>>>>>>>>>>>>> circle- free ...
He doesn't say the two machines generated by the >>>>>>>>>>>>>>>>>>>>>>> two problems are in any way equivalent, he says >>>>>>>>>>>>>>>>>>>>>>> that the PROBLEMS are equivalent, >>>>>>>>>>>>>>>>>>>>>>
No, he his saying the problems are equivalent as to >>>>>>>>>>>>>>>>>>>>> the nature
and if so could be used to enumerate the circle- >>>>>>>>>>>>>>>>>>>>>> free machines,
making the problem of enumerating the sets >>>>>>>>>>>>>>>>>>>>>> equivalent,
That isn't what "equivalent" means.
if problem A and B are equivalent:
then a solution to A can be used to produce a >>>>>>>>>>>>>>>>>>>> solution to B
AND
a solution to B can be used to produce a solution to A >>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all models >>>>>>>>>>>>>>>>>>> they produce the same "answer".
Since the problem is the question of "Can" you do >>>>>>>>>>>>>>>>>>> something,
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be used to >>>>>>>>>>>>>>>>>>>> produce a solution to enumerating computable >>>>>>>>>>>>>>>>>>>> numbers, but the reverse is *NOT* true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was >>>>>>>>>>>>>>>>> saying, and thus are trying to kill a strawman. >>>>>>>>>>>>>>>>>
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even whole >>>>>>>>>>>>>>>> paper, but at least the _section_ rick please... >>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON THE SAME >>>>>>>>>>>>>>>> FUCKING PAGES WE'VE BEEN TALKING ABOUT p246:
| The simplest and most direct proof of this is by >>>>>>>>>>>>>>>> showing that,
| if this general process exists [for circle-free machines] >>>>>>>>>>>>>>>> | then there is a machine which computes β
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>> <=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_ >>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a turing >>>>>>>>>>>>>>>> machine, produce a general process to output whether a >>>>>>>>>>>>>>>> machine is circle- free or not
the _first fallacy_ is that because that isn't actually >>>>>>>>>>>>>>>> equivalent to enumerating computable sequences (which is >>>>>>>>>>>>>>>> a lesser problem that only needs to recognize a subset >>>>>>>>>>>>>>>> of circle- free machines), ruling out a general process >>>>>>>>>>>>>>>> for deciding circle-free machine does _not_ actually >>>>>>>>>>>>>>>> rule out a general process for enumerating computable >>>>>>>>>>>>>>>> numbers
A fallacy in your mind, because you don't understand what >>>>>>>>>>>>>>> he means by equivalent.
how can computing a _subset_ of circle-free machines be >>>>>>>>>>>>>> equivalent to compute a _total_ set of circle-free >>>>>>>>>>>>>> machines...???
Who said they were equivalent COMPUTATIONS.
| 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
Right, equivLWNR PROBLEM, which means both problems are >>>>>>>>>>> either solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other... >>>>>>>>>
Where are you getting your definitions? Because you are using >>>>>>>>> the wrong ones.
All you are doing is proving your stubborn refusal to learn >>>>>>>>> what you are talking about, and that you don't care you are >>>>>>>>> ignorant.
Sure they are, you just don't know what that means as you
which is a fallacy in this case, they are not equivalent problems >>>>>>>>>
continue to hang on to your errors because you don't understand >>>>>>>>> the language you are reading.
IT seems you are just showing you don't know what the word >>>>>>>>>>> means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed out, >>>>>>>>> as it shows how much of an ass you are.
The problem of creating the computations are equivalent >>>>>>>>>>>>> PROBLEMS.
idk why ur gaslighting me about this, but it's pretty >>>>>>>>>>>> ridiculous richard
Because I am not, you are gaslighting yourself with your >>>>>>>>>>> false definitions that you try to insist on.
if problems are equivalent then a solution to A can be used >>>>>>>>>>>> to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck >>>>>>>>>>>> along with ur willful ignorance and gaslighting dick
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the >>>>>>>>>>> field, a field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better >>>>>>>>>>> than people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect. >>>>>>>>
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that
claims I am wrong?
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always
without agreeing more than a spattering of times over literally
months of engagement,
on top of heaps of unjustified insults completely unbecoming of
anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject stupidity
where u "win" when the opponent gives on u being and endless fucking
troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_
Nope, if the opponent is as brain dead as you show yourself, it isn't a
calling me brain dead is incredibly toxic,
No more than what you have called me.
It seems your nature is to insult and blame others for your own failings.
matter of convincing you, but protecting the naive from you lies.
this group is bunch of boomers who spent decades losing arguments
amongst themselves...
who in the fuck here is "naive"???
YOU.
The people here mostly know what they are talking about, because they
have studied it (some like Olcott and you are the exception).
It seems you are just admitting that you are stuck in your lies and
just can't think because, like Olcott, you have successfully gatlit
yourself into being convinced of your lies.
i demonstrated two distinct fallacies in turing's paper, that really
aren't the hard to understand,
No, you demonstrated that you do don't understand what he is saying,
regardless of whether turing's conclusions are ultimately correct or not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Yes, while the title of the paper uses the topic of "Computable
Numbers", and the part of the proof focuses on the related concept of machines that compute them, he DOES show a proof, that could be
similarly used to prove the uncomputablility of the computable numbers.
Your problem is you have such a wooden and limited knowledge of what you read, you can't understand what he is doing.
u've never "won" an argument here in the decades u wasted ur life here >>>>
get back to helping america bomb muzzies for their joo overlords,
that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but your bluster. >>>
Sorry, that won't cut it for begging people to give you money to carry
who am i begging??? who around here even has money to give??? 😂😂😂 >>
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your work so you
can complete the parts that you admit have holes in them.
You seem to expect that "someone" will like your crap work enough to pay
you to continue working on it with the hope that you can materialize
your unicorn, even though they have been proven to be impossible.
YOU are the idiot, with a foul mouth, that doesn't know what he is
talking about.
your ignorant ideas further, as you are just showing there isn't
anything to base going farther on.
If you can show an actual error I am making, with sources to back
up your claims, present them.
The problem is you KNOW that you don't know what you are talking
about because you have ADMITTED to not actually studing more that a >>>>> few papers, but you think you are smarter than the people who wrote >>>>> them.
YOU are the one flying to the grave in a crashing plane of ignorance. >>>>>
I will note, that just like with Peter Olcott, YOU are the one that >>>>> started the insults, showing whose mind is in the gutter.
On 15/03/2026 19:22, dart200 wrote:
that's a total L bro. if u fail to convince ur opponent, u _lose_
That's only in some fields of argumentation such as
- student union game/theatre debate,
- law courts, and
- dysfunctional marriages.
In engineering and science we either figure out an accurate description
of reality and both win, or we fail to do so and both lose.
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:calling me brain dead is incredibly toxic,
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:ur whole response is just a shitpile of insults and fallacies >>>>>>>>>
On 3/14/26 8:08 PM, Richard Damon wrote:Nope.
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:And B is the machine that computes the diagonals of the >>>>>>>>>>>>>>>> results of the enumeration of circle-free machines. >>>>>>>>>>>>>>>>
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 >>>>>>>>>>>>>>>>>>>>>>>>>> wrote:Where do you get that definition?
if problem A and B are equivalent:EQUIVALENT. Not the SAME.| the problem of enumerating computable sequences >>>>>>>>>>>>>>>>>>>>>>> is equivalentBut he didn't, that is just what your ignorant >>>>>>>>>>>>>>>>>>>>>>>> brain thinks he must have been talking about. >>>>>>>>>>>>>>>>>>>>>>>there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>But they’re a subset of the “ill-defined” set >>>>>>>>>>>>>>>>>>>>>>>>>> that Turing was
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 >>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
Because of this fallacy, the proof found on >>>>>>>>>>>>>>>>>>>>>>>>>>>>> the following p247,
where an ill-defined machine 𝓗 (which >>>>>>>>>>>>>>>>>>>>>>>>>>>>> attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is found to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> be undecidable in
respect to circle-free decider 𝓓; does not >>>>>>>>>>>>>>>>>>>>>>>>>>>>> then prove an
impossibility for enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequences.
But if the machine can be “ill-defined”, yet >>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable,
that must mean any “better-defined” machine >>>>>>>>>>>>>>>>>>>>>>>>>>>> that also satisfies
those “ill-defined” criteria must be >>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable.
the "better-defined" machine don't satisfy >>>>>>>>>>>>>>>>>>>>>>>>>>> the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>
considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines
*all* computable sequences
these sets are _not_ bijectable, and equating >>>>>>>>>>>>>>>>>>>>>>>>> the solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>>>>
| 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 >>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what EQUIVALENT means >>>>>>>>>>>>>>>>>>>>>> here.
After all Functional Equivalence doesn't mean the >>>>>>>>>>>>>>>>>>>>>> same machine or even using the same basic algorithm. >>>>>>>>>>>>>>>>>>>>>>
he's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>>>>>>> computable sequences, then u could use that >>>>>>>>>>>>>>>>>>>>>>> solution to determine whether any given machine >>>>>>>>>>>>>>>>>>>>>>> is circle- free ...
He doesn't say the two machines generated by the >>>>>>>>>>>>>>>>>>>>>>>> two problems are in any way equivalent, he says >>>>>>>>>>>>>>>>>>>>>>>> that the PROBLEMS are equivalent, >>>>>>>>>>>>>>>>>>>>>>>
No, he his saying the problems are equivalent as >>>>>>>>>>>>>>>>>>>>>> to the nature
and if so could be used to enumerate the circle- >>>>>>>>>>>>>>>>>>>>>>> free machines,
making the problem of enumerating the sets >>>>>>>>>>>>>>>>>>>>>>> equivalent,
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to produce a >>>>>>>>>>>>>>>>>>>>> solution to B
AND
a solution to B can be used to produce a solution to A >>>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all >>>>>>>>>>>>>>>>>>>> models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" you do >>>>>>>>>>>>>>>>>>>> something,
But it doesn't need to.
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be used to >>>>>>>>>>>>>>>>>>>>> produce a solution to enumerating computable >>>>>>>>>>>>>>>>>>>>> numbers, but the reverse is *NOT* true >>>>>>>>>>>>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was >>>>>>>>>>>>>>>>>> saying, and thus are trying to kill a strawman. >>>>>>>>>>>>>>>>>>
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even whole >>>>>>>>>>>>>>>>> paper, but at least the _section_ rick please... >>>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON THE SAME >>>>>>>>>>>>>>>>> FUCKING PAGES WE'VE BEEN TALKING ABOUT p246: >>>>>>>>>>>>>>>>>
| The simplest and most direct proof of this is by >>>>>>>>>>>>>>>>> showing that,
| if this general process exists [for circle-free >>>>>>>>>>>>>>>>> machines]
| then there is a machine which computes β
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>> <=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_ >>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a turing >>>>>>>>>>>>>>>>> machine, produce a general process to output whether a >>>>>>>>>>>>>>>>> machine is circle- free or not
the _first fallacy_ is that because that isn't actually >>>>>>>>>>>>>>>>> equivalent to enumerating computable sequences (which >>>>>>>>>>>>>>>>> is a lesser problem that only needs to recognize a >>>>>>>>>>>>>>>>> subset of circle- free machines), ruling out a general >>>>>>>>>>>>>>>>> process for deciding circle-free machine does _not_ >>>>>>>>>>>>>>>>> actually rule out a general process for enumerating >>>>>>>>>>>>>>>>> computable numbers
A fallacy in your mind, because you don't understand >>>>>>>>>>>>>>>> what he means by equivalent.
how can computing a _subset_ of circle-free machines be >>>>>>>>>>>>>>> equivalent to compute a _total_ set of circle-free >>>>>>>>>>>>>>> machines...???
Who said they were equivalent COMPUTATIONS.
| 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
Right, equivLWNR PROBLEM, which means both problems are >>>>>>>>>>>> either solvable or not (under all applicable models).
_because_ a solution to one leads to a solution for the other... >>>>>>>>>>
Where are you getting your definitions? Because you are using >>>>>>>>>> the wrong ones.
All you are doing is proving your stubborn refusal to learn >>>>>>>>>> what you are talking about, and that you don't care you are >>>>>>>>>> ignorant.
which is a fallacy in this case, they are not equivalent >>>>>>>>>>> problems
Sure they are, you just don't know what that means as you >>>>>>>>>> continue to hang on to your errors because you don't
understand the language you are reading.
IT seems you are just showing you don't know what the word >>>>>>>>>>>> means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed >>>>>>>>>> out, as it shows how much of an ass you are.
The problem of creating the computations are equivalent >>>>>>>>>>>>>> PROBLEMS.
idk why ur gaslighting me about this, but it's pretty >>>>>>>>>>>>> ridiculous richard
Because I am not, you are gaslighting yourself with your >>>>>>>>>>>> false definitions that you try to insist on.
if problems are equivalent then a solution to A can be used >>>>>>>>>>>>> to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck >>>>>>>>>>>>> along with ur willful ignorance and gaslighting dick >>>>>>>>>>>>>
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in the >>>>>>>>>>>> field, a field you have admitted being untrained in.
So you admit your ignorance, but insist you must know better >>>>>>>>>>>> than people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect. >>>>>>>>>
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that
claims I am wrong?
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always
without agreeing more than a spattering of times over literally >>>>>>> months of engagement,
on top of heaps of unjustified insults completely unbecoming of >>>>>>> anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject stupidity
where u "win" when the opponent gives on u being and endless
fucking troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_
Nope, if the opponent is as brain dead as you show yourself, it isn't a >>>
No more than what you have called me.
no more u say??? u can barely construct a single sentence without adding
an insult after it...
It seems your nature is to insult and blame others for your own failings.
matter of convincing you, but protecting the naive from you lies.
this group is bunch of boomers who spent decades losing arguments
amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own ideas???
😂😂😂
The people here mostly know what they are talking about, because they
have studied it (some like Olcott and you are the exception).
It seems you are just admitting that you are stuck in your lies and
just can't think because, like Olcott, you have successfully gatlit
yourself into being convinced of your lies.
i demonstrated two distinct fallacies in turing's paper, that really
aren't the hard to understand,
No, you demonstrated that you do don't understand what he is saying,
the fact u continually try to gaslight me into thinking i haven't
understood his argument well enough is not only incredibly toxic but
let's me know ur completely fine with blatantly lying at me to "win" an argument,
that's not what the side with truth does, or even remotely needs to do.
and if u can't recognize that, i'm sorry for all fallacy u've bought
into across all the things u believe
> take them to the grave bro
>
> #god
regardless of whether turing's conclusions are ultimately correct or
not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Yes, while the title of the paper uses the topic of "Computable
Numbers", and the part of the proof focuses on the related concept of
machines that compute them, he DOES show a proof, that could be
similarly used to prove the uncomputablility of the computable numbers.
Your problem is you have such a wooden and limited knowledge of what
you read, you can't understand what he is doing.
who am i begging??? who around here even has money to give??? 😂😂😂 >>>
u've never "won" an argument here in the decades u wasted ur life here >>>>>
get back to helping america bomb muzzies for their joo overlords,
that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but your
bluster.
Sorry, that won't cut it for begging people to give you money to carry >>>
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your work so
you can complete the parts that you admit have holes in them.
yes, certain further work would take time and therefore funds, and that
kind of work will remain out of scope of this discussion. that's a
statement of fact, not "begging" u sad dishonest old man
You seem to expect that "someone" will like your crap work enough to
pay you to continue working on it with the hope that you can
materialize your unicorn, even though they have been proven to be
impossible.
YOU are the idiot, with a foul mouth, that doesn't know what he is
talking about.
your ignorant ideas further, as you are just showing there isn't
anything to base going farther on.
If you can show an actual error I am making, with sources to back >>>>>> up your claims, present them.
The problem is you KNOW that you don't know what you are talking
about because you have ADMITTED to not actually studing more that >>>>>> a few papers, but you think you are smarter than the people who
wrote them.
YOU are the one flying to the grave in a crashing plane of ignorance. >>>>>>
I will note, that just like with Peter Olcott, YOU are the one
that started the insults, showing whose mind is in the gutter.
On 3/16/26 4:26 AM, Tristan Wibberley wrote:
On 15/03/2026 19:22, dart200 wrote:
that's a total L bro. if u fail to convince ur opponent, u _lose_
That's only in some fields of argumentation such as
- student union game/theatre debate,
- law courts, and
- dysfunctional marriages.
In engineering and science we either figure out an accurate description
of reality and both win, or we fail to do so and both lose.
consensus with ur peers is incredibly important with science,
what are you talking about???
On 3/16/26 1:11 PM, dart200 wrote:
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:ur whole response is just a shitpile of insults and fallacies >>>>>>>>>>
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:_because_ a solution to one leads to a solution for the >>>>>>>>>>>> other...
On 3/14/26 12:20 PM, Richard Damon wrote:
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote:And B is the machine that computes the diagonals of the >>>>>>>>>>>>>>>>> results of the enumeration of circle-free machines. >>>>>>>>>>>>>>>>>
On 3/13/26 6:12 PM, dart200 wrote:
On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote:
On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 >>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
if problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME.| the problem of enumerating computable >>>>>>>>>>>>>>>>>>>>>>>> sequences is equivalentBut he didn't, that is just what your ignorant >>>>>>>>>>>>>>>>>>>>>>>>> brain thinks he must have been talking about. >>>>>>>>>>>>>>>>>>>>>>>>there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>But they’re a subset of the “ill-defined” set
On Tue, 10 Mar 2026 09:51:43 -0700, dart200 >>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:the "better-defined" machine don't satisfy >>>>>>>>>>>>>>>>>>>>>>>>>>>> the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof found >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 (which >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is found >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; does not >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> then prove an
impossibility for enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequences.
But if the machine can be “ill-defined”, >>>>>>>>>>>>>>>>>>>>>>>>>>>>> yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better-defined” machine >>>>>>>>>>>>>>>>>>>>>>>>>>>>> that also satisfies
those “ill-defined” criteria must be >>>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
that Turing was
considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and equating >>>>>>>>>>>>>>>>>>>>>>>>>> the solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>>>>>
| 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 >>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what EQUIVALENT >>>>>>>>>>>>>>>>>>>>>>> means here.
After all Functional Equivalence doesn't mean the >>>>>>>>>>>>>>>>>>>>>>> same machine or even using the same basic algorithm. >>>>>>>>>>>>>>>>>>>>>>>
he's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>>>>>>>> computable sequences, then u could use that >>>>>>>>>>>>>>>>>>>>>>>> solution to determine whether any given machine >>>>>>>>>>>>>>>>>>>>>>>> is circle- free ...
He doesn't say the two machines generated by >>>>>>>>>>>>>>>>>>>>>>>>> the two problems are in any way equivalent, he >>>>>>>>>>>>>>>>>>>>>>>>> says that the PROBLEMS are equivalent, >>>>>>>>>>>>>>>>>>>>>>>>
No, he his saying the problems are equivalent as >>>>>>>>>>>>>>>>>>>>>>> to the nature
and if so could be used to enumerate the circle- >>>>>>>>>>>>>>>>>>>>>>>> free machines,
making the problem of enumerating the sets >>>>>>>>>>>>>>>>>>>>>>>> equivalent,
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to produce a >>>>>>>>>>>>>>>>>>>>>> solution to B
AND
a solution to B can be used to produce a solution >>>>>>>>>>>>>>>>>>>>>> to A
Where do you get that definition?
Two problems are logically eqivalent if in all >>>>>>>>>>>>>>>>>>>>> models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" you do >>>>>>>>>>>>>>>>>>>>> something,
But it doesn't need to.
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be used to >>>>>>>>>>>>>>>>>>>>>> produce a solution to enumerating computable >>>>>>>>>>>>>>>>>>>>>> numbers, but the reverse is *NOT* true >>>>>>>>>>>>>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was >>>>>>>>>>>>>>>>>>> saying, and thus are trying to kill a strawman. >>>>>>>>>>>>>>>>>>>
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even whole >>>>>>>>>>>>>>>>>> paper, but at least the _section_ rick please... >>>>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON THE SAME >>>>>>>>>>>>>>>>>> FUCKING PAGES WE'VE BEEN TALKING ABOUT p246: >>>>>>>>>>>>>>>>>>
| The simplest and most direct proof of this is by >>>>>>>>>>>>>>>>>> showing that,
| if this general process exists [for circle-free >>>>>>>>>>>>>>>>>> machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>>> <=> enumerating computable sequence
=> diagonal is computable
=> β is computable _contradiction_ >>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a turing >>>>>>>>>>>>>>>>>> machine, produce a general process to output whether a >>>>>>>>>>>>>>>>>> machine is circle- free or not
the _first fallacy_ is that because that isn't >>>>>>>>>>>>>>>>>> actually equivalent to enumerating computable >>>>>>>>>>>>>>>>>> sequences (which is a lesser problem that only needs >>>>>>>>>>>>>>>>>> to recognize a subset of circle- free machines), >>>>>>>>>>>>>>>>>> ruling out a general process for deciding circle-free >>>>>>>>>>>>>>>>>> machine does _not_ actually rule out a general process >>>>>>>>>>>>>>>>>> for enumerating computable numbers
A fallacy in your mind, because you don't understand >>>>>>>>>>>>>>>>> what he means by equivalent.
how can computing a _subset_ of circle-free machines be >>>>>>>>>>>>>>>> equivalent to compute a _total_ set of circle-free >>>>>>>>>>>>>>>> machines...???
Who said they were equivalent COMPUTATIONS.
| 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
Right, equivLWNR PROBLEM, which means both problems are >>>>>>>>>>>>> either solvable or not (under all applicable models). >>>>>>>>>>>>
Nope.
Where are you getting your definitions? Because you are using >>>>>>>>>>> the wrong ones.
All you are doing is proving your stubborn refusal to learn >>>>>>>>>>> what you are talking about, and that you don't care you are >>>>>>>>>>> ignorant.
which is a fallacy in this case, they are not equivalent >>>>>>>>>>>> problems
Sure they are, you just don't know what that means as you >>>>>>>>>>> continue to hang on to your errors because you don't
understand the language you are reading.
IT seems you are just showing you don't know what the word >>>>>>>>>>>>> means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed >>>>>>>>>>> out, as it shows how much of an ass you are.
The problem of creating the computations are equivalent >>>>>>>>>>>>>>> PROBLEMS.
idk why ur gaslighting me about this, but it's pretty >>>>>>>>>>>>>> ridiculous richard
Because I am not, you are gaslighting yourself with your >>>>>>>>>>>>> false definitions that you try to insist on.
if problems are equivalent then a solution to A can be >>>>>>>>>>>>>> used to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck >>>>>>>>>>>>>> along with ur willful ignorance and gaslighting dick >>>>>>>>>>>>>>
But where do you get your definition of equivalent.
As I have pointed out, that isn't the definition used in >>>>>>>>>>>>> the field, a field you have admitted being untrained in. >>>>>>>>>>>>>
So you admit your ignorance, but insist you must know >>>>>>>>>>>>> better than people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect. >>>>>>>>>>
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that >>>>>>>>> claims I am wrong?
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always >>>>>>>> without agreeing more than a spattering of times over literally >>>>>>>> months of engagement,
on top of heaps of unjustified insults completely unbecoming of >>>>>>>> anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject stupidity
where u "win" when the opponent gives on u being and endless
fucking troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_
Nope, if the opponent is as brain dead as you show yourself, it
isn't a
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence without
adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown increadible ignorance and failure in logic. You just have a foul mouth.
It seems your nature is to insult and blame others for your own
failings.
matter of convincing you, but protecting the naive from you lies.
this group is bunch of boomers who spent decades losing arguments
amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own ideas???
😂😂😂
Trying to. After all, I need to do something to wake you up to your own
lies that you have gaslit yourself with.
The people here mostly know what they are talking about, because they
have studied it (some like Olcott and you are the exception).
It seems you are just admitting that you are stuck in your lies and >>>>> just can't think because, like Olcott, you have successfully gatlit >>>>> yourself into being convinced of your lies.
i demonstrated two distinct fallacies in turing's paper, that really
aren't the hard to understand,
No, you demonstrated that you do don't understand what he is saying,
the fact u continually try to gaslight me into thinking i haven't
understood his argument well enough is not only incredibly toxic but
let's me know ur completely fine with blatantly lying at me to "win"
an argument,
But it isn't gaslighting, it is FACT.
that's not what the side with truth does, or even remotely needs to
do. and if u can't recognize that, i'm sorry for all fallacy u've
bought into across all the things u believe
Claiming truth is gaslighting is just gaslighting.
> take them to the grave bro
>
> #god
You got a source for that?
Or do you have the same "god" complex as Olcott?
regardless of whether turing's conclusions are ultimately correct or
not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Then show me your source.
Your claim is just the gaslighting you claim I am doing.
Yes, while the title of the paper uses the topic of "Computable
Numbers", and the part of the proof focuses on the related concept of
machines that compute them, he DOES show a proof, that could be
similarly used to prove the uncomputablility of the computable numbers.
Your problem is you have such a wooden and limited knowledge of what
you read, you can't understand what he is doing.
who am i begging??? who around here even has money to give??? 😂😂😂 >>>>
u've never "won" an argument here in the decades u wasted ur life >>>>>> here
get back to helping america bomb muzzies for their joo overlords, >>>>>> that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but your
bluster.
Sorry, that won't cut it for begging people to give you money to carry >>>>
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your work so
you can complete the parts that you admit have holes in them.
yes, certain further work would take time and therefore funds, and
that kind of work will remain out of scope of this discussion. that's
a statement of fact, not "begging" u sad dishonest old man
But, since you idea have been proven wrong, and you don't even try to
refute it, why should anyone support that work.
As I said, Your PRD couldn't have accepted ANY machine that generates
the same computable number as anti-fixed-H, and thus the enumeration it generates is NOT complete.
Thus, nothing you have done with it meets the requirements for the computation you talk about, as it, by definition, starts with a complete enumeration.
There is not contradiction in my anti-fixed-H if the enumeration isn't complete, but you also are proven to just be a liar about your claim of showing a way to compute a diagonal of a complete enumeration of the computable numbers.
You seem to expect that "someone" will like your crap work enough to
pay you to continue working on it with the hope that you can
materialize your unicorn, even though they have been proven to be
impossible.
YOU are the idiot, with a foul mouth, that doesn't know what he is
talking about.
your ignorant ideas further, as you are just showing there isn't
anything to base going farther on.
If you can show an actual error I am making, with sources to back >>>>>>> up your claims, present them.
The problem is you KNOW that you don't know what you are talking >>>>>>> about because you have ADMITTED to not actually studing more that >>>>>>> a few papers, but you think you are smarter than the people who >>>>>>> wrote them.
YOU are the one flying to the grave in a crashing plane of
ignorance.
I will note, that just like with Peter Olcott, YOU are the one
that started the insults, showing whose mind is in the gutter.
On 3/16/26 1:13 PM, dart200 wrote:
On 3/16/26 4:26 AM, Tristan Wibberley wrote:
On 15/03/2026 19:22, dart200 wrote:
that's a total L bro. if u fail to convince ur opponent, u _lose_
That's only in some fields of argumentation such as
- student union game/theatre debate,
- law courts, and
- dysfunctional marriages.
In engineering and science we either figure out an accurate description
of reality and both win, or we fail to do so and both lose.
consensus with ur peers is incredibly important with science,
what are you talking about???
But consensus only comes by showing convincing evidence.
Science will challenge new ideas, but once there is PROOF (and not just
wild conjecture) it will move to it.
But of course, since you can't actually build a proof of your ideas,
they will just keep getting rejected.
Maybe if you try to address the errors being pointed out, you might get
a bit more traction.
Just swearing at them won't get anyone to think your ideas are good, but just relegates them to the gutter they belong in.
On 3/16/26 6:50 PM, Richard Damon wrote:
On 3/16/26 1:11 PM, dart200 wrote:
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:Nope, if the opponent is as brain dead as you show yourself, it
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:
On 3/15/26 12:27 AM, dart200 wrote:ur whole response is just a shitpile of insults and fallacies >>>>>>>>>>>
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:_because_ a solution to one leads to a solution for the >>>>>>>>>>>>> other...
On 3/14/26 12:20 PM, Richard Damon wrote:Right, equivLWNR PROBLEM, which means both problems are >>>>>>>>>>>>>> either solvable or not (under all applicable models). >>>>>>>>>>>>>
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/26 6:12 PM, dart200 wrote:And B is the machine that computes the diagonals of >>>>>>>>>>>>>>>>>> the results of the enumeration of circle-free machines. >>>>>>>>>>>>>>>>>>
On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 >>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
if problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME.| the problem of enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>> sequences is equivalentBut he didn't, that is just what your ignorant >>>>>>>>>>>>>>>>>>>>>>>>>> brain thinks he must have been talking about. >>>>>>>>>>>>>>>>>>>>>>>>>there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro >>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:But they’re a subset of the “ill-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>> set that Turing was
the "better-defined" machine don't satisfy >>>>>>>>>>>>>>>>>>>>>>>>>>>>> the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote:
Because of this fallacy, the proof found >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 (which >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is found >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; does >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> not then prove an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility for enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequences.
But if the machine can be “ill-defined”, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine that also satisfies >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> those “ill-defined” criteria must be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and equating >>>>>>>>>>>>>>>>>>>>>>>>>>> the solution of them as the same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>>>>>>
| 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 >>>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what EQUIVALENT >>>>>>>>>>>>>>>>>>>>>>>> means here.
After all Functional Equivalence doesn't mean >>>>>>>>>>>>>>>>>>>>>>>> the same machine or even using the same basic >>>>>>>>>>>>>>>>>>>>>>>> algorithm.
he's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>>>>>>>>> computable sequences, then u could use that >>>>>>>>>>>>>>>>>>>>>>>>> solution to determine whether any given machine >>>>>>>>>>>>>>>>>>>>>>>>> is circle- free ...
He doesn't say the two machines generated by >>>>>>>>>>>>>>>>>>>>>>>>>> the two problems are in any way equivalent, he >>>>>>>>>>>>>>>>>>>>>>>>>> says that the PROBLEMS are equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>
No, he his saying the problems are equivalent as >>>>>>>>>>>>>>>>>>>>>>>> to the nature
and if so could be used to enumerate the >>>>>>>>>>>>>>>>>>>>>>>>> circle- free machines,
making the problem of enumerating the sets >>>>>>>>>>>>>>>>>>>>>>>>> equivalent,
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to produce a >>>>>>>>>>>>>>>>>>>>>>> solution to B
AND
a solution to B can be used to produce a solution >>>>>>>>>>>>>>>>>>>>>>> to A
Where do you get that definition?
Two problems are logically eqivalent if in all >>>>>>>>>>>>>>>>>>>>>> models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" you do >>>>>>>>>>>>>>>>>>>>>> something,
But it doesn't need to.
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be used to >>>>>>>>>>>>>>>>>>>>>>> produce a solution to enumerating computable >>>>>>>>>>>>>>>>>>>>>>> numbers, but the reverse is *NOT* true >>>>>>>>>>>>>>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was >>>>>>>>>>>>>>>>>>>> saying, and thus are trying to kill a strawman. >>>>>>>>>>>>>>>>>>>>
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even whole >>>>>>>>>>>>>>>>>>> paper, but at least the _section_ rick please... >>>>>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON THE SAME >>>>>>>>>>>>>>>>>>> FUCKING PAGES WE'VE BEEN TALKING ABOUT p246: >>>>>>>>>>>>>>>>>>>
| The simplest and most direct proof of this is by >>>>>>>>>>>>>>>>>>> showing that,
| if this general process exists [for circle-free >>>>>>>>>>>>>>>>>>> machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>>
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>>>> <=> enumerating computable sequence >>>>>>>>>>>>>>>>>>> => diagonal is computable
=> β is computable _contradiction_ >>>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a turing >>>>>>>>>>>>>>>>>>> machine, produce a general process to output whether >>>>>>>>>>>>>>>>>>> a machine is circle- free or not
A fallacy in your mind, because you don't understand >>>>>>>>>>>>>>>>>> what he means by equivalent.
the _first fallacy_ is that because that isn't >>>>>>>>>>>>>>>>>>> actually equivalent to enumerating computable >>>>>>>>>>>>>>>>>>> sequences (which is a lesser problem that only needs >>>>>>>>>>>>>>>>>>> to recognize a subset of circle- free machines), >>>>>>>>>>>>>>>>>>> ruling out a general process for deciding circle-free >>>>>>>>>>>>>>>>>>> machine does _not_ actually rule out a general >>>>>>>>>>>>>>>>>>> process for enumerating computable numbers >>>>>>>>>>>>>>>>>>
how can computing a _subset_ of circle-free machines be >>>>>>>>>>>>>>>>> equivalent to compute a _total_ set of circle-free >>>>>>>>>>>>>>>>> machines...???
Who said they were equivalent COMPUTATIONS.
| 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 >>>>>>>>>>>>>>
Nope.
Where are you getting your definitions? Because you are >>>>>>>>>>>> using the wrong ones.
All you are doing is proving your stubborn refusal to learn >>>>>>>>>>>> what you are talking about, and that you don't care you are >>>>>>>>>>>> ignorant.
which is a fallacy in this case, they are not equivalent >>>>>>>>>>>>> problems
Sure they are, you just don't know what that means as you >>>>>>>>>>>> continue to hang on to your errors because you don't
understand the language you are reading.
IT seems you are just showing you don't know what the word >>>>>>>>>>>>>> means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed >>>>>>>>>>>> out, as it shows how much of an ass you are.
The problem of creating the computations are equivalent >>>>>>>>>>>>>>>> PROBLEMS.
idk why ur gaslighting me about this, but it's pretty >>>>>>>>>>>>>>> ridiculous richard
Because I am not, you are gaslighting yourself with your >>>>>>>>>>>>>> false definitions that you try to insist on.
if problems are equivalent then a solution to A can be >>>>>>>>>>>>>>> used to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the fuck >>>>>>>>>>>>>>> along with ur willful ignorance and gaslighting dick >>>>>>>>>>>>>>>
But where do you get your definition of equivalent. >>>>>>>>>>>>>>
As I have pointed out, that isn't the definition used in >>>>>>>>>>>>>> the field, a field you have admitted being untrained in. >>>>>>>>>>>>>>
So you admit your ignorance, but insist you must know >>>>>>>>>>>>>> better than people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the effect. >>>>>>>>>>>
can't wait to see u take ur ignorance to the grave dick
So, what is my error, with actual SOURCES for your data that >>>>>>>>>> claims I am wrong?
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always >>>>>>>>> without agreeing more than a spattering of times over literally >>>>>>>>> months of engagement,
on top of heaps of unjustified insults completely unbecoming of >>>>>>>>> anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject stupidity >>>>>>> where u "win" when the opponent gives on u being and endless
fucking troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_ >>>>>>
isn't a
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence without
adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown increadible
ignorance and failure in logic. You just have a foul mouth.
AHAHAHAHA, how motherfucking toxic do u have to try to rationalize
insults as "factually based"????
😂😂😂
It seems your nature is to insult and blame others for your own
failings.
matter of convincing you, but protecting the naive from you lies.
this group is bunch of boomers who spent decades losing arguments
amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own ideas???
😂😂😂
Trying to. After all, I need to do something to wake you up to your
own lies that you have gaslit yourself with.
lol, i fixed turing's diagonal in a way that hasn't been addressed in literature before
(and before u try to make yet another baseless claim that it must have
been, show me the proof instead of baselessly just claiming u fucking twat)
The people here mostly know what they are talking about, because
they have studied it (some like Olcott and you are the exception).
It seems you are just admitting that you are stuck in your lies
and just can't think because, like Olcott, you have successfully
gatlit yourself into being convinced of your lies.
i demonstrated two distinct fallacies in turing's paper, that
really aren't the hard to understand,
No, you demonstrated that you do don't understand what he is saying,
the fact u continually try to gaslight me into thinking i haven't
understood his argument well enough is not only incredibly toxic but
let's me know ur completely fine with blatantly lying at me to "win"
an argument,
But it isn't gaslighting, it is FACT.
idk how u read everything i wrote and try to claim i just don't understand,
i have been manipulating his ideas in ways that have never been done
before, and u can't even acknowledge that i understand his ideas??? sheesh,
i'm swimming thru a swamp of endless gaslighting, fostered a toxic
mentality festering the fundamentals of math hostile to any sort of meaningful innovation at the core for some ungodly reason
that's not what the side with truth does, or even remotely needs to
do. and if u can't recognize that, i'm sorry for all fallacy u've
bought into across all the things u believe
Claiming truth is gaslighting is just gaslighting.
> take them to the grave bro
>
> #god
You got a source for that?
Or do you have the same "god" complex as Olcott?
oh does he also claim that _we are god_ ???
regardless of whether turing's conclusions are ultimately correct
or not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Then show me your source.
my source for fallacies that *i* pointed out???
i would be source for that... duh?
Your claim is just the gaslighting you claim I am doing.
Yes, while the title of the paper uses the topic of "Computable
Numbers", and the part of the proof focuses on the related concept
of machines that compute them, he DOES show a proof, that could be
similarly used to prove the uncomputablility of the computable numbers. >>>>
Your problem is you have such a wooden and limited knowledge of what
you read, you can't understand what he is doing.
u've never "won" an argument here in the decades u wasted ur life >>>>>>> here
get back to helping america bomb muzzies for their joo overlords, >>>>>>> that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but your
bluster.
Sorry, that won't cut it for begging people to give you money to
carry
who am i begging??? who around here even has money to give??? 😂😂😂
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your work so
you can complete the parts that you admit have holes in them.
yes, certain further work would take time and therefore funds, and
that kind of work will remain out of scope of this discussion. that's
a statement of fact, not "begging" u sad dishonest old man
But, since you idea have been proven wrong, and you don't even try to
refute it, why should anyone support that work.
As I said, Your PRD couldn't have accepted ANY machine that generates
the same computable number as anti-fixed-H, and thus the enumeration
it generates is NOT complete.
right that is a thorn in my side atm, and there are several avenues i'm exploring in thot on how to deal with that
- first of all i'm not entirely convinced there isn't some strategy
i'm missing that might still yet get it on the diagonal. we haven't even build an enumeration of computable numbers cause we haven't discussed
the dedpuing logic, and i don't know how that impact the current
predicament
- and even if so do we actually care about the computations being
done outside of the decidably enumerable set? can we build a way to
identify and classify what they do?
- and/or is this the wall i needed to justify the jump to
constructing RTMs, cause the "incompleteness" there happens due to
actual lies built into the computations...
look i acknowledged ur argument!
which is literally more generous than you've ever been towards me, cause u've never admitted an ounce of validity of my words, despite
understanding them well enough to pounce on any and all criticisms...
now *that* is fking toxic bro, and if u think i'm going to be swayed by
such toxicity, well i know some therapists u can talk to about that
kinda negative mindset rick, their teachings served me well thus far
Thus, nothing you have done with it meets the requirements for the
computation you talk about, as it, by definition, starts with a
complete enumeration.
There is not contradiction in my anti-fixed-H if the enumeration isn't
complete, but you also are proven to just be a liar about your claim
of showing a way to compute a diagonal of a complete enumeration of
the computable numbers.
You seem to expect that "someone" will like your crap work enough to
pay you to continue working on it with the hope that you can
materialize your unicorn, even though they have been proven to be
impossible.
YOU are the idiot, with a foul mouth, that doesn't know what he is
talking about.
your ignorant ideas further, as you are just showing there isn't
anything to base going farther on.
If you can show an actual error I am making, with sources to
back up your claims, present them.
The problem is you KNOW that you don't know what you are talking >>>>>>>> about because you have ADMITTED to not actually studing more
that a few papers, but you think you are smarter than the people >>>>>>>> who wrote them.
YOU are the one flying to the grave in a crashing plane of
ignorance.
I will note, that just like with Peter Olcott, YOU are the one >>>>>>>> that started the insults, showing whose mind is in the gutter.
On 3/16/26 6:54 PM, Richard Damon wrote:
On 3/16/26 1:13 PM, dart200 wrote:
On 3/16/26 4:26 AM, Tristan Wibberley wrote:
On 15/03/2026 19:22, dart200 wrote:
that's a total L bro. if u fail to convince ur opponent, u _lose_
That's only in some fields of argumentation such as
- student union game/theatre debate,
- law courts, and
- dysfunctional marriages.
In engineering and science we either figure out an accurate description >>>> of reality and both win, or we fail to do so and both lose.
consensus with ur peers is incredibly important with science,
what are you talking about???
But consensus only comes by showing convincing evidence.
Science will challenge new ideas, but once there is PROOF (and not
just wild conjecture) it will move to it.
yeah where's the PROOF for that rick??? lol
But of course, since you can't actually build a proof of your ideas,
they will just keep getting rejected.
Maybe if you try to address the errors being pointed out, you might
get a bit more traction.
Just swearing at them won't get anyone to think your ideas are good,
but just relegates them to the gutter they belong in.
people who confuse tact with correctness deserve to take that utter
nonsense to the grave, rick
i'm only reflecting the abject toxicity and unwillingness to cooperate
right back where it comes from: people like you
On 3/17/26 12:55 AM, dart200 wrote:
On 3/16/26 6:50 PM, Richard Damon wrote:
On 3/16/26 1:11 PM, dart200 wrote:
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:Nope, if the opponent is as brain dead as you show yourself, it >>>>>>> isn't a
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:So, what is my error, with actual SOURCES for your data that >>>>>>>>>>> claims I am wrong?
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:_because_ a solution to one leads to a solution for the >>>>>>>>>>>>>> other...
On 3/14/26 12:20 PM, Richard Damon wrote:Right, equivLWNR PROBLEM, which means both problems are >>>>>>>>>>>>>>> either solvable or not (under all applicable models). >>>>>>>>>>>>>>
On 3/14/26 1:29 PM, dart200 wrote:
On 3/14/26 4:48 AM, Richard Damon wrote:
On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:12 PM, dart200 wrote:And B is the machine that computes the diagonals of >>>>>>>>>>>>>>>>>>> the results of the enumeration of circle-free machines. >>>>>>>>>>>>>>>>>>>
On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, dart200 >>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
if problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME.there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:But they’re a subset of the “ill-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>> set that Turing was
the "better-defined" machine don't satisfy >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote:
Because of this fallacy, the proof found >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 (which >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is found >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; does >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> not then prove an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility for enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequences.
But if the machine can be “ill-defined”, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine that also satisfies >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> those “ill-defined” criteria must be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and >>>>>>>>>>>>>>>>>>>>>>>>>>>> equating the solution of them as the same is >>>>>>>>>>>>>>>>>>>>>>>>>>>> a _fallacy_
But he didn't, that is just what your >>>>>>>>>>>>>>>>>>>>>>>>>>> ignorant brain thinks he must have been >>>>>>>>>>>>>>>>>>>>>>>>>>> talking about.
| 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 >>>>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what EQUIVALENT >>>>>>>>>>>>>>>>>>>>>>>>> means here.
After all Functional Equivalence doesn't mean >>>>>>>>>>>>>>>>>>>>>>>>> the same machine or even using the same basic >>>>>>>>>>>>>>>>>>>>>>>>> algorithm.
No, he his saying the problems are equivalent >>>>>>>>>>>>>>>>>>>>>>>>> as to the naturehe's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences, then u could use that >>>>>>>>>>>>>>>>>>>>>>>>>> solution to determine whether any given >>>>>>>>>>>>>>>>>>>>>>>>>> machine is circle- free ... >>>>>>>>>>>>>>>>>>>>>>>>>
He doesn't say the two machines generated by >>>>>>>>>>>>>>>>>>>>>>>>>>> the two problems are in any way equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>>> he says that the PROBLEMS are equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>>
and if so could be used to enumerate the >>>>>>>>>>>>>>>>>>>>>>>>>> circle- free machines,
making the problem of enumerating the sets >>>>>>>>>>>>>>>>>>>>>>>>>> equivalent,
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>> solution to B
AND
a solution to B can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>> solution to A
Where do you get that definition? >>>>>>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all >>>>>>>>>>>>>>>>>>>>>>> models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" you do >>>>>>>>>>>>>>>>>>>>>>> something,
But it doesn't need to.
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be used to >>>>>>>>>>>>>>>>>>>>>>>> produce a solution to enumerating computable >>>>>>>>>>>>>>>>>>>>>>>> numbers, but the reverse is *NOT* true >>>>>>>>>>>>>>>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was >>>>>>>>>>>>>>>>>>>>> saying, and thus are trying to kill a strawman. >>>>>>>>>>>>>>>>>>>>>
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even whole >>>>>>>>>>>>>>>>>>>> paper, but at least the _section_ rick please... >>>>>>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON THE >>>>>>>>>>>>>>>>>>>> SAME FUCKING PAGES WE'VE BEEN TALKING ABOUT p246: >>>>>>>>>>>>>>>>>>>>
| The simplest and most direct proof of this is by >>>>>>>>>>>>>>>>>>>> showing that,
| if this general process exists [for circle-free >>>>>>>>>>>>>>>>>>>> machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>>>
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>>>>> <=> enumerating computable sequence >>>>>>>>>>>>>>>>>>>> => diagonal is computable
=> β is computable _contradiction_ >>>>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a turing >>>>>>>>>>>>>>>>>>>> machine, produce a general process to output whether >>>>>>>>>>>>>>>>>>>> a machine is circle- free or not
A fallacy in your mind, because you don't understand >>>>>>>>>>>>>>>>>>> what he means by equivalent.
the _first fallacy_ is that because that isn't >>>>>>>>>>>>>>>>>>>> actually equivalent to enumerating computable >>>>>>>>>>>>>>>>>>>> sequences (which is a lesser problem that only needs >>>>>>>>>>>>>>>>>>>> to recognize a subset of circle- free machines), >>>>>>>>>>>>>>>>>>>> ruling out a general process for deciding circle- >>>>>>>>>>>>>>>>>>>> free machine does _not_ actually rule out a general >>>>>>>>>>>>>>>>>>>> process for enumerating computable numbers >>>>>>>>>>>>>>>>>>>
how can computing a _subset_ of circle-free machines >>>>>>>>>>>>>>>>>> be equivalent to compute a _total_ set of circle-free >>>>>>>>>>>>>>>>>> machines...???
Who said they were equivalent COMPUTATIONS.
| 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 >>>>>>>>>>>>>>>
Nope.
Where are you getting your definitions? Because you are >>>>>>>>>>>>> using the wrong ones.
All you are doing is proving your stubborn refusal to learn >>>>>>>>>>>>> what you are talking about, and that you don't care you are >>>>>>>>>>>>> ignorant.
which is a fallacy in this case, they are not equivalent >>>>>>>>>>>>>> problems
Sure they are, you just don't know what that means as you >>>>>>>>>>>>> continue to hang on to your errors because you don't >>>>>>>>>>>>> understand the language you are reading.
IT seems you are just showing you don't know what the >>>>>>>>>>>>>>> word means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed >>>>>>>>>>>>> out, as it shows how much of an ass you are.
The problem of creating the computations are equivalent >>>>>>>>>>>>>>>>> PROBLEMS.
idk why ur gaslighting me about this, but it's pretty >>>>>>>>>>>>>>>> ridiculous richard
Because I am not, you are gaslighting yourself with your >>>>>>>>>>>>>>> false definitions that you try to insist on.
if problems are equivalent then a solution to A can be >>>>>>>>>>>>>>>> used to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the >>>>>>>>>>>>>>>> fuck along with ur willful ignorance and gaslighting dick >>>>>>>>>>>>>>>>
But where do you get your definition of equivalent. >>>>>>>>>>>>>>>
As I have pointed out, that isn't the definition used in >>>>>>>>>>>>>>> the field, a field you have admitted being untrained in. >>>>>>>>>>>>>>>
So you admit your ignorance, but insist you must know >>>>>>>>>>>>>>> better than people who actually know something.
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the >>>>>>>>>>>>> effect.
ur whole response is just a shitpile of insults and fallacies >>>>>>>>>>>>
can't wait to see u take ur ignorance to the grave dick >>>>>>>>>>>
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence always >>>>>>>>>> without agreeing more than a spattering of times over
literally months of engagement,
on top of heaps of unjustified insults completely unbecoming >>>>>>>>>> of anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject stupidity >>>>>>>> where u "win" when the opponent gives on u being and endless
fucking troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_ >>>>>>>
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence without
adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown increadible
ignorance and failure in logic. You just have a foul mouth.
AHAHAHAHA, how motherfucking toxic do u have to try to rationalize
insults as "factually based"????
😂😂😂
What is "toxic" about truth? It is only toxic to people who live on lies.
Shows how stupid you are.
Note, YOU are the one providing the evidence to prove my statements, by
just ignoring the facts.
Of course, what you are proving is you don't understand what truth or
facts actually are, since you logic is based on being able to presume something without proof.
It seems your nature is to insult and blame others for your own
failings.
matter of convincing you, but protecting the naive from you lies. >>>>>>this group is bunch of boomers who spent decades losing arguments >>>>>> amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own ideas???
😂😂😂
Trying to. After all, I need to do something to wake you up to your
own lies that you have gaslit yourself with.
lol, i fixed turing's diagonal in a way that hasn't been addressed in
literature before
No you haven't, because you don't understand the requirements.
The first problem is, you haven't created the enumeration required to compute the diagonal of.
We know this, because we can convert your fixed-H to be anti-fixed-H
that outputs the opposite digits that fixed-H does (using the trick of fixed-H, using the number of fixed-H, not anti-fixed-H), and thus shows
that if fixed-H is computing the diagonal, anti-fixed-H is computing the anti-diagonal, but we also see that this anti-diagonal isn't in the enumeration, and thus the enumeration can't be complete.
If it isn't complete, then the diagonal isn't the diagonal of the enumeration that Turing was talking about.
(and before u try to make yet another baseless claim that it must have
been, show me the proof instead of baselessly just claiming u fucking
twat)
And what is wrong about this proof.
Your enumeration generated by PRD just can not be COMPLETE, including at least one instance of EVERY computable number.
PROVE ME WRONG.
The people here mostly know what they are talking about, because
they have studied it (some like Olcott and you are the exception).
It seems you are just admitting that you are stuck in your lies >>>>>>> and just can't think because, like Olcott, you have successfully >>>>>>> gatlit yourself into being convinced of your lies.
i demonstrated two distinct fallacies in turing's paper, that
really aren't the hard to understand,
No, you demonstrated that you do don't understand what he is saying,
the fact u continually try to gaslight me into thinking i haven't
understood his argument well enough is not only incredibly toxic but
let's me know ur completely fine with blatantly lying at me to "win"
an argument,
But it isn't gaslighting, it is FACT.
idk how u read everything i wrote and try to claim i just don't
understand,
Because you keep claiming things that aren't there.
i have been manipulating his ideas in ways that have never been done
before, and u can't even acknowledge that i understand his ideas???
sheesh,
Right, by twisting words to not mean what they mean.
i'm swimming thru a swamp of endless gaslighting, fostered a toxic
mentality festering the fundamentals of math hostile to any sort of
meaningful innovation at the core for some ungodly reason
Yes, the endless gaslighting that you have done to yourself, causing you
to think that people point out truth to you are gaslighting you.
The fact you can't actually prove anything should be your first sign
that you have something wrong.
Your world is just built on your own lies and fantasies.
that's not what the side with truth does, or even remotely needs to
do. and if u can't recognize that, i'm sorry for all fallacy u've
bought into across all the things u believe
Claiming truth is gaslighting is just gaslighting.
> take them to the grave bro
>
> #god
You got a source for that?
Or do you have the same "god" complex as Olcott?
oh does he also claim that _we are god_ ???
No, just him.
It seems you like to "quote" things, claiming them to be from "god" (as
he signs them). That seems to imply you think you have a special link to him.
regardless of whether turing's conclusions are ultimately correct >>>>>> or not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Then show me your source.
my source for fallacies that *i* pointed out???
i would be source for that... duh?
So, what is that source?
That is your problem, you have no sources but your own misunderstanding.
That come from your gaslighting of yourself to brainwash you into
thinking you don't need sources.
Your claim is just the gaslighting you claim I am doing.
Yes, while the title of the paper uses the topic of "Computable
Numbers", and the part of the proof focuses on the related concept
of machines that compute them, he DOES show a proof, that could be
similarly used to prove the uncomputablility of the computable
numbers.
Your problem is you have such a wooden and limited knowledge of
what you read, you can't understand what he is doing.
u've never "won" an argument here in the decades u wasted ur
life here
get back to helping america bomb muzzies for their joo
overlords, that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but your >>>>>>> bluster.
Sorry, that won't cut it for begging people to give you money to >>>>>>> carry
who am i begging??? who around here even has money to give??? 😂😂😂
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your work so >>>>> you can complete the parts that you admit have holes in them.
yes, certain further work would take time and therefore funds, and
that kind of work will remain out of scope of this discussion.
that's a statement of fact, not "begging" u sad dishonest old man
But, since you idea have been proven wrong, and you don't even try to
refute it, why should anyone support that work.
As I said, Your PRD couldn't have accepted ANY machine that generates
the same computable number as anti-fixed-H, and thus the enumeration
it generates is NOT complete.
right that is a thorn in my side atm, and there are several avenues
i'm exploring in thot on how to deal with that
- first of all i'm not entirely convinced there isn't some strategy
i'm missing that might still yet get it on the diagonal. we haven't
even build an enumeration of computable numbers cause we haven't
discussed the dedpuing logic, and i don't know how that impact the
current predicament
- and even if so do we actually care about the computations being
done outside of the decidably enumerable set? can we build a way to
identify and classify what they do?
- and/or is this the wall i needed to justify the jump to
constructing RTMs, cause the "incompleteness" there happens due to
actual lies built into the computations...
look i acknowledged ur argument!
which is literally more generous than you've ever been towards me,
cause u've never admitted an ounce of validity of my words, despite
understanding them well enough to pounce on any and all criticisms...
So, if the number computed by anti-fixed-H isn't in the enumeration, how
can PRD, or ANY PRD that could exist (and then an anti-fixed-H be built
on it).
Your problem is you don't understand the fundamental nature of what a computation is.
now *that* is fking toxic bro, and if u think i'm going to be swayed
by such toxicity, well i know some therapists u can talk to about that
kinda negative mindset rick, their teachings served me well thus far
In other words, your whole plan is to hope that a magic fairy dust
powered unicorn can give you the answer.
So, you assume that rules must not apply, but then you don't even know
the basic definitions of what you are talking about, so of course you
can't understand the rules.
Your "logic" is based on the rules not applying and computations not
being actually computations. In part, because you don't actually
understand what a computation is, and thus you imagine things that
aren't computations but wnat to think of them as possibly being a computation.
Thus, nothing you have done with it meets the requirements for the
computation you talk about, as it, by definition, starts with a
complete enumeration.
There is not contradiction in my anti-fixed-H if the enumeration
isn't complete, but you also are proven to just be a liar about your
claim of showing a way to compute a diagonal of a complete
enumeration of the computable numbers.
You seem to expect that "someone" will like your crap work enough
to pay you to continue working on it with the hope that you can
materialize your unicorn, even though they have been proven to be
impossible.
YOU are the idiot, with a foul mouth, that doesn't know what he is
talking about.
your ignorant ideas further, as you are just showing there isn't >>>>>>> anything to base going farther on.
If you can show an actual error I am making, with sources to >>>>>>>>> back up your claims, present them.
The problem is you KNOW that you don't know what you are
talking about because you have ADMITTED to not actually studing >>>>>>>>> more that a few papers, but you think you are smarter than the >>>>>>>>> people who wrote them.
YOU are the one flying to the grave in a crashing plane of
ignorance.
I will note, that just like with Peter Olcott, YOU are the one >>>>>>>>> that started the insults, showing whose mind is in the gutter. >>>>>>>>
On 3/17/26 7:46 PM, Richard Damon wrote:
On 3/17/26 12:55 AM, dart200 wrote:
On 3/16/26 6:50 PM, Richard Damon wrote:
On 3/16/26 1:11 PM, dart200 wrote:AHAHAHAHA, how motherfucking toxic do u have to try to rationalize
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:Nope, if the opponent is as brain dead as you show yourself, it >>>>>>>> isn't a
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:So, what is my error, with actual SOURCES for your data that >>>>>>>>>>>> claims I am wrong?
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:_because_ a solution to one leads to a solution for the >>>>>>>>>>>>>>> other...
On 3/14/26 12:20 PM, Richard Damon wrote:Right, equivLWNR PROBLEM, which means both problems are >>>>>>>>>>>>>>>> either solvable or not (under all applicable models). >>>>>>>>>>>>>>>
On 3/14/26 1:29 PM, dart200 wrote:| 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 >>>>>>>>>>>>>>>>
On 3/14/26 4:48 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/26 12:43 AM, dart200 wrote:
On 3/13/26 4:28 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:12 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro >>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:And B is the machine that computes the diagonals of >>>>>>>>>>>>>>>>>>>> the results of the enumeration of circle-free machines. >>>>>>>>>>>>>>>>>>>>
if problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME.On Thu, 12 Mar 2026 00:41:06 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote:there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:But they’re a subset of the “ill-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> set that Turing was >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
the "better-defined" machine don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfy the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 (which >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; does >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> not then prove an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility for enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But if the machine can be “ill-defined”, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine that also satisfies >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> those “ill-defined” criteria must be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and >>>>>>>>>>>>>>>>>>>>>>>>>>>>> equating the solution of them as the same >>>>>>>>>>>>>>>>>>>>>>>>>>>>> is a _fallacy_
But he didn't, that is just what your >>>>>>>>>>>>>>>>>>>>>>>>>>>> ignorant brain thinks he must have been >>>>>>>>>>>>>>>>>>>>>>>>>>>> talking about.
| 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 >>>>>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what EQUIVALENT >>>>>>>>>>>>>>>>>>>>>>>>>> means here.
After all Functional Equivalence doesn't mean >>>>>>>>>>>>>>>>>>>>>>>>>> the same machine or even using the same basic >>>>>>>>>>>>>>>>>>>>>>>>>> algorithm.
No, he his saying the problems are equivalent >>>>>>>>>>>>>>>>>>>>>>>>>> as to the naturehe's literally saying that if u can enumerate >>>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences, then u could use that >>>>>>>>>>>>>>>>>>>>>>>>>>> solution to determine whether any given >>>>>>>>>>>>>>>>>>>>>>>>>>> machine is circle- free ... >>>>>>>>>>>>>>>>>>>>>>>>>>
He doesn't say the two machines generated by >>>>>>>>>>>>>>>>>>>>>>>>>>>> the two problems are in any way equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>>>> he says that the PROBLEMS are equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>>>
and if so could be used to enumerate the >>>>>>>>>>>>>>>>>>>>>>>>>>> circle- free machines,
making the problem of enumerating the sets >>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent,
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>> solution to B
AND
a solution to B can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>> solution to A
Where do you get that definition? >>>>>>>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all >>>>>>>>>>>>>>>>>>>>>>>> models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" you >>>>>>>>>>>>>>>>>>>>>>>> do something,
But it doesn't need to.
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be used >>>>>>>>>>>>>>>>>>>>>>>>> to produce a solution to enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>> numbers, but the reverse is *NOT* true >>>>>>>>>>>>>>>>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was >>>>>>>>>>>>>>>>>>>>>> saying, and thus are trying to kill a strawman. >>>>>>>>>>>>>>>>>>>>>>
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even whole >>>>>>>>>>>>>>>>>>>>> paper, but at least the _section_ rick please... >>>>>>>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON THE >>>>>>>>>>>>>>>>>>>>> SAME FUCKING PAGES WE'VE BEEN TALKING ABOUT p246: >>>>>>>>>>>>>>>>>>>>>
| The simplest and most direct proof of this is by >>>>>>>>>>>>>>>>>>>>> showing that,
| if this general process exists [for circle-free >>>>>>>>>>>>>>>>>>>>> machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>>>>
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>>>>>> <=> enumerating computable sequence >>>>>>>>>>>>>>>>>>>>> => diagonal is computable
=> β is computable _contradiction_ >>>>>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a turing >>>>>>>>>>>>>>>>>>>>> machine, produce a general process to output >>>>>>>>>>>>>>>>>>>>> whether a machine is circle- free or not >>>>>>>>>>>>>>>>>>>>
the _first fallacy_ is that because that isn't >>>>>>>>>>>>>>>>>>>>> actually equivalent to enumerating computable >>>>>>>>>>>>>>>>>>>>> sequences (which is a lesser problem that only >>>>>>>>>>>>>>>>>>>>> needs to recognize a subset of circle- free >>>>>>>>>>>>>>>>>>>>> machines), ruling out a general process for >>>>>>>>>>>>>>>>>>>>> deciding circle- free machine does _not_ actually >>>>>>>>>>>>>>>>>>>>> rule out a general process for enumerating >>>>>>>>>>>>>>>>>>>>> computable numbers
A fallacy in your mind, because you don't understand >>>>>>>>>>>>>>>>>>>> what he means by equivalent.
how can computing a _subset_ of circle-free machines >>>>>>>>>>>>>>>>>>> be equivalent to compute a _total_ set of circle-free >>>>>>>>>>>>>>>>>>> machines...???
Who said they were equivalent COMPUTATIONS. >>>>>>>>>>>>>>>>>
Nope.
Where are you getting your definitions? Because you are >>>>>>>>>>>>>> using the wrong ones.
All you are doing is proving your stubborn refusal to >>>>>>>>>>>>>> learn what you are talking about, and that you don't care >>>>>>>>>>>>>> you are ignorant.
which is a fallacy in this case, they are not equivalent >>>>>>>>>>>>>>> problems
Sure they are, you just don't know what that means as you >>>>>>>>>>>>>> continue to hang on to your errors because you don't >>>>>>>>>>>>>> understand the language you are reading.
IT seems you are just showing you don't know what the >>>>>>>>>>>>>>>> word means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being pointed >>>>>>>>>>>>>> out, as it shows how much of an ass you are.
The problem of creating the computations are >>>>>>>>>>>>>>>>>> equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's pretty >>>>>>>>>>>>>>>>> ridiculous richard
Because I am not, you are gaslighting yourself with your >>>>>>>>>>>>>>>> false definitions that you try to insist on.
if problems are equivalent then a solution to A can be >>>>>>>>>>>>>>>>> used to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the >>>>>>>>>>>>>>>>> fuck along with ur willful ignorance and gaslighting dick >>>>>>>>>>>>>>>>>
But where do you get your definition of equivalent. >>>>>>>>>>>>>>>>
As I have pointed out, that isn't the definition used in >>>>>>>>>>>>>>>> the field, a field you have admitted being untrained in. >>>>>>>>>>>>>>>>
So you admit your ignorance, but insist you must know >>>>>>>>>>>>>>>> better than people who actually know something. >>>>>>>>>>>>>>>>
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the >>>>>>>>>>>>>> effect.
ur whole response is just a shitpile of insults and fallacies >>>>>>>>>>>>>
can't wait to see u take ur ignorance to the grave dick >>>>>>>>>>>>
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence >>>>>>>>>>> always without agreeing more than a spattering of times over >>>>>>>>>>> literally months of engagement,
on top of heaps of unjustified insults completely unbecoming >>>>>>>>>>> of anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject stupidity >>>>>>>>> where u "win" when the opponent gives on u being and endless >>>>>>>>> fucking troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_ >>>>>>>>
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence without
adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown
increadible ignorance and failure in logic. You just have a foul mouth. >>>
insults as "factually based"????
😂😂😂
What is "toxic" about truth? It is only toxic to people who live on lies.
Shows how stupid you are.
Note, YOU are the one providing the evidence to prove my statements,
by just ignoring the facts.
Of course, what you are proving is you don't understand what truth or
facts actually are, since you logic is based on being able to presume
something without proof.
and literally nothing meaningful was said here
It seems your nature is to insult and blame others for your own
failings.
matter of convincing you, but protecting the naive from you lies. >>>>>>>this group is bunch of boomers who spent decades losing arguments >>>>>>> amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own ideas???
😂😂😂
Trying to. After all, I need to do something to wake you up to your
own lies that you have gaslit yourself with.
lol, i fixed turing's diagonal in a way that hasn't been addressed in
literature before
No you haven't, because you don't understand the requirements.
i certainly fixed it from stumbling on itself, which is a fix that has
not been addressed in any literature, until my paper
i have then encountered a new problem, which doesn't negate the fix i
did make u ungrateful tard
The first problem is, you haven't created the enumeration required to
compute the diagonal of.
We know this, because we can convert your fixed-H to be anti-fixed-H
that outputs the opposite digits that fixed-H does (using the trick of
fixed-H, using the number of fixed-H, not anti-fixed-H), and thus
shows that if fixed-H is computing the diagonal, anti-fixed-H is
computing the anti-diagonal, but we also see that this anti-diagonal
isn't in the enumeration, and thus the enumeration can't be complete.
that does not prove there exists no further tricks that might still get
it on the diagonal somehow,
If it isn't complete, then the diagonal isn't the diagonal of the
enumeration that Turing was talking about.
even if there doesn't,
there may be provable limits on what computations can and cannot be computably enumerated on the diagonal,
which is certainly a step up from the over-applied rice's theorem know- nothing nonsense u see with theorists today,
like what if those various pseudo-anti-diagonals (as they aren't true
total anti-diagonal) are the *only* set of computable numbers we can't computably enumerate on the list???
that would also be a huge win, cause those computations don't compute relationships we care about, so failing to enumerate them totally just doesn't really matter 🤷
(and before u try to make yet another baseless claim that it must
have been, show me the proof instead of baselessly just claiming u
fucking twat)
And what is wrong about this proof.
i don't have to be right about literally every possible future goal post
to right about one goal post in a unique way that's never been done
before. the fact i could even hit that goal post is to me a massive sign things have been missed in the fundamentals, rick
it should be to you as well, but my god are obsessed with clinging to certain uncertainty it's abcerd
Your enumeration generated by PRD just can not be COMPLETE, including
at least one instance of EVERY computable number.
PROVE ME WRONG.
it's really kind of sad how much toxicity and generally unwillingness to cooperate i've encounter when trying to explore these ideas,
i hope future academia may take heed from what i've had to endure thus
far, pretty much on my own. heck i hope current academic might too ...
but that might be asking for too much at this point eh???
the fact u continually try to gaslight me into thinking i haven't
The people here mostly know what they are talking about, because
they have studied it (some like Olcott and you are the exception). >>>>>>
It seems you are just admitting that you are stuck in your lies >>>>>>>> and just can't think because, like Olcott, you have successfully >>>>>>>> gatlit yourself into being convinced of your lies.
i demonstrated two distinct fallacies in turing's paper, that
really aren't the hard to understand,
No, you demonstrated that you do don't understand what he is saying, >>>>>
understood his argument well enough is not only incredibly toxic
but let's me know ur completely fine with blatantly lying at me to
"win" an argument,
But it isn't gaslighting, it is FACT.
idk how u read everything i wrote and try to claim i just don't
understand,
Because you keep claiming things that aren't there.
i have been manipulating his ideas in ways that have never been done
before, and u can't even acknowledge that i understand his ideas???
sheesh,
Right, by twisting words to not mean what they mean.
i'm swimming thru a swamp of endless gaslighting, fostered a toxic
mentality festering the fundamentals of math hostile to any sort of
meaningful innovation at the core for some ungodly reason
Yes, the endless gaslighting that you have done to yourself, causing
you to think that people point out truth to you are gaslighting you.
The fact you can't actually prove anything should be your first sign
that you have something wrong.
Your world is just built on your own lies and fantasies.
the fallacies i picked out are still fallacies
that's not what the side with truth does, or even remotely needs to >>>>> do. and if u can't recognize that, i'm sorry for all fallacy u've
bought into across all the things u believe
Claiming truth is gaslighting is just gaslighting.
> take them to the grave bro
>
> #god
You got a source for that?
Or do you have the same "god" complex as Olcott?
oh does he also claim that _we are god_ ???
No, just him.
It seems you like to "quote" things, claiming them to be from
"god" (as he signs them). That seems to imply you think you have a
special link to him.
like i said: _we are god_
> so anyone can do it eh???
>
> #god
if this "me" is particularly special, that is only due by seeding a
trend, if a trend even ever takes off, which is yet to be seen...
🤷🤷🤷
u can write it off as psycho-spiritual outburst of frustration from constantly banging my head against the various mental walls keeping us chained to acting _far_ less ethically than we should,
but don't count on me stopping. the grave we've been digging for our
species thru our systemic moral negligence is _deep_ ,
and it's gunna take some dramatic times to pull ourselves out
like how many more pedo islands do u think exist?
regardless of whether turing's conclusions are ultimately correct >>>>>>> or not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Then show me your source.
my source for fallacies that *i* pointed out???
i would be source for that... duh?
So, what is that source?
That is your problem, you have no sources but your own misunderstanding.
That come from your gaslighting of yourself to brainwash you into
thinking you don't need sources.
why would i need sources to justify novel arguments???
are you asking for me to repeat the arguments i generated? u can either reread the posts i made on usenet
or here for fallacy 1: https://www.academia.edu/165010519
Your claim is just the gaslighting you claim I am doing.
Yes, while the title of the paper uses the topic of "Computable
Numbers", and the part of the proof focuses on the related concept >>>>>> of machines that compute them, he DOES show a proof, that could be >>>>>> similarly used to prove the uncomputablility of the computable
numbers.
Your problem is you have such a wooden and limited knowledge of
what you read, you can't understand what he is doing.
u've never "won" an argument here in the decades u wasted ur >>>>>>>>> life here
get back to helping america bomb muzzies for their joo
overlords, that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but your >>>>>>>> bluster.
Sorry, that won't cut it for begging people to give you money to >>>>>>>> carry
who am i begging??? who around here even has money to give??? 😂😂😂
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your work
so you can complete the parts that you admit have holes in them.
yes, certain further work would take time and therefore funds, and
that kind of work will remain out of scope of this discussion.
that's a statement of fact, not "begging" u sad dishonest old man
But, since you idea have been proven wrong, and you don't even try
to refute it, why should anyone support that work.
As I said, Your PRD couldn't have accepted ANY machine that
generates the same computable number as anti-fixed-H, and thus the
enumeration it generates is NOT complete.
right that is a thorn in my side atm, and there are several avenues
i'm exploring in thot on how to deal with that
- first of all i'm not entirely convinced there isn't some
strategy i'm missing that might still yet get it on the diagonal. we
haven't even build an enumeration of computable numbers cause we
haven't discussed the dedpuing logic, and i don't know how that
impact the current predicament
- and even if so do we actually care about the computations being
done outside of the decidably enumerable set? can we build a way to
identify and classify what they do?
- and/or is this the wall i needed to justify the jump to
constructing RTMs, cause the "incompleteness" there happens due to
actual lies built into the computations...
look i acknowledged ur argument!
which is literally more generous than you've ever been towards me,
cause u've never admitted an ounce of validity of my words, despite
understanding them well enough to pounce on any and all criticisms...
So, if the number computed by anti-fixed-H isn't in the enumeration,
how can PRD, or ANY PRD that could exist (and then an anti-fixed-H be
built on it).
Your problem is you don't understand the fundamental nature of what a
computation is.
now *that* is fking toxic bro, and if u think i'm going to be swayed
by such toxicity, well i know some therapists u can talk to about
that kinda negative mindset rick, their teachings served me well thus
far
In other words, your whole plan is to hope that a magic fairy dust
powered unicorn can give you the answer.
So, you assume that rules must not apply, but then you don't even know
the basic definitions of what you are talking about, so of course you
can't understand the rules.
Your "logic" is based on the rules not applying and computations not
being actually computations. In part, because you don't actually
understand what a computation is, and thus you imagine things that
aren't computations but wnat to think of them as possibly being a
computation.
again, nothing was said here
Thus, nothing you have done with it meets the requirements for the
computation you talk about, as it, by definition, starts with a
complete enumeration.
There is not contradiction in my anti-fixed-H if the enumeration
isn't complete, but you also are proven to just be a liar about your
claim of showing a way to compute a diagonal of a complete
enumeration of the computable numbers.
You seem to expect that "someone" will like your crap work enough >>>>>> to pay you to continue working on it with the hope that you can
materialize your unicorn, even though they have been proven to be >>>>>> impossible.
YOU are the idiot, with a foul mouth, that doesn't know what he is >>>>>> talking about.
your ignorant ideas further, as you are just showing there isn't >>>>>>>> anything to base going farther on.
If you can show an actual error I am making, with sources to >>>>>>>>>> back up your claims, present them.
The problem is you KNOW that you don't know what you are
talking about because you have ADMITTED to not actually
studing more that a few papers, but you think you are smarter >>>>>>>>>> than the people who wrote them.
YOU are the one flying to the grave in a crashing plane of >>>>>>>>>> ignorance.
I will note, that just like with Peter Olcott, YOU are the one >>>>>>>>>> that started the insults, showing whose mind is in the gutter. >>>>>>>>>
you think CHANGING the requitrements is a valid process for meeting them.
On 3/18/26 3:13 AM, dart200 wrote:
On 3/17/26 7:46 PM, Richard Damon wrote:
On 3/17/26 12:55 AM, dart200 wrote:
On 3/16/26 6:50 PM, Richard Damon wrote:
On 3/16/26 1:11 PM, dart200 wrote:
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:Nope, if the opponent is as brain dead as you show yourself, it >>>>>>>>> isn't a
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:So, what is my error, with actual SOURCES for your data >>>>>>>>>>>>> that claims I am wrong?
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:_because_ a solution to one leads to a solution for the >>>>>>>>>>>>>>>> other...
On 3/14/26 12:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 3/14/26 1:29 PM, dart200 wrote:Right, equivLWNR PROBLEM, which means both problems are >>>>>>>>>>>>>>>>> either solvable or not (under all applicable models). >>>>>>>>>>>>>>>>
| 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 >>>>>>>>>>>>>>>>>On 3/14/26 4:48 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 3/14/26 12:43 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/26 4:28 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:12 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
how can computing a _subset_ of circle-free machines >>>>>>>>>>>>>>>>>>>> be equivalent to compute a _total_ set of circle- >>>>>>>>>>>>>>>>>>>> free machines...???And B is the machine that computes the diagonals of >>>>>>>>>>>>>>>>>>>>> the results of the enumeration of circle-free >>>>>>>>>>>>>>>>>>>>> machines.if problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME. >>>>>>>>>>>>>>>>>>>>>>>>>>>On Thu, 12 Mar 2026 00:41:06 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote:there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence D’Oliveiro >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:But they’re a subset of the “ill-defined”
the "better-defined" machine don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfy the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 (which >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; does >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> not then prove an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility for enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But if the machine can be “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined”, yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine that also satisfies >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> those “ill-defined” criteria must be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
set that Turing was >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with Turing. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equating the solution of them as the same >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is a _fallacy_
But he didn't, that is just what your >>>>>>>>>>>>>>>>>>>>>>>>>>>>> ignorant brain thinks he must have been >>>>>>>>>>>>>>>>>>>>>>>>>>>>> talking about.
| 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 >>>>>>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what EQUIVALENT >>>>>>>>>>>>>>>>>>>>>>>>>>> means here.
After all Functional Equivalence doesn't mean >>>>>>>>>>>>>>>>>>>>>>>>>>> the same machine or even using the same basic >>>>>>>>>>>>>>>>>>>>>>>>>>> algorithm.
No, he his saying the problems are equivalent >>>>>>>>>>>>>>>>>>>>>>>>>>> as to the nature
He doesn't say the two machines generated >>>>>>>>>>>>>>>>>>>>>>>>>>>>> by the two problems are in any way >>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent, he says that the PROBLEMS are >>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent,
he's literally saying that if u can >>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerate computable sequences, then u could >>>>>>>>>>>>>>>>>>>>>>>>>>>> use that solution to determine whether any >>>>>>>>>>>>>>>>>>>>>>>>>>>> given machine is circle- free ... >>>>>>>>>>>>>>>>>>>>>>>>>>>
and if so could be used to enumerate the >>>>>>>>>>>>>>>>>>>>>>>>>>>> circle- free machines, >>>>>>>>>>>>>>>>>>>>>>>>>>>>
making the problem of enumerating the sets >>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent,
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>>> solution to B
AND
a solution to B can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>>> solution to A
Where do you get that definition? >>>>>>>>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all >>>>>>>>>>>>>>>>>>>>>>>>> models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" you >>>>>>>>>>>>>>>>>>>>>>>>> do something,
But it doesn't need to.
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be used >>>>>>>>>>>>>>>>>>>>>>>>>> to produce a solution to enumerating >>>>>>>>>>>>>>>>>>>>>>>>>> computable numbers, but the reverse is *NOT* true >>>>>>>>>>>>>>>>>>>>>>>>>
yes it does, rick
WHY?
As I have said, you don't understand what he was >>>>>>>>>>>>>>>>>>>>>>> saying, and thus are trying to kill a strawman. >>>>>>>>>>>>>>>>>>>>>>>
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even >>>>>>>>>>>>>>>>>>>>>> whole paper, but at least the _section_ rick >>>>>>>>>>>>>>>>>>>>>> please...
i'm tired of answering questions that ARE ON THE >>>>>>>>>>>>>>>>>>>>>> SAME FUCKING PAGES WE'VE BEEN TALKING ABOUT p246: >>>>>>>>>>>>>>>>>>>>>>
| The simplest and most direct proof of this is by >>>>>>>>>>>>>>>>>>>>>> showing that,
| if this general process exists [for circle-free >>>>>>>>>>>>>>>>>>>>>> machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>>>>>
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>>>>>>> <=> enumerating computable sequence >>>>>>>>>>>>>>>>>>>>>> => diagonal is computable >>>>>>>>>>>>>>>>>>>>>> => β is computable _contradiction_ >>>>>>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a turing >>>>>>>>>>>>>>>>>>>>>> machine, produce a general process to output >>>>>>>>>>>>>>>>>>>>>> whether a machine is circle- free or not >>>>>>>>>>>>>>>>>>>>>
the _first fallacy_ is that because that isn't >>>>>>>>>>>>>>>>>>>>>> actually equivalent to enumerating computable >>>>>>>>>>>>>>>>>>>>>> sequences (which is a lesser problem that only >>>>>>>>>>>>>>>>>>>>>> needs to recognize a subset of circle- free >>>>>>>>>>>>>>>>>>>>>> machines), ruling out a general process for >>>>>>>>>>>>>>>>>>>>>> deciding circle- free machine does _not_ actually >>>>>>>>>>>>>>>>>>>>>> rule out a general process for enumerating >>>>>>>>>>>>>>>>>>>>>> computable numbers
A fallacy in your mind, because you don't >>>>>>>>>>>>>>>>>>>>> understand what he means by equivalent. >>>>>>>>>>>>>>>>>>>>
Who said they were equivalent COMPUTATIONS. >>>>>>>>>>>>>>>>>>
Nope.
Where are you getting your definitions? Because you are >>>>>>>>>>>>>>> using the wrong ones.
All you are doing is proving your stubborn refusal to >>>>>>>>>>>>>>> learn what you are talking about, and that you don't care >>>>>>>>>>>>>>> you are ignorant.
which is a fallacy in this case, they are not equivalent >>>>>>>>>>>>>>>> problems
Sure they are, you just don't know what that means as you >>>>>>>>>>>>>>> continue to hang on to your errors because you don't >>>>>>>>>>>>>>> understand the language you are reading.
IT seems you are just showing you don't know what the >>>>>>>>>>>>>>>>> word means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being >>>>>>>>>>>>>>> pointed out, as it shows how much of an ass you are. >>>>>>>>>>>>>>>
The problem of creating the computations are >>>>>>>>>>>>>>>>>>> equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's pretty >>>>>>>>>>>>>>>>>> ridiculous richard
Because I am not, you are gaslighting yourself with >>>>>>>>>>>>>>>>> your false definitions that you try to insist on. >>>>>>>>>>>>>>>>>
if problems are equivalent then a solution to A can be >>>>>>>>>>>>>>>>>> used to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the >>>>>>>>>>>>>>>>>> fuck along with ur willful ignorance and gaslighting dick >>>>>>>>>>>>>>>>>>
But where do you get your definition of equivalent. >>>>>>>>>>>>>>>>>
As I have pointed out, that isn't the definition used >>>>>>>>>>>>>>>>> in the field, a field you have admitted being untrained >>>>>>>>>>>>>>>>> in.
So you admit your ignorance, but insist you must know >>>>>>>>>>>>>>>>> better than people who actually know something. >>>>>>>>>>>>>>>>>
In a word, Dunning-Kruger
never seen anyone bring that up in good faith
Which is a response typical of those suffering from the >>>>>>>>>>>>>>> effect.
ur whole response is just a shitpile of insults and fallacies >>>>>>>>>>>>>>
can't wait to see u take ur ignorance to the grave dick >>>>>>>>>>>>>
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence >>>>>>>>>>>> always without agreeing more than a spattering of times over >>>>>>>>>>>> literally months of engagement,
on top of heaps of unjustified insults completely unbecoming >>>>>>>>>>>> of anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject
stupidity where u "win" when the opponent gives on u being and >>>>>>>>>> endless fucking troll???
that's a total L bro. if u fail to convince ur opponent, u _lose_ >>>>>>>>>
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence without >>>>>> adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown
increadible ignorance and failure in logic. You just have a foul
mouth.
AHAHAHAHA, how motherfucking toxic do u have to try to rationalize
insults as "factually based"????
😂😂😂
What is "toxic" about truth? It is only toxic to people who live on
lies.
Shows how stupid you are.
Note, YOU are the one providing the evidence to prove my statements,
by just ignoring the facts.
Of course, what you are proving is you don't understand what truth or
facts actually are, since you logic is based on being able to presume
something without proof.
and literally nothing meaningful was said here
If pointing out errors isn't meaningful, that points out your
fundamental error in logic.
I guess your world is just built on your own fantasies, and you just
don't care about what is actually true.
Maybe you will be able to just imagine the food and shelter you are
going to need to keep on living once your money runs out.
It seems your nature is to insult and blame others for your own >>>>>>> failings.
matter of convincing you, but protecting the naive from you lies. >>>>>>>>this group is bunch of boomers who spent decades losing
arguments amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own ideas??? >>>>>>
😂😂😂
Trying to. After all, I need to do something to wake you up to your >>>>> own lies that you have gaslit yourself with.
lol, i fixed turing's diagonal in a way that hasn't been addressed
in literature before
No you haven't, because you don't understand the requirements.
i certainly fixed it from stumbling on itself, which is a fix that has
not been addressed in any literature, until my paper
In other words, you think CHANGING the requitrements is a valid process
for meeting them.
i have then encountered a new problem, which doesn't negate the fix i
did make u ungrateful tard
Sure it does. You have presumed the enumeration that has been proved can
not be made.
There can not be an effective (computable) enumeration that includes all computable number, as any method that generates one allows the computing
of a number that doesn't exist in that set.
Thus, any computed enumeration of computable numbers is necessarily incomplete.
All you are doint is proving you don't understand what it means to be
able to compute something.
It seems your idea of computing allows an algorithm to assume that it
(or even an equivalent to it) can't be embedded into another algorithm, which makes your concept strictly weaker in the power to compute than
the methods used by Turing Machines and their equivalents.
The first problem is, you haven't created the enumeration required to
compute the diagonal of.
We know this, because we can convert your fixed-H to be anti-fixed-H
that outputs the opposite digits that fixed-H does (using the trick
of fixed-H, using the number of fixed-H, not anti-fixed-H), and thus
shows that if fixed-H is computing the diagonal, anti-fixed-H is
computing the anti-diagonal, but we also see that this anti-diagonal
isn't in the enumeration, and thus the enumeration can't be complete.
that does not prove there exists no further tricks that might still
get it on the diagonal somehow,
Sure it does.
*ANY* method to generate an enumeration of computable numbers allows the creation of a computation that computes a number not in the set that it generated.
Thus NO method to generate an enumeratio of computable numbers can
create a total enumeration.
Your assumption of trick is just depending on magic fairy dust from a unicorn to create the impossible.
If it isn't complete, then the diagonal isn't the diagonal of the
enumeration that Turing was talking about.
even if there doesn't,
there may be provable limits on what computations can and cannot be
computably enumerated on the diagonal,
But not that excludes the anti-diagonal, since we HAVE the description
of the algorithm that generates it, at least if we have an algorithm to generate the enumeration.
which is certainly a step up from the over-applied rice's theorem
know- nothing nonsense u see with theorists today,
You seem to have the problem of not understand what Rice proves.
There can be many machines that compute PARTIAL classifications or
decisions on machines, just not a TOTAL classification.
like what if those various pseudo-anti-diagonals (as they aren't true
total anti-diagonal) are the *only* set of computable numbers we can't
computably enumerate on the list???
But they aren't. There are other proof, well beyond your head, that show that other questions turn out to not be computable.
In fact, by simple "counting" we can tell that there are an infinite
number of uncomputable problems for every computable one.
that would also be a huge win, cause those computations don't compute
relationships we care about, so failing to enumerate them totally just
doesn't really matter 🤷
Sure it does. By knowing that TOTAL classification is impossible, we
know that we need to look at what classes of inputs a given algorithm
can work on.
Thus, like where you started with, because we KNOW we can't totally
solve the Halting Problem, we accept that we need to allow our algorithm
to decide that some cases might not be deciable, and work on the cases
we can decide on.
A correctness proving program doesn't need to prove EVERY program
correct or wrong, but can prove SOME programs correct, SOME programs it
can point out errors, and some it tells us they are too complicated for
it to process.
If it tells us it is too complicted, if we really need the proof, we
need to revise it to simplify it. (Or it may be that the problem we are working on is just uncomputable, so no program CAN be proven correct,
and we need to build a partial version that admits that there are cases
it can't get correct)
(and before u try to make yet another baseless claim that it must
have been, show me the proof instead of baselessly just claiming u
fucking twat)
And what is wrong about this proof.
i don't have to be right about literally every possible future goal
post to right about one goal post in a unique way that's never been
done before. the fact i could even hit that goal post is to me a
massive sign things have been missed in the fundamentals, rick
it should be to you as well, but my god are obsessed with clinging to
certain uncertainty it's abcerd
But the fact that you current claims are based on NO evidence, means you
are starting with nothing.
You claim that something might be possible, when it is shown that it
can't be.
Your world is just built on the assumption that the rules don't apply.
That is a world of fantasy and lies.
Your enumeration generated by PRD just can not be COMPLETE, including
at least one instance of EVERY computable number.
PROVE ME WRONG.
it's really kind of sad how much toxicity and generally unwillingness
to cooperate i've encounter when trying to explore these ideas,
The most toxic thing is to just lie to yourself about what can be done.
i hope future academia may take heed from what i've had to endure thus
far, pretty much on my own. heck i hope current academic might too ...
but that might be asking for too much at this point eh???
In other words, you hope academia might allow people to live in error
and self-deciet?
Your problem is you reject people pointing out the errors in your work,
because you assume you must be right, even when you admit you don't
really understand the field.
Your works is based on ignorant assumptions, not facts.
That is a dark world of lies.
the fact u continually try to gaslight me into thinking i haven't >>>>>> understood his argument well enough is not only incredibly toxic
The people here mostly know what they are talking about, because >>>>>>> they have studied it (some like Olcott and you are the exception). >>>>>>>
i demonstrated two distinct fallacies in turing's paper, that >>>>>>>> really aren't the hard to understand,
It seems you are just admitting that you are stuck in your lies >>>>>>>>> and just can't think because, like Olcott, you have
successfully gatlit yourself into being convinced of your lies. >>>>>>>>
No, you demonstrated that you do don't understand what he is saying, >>>>>>
but let's me know ur completely fine with blatantly lying at me to >>>>>> "win" an argument,
But it isn't gaslighting, it is FACT.
idk how u read everything i wrote and try to claim i just don't
understand,
Because you keep claiming things that aren't there.
i have been manipulating his ideas in ways that have never been done
before, and u can't even acknowledge that i understand his ideas???
sheesh,
Right, by twisting words to not mean what they mean.
i'm swimming thru a swamp of endless gaslighting, fostered a toxic
mentality festering the fundamentals of math hostile to any sort of
meaningful innovation at the core for some ungodly reason
Yes, the endless gaslighting that you have done to yourself, causing
you to think that people point out truth to you are gaslighting you.
The fact you can't actually prove anything should be your first sign
that you have something wrong.
Your world is just built on your own lies and fantasies.
the fallacies i picked out are still fallacies
No, because your claimed fallacies are based on you using a definist fallacy.
Starting with that error, NOTHING you have said has any basis to point
out error.
that's not what the side with truth does, or even remotely needs
to do. and if u can't recognize that, i'm sorry for all fallacy
u've bought into across all the things u believe
Claiming truth is gaslighting is just gaslighting.
> take them to the grave bro
>
> #god
You got a source for that?
Or do you have the same "god" complex as Olcott?
oh does he also claim that _we are god_ ???
No, just him.
It seems you like to "quote" things, claiming them to be from "god"
(as he signs them). That seems to imply you think you have a special
link to him.
like i said: _we are god_
> so anyone can do it eh???
>
> #god
if this "me" is particularly special, that is only due by seeding a
trend, if a trend even ever takes off, which is yet to be seen...
But "we" are not "god", and assuming you have the divine power of god is
the beginning of your own condemnation to a life of error and dispare.
🤷🤷🤷
u can write it off as psycho-spiritual outburst of frustration from
constantly banging my head against the various mental walls keeping us
chained to acting _far_ less ethically than we should,
Or, those "walls" are the boundries that aim us to what can be done.
but don't count on me stopping. the grave we've been digging for our
species thru our systemic moral negligence is _deep_ ,
But it seems, the grave you see, is just the grave for those that think
like you, and that the uncompuatable nature of some things means we
can't "do our best" and handle the cases we actually care about.
We CAN prove that some programs are correct. The cost is just too high
to be used everywhere, and many programs don't actually need to be
provably correct.
YOU are the one that rejects that this ACHEIVABLE (and ACHEIVED) goal is good enough, and seem to want that all work stops until we can do the impossible.
THAT is the truely toxic.
and it's gunna take some dramatic times to pull ourselves out
like how many more pedo islands do u think exist?
regardless of whether turing's conclusions are ultimately
correct or not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Then show me your source.
my source for fallacies that *i* pointed out???
i would be source for that... duh?
So, what is that source?
That is your problem, you have no sources but your own misunderstanding. >>>
That come from your gaslighting of yourself to brainwash you into
thinking you don't need sources.
why would i need sources to justify novel arguments???
To show that you claims are based on FACTS and not errors?
All you are doing is proving that you are ignorant and stupid.
Most "novel" arguement are just errors and fallacies.
are you asking for me to repeat the arguments i generated? u can
either reread the posts i made on usenet
or here for fallacy 1: https://www.academia.edu/165010519
Which is just repeating the error that thinking that equivalent problems
are solving the same problem.
Where is the definition that says that?
Equivalent problems are problems that are true/solvable or false/
unsolvable together.
Going to the fallacy of appeal to authority, using yourself as the
authority is just stupid.
It is also your definist fallacy, as you are trying to redefine the word "equivalent" as used as a modifier for problem.
Your claim is just the gaslighting you claim I am doing.
Yes, while the title of the paper uses the topic of "Computable >>>>>>> Numbers", and the part of the proof focuses on the related
concept of machines that compute them, he DOES show a proof, that >>>>>>> could be similarly used to prove the uncomputablility of the
computable numbers.
Your problem is you have such a wooden and limited knowledge of >>>>>>> what you read, you can't understand what he is doing.
u've never "won" an argument here in the decades u wasted ur >>>>>>>>>> life here
get back to helping america bomb muzzies for their joo
overlords, that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but your >>>>>>>>> bluster.
Sorry, that won't cut it for begging people to give you money >>>>>>>>> to carry
who am i begging??? who around here even has money to give??? 😂 >>>>>>>> 😂😂
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your work >>>>>>> so you can complete the parts that you admit have holes in them.
yes, certain further work would take time and therefore funds, and >>>>>> that kind of work will remain out of scope of this discussion.
that's a statement of fact, not "begging" u sad dishonest old man
But, since you idea have been proven wrong, and you don't even try
to refute it, why should anyone support that work.
As I said, Your PRD couldn't have accepted ANY machine that
generates the same computable number as anti-fixed-H, and thus the
enumeration it generates is NOT complete.
right that is a thorn in my side atm, and there are several avenues
i'm exploring in thot on how to deal with that
- first of all i'm not entirely convinced there isn't some
strategy i'm missing that might still yet get it on the diagonal. we
haven't even build an enumeration of computable numbers cause we
haven't discussed the dedpuing logic, and i don't know how that
impact the current predicament
- and even if so do we actually care about the computations being >>>> done outside of the decidably enumerable set? can we build a way to
identify and classify what they do?
- and/or is this the wall i needed to justify the jump to
constructing RTMs, cause the "incompleteness" there happens due to
actual lies built into the computations...
look i acknowledged ur argument!
which is literally more generous than you've ever been towards me,
cause u've never admitted an ounce of validity of my words, despite
understanding them well enough to pounce on any and all criticisms...
So, if the number computed by anti-fixed-H isn't in the enumeration,
how can PRD, or ANY PRD that could exist (and then an anti-fixed-H be
built on it).
Your problem is you don't understand the fundamental nature of what a
computation is.
now *that* is fking toxic bro, and if u think i'm going to be swayed
by such toxicity, well i know some therapists u can talk to about
that kinda negative mindset rick, their teachings served me well
thus far
In other words, your whole plan is to hope that a magic fairy dust
powered unicorn can give you the answer.
So, you assume that rules must not apply, but then you don't even
know the basic definitions of what you are talking about, so of
course you can't understand the rules.
Your "logic" is based on the rules not applying and computations not
being actually computations. In part, because you don't actually
understand what a computation is, and thus you imagine things that
aren't computations but wnat to think of them as possibly being a
computation.
again, nothing was said here
In part because there wasn't anything to reply to.
IT seems you have run out of ways to fabricate your ideas, so you just insult the messager pointing out your errors.
Thus, nothing you have done with it meets the requirements for the
computation you talk about, as it, by definition, starts with a
complete enumeration.
There is not contradiction in my anti-fixed-H if the enumeration
isn't complete, but you also are proven to just be a liar about
your claim of showing a way to compute a diagonal of a complete
enumeration of the computable numbers.
You seem to expect that "someone" will like your crap work enough >>>>>>> to pay you to continue working on it with the hope that you can >>>>>>> materialize your unicorn, even though they have been proven to be >>>>>>> impossible.
YOU are the idiot, with a foul mouth, that doesn't know what he >>>>>>> is talking about.
your ignorant ideas further, as you are just showing there
isn't anything to base going farther on.
If you can show an actual error I am making, with sources to >>>>>>>>>>> back up your claims, present them.
The problem is you KNOW that you don't know what you are >>>>>>>>>>> talking about because you have ADMITTED to not actually >>>>>>>>>>> studing more that a few papers, but you think you are smarter >>>>>>>>>>> than the people who wrote them.
YOU are the one flying to the grave in a crashing plane of >>>>>>>>>>> ignorance.
I will note, that just like with Peter Olcott, YOU are the >>>>>>>>>>> one that started the insults, showing whose mind is in the >>>>>>>>>>> gutter.
On 18/03/2026 11:32, Richard Damon wrote:
you think CHANGING the requitrements is a valid process for meeting them.
That's on the topic of project process, not any of the included groups.
The message body is Copyright (C) 2026 Tristan Wibberley except
citations and quotations noted. All Rights Reserved except that you
may,of course, cite it academically giving credit to me, distribute
it verbatim as part of a usenet system or its archives, and use it to
promote my greatness and general superiority without
misrepresentation of my opinions other than my opinion of my
greatness and general superiority which you _may_ misrepresent. You definitely MAY NOT train any production AI system with it but you may
train experimental AI that will only be used for evaluation of the AI
methods it implements.
On 3/18/26 4:32 AM, Richard Damon wrote:
On 3/18/26 3:13 AM, dart200 wrote:
and literally nothing meaningful was said here
If pointing out errors isn't meaningful, that points out your
fundamental error in logic.
I guess your world is just built on your own fantasies, and you just
don't care about what is actually true.
Maybe you will be able to just imagine the food and shelter you are
going to need to keep on living once your money runs out.
three more sentences of nothing
On 3/18/26 4:32 AM, Richard Damon wrote:
On 3/18/26 3:13 AM, dart200 wrote:
On 3/17/26 7:46 PM, Richard Damon wrote:
On 3/17/26 12:55 AM, dart200 wrote:
On 3/16/26 6:50 PM, Richard Damon wrote:
On 3/16/26 1:11 PM, dart200 wrote:
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:So, what is my error, with actual SOURCES for your data >>>>>>>>>>>>>> that claims I am wrong?
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:_because_ a solution to one leads to a solution for the >>>>>>>>>>>>>>>>> other...
On 3/14/26 12:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/26 1:29 PM, dart200 wrote:Right, equivLWNR PROBLEM, which means both problems >>>>>>>>>>>>>>>>>> are either solvable or not (under all applicable models). >>>>>>>>>>>>>>>>>
| 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 >>>>>>>>>>>>>>>>>>On 3/14/26 4:48 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/26 12:43 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 4:28 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:12 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
how can computing a _subset_ of circle-free >>>>>>>>>>>>>>>>>>>>> machines be equivalent to compute a _total_ set of >>>>>>>>>>>>>>>>>>>>> circle- free machines...???And B is the machine that computes the diagonals >>>>>>>>>>>>>>>>>>>>>> of the results of the enumeration of circle-free >>>>>>>>>>>>>>>>>>>>>> machines.if problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME. >>>>>>>>>>>>>>>>>>>>>>>>>>>>On Thu, 12 Mar 2026 00:41:06 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But they’re a subset of the “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined” set that Turing was >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>the "better-defined" machine don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfy the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 (which >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> does not then prove an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility for enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But if the machine can be “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined”, yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine that also satisfies >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> those “ill-defined” criteria must be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Turing.
there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equating the solution of them as the same >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is a _fallacy_
But he didn't, that is just what your >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ignorant brain thinks he must have been >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> talking about.
| 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 >>>>>>>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what EQUIVALENT >>>>>>>>>>>>>>>>>>>>>>>>>>>> means here.
After all Functional Equivalence doesn't >>>>>>>>>>>>>>>>>>>>>>>>>>>> mean the same machine or even using the same >>>>>>>>>>>>>>>>>>>>>>>>>>>> basic algorithm.
No, he his saying the problems are >>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent as to the nature >>>>>>>>>>>>>>>>>>>>>>>>>>>>
He doesn't say the two machines generated >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> by the two problems are in any way >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent, he says that the PROBLEMS are >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent,
he's literally saying that if u can >>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerate computable sequences, then u >>>>>>>>>>>>>>>>>>>>>>>>>>>>> could use that solution to determine >>>>>>>>>>>>>>>>>>>>>>>>>>>>> whether any given machine is circle- free ... >>>>>>>>>>>>>>>>>>>>>>>>>>>>
and if so could be used to enumerate the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> circle- free machines, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
making the problem of enumerating the sets >>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent,
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>>>> solution to B
AND
a solution to B can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>>>> solution to A
Where do you get that definition? >>>>>>>>>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in all >>>>>>>>>>>>>>>>>>>>>>>>>> models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" you >>>>>>>>>>>>>>>>>>>>>>>>>> do something,
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be used >>>>>>>>>>>>>>>>>>>>>>>>>>> to produce a solution to enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>> computable numbers, but the reverse is *NOT* >>>>>>>>>>>>>>>>>>>>>>>>>>> true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he was >>>>>>>>>>>>>>>>>>>>>>>> saying, and thus are trying to kill a strawman. >>>>>>>>>>>>>>>>>>>>>>>>
Where does he ACTUALLY SAY that the machine that >>>>>>>>>>>>>>>>>>>>>>>> generates circle- ftee machihes could be used to >>>>>>>>>>>>>>>>>>>>>>>> enumerate computable numbers.
my god rick, please fucking read the not even >>>>>>>>>>>>>>>>>>>>>>> whole paper, but at least the _section_ rick >>>>>>>>>>>>>>>>>>>>>>> please...
i'm tired of answering questions that ARE ON THE >>>>>>>>>>>>>>>>>>>>>>> SAME FUCKING PAGES WE'VE BEEN TALKING ABOUT p246: >>>>>>>>>>>>>>>>>>>>>>>
| The simplest and most direct proof of this is >>>>>>>>>>>>>>>>>>>>>>> by showing that,
| if this general process exists [for circle-free >>>>>>>>>>>>>>>>>>>>>>> machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>>>>>>
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>>>>>>>>>
Why doesn't the program do that?
ok ok i will even tho u will continue to disagree... >>>>>>>>>>>>>>>>>>>>>>>
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>>>>>>>> <=> enumerating computable sequence >>>>>>>>>>>>>>>>>>>>>>> => diagonal is computable >>>>>>>>>>>>>>>>>>>>>>> => β is computable _contradiction_ >>>>>>>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a >>>>>>>>>>>>>>>>>>>>>>> turing machine, produce a general process to >>>>>>>>>>>>>>>>>>>>>>> output whether a machine is circle- free or not >>>>>>>>>>>>>>>>>>>>>>
the _first fallacy_ is that because that isn't >>>>>>>>>>>>>>>>>>>>>>> actually equivalent to enumerating computable >>>>>>>>>>>>>>>>>>>>>>> sequences (which is a lesser problem that only >>>>>>>>>>>>>>>>>>>>>>> needs to recognize a subset of circle- free >>>>>>>>>>>>>>>>>>>>>>> machines), ruling out a general process for >>>>>>>>>>>>>>>>>>>>>>> deciding circle- free machine does _not_ actually >>>>>>>>>>>>>>>>>>>>>>> rule out a general process for enumerating >>>>>>>>>>>>>>>>>>>>>>> computable numbers
A fallacy in your mind, because you don't >>>>>>>>>>>>>>>>>>>>>> understand what he means by equivalent. >>>>>>>>>>>>>>>>>>>>>
Who said they were equivalent COMPUTATIONS. >>>>>>>>>>>>>>>>>>>
Nope.
Where are you getting your definitions? Because you are >>>>>>>>>>>>>>>> using the wrong ones.
All you are doing is proving your stubborn refusal to >>>>>>>>>>>>>>>> learn what you are talking about, and that you don't >>>>>>>>>>>>>>>> care you are ignorant.
which is a fallacy in this case, they are not >>>>>>>>>>>>>>>>> equivalent problems
Sure they are, you just don't know what that means as >>>>>>>>>>>>>>>> you continue to hang on to your errors because you don't >>>>>>>>>>>>>>>> understand the language you are reading.
IT seems you are just showing you don't know what the >>>>>>>>>>>>>>>>>> word means, because you are just ignornat.
ur an ass dick
No, you are. You just don't like your errors being >>>>>>>>>>>>>>>> pointed out, as it shows how much of an ass you are. >>>>>>>>>>>>>>>>
Which is a response typical of those suffering from the >>>>>>>>>>>>>>>> effect.
The problem of creating the computations are >>>>>>>>>>>>>>>>>>>> equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's pretty >>>>>>>>>>>>>>>>>>> ridiculous richard
Because I am not, you are gaslighting yourself with >>>>>>>>>>>>>>>>>> your false definitions that you try to insist on. >>>>>>>>>>>>>>>>>>
if problems are equivalent then a solution to A can >>>>>>>>>>>>>>>>>>> be used to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the >>>>>>>>>>>>>>>>>>> fuck along with ur willful ignorance and gaslighting >>>>>>>>>>>>>>>>>>> dick
But where do you get your definition of equivalent. >>>>>>>>>>>>>>>>>>
As I have pointed out, that isn't the definition used >>>>>>>>>>>>>>>>>> in the field, a field you have admitted being >>>>>>>>>>>>>>>>>> untrained in.
So you admit your ignorance, but insist you must know >>>>>>>>>>>>>>>>>> better than people who actually know something. >>>>>>>>>>>>>>>>>>
In a word, Dunning-Kruger
never seen anyone bring that up in good faith >>>>>>>>>>>>>>>>
ur whole response is just a shitpile of insults and >>>>>>>>>>>>>>> fallacies
can't wait to see u take ur ignorance to the grave dick >>>>>>>>>>>>>>
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll,
and have literally argued against every single sentence >>>>>>>>>>>>> always without agreeing more than a spattering of times >>>>>>>>>>>>> over literally months of engagement,
on top of heaps of unjustified insults completely
unbecoming of anyone engaged in serious discussion,
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject
stupidity where u "win" when the opponent gives on u being >>>>>>>>>>> and endless fucking troll???
that's a total L bro. if u fail to convince ur opponent, u >>>>>>>>>>> _lose_
Nope, if the opponent is as brain dead as you show yourself, >>>>>>>>>> it isn't a
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence without >>>>>>> adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown
increadible ignorance and failure in logic. You just have a foul
mouth.
AHAHAHAHA, how motherfucking toxic do u have to try to rationalize
insults as "factually based"????
😂😂😂
What is "toxic" about truth? It is only toxic to people who live on
lies.
Shows how stupid you are.
Note, YOU are the one providing the evidence to prove my statements,
by just ignoring the facts.
Of course, what you are proving is you don't understand what truth
or facts actually are, since you logic is based on being able to
presume something without proof.
and literally nothing meaningful was said here
If pointing out errors isn't meaningful, that points out your
fundamental error in logic.
I guess your world is just built on your own fantasies, and you just
don't care about what is actually true.
Maybe you will be able to just imagine the food and shelter you are
going to need to keep on living once your money runs out.
three more sentences of nothing
It seems your nature is to insult and blame others for your own >>>>>>>> failings.
matter of convincing you, but protecting the naive from you lies. >>>>>>>>>this group is bunch of boomers who spent decades losing
arguments amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own ideas??? >>>>>>>
😂😂😂
Trying to. After all, I need to do something to wake you up to
your own lies that you have gaslit yourself with.
lol, i fixed turing's diagonal in a way that hasn't been addressed
in literature before
No you haven't, because you don't understand the requirements.
i certainly fixed it from stumbling on itself, which is a fix that
has not been addressed in any literature, until my paper
In other words, you think CHANGING the requitrements is a valid
process for meeting them.
... there was no need for the diagonal to test itself ...
the fact u can't even acknowledge that as useful is incredibly toxic tbh
i have then encountered a new problem, which doesn't negate the fix i
did make u ungrateful tard
Sure it does. You have presumed the enumeration that has been proved
can not be made.
toxic toxic toxic i'm so very tired of being on a world surrounded by
toxic people, rick
There can not be an effective (computable) enumeration that includes
all computable number, as any method that generates one allows the
computing of a number that doesn't exist in that set.
Thus, any computed enumeration of computable numbers is necessarily
incomplete.
All you are doint is proving you don't understand what it means to be
able to compute something.
It seems your idea of computing allows an algorithm to assume that it
(or even an equivalent to it) can't be embedded into another
algorithm, which makes your concept strictly weaker in the power to
compute than the methods used by Turing Machines and their equivalents.
The first problem is, you haven't created the enumeration required
to compute the diagonal of.
We know this, because we can convert your fixed-H to be anti-fixed-H
that outputs the opposite digits that fixed-H does (using the trick
of fixed-H, using the number of fixed-H, not anti-fixed-H), and thus
shows that if fixed-H is computing the diagonal, anti-fixed-H is
computing the anti-diagonal, but we also see that this anti-diagonal
isn't in the enumeration, and thus the enumeration can't be complete.
that does not prove there exists no further tricks that might still
get it on the diagonal somehow,
Sure it does.
rick, the only reason i got to this new problem was by ignoring all the idiots telling me turing's proof was absolute
*ANY* method to generate an enumeration of computable numbers allows
the creation of a computation that computes a number not in the set
that it generated.
Thus NO method to generate an enumeratio of computable numbers can
create a total enumeration.
Your assumption of trick is just depending on magic fairy dust from a
unicorn to create the impossible.
If it isn't complete, then the diagonal isn't the diagonal of the
enumeration that Turing was talking about.
even if there doesn't,
there may be provable limits on what computations can and cannot be
computably enumerated on the diagonal,
But not that excludes the anti-diagonal, since we HAVE the description
of the algorithm that generates it, at least if we have an algorithm
to generate the enumeration.
which is certainly a step up from the over-applied rice's theorem
know- nothing nonsense u see with theorists today,
You seem to have the problem of not understand what Rice proves.
rice's theorem doesn't even apply to all semantics of a machine, just
those which are detectable from the output ... a phrase which ur gunna disagree which because ur kind of a moron rick, but ben at least did acknowledge
There can be many machines that compute PARTIAL classifications or
decisions on machines, just not a TOTAL classification.
like what if those various pseudo-anti-diagonals (as they aren't true
total anti-diagonal) are the *only* set of computable numbers we
can't computably enumerate on the list???
But they aren't. There are other proof, well beyond your head, that
show that other questions turn out to not be computable.
nah rick, i'm _never_ gunna accept anyone telling me anything is "beyond
my head" and the fact u even try to write that is _incredibly toxic_
i mean, the gall of u ever writing that out to someone is just
_incredibly anti-intellectual_ , why would u ever demand someone accept something that exist "beyond their understanding" like a fking a
religious nutjob???
my god rick, u are such an fucking gross hypocrite
the fact anyone in this group supports ur writing is just such a stain
on this group, and really demonstrates the hostility and toxicity being harbored at the core of computing (and really academia at large)
In fact, by simple "counting" we can tell that there are an infinite
number of uncomputable problems for every computable one.
ofc there are countable infinite variations on it. that doesn't mean
there aren't limits to the kinds of computations in that set
that would also be a huge win, cause those computations don't compute
relationships we care about, so failing to enumerate them totally
just doesn't really matter 🤷
Sure it does. By knowing that TOTAL classification is impossible, we
know that we need to look at what classes of inputs a given algorithm
can work on.
Thus, like where you started with, because we KNOW we can't totally
solve the Halting Problem, we accept that we need to allow our
algorithm to decide that some cases might not be deciable, and work on
the cases we can decide on.
that's already a huge step up from before where you were advocating for programs that we couldn't even generally decide on their decidability
A correctness proving program doesn't need to prove EVERY program
correct or wrong, but can prove SOME programs correct, SOME programs
it can point out errors, and some it tells us they are too complicated
for it to process.
"too complicated to process" is a different theory rick. that's
complexity theory not computability theory.
If it tells us it is too complicted, if we really need the proof, we
need to revise it to simplify it. (Or it may be that the problem we
are working on is just uncomputable, so no program CAN be proven correct,
u haven't demonstrated an actual machine we can't prove correct,
the machines PRD failed to classify are still provable in what they do
from our perspective (we both know the pseudo-anti-diagonals are circle- free and can prove it ... that's how we know PRD "missed" them),
regardless of whether PRD could classify them or not
see ur kinda stuck in a rut here. any circle-free machine can prove that
PRD fails to enumerate is still a machine that was proven as circle-
free ...
i feel this is going to end up in abandoning the ct-thesis rick. tm computability has limits due to self-referential weirdness, and they
aren't the same as the limits we 3rd party observers are bounded by
because we're aren't subject to that same self-referential weirdness
and we need to build a partial version that admits that there are
cases it can't get correct)
(and before u try to make yet another baseless claim that it must
have been, show me the proof instead of baselessly just claiming u
fucking twat)
And what is wrong about this proof.
i don't have to be right about literally every possible future goal
post to right about one goal post in a unique way that's never been
done before. the fact i could even hit that goal post is to me a
massive sign things have been missed in the fundamentals, rick
it should be to you as well, but my god are obsessed with clinging to
certain uncertainty it's abcerd
But the fact that you current claims are based on NO evidence, means
you are starting with nothing.
You claim that something might be possible, when it is shown that it
can't be.
Your world is just built on the assumption that the rules don't apply.
That is a world of fantasy and lies.
nothing was said here
Your enumeration generated by PRD just can not be COMPLETE,
including at least one instance of EVERY computable number.
PROVE ME WRONG.
it's really kind of sad how much toxicity and generally unwillingness
to cooperate i've encounter when trying to explore these ideas,
The most toxic thing is to just lie to yourself about what can be done.
i hope future academia may take heed from what i've had to endure
thus far, pretty much on my own. heck i hope current academic might
too ...
but that might be asking for too much at this point eh???
In other words, you hope academia might allow people to live in error
and self-deciet?
i would be nice if u could even read simple sentence accurately.
i said it was too much to ask for, in that i hope for it, but don't
expect it. not sure where u pulled hoping for opposite from... but i
never claimed that
Your problem is you reject people pointing out the errors in your work,
rick, u have problems reading simple sentences much of the time
because you assume you must be right, even when you admit you don't
really understand the field.
Your works is based on ignorant assumptions, not facts.
That is a dark world of lies.
The people here mostly know what they are talking about, because >>>>>>>> they have studied it (some like Olcott and you are the exception). >>>>>>>>
i demonstrated two distinct fallacies in turing's paper, that >>>>>>>>> really aren't the hard to understand,
It seems you are just admitting that you are stuck in your >>>>>>>>>> lies and just can't think because, like Olcott, you have
successfully gatlit yourself into being convinced of your lies. >>>>>>>>>
No, you demonstrated that you do don't understand what he is
saying,
the fact u continually try to gaslight me into thinking i haven't >>>>>>> understood his argument well enough is not only incredibly toxic >>>>>>> but let's me know ur completely fine with blatantly lying at me >>>>>>> to "win" an argument,
But it isn't gaslighting, it is FACT.
idk how u read everything i wrote and try to claim i just don't
understand,
Because you keep claiming things that aren't there.
i have been manipulating his ideas in ways that have never been
done before, and u can't even acknowledge that i understand his
ideas??? sheesh,
Right, by twisting words to not mean what they mean.
i'm swimming thru a swamp of endless gaslighting, fostered a toxic
mentality festering the fundamentals of math hostile to any sort of >>>>> meaningful innovation at the core for some ungodly reason
Yes, the endless gaslighting that you have done to yourself, causing
you to think that people point out truth to you are gaslighting you.
The fact you can't actually prove anything should be your first sign
that you have something wrong.
Your world is just built on your own lies and fantasies.
the fallacies i picked out are still fallacies
No, because your claimed fallacies are based on you using a definist
fallacy.
fallacy 1) identifying a subset is _NOT_ the same problem as identify
the entirety of a set,
fallacy 2) computing the diagonal does _NOT_ then grant an able to
computing an anti-diagonal
i'm not redefining terms in either (computable numbers *ARE* a subset of circle-machines), ur claims of fallacy are incorrect
in a reasonable debate, that should suffice, but u have been anything
but reasonable with me rick
Starting with that error, NOTHING you have said has any basis to point
out error.
that's not what the side with truth does, or even remotely needs >>>>>>> to do. and if u can't recognize that, i'm sorry for all fallacy >>>>>>> u've bought into across all the things u believe
Claiming truth is gaslighting is just gaslighting.
> take them to the grave bro
>
> #god
You got a source for that?
Or do you have the same "god" complex as Olcott?
oh does he also claim that _we are god_ ???
No, just him.
It seems you like to "quote" things, claiming them to be from
"god" (as he signs them). That seems to imply you think you have a
special link to him.
like i said: _we are god_
> so anyone can do it eh???
>
> #god
if this "me" is particularly special, that is only due by seeding a
trend, if a trend even ever takes off, which is yet to be seen...
But "we" are not "god", and assuming you have the divine power of god
is the beginning of your own condemnation to a life of error and dispare.
i never said _i_ had that ...
🤷🤷🤷
u can write it off as psycho-spiritual outburst of frustration from
constantly banging my head against the various mental walls keeping
us chained to acting _far_ less ethically than we should,
Or, those "walls" are the boundries that aim us to what can be done.
rick u would be the kinda unethical fuck "aiming" at people to bomb, eh???
but don't count on me stopping. the grave we've been digging for our
species thru our systemic moral negligence is _deep_ ,
But it seems, the grave you see, is just the grave for those that think
that grave is far deeper than mistakes in the fundamentals of computing
like you, and that the uncompuatable nature of some things means we
can't "do our best" and handle the cases we actually care about.
We CAN prove that some programs are correct. The cost is just too high
to be used everywhere, and many programs don't actually need to be
provably correct.
muh econobabble, right ...
YOU are the one that rejects that this ACHEIVABLE (and ACHEIVED) goal
is good enough, and seem to want that all work stops until we can do
the impossible.
THAT is the truely toxic.
and it's gunna take some dramatic times to pull ourselves out
like how many more pedo islands do u think exist?
regardless of whether turing's conclusions are ultimately
correct or not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Then show me your source.
my source for fallacies that *i* pointed out???
i would be source for that... duh?
So, what is that source?
That is your problem, you have no sources but your own
misunderstanding.
That come from your gaslighting of yourself to brainwash you into
thinking you don't need sources.
why would i need sources to justify novel arguments???
To show that you claims are based on FACTS and not errors?
All you are doing is proving that you are ignorant and stupid.
Most "novel" arguement are just errors and fallacies.
are you asking for me to repeat the arguments i generated? u can
either reread the posts i made on usenet
or here for fallacy 1: https://www.academia.edu/165010519
Which is just repeating the error that thinking that equivalent
problems are solving the same problem.
Where is the definition that says that?
Equivalent problems are problems that are true/solvable or false/
unsolvable together.
... right and you prove that relationship by demonstrating that a
solution to one problem causally leads to a solution for the other and vise-versa, which is _WHY_ they are solvable/unsolvable together ...
which ur never going to accept, because then u'd have to admit i did
even /one thing/ correct,
> which you are too toxic a person to ever admit
>
> #god
may u someday repent for the endless sinning u've committed upon me 🙏
Going to the fallacy of appeal to authority, using yourself as the
authority is just stupid.
It is also your definist fallacy, as you are trying to redefine the
word "equivalent" as used as a modifier for problem.
So, if the number computed by anti-fixed-H isn't in the enumeration,
Your claim is just the gaslighting you claim I am doing.
yes, certain further work would take time and therefore funds,
Yes, while the title of the paper uses the topic of "Computable >>>>>>>> Numbers", and the part of the proof focuses on the related
concept of machines that compute them, he DOES show a proof,
that could be similarly used to prove the uncomputablility of >>>>>>>> the computable numbers.
Your problem is you have such a wooden and limited knowledge of >>>>>>>> what you read, you can't understand what he is doing.
u've never "won" an argument here in the decades u wasted ur >>>>>>>>>>> life here
get back to helping america bomb muzzies for their joo
overlords, that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but >>>>>>>>>> your bluster.
Sorry, that won't cut it for begging people to give you money >>>>>>>>>> to carry
who am i begging??? who around here even has money to give??? >>>>>>>>> 😂 😂😂
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your work >>>>>>>> so you can complete the parts that you admit have holes in them. >>>>>>>
and that kind of work will remain out of scope of this
discussion. that's a statement of fact, not "begging" u sad
dishonest old man
But, since you idea have been proven wrong, and you don't even try >>>>>> to refute it, why should anyone support that work.
As I said, Your PRD couldn't have accepted ANY machine that
generates the same computable number as anti-fixed-H, and thus the >>>>>> enumeration it generates is NOT complete.
right that is a thorn in my side atm, and there are several avenues >>>>> i'm exploring in thot on how to deal with that
- first of all i'm not entirely convinced there isn't some
strategy i'm missing that might still yet get it on the diagonal.
we haven't even build an enumeration of computable numbers cause we >>>>> haven't discussed the dedpuing logic, and i don't know how that
impact the current predicament
- and even if so do we actually care about the computations
being done outside of the decidably enumerable set? can we build a
way to identify and classify what they do?
- and/or is this the wall i needed to justify the jump to
constructing RTMs, cause the "incompleteness" there happens due to
actual lies built into the computations...
look i acknowledged ur argument!
which is literally more generous than you've ever been towards me,
cause u've never admitted an ounce of validity of my words, despite >>>>> understanding them well enough to pounce on any and all criticisms... >>>>
how can PRD, or ANY PRD that could exist (and then an anti-fixed-H
be built on it).
Your problem is you don't understand the fundamental nature of what
a computation is.
now *that* is fking toxic bro, and if u think i'm going to be
swayed by such toxicity, well i know some therapists u can talk to
about that kinda negative mindset rick, their teachings served me
well thus far
In other words, your whole plan is to hope that a magic fairy dust
powered unicorn can give you the answer.
So, you assume that rules must not apply, but then you don't even
know the basic definitions of what you are talking about, so of
course you can't understand the rules.
Your "logic" is based on the rules not applying and computations not
being actually computations. In part, because you don't actually
understand what a computation is, and thus you imagine things that
aren't computations but wnat to think of them as possibly being a
computation.
again, nothing was said here
In part because there wasn't anything to reply to.
IT seems you have run out of ways to fabricate your ideas, so you just
insult the messager pointing out your errors.
Thus, nothing you have done with it meets the requirements for the >>>>>> computation you talk about, as it, by definition, starts with a
complete enumeration.
There is not contradiction in my anti-fixed-H if the enumeration
isn't complete, but you also are proven to just be a liar about
your claim of showing a way to compute a diagonal of a complete
enumeration of the computable numbers.
You seem to expect that "someone" will like your crap work
enough to pay you to continue working on it with the hope that >>>>>>>> you can materialize your unicorn, even though they have been
proven to be impossible.
YOU are the idiot, with a foul mouth, that doesn't know what he >>>>>>>> is talking about.
your ignorant ideas further, as you are just showing there >>>>>>>>>> isn't anything to base going farther on.
If you can show an actual error I am making, with sources to >>>>>>>>>>>> back up your claims, present them.
The problem is you KNOW that you don't know what you are >>>>>>>>>>>> talking about because you have ADMITTED to not actually >>>>>>>>>>>> studing more that a few papers, but you think you are >>>>>>>>>>>> smarter than the people who wrote them.
YOU are the one flying to the grave in a crashing plane of >>>>>>>>>>>> ignorance.
I will note, that just like with Peter Olcott, YOU are the >>>>>>>>>>>> one that started the insults, showing whose mind is in the >>>>>>>>>>>> gutter.
On 3/18/26 1:32 PM, dart200 wrote:
On 3/18/26 4:32 AM, Richard Damon wrote:
On 3/18/26 3:13 AM, dart200 wrote:
On 3/17/26 7:46 PM, Richard Damon wrote:
On 3/17/26 12:55 AM, dart200 wrote:
On 3/16/26 6:50 PM, Richard Damon wrote:
On 3/16/26 1:11 PM, dart200 wrote:
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:So, what is my error, with actual SOURCES for your data >>>>>>>>>>>>>>> that claims I am wrong?
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote:
On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 3/14/26 1:29 PM, dart200 wrote:Right, equivLWNR PROBLEM, which means both problems >>>>>>>>>>>>>>>>>>> are either solvable or not (under all applicable >>>>>>>>>>>>>>>>>>> models).
| 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 >>>>>>>>>>>>>>>>>>>On 3/14/26 4:48 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 12:43 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 4:28 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:12 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence D’Oliveiro >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
how can computing a _subset_ of circle-free >>>>>>>>>>>>>>>>>>>>>> machines be equivalent to compute a _total_ set of >>>>>>>>>>>>>>>>>>>>>> circle- free machines...???And B is the machine that computes the diagonals >>>>>>>>>>>>>>>>>>>>>>> of the results of the enumeration of circle-free >>>>>>>>>>>>>>>>>>>>>>> machines.my god rick, please fucking read the not even >>>>>>>>>>>>>>>>>>>>>>>> whole paper, but at least the _section_ rick >>>>>>>>>>>>>>>>>>>>>>>> please...if problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>But he didn't, that is just what your >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ignorant brain thinks he must have been >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> talking about.On Thu, 12 Mar 2026 00:41:06 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On 3/12/26 12:17 AM, Lawrence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But they’re a subset of the “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined” set that Turing was >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>the "better-defined" machine don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfy the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> (which attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> does not then prove an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility for enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But if the machine can be “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined”, yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine that also satisfies >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> those “ill-defined” criteria must be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Turing.
there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equating the solution of them as the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
| 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 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what >>>>>>>>>>>>>>>>>>>>>>>>>>>>> EQUIVALENT means here. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
After all Functional Equivalence doesn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>> mean the same machine or even using the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> same basic algorithm. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
No, he his saying the problems are >>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent as to the nature >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
He doesn't say the two machines generated >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> by the two problems are in any way >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent, he says that the PROBLEMS are >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent,
he's literally saying that if u can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerate computable sequences, then u >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> could use that solution to determine >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> whether any given machine is circle- free ... >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
and if so could be used to enumerate the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> circle- free machines, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
making the problem of enumerating the sets >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent,
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to produce >>>>>>>>>>>>>>>>>>>>>>>>>>>> a solution to B
AND
a solution to B can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>>>>> solution to A
Where do you get that definition? >>>>>>>>>>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in >>>>>>>>>>>>>>>>>>>>>>>>>>> all models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" >>>>>>>>>>>>>>>>>>>>>>>>>>> you do something,
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be >>>>>>>>>>>>>>>>>>>>>>>>>>>> used to produce a solution to enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>> computable numbers, but the reverse is *NOT* >>>>>>>>>>>>>>>>>>>>>>>>>>>> true
But it doesn't need to.
yes it does, rick
WHY?
As I have said, you don't understand what he >>>>>>>>>>>>>>>>>>>>>>>>> was saying, and thus are trying to kill a >>>>>>>>>>>>>>>>>>>>>>>>> strawman.
Where does he ACTUALLY SAY that the machine >>>>>>>>>>>>>>>>>>>>>>>>> that generates circle- ftee machihes could be >>>>>>>>>>>>>>>>>>>>>>>>> used to enumerate computable numbers. >>>>>>>>>>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON THE >>>>>>>>>>>>>>>>>>>>>>>> SAME FUCKING PAGES WE'VE BEEN TALKING ABOUT p246: >>>>>>>>>>>>>>>>>>>>>>>>
| The simplest and most direct proof of this is >>>>>>>>>>>>>>>>>>>>>>>> by showing that,
| if this general process exists [for circle- >>>>>>>>>>>>>>>>>>>>>>>> free machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>>>>>>>
do i need to spell out why with even more detail??? >>>>>>>>>>>>>>>>>>>>>>>
Why doesn't the program do that? >>>>>>>>>>>>>>>>>>>>>>>
ok ok i will even tho u will continue to >>>>>>>>>>>>>>>>>>>>>>>> disagree...
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>>>>>>>>> <=> enumerating computable sequence >>>>>>>>>>>>>>>>>>>>>>>> => diagonal is computable >>>>>>>>>>>>>>>>>>>>>>>> => β is computable _contradiction_ >>>>>>>>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a >>>>>>>>>>>>>>>>>>>>>>>> turing machine, produce a general process to >>>>>>>>>>>>>>>>>>>>>>>> output whether a machine is circle- free or not >>>>>>>>>>>>>>>>>>>>>>>
the _first fallacy_ is that because that isn't >>>>>>>>>>>>>>>>>>>>>>>> actually equivalent to enumerating computable >>>>>>>>>>>>>>>>>>>>>>>> sequences (which is a lesser problem that only >>>>>>>>>>>>>>>>>>>>>>>> needs to recognize a subset of circle- free >>>>>>>>>>>>>>>>>>>>>>>> machines), ruling out a general process for >>>>>>>>>>>>>>>>>>>>>>>> deciding circle- free machine does _not_ >>>>>>>>>>>>>>>>>>>>>>>> actually rule out a general process for >>>>>>>>>>>>>>>>>>>>>>>> enumerating computable numbers
A fallacy in your mind, because you don't >>>>>>>>>>>>>>>>>>>>>>> understand what he means by equivalent. >>>>>>>>>>>>>>>>>>>>>>
Who said they were equivalent COMPUTATIONS. >>>>>>>>>>>>>>>>>>>>
_because_ a solution to one leads to a solution for >>>>>>>>>>>>>>>>>> the other...
Nope.
Where are you getting your definitions? Because you are >>>>>>>>>>>>>>>>> using the wrong ones.
All you are doing is proving your stubborn refusal to >>>>>>>>>>>>>>>>> learn what you are talking about, and that you don't >>>>>>>>>>>>>>>>> care you are ignorant.
which is a fallacy in this case, they are not >>>>>>>>>>>>>>>>>> equivalent problems
Sure they are, you just don't know what that means as >>>>>>>>>>>>>>>>> you continue to hang on to your errors because you >>>>>>>>>>>>>>>>> don't understand the language you are reading. >>>>>>>>>>>>>>>>>
ur an ass dick
IT seems you are just showing you don't know what the >>>>>>>>>>>>>>>>>>> word means, because you are just ignornat. >>>>>>>>>>>>>>>>>>
No, you are. You just don't like your errors being >>>>>>>>>>>>>>>>> pointed out, as it shows how much of an ass you are. >>>>>>>>>>>>>>>>>
Which is a response typical of those suffering from the >>>>>>>>>>>>>>>>> effect.
The problem of creating the computations are >>>>>>>>>>>>>>>>>>>>> equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's >>>>>>>>>>>>>>>>>>>> pretty ridiculous richard
Because I am not, you are gaslighting yourself with >>>>>>>>>>>>>>>>>>> your false definitions that you try to insist on. >>>>>>>>>>>>>>>>>>>
if problems are equivalent then a solution to A can >>>>>>>>>>>>>>>>>>>> be used to solve B and vise versa ...
Says who?
if u don't agree with this then u can move right the >>>>>>>>>>>>>>>>>>>> fuck along with ur willful ignorance and gaslighting >>>>>>>>>>>>>>>>>>>> dick
But where do you get your definition of equivalent. >>>>>>>>>>>>>>>>>>>
As I have pointed out, that isn't the definition used >>>>>>>>>>>>>>>>>>> in the field, a field you have admitted being >>>>>>>>>>>>>>>>>>> untrained in.
So you admit your ignorance, but insist you must know >>>>>>>>>>>>>>>>>>> better than people who actually know something. >>>>>>>>>>>>>>>>>>>
In a word, Dunning-Kruger
never seen anyone bring that up in good faith >>>>>>>>>>>>>>>>>
ur whole response is just a shitpile of insults and >>>>>>>>>>>>>>>> fallacies
can't wait to see u take ur ignorance to the grave dick >>>>>>>>>>>>>>>
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll, >>>>>>>>>>>>>>
and have literally argued against every single sentence >>>>>>>>>>>>>> always without agreeing more than a spattering of times >>>>>>>>>>>>>> over literally months of engagement,
on top of heaps of unjustified insults completely >>>>>>>>>>>>>> unbecoming of anyone engaged in serious discussion, >>>>>>>>>>>>>>
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject >>>>>>>>>>>> stupidity where u "win" when the opponent gives on u being >>>>>>>>>>>> and endless fucking troll???
that's a total L bro. if u fail to convince ur opponent, u >>>>>>>>>>>> _lose_
Nope, if the opponent is as brain dead as you show yourself, >>>>>>>>>>> it isn't a
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence
without adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown
increadible ignorance and failure in logic. You just have a foul >>>>>>> mouth.
AHAHAHAHA, how motherfucking toxic do u have to try to rationalize >>>>>> insults as "factually based"????
😂😂😂
What is "toxic" about truth? It is only toxic to people who live on >>>>> lies.
Shows how stupid you are.
Note, YOU are the one providing the evidence to prove my
statements, by just ignoring the facts.
Of course, what you are proving is you don't understand what truth
or facts actually are, since you logic is based on being able to
presume something without proof.
and literally nothing meaningful was said here
If pointing out errors isn't meaningful, that points out your
fundamental error in logic.
I guess your world is just built on your own fantasies, and you just
don't care about what is actually true.
Maybe you will be able to just imagine the food and shelter you are
going to need to keep on living once your money runs out.
three more sentences of nothing
And one for you too.
I guess the actual pointing out of the errors of your "logic" is
meaningless to you, as truth and reality have no meaning to you.
You will problem call this meaningless too, just proving that you don't understand what is being said, and things you don't understand are
"nothing" to you.
It seems your nature is to insult and blame others for your own >>>>>>>>> failings.
matter of convincing you, but protecting the naive from you >>>>>>>>>>> lies.
this group is bunch of boomers who spent decades losing
arguments amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own ideas??? >>>>>>>>
😂😂😂
Trying to. After all, I need to do something to wake you up to
your own lies that you have gaslit yourself with.
lol, i fixed turing's diagonal in a way that hasn't been addressed >>>>>> in literature before
No you haven't, because you don't understand the requirements.
i certainly fixed it from stumbling on itself, which is a fix that
has not been addressed in any literature, until my paper
In other words, you think CHANGING the requitrements is a valid
process for meeting them.
... there was no need for the diagonal to test itself ...
But only if it IS in the enumeration at that point.
Note, your "fixed-anti-fixed-H" doesn't actually compute the actual anti-diagonal.
the fact u can't even acknowledge that as useful is incredibly toxic tbh
Lying is not useful.
The problem is your fixed-anti-fixed-H doesn't fix the problem, as we
still have that anti-fixed-H shows that yoru PRD doesn't meet its requirements.
If PRD only accepts circle-free machines, it WILL be a circle-free
machine, and thus the results it generates WILL be a computable number,
but that number can not exist at any finite point in the enumartion generated by PRD.
Thus, PRD fails to generate a COMPLETE enumeration, and thus your fixed-
H does not compute the diagonal of such a complete enumeration, because
the enumeration it computed the diagonal of was not complete.
This seems beyond your ability to understand, so I wonder what fallacy
you are going to use to try to refute this. or perhaps you just ignore
it, and thus admit you don't have an answer.
i have then encountered a new problem, which doesn't negate the fix
i did make u ungrateful tard
Sure it does. You have presumed the enumeration that has been proved
can not be made.
toxic toxic toxic i'm so very tired of being on a world surrounded by
toxic people, rick
LYING is toxic.
Your enumeration is not what you claim, and thus your claims are the
toxic part.
It seems you like to call as "toxic" anything you can't deal with, as
you method of handling the ERRORS in you logic.
YOU are the toxic one.
There can not be an effective (computable) enumeration that includes
all computable number, as any method that generates one allows the
computing of a number that doesn't exist in that set.
Thus, any computed enumeration of computable numbers is necessarily
incomplete.
All you are doint is proving you don't understand what it means to be
able to compute something.
It seems your idea of computing allows an algorithm to assume that it
(or even an equivalent to it) can't be embedded into another
algorithm, which makes your concept strictly weaker in the power to
compute than the methods used by Turing Machines and their equivalents.
The first problem is, you haven't created the enumeration required
to compute the diagonal of.
We know this, because we can convert your fixed-H to be anti-fixed- >>>>> H that outputs the opposite digits that fixed-H does (using the
trick of fixed-H, using the number of fixed-H, not anti-fixed-H),
and thus shows that if fixed-H is computing the diagonal, anti-
fixed-H is computing the anti-diagonal, but we also see that this
anti-diagonal isn't in the enumeration, and thus the enumeration
can't be complete.
that does not prove there exists no further tricks that might still
get it on the diagonal somehow,
Sure it does.
rick, the only reason i got to this new problem was by ignoring all
the idiots telling me turing's proof was absolute
So, you jumped out of the frying pan into the flames and are burning yourself to death.
Your "logic" is based on ignoring FACTS and definitions, and assuming
that magic fairy dust powered unicorns can make you impossible ideas work.
This is the work of a mind destroyed by a steady diet of your own toxic
lies that have eaten out every ounce of reasoning you might have had.
*ANY* method to generate an enumeration of computable numbers allows
the creation of a computation that computes a number not in the set
that it generated.
Thus NO method to generate an enumeratio of computable numbers can
create a total enumeration.
Your assumption of trick is just depending on magic fairy dust from a
unicorn to create the impossible.
If it isn't complete, then the diagonal isn't the diagonal of the
enumeration that Turing was talking about.
even if there doesn't,
there may be provable limits on what computations can and cannot be
computably enumerated on the diagonal,
But not that excludes the anti-diagonal, since we HAVE the
description of the algorithm that generates it, at least if we have
an algorithm to generate the enumeration.
which is certainly a step up from the over-applied rice's theorem
know- nothing nonsense u see with theorists today,
You seem to have the problem of not understand what Rice proves.
rice's theorem doesn't even apply to all semantics of a machine, just
those which are detectable from the output ... a phrase which ur gunna
disagree which because ur kind of a moron rick, but ben at least did
acknowledge
So, your sub-machine doesn't generate that as part of its output?
Then how does the outer machine know the answer it gabe?
There can be many machines that compute PARTIAL classifications or
decisions on machines, just not a TOTAL classification.
like what if those various pseudo-anti-diagonals (as they aren't
true total anti-diagonal) are the *only* set of computable numbers
we can't computably enumerate on the list???
But they aren't. There are other proof, well beyond your head, that
show that other questions turn out to not be computable.
nah rick, i'm _never_ gunna accept anyone telling me anything is
"beyond my head" and the fact u even try to write that is _incredibly
toxic_
DUNNING-KRUGER in action,
i mean, the gall of u ever writing that out to someone is just
_incredibly anti-intellectual_ , why would u ever demand someone
accept something that exist "beyond their understanding" like a fking
a religious nutjob???
But you PROVE that it is beyond your understanding.
my god rick, u are such an fucking gross hypocrite
the fact anyone in this group supports ur writing is just such a stain
on this group, and really demonstrates the hostility and toxicity
being harbored at the core of computing (and really academia at large)
So, if you think you are so smart, submit your work to a real peer-
reviewed journal and see how fast it is shot down.
The "hostility" you perceive is people pointing out your errors that you refuse to accept, because "Truth" isn't something you world can handle.
In fact, by simple "counting" we can tell that there are an infinite
number of uncomputable problems for every computable one.
ofc there are countable infinite variations on it. that doesn't mean
there aren't limits to the kinds of computations in that set
I guess you haven't read any of the papers of the other kinds of uncomputable problems, one NOT based on a "self-reference".
Yes, many of them allow you to, as ONE of the ways, to prove them uncomputable, show that them being computable would allow you to compute
the answer to the uncomputable problems due to self-reference.
But, they don't themselves use that sort of self-reference.
that would also be a huge win, cause those computations don't
compute relationships we care about, so failing to enumerate them
totally just doesn't really matter 🤷
Sure it does. By knowing that TOTAL classification is impossible, we
know that we need to look at what classes of inputs a given algorithm
can work on.
Thus, like where you started with, because we KNOW we can't totally
solve the Halting Problem, we accept that we need to allow our
algorithm to decide that some cases might not be deciable, and work
on the cases we can decide on.
that's already a huge step up from before where you were advocating
for programs that we couldn't even generally decide on their decidability
But that is still true, and not contradictory with the above.
There are many programs that we can decide on.
But there are also some that we can't, and some we can't even decide
that there behavior is unknowable.
A correctness proving program doesn't need to prove EVERY program
correct or wrong, but can prove SOME programs correct, SOME programs
it can point out errors, and some it tells us they are too
complicated for it to process.
"too complicated to process" is a different theory rick. that's
complexity theory not computability theory.
No, this isn't "complexity" as in O notation complexity, but that our processing, but necessity, can't try to handle all cases to all depths,
but, to avoid getting stuck and not answering, exstablishes finite
limits on resources that can be expended on the various parts of the analysis, and if the analysis of the program hits one of these limits,
we classify it as "too complex".
If it tells us it is too complicted, if we really need the proof, we
need to revise it to simplify it. (Or it may be that the problem we
are working on is just uncomputable, so no program CAN be proven
correct,
u haven't demonstrated an actual machine we can't prove correct,
the machines PRD failed to classify are still provable in what they do
from our perspective (we both know the pseudo-anti-diagonals are
circle- free and can prove it ... that's how we know PRD "missed"
them), regardless of whether PRD could classify them or not
And thus, PRD can not be "correct" to its specification, as one of the REQUIREMENTS was that it would accept at least one machine that
generates EVERY computable number.
the "anti-diagonal" anti-fixed=H is not "pseudo" anything, given your claimed PRD, it is a REAL machine, that computes a REAL computable
number that no machine in your enumerate generates.
see ur kinda stuck in a rut here. any circle-free machine can prove
that PRD fails to enumerate is still a machine that was proven as
circle- free ...
So, you are forgetting that for your claim was that fixed-H generates an computable diagonal of a set that is a enumeration of a set of amchines
that contains EVERY computable number.
You forgot that requirement, as you went off on your strawman fallacy.
i feel this is going to end up in abandoning the ct-thesis rick. tm
computability has limits due to self-referential weirdness, and they
aren't the same as the limits we 3rd party observers are bounded by
because we're aren't subject to that same self-referential weirdness
feelings don't generate proofs.
and we need to build a partial version that admits that there are
cases it can't get correct)
(and before u try to make yet another baseless claim that it must >>>>>> have been, show me the proof instead of baselessly just claiming u >>>>>> fucking twat)
And what is wrong about this proof.
i don't have to be right about literally every possible future goal
post to right about one goal post in a unique way that's never been
done before. the fact i could even hit that goal post is to me a
massive sign things have been missed in the fundamentals, rick
it should be to you as well, but my god are obsessed with clinging
to certain uncertainty it's abcerd
But the fact that you current claims are based on NO evidence, means
you are starting with nothing.
You claim that something might be possible, when it is shown that it
can't be.
Your world is just built on the assumption that the rules don't
apply. That is a world of fantasy and lies.
nothing was said here
Sure there was, you just can't understand it, as your world is built on
that lie.
Your enumeration generated by PRD just can not be COMPLETE,
including at least one instance of EVERY computable number.
PROVE ME WRONG.
it's really kind of sad how much toxicity and generally
unwillingness to cooperate i've encounter when trying to explore
these ideas,
The most toxic thing is to just lie to yourself about what can be done.
i hope future academia may take heed from what i've had to endure
thus far, pretty much on my own. heck i hope current academic might
too ...
but that might be asking for too much at this point eh???
In other words, you hope academia might allow people to live in error
and self-deciet?
i would be nice if u could even read simple sentence accurately.
i said it was too much to ask for, in that i hope for it, but don't
expect it. not sure where u pulled hoping for opposite from... but i
never claimed that
No, the problem is you think you are being treated unfairly, but you are not, you are treating truth unfairly.
You ARE living a life of lies, based on the ignoring of basic principles.
Your hope is for a world where error is just tolerated under some guise
of acceptance.
Your problem is you reject people pointing out the errors in your work,
rick, u have problems reading simple sentences much of the time
Less than you do.
You don't even know what an "equivalent problem" is.
Or what a "computation" is.
because you assume you must be right, even when you admit you don't
really understand the field.
Your works is based on ignorant assumptions, not facts.
That is a dark world of lies.
The people here mostly know what they are talking about,
because they have studied it (some like Olcott and you are the >>>>>>>>> exception).
i demonstrated two distinct fallacies in turing's paper, that >>>>>>>>>> really aren't the hard to understand,
It seems you are just admitting that you are stuck in your >>>>>>>>>>> lies and just can't think because, like Olcott, you have >>>>>>>>>>> successfully gatlit yourself into being convinced of your lies. >>>>>>>>>>
No, you demonstrated that you do don't understand what he is >>>>>>>>> saying,
the fact u continually try to gaslight me into thinking i
haven't understood his argument well enough is not only
incredibly toxic but let's me know ur completely fine with
blatantly lying at me to "win" an argument,
But it isn't gaslighting, it is FACT.
idk how u read everything i wrote and try to claim i just don't
understand,
Because you keep claiming things that aren't there.
i have been manipulating his ideas in ways that have never been
done before, and u can't even acknowledge that i understand his
ideas??? sheesh,
Right, by twisting words to not mean what they mean.
i'm swimming thru a swamp of endless gaslighting, fostered a toxic >>>>>> mentality festering the fundamentals of math hostile to any sort
of meaningful innovation at the core for some ungodly reason
Yes, the endless gaslighting that you have done to yourself,
causing you to think that people point out truth to you are
gaslighting you.
The fact you can't actually prove anything should be your first
sign that you have something wrong.
Your world is just built on your own lies and fantasies.
the fallacies i picked out are still fallacies
No, because your claimed fallacies are based on you using a definist
fallacy.
fallacy 1) identifying a subset is _NOT_ the same problem as identify
the entirety of a set,
No, but might be an equivalent problem.
fallacy 2) computing the diagonal does _NOT_ then grant an able to
computing an anti-diagonal
Why not?
WHy can you NOT just change that program to reverse the value written to
the perminante cells, and any decision based on reading one of those cells?
How does that NOT result in that result?
i'm not redefining terms in either (computable numbers *ARE* a subset
of circle-machines), ur claims of fallacy are incorrect
No they are not.
Computable numbers are numbers, that are computable by machines.
Circle-free machine generate computab;e numbers
in a reasonable debate, that should suffice, but u have been anything
but reasonable with me rick
Why? It is just a stupid category error.
Starting with that error, NOTHING you have said has any basis to
point out error.
that's not what the side with truth does, or even remotely needs >>>>>>>> to do. and if u can't recognize that, i'm sorry for all fallacy >>>>>>>> u've bought into across all the things u believe
Claiming truth is gaslighting is just gaslighting.
> take them to the grave bro
>
> #god
You got a source for that?
Or do you have the same "god" complex as Olcott?
oh does he also claim that _we are god_ ???
No, just him.
It seems you like to "quote" things, claiming them to be from "god" >>>>> (as he signs them). That seems to imply you think you have a
special link to him.
like i said: _we are god_
> so anyone can do it eh???
>
> #god
if this "me" is particularly special, that is only due by seeding a
trend, if a trend even ever takes off, which is yet to be seen...
But "we" are not "god", and assuming you have the divine power of god
is the beginning of your own condemnation to a life of error and
dispare.
i never said _i_ had that ...
YOU are quoting what you think "god" has said.
Which sort of implies you think you know what he is saying.
🤷🤷🤷
u can write it off as psycho-spiritual outburst of frustration from
constantly banging my head against the various mental walls keeping
us chained to acting _far_ less ethically than we should,
Or, those "walls" are the boundries that aim us to what can be done.
rick u would be the kinda unethical fuck "aiming" at people to bomb,
eh???
but don't count on me stopping. the grave we've been digging for our
species thru our systemic moral negligence is _deep_ ,
But it seems, the grave you see, is just the grave for those that think
that grave is far deeper than mistakes in the fundamentals of computing
Yep, your stupidity runs very deep.
like you, and that the uncompuatable nature of some things means we
can't "do our best" and handle the cases we actually care about.
We CAN prove that some programs are correct. The cost is just too
high to be used everywhere, and many programs don't actually need to
be provably correct.
muh econobabble, right ...
YOU are the one that rejects that this ACHEIVABLE (and ACHEIVED) goal
is good enough, and seem to want that all work stops until we can do
the impossible.
THAT is the truely toxic.
and it's gunna take some dramatic times to pull ourselves out
like how many more pedo islands do u think exist?
regardless of whether turing's conclusions are ultimately >>>>>>>>>> correct or not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Then show me your source.
my source for fallacies that *i* pointed out???
i would be source for that... duh?
So, what is that source?
That is your problem, you have no sources but your own
misunderstanding.
That come from your gaslighting of yourself to brainwash you into
thinking you don't need sources.
why would i need sources to justify novel arguments???
To show that you claims are based on FACTS and not errors?
All you are doing is proving that you are ignorant and stupid.
Most "novel" arguement are just errors and fallacies.
are you asking for me to repeat the arguments i generated? u can
either reread the posts i made on usenet
or here for fallacy 1: https://www.academia.edu/165010519
Which is just repeating the error that thinking that equivalent
problems are solving the same problem.
Where is the definition that says that?
Equivalent problems are problems that are true/solvable or false/
unsolvable together.
... right and you prove that relationship by demonstrating that a
solution to one problem causally leads to a solution for the other and
vise-versa, which is _WHY_ they are solvable/unsolvable together ...
Right. SO it isn't that the solution for one DIRECTLY solves the other,
but something about the nature of one solution leads to the other.
Admittedly, Turing did not present a proof of their equivalence here,
but that failure does not make it a fallacy of the arguement.
You need to prove that it wasn't established, or at least been accepted elsewhere.
which ur never going to accept, because then u'd have to admit i did
even /one thing/ correct,
> which you are too toxic a person to ever admit
>
> #god
You don't understand what he is saying, do you.
YOU are the one that is full of toxic waste.
may u someday repent for the endless sinning u've committed upon me 🙏
Telling the truth is NOT a sin.
LYING, as you do IS.
As is claiming revelation from god that he didn't actually give you.
Going to the fallacy of appeal to authority, using yourself as the
authority is just stupid.
It is also your definist fallacy, as you are trying to redefine the
word "equivalent" as used as a modifier for problem.
Your claim is just the gaslighting you claim I am doing.
Yes, while the title of the paper uses the topic of "Computable >>>>>>>>> Numbers", and the part of the proof focuses on the related
concept of machines that compute them, he DOES show a proof, >>>>>>>>> that could be similarly used to prove the uncomputablility of >>>>>>>>> the computable numbers.
Your problem is you have such a wooden and limited knowledge of >>>>>>>>> what you read, you can't understand what he is doing.
u've never "won" an argument here in the decades u wasted ur >>>>>>>>>>>> life here
get back to helping america bomb muzzies for their joo >>>>>>>>>>>> overlords, that's all ur good for
So, you still can't point out any error with a source!
So you are just admitting that you don't have anything but >>>>>>>>>>> your bluster.
Sorry, that won't cut it for begging people to give you money >>>>>>>>>>> to carry
who am i begging??? who around here even has money to give??? >>>>>>>>>> 😂 😂😂
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your >>>>>>>>> work so you can complete the parts that you admit have holes in >>>>>>>>> them.
yes, certain further work would take time and therefore funds, >>>>>>>> and that kind of work will remain out of scope of this
discussion. that's a statement of fact, not "begging" u sad
dishonest old man
But, since you idea have been proven wrong, and you don't even
try to refute it, why should anyone support that work.
As I said, Your PRD couldn't have accepted ANY machine that
generates the same computable number as anti-fixed-H, and thus
the enumeration it generates is NOT complete.
right that is a thorn in my side atm, and there are several
avenues i'm exploring in thot on how to deal with that
- first of all i'm not entirely convinced there isn't some
strategy i'm missing that might still yet get it on the diagonal. >>>>>> we haven't even build an enumeration of computable numbers cause
we haven't discussed the dedpuing logic, and i don't know how that >>>>>> impact the current predicament
- and even if so do we actually care about the computations
being done outside of the decidably enumerable set? can we build a >>>>>> way to identify and classify what they do?
- and/or is this the wall i needed to justify the jump to
constructing RTMs, cause the "incompleteness" there happens due to >>>>>> actual lies built into the computations...
look i acknowledged ur argument!
which is literally more generous than you've ever been towards me, >>>>>> cause u've never admitted an ounce of validity of my words,
despite understanding them well enough to pounce on any and all
criticisms...
So, if the number computed by anti-fixed-H isn't in the
enumeration, how can PRD, or ANY PRD that could exist (and then an
anti-fixed-H be built on it).
Your problem is you don't understand the fundamental nature of what >>>>> a computation is.
now *that* is fking toxic bro, and if u think i'm going to be
swayed by such toxicity, well i know some therapists u can talk to >>>>>> about that kinda negative mindset rick, their teachings served me >>>>>> well thus far
In other words, your whole plan is to hope that a magic fairy dust
powered unicorn can give you the answer.
So, you assume that rules must not apply, but then you don't even
know the basic definitions of what you are talking about, so of
course you can't understand the rules.
Your "logic" is based on the rules not applying and computations
not being actually computations. In part, because you don't
actually understand what a computation is, and thus you imagine
things that aren't computations but wnat to think of them as
possibly being a computation.
again, nothing was said here
In part because there wasn't anything to reply to.
IT seems you have run out of ways to fabricate your ideas, so you
just insult the messager pointing out your errors.
Thus, nothing you have done with it meets the requirements for
the computation you talk about, as it, by definition, starts with >>>>>>> a complete enumeration.
There is not contradiction in my anti-fixed-H if the enumeration >>>>>>> isn't complete, but you also are proven to just be a liar about >>>>>>> your claim of showing a way to compute a diagonal of a complete >>>>>>> enumeration of the computable numbers.
You seem to expect that "someone" will like your crap work
enough to pay you to continue working on it with the hope that >>>>>>>>> you can materialize your unicorn, even though they have been >>>>>>>>> proven to be impossible.
YOU are the idiot, with a foul mouth, that doesn't know what he >>>>>>>>> is talking about.
your ignorant ideas further, as you are just showing there >>>>>>>>>>> isn't anything to base going farther on.
If you can show an actual error I am making, with sources >>>>>>>>>>>>> to back up your claims, present them.
The problem is you KNOW that you don't know what you are >>>>>>>>>>>>> talking about because you have ADMITTED to not actually >>>>>>>>>>>>> studing more that a few papers, but you think you are >>>>>>>>>>>>> smarter than the people who wrote them.
YOU are the one flying to the grave in a crashing plane of >>>>>>>>>>>>> ignorance.
I will note, that just like with Peter Olcott, YOU are the >>>>>>>>>>>>> one that started the insults, showing whose mind is in the >>>>>>>>>>>>> gutter.
On 3/18/26 8:14 PM, Richard Damon wrote:
On 3/18/26 1:32 PM, dart200 wrote:
On 3/18/26 4:32 AM, Richard Damon wrote:
On 3/18/26 3:13 AM, dart200 wrote:
On 3/17/26 7:46 PM, Richard Damon wrote:
On 3/17/26 12:55 AM, dart200 wrote:
On 3/16/26 6:50 PM, Richard Damon wrote:
On 3/16/26 1:11 PM, dart200 wrote:
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:So, what is my error, with actual SOURCES for your data >>>>>>>>>>>>>>>> that claims I am wrong?
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/26 1:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 4:48 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 12:43 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 4:28 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:12 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Right, equivLWNR PROBLEM, which means both problems >>>>>>>>>>>>>>>>>>>> are either solvable or not (under all applicable >>>>>>>>>>>>>>>>>>>> models).
| 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 >>>>>>>>>>>>>>>>>>>>how can computing a _subset_ of circle-free >>>>>>>>>>>>>>>>>>>>>>> machines be equivalent to compute a _total_ set >>>>>>>>>>>>>>>>>>>>>>> of circle- free machines...???my god rick, please fucking read the not even >>>>>>>>>>>>>>>>>>>>>>>>> whole paper, but at least the _section_ rick >>>>>>>>>>>>>>>>>>>>>>>>> please...yes it does, rickif problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But he didn't, that is just what your >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ignorant brain thinks he must have been >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> talking about.On 3/12/26 12:17 AM, Lawrence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But they’re a subset of the “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined” set that Turing was >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>the "better-defined" machine don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfy the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> (which attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> does not then prove an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility for enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But if the machine can be “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined”, yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine that also satisfies >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> those “ill-defined” criteria must be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Turing.
there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equating the solution of them as the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
| 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 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> EQUIVALENT means here. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
After all Functional Equivalence doesn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mean the same machine or even using the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> same basic algorithm. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
he's literally saying that if u can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerate computable sequences, then u >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> could use that solution to determine >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> whether any given machine is circle- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> free ...
He doesn't say the two machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> generated by the two problems are in any >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> way equivalent, he says that the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> PROBLEMS are equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
No, he his saying the problems are >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent as to the nature >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
and if so could be used to enumerate the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> circle- free machines, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
making the problem of enumerating the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sets equivalent,
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to produce >>>>>>>>>>>>>>>>>>>>>>>>>>>>> a solution to B
AND
a solution to B can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>>>>>> solution to A
Where do you get that definition? >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in >>>>>>>>>>>>>>>>>>>>>>>>>>>> all models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" >>>>>>>>>>>>>>>>>>>>>>>>>>>> you do something,
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be >>>>>>>>>>>>>>>>>>>>>>>>>>>>> used to produce a solution to enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable numbers, but the reverse is >>>>>>>>>>>>>>>>>>>>>>>>>>>>> *NOT* true
But it doesn't need to. >>>>>>>>>>>>>>>>>>>>>>>>>>>
WHY?
As I have said, you don't understand what he >>>>>>>>>>>>>>>>>>>>>>>>>> was saying, and thus are trying to kill a >>>>>>>>>>>>>>>>>>>>>>>>>> strawman.
Where does he ACTUALLY SAY that the machine >>>>>>>>>>>>>>>>>>>>>>>>>> that generates circle- ftee machihes could be >>>>>>>>>>>>>>>>>>>>>>>>>> used to enumerate computable numbers. >>>>>>>>>>>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON >>>>>>>>>>>>>>>>>>>>>>>>> THE SAME FUCKING PAGES WE'VE BEEN TALKING ABOUT >>>>>>>>>>>>>>>>>>>>>>>>> p246:
| The simplest and most direct proof of this is >>>>>>>>>>>>>>>>>>>>>>>>> by showing that,
| if this general process exists [for circle- >>>>>>>>>>>>>>>>>>>>>>>>> free machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>>>>>>>>
do i need to spell out why with even more >>>>>>>>>>>>>>>>>>>>>>>>> detail???
And B is the machine that computes the diagonals >>>>>>>>>>>>>>>>>>>>>>>> of the results of the enumeration of circle-free >>>>>>>>>>>>>>>>>>>>>>>> machines.
Why doesn't the program do that? >>>>>>>>>>>>>>>>>>>>>>>>
ok ok i will even tho u will continue to >>>>>>>>>>>>>>>>>>>>>>>>> disagree...
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>> <=> enumerating computable sequence >>>>>>>>>>>>>>>>>>>>>>>>> => diagonal is computable >>>>>>>>>>>>>>>>>>>>>>>>> => β is computable _contradiction_ >>>>>>>>>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a >>>>>>>>>>>>>>>>>>>>>>>>> turing machine, produce a general process to >>>>>>>>>>>>>>>>>>>>>>>>> output whether a machine is circle- free or not >>>>>>>>>>>>>>>>>>>>>>>>
A fallacy in your mind, because you don't >>>>>>>>>>>>>>>>>>>>>>>> understand what he means by equivalent. >>>>>>>>>>>>>>>>>>>>>>>
the _first fallacy_ is that because that isn't >>>>>>>>>>>>>>>>>>>>>>>>> actually equivalent to enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>> sequences (which is a lesser problem that only >>>>>>>>>>>>>>>>>>>>>>>>> needs to recognize a subset of circle- free >>>>>>>>>>>>>>>>>>>>>>>>> machines), ruling out a general process for >>>>>>>>>>>>>>>>>>>>>>>>> deciding circle- free machine does _not_ >>>>>>>>>>>>>>>>>>>>>>>>> actually rule out a general process for >>>>>>>>>>>>>>>>>>>>>>>>> enumerating computable numbers >>>>>>>>>>>>>>>>>>>>>>>>
Who said they were equivalent COMPUTATIONS. >>>>>>>>>>>>>>>>>>>>>
_because_ a solution to one leads to a solution for >>>>>>>>>>>>>>>>>>> the other...
Nope.
Where are you getting your definitions? Because you >>>>>>>>>>>>>>>>>> are using the wrong ones.
All you are doing is proving your stubborn refusal to >>>>>>>>>>>>>>>>>> learn what you are talking about, and that you don't >>>>>>>>>>>>>>>>>> care you are ignorant.
which is a fallacy in this case, they are not >>>>>>>>>>>>>>>>>>> equivalent problems
Sure they are, you just don't know what that means as >>>>>>>>>>>>>>>>>> you continue to hang on to your errors because you >>>>>>>>>>>>>>>>>> don't understand the language you are reading. >>>>>>>>>>>>>>>>>>
ur an ass dick
IT seems you are just showing you don't know what >>>>>>>>>>>>>>>>>>>> the word means, because you are just ignornat. >>>>>>>>>>>>>>>>>>>
No, you are. You just don't like your errors being >>>>>>>>>>>>>>>>>> pointed out, as it shows how much of an ass you are. >>>>>>>>>>>>>>>>>>
Which is a response typical of those suffering from >>>>>>>>>>>>>>>>>> the effect.
The problem of creating the computations are >>>>>>>>>>>>>>>>>>>>>> equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's >>>>>>>>>>>>>>>>>>>>> pretty ridiculous richard
Because I am not, you are gaslighting yourself with >>>>>>>>>>>>>>>>>>>> your false definitions that you try to insist on. >>>>>>>>>>>>>>>>>>>>
Says who?
if problems are equivalent then a solution to A can >>>>>>>>>>>>>>>>>>>>> be used to solve B and vise versa ... >>>>>>>>>>>>>>>>>>>>
if u don't agree with this then u can move right >>>>>>>>>>>>>>>>>>>>> the fuck along with ur willful ignorance and >>>>>>>>>>>>>>>>>>>>> gaslighting dick
But where do you get your definition of equivalent. >>>>>>>>>>>>>>>>>>>>
As I have pointed out, that isn't the definition >>>>>>>>>>>>>>>>>>>> used in the field, a field you have admitted being >>>>>>>>>>>>>>>>>>>> untrained in.
So you admit your ignorance, but insist you must >>>>>>>>>>>>>>>>>>>> know better than people who actually know something. >>>>>>>>>>>>>>>>>>>>
In a word, Dunning-Kruger
never seen anyone bring that up in good faith >>>>>>>>>>>>>>>>>>
ur whole response is just a shitpile of insults and >>>>>>>>>>>>>>>>> fallacies
can't wait to see u take ur ignorance to the grave dick >>>>>>>>>>>>>>>>
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll, >>>>>>>>>>>>>>>
and have literally argued against every single sentence >>>>>>>>>>>>>>> always without agreeing more than a spattering of times >>>>>>>>>>>>>>> over literally months of engagement,
on top of heaps of unjustified insults completely >>>>>>>>>>>>>>> unbecoming of anyone engaged in serious discussion, >>>>>>>>>>>>>>>
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject >>>>>>>>>>>>> stupidity where u "win" when the opponent gives on u being >>>>>>>>>>>>> and endless fucking troll???
that's a total L bro. if u fail to convince ur opponent, u >>>>>>>>>>>>> _lose_
Nope, if the opponent is as brain dead as you show yourself, >>>>>>>>>>>> it isn't a
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence
without adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown
increadible ignorance and failure in logic. You just have a foul >>>>>>>> mouth.
AHAHAHAHA, how motherfucking toxic do u have to try to
rationalize insults as "factually based"????
😂😂😂
What is "toxic" about truth? It is only toxic to people who live
on lies.
Shows how stupid you are.
Note, YOU are the one providing the evidence to prove my
statements, by just ignoring the facts.
Of course, what you are proving is you don't understand what truth >>>>>> or facts actually are, since you logic is based on being able to
presume something without proof.
and literally nothing meaningful was said here
If pointing out errors isn't meaningful, that points out your
fundamental error in logic.
I guess your world is just built on your own fantasies, and you just
don't care about what is actually true.
Maybe you will be able to just imagine the food and shelter you are
going to need to keep on living once your money runs out.
three more sentences of nothing
And one for you too.
I guess the actual pointing out of the errors of your "logic" is
meaningless to you, as truth and reality have no meaning to you.
You will problem call this meaningless too, just proving that you
don't understand what is being said, and things you don't understand
are "nothing" to you.
u really do like inflating ur replies with a bunch of nothing eh???
It seems your nature is to insult and blame others for your >>>>>>>>>> own failings.
matter of convincing you, but protecting the naive from you >>>>>>>>>>>> lies.
this group is bunch of boomers who spent decades losing >>>>>>>>>>> arguments amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own ideas??? >>>>>>>>>
😂😂😂
Trying to. After all, I need to do something to wake you up to >>>>>>>> your own lies that you have gaslit yourself with.
lol, i fixed turing's diagonal in a way that hasn't been
addressed in literature before
No you haven't, because you don't understand the requirements.
i certainly fixed it from stumbling on itself, which is a fix that
has not been addressed in any literature, until my paper
In other words, you think CHANGING the requitrements is a valid
process for meeting them.
... there was no need for the diagonal to test itself ...
But only if it IS in the enumeration at that point.
... right but that problem was only discovered *after* i figured out
there was no need for the diagonal to test itself ...
i love how just discredit actual innovation because of the fact it lead
to a different problem arising, fucking toxic as shit
Note, your "fixed-anti-fixed-H" doesn't actually compute the actual
anti-diagonal.
the fact u can't even acknowledge that as useful is incredibly toxic tbh
Lying is not useful.
it not lying to make a fix and then discover a *new* problem ???
The problem is your fixed-anti-fixed-H doesn't fix the problem, as we
still have that anti-fixed-H shows that yoru PRD doesn't meet its
requirements.
what i have done is shown the proof turing made as silly (to anyone who
can reason),
notwithstanding ur aggressively defeatist comments, the new problem that arose has not been analyzed sufficiently
If PRD only accepts circle-free machines, it WILL be a circle-free
machine, and thus the results it generates WILL be a computable
number, but that number can not exist at any finite point in the
enumartion generated by PRD.
Thus, PRD fails to generate a COMPLETE enumeration, and thus your
fixed- H does not compute the diagonal of such a complete enumeration,
because the enumeration it computed the diagonal of was not complete.
This seems beyond your ability to understand, so I wonder what fallacy
you are going to use to try to refute this. or perhaps you just ignore
it, and thus admit you don't have an answer.
i have then encountered a new problem, which doesn't negate the fix >>>>> i did make u ungrateful tard
Sure it does. You have presumed the enumeration that has been proved
can not be made.
toxic toxic toxic i'm so very tired of being on a world surrounded by
toxic people, rick
LYING is toxic.
calling me a liar is toxic, as i'm not intentionally misleading, i'm
just exploring the bounds of computability, and i shouldn't have to be running up against such toxicity in order to do so
clearly the consensus is defensive over being so wrong for so long, i
don't even get the need to be defensive here. i haven't the foggiest
clue what wrongs u think could be committed over exploring these issues,
cause the way we go about computing irl is already so batshit insane unconscious AI does it better probabilisticly than ur average corpo
coder...
Your enumeration is not what you claim, and thus your claims are the
toxic part.
It seems you like to call as "toxic" anything you can't deal with, as
you method of handling the ERRORS in you logic.
YOU are the toxic one.
There can not be an effective (computable) enumeration that includes
all computable number, as any method that generates one allows the
computing of a number that doesn't exist in that set.
Thus, any computed enumeration of computable numbers is necessarily
incomplete.
All you are doint is proving you don't understand what it means to
be able to compute something.
It seems your idea of computing allows an algorithm to assume that
it (or even an equivalent to it) can't be embedded into another
algorithm, which makes your concept strictly weaker in the power to
compute than the methods used by Turing Machines and their equivalents. >>>>
The first problem is, you haven't created the enumeration required >>>>>> to compute the diagonal of.
We know this, because we can convert your fixed-H to be anti-
fixed- H that outputs the opposite digits that fixed-H does (using >>>>>> the trick of fixed-H, using the number of fixed-H, not anti-fixed- >>>>>> H), and thus shows that if fixed-H is computing the diagonal,
anti- fixed-H is computing the anti-diagonal, but we also see that >>>>>> this anti-diagonal isn't in the enumeration, and thus the
enumeration can't be complete.
that does not prove there exists no further tricks that might still >>>>> get it on the diagonal somehow,
Sure it does.
rick, the only reason i got to this new problem was by ignoring all
the idiots telling me turing's proof was absolute
So, you jumped out of the frying pan into the flames and are burning
yourself to death.
again, why do u have a stick lodged so far up your asshole, dick?
Your "logic" is based on ignoring FACTS and definitions, and assuming
that magic fairy dust powered unicorns can make you impossible ideas
work.
This is the work of a mind destroyed by a steady diet of your own
toxic lies that have eaten out every ounce of reasoning you might have
had.
incredibly toxic thing to state, what are you hiding?
*ANY* method to generate an enumeration of computable numbers allows
the creation of a computation that computes a number not in the set
that it generated.
Thus NO method to generate an enumeratio of computable numbers can
create a total enumeration.
Your assumption of trick is just depending on magic fairy dust from
a unicorn to create the impossible.
If it isn't complete, then the diagonal isn't the diagonal of the >>>>>> enumeration that Turing was talking about.
even if there doesn't,
there may be provable limits on what computations can and cannot be >>>>> computably enumerated on the diagonal,
But not that excludes the anti-diagonal, since we HAVE the
description of the algorithm that generates it, at least if we have
an algorithm to generate the enumeration.
which is certainly a step up from the over-applied rice's theorem
know- nothing nonsense u see with theorists today,
You seem to have the problem of not understand what Rice proves.
rice's theorem doesn't even apply to all semantics of a machine, just
those which are detectable from the output ... a phrase which ur
gunna disagree which because ur kind of a moron rick, but ben at
least did acknowledge
So, your sub-machine doesn't generate that as part of its output?
Then how does the outer machine know the answer it gabe?
There can be many machines that compute PARTIAL classifications or
decisions on machines, just not a TOTAL classification.
like what if those various pseudo-anti-diagonals (as they aren't
true total anti-diagonal) are the *only* set of computable numbers
we can't computably enumerate on the list???
But they aren't. There are other proof, well beyond your head, that
show that other questions turn out to not be computable.
nah rick, i'm _never_ gunna accept anyone telling me anything is
"beyond my head" and the fact u even try to write that is _incredibly
toxic_
DUNNING-KRUGER in action,
AD HOMINEM in action,
the term "dunning-kruger" serves no purpose to convey good information.
it's only used to convey bad information by lazy people who act in bad
faith towards other
i mean, the gall of u ever writing that out to someone is just
_incredibly anti-intellectual_ , why would u ever demand someone
accept something that exist "beyond their understanding" like a fking
a religious nutjob???
But you PROVE that it is beyond your understanding.
i can understand something without accepting it, u dunce
my god rick, u are such an fucking gross hypocrite
the fact anyone in this group supports ur writing is just such a
stain on this group, and really demonstrates the hostility and
toxicity being harbored at the core of computing (and really academia
at large)
So, if you think you are so smart, submit your work to a real peer-
reviewed journal and see how fast it is shot down.
i already know there's a large bandwagon rick. i also know the bandwagon fallacies are a thing because something large groups of people are all
wrong in the same manner.
the two fallacies i spotted i fully intend to get published. they don't prove turing wrong, but they do warrant revisiting the arguments. who
knows what other fallacies are lurking that i haven't spotted yet
The "hostility" you perceive is people pointing out your errors that
you refuse to accept, because "Truth" isn't something you world can
handle.
actually i've never had them "point out" errors. their negligence is to
the point of literally not even reading the submissions because they
trust their internalized hubris *that* strongly
In fact, by simple "counting" we can tell that there are an infinite
number of uncomputable problems for every computable one.
ofc there are countable infinite variations on it. that doesn't mean
there aren't limits to the kinds of computations in that set
I guess you haven't read any of the papers of the other kinds of
uncomputable problems, one NOT based on a "self-reference".
more than half are "proven" thru a reduction to the halting problem, and
tbh that's where my focus lies: decision problems with computing
honestly i don't even need to compute a full diagonal to throw a wrench
into much of this. if can prove which computations belong on the PRD diagonal vs not ...
then we would need to revist those proofs to ensure the problem is at
least computed by a machine proven to not exist on the enumerable diagonal... otherwise why should we believe it to be uncomputable???
there's just so many angles here that just haven't been worked. u poo- pooing me about literally all of them is just laziness that has driven
deep into the territory of blatantly intellectual negligent,
tbh yes:
*i'm calling the entirety of CS academia intellectually negligent*
Yes, many of them allow you to, as ONE of the ways, to prove them
uncomputable, show that them being computable would allow you to
compute the answer to the uncomputable problems due to self-reference.
But, they don't themselves use that sort of self-reference.
that would also be a huge win, cause those computations don't
compute relationships we care about, so failing to enumerate them
totally just doesn't really matter 🤷
Sure it does. By knowing that TOTAL classification is impossible, we
know that we need to look at what classes of inputs a given
algorithm can work on.
Thus, like where you started with, because we KNOW we can't totally
solve the Halting Problem, we accept that we need to allow our
algorithm to decide that some cases might not be deciable, and work
on the cases we can decide on.
that's already a huge step up from before where you were advocating
for programs that we couldn't even generally decide on their
decidability
But that is still true, and not contradictory with the above.
There are many programs that we can decide on.
But there are also some that we can't, and some we can't even decide
that there behavior is unknowable.
this part i still entirely disagree with. we proved what the anti-
diagonal does even if it wasn't on PRDs diagonal...
how is that???
(because undecidability in computing _only_ exists between a machine and
the *specific* classifiers it paradoxes _not_ generally)
to prove a machine with complete unknowable decidability i think u'd
need to show a machine that exists on _no_ possible diagonal ... which i
do _not_ think is possible
A correctness proving program doesn't need to prove EVERY program
correct or wrong, but can prove SOME programs correct, SOME programs
it can point out errors, and some it tells us they are too
complicated for it to process.
"too complicated to process" is a different theory rick. that's
complexity theory not computability theory.
No, this isn't "complexity" as in O notation complexity, but that our
processing, but necessity, can't try to handle all cases to all
depths, but, to avoid getting stuck and not answering, exstablishes
finite limits on resources that can be expended on the various parts
of the analysis, and if the analysis of the program hits one of these
limits, we classify it as "too complex".
If it tells us it is too complicted, if we really need the proof, we
need to revise it to simplify it. (Or it may be that the problem we
are working on is just uncomputable, so no program CAN be proven
correct,
u haven't demonstrated an actual machine we can't prove correct,
the machines PRD failed to classify are still provable in what they
do from our perspective (we both know the pseudo-anti-diagonals are
circle- free and can prove it ... that's how we know PRD "missed"
them), regardless of whether PRD could classify them or not
And thus, PRD can not be "correct" to its specification, as one of the
REQUIREMENTS was that it would accept at least one machine that
generates EVERY computable number.
that was what i thought it could do, i'm unsure as of right now
the "anti-diagonal" anti-fixed=H is not "pseudo" anything, given your
claimed PRD, it is a REAL machine, that computes a REAL computable
number that no machine in your enumerate generates.
see ur kinda stuck in a rut here. any circle-free machine can prove
that PRD fails to enumerate is still a machine that was proven as
circle- free ...
So, you are forgetting that for your claim was that fixed-H generates
an computable diagonal of a set that is a enumeration of a set of
amchines that contains EVERY computable number.
You forgot that requirement, as you went off on your strawman fallacy.
i feel this is going to end up in abandoning the ct-thesis rick. tm
computability has limits due to self-referential weirdness, and they
aren't the same as the limits we 3rd party observers are bounded by
because we're aren't subject to that same self-referential weirdness
feelings don't generate proofs.
quite the opposite, really
one _must_ have the feeling in order to motivate themselves to produce
the proof ya dingdong
and we need to build a partial version that admits that there are
cases it can't get correct)
(and before u try to make yet another baseless claim that it must >>>>>>> have been, show me the proof instead of baselessly just claiming >>>>>>> u fucking twat)
And what is wrong about this proof.
i don't have to be right about literally every possible future goal >>>>> post to right about one goal post in a unique way that's never been >>>>> done before. the fact i could even hit that goal post is to me a
massive sign things have been missed in the fundamentals, rick
it should be to you as well, but my god are obsessed with clinging
to certain uncertainty it's abcerd
But the fact that you current claims are based on NO evidence, means
you are starting with nothing.
You claim that something might be possible, when it is shown that it
can't be.
Your world is just built on the assumption that the rules don't
apply. That is a world of fantasy and lies.
nothing was said here
Sure there was, you just can't understand it, as your world is built
on that lie.
Your enumeration generated by PRD just can not be COMPLETE,
including at least one instance of EVERY computable number.
PROVE ME WRONG.
it's really kind of sad how much toxicity and generally
unwillingness to cooperate i've encounter when trying to explore
these ideas,
The most toxic thing is to just lie to yourself about what can be done. >>>>
i hope future academia may take heed from what i've had to endure
thus far, pretty much on my own. heck i hope current academic might >>>>> too ...
but that might be asking for too much at this point eh???
In other words, you hope academia might allow people to live in
error and self-deciet?
i would be nice if u could even read simple sentence accurately.
i said it was too much to ask for, in that i hope for it, but don't
expect it. not sure where u pulled hoping for opposite from... but i
never claimed that
No, the problem is you think you are being treated unfairly, but you
are not, you are treating truth unfairly.
You ARE living a life of lies, based on the ignoring of basic principles.
Your hope is for a world where error is just tolerated under some
guise of acceptance.
rick, u have problems reading simple sentences much of the time
Your problem is you reject people pointing out the errors in your work, >>>
Less than you do.
You don't even know what an "equivalent problem" is.
Or what a "computation" is.
because you assume you must be right, even when you admit you don't
really understand the field.
Your works is based on ignorant assumptions, not facts.
That is a dark world of lies.
The people here mostly know what they are talking about,
because they have studied it (some like Olcott and you are the >>>>>>>>>> exception).
i demonstrated two distinct fallacies in turing's paper, that >>>>>>>>>>> really aren't the hard to understand,
It seems you are just admitting that you are stuck in your >>>>>>>>>>>> lies and just can't think because, like Olcott, you have >>>>>>>>>>>> successfully gatlit yourself into being convinced of your lies. >>>>>>>>>>>
No, you demonstrated that you do don't understand what he is >>>>>>>>>> saying,
the fact u continually try to gaslight me into thinking i
haven't understood his argument well enough is not only
incredibly toxic but let's me know ur completely fine with
blatantly lying at me to "win" an argument,
But it isn't gaslighting, it is FACT.
idk how u read everything i wrote and try to claim i just don't >>>>>>> understand,
Because you keep claiming things that aren't there.
i have been manipulating his ideas in ways that have never been >>>>>>> done before, and u can't even acknowledge that i understand his >>>>>>> ideas??? sheesh,
Right, by twisting words to not mean what they mean.
Yes, the endless gaslighting that you have done to yourself,
i'm swimming thru a swamp of endless gaslighting, fostered a
toxic mentality festering the fundamentals of math hostile to any >>>>>>> sort of meaningful innovation at the core for some ungodly reason >>>>>>
causing you to think that people point out truth to you are
gaslighting you.
The fact you can't actually prove anything should be your first
sign that you have something wrong.
Your world is just built on your own lies and fantasies.
the fallacies i picked out are still fallacies
No, because your claimed fallacies are based on you using a definist
fallacy.
fallacy 1) identifying a subset is _NOT_ the same problem as identify
the entirety of a set,
No, but might be an equivalent problem.
MIGHT is not a proof
fallacy 2) computing the diagonal does _NOT_ then grant an able to
computing an anti-diagonal
Why not?
WHy can you NOT just change that program to reverse the value written
to the perminante cells, and any decision based on reading one of
those cells?
How does that NOT result in that result?
because u can't then put anti-diagonal machine on the diagonal!
so it's never going to be a true anti-diagonal, just for everything but itself ... that's literally _why_ PRD has problems with anti_H atm, _why
do i need to explain this_ ???
what u *can* do is mechanically feed the output of the diagonal to
another machine as input ... and it can reverse the digits one by one without knowing with the input was. but u _cannot_ represent that computation entirely within the scope of recursive turing machine definitions
hence why u can take the ct-thesis to grave with u, rick
i'm not redefining terms in either (computable numbers *ARE* a subset
of circle-machines), ur claims of fallacy are incorrect
No they are not.
Computable numbers are numbers, that are computable by machines.
Circle-free machine generate computab;e numbers
but we don't need *all* circle-free machines, just one for each
computable number
in a reasonable debate, that should suffice, but u have been anything
but reasonable with me rick
Why? It is just a stupid category error.
Starting with that error, NOTHING you have said has any basis to
point out error.
that's not what the side with truth does, or even remotely
needs to do. and if u can't recognize that, i'm sorry for all >>>>>>>>> fallacy u've bought into across all the things u believe
Claiming truth is gaslighting is just gaslighting.
> take them to the grave bro
>
> #god
You got a source for that?
Or do you have the same "god" complex as Olcott?
oh does he also claim that _we are god_ ???
No, just him.
It seems you like to "quote" things, claiming them to be from
"god" (as he signs them). That seems to imply you think you have a >>>>>> special link to him.
like i said: _we are god_
> so anyone can do it eh???
>
> #god
if this "me" is particularly special, that is only due by seeding a >>>>> trend, if a trend even ever takes off, which is yet to be seen...
But "we" are not "god", and assuming you have the divine power of
god is the beginning of your own condemnation to a life of error and
dispare.
i never said _i_ had that ...
YOU are quoting what you think "god" has said.
🎶🎶🎶 what if god was one of us??? 🎶🎶🎶
Which sort of implies you think you know what he is saying.
i don't need to explain my moral frustrations to anyone
> take it or leave it 🤷🤷🤷
>
> #god
🤷🤷🤷
u can write it off as psycho-spiritual outburst of frustration from >>>>> constantly banging my head against the various mental walls keeping >>>>> us chained to acting _far_ less ethically than we should,
Or, those "walls" are the boundries that aim us to what can be done.
rick u would be the kinda unethical fuck "aiming" at people to bomb,
eh???
that grave is far deeper than mistakes in the fundamentals of computing
but don't count on me stopping. the grave we've been digging for
our species thru our systemic moral negligence is _deep_ ,
But it seems, the grave you see, is just the grave for those that think >>>
Yep, your stupidity runs very deep.
like you, and that the uncompuatable nature of some things means we
can't "do our best" and handle the cases we actually care about.
We CAN prove that some programs are correct. The cost is just too
high to be used everywhere, and many programs don't actually need to
be provably correct.
muh econobabble, right ...
YOU are the one that rejects that this ACHEIVABLE (and ACHEIVED)
goal is good enough, and seem to want that all work stops until we
can do the impossible.
THAT is the truely toxic.
and it's gunna take some dramatic times to pull ourselves out
like how many more pedo islands do u think exist?
regardless of whether turing's conclusions are ultimately >>>>>>>>>>> correct or not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Then show me your source.
my source for fallacies that *i* pointed out???
i would be source for that... duh?
So, what is that source?
That is your problem, you have no sources but your own
misunderstanding.
That come from your gaslighting of yourself to brainwash you into >>>>>> thinking you don't need sources.
why would i need sources to justify novel arguments???
To show that you claims are based on FACTS and not errors?
All you are doing is proving that you are ignorant and stupid.
Most "novel" arguement are just errors and fallacies.
are you asking for me to repeat the arguments i generated? u can
either reread the posts i made on usenet
or here for fallacy 1: https://www.academia.edu/165010519
Which is just repeating the error that thinking that equivalent
problems are solving the same problem.
Where is the definition that says that?
Equivalent problems are problems that are true/solvable or false/
unsolvable together.
... right and you prove that relationship by demonstrating that a
solution to one problem causally leads to a solution for the other
and vise-versa, which is _WHY_ they are solvable/unsolvable together ...
Right. SO it isn't that the solution for one DIRECTLY solves the
other, but something about the nature of one solution leads to the other.
Admittedly, Turing did not present a proof of their equivalence here,
but that failure does not make it a fallacy of the arguement.
You need to prove that it wasn't established, or at least been
accepted elsewhere.
which ur never going to accept, because then u'd have to admit i did
even /one thing/ correct,
> which you are too toxic a person to ever admit
>
> #god
You don't understand what he is saying, do you.
YOU are the one that is full of toxic waste.
may u someday repent for the endless sinning u've committed upon me 🙏
Telling the truth is NOT a sin.
LYING, as you do IS.
As is claiming revelation from god that he didn't actually give you.
so now ur claiming to speak for god, eh???
Going to the fallacy of appeal to authority, using yourself as the
authority is just stupid.
It is also your definist fallacy, as you are trying to redefine the
word "equivalent" as used as a modifier for problem.
Your claim is just the gaslighting you claim I am doing.
Yes, while the title of the paper uses the topic of
"Computable Numbers", and the part of the proof focuses on the >>>>>>>>>> related concept of machines that compute them, he DOES show a >>>>>>>>>> proof, that could be similarly used to prove the
uncomputablility of the computable numbers.
Your problem is you have such a wooden and limited knowledge >>>>>>>>>> of what you read, you can't understand what he is doing.
u've never "won" an argument here in the decades u wasted >>>>>>>>>>>>> ur life here
get back to helping america bomb muzzies for their joo >>>>>>>>>>>>> overlords, that's all ur good for
So, you still can't point out any error with a source! >>>>>>>>>>>>
So you are just admitting that you don't have anything but >>>>>>>>>>>> your bluster.
Sorry, that won't cut it for begging people to give you >>>>>>>>>>>> money to carry
who am i begging??? who around here even has money to give??? >>>>>>>>>>> 😂 😂😂
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your >>>>>>>>>> work so you can complete the parts that you admit have holes >>>>>>>>>> in them.
yes, certain further work would take time and therefore funds, >>>>>>>>> and that kind of work will remain out of scope of this
discussion. that's a statement of fact, not "begging" u sad >>>>>>>>> dishonest old man
But, since you idea have been proven wrong, and you don't even >>>>>>>> try to refute it, why should anyone support that work.
As I said, Your PRD couldn't have accepted ANY machine that
generates the same computable number as anti-fixed-H, and thus >>>>>>>> the enumeration it generates is NOT complete.
right that is a thorn in my side atm, and there are several
avenues i'm exploring in thot on how to deal with that
- first of all i'm not entirely convinced there isn't some >>>>>>> strategy i'm missing that might still yet get it on the diagonal. >>>>>>> we haven't even build an enumeration of computable numbers cause >>>>>>> we haven't discussed the dedpuing logic, and i don't know how
that impact the current predicament
- and even if so do we actually care about the computations >>>>>>> being done outside of the decidably enumerable set? can we build >>>>>>> a way to identify and classify what they do?
- and/or is this the wall i needed to justify the jump to
constructing RTMs, cause the "incompleteness" there happens due >>>>>>> to actual lies built into the computations...
look i acknowledged ur argument!
which is literally more generous than you've ever been towards
me, cause u've never admitted an ounce of validity of my words, >>>>>>> despite understanding them well enough to pounce on any and all >>>>>>> criticisms...
So, if the number computed by anti-fixed-H isn't in the
enumeration, how can PRD, or ANY PRD that could exist (and then an >>>>>> anti-fixed-H be built on it).
Your problem is you don't understand the fundamental nature of
what a computation is.
now *that* is fking toxic bro, and if u think i'm going to be
swayed by such toxicity, well i know some therapists u can talk >>>>>>> to about that kinda negative mindset rick, their teachings served >>>>>>> me well thus far
In other words, your whole plan is to hope that a magic fairy dust >>>>>> powered unicorn can give you the answer.
So, you assume that rules must not apply, but then you don't even >>>>>> know the basic definitions of what you are talking about, so of
course you can't understand the rules.
Your "logic" is based on the rules not applying and computations
not being actually computations. In part, because you don't
actually understand what a computation is, and thus you imagine
things that aren't computations but wnat to think of them as
possibly being a computation.
again, nothing was said here
In part because there wasn't anything to reply to.
IT seems you have run out of ways to fabricate your ideas, so you
just insult the messager pointing out your errors.
Thus, nothing you have done with it meets the requirements for >>>>>>>> the computation you talk about, as it, by definition, starts
with a complete enumeration.
There is not contradiction in my anti-fixed-H if the enumeration >>>>>>>> isn't complete, but you also are proven to just be a liar about >>>>>>>> your claim of showing a way to compute a diagonal of a complete >>>>>>>> enumeration of the computable numbers.
You seem to expect that "someone" will like your crap work >>>>>>>>>> enough to pay you to continue working on it with the hope that >>>>>>>>>> you can materialize your unicorn, even though they have been >>>>>>>>>> proven to be impossible.
YOU are the idiot, with a foul mouth, that doesn't know what >>>>>>>>>> he is talking about.
your ignorant ideas further, as you are just showing there >>>>>>>>>>>> isn't anything to base going farther on.
If you can show an actual error I am making, with sources >>>>>>>>>>>>>> to back up your claims, present them.
The problem is you KNOW that you don't know what you are >>>>>>>>>>>>>> talking about because you have ADMITTED to not actually >>>>>>>>>>>>>> studing more that a few papers, but you think you are >>>>>>>>>>>>>> smarter than the people who wrote them.
YOU are the one flying to the grave in a crashing plane of >>>>>>>>>>>>>> ignorance.
I will note, that just like with Peter Olcott, YOU are the >>>>>>>>>>>>>> one that started the insults, showing whose mind is in the >>>>>>>>>>>>>> gutter.
On 3/19/26 1:35 AM, dart200 wrote:
On 3/18/26 8:14 PM, Richard Damon wrote:
On 3/18/26 1:32 PM, dart200 wrote:
On 3/18/26 4:32 AM, Richard Damon wrote:
On 3/18/26 3:13 AM, dart200 wrote:
On 3/17/26 7:46 PM, Richard Damon wrote:
On 3/17/26 12:55 AM, dart200 wrote:
On 3/16/26 6:50 PM, Richard Damon wrote:
On 3/16/26 1:11 PM, dart200 wrote:
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote:So, what is my error, with actual SOURCES for your data >>>>>>>>>>>>>>>>> that claims I am wrong?
On 3/15/26 12:27 AM, dart200 wrote:
On 3/14/26 8:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 3/14/26 3:56 PM, dart200 wrote:
On 3/14/26 12:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 1:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 4:48 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 12:43 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 4:28 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:12 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Right, equivLWNR PROBLEM, which means both problems >>>>>>>>>>>>>>>>>>>>> are either solvable or not (under all applicable >>>>>>>>>>>>>>>>>>>>> models).
| the problem of enumerating computable sequences is >>>>>>>>>>>>>>>>>>>>>> | /equivalent/ to the problem of finding out >>>>>>>>>>>>>>>>>>>>>> whether ahow can computing a _subset_ of circle-free >>>>>>>>>>>>>>>>>>>>>>>> machines be equivalent to compute a _total_ set >>>>>>>>>>>>>>>>>>>>>>>> of circle- free machines...???my god rick, please fucking read the not even >>>>>>>>>>>>>>>>>>>>>>>>>> whole paper, but at least the _section_ rick >>>>>>>>>>>>>>>>>>>>>>>>>> please...yes it does, rickif problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>| 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 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But he didn't, that is just what your >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ignorant brain thinks he must have been >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> talking about. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>On 3/12/26 12:17 AM, Lawrence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But they’re a subset of the “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined” set that Turing was >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>the "better-defined" machine don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfy the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the proof >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> (which attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> found to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> does not then prove an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility for enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But if the machine can be “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined”, yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better-defined” >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> machine that also satisfies >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> those “ill-defined” criteria must >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> be provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an entirely >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Turing.
there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equating the solution of them as the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> EQUIVALENT means here. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
After all Functional Equivalence doesn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mean the same machine or even using the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> same basic algorithm. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
he's literally saying that if u can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerate computable sequences, then u >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> could use that solution to determine >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> whether any given machine is circle- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> free ...
He doesn't say the two machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> generated by the two problems are in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> any way equivalent, he says that the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> PROBLEMS are equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
No, he his saying the problems are >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent as to the nature >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
and if so could be used to enumerate the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> circle- free machines, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
making the problem of enumerating the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sets equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> produce a solution to B >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
AND
a solution to B can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> solution to A
Where do you get that definition? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in >>>>>>>>>>>>>>>>>>>>>>>>>>>>> all models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" >>>>>>>>>>>>>>>>>>>>>>>>>>>>> you do something,
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> used to produce a solution to enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable numbers, but the reverse is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *NOT* true
But it doesn't need to. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
WHY?
As I have said, you don't understand what he >>>>>>>>>>>>>>>>>>>>>>>>>>> was saying, and thus are trying to kill a >>>>>>>>>>>>>>>>>>>>>>>>>>> strawman.
Where does he ACTUALLY SAY that the machine >>>>>>>>>>>>>>>>>>>>>>>>>>> that generates circle- ftee machihes could be >>>>>>>>>>>>>>>>>>>>>>>>>>> used to enumerate computable numbers. >>>>>>>>>>>>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON >>>>>>>>>>>>>>>>>>>>>>>>>> THE SAME FUCKING PAGES WE'VE BEEN TALKING >>>>>>>>>>>>>>>>>>>>>>>>>> ABOUT p246:
| The simplest and most direct proof of this >>>>>>>>>>>>>>>>>>>>>>>>>> is by showing that,
| if this general process exists [for circle- >>>>>>>>>>>>>>>>>>>>>>>>>> free machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>>>>>>>>>
do i need to spell out why with even more >>>>>>>>>>>>>>>>>>>>>>>>>> detail???
And B is the machine that computes the >>>>>>>>>>>>>>>>>>>>>>>>> diagonals of the results of the enumeration of >>>>>>>>>>>>>>>>>>>>>>>>> circle-free machines.
Why doesn't the program do that? >>>>>>>>>>>>>>>>>>>>>>>>>
ok ok i will even tho u will continue to >>>>>>>>>>>>>>>>>>>>>>>>>> disagree...
turing's logic is:
general process to decide on circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>> <=> enumerating computable sequence >>>>>>>>>>>>>>>>>>>>>>>>>> => diagonal is computable >>>>>>>>>>>>>>>>>>>>>>>>>> => β is computable _contradiction_ >>>>>>>>>>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a >>>>>>>>>>>>>>>>>>>>>>>>>> turing machine, produce a general process to >>>>>>>>>>>>>>>>>>>>>>>>>> output whether a machine is circle- free or not >>>>>>>>>>>>>>>>>>>>>>>>>
A fallacy in your mind, because you don't >>>>>>>>>>>>>>>>>>>>>>>>> understand what he means by equivalent. >>>>>>>>>>>>>>>>>>>>>>>>
the _first fallacy_ is that because that isn't >>>>>>>>>>>>>>>>>>>>>>>>>> actually equivalent to enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>>> sequences (which is a lesser problem that only >>>>>>>>>>>>>>>>>>>>>>>>>> needs to recognize a subset of circle- free >>>>>>>>>>>>>>>>>>>>>>>>>> machines), ruling out a general process for >>>>>>>>>>>>>>>>>>>>>>>>>> deciding circle- free machine does _not_ >>>>>>>>>>>>>>>>>>>>>>>>>> actually rule out a general process for >>>>>>>>>>>>>>>>>>>>>>>>>> enumerating computable numbers >>>>>>>>>>>>>>>>>>>>>>>>>
Who said they were equivalent COMPUTATIONS. >>>>>>>>>>>>>>>>>>>>>>
| given number is the D.N of a circle-free machine >>>>>>>>>>>>>>>>>>>>>
_because_ a solution to one leads to a solution for >>>>>>>>>>>>>>>>>>>> the other...
Nope.
Where are you getting your definitions? Because you >>>>>>>>>>>>>>>>>>> are using the wrong ones.
All you are doing is proving your stubborn refusal to >>>>>>>>>>>>>>>>>>> learn what you are talking about, and that you don't >>>>>>>>>>>>>>>>>>> care you are ignorant.
which is a fallacy in this case, they are not >>>>>>>>>>>>>>>>>>>> equivalent problems
Sure they are, you just don't know what that means as >>>>>>>>>>>>>>>>>>> you continue to hang on to your errors because you >>>>>>>>>>>>>>>>>>> don't understand the language you are reading. >>>>>>>>>>>>>>>>>>>
ur an ass dick
IT seems you are just showing you don't know what >>>>>>>>>>>>>>>>>>>>> the word means, because you are just ignornat. >>>>>>>>>>>>>>>>>>>>
No, you are. You just don't like your errors being >>>>>>>>>>>>>>>>>>> pointed out, as it shows how much of an ass you are. >>>>>>>>>>>>>>>>>>>
Which is a response typical of those suffering from >>>>>>>>>>>>>>>>>>> the effect.
The problem of creating the computations are >>>>>>>>>>>>>>>>>>>>>>> equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's >>>>>>>>>>>>>>>>>>>>>> pretty ridiculous richard
Because I am not, you are gaslighting yourself with >>>>>>>>>>>>>>>>>>>>> your false definitions that you try to insist on. >>>>>>>>>>>>>>>>>>>>>
Says who?
if problems are equivalent then a solution to A >>>>>>>>>>>>>>>>>>>>>> can be used to solve B and vise versa ... >>>>>>>>>>>>>>>>>>>>>
if u don't agree with this then u can move right >>>>>>>>>>>>>>>>>>>>>> the fuck along with ur willful ignorance and >>>>>>>>>>>>>>>>>>>>>> gaslighting dick
But where do you get your definition of equivalent. >>>>>>>>>>>>>>>>>>>>>
As I have pointed out, that isn't the definition >>>>>>>>>>>>>>>>>>>>> used in the field, a field you have admitted being >>>>>>>>>>>>>>>>>>>>> untrained in.
So you admit your ignorance, but insist you must >>>>>>>>>>>>>>>>>>>>> know better than people who actually know something. >>>>>>>>>>>>>>>>>>>>>
In a word, Dunning-Kruger
never seen anyone bring that up in good faith >>>>>>>>>>>>>>>>>>>
ur whole response is just a shitpile of insults and >>>>>>>>>>>>>>>>>> fallacies
can't wait to see u take ur ignorance to the grave dick >>>>>>>>>>>>>>>>>
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll, >>>>>>>>>>>>>>>>
and have literally argued against every single sentence >>>>>>>>>>>>>>>> always without agreeing more than a spattering of times >>>>>>>>>>>>>>>> over literally months of engagement,
on top of heaps of unjustified insults completely >>>>>>>>>>>>>>>> unbecoming of anyone engaged in serious discussion, >>>>>>>>>>>>>>>>
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject >>>>>>>>>>>>>> stupidity where u "win" when the opponent gives on u being >>>>>>>>>>>>>> and endless fucking troll???
that's a total L bro. if u fail to convince ur opponent, u >>>>>>>>>>>>>> _lose_
Nope, if the opponent is as brain dead as you show
yourself, it isn't a
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence >>>>>>>>>> without adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown
increadible ignorance and failure in logic. You just have a >>>>>>>>> foul mouth.
AHAHAHAHA, how motherfucking toxic do u have to try to
rationalize insults as "factually based"????
😂😂😂
What is "toxic" about truth? It is only toxic to people who live >>>>>>> on lies.
Shows how stupid you are.
Note, YOU are the one providing the evidence to prove my
statements, by just ignoring the facts.
Of course, what you are proving is you don't understand what
truth or facts actually are, since you logic is based on being
able to presume something without proof.
and literally nothing meaningful was said here
If pointing out errors isn't meaningful, that points out your
fundamental error in logic.
I guess your world is just built on your own fantasies, and you
just don't care about what is actually true.
Maybe you will be able to just imagine the food and shelter you are >>>>> going to need to keep on living once your money runs out.
three more sentences of nothing
And one for you too.
I guess the actual pointing out of the errors of your "logic" is
meaningless to you, as truth and reality have no meaning to you.
You will problem call this meaningless too, just proving that you
don't understand what is being said, and things you don't understand
are "nothing" to you.
u really do like inflating ur replies with a bunch of nothing eh???
It seems your nature is to insult and blame others for your >>>>>>>>>>> own failings.
matter of convincing you, but protecting the naive from you >>>>>>>>>>>>> lies.
this group is bunch of boomers who spent decades losing >>>>>>>>>>>> arguments amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own >>>>>>>>>> ideas???
😂😂😂
Trying to. After all, I need to do something to wake you up to >>>>>>>>> your own lies that you have gaslit yourself with.
lol, i fixed turing's diagonal in a way that hasn't been
addressed in literature before
No you haven't, because you don't understand the requirements.
i certainly fixed it from stumbling on itself, which is a fix that >>>>>> has not been addressed in any literature, until my paper
In other words, you think CHANGING the requitrements is a valid
process for meeting them.
... there was no need for the diagonal to test itself ...
But only if it IS in the enumeration at that point.
... right but that problem was only discovered *after* i figured out
there was no need for the diagonal to test itself ...
But the Turing_H still exist, and D can't answer about IT.
The fact you can make an alternate machine that it can answer about
doesn't handle the problem it has with Turing's version of it.
After all, "D" needs to be able to answer about ALL inputs.
I guess you you, a requriement for ALL inputs only needs to work for
many inputs. Thus, it is easy to prove that programs are correct, if
they only need to work for MANY inputs, and not all.
And your PRD fails to accept some machine for EVERY computaable number,
as nowhere in its set of accepted machines is one that computes the same number as my anti-fixed_H, which *IS* a circle-free machine since it
onlyy simulates machines that PRD decides are circle-free.
i love how just discredit actual innovation because of the fact it
lead to a different problem arising, fucking toxic as shit
You mean your LIES that are based on ignoring the errors pointed out in them, because you work off of strawmen.
It doesn't matter that D can answer about your fixed-H, creating that
still doesn't handle Turing's H which is still a valid machine.
And your PRD doesn't meet its requirement, since no machine it accepts computes the same value as anti-fixed-H.
Your problems seems to be that you don't understand what REQUIREMENTS
are and that you actually need to meet them.
Note, your "fixed-anti-fixed-H" doesn't actually compute the actual
anti-diagonal.
the fact u can't even acknowledge that as useful is incredibly toxic
tbh
Lying is not useful.
it not lying to make a fix and then discover a *new* problem ???
But you didn't "fix" D, you made a strawman that it get correct, like it doees for so many other machines.
You don't seem to understand the nature of problems.
The problem is your fixed-anti-fixed-H doesn't fix the problem, as we
still have that anti-fixed-H shows that yoru PRD doesn't meet its
requirements.
what i have done is shown the proof turing made as silly (to anyone
who can reason),
No, you have shown that you don't know what you are talking about.
How does showing a case that the decider gets right negate the problem
of the input that it can't get right?
notwithstanding ur aggressively defeatist comments, the new problem
that arose has not been analyzed sufficiently
Sure it has. I have shown an machine that computes a number that PRD can
not accept a machine that compute that same number.
You are just stuck with your head in the sand (or up your a**) refusing
to even look at the problem, because you think the magic unicorns can
make it go away.
If PRD only accepts circle-free machines, it WILL be a circle-free
machine, and thus the results it generates WILL be a computable
number, but that number can not exist at any finite point in the
enumartion generated by PRD.
Thus, PRD fails to generate a COMPLETE enumeration, and thus your
fixed- H does not compute the diagonal of such a complete
enumeration, because the enumeration it computed the diagonal of was
not complete.
This seems beyond your ability to understand, so I wonder what
fallacy you are going to use to try to refute this. or perhaps you
just ignore it, and thus admit you don't have an answer.
i have then encountered a new problem, which doesn't negate the
fix i did make u ungrateful tard
Sure it does. You have presumed the enumeration that has been
proved can not be made.
toxic toxic toxic i'm so very tired of being on a world surrounded
by toxic people, rick
LYING is toxic.
calling me a liar is toxic, as i'm not intentionally misleading, i'm
just exploring the bounds of computability, and i shouldn't have to be
running up against such toxicity in order to do so
No, it is TRUTH.
Calling TRUTH toxic is just more toxicity.
clearly the consensus is defensive over being so wrong for so long, i
don't even get the need to be defensive here. i haven't the foggiest
clue what wrongs u think could be committed over exploring these issues,
Right, you don't need to be defensive as you just let the errors sit
there proving your stupidity.
cause the way we go about computing irl is already so batshit insane
unconscious AI does it better probabilisticly than ur average corpo
coder...
Maybe your problem is that you don't understand what Computation theory
is about.
And if you think AI does a good job, then you are just admititng that
you think lies are valid, as AI is based on just sounding good, not
making sound judgements.
Your enumeration is not what you claim, and thus your claims are the
toxic part.
It seems you like to call as "toxic" anything you can't deal with, as
you method of handling the ERRORS in you logic.
YOU are the toxic one.
There can not be an effective (computable) enumeration that
includes all computable number, as any method that generates one
allows the computing of a number that doesn't exist in that set.
Thus, any computed enumeration of computable numbers is necessarily >>>>> incomplete.
All you are doint is proving you don't understand what it means to
be able to compute something.
It seems your idea of computing allows an algorithm to assume that
it (or even an equivalent to it) can't be embedded into another
algorithm, which makes your concept strictly weaker in the power to >>>>> compute than the methods used by Turing Machines and their
equivalents.
The first problem is, you haven't created the enumeration
required to compute the diagonal of.
We know this, because we can convert your fixed-H to be anti-
fixed- H that outputs the opposite digits that fixed-H does
(using the trick of fixed-H, using the number of fixed-H, not
anti-fixed- H), and thus shows that if fixed-H is computing the >>>>>>> diagonal, anti- fixed-H is computing the anti-diagonal, but we
also see that this anti-diagonal isn't in the enumeration, and
thus the enumeration can't be complete.
that does not prove there exists no further tricks that might
still get it on the diagonal somehow,
Sure it does.
rick, the only reason i got to this new problem was by ignoring all
the idiots telling me turing's proof was absolute
So, you jumped out of the frying pan into the flames and are burning
yourself to death.
again, why do u have a stick lodged so far up your asshole, dick?
Just trying to get you to take yours out so you can see the truth.
Your "logic" is based on ignoring FACTS and definitions, and assuming
that magic fairy dust powered unicorns can make you impossible ideas
work.
This is the work of a mind destroyed by a steady diet of your own
toxic lies that have eaten out every ounce of reasoning you might
have had.
incredibly toxic thing to state, what are you hiding?
Nothing.
What are YOU hiding behind your need to lie and believe in the impossible?
*ANY* method to generate an enumeration of computable numbers
allows the creation of a computation that computes a number not in
the set that it generated.
Thus NO method to generate an enumeratio of computable numbers can
create a total enumeration.
Your assumption of trick is just depending on magic fairy dust from >>>>> a unicorn to create the impossible.
If it isn't complete, then the diagonal isn't the diagonal of the >>>>>>> enumeration that Turing was talking about.
even if there doesn't,
there may be provable limits on what computations can and cannot
be computably enumerated on the diagonal,
But not that excludes the anti-diagonal, since we HAVE the
description of the algorithm that generates it, at least if we have >>>>> an algorithm to generate the enumeration.
which is certainly a step up from the over-applied rice's theorem >>>>>> know- nothing nonsense u see with theorists today,
You seem to have the problem of not understand what Rice proves.
rice's theorem doesn't even apply to all semantics of a machine,
just those which are detectable from the output ... a phrase which
ur gunna disagree which because ur kind of a moron rick, but ben at
least did acknowledge
So, your sub-machine doesn't generate that as part of its output?
Then how does the outer machine know the answer it gabe?
There can be many machines that compute PARTIAL classifications or
decisions on machines, just not a TOTAL classification.
like what if those various pseudo-anti-diagonals (as they aren't
true total anti-diagonal) are the *only* set of computable numbers >>>>>> we can't computably enumerate on the list???
But they aren't. There are other proof, well beyond your head, that >>>>> show that other questions turn out to not be computable.
nah rick, i'm _never_ gunna accept anyone telling me anything is
"beyond my head" and the fact u even try to write that is
_incredibly toxic_
DUNNING-KRUGER in action,
AD HOMINEM in action,
Nope. I have never said your arguement as invalid BECAUSE it was you
that said it, which is the basis of "Ad Hominem". I have pointed out
YOUR errors, and you just refuse to look at them.
the term "dunning-kruger" serves no purpose to convey good
information. it's only used to convey bad information by lazy people
who act in bad faith towards other
Sure it does, it explains your statement. You can't accept the truth, because you are so certain you must be right, that you just blantently ignore anything that might prove you wrong.
That is just you being toxic.
i mean, the gall of u ever writing that out to someone is just
_incredibly anti-intellectual_ , why would u ever demand someone
accept something that exist "beyond their understanding" like a
fking a religious nutjob???
But you PROVE that it is beyond your understanding.
i can understand something without accepting it, u dunce
Then you admit that you are INTENTIONALLY lying, and that you are just patholg]ogicallt a toxic liar.
my god rick, u are such an fucking gross hypocrite
the fact anyone in this group supports ur writing is just such a
stain on this group, and really demonstrates the hostility and
toxicity being harbored at the core of computing (and really
academia at large)
So, if you think you are so smart, submit your work to a real peer-
reviewed journal and see how fast it is shot down.
i already know there's a large bandwagon rick. i also know the
bandwagon fallacies are a thing because something large groups of
people are all wrong in the same manner.
In other words, The "world" is just wrong and your imaginary world is correct.
Sorry, that is just demonstrating your insanity.
the two fallacies i spotted i fully intend to get published. they
don't prove turing wrong, but they do warrant revisiting the
arguments. who knows what other fallacies are lurking that i haven't
spotted yet
Except they are fallacies, but you not understanding the meaning of his words because you have INTENTIONALLY (it seems) chosen to be ignorant so
you can lie about it.
The "hostility" you perceive is people pointing out your errors that
you refuse to accept, because "Truth" isn't something you world can
handle.
actually i've never had them "point out" errors. their negligence is
to the point of literally not even reading the submissions because
they trust their internalized hubris *that* strongly
Probably because it is SO bad it isn't worth their time.
I will point out, that you are proving it is not worth pointing out your errors because you just ignore the advice that you asked for.
In fact, by simple "counting" we can tell that there are an
infinite number of uncomputable problems for every computable one.
ofc there are countable infinite variations on it. that doesn't mean
there aren't limits to the kinds of computations in that set
I guess you haven't read any of the papers of the other kinds of
uncomputable problems, one NOT based on a "self-reference".
more than half are "proven" thru a reduction to the halting problem,
and tbh that's where my focus lies: decision problems with computing
But most of those have OTHER proofs that don't reduce to the halting problem.
And, since you admit that we can't actually solve the halting problem, having a proof that reduces to it is valid.
honestly i don't even need to compute a full diagonal to throw a
wrench into much of this. if can prove which computations belong on
the PRD diagonal vs not ...
In other words you are ADMITTING that you claim that your PRD accepts at least one machine for every computable number ois just a lie.
then we would need to revist those proofs to ensure the problem is at
least computed by a machine proven to not exist on the enumerable
diagonal... otherwise why should we believe it to be uncomputable???
So, where is anti-fixed-H in your enumeration?
Computing a partial enumeration was never said to be a problem.
Again, you don't seem to understand that *ALL* does mean ALL with no exceptions.
there's just so many angles here that just haven't been worked. u poo-
pooing me about literally all of them is just laziness that has driven
deep into the territory of blatantly intellectual negligent,
Yes, you deflect yourself with all your strawmen, so you can ignore the errors pointed out, and you keep on hoping that you can find that magic unicorn that can make the problem machines just disappear.
tbh yes:
*i'm calling the entirety of CS academia intellectually negligent*
No, you are just admitting your own intentional ignorance so you can lie about not seeing the problems.
Yes, many of them allow you to, as ONE of the ways, to prove them
uncomputable, show that them being computable would allow you to
compute the answer to the uncomputable problems due to self-reference.
But, they don't themselves use that sort of self-reference.
that would also be a huge win, cause those computations don't
compute relationships we care about, so failing to enumerate them >>>>>> totally just doesn't really matter 🤷
Sure it does. By knowing that TOTAL classification is impossible,
we know that we need to look at what classes of inputs a given
algorithm can work on.
Thus, like where you started with, because we KNOW we can't totally >>>>> solve the Halting Problem, we accept that we need to allow our
algorithm to decide that some cases might not be deciable, and work >>>>> on the cases we can decide on.
that's already a huge step up from before where you were advocating
for programs that we couldn't even generally decide on their
decidability
But that is still true, and not contradictory with the above.
There are many programs that we can decide on.
But there are also some that we can't, and some we can't even decide
that there behavior is unknowable.
this part i still entirely disagree with. we proved what the anti-
diagonal does even if it wasn't on PRDs diagonal...
Then it isn't an "anti-diagonal", but just a strawman.
how is that???
Because it is just a strawman, that doesn't negate the problem of the
actual anti-diagonal program that proves that PRD doesn't do a complete enumeration.
(because undecidability in computing _only_ exists between a machine
and the *specific* classifiers it paradoxes _not_ generally)
Nope. And that is one of the roots of your problem. "Udecidability" is
about a "problem", a defined "classification". It has nothing about a specific machine.
Of course, if you admit that you are in a totally new field with new definitions, go through and define EVERYTHING it needs and show it is useful.
Of course, you are unlikely to find anyone willing to fund that research since you show so little understanding of the system you claimed to have been talking about.
to prove a machine with complete unknowable decidability i think u'd
need to show a machine that exists on _no_ possible diagonal ... which
i do _not_ think is possible
You just are proving you don't understand what you are talking about.
A correctness proving program doesn't need to prove EVERY program
correct or wrong, but can prove SOME programs correct, SOME
programs it can point out errors, and some it tells us they are too >>>>> complicated for it to process.
"too complicated to process" is a different theory rick. that's
complexity theory not computability theory.
No, this isn't "complexity" as in O notation complexity, but that our
processing, but necessity, can't try to handle all cases to all
depths, but, to avoid getting stuck and not answering, exstablishes
finite limits on resources that can be expended on the various parts
of the analysis, and if the analysis of the program hits one of these
limits, we classify it as "too complex".
If it tells us it is too complicted, if we really need the proof,
we need to revise it to simplify it. (Or it may be that the problem >>>>> we are working on is just uncomputable, so no program CAN be proven >>>>> correct,
u haven't demonstrated an actual machine we can't prove correct,
the machines PRD failed to classify are still provable in what they
do from our perspective (we both know the pseudo-anti-diagonals are
circle- free and can prove it ... that's how we know PRD "missed"
them), regardless of whether PRD could classify them or not
And thus, PRD can not be "correct" to its specification, as one of
the REQUIREMENTS was that it would accept at least one machine that
generates EVERY computable number.
that was what i thought it could do, i'm unsure as of right now
Right, you "logic" isn't based on being right, but sounding sort of good.
Just like AI. Only AI was trained on a lot of material, It seems you
know very little.
the "anti-diagonal" anti-fixed=H is not "pseudo" anything, given your
claimed PRD, it is a REAL machine, that computes a REAL computable
number that no machine in your enumerate generates.
see ur kinda stuck in a rut here. any circle-free machine can prove
that PRD fails to enumerate is still a machine that was proven as
circle- free ...
So, you are forgetting that for your claim was that fixed-H generates
an computable diagonal of a set that is a enumeration of a set of
amchines that contains EVERY computable number.
You forgot that requirement, as you went off on your strawman fallacy.
i feel this is going to end up in abandoning the ct-thesis rick. tm
computability has limits due to self-referential weirdness, and they
aren't the same as the limits we 3rd party observers are bounded by
because we're aren't subject to that same self-referential weirdness
feelings don't generate proofs.
quite the opposite, really
one _must_ have the feeling in order to motivate themselves to produce
the proof ya dingdong
In other words, you don't understand how logic works.
Yes, feelings can provide impetus and motivation, but they do not
themselves generate a proof.
You need to start from know truths to build a proof, which means you
need to start knowing something and not just working from feelings.
Your world is just a giant fallacy, so that is what you see in others, because you just don't know better.
and we need to build a partial version that admits that there are
cases it can't get correct)
(and before u try to make yet another baseless claim that it
must have been, show me the proof instead of baselessly just
claiming u fucking twat)
And what is wrong about this proof.
i don't have to be right about literally every possible future
goal post to right about one goal post in a unique way that's
never been done before. the fact i could even hit that goal post
is to me a massive sign things have been missed in the
fundamentals, rick
it should be to you as well, but my god are obsessed with clinging >>>>>> to certain uncertainty it's abcerd
But the fact that you current claims are based on NO evidence,
means you are starting with nothing.
You claim that something might be possible, when it is shown that
it can't be.
Your world is just built on the assumption that the rules don't
apply. That is a world of fantasy and lies.
nothing was said here
Sure there was, you just can't understand it, as your world is built
on that lie.
Your enumeration generated by PRD just can not be COMPLETE,
including at least one instance of EVERY computable number.
PROVE ME WRONG.
it's really kind of sad how much toxicity and generally
unwillingness to cooperate i've encounter when trying to explore
these ideas,
The most toxic thing is to just lie to yourself about what can be
done.
i hope future academia may take heed from what i've had to endure >>>>>> thus far, pretty much on my own. heck i hope current academic
might too ...
but that might be asking for too much at this point eh???
In other words, you hope academia might allow people to live in
error and self-deciet?
i would be nice if u could even read simple sentence accurately.
i said it was too much to ask for, in that i hope for it, but don't
expect it. not sure where u pulled hoping for opposite from... but i
never claimed that
No, the problem is you think you are being treated unfairly, but you
are not, you are treating truth unfairly.
You ARE living a life of lies, based on the ignoring of basic
principles.
Your hope is for a world where error is just tolerated under some
guise of acceptance.
Your problem is you reject people pointing out the errors in your
work,
rick, u have problems reading simple sentences much of the time
Less than you do.
You don't even know what an "equivalent problem" is.
Or what a "computation" is.
because you assume you must be right, even when you admit you don't >>>>> really understand the field.
Your works is based on ignorant assumptions, not facts.
That is a dark world of lies.
The people here mostly know what they are talking about, >>>>>>>>>>> because they have studied it (some like Olcott and you are >>>>>>>>>>> the exception).
It seems you are just admitting that you are stuck in your >>>>>>>>>>>>> lies and just can't think because, like Olcott, you have >>>>>>>>>>>>> successfully gatlit yourself into being convinced of your >>>>>>>>>>>>> lies.
i demonstrated two distinct fallacies in turing's paper, >>>>>>>>>>>> that really aren't the hard to understand,
No, you demonstrated that you do don't understand what he is >>>>>>>>>>> saying,
the fact u continually try to gaslight me into thinking i >>>>>>>>>> haven't understood his argument well enough is not only
incredibly toxic but let's me know ur completely fine with >>>>>>>>>> blatantly lying at me to "win" an argument,
But it isn't gaslighting, it is FACT.
idk how u read everything i wrote and try to claim i just don't >>>>>>>> understand,
Because you keep claiming things that aren't there.
i have been manipulating his ideas in ways that have never been >>>>>>>> done before, and u can't even acknowledge that i understand his >>>>>>>> ideas??? sheesh,
Right, by twisting words to not mean what they mean.
i'm swimming thru a swamp of endless gaslighting, fostered a
toxic mentality festering the fundamentals of math hostile to >>>>>>>> any sort of meaningful innovation at the core for some ungodly >>>>>>>> reason
Yes, the endless gaslighting that you have done to yourself,
causing you to think that people point out truth to you are
gaslighting you.
The fact you can't actually prove anything should be your first >>>>>>> sign that you have something wrong.
Your world is just built on your own lies and fantasies.
the fallacies i picked out are still fallacies
No, because your claimed fallacies are based on you using a
definist fallacy.
fallacy 1) identifying a subset is _NOT_ the same problem as
identify the entirety of a set,
No, but might be an equivalent problem.
MIGHT is not a proof
I have agreed that he didn't present the proof here, but that doesn't
mean it wasn't a fact proven before and was just known to his intended audience.
fallacy 2) computing the diagonal does _NOT_ then grant an able to
computing an anti-diagonal
Why not?
WHy can you NOT just change that program to reverse the value written
to the perminante cells, and any decision based on reading one of
those cells?
How does that NOT result in that result?
because u can't then put anti-diagonal machine on the diagonal!
But, if the enumeration is computable, you can compute the diagonal, and
the anti-diagonal, and since the anti-diagonal isn't in the enumeration,
it shows your computed enumeration couldn't have been complete.
so it's never going to be a true anti-diagonal, just for everything
but itself ... that's literally _why_ PRD has problems with anti_H
atm, _why do i need to explain this_ ???
It *IS* the true anti-diagonal of the partial enumeration generated by
your PRD.
And it proves that PRD can't do a complete enumeration.
what u *can* do is mechanically feed the output of the diagonal to
another machine as input ... and it can reverse the digits one by one
without knowing with the input was. but u _cannot_ represent that
computation entirely within the scope of recursive turing machine
definitions
Sure you can. What was wrong with my program?
hence why u can take the ct-thesis to grave with u, rick
It seems you have taken your reasoning to the grave and buried it so you can't find it any more.
Your world is populated by imaginary fairy dust powered magic unicorns
that let you do what every you want them to, even if that world doesn't exist.
i'm not redefining terms in either (computable numbers *ARE* a
subset of circle-machines), ur claims of fallacy are incorrect
No they are not.
Computable numbers are numbers, that are computable by machines.
Circle-free machine generate computab;e numbers
but we don't need *all* circle-free machines, just one for each
computable number
RIGHT, so where was the one for the number that anti-fixed-H computed?
It can't exist, as if it was the k'th number accepted by PRD, then then
k'th digit of anti-fixed-H's number differs from it.
in a reasonable debate, that should suffice, but u have been
anything but reasonable with me rick
Why? It is just a stupid category error.
Starting with that error, NOTHING you have said has any basis to
point out error.
But "we" are not "god", and assuming you have the divine power of
that's not what the side with truth does, or even remotely >>>>>>>>>> needs to do. and if u can't recognize that, i'm sorry for all >>>>>>>>>> fallacy u've bought into across all the things u believe
Claiming truth is gaslighting is just gaslighting.
> take them to the grave bro
>
> #god
You got a source for that?
Or do you have the same "god" complex as Olcott?
oh does he also claim that _we are god_ ???
No, just him.
It seems you like to "quote" things, claiming them to be from
"god" (as he signs them). That seems to imply you think you have >>>>>>> a special link to him.
like i said: _we are god_
> so anyone can do it eh???
>
> #god
if this "me" is particularly special, that is only due by seeding >>>>>> a trend, if a trend even ever takes off, which is yet to be seen... >>>>>
god is the beginning of your own condemnation to a life of error
and dispare.
i never said _i_ had that ...
YOU are quoting what you think "god" has said.
🎶🎶🎶 what if god was one of us??? 🎶🎶🎶
Which means you don't understand who God is.
Which sort of implies you think you know what he is saying.
i don't need to explain my moral frustrations to anyone
> take it or leave it 🤷🤷🤷
>
> #god
But you are just showing how STUPID, IGNORANT, and AROGANT you are
spreading your toxic lies.
Yep, your stupidity runs very deep.
🤷🤷🤷
u can write it off as psycho-spiritual outburst of frustration
from constantly banging my head against the various mental walls
keeping us chained to acting _far_ less ethically than we should,
Or, those "walls" are the boundries that aim us to what can be done.
rick u would be the kinda unethical fuck "aiming" at people to bomb,
eh???
but don't count on me stopping. the grave we've been digging for
our species thru our systemic moral negligence is _deep_ ,
But it seems, the grave you see, is just the grave for those that
think
that grave is far deeper than mistakes in the fundamentals of computing >>>
like you, and that the uncompuatable nature of some things means we >>>>> can't "do our best" and handle the cases we actually care about.
We CAN prove that some programs are correct. The cost is just too
high to be used everywhere, and many programs don't actually need
to be provably correct.
muh econobabble, right ...
YOU are the one that rejects that this ACHEIVABLE (and ACHEIVED)
goal is good enough, and seem to want that all work stops until we
can do the impossible.
THAT is the truely toxic.
and it's gunna take some dramatic times to pull ourselves out
like how many more pedo islands do u think exist?
regardless of whether turing's conclusions are ultimately >>>>>>>>>>>> correct or not:
the fallacies are still fallacies
No, the fallacies are mostly in your own understanding.
Then show me your source.
my source for fallacies that *i* pointed out???
i would be source for that... duh?
So, what is that source?
That is your problem, you have no sources but your own
misunderstanding.
That come from your gaslighting of yourself to brainwash you into >>>>>>> thinking you don't need sources.
why would i need sources to justify novel arguments???
To show that you claims are based on FACTS and not errors?
All you are doing is proving that you are ignorant and stupid.
Most "novel" arguement are just errors and fallacies.
are you asking for me to repeat the arguments i generated? u can
either reread the posts i made on usenet
or here for fallacy 1: https://www.academia.edu/165010519
Which is just repeating the error that thinking that equivalent
problems are solving the same problem.
Where is the definition that says that?
Equivalent problems are problems that are true/solvable or false/
unsolvable together.
... right and you prove that relationship by demonstrating that a
solution to one problem causally leads to a solution for the other
and vise-versa, which is _WHY_ they are solvable/unsolvable
together ...
Right. SO it isn't that the solution for one DIRECTLY solves the
other, but something about the nature of one solution leads to the
other.
Admittedly, Turing did not present a proof of their equivalence here,
but that failure does not make it a fallacy of the arguement.
You need to prove that it wasn't established, or at least been
accepted elsewhere.
which ur never going to accept, because then u'd have to admit i did
even /one thing/ correct,
> which you are too toxic a person to ever admit
>
> #god
You don't understand what he is saying, do you.
YOU are the one that is full of toxic waste.
Telling the truth is NOT a sin.;may u someday repent for the endless sinning u've committed upon me 🙏 >>>
LYING, as you do IS.
As is claiming revelation from god that he didn't actually give you.
so now ur claiming to speak for god, eh???
Going to the fallacy of appeal to authority, using yourself as the
authority is just stupid.
It is also your definist fallacy, as you are trying to redefine the >>>>> word "equivalent" as used as a modifier for problem.
Your claim is just the gaslighting you claim I am doing.
Yes, while the title of the paper uses the topic of
"Computable Numbers", and the part of the proof focuses on >>>>>>>>>>> the related concept of machines that compute them, he DOES >>>>>>>>>>> show a proof, that could be similarly used to prove the >>>>>>>>>>> uncomputablility of the computable numbers.
Your problem is you have such a wooden and limited knowledge >>>>>>>>>>> of what you read, you can't understand what he is doing. >>>>>>>>>>>
u've never "won" an argument here in the decades u wasted >>>>>>>>>>>>>> ur life here
get back to helping america bomb muzzies for their joo >>>>>>>>>>>>>> overlords, that's all ur good for
So, you still can't point out any error with a source! >>>>>>>>>>>>>
So you are just admitting that you don't have anything but >>>>>>>>>>>>> your bluster.
Sorry, that won't cut it for begging people to give you >>>>>>>>>>>>> money to carry
who am i begging??? who around here even has money to >>>>>>>>>>>> give??? 😂 😂😂
dicks and making a random ass accusations:
name a more iconic duo
You previously have said that you need someone to fund your >>>>>>>>>>> work so you can complete the parts that you admit have holes >>>>>>>>>>> in them.
yes, certain further work would take time and therefore funds, >>>>>>>>>> and that kind of work will remain out of scope of this
discussion. that's a statement of fact, not "begging" u sad >>>>>>>>>> dishonest old man
But, since you idea have been proven wrong, and you don't even >>>>>>>>> try to refute it, why should anyone support that work.
As I said, Your PRD couldn't have accepted ANY machine that >>>>>>>>> generates the same computable number as anti-fixed-H, and thus >>>>>>>>> the enumeration it generates is NOT complete.
right that is a thorn in my side atm, and there are several
avenues i'm exploring in thot on how to deal with that
- first of all i'm not entirely convinced there isn't some >>>>>>>> strategy i'm missing that might still yet get it on the
diagonal. we haven't even build an enumeration of computable
numbers cause we haven't discussed the dedpuing logic, and i
don't know how that impact the current predicament
- and even if so do we actually care about the computations >>>>>>>> being done outside of the decidably enumerable set? can we build >>>>>>>> a way to identify and classify what they do?
- and/or is this the wall i needed to justify the jump to >>>>>>>> constructing RTMs, cause the "incompleteness" there happens due >>>>>>>> to actual lies built into the computations...
look i acknowledged ur argument!
which is literally more generous than you've ever been towards >>>>>>>> me, cause u've never admitted an ounce of validity of my words, >>>>>>>> despite understanding them well enough to pounce on any and all >>>>>>>> criticisms...
So, if the number computed by anti-fixed-H isn't in the
enumeration, how can PRD, or ANY PRD that could exist (and then >>>>>>> an anti-fixed-H be built on it).
Your problem is you don't understand the fundamental nature of
what a computation is.
now *that* is fking toxic bro, and if u think i'm going to be >>>>>>>> swayed by such toxicity, well i know some therapists u can talk >>>>>>>> to about that kinda negative mindset rick, their teachings
served me well thus far
In other words, your whole plan is to hope that a magic fairy
dust powered unicorn can give you the answer.
So, you assume that rules must not apply, but then you don't even >>>>>>> know the basic definitions of what you are talking about, so of >>>>>>> course you can't understand the rules.
Your "logic" is based on the rules not applying and computations >>>>>>> not being actually computations. In part, because you don't
actually understand what a computation is, and thus you imagine >>>>>>> things that aren't computations but wnat to think of them as
possibly being a computation.
again, nothing was said here
In part because there wasn't anything to reply to.
IT seems you have run out of ways to fabricate your ideas, so you
just insult the messager pointing out your errors.
Thus, nothing you have done with it meets the requirements for >>>>>>>>> the computation you talk about, as it, by definition, starts >>>>>>>>> with a complete enumeration.
There is not contradiction in my anti-fixed-H if the
enumeration isn't complete, but you also are proven to just be >>>>>>>>> a liar about your claim of showing a way to compute a diagonal >>>>>>>>> of a complete enumeration of the computable numbers.
You seem to expect that "someone" will like your crap work >>>>>>>>>>> enough to pay you to continue working on it with the hope >>>>>>>>>>> that you can materialize your unicorn, even though they have >>>>>>>>>>> been proven to be impossible.
YOU are the idiot, with a foul mouth, that doesn't know what >>>>>>>>>>> he is talking about.
your ignorant ideas further, as you are just showing there >>>>>>>>>>>>> isn't anything to base going farther on.
If you can show an actual error I am making, with sources >>>>>>>>>>>>>>> to back up your claims, present them.
The problem is you KNOW that you don't know what you are >>>>>>>>>>>>>>> talking about because you have ADMITTED to not actually >>>>>>>>>>>>>>> studing more that a few papers, but you think you are >>>>>>>>>>>>>>> smarter than the people who wrote them.
YOU are the one flying to the grave in a crashing plane >>>>>>>>>>>>>>> of ignorance.
I will note, that just like with Peter Olcott, YOU are >>>>>>>>>>>>>>> the one that started the insults, showing whose mind is >>>>>>>>>>>>>>> in the gutter.
On 3/19/26 4:52 PM, Richard Damon wrote:
On 3/19/26 1:35 AM, dart200 wrote:
On 3/18/26 8:14 PM, Richard Damon wrote:
On 3/18/26 1:32 PM, dart200 wrote:
On 3/18/26 4:32 AM, Richard Damon wrote:
On 3/18/26 3:13 AM, dart200 wrote:
On 3/17/26 7:46 PM, Richard Damon wrote:
On 3/17/26 12:55 AM, dart200 wrote:
On 3/16/26 6:50 PM, Richard Damon wrote:
On 3/16/26 1:11 PM, dart200 wrote:
On 3/16/26 3:51 AM, Richard Damon wrote:
On 3/15/26 8:28 PM, dart200 wrote:
On 3/15/26 4:41 PM, Richard Damon wrote:
On 3/15/26 3:22 PM, dart200 wrote:
On 3/15/26 12:06 PM, Richard Damon wrote:
On 3/15/26 2:31 PM, dart200 wrote:
On 3/15/26 11:12 AM, Richard Damon wrote:
On 3/15/26 12:05 PM, dart200 wrote:
On 3/15/26 3:48 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/15/26 12:27 AM, dart200 wrote:So, what is my error, with actual SOURCES for your >>>>>>>>>>>>>>>>>> data that claims I am wrong?
On 3/14/26 8:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/26 3:56 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 12:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 1:29 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 4:48 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/26 12:43 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 4:28 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:12 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 11:47 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 1:33 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 10:11 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 12:41 PM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 6:52 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/26 3:30 AM, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/26 10:53 PM, Lawrence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On Thu, 12 Mar 2026 00:41:06 -0700, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Right, equivLWNR PROBLEM, which means both >>>>>>>>>>>>>>>>>>>>>> problems are either solvable or not (under all >>>>>>>>>>>>>>>>>>>>>> applicable models).| the problem of enumerating computable sequences is >>>>>>>>>>>>>>>>>>>>>>> | /equivalent/ to the problem of finding out >>>>>>>>>>>>>>>>>>>>>>> whether aWho said they were equivalent COMPUTATIONS. >>>>>>>>>>>>>>>>>>>>>>>how can computing a _subset_ of circle-free >>>>>>>>>>>>>>>>>>>>>>>>> machines be equivalent to compute a _total_ set >>>>>>>>>>>>>>>>>>>>>>>>> of circle- free machines...??? >>>>>>>>>>>>>>>>>>>>>>>>my god rick, please fucking read the not even >>>>>>>>>>>>>>>>>>>>>>>>>>> whole paper, but at least the _section_ rick >>>>>>>>>>>>>>>>>>>>>>>>>>> please...yes it does, rickif problem A and B are equivalent: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>EQUIVALENT. Not the SAME. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>| 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 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But he didn't, that is just what your >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ignorant brain thinks he must have >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> been talking about. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>there are two sets being conflated here: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>On 3/12/26 12:17 AM, Lawrence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> D’Oliveiro wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But they’re a subset of the “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined” set that Turing was >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> considering, are they not? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
On Tue, 10 Mar 2026 09:51:43 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> -0700, dart200 wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>the "better-defined" machine don't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfy the criteria to be undecidable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because of this fallacy, the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof found on the following p247, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> where an ill-defined machine 𝓗 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> (which attempts and fails to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> compute the direct diagonal β’) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is found to be undecidable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> respect to circle-free decider 𝓓; >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> does not then prove an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility for enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But if the machine can be “ill- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined”, yet provably undecidable, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that must mean any “better- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> defined” machine that also satisfies >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> those “ill-defined” criteria must >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> be provably undecidable. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Unless you’re considering an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> entirely different set, in which case >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> your argument has nothing to do with >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Turing. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* circle-free machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*all* computable sequences >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
these sets are _not_ bijectable, and >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equating the solution of them as the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> same is a _fallacy_ >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I guess you don't understand what >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> EQUIVALENT means here. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
After all Functional Equivalence doesn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mean the same machine or even using the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> same basic algorithm. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
he's literally saying that if u can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerate computable sequences, then u >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> could use that solution to determine >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> whether any given machine is circle- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> free ...
He doesn't say the two machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> generated by the two problems are in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> any way equivalent, he says that the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> PROBLEMS are equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
No, he his saying the problems are >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent as to the nature >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
That isn't what "equivalent" means. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
and if so could be used to enumerate >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the circle- free machines, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
making the problem of enumerating the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sets equivalent, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
then a solution to A can be used to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> produce a solution to B >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
AND
a solution to B can be used to produce a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> solution to A
Where do you get that definition? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Two problems are logically eqivalent if in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> all models they produce the same "answer". >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since the problem is the question of "Can" >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> you do something,
turing is wrong about this. a solution to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> enumerating circle- free machines can be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> used to produce a solution to enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> computable numbers, but the reverse is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *NOT* true
But it doesn't need to. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
WHY?
As I have said, you don't understand what he >>>>>>>>>>>>>>>>>>>>>>>>>>>> was saying, and thus are trying to kill a >>>>>>>>>>>>>>>>>>>>>>>>>>>> strawman.
Where does he ACTUALLY SAY that the machine >>>>>>>>>>>>>>>>>>>>>>>>>>>> that generates circle- ftee machihes could >>>>>>>>>>>>>>>>>>>>>>>>>>>> be used to enumerate computable numbers. >>>>>>>>>>>>>>>>>>>>>>>>>>>
i'm tired of answering questions that ARE ON >>>>>>>>>>>>>>>>>>>>>>>>>>> THE SAME FUCKING PAGES WE'VE BEEN TALKING >>>>>>>>>>>>>>>>>>>>>>>>>>> ABOUT p246:
| The simplest and most direct proof of this >>>>>>>>>>>>>>>>>>>>>>>>>>> is by showing that,
| if this general process exists [for circle- >>>>>>>>>>>>>>>>>>>>>>>>>>> free machines]
| then there is a machine which computes β >>>>>>>>>>>>>>>>>>>>>>>>>>>
do i need to spell out why with even more >>>>>>>>>>>>>>>>>>>>>>>>>>> detail???
And B is the machine that computes the >>>>>>>>>>>>>>>>>>>>>>>>>> diagonals of the results of the enumeration of >>>>>>>>>>>>>>>>>>>>>>>>>> circle-free machines.
Why doesn't the program do that? >>>>>>>>>>>>>>>>>>>>>>>>>>
ok ok i will even tho u will continue to >>>>>>>>>>>>>>>>>>>>>>>>>>> disagree...
turing's logic is:
general process to decide on circle-free >>>>>>>>>>>>>>>>>>>>>>>>>>> machines
<=> enumerating computable sequence >>>>>>>>>>>>>>>>>>>>>>>>>>> => diagonal is computable >>>>>>>>>>>>>>>>>>>>>>>>>>> => β is computable _contradiction_ >>>>>>>>>>>>>>>>>>>>>>>>>>
sure he demonstrated that we cannot, with a >>>>>>>>>>>>>>>>>>>>>>>>>>> turing machine, produce a general process to >>>>>>>>>>>>>>>>>>>>>>>>>>> output whether a machine is circle- free or not >>>>>>>>>>>>>>>>>>>>>>>>>>
the _first fallacy_ is that because that >>>>>>>>>>>>>>>>>>>>>>>>>>> isn't actually equivalent to enumerating >>>>>>>>>>>>>>>>>>>>>>>>>>> computable sequences (which is a lesser >>>>>>>>>>>>>>>>>>>>>>>>>>> problem that only needs to recognize a subset >>>>>>>>>>>>>>>>>>>>>>>>>>> of circle- free machines), ruling out a >>>>>>>>>>>>>>>>>>>>>>>>>>> general process for deciding circle- free >>>>>>>>>>>>>>>>>>>>>>>>>>> machine does _not_ actually rule out a >>>>>>>>>>>>>>>>>>>>>>>>>>> general process for enumerating computable >>>>>>>>>>>>>>>>>>>>>>>>>>> numbers
A fallacy in your mind, because you don't >>>>>>>>>>>>>>>>>>>>>>>>>> understand what he means by equivalent. >>>>>>>>>>>>>>>>>>>>>>>>>
| given number is the D.N of a circle-free machine >>>>>>>>>>>>>>>>>>>>>>
_because_ a solution to one leads to a solution for >>>>>>>>>>>>>>>>>>>>> the other...
Nope.
Where are you getting your definitions? Because you >>>>>>>>>>>>>>>>>>>> are using the wrong ones.
All you are doing is proving your stubborn refusal >>>>>>>>>>>>>>>>>>>> to learn what you are talking about, and that you >>>>>>>>>>>>>>>>>>>> don't care you are ignorant.
which is a fallacy in this case, they are not >>>>>>>>>>>>>>>>>>>>> equivalent problems
Sure they are, you just don't know what that means >>>>>>>>>>>>>>>>>>>> as you continue to hang on to your errors because >>>>>>>>>>>>>>>>>>>> you don't understand the language you are reading. >>>>>>>>>>>>>>>>>>>>
ur an ass dick
IT seems you are just showing you don't know what >>>>>>>>>>>>>>>>>>>>>> the word means, because you are just ignornat. >>>>>>>>>>>>>>>>>>>>>
No, you are. You just don't like your errors being >>>>>>>>>>>>>>>>>>>> pointed out, as it shows how much of an ass you are. >>>>>>>>>>>>>>>>>>>>
Which is a response typical of those suffering from >>>>>>>>>>>>>>>>>>>> the effect.
The problem of creating the computations are >>>>>>>>>>>>>>>>>>>>>>>> equivalent PROBLEMS.
idk why ur gaslighting me about this, but it's >>>>>>>>>>>>>>>>>>>>>>> pretty ridiculous richard
Because I am not, you are gaslighting yourself >>>>>>>>>>>>>>>>>>>>>> with your false definitions that you try to insist >>>>>>>>>>>>>>>>>>>>>> on.
Says who?
if problems are equivalent then a solution to A >>>>>>>>>>>>>>>>>>>>>>> can be used to solve B and vise versa ... >>>>>>>>>>>>>>>>>>>>>>
if u don't agree with this then u can move right >>>>>>>>>>>>>>>>>>>>>>> the fuck along with ur willful ignorance and >>>>>>>>>>>>>>>>>>>>>>> gaslighting dick
But where do you get your definition of equivalent. >>>>>>>>>>>>>>>>>>>>>>
As I have pointed out, that isn't the definition >>>>>>>>>>>>>>>>>>>>>> used in the field, a field you have admitted being >>>>>>>>>>>>>>>>>>>>>> untrained in.
So you admit your ignorance, but insist you must >>>>>>>>>>>>>>>>>>>>>> know better than people who actually know something. >>>>>>>>>>>>>>>>>>>>>>
In a word, Dunning-Kruger
never seen anyone bring that up in good faith >>>>>>>>>>>>>>>>>>>>
ur whole response is just a shitpile of insults and >>>>>>>>>>>>>>>>>>> fallacies
can't wait to see u take ur ignorance to the grave dick >>>>>>>>>>>>>>>>>>
bro, i'm done arguing with u
u've got the ethical credibility of a fking troll, >>>>>>>>>>>>>>>>>
and have literally argued against every single sentence >>>>>>>>>>>>>>>>> always without agreeing more than a spattering of times >>>>>>>>>>>>>>>>> over literally months of engagement,
on top of heaps of unjustified insults completely >>>>>>>>>>>>>>>>> unbecoming of anyone engaged in serious discussion, >>>>>>>>>>>>>>>>>
take ur ignorance down into the grave dick
> that's all it's good for
>
> #god
I.E, I got you good and you can't handle it.
u got me good???
is that how u see this as??? a teenage game of abject >>>>>>>>>>>>>>> stupidity where u "win" when the opponent gives on u >>>>>>>>>>>>>>> being and endless fucking troll???
that's a total L bro. if u fail to convince ur opponent, >>>>>>>>>>>>>>> u _lose_
Nope, if the opponent is as brain dead as you show >>>>>>>>>>>>>> yourself, it isn't a
calling me brain dead is incredibly toxic,
No more than what you have called me.
no more u say??? u can barely construct a single sentence >>>>>>>>>>> without adding an insult after it...
As I said, you started it.
Note, my "insults" are factually based, as you have shown >>>>>>>>>> increadible ignorance and failure in logic. You just have a >>>>>>>>>> foul mouth.
AHAHAHAHA, how motherfucking toxic do u have to try to
rationalize insults as "factually based"????
😂😂😂
What is "toxic" about truth? It is only toxic to people who live >>>>>>>> on lies.
Shows how stupid you are.
Note, YOU are the one providing the evidence to prove my
statements, by just ignoring the facts.
Of course, what you are proving is you don't understand what
truth or facts actually are, since you logic is based on being >>>>>>>> able to presume something without proof.
and literally nothing meaningful was said here
If pointing out errors isn't meaningful, that points out your
fundamental error in logic.
I guess your world is just built on your own fantasies, and you
just don't care about what is actually true.
Maybe you will be able to just imagine the food and shelter you
are going to need to keep on living once your money runs out.
three more sentences of nothing
And one for you too.
I guess the actual pointing out of the errors of your "logic" is
meaningless to you, as truth and reality have no meaning to you.
You will problem call this meaningless too, just proving that you
don't understand what is being said, and things you don't understand
are "nothing" to you.
u really do like inflating ur replies with a bunch of nothing eh???
It seems your nature is to insult and blame others for your >>>>>>>>>>>> own failings.
matter of convincing you, but protecting the naive from >>>>>>>>>>>>>> you lies.
this group is bunch of boomers who spent decades losing >>>>>>>>>>>>> arguments amongst themselves...
who in the fuck here is "naive"???
YOU.
lol, ur going to insult me into protecting me from my own >>>>>>>>>>> ideas???
😂😂😂
Trying to. After all, I need to do something to wake you up to >>>>>>>>>> your own lies that you have gaslit yourself with.
lol, i fixed turing's diagonal in a way that hasn't been
addressed in literature before
No you haven't, because you don't understand the requirements.
i certainly fixed it from stumbling on itself, which is a fix
that has not been addressed in any literature, until my paper
In other words, you think CHANGING the requitrements is a valid
process for meeting them.
... there was no need for the diagonal to test itself ...
But only if it IS in the enumeration at that point.
... right but that problem was only discovered *after* i figured out
there was no need for the diagonal to test itself ...
But the Turing_H still exist, and D can't answer about IT.
it is a bizarre contradiction to claim those exist after using them in a proof that proves they can't exist
the reason they don't exist is not due to some "limit to computability", it's because D is under specified and does not handle /undecidable input/
The fact you can make an alternate machine that it can answer about
doesn't handle the problem it has with Turing's version of it.
After all, "D" needs to be able to answer about ALL inputs.
I guess you you, a requriement for ALL inputs only needs to work for
many inputs. Thus, it is easy to prove that programs are correct, if
they only need to work for MANY inputs, and not all.
And your PRD fails to accept some machine for EVERY computaable
number, as nowhere in its set of accepted machines is one that
computes the same number as my anti-fixed_H, which *IS* a circle-free
machine since it onlyy simulates machines that PRD decides are circle-
free.
right, the fact a _specific_ decider fails to classify a _specific_
input does not then mean we cannot prove what it does
our proofs are not subject to the same self-referential weirdness as TM computed proofs are
i love how just discredit actual innovation because of the fact it
lead to a different problem arising, fucking toxic as shit
You mean your LIES that are based on ignoring the errors pointed out
in them, because you work off of strawmen.
It doesn't matter that D can answer about your fixed-H, creating that
still doesn't handle Turing's H which is still a valid machine.
turing_H is not a real machine that can exist in the enumeration,
because D is not a real machine that can exist in the enumeration,
as D is too under specified to be actually implemented as a real
machine, therefore PRD does _not_ need to handle turing_H
the _only_ angle you have on me right now is the fact idk how to get anti_fixed_H on PRDs diagonal, and that's probably not as good an angle
as u would hope
And your PRD doesn't meet its requirement, since no machine it accepts
computes the same value as anti-fixed-H.
Your problems seems to be that you don't understand what REQUIREMENTS
are and that you actually need to meet them.
unfortunately ur not my boss, nor an arbiter of trust, nor even just
like an honest person ...
so i'm not just not required to stick to the "requirements". this is *exploring* the space, meaning i'm gunna have to go down wrong paths for
a bit to figure out precisely what i'm trying to uncover
Note, your "fixed-anti-fixed-H" doesn't actually compute the actual
anti-diagonal.
the fact u can't even acknowledge that as useful is incredibly
toxic tbh
Lying is not useful.
it not lying to make a fix and then discover a *new* problem ???
But you didn't "fix" D, you made a strawman that it get correct, like
it doees for so many other machines.
i produced a diagonal of _almost_ all computable numbers
and there are at least a few paths forward
1) some trick that can still put anti_fixed_H on the computable diagonal
2) demonstrating that we can classify what can and cannot be on the computable diagonal, restricting the numbers that aren't computably enumerable
3) using this a stepping stone to RTMs where the anti-diagonal trick
doesn't work, philosophical question of whether these machines are
important becomes a lot more clear
4) follow up discussion on what undecidability really is
You don't seem to understand the nature of problems.
The problem is your fixed-anti-fixed-H doesn't fix the problem, as
we still have that anti-fixed-H shows that yoru PRD doesn't meet its
requirements.
what i have done is shown the proof turing made as silly (to anyone
who can reason),
No, you have shown that you don't know what you are talking about.
there is no way to form an anti-diagonal even with a computable diagonal
How does showing a case that the decider gets right negate the problem
of the input that it can't get right?
turing_H isn't possible input, because D isn't a possible machine
notwithstanding ur aggressively defeatist comments, the new problem
that arose has not been analyzed sufficiently
Sure it has. I have shown an machine that computes a number that PRD
can not accept a machine that compute that same number.
actually i showed that machine, first
You are just stuck with your head in the sand (or up your a**)
refusing to even look at the problem, because you think the magic
unicorns can make it go away.
If PRD only accepts circle-free machines, it WILL be a circle-free
machine, and thus the results it generates WILL be a computable
number, but that number can not exist at any finite point in the
enumartion generated by PRD.
Thus, PRD fails to generate a COMPLETE enumeration, and thus your
fixed- H does not compute the diagonal of such a complete
enumeration, because the enumeration it computed the diagonal of was
not complete.
This seems beyond your ability to understand, so I wonder what
fallacy you are going to use to try to refute this. or perhaps you
just ignore it, and thus admit you don't have an answer.
i have then encountered a new problem, which doesn't negate the >>>>>>> fix i did make u ungrateful tard
Sure it does. You have presumed the enumeration that has been
proved can not be made.
toxic toxic toxic i'm so very tired of being on a world surrounded
by toxic people, rick
LYING is toxic.
calling me a liar is toxic, as i'm not intentionally misleading, i'm
just exploring the bounds of computability, and i shouldn't have to
be running up against such toxicity in order to do so
No, it is TRUTH.
Calling TRUTH toxic is just more toxicity.
i just don't care what u think here.
very little was actually built on this "truth" u seemingly hold so dear,
overturning it doesn't upset much about what we already know about
computing so i really don't feel bad about exploring in how correct they actually are,
u can cry about limits being important all u want (not that u know what
they are since they are unknowable limits to you), probably including something about wasting time if we don't recognize them (except how can
we if we can't even _know_ them???) ...
but i can't take anything like that serious because the way we go about computing is ungodly inefficient due to severe lack of overall
cooperation that such arguments seem incredibly detached from reality
if there are limits to tm computing, i should think they are at least *knowable*
clearly the consensus is defensive over being so wrong for so long, i
don't even get the need to be defensive here. i haven't the foggiest
clue what wrongs u think could be committed over exploring these issues,
Right, you don't need to be defensive as you just let the errors sit
there proving your stupidity.
cause the way we go about computing irl is already so batshit insane
unconscious AI does it better probabilisticly than ur average corpo
coder...
Maybe your problem is that you don't understand what Computation
theory is about.
And if you think AI does a good job, then you are just admititng that
you think lies are valid, as AI is based on just sounding good, not
making sound judgements.
i don't. it's just that corpo code is so bad that corpo coders don't
even notice the difference
Your enumeration is not what you claim, and thus your claims are the
toxic part.
It seems you like to call as "toxic" anything you can't deal with,
as you method of handling the ERRORS in you logic.
YOU are the toxic one.
There can not be an effective (computable) enumeration that
includes all computable number, as any method that generates one
allows the computing of a number that doesn't exist in that set.
Thus, any computed enumeration of computable numbers is
necessarily incomplete.
All you are doint is proving you don't understand what it means to >>>>>> be able to compute something.
It seems your idea of computing allows an algorithm to assume that >>>>>> it (or even an equivalent to it) can't be embedded into another
algorithm, which makes your concept strictly weaker in the power
to compute than the methods used by Turing Machines and their
equivalents.
The first problem is, you haven't created the enumeration
required to compute the diagonal of.
We know this, because we can convert your fixed-H to be anti- >>>>>>>> fixed- H that outputs the opposite digits that fixed-H does
(using the trick of fixed-H, using the number of fixed-H, not >>>>>>>> anti-fixed- H), and thus shows that if fixed-H is computing the >>>>>>>> diagonal, anti- fixed-H is computing the anti-diagonal, but we >>>>>>>> also see that this anti-diagonal isn't in the enumeration, and >>>>>>>> thus the enumeration can't be complete.
that does not prove there exists no further tricks that might
still get it on the diagonal somehow,
Sure it does.
rick, the only reason i got to this new problem was by ignoring all >>>>> the idiots telling me turing's proof was absolute
So, you jumped out of the frying pan into the flames and are burning
yourself to death.
again, why do u have a stick lodged so far up your asshole, dick?
Just trying to get you to take yours out so you can see the truth.
Your "logic" is based on ignoring FACTS and definitions, and
assuming that magic fairy dust powered unicorns can make you
impossible ideas work.
This is the work of a mind destroyed by a steady diet of your own
toxic lies that have eaten out every ounce of reasoning you might
have had.
incredibly toxic thing to state, what are you hiding?
Nothing.
What are YOU hiding behind your need to lie and believe in the
impossible?
*ANY* method to generate an enumeration of computable numbers
allows the creation of a computation that computes a number not in >>>>>> the set that it generated.
Thus NO method to generate an enumeratio of computable numbers can >>>>>> create a total enumeration.
Your assumption of trick is just depending on magic fairy dust
from a unicorn to create the impossible.
If it isn't complete, then the diagonal isn't the diagonal of >>>>>>>> the enumeration that Turing was talking about.
even if there doesn't,
there may be provable limits on what computations can and cannot >>>>>>> be computably enumerated on the diagonal,
But not that excludes the anti-diagonal, since we HAVE the
description of the algorithm that generates it, at least if we
have an algorithm to generate the enumeration.
which is certainly a step up from the over-applied rice's theorem >>>>>>> know- nothing nonsense u see with theorists today,
You seem to have the problem of not understand what Rice proves.
rice's theorem doesn't even apply to all semantics of a machine,
just those which are detectable from the output ... a phrase which
ur gunna disagree which because ur kind of a moron rick, but ben at >>>>> least did acknowledge
So, your sub-machine doesn't generate that as part of its output?
Then how does the outer machine know the answer it gabe?
There can be many machines that compute PARTIAL classifications or >>>>>> decisions on machines, just not a TOTAL classification.
like what if those various pseudo-anti-diagonals (as they aren't >>>>>>> true total anti-diagonal) are the *only* set of computable
numbers we can't computably enumerate on the list???
But they aren't. There are other proof, well beyond your head,
that show that other questions turn out to not be computable.
nah rick, i'm _never_ gunna accept anyone telling me anything is
"beyond my head" and the fact u even try to write that is
_incredibly toxic_
DUNNING-KRUGER in action,
AD HOMINEM in action,
Nope. I have never said your arguement as invalid BECAUSE it was you
that said it, which is the basis of "Ad Hominem". I have pointed out
YOUR errors, and you just refuse to look at them.
the term "dunning-kruger" serves no purpose to convey good
information. it's only used to convey bad information by lazy people
who act in bad faith towards other
Sure it does, it explains your statement. You can't accept the truth,
because you are so certain you must be right, that you just blantently
ignore anything that might prove you wrong.
clearly i don't just blatantly ignore it, given that i've _acknowledged_
the thorn in my side as of now
That is just you being toxic.
keep up the good work! 👍
i mean, the gall of u ever writing that out to someone is just
_incredibly anti-intellectual_ , why would u ever demand someone
accept something that exist "beyond their understanding" like a
fking a religious nutjob???
But you PROVE that it is beyond your understanding.
i can understand something without accepting it, u dunce
Then you admit that you are INTENTIONALLY lying, and that you are just
patholg]ogicallt a toxic liar.
bro i am honest to fault much of the time because i generally can't
stand lying, it is so bizarre to see u try to paint me as intentionally lying
my god rick, u are such an fucking gross hypocrite
the fact anyone in this group supports ur writing is just such a
stain on this group, and really demonstrates the hostility and
toxicity being harbored at the core of computing (and really
academia at large)
So, if you think you are so smart, submit your work to a real peer-
reviewed journal and see how fast it is shot down.
i already know there's a large bandwagon rick. i also know the
bandwagon fallacies are a thing because something large groups of
people are all wrong in the same manner.
In other words, The "world" is just wrong and your imaginary world is
correct.
idk who has more hubris:
someone who thinks the consensus will be overturned...
or something who thinks the consensus will never be overturned for all eternity...
Sorry, that is just demonstrating your insanity.
the two fallacies i spotted i fully intend to get published. they
don't prove turing wrong, but they do warrant revisiting the
arguments. who knows what other fallacies are lurking that i haven't
spotted yet
Except they are fallacies, but you not understanding the meaning of
his words because you have INTENTIONALLY (it seems) chosen to be
ignorant so you can lie about it.
The "hostility" you perceive is people pointing out your errors that
you refuse to accept, because "Truth" isn't something you world can
handle.
actually i've never had them "point out" errors. their negligence is
to the point of literally not even reading the submissions because
they trust their internalized hubris *that* strongly
Probably because it is SO bad it isn't worth their time.
well that's the kind of consensus that will invariably miss something in their hubris 🤷
I will point out, that you are proving it is not worth pointing out
your errors because you just ignore the advice that you asked for.
In fact, by simple "counting" we can tell that there are an
infinite number of uncomputable problems for every computable one.
ofc there are countable infinite variations on it. that doesn't
mean there aren't limits to the kinds of computations in that set
I guess you haven't read any of the papers of the other kinds of
uncomputable problems, one NOT based on a "self-reference".
more than half are "proven" thru a reduction to the halting problem,
and tbh that's where my focus lies: decision problems with computing
But most of those have OTHER proofs that don't reduce to the halting
problem.
And, since you admit that we can't actually solve the halting problem,
having a proof that reduces to it is valid.
honestly i don't even need to compute a full diagonal to throw a
wrench into much of this. if can prove which computations belong on
the PRD diagonal vs not ...
In other words you are ADMITTING that you claim that your PRD accepts
at least one machine for every computable number ois just a lie.
then we would need to revist those proofs to ensure the problem is at
least computed by a machine proven to not exist on the enumerable
diagonal... otherwise why should we believe it to be uncomputable???
So, where is anti-fixed-H in your enumeration?
Computing a partial enumeration was never said to be a problem.
Again, you don't seem to understand that *ALL* does mean ALL with no
exceptions.
there's just so many angles here that just haven't been worked. u
poo- pooing me about literally all of them is just laziness that has
driven deep into the territory of blatantly intellectual negligent,
Yes, you deflect yourself with all your strawmen, so you can ignore
the errors pointed out, and you keep on hoping that you can find that
magic unicorn that can make the problem machines just disappear.
tbh yes:
*i'm calling the entirety of CS academia intellectually negligent*
No, you are just admitting your own intentional ignorance so you can
lie about not seeing the problems.
Yes, many of them allow you to, as ONE of the ways, to prove them
uncomputable, show that them being computable would allow you to
compute the answer to the uncomputable problems due to self-reference. >>>>
But, they don't themselves use that sort of self-reference.
that would also be a huge win, cause those computations don't
compute relationships we care about, so failing to enumerate them >>>>>>> totally just doesn't really matter 🤷
Sure it does. By knowing that TOTAL classification is impossible, >>>>>> we know that we need to look at what classes of inputs a given
algorithm can work on.
Thus, like where you started with, because we KNOW we can't
totally solve the Halting Problem, we accept that we need to allow >>>>>> our algorithm to decide that some cases might not be deciable, and >>>>>> work on the cases we can decide on.
that's already a huge step up from before where you were advocating >>>>> for programs that we couldn't even generally decide on their
decidability
But that is still true, and not contradictory with the above.
There are many programs that we can decide on.
But there are also some that we can't, and some we can't even decide
that there behavior is unknowable.
this part i still entirely disagree with. we proved what the anti-
diagonal does even if it wasn't on PRDs diagonal...
Then it isn't an "anti-diagonal", but just a strawman.
lol ur just trying to focus on failures, but this is *exploring* the
bounds. so will accept any given failure can still lead to new results
in different aspects. idk why i need to explain this other than academia
is mostly a dribbling shitshow focused on regurgitation for kudos rather than curiosity at the bounds... or maybe that's just you eh???
I'M RESPONDING TO THIS CLAIM: but there are also some that we can't, and some we can't even decide that there behavior is unknowable
again: the number computed by anti_H was not on PRDs diagonal, yet we
still know it's circle free...
which means if we manually compute PRDs diagonal, but just injected any computable numbers it missed along the way... we could manually compute
the full diagonal even if a TM couldn't
oh dear lord: the ct-thesis is so damn cooked
how is that???
Because it is just a strawman, that doesn't negate the problem of the
actual anti-diagonal program that proves that PRD doesn't do a
complete enumeration.
(because undecidability in computing _only_ exists between a machine
and the *specific* classifiers it paradoxes _not_ generally)
Nope. And that is one of the roots of your problem. "Udecidability" is
about a "problem", a defined "classification". It has nothing about a
specific machine.
i full-heatedly disagree that the consensus has done a good job as
defining undecidability
Of course, if you admit that you are in a totally new field with new
definitions, go through and define EVERYTHING it needs and show it is
useful.
red herring with gish gallop mixed in. it is totally bizarre to suggest
i need to redefine *everything* to change small aspects. lol
Of course, you are unlikely to find anyone willing to fund that
research since you show so little understanding of the system you
claimed to have been talking about.
see what i'm saying: not even you think it's genuine advice, so why
bother suggesting it?
to prove a machine with complete unknowable decidability i think u'd
need to show a machine that exists on _no_ possible diagonal ...
which i do _not_ think is possible
You just are proving you don't understand what you are talking about.
not a response, rick
A correctness proving program doesn't need to prove EVERY program >>>>>> correct or wrong, but can prove SOME programs correct, SOME
programs it can point out errors, and some it tells us they are
too complicated for it to process.
"too complicated to process" is a different theory rick. that's
complexity theory not computability theory.
No, this isn't "complexity" as in O notation complexity, but that
our processing, but necessity, can't try to handle all cases to all
depths, but, to avoid getting stuck and not answering, exstablishes
finite limits on resources that can be expended on the various parts
of the analysis, and if the analysis of the program hits one of
these limits, we classify it as "too complex".
If it tells us it is too complicted, if we really need the proof, >>>>>> we need to revise it to simplify it. (Or it may be that the
problem we are working on is just uncomputable, so no program CAN >>>>>> be proven correct,
u haven't demonstrated an actual machine we can't prove correct,
the machines PRD failed to classify are still provable in what they >>>>> do from our perspective (we both know the pseudo-anti-diagonals are >>>>> circle- free and can prove it ... that's how we know PRD "missed"
them), regardless of whether PRD could classify them or not
And thus, PRD can not be "correct" to its specification, as one of
the REQUIREMENTS was that it would accept at least one machine that
generates EVERY computable number.
that was what i thought it could do, i'm unsure as of right now
Right, you "logic" isn't based on being right, but sounding sort of good.
Just like AI. Only AI was trained on a lot of material, It seems you
know very little.
the "anti-diagonal" anti-fixed=H is not "pseudo" anything, given
your claimed PRD, it is a REAL machine, that computes a REAL
computable number that no machine in your enumerate generates.
see ur kinda stuck in a rut here. any circle-free machine can prove >>>>> that PRD fails to enumerate is still a machine that was proven as
circle- free ...
So, you are forgetting that for your claim was that fixed-H
generates an computable diagonal of a set that is a enumeration of a
set of amchines that contains EVERY computable number.
You forgot that requirement, as you went off on your strawman fallacy. >>>>
i feel this is going to end up in abandoning the ct-thesis rick. tm >>>>> computability has limits due to self-referential weirdness, and
they aren't the same as the limits we 3rd party observers are
bounded by because we're aren't subject to that same self-
referential weirdness
feelings don't generate proofs.
quite the opposite, really
one _must_ have the feeling in order to motivate themselves to
produce the proof ya dingdong
In other words, you don't understand how logic works.
Yes, feelings can provide impetus and motivation, but they do not
themselves generate a proof.
You need to start from know truths to build a proof, which means you
need to start knowing something and not just working from feelings.
Your world is just a giant fallacy, so that is what you see in others,
because you just don't know better.
and we need to build a partial version that admits that there are >>>>>> cases it can't get correct)
(and before u try to make yet another baseless claim that it >>>>>>>>> must have been, show me the proof instead of baselessly just >>>>>>>>> claiming u fucking twat)
And what is wrong about this proof.
i don't have to be right about literally every possible future
goal post to right about one goal post in a unique way that's
never been done before. the fact i could even hit that goal post >>>>>>> is to me a massive sign things have been missed in the
fundamentals, rick
it should be to you as well, but my god are obsessed with
clinging to certain uncertainty it's abcerd
But the fact that you current claims are based on NO evidence,
means you are starting with nothing.
You claim that something might be possible, when it is shown that >>>>>> it can't be.
Your world is just built on the assumption that the rules don't
apply. That is a world of fantasy and lies.
nothing was said here
Sure there was, you just can't understand it, as your world is built
on that lie.
Your enumeration generated by PRD just can not be COMPLETE,
including at least one instance of EVERY computable number.
PROVE ME WRONG.
it's really kind of sad how much toxicity and generally
unwillingness to cooperate i've encounter when trying to explore >>>>>>> these ideas,
The most toxic thing is to just lie to yourself about what can be >>>>>> done.
i hope future academia may take heed from what i've had to endure >>>>>>> thus far, pretty much on my own. heck i hope current academic
might too ...
but that might be asking for too much at this point eh???
In other words, you hope academia might allow people to live in
error and self-deciet?
i would be nice if u could even read simple sentence accurately.
i said it was too much to ask for, in that i hope for it, but don't >>>>> expect it. not sure where u pulled hoping for opposite from... but
i never claimed that
No, the problem is you think you are being treated unfairly, but you
are not, you are treating truth unfairly.
You ARE living a life of lies, based on the ignoring of basic
principles.
Your hope is for a world where error is just tolerated under some
guise of acceptance.
Your problem is you reject people pointing out the errors in your >>>>>> work,
rick, u have problems reading simple sentences much of the time
Less than you do.
You don't even know what an "equivalent problem" is.
Or what a "computation" is.
because you assume you must be right, even when you admit you
don't really understand the field.
Your works is based on ignorant assumptions, not facts.
That is a dark world of lies.
The people here mostly know what they are talking about, >>>>>>>>>>>> because they have studied it (some like Olcott and you are >>>>>>>>>>>> the exception).
It seems you are just admitting that you are stuck in your >>>>>>>>>>>>>> lies and just can't think because, like Olcott, you have >>>>>>>>>>>>>> successfully gatlit yourself into being convinced of your >>>>>>>>>>>>>> lies.
i demonstrated two distinct fallacies in turing's paper, >>>>>>>>>>>>> that really aren't the hard to understand,
No, you demonstrated that you do don't understand what he is >>>>>>>>>>>> saying,
the fact u continually try to gaslight me into thinking i >>>>>>>>>>> haven't understood his argument well enough is not only >>>>>>>>>>> incredibly toxic but let's me know ur completely fine with >>>>>>>>>>> blatantly lying at me to "win" an argument,
But it isn't gaslighting, it is FACT.
idk how u read everything i wrote and try to claim i just don't >>>>>>>>> understand,
Because you keep claiming things that aren't there.
i have been manipulating his ideas in ways that have never been >>>>>>>>> done before, and u can't even acknowledge that i understand his >>>>>>>>> ideas??? sheesh,
Right, by twisting words to not mean what they mean.
i'm swimming thru a swamp of endless gaslighting, fostered a >>>>>>>>> toxic mentality festering the fundamentals of math hostile to >>>>>>>>> any sort of meaningful innovation at the core for some ungodly >>>>>>>>> reason
Yes, the endless gaslighting that you have done to yourself,
causing you to think that people point out truth to you are
gaslighting you.
The fact you can't actually prove anything should be your first >>>>>>>> sign that you have something wrong.
Your world is just built on your own lies and fantasies.
the fallacies i picked out are still fallacies
No, because your claimed fallacies are based on you using a
definist fallacy.
fallacy 1) identifying a subset is _NOT_ the same problem as
identify the entirety of a set,
No, but might be an equivalent problem.
MIGHT is not a proof
I have agreed that he didn't present the proof here, but that doesn't
mean it wasn't a fact proven before and was just known to his intended
audience.
i love how you also just assume the proof must exist
fallacy 2) computing the diagonal does _NOT_ then grant an able to
computing an anti-diagonal
Why not?
WHy can you NOT just change that program to reverse the value
written to the perminante cells, and any decision based on reading
one of those cells?
How does that NOT result in that result?
because u can't then put anti-diagonal machine on the diagonal!
But, if the enumeration is computable, you can compute the diagonal,
and the anti-diagonal, and since the anti-diagonal isn't in the
enumeration, it shows your computed enumeration couldn't have been
complete.
this doesn't change the fact one cannot use an enumeration to compute a
true anti-diagonal.
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