• Re: Kaz does not understand his own code --- I AM PROVED EXACTLYCORRECT

    From olcott@polcott333@gmail.com to comp.theory,comp.lang.c++,comp.lang.c,comp.ai.philosophy on Tue Nov 4 21:51:10 2025
    From Newsgroup: comp.ai.philosophy

    On 11/4/2025 8:43 PM, Kaz Kylheku wrote:
    On 2025-11-05, olcott <polcott333@gmail.com> wrote:
    On 11/4/2025 7:20 PM, Mike Terry wrote:

    I don't really see that PO has declined recently - It seems to me he's
    been the way he is for as long as I recall.  I would say he's become
    markedly ruder of late, though!  Maybe that's all related to his LLMs
    pandering to his delusions and giving him a renewed confidence.


    Mike.



    news://news.eternal-september.org/20251103195844.661@kylheku.com

    On 11/3/2025 10:28 PM, Kaz Kylheku wrote:

    How about this minimal viable H:

    #include <interpret.h> // C interpreter's own API

    bool H(fptr P)
    {
    interp *s = interp_init(P);

    for (int i = 0; i < 3; i++) {
    if (interp_step(s))
    return true;
    }

    return false;
    }

    H initializes an interpreter for its argument P.
    Then it applies a very simple abort logic: it
    steps the interpreter state three times. If
    during those three steps, P terminates, it returns
    true. Otherwise it assumes P is nonterminating and
    returns false.

    (Pretend that more complicated abort criteria are there.)

    The interpreter API consists of primitives built
    into the system, so it isn't traced.

    So then we have D:

    void D(void)
    {
    if (H(D)) { for (;;); }
    return;
    }

    Let's trace H(D). We indicate the simulation levels from 0,
    step numbers from 1 within each level, with a bit of indentation
    to tell apart the levels:


    *This is the part that Kaz does not understand*

    i == 0 reaches if (interp_step(s))
    i == 1 reaches if (interp_step(s))
    i == 2 reaches if (interp_step(s))
    i == 3 NEVER reaches if (interp_step(s))

    The whole point is that D simulated by H
    cannot possbly reach its own simulated
    "return" statement no matter what H does.

    Yes; this doesn't happen while H is running.


    Then that all by itself conclusively proves that
    the input to H(D) specifies non-halting behavior
    when halting behavior is stipulated to mean behavior
    that cannot possibly reach its own final halt state
    as measured by D simulated by H.

    Within the 100% perfectly exact meaning of those
    words without the tiniest little change
    I AM PROVED EXACTLY CORRECT.
    --
    Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
    hits a target no one else can see." Arthur Schopenhauer
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