• Proof that D simulated by H never reaches its own simulated "return" statement

    From olcott@[email protected] to comp.theory,comp.lang.c++,comp.lang.c,comp.ai.philosophy on Fri Nov 7 08:04:22 2025
    From Newsgroup: comp.ai.philosophy

    On 11/7/2025 2:26 AM, Mikko wrote:
    On 2025-11-06 10:54:43 +0000, olcott said:

    On 11/6/2025 2:55 AM, Mikko wrote:
    On 2025-11-05 11:37:15 +0000, olcott said:

    On 11/5/2025 4:53 AM, Mikko wrote:
    On 2025-11-04 13:56:17 +0000, olcott said:

    On 11/4/2025 4:10 AM, Mikko wrote:
    On 2025-11-03 23:28:03 +0000, olcott said:

    On 11/2/2025 6:28 AM, Mikko wrote:

    Irrelevant here. I said "Perhaps you should try to improve your >>>>>>>>> perfirmance in the art of excution traces". You didn't ppint out >>>>>>>>> any error or incoherence in that sentence. Instead you said
    "There is no mistake in the essence of above."


    *Claude, ChatGPGT, Gemini and Grok all agree on this*
    *links provided below*

    int D()
    {
       int Halt_Status = H(D);
       if (Halt_Status)
         HERE: goto HERE;
       return Halt_Status;
    }

    The function H is a simulating termination analyzer:
    (a) Detects a non-terminating behavior pattern:
         abort simulation and return 0.
    (b) Simulated input reaches its simulated
         "return" statement: return 1.

    When given a function P, it literally simulates each
    step of executing P() to *see whether that simulated*
    *execution ever reaches a return statement* Now let H
    simulate D. Based only on the outcome of that literal
    simulation (not on reasoning about what should happen),
    what result should H(D) produce?

    That is irrelevant to anything in the quoted text.

    You didn't even pay attention to the words the last paragraph.

    Everything in the last paragraph, including the question, is
    as ifrrelevant as the text before.

    One should also note that you didn't identify any relevance.

    It forms a proof that H(D) is correct to reject (D).
    That you didn't see that proves that you are not
    paying close enough attention.

    If does not form a proof of anything.

    The proof would be my reviewers actually knowing C
    programming and being able to do their own execution
    trace.

    No, that would not be a proof. A proof is a seuence of sentences,

    A proof is any damn thing that shows a conclusion
    is necessarily true.

    When we start with the semantics of the C programming
    language and this finite string of ASCII characters
    defining the C function D:

    int D()
    {
    int Halt_Status = H(D);
    if (Halt_Status)
    HERE: goto HERE;
    return Halt_Status;
    }

    And the knowledge that H is a C interpreter
    that takes the above test.c file as an input
    and knows that the call to H(D) in D calls
    itself to simulate the text body of D then
    the execution trace proves that D simulated
    by H cannot possibly reach its own simulated
    "return" statement final halt state.

    each of which either is a premise or follows from earlier sentences
    in the proof by a truth-preserving transformatios. The proof proves
    its last sentence, which is called the concluson of the proof.

    --
    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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