• Re: Concise refutation of halting problem proofs V21 [ Richard is not t

    From olcott@21:1/5 to Richard Damon on Mon Nov 22 10:37:48 2021
    XPost: comp.theory, sci.logic, sci.math

    On 11/22/2021 10:05 AM, Richard Damon wrote:
    On 11/22/21 10:28 AM, olcott wrote:
    On 11/22/2021 6:20 AM, Python wrote:
    olcott wrote:
    #include <stdint.h>
    #include <stdio.h>
    typedef int (*ptr)();

    int H(ptr x, ptr y)
    {
       x(y); // direct execution of P(P)
       return 1;
    }

    // Minimal essence of Linz(1990) Ĥ
    // and Strachey(1965) P
    int P(ptr x)
    {
       H(x, x);
       return 1; // Give P a last instruction at the "c" level
    }

    int main(void)
    {
       H(P, P);
    }

    $ ./olcott
    Segmentation fault: 11


    Thus proving that H(P,P) never halts.


    And if H(P,P) never halts, it isn't a valid decider.

    FAIL.

    I take back that you are sufficiently technically competent to
    comprehend any of the following, not even the Linz quote:

    Computation that halts
    a computation is said to halt whenever it enters a final state.
    (Linz:1990:234)

    [PSR set] Combinations of H/P having pathological self-reference
    Every H of H(P,P) invoked from main() where P(P) calls this same H(P,P)
    and H simulates or executes its input and aborts or does not abort its
    input P never reaches its last instruction.

    [PSR subset] Because we know that the input to H(P,P) never halts for
    the whole PSR set and a subset of these H/P combinations aborts the
    execution or simulation of its input then we know that for this entire
    PSR subset the input to H(P,P) never halts and H(P,P) halts.



    --
    Copyright 2021 Pete 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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