• Concise refutation of halting problem proofs V33

    From olcott@21:1/5 to All on Sat Nov 27 08:44:46 2021
    XPost: comp.theory, sci.logic, sci.math

    #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);
    }

    The above program is obviously infinitely recursive. It is self evident
    that when 0 to ∞ steps of the input to H(P,P) are directly executed or correctly simulated that the input to H(P,P) never reaches its final instruction.

    computation that halts a computation halts whenever it enters a final
    state (Linz:1990:234) thus none of the simulated or executed 0 to ∞
    steps of the input to H(P,P) ever halt.

    PSR set (pathological self-reference)
    H1(P1,P1) Is the above code.
    H2(P2,P2) Is the above code where H2 simulates rather than directly
    executes its input.
    H3(P3,P3) Is the execution of N steps of the input of H1(P1,P1).
    H4(P4,P4) Is the simulation of N steps of the input of H2(P2,P2).

    Every Hn(Px,Py) that returns a value returns 1 except for instances of
    {H3, H4} that determine whether or not to return {0,1} on the basis of
    the behavior of their input.

    The correct pure simulation of N steps of the input to H(P,P) by H is
    always a correct halt deciding basis where P has reached its final state
    or H has correctly detected that P would never reach its final state.

    The point in the sequence of N steps where the execution trace of the simulation of P shows that P is about to call H(P,P) again with the same
    input that H was called with provides conclusive proof that P would be infinitely recursive unless H aborted its simulation.
    *In this H4(P4,P4)==0 computation P4 is dependent on H4 altering the
    behavior of P4.*

    When directly executed P(P) calls H(P,P) and the simulated P(P) reaches
    the point where it would call H(P,P) with the same parameters that H was
    called with H returns 0 to this directly executed P.
    *In this H1(P4,P4)==1 computation P4 is independent of H1.*

    H is a computable function that accepts or rejects inputs in its domain
    on the basis that these inputs specify a sequence of configurations that
    reach their final state.




    Halting problem undecidability and infinitely nested simulation (V2)

    https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2

    --
    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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