• H(D,D)==0 is proven to be correct [V7]

    From olcott@21:1/5 to All on Mon Mar 6 18:26:12 2023
    XPost: comp.theory, sci.logic, sci.math

    https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X


    MIT Professor Michael Sipser has agreed that this verbatim paragraph is
    correct (he has neither reviewed nor agreed to anything else):

    (a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running
    unless aborted then (b) H can abort its simulation of D and correctly
    report that D specifies a non-halting sequence of configurations.

    To the best of my understanding (a) proves (b) is simply a tautology.
    (proven true entirely on the basis of the meaning of its words)

    01 int D(int (*x)())
    02 {
    03 int Halt_Status = H(x, x);
    04 if (Halt_Status)
    05 HERE: goto HERE;
    06 return Halt_Status;
    07 }
    08
    09 void main()
    10 {
    11 D(D);
    12 }

    Here is the sequence when H never aborts it simulation
    main() calls D(D) at line 11
    that calls H(D,D) that simulates D(D) at line 03
    that calls H(D,D) that simulates D(D) at line 03
    that calls H(D,D) that simulates D(D) ... repeating ...

    Because it is an easily verified fact that D(D) would never stop running
    unless H aborts its simulation of D, H is necessarily correct to return
    0 indicating this verified fact.

    This is every aspect of my whole proof

    (a) (proven above)
    (a) proves (b) (tautology)
    ---------------------------------
    ∴ H(D,D)==0 is correct


    --
    Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
    hits a target no one else can see." Arthur Schopenhauer

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)