Re: Can you see that D correctly simulated by H remains stuck in recursive simulation?

Op 23.mei.2024 om 18:52 schreef olcott:

typedef int (*ptr)();  // ptr is pointer to int function in C
00       int H(ptr p, ptr i);
01       int D(ptr p)
02       {
03         int Halt_Status = H(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
08
09       int main()
10       {
11         H(D,D);
12         return 0;
13       }
 The above template refers to an infinite set of H/D pairs where D is
correctly simulated by pure function H. This was done because many
reviewers used the shell game ploy to endlessly switch which H/D was
being referred to.
 *Correct Simulation Defined*
This is provided because every reviewer had a different notion of
correct simulation that diverges from this notion.
 In the above case a simulator is an x86 emulator that correctly emulates
at least one of the x86 instructions of D in the order specified by the
x86 instructions of D.
 This may include correctly emulating the x86 instructions of H in the
order specified by the x86 instructions of H thus calling H(D,D) in
recursive simulation.
 *Execution Trace*
Line 11: main() invokes H(D,D); H(D,D) simulates lines 01, 02, and 03 of
D. This invokes H(D,D) again to repeat the process in endless recursive
simulation.

The case can be simplified even more (D is not needed):
typedef int (*ptr)();  // ptr is pointer to int function in C
00       int H(ptr p, ptr i);
01       int main()
02       {
03         H(H,H);
04         return 0;
05       }
If olcott's claim is true, then also main will never reach line 04. This would prove that H is non-halting. This would prove that a simulating halt-decider cannot be used as a halt-decider, because it does not halt.