Re: 195 page execution trace of DDD correctly simulated by HH0

Am Thu, 27 Jun 2024 12:30:38 -0500 schrieb olcott:

void DDD()
{
   H0(DDD);
}
 
int main()
{
   H0(Infinite_Loop);
   H0(Infinite_Recursion);
   H0(DDD);
}

When this is construed as non-halting criteria then simulating
termination analyzer H0 is correct to reject these inputs as non-halting
by returning 0 to its caller.

To the caller DDD, which then returns to its own caller H0, which returns
„halting” to main… hold on.

Simulating termination analyzers must report on the behavior that their
finite string input specifies thus H0 must report that DDD correctly
emulated by H0 remains stuck in recursive simulation.

H0 must not report on itself, only on DDD. Which you’ve proven halts.
We don’t care how H0 deviates (i.e. is incorrect) in its simulation.
That would be main {H0(H0(DDD))}.