Re: How do simulating termination analyzers work?

On 6/17/25 4:34 PM, olcott wrote:

void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}
 void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
 void DDD()
{
   HHH(DDD);
   return;
}
 When it is understood that HHH does simulate itself
simulating DDD then any first year CS student knows
that when each of the above are correctly simulated
by HHH that none of them ever stop running unless aborted.

WHich means that the code for HHH is part of the input, and thus there is just ONE HHH in existance at this time.
Since that code aborts its simulation to return the answer that you claim, you are just lying that it did a correct simulation (which in this context means complete)
Partial simulation NEVER "define" an input to be non-halting, and the only way for it to give the answer that you claim means that since it *DOES* abort, the "hypothetical" HHH that you are thinking of that doesn't, isn't actually the HHH that is here, and the input that it would look at is different, so irrelevent,.

 The same thing applies to these two, yet they may be
too difficult for a first year CS student.
 int Sipser_D()
{
   if (HHH(Sipser_D) == 1)
     return 0;
   return 1;
}
 int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
 

And the same comment applies here.
Since you are admitting that you HHH returns 0 in all these cases, *ALL* of these input are halting (and Sipser_D returns 1) so your HHH is just incorrect in every case.
All you are doing is proving that you are nothing but a stupid liar.