Re: DDD correctly simulated by HHH can't possibly reach its own "return" statement --- Liars on comp.theory

On 27/06/2025 20:36, olcott wrote:

I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
 I have included proof that the people on comp.theory
lied about this at the bottom.
 typedef void (*ptr)();
int HHH(ptr P);
 void DDD()
{
   HHH(DDD);
   return;
}
 int main()
{
   HHH(DDD);
   DDD();
}
 Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
 On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
 > In comp.theory olcott <polcott333@gmail.com
 > wrote:
 >
 >> I know that DDD .... simulated by HHH cannot
 >> possibly reach its own simulated "return" statement
 >> final halt state because the execution trace
 >> conclusively proves this.
 >
 > Everybody else knows this, too, and nobody has
 > said otherwise. *The conclusion is that the*
 > *simulation by HHH is incorrect*
 >
 *That last sentence is an intentional falsehood*

Well, people here use the term "simulation" in a number of ways, right?  Like I've pointed out to you on several occasions.  Maybe you should consider a simple misunderstanding over terminology before assuming bad intentions.
So what does Alan mean by "the simulation by HHH is incorrect", exactly?  (And why do you think it is incorrect?)
That is a question for PO, rather than Alan, since it is PO who claims Alan is lying...
Mike.