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

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*
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer