DD simulated by HHH cannot possibly halt (Halting Problem)

On 4/8/2025 5:53 PM, Richard Damon wrote:

On 4/8/25 11:17 AM, olcott wrote:

On 4/8/2025 2:51 AM, Mikko wrote:

On 2025-04-08 04:33:57 +0000, olcott said:
>

>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   HHH(DD);
}
>
*Simulating termination analyzer Principle*
It is always correct for any simulating termination
analyzer to stop simulating and reject any input that
would otherwise prevent its own termination.
>
Rational minds would agree that the above principle
is correct and directly applies to HHH(DD) rejecting
its input.

>
 From the meaning of the word "correct" obviously follows that it is
never correct to reject a terminating input as non-terminating.
>

>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
This stuff is simply over-your-head.
HHH(DD) meets the above: *Simulating termination analyzer Principle*
Anyone with sufficient competence with the C programming language
will understand this.
>

 NO, because either:
 

Yes because this is A VERIFIED FACT:
DD simulated by HHH cannot possibly halt** (Halting Problem)
** Reach its own "return" instruction final halt state no
matter how many straw-man deceptions you try to get away with
by changing the subject to DD NOT simulated by HHH.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer