Re: HHH(DD) does correctly reject its input as non-halting --- VERIFIED FACT

Op 14.jun.2025 om 15:38 schreef olcott:

On 6/14/2025 4:10 AM, Fred. Zwarts wrote:

Op 13.jun.2025 om 17:53 schreef olcott:

On 6/13/2025 5:51 AM, Mikko wrote:

On 2025-06-12 15:30:05 +0000, olcott said:
>

int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
It is a verified fact that DD() *is* one of the forms
of the counter-example input as such an input would
be encoded in C. Christopher Strachey wrote his in CPL.
>
// rec routine P
//   §L :if T[P] go to L
//     Return §
// https://academic.oup.com/comjnl/article/7/4/313/354243
void Strachey_P()
{
   L: if (HHH(Strachey_P)) goto L;
   return;
}
>
https://academic.oup.com/comjnl/article-abstract/7/4/313/354243? redirectedFrom=fulltext

>
Strachey only informally presents the idea of the proof. Formalism
and details needed in a rigorous proof is not shown.
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
_DDD()
[00002192] 55             push ebp
[00002193] 8bec           mov ebp,esp
[00002195] 6892210000     push 00002192
[0000219a] e833f4ffff     call 000015d2  // call HHH
[0000219f] 83c404         add esp,+04
[000021a2] 5d             pop ebp
[000021a3] c3             ret
Size in bytes:(0018) [000021a3]
>
Exactly how would DDD correctly emulated by HHH
reach its own "ret" instruction final halt state?

>
Indeed, HHH fails where other world-class simulators have no problem to simulate the program specified in the input.
>

 So you still don't understand what recursive simulation is?
 

It seems I understand it better than you do. You seem to think that every recursion is a infinite recursion. As soon as you see a recursion, you think it has been proven that it is an infinite recursion, even if the code specifies an abort and halt.