Re: How do simulating termination analyzers work?

On 6/18/25 9:46 AM, olcott wrote:

On 6/18/2025 5:12 AM, Fred. Zwarts wrote:

Op 18.jun.2025 om 03:54 schreef olcott:

On 6/17/2025 8:19 PM, Richard Damon wrote:

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)
>

>
*none of them ever stop running unless aborted*

>
All of them do abort and their simulation does not need an abort.
>

 *It is not given that any of them abort*
 

But it either does or it doesn't, and different HHHs give different DDD so you can't compare their behavior.
The HHH that aborts, doesn't actually have proof that its input doesn't halt, and it has been shown that it does.
The HHH that doesn't abort, never answers.
So both of them are just wrong.
Also, youj *HAVE* defined what HHH *IS* as you claim Halt7.c is part of the memory, and that DEFINES what HHH MUST BE.
Your logic is just like saying a one mile road is infinite in length, just because you turned off it after walking on it 100 feet and then getting off it.
Sorry, but you are just showing that you are a stupid liar.