Re: How do simulating termination analyzers work?

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.

*none of them ever stop running unless aborted*

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

*none of them ever stop running unless aborted*

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

 Do you agree or can you refute THIS EXACT POINT?
Do you agree or can you refute THIS EXACT POINT?
Do you agree or can you refute THIS EXACT POINT?
 

We agree that the non-existing hypothetical HHH seen only in your dreams that does not abort will never stop running.
But simulation of the actual HHH specified in the input for the simulator, that has code to abort and halt does not need to be aborted.
Sum(2,3) should calculate the sum of its input 2 and 3, not of the non-existing hypothetical inputs 7 and 9.