Re: How do simulating termination analyzers work?

On 6/19/25 2:59 AM, olcott wrote:

On 6/19/2025 1:17 AM, Mikko wrote:

On 2025-06-18 13:46:16 +0000, olcott said:
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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;
}
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void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
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void DDD()
{
   HHH(DDD);
   return;
}
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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.
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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)
>

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*none of them ever stop running unless aborted*

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

>
*It is not given that any of them abort*

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It is known a priori that HHH either does or does not abort.

 Very good.
 

If HHH does
not abort it does not terminate the simulation of DDD and therefore

 DDD never stops running.

Sure it does, if the HHH that it is built on gives an answer.
Your problem is you think that the HHH that does not terminate its simulation can be the same program as the one that gives the answer, proving your utter stupidity.

 

does
not report correctly. If HHH does abort it reports that DDD does not
halt, which is incorrect as in that case DDD does halt. HHH is correct
about DDD only if it does abort its simulation and reports "halts".
But you HHH does not do that.
>