Re: How do simulating termination analyzers work?

Am Wed, 18 Jun 2025 09:50:18 -0500 schrieb olcott:

On 6/18/2025 9:05 AM, joes wrote:

Am Wed, 18 Jun 2025 08:46:16 -0500 schrieb olcott:

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:

 

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*

Huh? They contain the code to abort, even if it is not simulated.
 

*none of them ever stop running unless aborted* yes or no?

If HHH(DDD), where DDD() only calls HHH(DDD), is simulated by a pure
simulator such as HHH1 (not by itself, which aborts), it stops running
by aborting (the simulator also terminates). All inner invocations of
HHH would have stopped running had the outermost one (which is still
only simulated by the real HHH1) not stopped simulating.
HHH1 simulating HHH1 of course doesn't stop running, but we don't have
HHH1 involved in either role here. (For completeness, HHH1(HHH) and
HHH(HHH1) also halt.)

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.