Sujet : Re: How do simulating termination analyzers work?
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 18. Jun 2025, 15:05:57
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <dc7254ded3893833b22f666b467c70c5b5832ef9@i2pn2.org>
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User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
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.
-- 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.
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