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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as
 non-halting V2
Date: Sat, 13 Jul 2024 13:17:23 -0000 (UTC)
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Am Sat, 13 Jul 2024 06:39:31 -0500 schrieb olcott:
> On 7/13/2024 3:15 AM, Fred. Zwarts wrote:
>> Op 13.jul.2024 om 01:19 schreef olcott:
>>> On 7/12/2024 5:56 PM, Richard Damon wrote:
>>>> On 7/12/24 10:56 AM, olcott wrote:
>>>> Which means the only "correct emulation" that tells the behavior of
>>>> the program at the input is a non-aborted one.

>>>> And thus HHH that do that know only the first N steps of the behavior
>>>> of DDD, which continues per the definition of the x86 instruction set
>>>> until the COMPLETE emulation (or direct execution) reaches a terminal
>>>> instruction.

>>>> And thus, the subset that only did a finite number of steps and
>>>> aborted its emulation on a non-terminal instrucition only have
>>>> partial knowledge of the behavior of their DDD, and by returning to
>>>> their caller, they establish that behavior for ALL copies of that
>>>> HHH, even the one that DDD calls, which shows that DDD will be
>>>> halting, even though HHH stopped its observation of the input before
>>>> it gets to that point.
This is the crux. You think that the nested HHH runs forever. But it
aborts just the same as the outer one.

>>>> Wrong. EVERY DDD of an HHH that simulated its input for only a finite
>>>> number of steps WILL halt becuase it will reach its final return.
>>>> The HHH that simulated it for only a finite number of steps, only
>>>> learned that finite number of steps of the behaivor, and in EVERY
>>>> case, when we look at the behavior past that point, which DOES occur
>>>> per the definition of the x86 instruction set, as we have not reached
>>>> a "termial" instruction that stops behavior, will see the HHH(DDD)
>>>> that DDD called continuing to simulate its input to the point that
>>>> this one was defined to stop, and then returns 0 to DDDD and then DDD
>>>> returning and ending the behavior.
>>>> You continue to stupidly confuse the PARTIAL observation that HHH
>>>> does of the behavior of DDD by its PARTIAL emulation with the ACTUAL
>>>> FULL behavior of DDD as defined by the full definition of the x86
>>>> insttuction set.
Worth requoting.

>>>> NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
>>>> shown to be ignorant of what you are talking about.
>>>> The HHH that did a partial emulation got the wrong answer, because
>>>> THEIR DDD will halt. and the HHH that doen't abort never get around
>>>> to rejecting its DDD as non-halting.
>>>
>>> When no DDD of every HHH/DDD that can possibly exist halts then each
>>> HHH that rejects its DDD as non-halting is necessarily correct.
>>>
>> This is double talk, because no HHH can possibly exist that simulates
>> itself correctly.
> Your definition of correct contradicts the semantics of the x86 language
> making it wrong.
What part of it contradicts which semantics?
What are the doubletalk and weasel words?
What are the shells of the game?

-- 
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.