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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as
 non-halting V2
Date: Sun, 14 Jul 2024 09:49:30 -0500
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On 7/14/2024 3:58 AM, Mikko wrote:
> On 2024-07-13 13:04:01 +0000, olcott said:
> 
>> On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
>>> Op 13.jul.2024 om 13:39 schreef 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:
>>>>>>>> We stipulate that the only measure of a correct emulation is the
>>>>>>>> semantics of the x86 programming language.
>>>>>>>
>>>>>>> Which means the only "correct emulation" that tells the behavior 
>>>>>>> of the program at the input is a non-aborted one.
>>>>>>>
>>>>>>>>
>>>>>>>> _DDD()
>>>>>>>> [00002163] 55         push ebp      ; housekeeping
>>>>>>>> [00002164] 8bec       mov ebp,esp   ; housekeeping
>>>>>>>> [00002166] 6863210000 push 00002163 ; push DDD
>>>>>>>> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
>>>>>>>> [00002170] 83c404     add esp,+04
>>>>>>>> [00002173] 5d         pop ebp
>>>>>>>> [00002174] c3         ret
>>>>>>>> Size in bytes:(0018) [00002174]
>>>>>>>>
>>>>>>>> When N steps of DDD are emulated by HHH according to the
>>>>>>>> semantics of the x86 language then N steps are emulated correctly.
>>>>>>>
>>>>>>> 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.
>>>>>>>
>>>>>>>>
>>>>>>>> When we examine the infinite set of every HHH/DDD pair such that:
>>>>>>>> HHH₁ one step of DDD is correctly emulated by HHH.
>>>>>>>> HHH₂ two steps of DDD are correctly emulated by HHH.
>>>>>>>> HHH₃ three steps of DDD are correctly emulated by HHH.
>>>>>>>> ...
>>>>>>>> HHH∞ The emulation of DDD by HHH never stops running.
>>>>>>>
>>>>>>> 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.
>>>>>>>
>>>>>>>>
>>>>>>>> The above specifies the infinite set of every HHH/DDD pair
>>>>>>>> where 1 to infinity steps of DDD are correctly emulated by HHH.
>>>>>>>>
>>>>>>>> No DDD instance of each HHH/DDD pair ever reaches past its
>>>>>>>> own machine address of 0000216b and halts.
>>>>>>>
>>>>>>> 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.
>>>>>>>
>>>>>>>
>>>>>>>>
>>>>>>>> Thus each HHH element of the above infinite set of HHH/DDD
>>>>>>>> pairs is necessarily correct to reject its DDD as non-halting.
>>>>>>>>
>>>>>>>
>>>>>>> Nope.
>>>>>>>
>>>>>>> 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.
>>>>>>
>>>>>> *Here is the gist of my proof it is irrefutable*
>>>>>> 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.
>>>>>>
>>>>>> *No double-talk and weasel words can overcome that*
>>>>>>
>>>>>
>>>>> 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.
>>>>
>>>
>>> You have a wrong understanding of the semantics of the x86 language. 
>>> You think that the x86 language specifies that skipping instructions 
>>> do not change the behaviour of a program.
>>
>> You have the wrong understanding of a decider.
> 
> You seem to have a wrong understanding of a decider.
> Calling a program a decider does not make it halt.
> 

Calling a program a decider places a requirement
on its behavior: that it must halt.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer