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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: Mon, 15 Jul 2024 08:41:17 -0500
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On 7/15/2024 3:15 AM, Mikko wrote:
> On 2024-07-14 14:49:30 +0000, olcott said:
> 
>> 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.
> 
> Placing a requirement that the program must halt does not make it halt.
> 

It requires it to halt on inputs in its domain or it does
not meet its own design spec.

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