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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is correctly rejected as non-halting V2
Date: Fri, 2 Aug 2024 09:32:25 +0300
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On 2024-08-01 11:56:31 +0000, olcott said:

> On 8/1/2024 2:49 AM, Mikko wrote:
>> On 2024-07-31 17:28:38 +0000, olcott said:
>> 
>>> On 7/31/2024 2:36 AM, Mikko wrote:
>>>> On 2024-07-16 18:18:07 +0000, olcott said:
>>>> 
>>>>> On 7/16/2024 2:57 AM, Mikko wrote:
>>>>>> On 2024-07-15 13:43:34 +0000, olcott said:
>>>>>> 
>>>>>>> On 7/15/2024 3:17 AM, Mikko wrote:
>>>>>>>> On 2024-07-14 14:50:47 +0000, olcott said:
>>>>>>>> 
>>>>>>>>> On 7/14/2024 5:09 AM, Mikko wrote:
>>>>>>>>>> On 2024-07-12 14:56:05 +0000, olcott said:
>>>>>>>>>> 
>>>>>>>>>>> We stipulate that the only measure of a correct emulation is the
>>>>>>>>>>> semantics of the x86 programming language.
>>>>>>>>>>> 
>>>>>>>>>>> _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.
>>>>>>>>>>> 
>>>>>>>>>>> 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.
>>>>>>>>>>> 
>>>>>>>>>>> The above specifies the infinite set of every HHH/DDD pair
>>>>>>>>>>> where 1 to infinity steps of DDD are correctly emulated by HHH.
>>>>>>>>>> 
>>>>>>>>>> You should use the indices here, too, e.g., "where 1 to infinity steps of
>>>>>>>>>> DDD₁ are correctly emulated by HHH₃" or whatever you mean.
>>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> DDD is the exact same fixed constant finite string that
>>>>>>>>> always calls HHH at the same fixed constant machine
>>>>>>>>> address.
>>>>>>>> 
>>>>>>>> If the function called by DDD is not part of the input then the input does
>>>>>>>> not specify a behaviour and the question whether DDD halts is ill-posed.
>>>>>>>> 
>>>>>>> 
>>>>>>> We don't care about whether HHH halts. We know that
>>>>>>> HHH halts or fails to meet its design spec.
>>>>>>> 
>>>>>>> We are only seeing if DDD correctly emulated by HHH
>>>>>>> can can possibly reach its own final state.
>>>>>> 
>>>>>> HHH does not see even that. It only sees whther that it does not emulate
>>>>>> DDD to its final state.
>>>>> 
>>>>> No. HHH is not judging whether or not itself is a correct
>>>>> emulator. The semantics of the x86 instructions that emulates
>>>>> prove that its emulation is correct.
>>>> 
>>>> Semantics of x86 language alone doesn't prove anything. Only a detailed
>>>> comparison of the emulator code to the x86 semantics may prove that.
>>> 
>>> A proof is any sequence of steps such that the conclusion
>>> is a necessary consequence of its basis.
>> 
>> Only if every "step" is a sentence.
> 
> Not at all.

From the meaning of "proof" directly follows that every proof is
a sequence of sentences.

-- 
Mikko