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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD incorrectly emulated by HHH is incorrectly rejected as
 non-halting V2
Date: Wed, 17 Jul 2024 19:56:23 -0400
Organization: i2pn2 (i2pn.org)
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On 7/17/24 9:04 AM, olcott wrote:
> On 7/17/2024 1:52 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.
>>
>> The semantics does not prove. Only a proof would prove.
>>
> 
> Nothing besides the semantics of English proves that
> a kitten is not any type of 15 story office building.
> 
> _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]
> 
> DDD emulated by HHH according to the semantic meaning of
> its x86 instructions never stop running unless aborted.
> 
> 

Wrong, EVERY DDD that is calls an HHH that any copy of it returns from 
the call HHH(DDD) to any caller makes a DDD that halts.

Yes, the HHH can never reach that state in its emulation, but DDD gets 
there.

You are just proving you are a stupid liar that doesn't know what he is 
talking about.