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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: People are still trying to get away with disagreeing with the
 semantics of the x86 language
Date: Wed, 3 Jul 2024 17:59:34 +0200
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Op 03.jul.2024 om 15:21 schreef olcott:
> On 7/3/2024 3:26 AM, Fred. Zwarts wrote:
>> Op 02.jul.2024 om 21:48 schreef olcott:
>>> On 7/2/2024 2:22 PM, Fred. Zwarts wrote:
>>>> Op 02.jul.2024 om 20:43 schreef olcott:
>>>>> On 7/2/2024 1:59 AM, Mikko wrote:
>>>>>> On 2024-07-01 12:44:57 +0000, olcott said:
>>>>>>
>>>>>>> On 7/1/2024 1:05 AM, Mikko wrote:
>>>>>>>> On 2024-06-30 17:18:09 +0000, olcott said:
>>>>>>>>>
>>>>>>>>> Richard just said that he affirms that when DDD correctly
>>>>>>>>> simulated by HHH calls HHH(DDD) that this call returns even
>>>>>>>>> though the semantics of the x86 language disagrees.
>>>>>>>>>
>>>>>>>>> On 6/30/2024 7:34 AM, Richard Damon wrote:
>>>>>>>>>  > It is still true that the xemantics of the x86
>>>>>>>>>  > language define the behavior of a set of bytes,
>>>>>>>>>  > as the behavior when you ACTUALLY RUN THEM,
>>>>>>>>>  > and nothing else.
>>>>>>>>>
>>>>>>>>> _DDD()
>>>>>>>>> [00002172] 55               push ebp      ; housekeeping
>>>>>>>>> [00002173] 8bec             mov ebp,esp   ; housekeeping
>>>>>>>>> [00002175] 6872210000       push 00002172 ; push DDD
>>>>>>>>> [0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
>>>>>>>>> [0000217f] 83c404           add esp,+04
>>>>>>>>> [00002182] 5d               pop ebp
>>>>>>>>> [00002183] c3               ret
>>>>>>>>> Size in bytes:(0018) [00002183]
>>>>>>>>>
>>>>>>>>> Richard thinks that he can get away with disagreeing with this
>>>>>>>>> verified fact:
>>>>>>>>>
>>>>>>>>> The call from DDD to HHH(DDD) when N steps of DDD are correctly
>>>>>>>>> emulated by any pure function x86 emulator HHH cannot possibly
>>>>>>>>> return.
>>>>>>>>
>>>>>>>> It is your HHH so you should know whether it returns. Others may
>>>>>>>> have wrong impression about it if they have trusted your lies.
>>>>>>>
>>>>>>> I have never lied about this.
>>>>>>
>>>>>> At least you have claimed more than proven.
>>>>>>
>>>>>>> _DDD()
>>>>>>> [00002172] 55               push ebp      ; housekeeping
>>>>>>> [00002173] 8bec             mov ebp,esp   ; housekeeping
>>>>>>> [00002175] 6872210000       push 00002172 ; push DDD
>>>>>>> [0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
>>>>>>> [0000217f] 83c404           add esp,+04
>>>>>>> [00002182] 5d               pop ebp
>>>>>>> [00002183] c3               ret
>>>>>>> Size in bytes:(0018) [00002183]
>>>>>>>
>>>>>>> DDD is correctly emulated by HHH which calls an
>>>>>>> emulated HHH(DDD) to repeat the process until aborted.
>>>>>>
>>>>>> The correctness remain unproven.
>>>>>>
>>>>>
>>>>> IT IS PROVEN BY THE SEMANTICS OF THE X86 LANGUAGE
>>>>> THAT YOU REMAIN WILLFULLY IGNORANT OF SEMANTICS OF
>>>>> THE X86 LANGUAGE DOES NOT MEAN IT HAS NOT BEEN PROVEN.
>>>>>
>>>>
>>>> Please, point to the paragraph in the specification of the X86 
>>>> language that says that a two cycle recursion should be aborted 
>>>> after one cycle.
>>>> Claiming that the abort is related to the x86 language is apparently 
>>>> wilfully incorrect.
>>>>
>>>>
>>>
>>> I am not going to show you the trace of the Peano axioms
>>> that prove the 2 + 3 = 5, if you disagree you are a liar
>>> or an ignoramus.
>>>
>>
>> This change of subject does not hide that your claim that the x86 
>> language proves your claim is incorrect as a verified fact.
>>
> 
> So you opted for liar then:
> 
> _DDD()
> [00002172] 55               push ebp      ; housekeeping
> [00002173] 8bec             mov ebp,esp   ; housekeeping
> [00002175] 6872210000       push 00002172 ; push DDD
> [0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
> [0000217f] 83c404           add esp,+04
> [00002182] 5d               pop ebp
> [00002183] c3               ret
> Size in bytes:(0018) [00002183]
> 
> DDD correctly emulated by HHH calls an emulated HHH(DDD)
> that emulates DDD that calls an emulated HHH(DDD)
> in a cycle that cannot end unless aborted.
> 

HHH aborts after two cycles. So, its simulation needs only two cycles to 
make it return. However, HHH is not able to simulate two cycles. 
Therefore, it aborts one cycle too soon. This shows that HHH cannot 
correctly simulate itself.
The 'unless' phrase is misleading, because it suggests that HHH will not 
abort. But dreaming of a HHH that does not abort is irrelevant. HHH 
*does* abort. Therefore, the simulation of HHH does not need to be 
aborted. This is shown to be correct when HHH is simulated by a 
simulator that does not abort too soon.
Therefore, you only prove that HHH cannot be simulated correctly by itself.
Don't start talking about the x86 language, because the x86 language 
does not define when an abort should occur.