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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: People are still trying to get away with disagreeing with the
 semantics of the x86 language
Date: Thu, 4 Jul 2024 12:52:17 -0400
Organization: i2pn2 (i2pn.org)
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On 7/4/24 12:31 PM, olcott wrote:
> On 7/4/2024 11:26 AM, joes wrote:
>> Am Thu, 04 Jul 2024 11:03:09 -0500 schrieb olcott:
>>
>>> On 7/4/2024 10:06 AM, joes wrote:
>>>> Am Thu, 04 Jul 2024 08:41:22 -0500 schrieb olcott:
>>>>> On 7/4/2024 8:26 AM, joes wrote:
>>>>>> Am Thu, 04 Jul 2024 07:46:15 -0500 schrieb olcott:
>>>>>>> On 7/4/2024 5:15 AM, joes wrote:
>>>>>>>> Am Wed, 03 Jul 2024 09:45:57 -0500 schrieb olcott:
>>>>>>>>> On 7/3/2024 9:39 AM, joes wrote:
>>>>>>>>>> Am Wed, 03 Jul 2024 08:21:40 -0500 schrieb 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.
>>>>>>>>>> Which semantics?
>>>>>>>> I repeat.
>>>>>> What x86 semantics say that HHH can’t return?
>>>> Hello?
>>
>>>>>>>>>>> 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.
>>>>>>>>>> But HHH aborts, so the cycle does end.
>>>>>>>>> As long as it is impossible for DDD correctly emulated by HHH to
>>>>>>>>> reach its own ret instruction then DDD never halts even when its
>>>>>>>>> stops running because its emulation was aborted.
>>>>>>>> HHH halts by definition. Why can’t DDD?
>>>>>>> By definition DDD calls its simulator.
>>>>>> Yes, and nothing else. So when HHH returns, so does DDD.
>>>>> *Machine address 00002174 of DDD is never reached*
>>>> Why not? Clearly HHH halts. Does it not return or what?
>>> The semantics of the x86 language proves that DDD correctly emulated by
>>> HHH cannot possibly reach its own machine address 00002183.
>> What semantics am I disagreeing with? Doesn’t HHH halt?
>>
> 
> The DDD correctly emulated by HHH in its own process
> context cannot possibly halt even if another entirely
> different instance of DDD does halt.
> 

Which just means that HHH can not emulate DDD to a final state.

That doesn't mean that the DDD being emulated can't be a program that 
reaches a final state, just that HHH doesn't emulate it far enough.

Partial Emulaitons do not directly provide evidence of future behavior, 
only the behavior of the steps that were emulated.