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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 11:25:25 -0400
Organization: i2pn2 (i2pn.org)
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On 7/4/24 8:46 AM, olcott wrote:
> 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.
>>
>>>>> 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 it simulator.
> 

No, it calls a simulating decider.

If HHH IS a decider, then DDD Halts.

Your problem is you LIE that the two HHHs are the same, because they 
can't be and meet your logic, or if they are, your logic is broken 
because it assumes the HHH that DDD calls acts diffferently then the HHH 
that main calls.

You, you logic is just based on illogical lies.