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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: This function proves that only the outermost HHH examines the execution trace
Date: Thu, 1 Aug 2024 10:52:39 +0300
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On 2024-07-31 17:33:38 +0000, olcott said:

> On 7/31/2024 4:15 AM, Mikko wrote:
>> On 2024-07-30 23:40:21 +0000, olcott said:
>> 
>>> On 7/30/2024 2:00 AM, Mikko wrote:
>>>> On 2024-07-29 16:50:53 +0000, olcott said:
>>>> 
>>>>> On 7/28/2024 3:59 AM, Mikko wrote:
>>>>>> On 2024-07-27 20:05:31 +0000, olcott said:
>>>>>>> If you had sufficient understanding of the x86 language
>>>>>>> you would know that DDD is correctly emulated by HHH.
>>>>>> 
>>>>>> If you had suffient understanding of x86 language and correctness
>>>>>> you would know that DDD is incorrectly emnulated by HHH.
>>>>> 
>>>>> This is only seems that way because every reviewer makes sure
>>>>> to ignore one aspect of the basis of another.
>>>> 
>>>> It is perfectly OK to ignore irrelevant details. A relevant detail
>>>> is the meaning of the word "emulate" as that determines what is a
>>>> correct emulation and what is not.
>>> 
>>> *It is not OK to ignore*
>>> 
>>> This algorithm is used by all the simulating termination analyzers:
>>> <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>>>      If simulating halt decider *H correctly simulates its input D*
>>>      *until H correctly determines that its simulated D would never*
>>>      *stop running unless aborted* then
>>> 
>>>      H can abort its simulation of D and correctly report that D
>>>      specifies a non-halting sequence of configurations.
>>> </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>>> 
>>> for DDD correctly emulated by HHH until...
>> 
>> It is as Sipser does not say whether DDD is correctly simulated by HHH
>> or what would constitute a correct simulation.
>> 
> 
> This has already been fully established elsewhere.

You have never shown any proof about either "correctly".

-- 
Mikko