Re: This function proves that only the outermost HHH examines the execution trace

Op 01.aug.2024 om 14:03 schreef olcott:

On 8/1/2024 2:52 AM, Mikko wrote:

On 2024-07-31 17:33:38 +0000, olcott said:
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On 7/31/2024 4:15 AM, Mikko wrote:

On 2024-07-30 23:40:21 +0000, olcott said:
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On 7/30/2024 2:00 AM, Mikko wrote:

On 2024-07-29 16:50:53 +0000, olcott said:
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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.

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If you had suffient understanding of x86 language and correctness
you would know that DDD is incorrectly emnulated by HHH.

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This is only seems that way because every reviewer makes sure
to ignore one aspect of the basis of another.

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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.

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*It is not OK to ignore*
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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
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     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>
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for DDD correctly emulated by HHH until...

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It is as Sipser does not say whether DDD is correctly simulated by HHH
or what would constitute a correct simulation.
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This has already been fully established elsewhere.

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You have never shown any proof about either "correctly".
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When instructions are executed/emulated according to the
semantics of the x86 language then they are executed/emulated
correctly.
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 But only those instructions. A halting program is simulated correctly if no instructions are skipped.