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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Who knows that DDD correctly simulated by HHH cannot possibly
 reach its own return instruction final state?
Date: Sat, 3 Aug 2024 16:00:01 -0400
Organization: i2pn2 (i2pn.org)
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On 8/3/24 3:06 PM, olcott wrote:
> On 8/3/2024 1:58 PM, Richard Damon wrote:
>> On 8/3/24 2:33 PM, olcott wrote:
>>> On 8/3/2024 1:09 PM, Richard Damon wrote:
>>>> On 8/3/24 1:58 PM, olcott wrote:
>>>>> Every DDD correctly emulated by any HHH for a finite or
>>>>> infinite number of steps never reaches its own "return"
>>>>> halt state.
>>>>>
>>>>
>>>> Nope. And you statment is just a incoherent statement, as no partial 
>>>> simulaitoni for a finite number of steps is "correct".
>>>>
>>>
>>> In other words you are trying to get away with saying that
>>> when N instructions are correctly emulated by HHH that none
>>> of these correctly emulated instructions were correctly emulated.
>>
>> No, I am saying that the result is NOT the final result that the x86 
>> semantics says will happen, because the x86 semantics says it does not 
>> stop therme
>>
> 
> The x86 semantics says that DDD correctly emulated by HHH
> never reaches its own halt state of "return" in any finite
> or infinite number of steps.
> 
> 

But only if HHH DOES correct emulation that never aborts.

If HHH does abort and return, then the x86 semantics says that DDD will 
reach its return statement.

You need to look at the whole program DDD to determine its behavior, and 
thus can't change HHH in the middle.

You have to start from truths, and not lies. You can't say that HHH does 
a correct emulation when it doesn't, and you have to use the definition 
that apply, and thus, correct simulation means one that doesn't abort.

It is impossible to have an HHH that both does a correct simulation and 
also aborts. Your assumption of this just proves your logic is flawed.