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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Who knows that DDD correctly simulated by HHH cannot possibly
 reach its own return instruction final state? BUT ONLY that DDD
Date: Sat, 3 Aug 2024 16:56:59 -0500
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On 8/3/2024 4:48 PM, Richard Damon wrote:
> On 8/3/24 4:52 PM, olcott wrote:
>> On 8/3/2024 3:45 PM, Richard Damon wrote:
>>> On 8/3/24 4:14 PM, olcott wrote:
>>>> On 8/3/2024 3:00 PM, Richard Damon wrote:
>>>>> 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.
>>>>>
>>>>
>>>> The x86 semantics says that
>>>> *DDD correctly emulated by HHH*
>>> ...
>>>> *DDD correctly emulated by HHH*
>>>> never reaches its own halt state of
>>>> "return" in any finite
>>>> or infinite number of steps.
>>>
>>> Yes, but only for an HHH that corectly emulates its input, which 
>>> means it never aborts, and only for the DDD that calls THAT HHH.
>>
>> *No you damned liar it does not mean that*
> 
> Why not? That *IS* the definition of correct emulation as defined by the 
> concept of a UTM, which is the only definition of emulation that lets 
> you get the actual full behavior of the program given.
> 
> 
>> It means that when 0 to infinity steps of DDD are
>> correctly emulated by its corresponding HHH not a
>> single DDD ever reaches its own halt state of "return".
> 
> Where do you get that from?
> 
> is it another lie that you took out of your ass?
> 
> Partial emulation is not correct emulation for the purposes of 
> determining full behavior of a program.
> 

When every possible finite or infinite emulation of DDD
by HHH according to the semantics of the x86 language
never reaches the final state of DDD then each HHH can
correctly take a wild guess that DDD never halts.


-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer