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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sat, 29 Mar 2025 11:45:16 -0500
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On 3/29/2025 5:55 AM, Richard Damon wrote:
> On 3/28/25 11:41 PM, olcott wrote:
>> On 3/28/2025 9:31 PM, Richard Damon wrote:
>>> On 3/28/25 10:08 PM, olcott wrote:
>>>> On 3/28/2025 8:05 PM, Richard Damon wrote:
>>>>> On 3/28/25 6:41 PM, olcott wrote:
>>>>>> On 3/28/2025 4:58 PM, Richard Damon wrote:
>>>>>>> On 3/28/25 2:13 PM, olcott wrote:
>>>>>>>> On 3/28/2025 8:50 AM, Richard Damon wrote:
>>>>>>>>> On 3/27/25 10:11 PM, olcott wrote:
>>>>>>>>>> On 3/27/2025 9:02 PM, Richard Damon wrote:
>>>>>>>>>>> On 3/27/25 9:10 PM, olcott wrote:
>>>>>>>>>>>> On 3/27/2025 7:47 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 3/27/25 8:11 PM, olcott wrote:
>>>>>>>>>>>>>> On 3/27/2025 4:56 PM, joes wrote:
>>>>>>>>>>>>>>> Am Thu, 27 Mar 2025 13:10:46 -0500 schrieb olcott:
>>>>>>>>>>>>>>>> On 3/27/2025 6:02 AM, Richard Damon wrote:
>>>>>>>>>>>>>>>>> On 3/26/25 11:47 PM, olcott wrote:
>>>>>>>>>>>>>>>>>> On 3/26/2025 10:28 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>> On 3/26/25 11:09 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>> On 3/26/2025 8:22 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Non-Halting is that the machine won't reach its 
>>>>>>>>>>>>>>>>>>>>> final staste even
>>>>>>>>>>>>>>>>>>>>> if an unbounded number of steps are emulated. Since 
>>>>>>>>>>>>>>>>>>>>> HHH doesn't do
>>>>>>>>>>>>>>>>>>>>> that, it isn't showing non-halting.
>>>>>>>>>>>>>>>>>>>> DDD emulated by any HHH will never reach its final 
>>>>>>>>>>>>>>>>>>>> state in an
>>>>>>>>>>>>>>>>>>>> unbounded number of steps.
>>>>>>>>>>>>>>>>>>> But DDD emulated by an actually correct emulator will,
>>>>>>>>>>>>>>>>>> If you were not intentionally persisting in a lie you 
>>>>>>>>>>>>>>>>>> would
>>>>>>>>>>>>>>>>>> acknowledge the dead obvious that DDD emulated by HHH 
>>>>>>>>>>>>>>>>>> according to the
>>>>>>>>>>>>>>>>>> semantics of the x86 language cannot possibly 
>>>>>>>>>>>>>>>>>> correctly reach its
>>>>>>>>>>>>>>>>>> final halt state.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> _DDD()
>>>>>>>>>>>>>> [00002172] 55         push ebp      ; housekeeping
>>>>>>>>>>>>>> [00002173] 8bec       mov  ebp,esp  ; housekeeping
>>>>>>>>>>>>>> [00002175] 6872210000 push 00002172 ; push DDD
>>>>>>>>>>>>>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>>>>>>>>>>>>>> [0000217f] 83c404     add  esp,+04
>>>>>>>>>>>>>> [00002182] 5d         pop  ebp
>>>>>>>>>>>>>> [00002183] c3         ret
>>>>>>>>>>>>>> Size in bytes:(0018) [00002183]
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Yes, HHH is not a correct simulator.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You say that it is not a correct simulator on the basis
>>>>>>>>>>>>>> of your ignorance of the x86 language that conclusively
>>>>>>>>>>>>>> proves that HHH does correctly simulate the first four
>>>>>>>>>>>>>> instructions of DDD and correctly simulates itself
>>>>>>>>>>>>>> simulating the first four instructions of DDD.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> It isn't a correct simulator, 
>>>>>>>>>>>>
>>>>>>>>>>>> You know that you are lying about this or you would
>>>>>>>>>>>> show how DDD emulated by HHH would reach its final state
>>>>>>>>>>>> ACCORDING TO THE SEMANTICS OF THE X86 LANGUAGE.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> It can't be, because your HHH doesn't meet your requirement.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> You cannot show that because you know you are lying about that.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Sure we can, make a main that directly calls HHH and then DDD, 
>>>>>>>>> then call HHH1(DDD)
>>>>>>>>>
>>>>>>>>> That HHH will return 0, saying that DDD is non-halting, but the 
>>>>>>>>> DDD wll return, showing that DDD is halting.
>>>>>>>>>
>>>>>>>>> Look at the trace that HHH generates, and that HHH1 generates, 
>>>>>>>>> HHH's will be a subset of the trace that HHH1 generates, 
>>>>>>>>> showing that it is NOT proof that this program is non-halting 
>>>>>>>>> as that exact same initial segment halts.
>>>>>>>>>
>>>>>>>>> Your argument about changing HHH shows that it doesn't halt is 
>>>>>>>>> just invalid, as then you either changed the input, or 
>>>>>>>>> demonstrated that you input was a class error as it didn't 
>>>>>>>>> contain the COMPLETE representation of the code of DDD.
>>>>>>>>>
>>>>>>>>> Sorry, This is what you have been told for years, but you 
>>>>>>>>> refuse to look at the truth, because you have been brainwashed 
>>>>>>>>> by your lies.
>>>>>>>>>
>>>>>>>>> Look
>>>>>>>>
>>>>>>>> I can't understand how that confused mess addresses
>>>>>>>> the point of this thread:
>>>>>>>>
>>>>>>>> It is a verified fact that the finite string of machine
>>>>>>>> code of DDD emulated by HHH according to the semantics of
>>>>>>>> the x86 language has different behavior than DDD emulated
>>>>>>>> by HHH1 according to the semantics of the x86 language.
>>>>>>>>
>>>>>>>
>>>>>>> Where did you "verify" that LIE.
>>>>>>>
>>>>>>> You claim fails the simple test:
>>>>>>>
>>>>>>> What is the first instruction actually correctly emulated by the 
>>>>>>> rules of the x86 language by HHH and HHH1 that had a different 
>>>>>>> result.
>>>>>>>
>>>>>>
>>>>>> When DDD emulated by HHH calls HHH(DDD) this call NEVER returns.
>>>>>
>>>>> Only because your HHH doesn't do a correct emulation. PERIOD,
>>>>>
>>>>
>>>> I am defining a correct emulation as obeying the semantics
>>>> of the x86 language and you are defining it to disagree
>>>> with this semantics thus proving that you know you are lying.
>>>>
>>>>
>>>
>>> Right, 
>>
>> I am glad that you finally admitted that you are lying.
> 
> Sorry, but you just demonstrated that you are just a liar.
> 

My whole sentence claimed that "you know you are lying"
and you said "right".


> I didn't agree to your actual claim, just your stated defintion that you 
> don't hold to. I said:
> 
> 
> Right, and that means you don't stop until the process does.
> 

You already know that DDD emulated by HHH according to the
semantics of the x86 language can not possibly reach its
final halt state and halt.

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer