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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: Fri, 28 Mar 2025 22:41:48 -0500
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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.

> and that means you don't stop until the process does.
> 
> Which your program doesn't do.
> 
> Sorry, but you are just admitting to being a SRUPID LIAR.
> 
> 


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