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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sat, 29 Mar 2025 06:55:10 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <b76458c4190da6acac7bfb38487104022fbda2f7@i2pn2.org>
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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.

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.

Which your program doesn't do.

Sorry, but you are just admitting to being a SRUPID LIAR.


Since your program doesn't meet your own requirements as stated, you are 
just a liar. Your problem seems to be that you don't understand many of 
the basic words you use like "correct" or "obeying", or what the x86 
language actually is.

Sorry, you are just confirming your reservation at the Lake of Fire hotel.

> 
>> 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.
>>
>>
> 
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