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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 14:32:46 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <f8c3f661b1ca19a7e4ea7837aaa161f8ad651a11@i2pn2.org>
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On 3/29/25 12:45 PM, olcott wrote:
> 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".

No, that is just a lie.

You said:
> 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. 

And 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.

Note, you are just attempting to LIE by taking statements out of contest.

My "Right" was an agreement to your definition, that it *IS* to agree to 
the semantics of the x86 language, I then clarify what that means, and 
the part which you don't follow, and then point out that your claim is 
just a lie.
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