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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Anyone that disagrees with this is not telling the truth --- V5
 --- Professor Sipser
Date: Wed, 21 Aug 2024 07:35:32 -0500
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On 8/21/2024 6:11 AM, Richard Damon wrote:
> On 8/20/24 11:01 PM, olcott wrote:
>> On 8/20/2024 9:53 PM, Richard Damon wrote:
>>> On 8/20/24 10:44 PM, olcott wrote:
>>>> On 8/20/2024 8:56 PM, Richard Damon wrote:
>>>>> On 8/20/24 9:17 PM, olcott wrote:
>>>>>> On 8/20/2024 7:50 PM, Richard Damon wrote:
>>>>>>> On 8/20/24 7:28 PM, olcott wrote:
>>>>>>>> On 8/20/2024 6:18 PM, Richard Damon wrote:
>>>>>>>>> On 8/20/24 9:09 AM, olcott wrote:
>>>>>>>>>> On 8/19/2024 11:02 PM, Richard Damon wrote:
>>>>>>>>>>> On 8/19/24 11:50 PM, olcott wrote:
>>>>>>>>>>>> On 8/19/2024 10:32 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 8/19/24 10:47 PM, olcott wrote:
>>>>>>>>>>>>>> *Everything that is not expressly stated below is*
>>>>>>>>>>>>>> *specified as unspecified*
>>>>>>>>>>>>>
>>>>>>>>>>>>> Looks like you still have this same condition.
>>>>>>>>>>>>>
>>>>>>>>>>>>> I thought you said you removed it.
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> void DDD()
>>>>>>>>>>>>>> {
>>>>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>>>>    return;
>>>>>>>>>>>>>> }
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> _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]
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> *It is a basic fact that DDD emulated by HHH according to*
>>>>>>>>>>>>>> *the semantics of the x86 language cannot possibly stop*
>>>>>>>>>>>>>> *running unless aborted* (out of memory error excluded)
>>>>>>>>>>>>>
>>>>>>>>>>>>> But it can't emulate DDD correctly past 4 instructions, 
>>>>>>>>>>>>> since the 5th instruciton to emulate doesn't exist.
>>>>>>>>>>>>>
>>>>>>>>>>>>> And, you can't include the memory that holds HHH, as you 
>>>>>>>>>>>>> mention HHHn below, so that changes, but DDD, so the input 
>>>>>>>>>>>>> doesn't and thus is CAN'T be part of the input.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> X = DDD emulated by HHH∞ according to the semantics of the 
>>>>>>>>>>>>>> x86 language
>>>>>>>>>>>>>> Y = HHH∞ never aborts its emulation of DDD
>>>>>>>>>>>>>> Z = DDD never stops running
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The above claim boils down to this: (X ∧ Y) ↔ Z
>>>>>>>>>>>>>
>>>>>>>>>>>>> And neither X or Y are possible.
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> x86utm takes the compiled Halt7.obj file of this c program
>>>>>>>>>>>>>> https://github.com/plolcott/x86utm/blob/master/Halt7.c
>>>>>>>>>>>>>> Thus making all of the code of HHH directly available to
>>>>>>>>>>>>>> DDD and itself. HHH emulates itself emulating DDD.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Which is irrelevent and a LIE as if HHHn is part of the 
>>>>>>>>>>>>> input, that input needs to be DDDn
>>>>>>>>>>>>>
>>>>>>>>>>>>> And, in fact,
>>>>>>>>>>>>>
>>>>>>>>>>>>> Since, you have just explicitly introduced that all of HHHn 
>>>>>>>>>>>>> is available to HHHn when it emulates its input, that DDD 
>>>>>>>>>>>>> must actually be DDDn as it changes.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Thus, your ACTUAL claim needs to be more like:
>>>>>>>>>>>>>
>>>>>>>>>>>>> X = DDD∞ emulated by HHH∞ according to the semantics of the 
>>>>>>>>>>>>> x86 language
>>>>>>>>>>>>> Y = HHH∞ never aborts its emulation of DDD∞
>>>>>>>>>>>>> Z = DDD∞ never stops running
>>>>>>>>>>>>>
>>>>>>>>>>>>> The above claim boils down to this: (X ∧ Y) ↔ Z
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Yes that is correct.
>>>>>>>>>>>
>>>>>>>>>>> So, you only prove that the DDD∞ that calls the HHH∞ is non- 
>>>>>>>>>>> halting.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Not any of the other DDDn
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>> Your problem is that for any other DDDn / HHHn, you don't 
>>>>>>>>>>>>> have Y so you don't have Z.
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> void EEE()
>>>>>>>>>>>>>> {
>>>>>>>>>>>>>>    HERE: goto HERE;
>>>>>>>>>>>>>> }
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> HHHn correctly predicts the behavior of DDD the same
>>>>>>>>>>>>>> way that HHHn correctly predicts the behavior of EEE.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Nope, HHHn can form a valid inductive proof of the input.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>> It can't for DDDn, since when we move to HHHn+1 we no 
>>>>>>>>>>>>> longer have DDDn but DDDn+1, which is a different input.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> You already agreed that (X ∧ Y) ↔ Z is correct.
>>>>>>>>>>>> Did you do an infinite trace in your mind?
>>>>>>>>>>>
>>>>>>>>>>> But only for DDD∞, not any of the other ones.
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> If you can do it and I can do it then HHH can
>>>>>>>>>>>> do this same sort of thing. Computations are
>>>>>>>>>>>> not inherently dumber than human minds.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> But HHHn isn't given DDD∞ as its input, so that doesn't matter.
>>>>>>>>>>>
>>>>>>>>>>> HHHn is given DDDn as its input,
>>>>>>>>>>>
>>>>>>>>>>> Remeber, since you said that the input to HHH includes all 
>>>>>>>>>>> the memory, if that differs, it is a DIFFERENT input, and 
>>>>>>>>>>> needs to be so marked.
>>>>>>>>>>>
>>>>>>>>>>> You are just admittig that you are just stupid and think two 
>>>>>>>>>>> things that are different are the same.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> *attempts to use misdirection to weasel word around this are 
>>>>>>>>>> dismissed*
>>>>>>>>>> *attempts to use misdirection to weasel word around this are 
>>>>>>>>>> dismissed*
>>>>>>>>>> *attempts to use misdirection to weasel word around this are 
>>>>>>>>>> dismissed*
>>>>>>>>>>
>>>>>>>>>> <MIT Professor Sipser agreed to ONLY these verbatim words 
>>>>>>>>>> 10/13/2022>
>>>>>>>>>>      If simulating halt decider H correctly simulates its input D
>>>>>>>>>>      until H correctly determines that its simulated D would 
>>>>>>>>>> never
>>>>>>>>>>      stop running unless aborted then
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Right, so the decider needs top be able to show that its exact 
>>>>>>>>> input will not halt.
>>>>>>>>
>>>>>>>> No it cannot possibly mean that or professor Sipser
>>>>>>>> would not agreed to the second half:
>>>>>>>>
>>>>>>>>      H can abort its simulation of D and correctly report that D
>>>>>>>>      specifies a non-halting sequence of configurations.
>>>>>>>> </MIT Professor Sipser agreed to ONLY these verbatim words 
>>>>>>>> 10/13/2022>
>>>>>>>>
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