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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: Tue, 20 Aug 2024 21:44:28 -0500
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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>
>>>>
>>>>
>>>
>>> Of course it means that, because Professoer Sipser would have 
>>> presumed that you built the machines PROPERLY, so that you COULD 
>>> think of changing THIS H to be non-aborting, while the input still 
>>> used the final version that it always uses,
>>>
>>
>> A machine cannot both abort and fail to abort an input
>> unless it modifies its own code dynamically.
> 
> 
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