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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 22:01:38 -0500
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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>
>>>>>>
>>>>>>
>>>>>
>>>>> 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 
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