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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Anyone that disagrees with this is not telling the truth --- V5
 --- Professor Sipser
Date: Tue, 20 Aug 2024 22:53:16 -0400
Organization: i2pn2 (i2pn.org)
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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 
>>>> used the final version that it always uses,
>>>>
========== REMAINDER OF ARTICLE TRUNCATED ==========