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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Anyone that disagrees with this is not telling the truth --- V5 --- Professor Sipser
Date: Wed, 21 Aug 2024 11:01:17 +0300
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On 2024-08-21 03:01:38 +0000, olcott said:

> 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 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.
>>>> 
>>>> 
>>>> Right, so HHH must do just one of them, and the DDD that calls it will 
>>>> act differently based on which one it is.
>>>> 
>>>>> 
>>>>> Professor Sipser would not have construed that I am referring
>>>>> to self-modifying code.
>>>> 
>>>> Nope, just giving the exact same input to two different version of the 
>>>> decider, the one that doesn't abort as the hypothetical, and the one 
>>>> that does as the actual (if the hypothical one doesn't halt).
>>>> 
>>>> The key is that both get the EXACT SAME input, the DDD that calls the 
>>>> HHH that does abort.
>>>> 
========== REMAINDER OF ARTICLE TRUNCATED ==========