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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: Wed, 21 Aug 2024 07:11:06 -0400
Organization: i2pn2 (i2pn.org)
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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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