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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 18:28:16 -0500
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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>


-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer