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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
Date: Tue, 20 Aug 2024 08:09:09 -0500
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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


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