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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 08:13:57 -0500
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On 8/20/2024 3:45 AM, joes wrote:
> Am Mon, 19 Aug 2024 23:33:52 -0500 schrieb olcott:
>> 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.
> 
>>>>>> _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]
>>>>> 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.
> Changing the code, but not the address, constitutes a change.
> 
>>>>>> 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.
> 
>>>>>> 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.
>>>
>> All of the DDD have identical bytes it is only the HHH that varies.
>> HHHn(DDD) predicts the behavior of HHH∞(DDD).
>> It does this same same way that HHHn(EEE)
>> predicts the behavior of HHH∞(EEE).
> The bytes of HHH are part of DDD.
> 

*The following criteria only means*
HHHn(DDD) correctly predicts the behavior of HHH∞(EEE) and HHH∞(DDD)

<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