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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: Tue, 27 Aug 2024 10:39:40 +0300
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On 2024-08-21 01:17:44 +0000, olcott said:

> 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.
> 
> Professor Sipser would not have construed that I am referring
> to self-modifying code.

That does not really matter. Everything that can be computed with
self-modifying code can be computed without self-modifying code.

-- 
Mikko