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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 20:00:55 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <dd8d9a0ef0fd3b97d7c9ac8555f2a5e5b733d9e7@i2pn2.org>
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On 8/21/24 8:35 AM, olcott wrote:
> On 8/21/2024 6:11 AM, Richard Damon wrote:
>> 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
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