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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 10:36:33 +0300
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On 2024-08-20 23:28:16 +0000, olcott said:

> 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:

Professor Sipser did not agree with any half, only the complete sentence.

>      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>


-- 
Mikko