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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
Date: Wed, 21 Aug 2024 10:28:31 +0300
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On 2024-08-20 13:13:57 +0000, olcott said:

> 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

No, that is not what they mean. The criteria don't say anything about
predicting the behaviour of HHH∞.

-- 
Mikko