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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Anyone that disagrees with this is not telling the truth --- V5
Date: Tue, 20 Aug 2024 08:45:25 -0000 (UTC)
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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.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.