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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Anyone that disagrees with this is not telling the truth --- V5
Date: Tue, 20 Aug 2024 12:29:31 +0200
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Op 20.aug.2024 om 06:33 schreef 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.
>>>>
>>>>>
>>>>> 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.
>>
> 
> All of the DDD have identical bytes it is only the HHH that varies.
> HHHn(DDD) predicts the behavior of HHH∞(DDD).
Not all HHH can be at the same memory at the same time. When HHHn is in 
the memory, then DDD calls HHHn, not HHH∞.
When HHHn is doing the simulation, HHHn is in that memory, therefore, it 
should simulate HHHn, not HHH∞.
They cannot be at the same memory location at the same time, unless you 
are cheating with the Root variable to switch between HHHn and HHH∞, 
which causes HHHn to process the non-input HHH∞ instead of the input HHHn.