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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Anyone that disagrees with this is not telling the truth --- V5
Date: Mon, 19 Aug 2024 22:50:16 -0500
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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.

> 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?

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.



-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer