Re: Anyone that disagrees with this is not telling the truth --- V3

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Sujet : Re: Anyone that disagrees with this is not telling the truth --- V3
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 19. Aug 2024, 12:50:30
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v9vbi6$2rjt1$1@dont-email.me>
References : 1 2 3 4
User-Agent : Mozilla Thunderbird
On 8/18/2024 8:15 PM, Richard Damon wrote:
On 8/18/24 9:02 PM, olcott wrote:
On 8/18/2024 7:51 AM, olcott wrote:
On 8/17/2024 7:29 AM, olcott wrote:
void DDD()
{
   HHH(DDD);
}
>
_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)
>
>
*The essence of the above verbal claim is translated into logic*
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
(X ∧ Y) ↔ Z
>
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.
>
>
>
All attempted rebuttals must take the form
of proving that (X ∧ Y) ↔ Z is not true
>
 No, it is pointing out that by the precise terms of your claims, X can't happen, as there isn't a full DDD to emulate.
 
You are simply ignoring one of the premises.

And, if you fix that, that the results only apply if in fact, *THE* HHH never aborts its simulation, and we can't be thinking about two different HHHs at a time, since they are in the same location of menory,
 
Yes this is the HHH.

Thus, HHH can't claim that if was something it wasn't the input wouldn't halt, as the input is only this input becuase HHH is what it is.
 
void EEE()
{
   HERE: goto HERE;
}
HHHn predicts the behavior of DDD∞ the same
way that HHHn predicts the behavior of EEE∞.

That this breaks your next steps, and thus you can't accept the limitations just shows how broken your proof is,
 IF we take you fix, which does require you rewording your claims, and you can't accept the second restriction, that just proves that your full "proof" is just based on some invalid steps.
 That we can be smart enough to see your error before you try to show it, just shows how bad your lies are.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Date Sujet#  Auteur
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