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 : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theory
Date : 20. Aug 2024, 10:15:12
Autres entêtes
Organisation : -
Message-ID : <va1mr0$3b0t9$2@dont-email.me>
References : 1 2 3 4
User-Agent : Unison/2.2
On 2024-08-19 01:15:01 +0000, Richard Damon said:

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.
Nvertheless, it Z is false then (X ∧ Y) ↔ Z is still true.
--
Mikko

Date Sujet#  Auteur
7 Jul 25 o 

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