Re: Anyone that disagrees with this is not telling the truth --- V5

On 2024-08-20 02:47:49 +0000, olcott said:

*Everything that is not expressly stated below is*
*specified as unspecified*
 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)
 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

No, it does not. Above X is not a truth bearer (no verb).
Y and Z are distinct propositions abot different things,
Y about HHH∞ and Z about DDD with no obvious connection.
In addition your "basic fact" menstions HHH but not HHH∞.
It is unspecifend whether HHH or HHH∞ ever emulate or
ever abort their emulation or have aborted their emulation
and therefore whether "DDD emulated by HHH" and "DDD
emulated by HHH∞" denote anything. As DDD only calls HHH
but not HHH∞ there is no connection between X and 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.
 void EEE()
{
   HERE: goto HERE;
}
 HHHn correctly predicts the behavior of DDD the same
way that HHHn correctly predicts the behavior of EEE.

It cannot correctly predicted the same if the behavours
of DDD and EEE are different.
--
Mikko