Re: Anyone that disagrees with this is not telling the truth ---- V4

On 8/19/2024 5:17 AM, Fred. Zwarts wrote:
*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
My claim boils down to this: (X ∧ Y) ↔ Z
void EEE()
{
   HERE: goto HERE;
}
HHHn correctly predicts the behavior of DDD the same
way that HHHn correctly predicts the behavior of EEE.

>

 Which proves that the simulation failed to reach the end. This makes the simulation incomplete and therefore incorrect.
The simulating HHH is programmed to abort and halt. The simulated HHH should behave exactly in the same way, so no cheating with the Root variable is allowed.
The the simulating HHH aborts when the simulated HHH has only one cycle to go, after which it would also abort and halt, but the simulating HHH failed to reach this end.

I made my claim more precise.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer