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

On 2024-08-19 12:34:54 +0000, olcott said:

On 8/19/2024 2:26 AM, Mikko wrote:

On 2024-08-18 12:48:32 +0000, olcott said:

 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.

 It is not an emulation of DDD if the execution differs from a
real execution of DDD.
 

 *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.

That you add "V4" in to the subject line means that your earlier
presentations were not good enough. However, there is nothing
above that would give any reason to expect anything better.
The most important undefined thing above is the problem.
There is only an attempt of a solution without a problem.
--
Mikko