Re: Proof that DDD specifies non-halting behavior --- point by point

On 8/13/2024 9:29 PM, Richard Damon wrote:

On 8/13/24 8:52 PM, olcott wrote:

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]
>
A simulation of N instructions of DDD by HHH according to
the semantics of the x86 language is necessarily correct.
>

 Nope, it is just the correct PARTIAL emulation of the first N instructions of DDD, and not of all of DDD,

That is what I said dufuss.

 

A correct simulation of N instructions of DDD by HHH is
sufficient to correctly predict the behavior of an unlimited
simulation.

 Nope, if a HHH returns to its caller,

*Try to show exactly how DDD emulated by HHH returns to its caller*
(the first one doesn't even have a caller)
Use the above machine language instructions to show this.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer