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

Op 14.aug.2024 om 15:18 schreef olcott:

On 8/14/2024 4:09 AM, Fred. Zwarts wrote:

Op 14.aug.2024 om 02:52 schreef olcott:

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]
>

>
Again the same joke? It seems you are short of memory.
>

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

>
It is only a correct start of an incomplete simulation.
>

 When one instruction is emulated completely then this
is a complete emulation of one instruction.

That is what I said: It is the correct start of the simulation of a program. But for the correct simulation of the program we need more.

 

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

>
It is not,

 When one instruction is emulated completely then this
is a complete emulation of one instruction.
Until you agree with that we are dead in the water.

I agreed already that the correct simulation of one instruction is the correct start of the simulation of a program. But for the correct simulation of the program we need more.