Re: D correctly simulated by H proved for THREE YEARS --- rewritten

On 6/11/24 11:34 PM, olcott wrote:

On 6/11/2024 9:32 PM, Richard Damon wrote:

On 6/11/24 8:57 PM, olcott wrote:

On 6/11/2024 7:20 PM, Python wrote:

Le 12/06/2024 à 01:23, olcott a écrit :
...

It turns out that by the generic definition of a decider
what the directly executed D(D) does is not any of the
business of H.

>
LOL
>

>
There are no finite string transformations from the input
to H to the behavior of D(D), thus the behavior of D(D)
is irrelevant.
>

>
Of course there is.
>
That is exactly what the definition of a UTM is.
>

 Unless we are as concrete as the x86 language truth slips
though the cracks of vagueness.
 Show each step of DDD correctly simulated by HH such that
DDD terminates normally.

WHy? I never claimed that to be true.
The lack of finding a couter example doesn't prove that no counter example exists, it might just not be discovered.
Your question shows that you don't understand that essential property of logic.
Your reliance on Strawmen, Red Herrings, and other dishonest dodges just shows that you don't have any actual logical argument foryou poosition, so you are going to try deceit.

 typedef void (*ptr)();
u32 HH(ptr P, ptr I);
 void DDD(int (*x)())
{
   HH(x, x);
}
 int main()
{
   HH(DDD,DDD);
}
 _DDD()
[00001df3] 55               push ebp
[00001df4] 8bec             mov ebp,esp
[00001df6] 8b4508           mov eax,[ebp+08]
[00001df9] 50               push eax
[00001dfa] 8b4d08           mov ecx,[ebp+08]
[00001dfd] 51               push ecx
[00001dfe] e830f5ffff       call 00001333
[00001e03] 83c408           add esp,+08
[00001e06] 5d               pop ebp
[00001e07] c3               ret
Size in bytes:(0021) [00001e07]