Re: Everyone on this forum besides Keith has been a damned liar about this point

On 6/10/25 4:07 PM, olcott wrote:

On 6/10/2025 2:32 PM, joes wrote:

Am Tue, 10 Jun 2025 12:27:48 -0500 schrieb olcott:

On 6/10/2025 4:00 AM, Fred. Zwarts wrote:

Op 09.jun.2025 om 16:43 schreef olcott:

On 6/9/2025 5:31 AM, Fred. Zwarts wrote:

Op 09.jun.2025 om 06:15 schreef olcott:

On 6/8/2025 10:42 PM, dbush wrote:

>

And because your HHH does not work with the description/
specification of an algorithm, by your own admission, you're not
working on the halting problem.
>

HHH(DDD) takes a finite string of x86 instructions that specify that
HHH simulates itself simulating DDD.

>
DDD does not specify that HHH should simulate itself. It could be
simulated by HHH1, which would (as you point out) not simulate itself.
>

And HHH fails to see the specification of the x86 instructions. It
aborts before it can see how the program ends.
>

It is a verified fact that unless the outer HHH aborts its simulation
of DDD that DDD simulated by HHH the directly executed DDD() and the
directly executed HHH() would never stop running.

>
But the abort is coded in the input.

>
I corrected you on this too many times. Stopping running is not halting.
Only reaching a final halt state is halting.
That I had to tell you this several times seems to prove that you are
dishonest.

>
No, the *input* DDD calls HHH, which contains an abort, but the outer
HHH doesn't simulate it up to that point.
>

 Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
 Likewise Infinite_Loop() is never simulated to
completion BECAUSE THERE IS NO COMPLETION.
 

Sure there is, it is just unbounded.
It is just like the Natural Numbers are a complete set.
You don't understand what "complete" means here, because you mind is just too small, and you don't understand the mathematics of infinite sets.