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

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.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer