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

On 6/13/2025 4:16 AM, Fred. Zwarts wrote:

Op 12.jun.2025 om 17:54 schreef olcott:

On 6/12/2025 4:51 AM, Fred. Zwarts wrote:

Op 11.jun.2025 om 17:34 schreef olcott:

On 6/11/2025 10:04 AM, joes wrote:

Am Wed, 11 Jun 2025 09:59:46 -0500 schrieb olcott:

On 6/11/2025 9:42 AM, joes wrote:

Am Wed, 11 Jun 2025 09:11:32 -0500 schrieb olcott:

On 6/11/2025 3:29 AM, Mikko wrote:

On 2025-06-10 16:10:49 +0000, olcott said:

On 6/10/2025 7:01 AM, Mikko wrote:

On 2025-06-09 14:46:30 +0000, olcott said:

>

I am no so stupid that I require a complete simulation of a
non-terminating input.

>
Yes you are. You just express your stupidity in another way.
>

It only takes two simulations of DDD by HHH for HHH to correctly
recognize a non-halting behavior pattern.

>
Either the pattern or the recognition is incorrect.

>
DDD correctly simulated by HHH cannot possibly reach its own "return"
statement final halt state. This by itself *is* complete proof that
the input to HHH(DDD) specifies non-halting behavior.

>
It is also proof that HHH doesn't terminate.
>

The fact that HHH reaches its own "return" statement final halt state
proves that you are incorrect.

Not when simulated.
>

>
DDD correctly simulated by HHH proves beyond all
possible doubt the exact behavior that the input
to HHH(DDD) specifies.

>
No. hHH aborts prematurely.

*Counter-factual and apparently over-your-head*
>
When HHH aborts after it emulates N instructions of DDD,
this same correctly emulated DDD has never reached its
own "return" statement final halt state.
>
>

 Wrong. The input for HHH is a pointer to code, including the part specifying the abort and halt.
World-class simulators show that more then N instructions must be simulated, including those within HHH. That HHH simulates only N instructions means that the abort is premature.

It is ridiculously stupid to require a simulating termination
analyzer to simulating a non terminating input to its non-existent
end.

That HHH never reaches the end of its simulation is a failure of HHH, not a property specified in the input.

Then show ALL of the details of exactly how DDD correctly
simulated by HHH reaches its simulated "return" statement
final halt state.
It is like you are claiming that square circles exist and
refuse to draw one.
void DDD()
{
   HHH(DDD);
   return;
}
The input to HHH(DDD) specifies non-halting behavior in
that DDD correctly simulated by HHH cannot possibly reach
its own simulated "return" statement final halt state.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer