Re: How do simulating termination analyzers work? ---Truth Maker Maximalism

On 7/5/2025 2:26 AM, Fred. Zwarts wrote:

Op 05.jul.2025 om 00:26 schreef olcott:

On 7/4/2025 3:45 PM, Richard Damon wrote:

On 7/4/25 3:15 PM, olcott wrote:

On 7/4/2025 2:09 PM, joes wrote:

Am Fri, 04 Jul 2025 13:37:25 -0500 schrieb olcott:

On 7/4/2025 1:23 PM, joes wrote:

Am Fri, 04 Jul 2025 12:30:43 -0500 schrieb olcott:

On 7/4/2025 8:37 AM, joes wrote:

Am Fri, 04 Jul 2025 07:16:23 -0500 schrieb olcott:

On 7/4/2025 3:55 AM, joes wrote:

>

You are effectively saying that all programs that start with a call
to HHH are the same.

The nesting is too deep to see what you are responding to.

Lol, you could have responded immediately. You know how to look up posts.
>

Yes it is, HHH should compute whether the code of DD halts when run.
You can't be thinking that is uncomputable.

Likewise we should also compute the area of a square circle with a
radius of 2.

Are you seriously suggesting that you can't compute what the code of
DDD does when executed?

Don't complain later.
>

Partial halt deciders have never been allowed to report on the
behavior of any directly executed Turing machine. Instead of this they
have used the behavior that their input machine description specifies
as a proxy.

And you think that DDD's direct execution is not specified by its
description?

I HAVE PROVEN THAT DDD CORRECTLY SIMULATED BY HHH DOES NOT HAVE THE SAME
BEHAVIOR AS DDD() THOUSANDS OF TIMES IN THE LAST THREE YEARS

No disagreement; not my question.
>

Now for the first time we see that DDD correctly simulated by HHH *IS
NOT A PROXY* for the behavior of the directly executed DDD().

Indeed, HHH does not simulate it correctly. (You can't mean that DDD is
*executed* incorrectly.)

You are using the wrong measure of correct.

So DDD specifies at least two different behaviours?
>

>
*Yes. This sums it up quite well* (its only 1.5 pages long)
https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c
>
>

>
Then your system LIES and is based on lies.
>
"Code" is deterministic, and thus every instruction when starting from the same state will always do the same thing.
>

>
That you are not bright enough to detect the recursive
simulation non terminating behavior pattern is no rebuttal
at all.
>

 There is only *finite* recursive simulation, so everybody bright enough understands that there is no non-terminating behaviour.

*It has never been whether it is finite or infinite*
It has always been: Would DDD simulated by HHH reach
its simulated final halt state in an infinite simulation?

Not understanding the difference between *finite* recursion and *infinite* recursion shows who is not bright enough.

No it shows that you are not bothering to pay close attention
or like Richard cannot remember anything that I ever said in
prior posts.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer