Re: D simulated by H unproved for THREE YEARS ---

On 6/10/2024 12:06 PM, joes wrote:

Am Mon, 10 Jun 2024 10:36:44 -0500 schrieb olcott:

On 6/10/2024 10:25 AM, joes wrote:

Am Mon, 10 Jun 2024 09:36:09 -0500 schrieb olcott:

On 6/10/2024 3:35 AM, joes wrote:

Am Sun, 09 Jun 2024 22:54:52 -0500 schrieb olcott:

On 5/29/2021 2:26 PM, olcott wrote:

>

THE ONLY POSSIBLE WAY for D simulated by H to have the same behavior
as the directly executed D(D) is for the instructions of D to be
incorrectly simulated by H (details provided below).

The only correct simulation must simulate incorrectly? Wat.

Try carefully studying those words again and again until you see how
your paraphrase is wrong.

Perhaps you can paraphrase it better?

A simulator MUST have the same behaviour.
 

That you do not look at my proof or have insufficient understanding
of the ax8 language counts for no rebuttal at all.
Saying that I am wrong without looking at my proof or failing
to understand the x86 language is less than no rebuttal at all.

H does not ignore that instruction and simulates itself simulating D.
The directly executed D(D) reaps the benefit of D simulated by H
proving that *its input never halts*

If that simulation is right, D(D) never halts.
 

Because the H(D,D) that D(D) calls recognizes the its input DOES NOT
HALT, it aborts the simulation of this input causing the directly
executed D(D) to halt.

Simulating something most definitely does NOT cause any change in
its behaviour.
If the abortion causes D to halt, how could H detect D not to?
If D halts, it does so whether it is simulated or not.

It is two different levels of what is essentially infinite
recursion that have different behavior.
If you don't understand infinite recursion then you will
never understand that:
when the second recursive call of infinite recursion must
be aborted to prevent infinite execution then first recursive
call will halt.
Failing to understand infinite recursion is no rebuttal at all.

How can H report "non-halting" when D(D) halts?

 

It might simply be over your head.
That is why I asked if you understand infinite recursion.

I proved that D simulated by H can only have the same behavior as the
directly executed D(D) when D is simulated by H incorrectly.

You've got it the wrong way around. A simulation must have the same
behaviour.

I proved otherwise. Maybe the proof is simply over-your-head?

Your simulator does not simulate if it has different behaviour.
 

It might simply be over your head.
That is why I asked if you understand infinite recursion.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer