Re: Boilerplate Reply -- different simulation

Am Sat, 22 Jun 2024 08:08:51 -0500 schrieb olcott:

On 6/22/2024 3:47 AM, joes wrote:

Am Fri, 21 Jun 2024 23:18:50 -0500 schrieb olcott:

On 6/21/2024 11:09 PM, joes wrote:

Am Fri, 21 Jun 2024 15:52:21 -0500 schrieb olcott:

On 6/21/2024 3:00 PM, Richard Damon wrote:

On 6/21/24 3:45 PM, olcott wrote:

On 6/21/2024 2:33 PM, Richard Damon wrote:

On 6/21/24 3:19 PM, olcott wrote:

No you cannot show that the mapping for the input to H(D,D) maps to
the behavior of D(D).

If it doesn't, H is not a simulator.
The input D(D) absolutely describes the behaviour of that machine.
H just can't map it.
Either H is not a decider or it returns.

H(D,D) is not free to simply assume that the call from D(D) to
H(D,D) will return.

Yes it is, because it is a decider. It (incorrectly) aborts
nonterminating inputs.

The behavior of D correctly simulated by H1 is the same as the
behavior of the directly executed D(D) because D does not call H1(D,D)
in recursive simulation.

D1 however, which calls H1(D1, D1), can't be decided by H1.

When you change the subject rather than address the point at hand I take
this to mean that you do not want an honest dialogue.

Unlike you, I replied to the points at hand. See below.

The behavior of D correctly simulated by H is NOT the same as the
behavior of D correctly simulated by H1 because D DOES call H(D,D) in
recursive simulation.

The simulation by H is then of course not correct.

What about the other points above?

Yeah, what about them?

--
Man kann mit dunklen Zahlen nicht rechnen. Für die eigentliche Mathematik
sind sie vollkommen nutzlos. --Wolfgang Mückenheim