Re: A computable function that reports on the behavior of its actual self is not allowed

On 5/13/24 9:37 AM, olcott wrote:

On 5/13/2024 3:22 AM, joes wrote:

Am Sun, 12 May 2024 18:19:12 -0500 schrieb olcott:

On 5/12/2024 5:40 PM, Richard Damon wrote:

On 5/12/24 6:21 PM, olcott wrote:

>

No decider is ever allowed to report on its own behavior thus
embedded_H as a simulating partial halt decider is NOT ALLOWED to
report on the direct execution of Ĥ ⟨Ĥ⟩ because this IS REPORTING ON
ITS OWN BEHAVIOR.

>
WHO SAYS THIS?

>
A decider must compute the mapping from an input.
Its actual self cannot possibly be an input.

Oh, it definitely can. It must decide for every machine, including
itself. How would you even detect that?
>

No decider takes an actual Turing machine as input thus no decider can
possibly take its actual self as input.

Well, an encoding of one.
>

 That is not the same, as I show below the behavior of the executed
Turing machine that contains the embedded partial halt decider is
different than the behavior of its simulated Turing Machine description.

Then your simulator is BROKEN.
Note, you have NEVER been able to show what instruction CORRECTLY SIMULSTED by your simulator differed from the actual execution, so your statement is just proven to be a known LIE.

 These two machines are one recursive simulation out-of-sync with each
other This changes their behavior relative to each other.

Nope.
Try to show it.

 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by embedded_H cannot possibly stop running
unless aborted. Whereas the behavior of Ĥ ⟨Ĥ⟩ is AFTER the
simulated input has already been aborted. These two are always
out-of-sync whenever there is a pathological relationship
between the partial halt decider and its input.

Excpet that statement is a "nonse" statement, as it is based on a condition that just doesn't happen. It doesn't matter what some other H^ using some other embedded_H does or doesn't do. What matters is what THIS H^ does

 The behavior of the directly execution machine and the correct
simulation of the machine description of this same machine is
always exactly the same:
*UNLESS THEY HAVE A PATHOLOGICAL RELATIONSHIP TO EACH OTHER*

Nope, just more of your lies.
Where is that in the DEFINITION of correct simulation?
You are just proving you don't care about the truth because you are just a pathological liar.

 

(a) The behavior of the directly executed Ĥ ⟨Ĥ⟩ is after embedded_H ⟨Ĥ⟩
⟨Ĥ⟩ has already aborted its simulation.
>
(b) The behavior of the simulated input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
is before embedded_H has aborted its simulation.
>
(c) These two behaviors (a) and (b) ARE NOT THE SAME. (a) will stop
running on its own (b) will never stop running unless aborted.

Then they are either not the same machine, or simulated incorrectly.
>