Re: Proof that DD correctly simulated by HH has different behavior than DD(DD) STEP(1)

On 6/9/2024 1:33 AM, Fred. Zwarts wrote:

Op 08.jun.2024 om 20:47 schreef olcott:

Before we can get to the behavior of the directly executed
DD(DD) we must first see that the Sipser approved criteria
have been met:
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D
until H correctly determines that its simulated D would never
stop running unless aborted then
>
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words10/13/2022>
>
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > I don't think that is the shell game. PO really /has/ an H
 > (it's trivial to do for this one case) that correctly determines
 > that P(P) *would* never stop running *unless* aborted.
>
Try to show how this DD correctly simulated by any HH ever
stops running without having its simulation aborted by HH.

 Stopping at your first error. So, we can focus on it. Your are asking a question that contradicts itself.

In other words the best selling author of theory of
computation textbooks doesn't have a clue?

A correct simulation of HH that aborts itself,
should simulate up to the point where the simulated HH aborts.

The outer HH always sees a longer execution trace than
the inner ones. Unless the outer one aborts none of them
abort.

That is logically impossible. So, either it is a correct simulation and then we see that the simulated HH aborts and returns, or the simulation is incorrect, because it assumes incorrectly that things that happen (abort) do not happen.
A premature conclusion.
 

_DD()
[00001e32] 55               push ebp      ; Begin DD
[00001e33] 8bec             mov ebp,esp
[00001e35] 51               push ecx
[00001e36] 8b4508           mov eax,[ebp+08]
[00001e39] 50               push eax      ; push DD
[00001e3a] 8b4d08           mov ecx,[ebp+08]
[00001e3d] 51               push ecx      ; push DD
[00001e3e] e83ff5ffff       call 00001382 ; call DD
If you can't see this then you seem to just not have
enough technical skill:
The first seven steps of DD correctly simulated by HH call
HH(DD,DD) to repeat these first seven steps.
HH then simulates itself simulating DD until this second
instance of DD calls another HH(DD,DD) to repeat these first
seven steps again.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer