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

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