Re: How Partial Simulations correctly determine non-halting ---Should I quit Richard at this point?

Am Sat, 08 Jun 2024 12:10:33 -0500 schrieb olcott:

On 6/8/2024 11:03 AM, Richard Damon wrote:

On 6/8/24 11:32 AM, olcott wrote:

On 6/8/2024 10:15 AM, Richard Damon wrote:

On 6/8/24 11:07 AM, olcott wrote:

On 6/8/2024 9:54 AM, Richard Damon wrote:

On 6/8/24 10:20 AM, olcott wrote:

On 6/8/2024 9:10 AM, Richard Damon wrote:

By your code, the simulator will "Debug Step" those instructions.
>

The underlying details of one HH are irrelevant when I reference an
infinite set:\

What are all the other HH?

*Ben already agreed that the criteria that*
*Professor Sipser agreed to have been met*
 
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.

What do I care what Ben thinks (sorry).
And Sipser's "criterion" is not at all disputed.

Yes, P(P) won't halt if the code of H won't abort it.
 

thus meeting the Sipser approved criteria.
 

But if the code of H does abort it, it will halt.
 

D is accountable for the behavior of its actual input and not freaking
accountable for any damn thing else in the whole freaking universe.

Halt deciders compute the mapping FROM THEIR INPUTS BASED ON THE ACTUAL
BEHAVIOR THAT THIS INPUT SPECIFIES.
This input specifies the behavior of DD correctly simulated by HH.

That everyone wants to avoid looking at my conclusive proof that DD
correctly simulated by HH has provably different behavior than the
direct executed DD(DD) is dishonest.

Dishonest is claiming a simulator gets to a different conclusion.

Thus, if the question actually WAS "Will the template P(P) Halt if the
decider it is pair with doesn't abort its simulation", then he agreed
you could answer No.
But that isn't the Halting Question, so it isn't the correct answer to
the question that H is supposed to answer.

--
joes