Re: Philosophy of Computation: Three seem to agree how emulating termination analyzers are supposed to work

Am Sun, 10 Nov 2024 15:45:37 -0600 schrieb olcott:

On 11/10/2024 3:02 PM, Richard Damon wrote:

On 11/10/24 2:28 PM, olcott wrote:

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

Right, if the correct (and thus complete) emulation of this precise
input would not halt.

That is what I have been saying for years.

If.

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 words 10/13/2022>

Which your H doesn't do.

It is a matter of objective fact H does abort its emulation and it does
reject its input D as non-halting.

And then it returns to the D that called it, which then halts anyway.
This H is the same as that one. They are the same. There is only one.
If not, H is not simulating the counterexample that calls H.

Correct simulation is defined as D is emulated by H according to the
semantics of the x86 language thus includes H emulating itself
emulating D.

And also means that it can not be aborted, as "stopping" in the middle
is not to the semantics of the x86 language.

Every H, HH, HHH, H1, HH1, and HHH1 (a) Predicts that its input would
not stop running unless aborted.
(b) Lets its input continue to run until completion.

No, it aborts.

But DDD doesn't call an "ideaized" version of HHH,

     *simulated D would never stop running unless aborted*
     has ALWAYS been this idealized input.

Ah no, that is not the counterexample. No disagreement with that.
We are not talking about the "idealised" version, but about
the actual, literal description of D, which calls H, which DOES
IN FACT abort.

*Breaking that down into its key element*
 > [This bounded HHH] must CORRECTLY determine what an unbounded
 > emulation of that input would do...

Problem is, the input under scrutiny changes along with HHH.

When that input is unbounded that means it is never aborted at any
level, otherwise it is bounded at some level thus not unbounded.

No, because there aren't "levels" of emulation under consideration
here.

     *simulated D would never stop running unless aborted*
Has always involved levels of simulation when H emulates itself
emulating D

Apparently D doesn’t call (or H doesn’t simulate) H, which aborts,
but rather H1, which doesn’t. The interesting case is H simulating
the actual input, which includes simulating the same abort check
itself is using.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.