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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Philosophy of Computation: Three seem to agree how emulating
 termination analyzers are supposed to work
Date: Sun, 10 Nov 2024 22:53:53 -0000 (UTC)
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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.