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From: joes <noreply@example.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Simulating termination analyzers by dummies --- What does halting
 mean?
Date: Tue, 18 Jun 2024 17:57:14 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Tue, 18 Jun 2024 12:25:44 -0500 schrieb olcott:
> On 6/18/2024 12:06 PM, joes wrote:
> void DDD()
> {
>    H0(DDD);
> }
> DDD correctly simulated by any H0 cannot possibly halt.
>> DDD halts iff H0 halts.
So H0 returns "doesn't halt" to DDD, which then stops running,
so H0 should have returned "halts".

> Some TM's loop and thus never stop running, this is classical
> non-halting behavior. UTM's simulate Turing machine descriptions.
> This is the same thing as an interpreter interpreting the source-code of
> a program.
Some TMs do not loop and do not halt.

> A UTM can be adapted so that it only simulates a fixed number of
> iterations of an input that loops. When this UTM stops simulating this
> Turing machine description we cannot correctly say that this looping
> input halted.
Yes. We also cannot say that that input was simulated correctly.

-- 
joes