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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic
Subject: Re: Simulating termination analyzers by dummies --- What does halting
 mean?
Date: Wed, 19 Jun 2024 10:18:24 +0200
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Op 18.jun.2024 om 19:25 schreef 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.
> 
> Halting is a technical term-of-the-art that corresponds
> to terminates normally. Because Turing machines are
> abstract mathematical objects there has been no notion
> of abnormal termination for a Turing machine.
> 
> We can derive a notion of abnormal termination for Turing
> machines from the standard terms-of-the-art.
> 
> 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.
> 
> 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.
> 

If the code specifies 5 iterations and the simulator simulates only 3 
iterations, it is incorrect to conclude that the repetition show 
non-halting behaviour.
Similarly, when your H, H0, or other H simulates itself, its simulation 
aborts one cycle too early and therefore the non-halting conclusion is 
incorrect.