Re: DDD correctly emulated by H0

On 25/06/2024 14:46, Alan Mackenzie wrote:

[ Followup-To: set ]
 In comp.theory olcott <polcott333@gmail.com> wrote:

On 6/25/2024 4:22 AM, joes wrote:

Am Sat, 22 Jun 2024 13:47:24 -0500 schrieb olcott:

On 6/22/2024 1:39 PM, Fred. Zwarts wrote:

Op 21.jun.2024 om 15:21 schreef olcott:

 

When we stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language then we see that when DDD is
correctly emulated by H0 that its call to H0(DDD) cannot possibly
return.

Yes. Which is wrong, because H0 should terminate.

 [ .... ]
 

The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.

 

Until you acknowledge this is true, this is the
only thing that I am willing to talk to you about.

 I think you are talking at cross purposes.  Joes's point is that H0
should terminate because it's a decider.  You're saying that when H0 is
"correctly" emulating, it won't terminate.  I don't recall seeing anybody
arguing against that.

Hehe, everyone has an opinion on what PO is saying! :)
So here's mine:  I think PO is saying that when DDD is correctly *emulated* [not "correctly /emulating/"] then "it" will not return.
To be clear, that's saying that the /emulation/ does not get as far as the final return, i.e. outer H0 will stop emulating (aka "abort") without having emulated DDD's return statement.  PO has acknowledged that the outer H0 will return after aborting its emulation.
So what? I hear everyone asking.  So what indeed! The "behaviour" of a partial emulation is not the behaviour of the computation itself [trivially] and not what the HP is about. But PO then gets hopelessly muddled, thinking at least two wrong things:
1)  That DDD would /never/ have terminated in "1-oo steps of emulation", i.e. that DDD really doesn't halt, not simply that H0 aborted it before it returned.  (Thats muddling all the different examples of H in his head each with their own personalised (D,D) inputs into one single super-H/super-D.  Of course, the FIXED DDD under discussion /does/ in fact return in some finite number of steps.  H0 is also fixed and is coded in a way that aborts /before/ that number of steps is emulated.)
2)  PO knows that the /reason/ H0 decided to abort was that it matched his "infinite recursive emulation" pattern in the emulation trace - therefore he believes it's correct to decide non-halting because his pattern proves that.  (That's just Wrong, but PO really really really believes the pattern is sound, so that's the end of it.  He has no intention or capability of ever attempting to prove his rule is sound.)
Mike.