Re: Proof that DDD specifies non-halting behavior --- Mike correcting Joes

Op 15.aug.2024 om 14:12 schreef olcott:

On 8/15/2024 2:00 AM, joes wrote:

Am Wed, 14 Aug 2024 16:07:43 +0100 schrieb Mike Terry:

On 14/08/2024 08:43, joes wrote:

Am Tue, 13 Aug 2024 21:38:07 -0500 schrieb olcott:

On 8/13/2024 9:29 PM, Richard Damon wrote:

On 8/13/24 8:52 PM, olcott wrote:

>

A simulation of N instructions of DDD by HHH according to the
semantics of the x86 language is necessarily correct.

Nope, it is just the correct PARTIAL emulation of the first N
instructions of DDD, and not of all of DDD,

That is what I said dufuss.

You were trying to label an incomplete/partial/aborted simulation as
correct.
>

A correct simulation of N instructions of DDD by HHH is sufficient
to correctly predict the behavior of an unlimited simulation.

Nope, if a HHH returns to its caller,

*Try to show exactly how DDD emulated by HHH returns to its caller*

how *HHH* returns

>

HHH simulates DDD    enter the matrix
    DDD calls HHH(DDD)    Fred: could be eliminated HHH simulates

DDD

    second level
      DDD calls HHH(DDD)    recursion detected
    HHH aborts, returns    outside interference DDD halts

voila

HHH halts

>
You're misunderstanding the scenario?  If your simulated HHH aborts its
simulation [line 5 above],
then the outer level H would have aborted its identical simulation
earlier.  You know that, right?

 

Of course. I made it only to illustrate one step in the paradoxical
reasoning, as long as we're calling programs that do or don't abort
the same.
>

 It is like I always pointed out. The outer HHH cannot
wait for the inner ones to abort because it would be
waiting forever.

Exactly. And when it aborts, it aborts too soon, one cycle before the simulated HHH would abort and halt.
HHH cannot possibly simulate itself correctly.