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

On 8/15/24 8:12 AM, olcott wrote:

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?