Re: Proof that DDD specifies non-halting behavior --- Mike correcting Joes and thus Fred too

Op 15.aug.2024 om 21:39 schreef olcott:

On 8/15/2024 1:35 PM, Mike Terry wrote:

On 15/08/2024 17:30, olcott wrote:

On 8/15/2024 10:40 AM, Fred. Zwarts wrote:

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.

>
Mike corrected you on this. You are wrong.

>
For the record, I did no such thing and Fred is correct.
>

 *Fred has the same incorrect views as joes*
*Here is where you agreed that Fred is wrong*
*when replying to joes*
 On 8/14/2024 10:07 AM, Mike Terry wrote:
 > On 14/08/2024 08:43, joes wrote:
 >> Am Tue, 13 Aug 2024 21:38:07 -0500 schrieb olcott:
 >>> *Try to show exactly how DDD emulated by HHH
 >>>  returns to its caller*>>
 >>> (the first one doesn't even have a caller)
 >>> Use the above machine language instructions to show this.
 >> 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?
 > [It's what people have been discussing here endlessly
 > for the last few months! :) ]
 >
 > So your trace is impossible...
 >
 

It is clear that olcott does not really read what I write. (Or is very short of memory.)
I never said such a thing.
I repeatedly told that the simulating HHH aborted when the simulated HHH had only one cycle to go. I never said that the simulated HHH reached it abort and halted.
In fact, I said that the fact that the simulation fails to reach the abort and halt of the simulated HHH proves that the simulation is incomplete and incorrect, because a complete simulation (such as by HHH1) shows that the simulated HHH would abort and halt.
It now becomes clear that you either never understood what I said, or your memory is indeed very short.
Give it some time to think about what I say, try to escape from rebuttal mode, instead of ignoring it immediately.