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

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Sujet : Re: Proof that DDD specifies non-halting behavior --- Mike correcting Joes
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory
Date : 16. Aug 2024, 02:57:45
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <8b56eba0ec44b78d23a1029236e2c22734d48ae9@i2pn2.org>
References : 1 2 3 4 5 6 7
User-Agent : Mozilla Thunderbird
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?
 
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.
 You either understand this or do not understand Mike's
correction.
But that is not an excuess to give the wrong answer.
That just points to the dilemma that it must try to solve, but turns out to be unsolvable, which is why the problem is uncomputable.
Something you don't seem able to understand, perhaps because you just don't understand the technical meaning of the words (just like you miss the technical meaning of so many words because you refuse to study them).

 
So your trace is impossible...
Just like all the others are wrong.
>
 

Date Sujet#  Auteur
5 Jul 25 o 

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