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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Proof that DDD specifies non-halting behavior --- Mike correcting
 Joes
Date: Thu, 15 Aug 2024 21:57:45 -0400
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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.
>>
> 
>