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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: Fri, 16 Aug 2024 10:18:43 -0400
Organization: i2pn2 (i2pn.org)
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On 8/16/24 8:51 AM, olcott wrote:
> On 8/16/2024 1:42 AM, Fred. Zwarts wrote:
>> Op 16.aug.2024 om 04:24 schreef olcott:
>>> On 8/15/2024 8:57 PM, Richard Damon wrote:
>>>> 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?
>>>
>>> That is the part that Joes and Fred do not understand.
>>>
>>
>> You do not understand what we say. I have repeated many times that the 
>> simulated HHH is aborted before it would halt by itself.
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> DDD emulated by HHH according to the semantics of the
> x86 language would never halt by itself.

But it does if the HHH that DDD calls is the HHH that you actually have 
that does abort its simulation.

You just LIE to everyone by pretending that HHH can somehow do this 
magical compete simulate and still get an answer.

Yes, *IF* your HHH actually does that complete emulation by the xx86 
language, then DDD will not return, but that is just a fantasy, because 
that isn't the HHH that you supply, and changing HHH changes the program 
DDD so you just compound your lies.

> 
>> The simulating HHH fails to reach that point, proving an incomplete 
>> simulation.
> 
> void Infinite_Recursion()
> {
>    Infinite_Recursion();
>    OutputString("I never make it here!\n");
> }
> 
> Unreachable code is unreachable code.

Right, and reachable code is reachable code, becuase HHH is the program 
that it is.

Since you claim HHH will "correctly" abort its simulation, we can know 
that HHH will abort its simulation and see if it was correct about it. 
SInce it has been stipulated that HHH does abort its simulation here, 
then it WILL return to DDD and DDD WILL Halt, and thus the CORRECT 
answer that HHH needed to have returned to be correct was Halting, and 
since that isn't what it did, it just isn't correct and your claim it 
was is just a LIE.


> 
>> That is what you do not understand, either because of incompetence, or 
>> unwillingness.
> 
> You might not even understand the above simple example.
> 

Sure I do, but it seems you don't.