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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is correctly rejected as
 non-halting.
Date: Thu, 11 Jul 2024 22:08:27 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <b61ed323c69ac7f8f7af365e7a7aba5085407d67@i2pn2.org>
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On 7/11/24 11:05 AM, olcott wrote:
> On 7/11/2024 9:25 AM, joes wrote:
>> Am Thu, 11 Jul 2024 09:10:24 -0500 schrieb olcott:
>>> On 7/11/2024 1:25 AM, Mikko wrote:
>>>> On 2024-07-10 17:53:38 +0000, olcott said:
>>>>> On 7/10/2024 12:45 PM, Fred. Zwarts wrote:
>>>>>> Op 10.jul.2024 om 17:03 schreef olcott:
>>
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>     HHH(DDD);
>>>>>>> }
>>>>>>> int main()
>>>>>>> {
>>>>>>>     HHH(DDD);
>>>>>>> }
>>>>>> Unneeded complexity. It is equivalent to:
>>>>>>         int main()
>>>>>>         {
>>>>>>           return HHH(main);
>>>>>>         }
>>>>> Every time any HHH correctly emulates DDD it calls the x86utm
>>>>> operating system to create a separate process context with its own
>>>>> memory virtual registers and stack, thus each recursively emulated DDD
>>>>> is a different instance.
>>>>
>>>> However, each of those instances has the same sequence of instructions
>>>> that the x86 language specifies the same operational meaning.
>>>>
>>> *That is counter-factual*
>> Contradicting yourself? "Counterfactual" usually means "if it were
>> different".
>>
>>> When DDD is correctly emulated by HHH according to the semantics of the
>>> x86 programming language HHH must abort its emulation of DDD or both HHH
>>> and DDD never halt.
> 
>> If the recursive call to HHH from DDD halts, the outer HHH doesn't need
>> to abort. 
> 
> Sure and when squares are round you can measure the radius of a square.
> 
>> DDD depends totally on HHH; it halts exactly when HHH does.
>> Which it does, because it aborts.
>>
> 
> Halting means reaching its own last instruction and
> terminating normally.
> 
>>> When DDD is correctly emulated by HHH1 according to the semantics of the
>>> x86 programming language HHH1 need not abort its emulation of DDD
>>> because HHH has already done this.
>> Where does HHH figure into this? It is not the simulator here.
>>
>>> The behavior of DDD emulated by HHH1 is identical to the behavior of the
>>> directly executed DDD().
>> At last!
>>
> 
> HHH must abort its simulation. HHH1 does not need to
> do that because HHH has already done this.
> 
> DDD correctly simulated by HHH has provably different
> behavior than DDD correctly simulated by HHH1.
> 

Which just shows that YOUR definition of "Correctly Simulated" that you 
are trying to use can't actually be a correct definition of an objective 
property.