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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is correctly rejected as non-halting.
Date: Sat, 13 Jul 2024 11:00:37 +0300
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On 2024-07-12 13:20:53 +0000, olcott said:

> On 7/12/2024 3:03 AM, Mikko wrote:
>> On 2024-07-11 14:10:24 +0000, olcott said:
>> 
>>> 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:
>>>>>>> typedef void (*ptr)();
>>>>>>> int HHH(ptr P);
>>>>>>> 
>>>>>>> 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*
>>> 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.
>> 
>> There is not "must" anywhere in the semantics of the programming language.
>> 
> 
> The semantics of the language specifies the behavior of
> the machine code thus deriving the must.

How can one derive "must" from the semantics of the machine code?

-- 
Mikko