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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Every sufficiently competent C programmer knows --- Very Stupid
 Mistake or Liars
Date: Fri, 14 Mar 2025 10:10:35 -0400
Organization: i2pn2 (i2pn.org)
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On 3/13/25 11:53 PM, olcott wrote:
> On 3/13/2025 10:03 PM, Richard Damon wrote:
>> On 3/13/25 10:07 PM, olcott wrote:
>>> On 3/13/2025 6:09 PM, Richard Damon wrote:
>>>> On 3/13/25 9:41 AM, olcott wrote:
>>>>> On 3/13/2025 6:18 AM, Dan Cross wrote:
>>>>>> In article <vqud4e$36e14$3@dont-email.me>,
>>>>>> Fred. Zwarts <F.Zwarts@HetNet.nl> wrote:
>>>>>>> Op 12.mrt.2025 om 16:31 schreef olcott:
>>>>>>>> [snip]
>>>>>>>> When N steps of DDD are correctly emulated by every element
>>>>>>>> of the set of C functions named HHH that do x86 emulation and
>>>>>>>>
>>>>>>>> N is each element of the set of natural numbers
>>>>>>>>
>>>>>>>> then no DDD of the set of HHH/DDD pairs ever reaches its
>>>>>>>> "return" instruction and terminates normally.
>>>>>>>
>>>>>>> In other words no HHH of the set of HHH/DDD pairs ever succeeds to
>>>>>>> complete the simulation of a halting program. Failure to reach 
>>>>>>> the end
>>>>>>> of a halting program is not a great success. If all HHH in this set
>>>>>>> fail, it would be better to change your mind and start working on
>>>>>>> something more useful.
>>>>>>
>>>>>> He seems to think that he's written a program that detects that
>>>>>> his thing hasn't 'reached its "return" instruction and
>>>>>> terminate[d] normally', given some number of steps, where that
>>>>>> number is ... the cardinality of the natural numbers.
>>>>>>
>>>>>> I wonder if he knows that the set of natural numbers is
>>>>>> infintite, though I suspect he'd say something like, "but it's
>>>>>> countable!"  To which I'd surmise that he has no idea what that
>>>>>> means.
>>>>>>
>>>>>
>>>>> void DDD()
>>>>> {
>>>>>    HHH(DDD);
>>>>>    return;
>>>>> }
>>>>>
>>>>> Everyone here knows that when N steps of DDD are correctly
>>>>> simulated by HHH that DDD never reaches its own "return"
>>>>> instruction and terminates normally thus never halts.
>>>>> *AND THEY LIE ABOUT IT BY ENDLESSLY CHANGING THE SUBJECT*
>>>>
>>>>
>>>> No, the PARTIAL EMULATION done by HHH can't reach that point,
>>>
>>> But a complete emulation can?
>>>
>>
>> Yes, but an HHH that gives an answer doesn't do one, due to the 
>> pathological design of the template used to build DD to the HHH it 
>> calls (which is the only HHH that can exist, or you have violated the 
>> basic rules of programing and logic).
>>
>> We have two basic cases,
>>
>> 1) if HHH does the partial emulation you describe, then the complete 
>> emulation of DD will see that DD call HHH, and it will emulate its 
>> input for a while, then abort and theu return 0 to DD which will then 
>> halt.
>>
> 
> int main()
> {
>    HHH(DDD); // No DDD can possibly ever return.
> }
> 

Since HHH doesn;t call DDD, the statement is vacuous and shows a 
fundamental ignorance of what is being talked about.

Yes, No HHH can emulated DDD to the end, but since halting is DEFINED by 
the behavior of the program, and for every HHH that aborts and returns, 
the program of DDD, as tested with:

int main()
{
    DDD()
}

will return to main, that shows that every HHH that returns 0 fails to 
be a Halt Decider or Termination Analyzer. PERIOD.

The fact that it may meet your defintion of POOP says maybe you have 
solvec the POOP problem, but it seems you can't even properly define it 
in the nornal field of Computation Theory.