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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: HHH(DDD)==0 is correct
Date: Mon, 7 Jul 2025 09:02:49 -0500
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On 7/7/2025 3:25 AM, Mikko wrote:
> On 2025-07-06 15:00:25 +0000, olcott said:
> 
>> On 7/6/2025 4:19 AM, Mikko wrote:
>>> On 2025-07-05 15:59:50 +0000, olcott said:
>>>
>>>> On 7/5/2025 2:30 AM, Fred. Zwarts wrote:
>>>>> Op 04.jul.2025 om 14:57 schreef olcott:
>>>>>> On 7/4/2025 2:42 AM, Mikko wrote:
>>>>>>> On 2025-07-03 15:17:53 +0000, olcott said:
>>>>>>>
>>>>>>>> On 7/3/2025 9:50 AM, Richard Damon wrote:
>>>>>>>>> On 7/3/25 10:39 AM, olcott wrote:
>>>>>>>>>> On 7/3/2025 9:16 AM, Richard Damon wrote:
>>>>>>>>>>> On 7/2/25 10:50 PM, olcott wrote:
>>>>>>>>>>>> On 7/1/2025 11:37 AM, Mr Flibble wrote:
>>>>>>>>>>>>> On Mon, 30 Jun 2025 21:12:48 -0400, Richard Damon wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 6/30/25 2:30 PM, Mr Flibble wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> PO just works off the lie that a correct simulation of the 
>>>>>>>>>>>>>> input is
>>>>>>>>>>>>>> different than the direct execution, even though he can't 
>>>>>>>>>>>>>> show the
>>>>>>>>>>>>>> instruction actually correctly simulated where they 
>>>>>>>>>>>>>> differ, and thus
>>>>>>>>>>>>>> proves he is lying.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The closest he comes is claiming that the simulation of 
>>>>>>>>>>>>>> the "Call HHH"
>>>>>>>>>>>>>> must be different when simulated then when executed, as 
>>>>>>>>>>>>>> for "some
>>>>>>>>>>>>>> reason" it must be just because otherwise HHH can't do the 
>>>>>>>>>>>>>> simulation.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Sorry, not being able to do something doesn't mean you get 
>>>>>>>>>>>>>> to redefine
>>>>>>>>>>>>>> it,
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You ar4e just showing you are as stupid as he is.
>>>>>>>>>>>>>
>>>>>>>>>>>>> No. A simulator does not have to run a simulation to 
>>>>>>>>>>>>> completion if it can
>>>>>>>>>>>>> determine that the input, A PROGRAM, never halts.
>>>>>>>>>>>>>
>>>>>>>>>>>>> /Flibble
>>>>>>>>>>>>
>>>>>>>>>>>> The most direct way to analyze this is that
>>>>>>>>>>>> HHH(DDD)==0 and HHH1(DDD)==1 are both correct
>>>>>>>>>>>> because DDD calls HHH(DDD) in recursive simulation and
>>>>>>>>>>>> DDD does not call HHH1(DDD) in recursive simulation.
>>>>>>>>>>>
>>>>>>>>>>> Nope. It seems you don't understand what the question 
>>>>>>>>>>> actually IS because you have just lied to yourself so much 
>>>>>>>>>>> that you lost the understanding of the queiston.
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> *I can't imagine how Mike does not get this*
>>>>>>>>>>>
>>>>>>>>>>> I can't understand
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> *Context of above dialogue*
>>>>>>>>>>>> *Context of above dialogue*
>>>>>>>>>>>> *Context of above dialogue*
>>>>>>>>>>>
>>>>>>>>>>> Context of your context:
>>>>>>>>>>>
>>>>>>>>>>> A Halt Decider is supposed to decide if the program given to 
>>>>>>>>>>> it (via some correct representation) will halt when run.
>>>>>>>>>>>
>>>>>>>>>>> Thus, "the input" needs to represent a program
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> typedef void (*ptr)();
>>>>>>>>>>>> int HHH(ptr P);
>>>>>>>>>>>>
>>>>>>>>>>>> void DDD()
>>>>>>>>>>>> {
>>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>>    return;
>>>>>>>>>>>> }
>>>>>>>>>>>>
>>>>>>>>>>>> int main()
>>>>>>>>>>>> {
>>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>> Which, by itself, isn't a valid input, or program. as HHH is 
>>>>>>>>>>> undefined.
>>>>>>>>>>>
>>>>>>>>>>> Each different definition of HHH, gives a different problem.
>>>>>>>>>>>
>>>>>>>>>>> Your "logic" seems to be based on trying to re-define what a 
>>>>>>>>>>> program is, which just makes it a lie.
>>>>>>>>>>>
>>>>>>>>>>> "Programs" must be complete and self-contained in the field 
>>>>>>>>>>> of computability theory, something you don't seem to understand.
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Termination Analyzer HHH simulates its input until
>>>>>>>>>>>> it detects a non-terminating behavior pattern. When
>>>>>>>>>>>> HHH detects such a pattern it aborts its simulation
>>>>>>>>>>>> and returns 0. (HHH1 has identical code)
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> But it CAN'T simulate the above input. as it isn't valid.
>>>>>>>>>>>
>>>>>>>>>>> You need to add the code of HHH to the input to let HHH 
>>>>>>>>>>> simulate "the input" to get anything.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> No I do not. The above paragraph has every detail that is needed.
>>>>>>>>>
>>>>>>>>> Then how do you correctly simulate something you do not have.
>>>>>>>>>
>>>>>>>>> Note, your "description" of HHH is just incorrect, as it is 
>>>>>>>>> also incomplete.
>>>>>>>>>
>>>>>>>>> Simulating a LIE just gives you a lie.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>> And at that point, you have different inputs for different 
>>>>>>>>>>> HHHs, and possibly different behaviors, which you logic 
>>>>>>>>>>> forgets to take into account, which just breaks it.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Wrong.
>>>>>>>>>> It is because the what I specified does take this
>>>>>>>>>> into account that HHH(DDD)==0 and HHH1(DDD)==1 are correct.
>>>>>>>>>
>>>>>>>>> Nope, becausee it violates the DEFINITION of what it means to 
>>>>>>>>> simulate something.
>>>>>>>>
>>>>>>>> *You don't even know what you mean by this*
>>>>>>>> What I mean is the execution trace that is derived
>>>>>>>> within the semantics of the C programming language.
>>>>>>>
>>>>>>> C lanbuage definition does not specifiy the senatics of the non- 
>>>>>>> standard
>>>>>>> lanugage extension that your HHH and HHH1 use.
>>>>>>
>>>>>>
>>>>>> *This is the ONLY specification of HHH that chatbots see*
>>>>>> Termination Analyzer HHH simulates its input until
>>>>>> it detects a non-terminating behavior pattern. When
>>>>>> HHH detects such a pattern it aborts its simulation
>>>>>> and returns 0.
>>>>>
>>>>> There is no non-termination behaviour to detect, because the input 
>>>>> specifies only a *finite* recursion.
>>>>
>>>> When DDD is infinitely simulated by HHH it never reaches
>>>
>>> When DDD is infinitely simulated by HHH that HHH never answers correctly
>>> (or otherwise) and therefore is not a halting decider.
>>
>> So by a kind of mathematical induction HHH correctly
>> predicts that DDD simulated by HHH (according to the
>> semantics of the C programming language) cannot possibly
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