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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: HHH(DD) --- COMPUTE ACTUAL MAPPING FROM INPUT TO OUTPUT ---
 Ignoramus !!!
Date: Wed, 30 Apr 2025 10:28:33 -0500
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On 4/29/2025 4:49 AM, Mikko wrote:
> On 2025-04-28 15:52:13 +0000, olcott said:
> 
>> On 4/28/2025 4:01 AM, Mikko wrote:
>>> On 2025-04-16 17:36:31 +0000, olcott said:
>>>
>>>> On 4/16/2025 7:29 AM, Richard Heathfield wrote:
>>>>> On 16/04/2025 12:40, olcott wrote:
>>>>>> sum(3,2) IS NOT THE SAME AS sum(5,2).
>>>>>> IT IS EITHER STUPID OR DISHONEST FOR YOU TO TRY TO
>>>>>> GET AWAY FOR CLAIMING THIS USING THE STRAW DECEPTION
>>>>>> INTENTIONALLY INCORRECT PARAPHRASE OF MY WORDS.
>>>>>
>>>>> Whether sum(3,2) is or is not the same as sum(5,2) is not the 
>>>>> question. The question is whether a universal termination analyser 
>>>>> can be constructed, and the answer is that it can't.
>>>>>
>>>>> This has been rigorously proved. If you want to overturn the proof 
>>>>> you've got your work cut out to persuade anyone to listen, not 
>>>>> least because anyone who tries to enter into a dialogue with you is 
>>>>> met with contempt and scorn.
>>>>>
>>>>> The proof stands.
>>>>>
>>>>
>>>> *corresponding output to the input*
>>>> *corresponding output to the input*
>>>> *corresponding output to the input*
>>>> *corresponding output to the input*
>>>> *corresponding output to the input*
>>>>
>>>> Not freaking allowed to look at any damn thing
>>>> else besides the freaking input. Must compute whatever
>>>> mapping ACTUALLY EXISTS FROM THIS INPUT.
>>>
>>> A halt decider is is not allowed to compute "whatever" mapping. It is
>>> required to compute one specific mapping: to "no" if the computation
>>> described by the input can be continesd forever without halting, to
>>> "no" otherwise.
>>
>> It must do this by applying the finite string transformation
>> rules specified by the x86 language to the input to HHH(DD).
> 
> No, it needn't. A halt decider cannot do other than certain finite string
> operations. No relation to x86 language is required.
> 
>> This DOES NOT DERIVE THE BEHAVIOR OF THE DIRECTLY EXECUTED DD.
> 
> Whether the execution is "direct" or otherwise is irrelevant. A computation
> either halts or not. A halt decider must just tell whether the somputation
> halts. It is true that no Turing machine can determine this about every
> computation, i.e., no Turing machine is a halt decider.
> 
>> It DOES DERIVE DD EMULATED BY HHH AND ALSO DERIVES THE RECURSIVE
>> EMULATION OF HHH EMULATING ITSELF EMULATING DD.
> 
> Which are not mentioned in the halting problem.
> 

When understand rather than simply ignore the HHH/DD
example it can be seen that every conventional halting
problem proof suffers the same fate. The contradictory
part of the "impossible" input IS NEVER REACHABLE.

int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}

It took several C programmers only a few minutes to see this.
The same thing applies to the Peter Linz proof.

Unlike the HHH/DD example the Linz proof lacks a fully
specified termination analyzer written in a language that
has a complete definition. Disagreeing with my 100%
complete proof is like disagreeing with arithmetic.

There are no weasel words or double-talk that can possibly
get around this:

DD emulated by HHH according to the finite string
transformation rules of the x86 language DOES NOT HALT!


-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer