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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Turing Machine computable functions apply finite string
 transformations to inputs VERIFIED FACT
Date: Wed, 7 May 2025 14:18:11 -0500
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On 5/1/2025 10:32 AM, Ben Bacarisse wrote:
> Richard Heathfield <rjh@cpax.org.uk> writes:
> 
>> On 30/04/2025 19:30, Mike Terry wrote:
>>> On 30/04/2025 16:46, Richard Heathfield wrote:
>>>> On 30/04/2025 16:15, olcott wrote:
>>>>> On 4/29/2025 5:03 PM, Richard Heathfield wrote:
>>>>>> On 29/04/2025 22:38, olcott wrote:
>>>>>>
>>>>>> <snip>
>>>>>>
>>>>>>>
>>>>>>> int DD()
>>>>>>> {
>>>>>>>     int Halt_Status = HHH(DD);
>>>>>>>     if (Halt_Status)
>>>>>>>       HERE: goto HERE;
>>>>>>>     return Halt_Status;
>>>>>>> }
>>>>>>>
>>>>>>> HHH is correct DD as non-halting BECAUSE THAT IS
>>>>>>> WHAT THE INPUT TO HHH(DD) SPECIFIES.
>>>>>>
>>>>>> You're going round the same loop again.
>>>>>>
>>>>>> Either your HHH() is a universal termination analyser or it isn't.
>>>>>
>>>>> The domain of HHH is DD.
>>>>
>>>> Then it is attacking not the Halting Problem but the Olcott Problem,
>>>> which is of interest to nobody but you.
>>> It would be (if correct) attacking the common proof for HP theorem as it
>>> occurs for instance in the Linz book which PO links to from time to time.
>>
>> Yes. That's what I call the Olcott Problem.
>>
>> De gustibus non est disputandum, but I venture to suggest that (correctly)
>> overturning Turing's proof would be of cosmos-rocking interest to the world
>> of computer science, compared to which pointing out a minor flaw in a
>> minor[1] proof would, even if correct, have no more effect on our field
>> than lobbing a pebble into the swash at high tide.
>>
>> I suspect that the only reason we bother to argue with Mr Olcott so much is
>> because (even if he does so unwittingly) he manages to convey the
>> appearance of attacking the Halting Problem, and arguing about the Halting
>> Problem is a lot more fun than arguing about the Olcott Problem.
>>
>> To be of any interest, solving the Olcott Problem would have to have
>> important consequences. But does it? Let's see.
>>
>> Dr Linz Theorem 12.1 (Halting Problem is Undecidable): There does not exist
>> any Turing machine H that behaves as required by Linz Definition 12.1. Thus
>> the halting problem is undecidable.
>>
>> Dr Linz has a proof for this claim, which can be found here:
>> <https://john.cs.olemiss.edu/~hcc/csci311/notes/chap12/ch12.pdf>
>>
>> If the proof is flawless, the conclusion stands and Mr Olcott is simply
>> wrong.
> 
> There is peculiar irony here.  The proof is /not/ flawless.  It has, in
> fact, a flaw that PO pointed out (although in passing).  PO does not
> care about the flaw because it is easily fixed, but it's there none the
> less[1].
> 
> Anyway, Linz only gives this argument because it is of historical
> interest.  His "real" proof immediately follows this argument (in the
> book) and is a trivial corollary of the fact, proved in chapter 11, that
> not all recursively enumerable languages are recursive.  But no crank
> ever addresses that proof.  I wonder why...
> 
>> If the proof is flawed through some error of reasoning, *either* it merely
>> fails to correctly support its conclusion *or* a duly corrected proof
>> /overturns/ the conclusion.
> 
> ... or a duly corrected proof /supports/ the conclusion.  Maybe you
> intended this possibility to be include in your first "merely fails to
> support" alternative.
> 
>> The latter would be /extremely/ interesting, but it would also mean that we
>> have two proofs proving opposite things, and so it would effectively be a
>> cataclysmic sideways attack on Turing's reasoning.
> 
> It would be a cataclysmic attack on all of conventional mathematics and
> logic as two proofs, one of T and another of ~T, makes the system of
> proof oneself inconsistent (and everything becomes a theorem).
> 
> [1] It centres on the naming of states in H and the derived machines H'
> and H^.  Details available if you really care.
> 

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

When the mapping from the input to the behavior
that this input specifies is used as the basis
then embedded_H correctly rejects ⟨Ĥ⟩ ⟨Ĥ⟩ because
recursive simulation prevents the correctly
simulated ⟨Ĥ⟩ from ever reaching its final state
of ⟨Ĥ.qn⟩

When we simply FALSELY ASSUME that ⟨Ĥ⟩ is able to
actually do the opposite of whatever embedded_H
reports then Linz would be (yet is not) correct.
This false conclusion is based on a false assumption.


-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer