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From: olcott <polcott2@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: ZFC solution to incorrect questions: reject them
Date: Tue, 12 Mar 2024 22:55:33 -0500
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On 3/12/2024 10:49 PM, Richard Damon wrote:
> On 3/12/24 7:41 PM, olcott wrote:
>> On 3/12/2024 9:17 PM, immibis wrote:
>>> On 13/03/24 02:47, olcott wrote:
>>>> On 3/12/2024 8:05 PM, immibis wrote:
>>>>> On 13/03/24 01:18, olcott wrote:
>>>>>> On 3/12/2024 7:10 PM, immibis wrote:
>>>>>>> So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
>>>>>> Exactly one element of Q differs by writing a 1 instead of a 0.
>>>>>
>>>>> That's part of δ but this mistake doesn't matter.
>>>>>
>>>>> It wasn't clear whether you were talking about a Turing machine 
>>>>> that was somehow identical but gave a different return value, or 
>>>>> one that was not identical. Now you have explained it is not 
>>>>> identical.
>>>>>
>>>> They are identical except for their return value that is specified
>>>> in a single state that is different.
>>>>
>>>> *This means that they implement the exact same algorithm*
>>>
>>> OK. Well, one of them gets the right answer and one of them gets the 
>>> wrong answer. What is the confusion?
>>
>> The Linz Ĥ.H machine gets the wrong answer on its own
>> machine description no matter how its Linz H is defined.
>>
>> This means that it gets the wrong answer on YES and the
>> wrong answer on NO.
>>
> 
> Not quite. It always gets the wrong answer, but only one of them for 
> each quesiton.
> 

They all gets the wrong answer on a whole class of questions because
epistemological antinomies are not rejected as semantically invalid input.

> For EACH SEPARATE definition of H, and thus H^, we have a different 
> question.
> 

They are all the same epistemological antinomy category of question.

> Note, the machine H^ isn't DEFINED to just get H^ as an input.
> 
> H^ is defined to get as an input, the description of ANY Turing Machine, 
> and to ask H what that machine applied to its description will do, and 
> then it does the opposite.
> 
> Thus, for every different H we go to test, we get a DIFFERENT H^ 
> machine. and when we look at the question to H (or H^.H) about the 
> description (H^) (H^),
> 
> If H (H^) (H^) goes to qn, then H^ (H^) goes to qn too and halts, so the 
> correct answer would have been to go to qy.
> 
> If H (H^) (H^) goes to qy, then H^ (H^) goes to qy too, and loops, so 
> the correct answer would have been to go to qn.
> 
> So, each case HAS a correct answer, just not the one that H (or H^.H) 
> goes to,
> 
> So yes, which ever one it goes to (and a given machine will only go to 
> one with this input) will be wrong, but the other one would have been 
> right, and an H* machine that answer the opposite of H would have been 
> correct.

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