Re: Bad faith and dishonesty

On 6/1/2025 11:22 AM, Mike Terry wrote:

On 01/06/2025 10:48, Mikko wrote:

On 2025-05-31 15:12:57 +0000, olcott said:
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On 5/31/2025 8:37 AM, Mike Terry wrote:

On 31/05/2025 11:03, Richard Heathfield wrote:

On 31/05/2025 10:42, Mikko wrote:

On 2025-05-30 17:27:05 +0000, olcott said:
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On 5/30/2025 12:06 PM, Richard Heathfield wrote:

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<snip>
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As far as I can recall, Olcott's ramblings never go within discus- throwing distance of a potentially erroneous step.

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There is no *INPUT* D to termination analyzer H
that can possibly do the opposite of whatever
value that H returns.

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We can apply Truing'c construction to every Truing machine and
 prove that no Turing machine is a halting decider. With a
similar construction many other problems about Turing machine
behaviour can be proven uncomputable, too.

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Indeed. In fact, if I'm reading Mr Olcott's retort correctly...
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Turing: imagine a program P...
Turing: ...argument goes here...
Turing: ...and it turns out that P can't exist.
Olcott: Turing is wrong, because P can't exist.
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I /seeeeee.../
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More like:
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Turing: imagine a program P...
Turing: ...modify P as follows to create D...
Turing: ...argument goes here...
Turing: ...so P decides D's halting incorrectly
Turing: ...and so P is not a halt decider.
Olcott: Turing is wrong, because *D* can't exist.
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I /seeeeee.../
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Mike.

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Turing wasn't even dealing with halting.

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No, but he proved that there are problems that are not Turing decidable.
That halting is one of them can be proven with an adaptation of Turing's
proof.
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The question Turing wanted to solve is whether a computation writes
infinitely many unerasable symbols. A computation that halts doesn't.
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*Here is the orignal Linz proof*
https://www.liarparadox.org/Linz_Proof.pdf

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Linz' presentation of the proof is not as good as it could be.
Linz is not careful with details. For example, the formulas in
the definition 12.1 specify to configurations (and a relation
between them) but does not specify which tape location is under
the head.

 Linz's notation does specify the tape head location. In his notation
     aQb,
 where a,b are strings of the tape alphabet, and Q is a TM state, the meaning is that the TM is in state Q, and the tape contains string ab (concatenation of a and b), and the tape head is about to read the first character of b.
 You're right that Linz's presentation is not completely careful.  He is aiming at CS students specifically without any special mathematical background.  Where I see lack of care, I also see that it would be clear how to address that issue if required, so it is not a big issue for me.
  Mike.
 

He made the big mistake of have two start states
in the same sequence of configurations.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer