Re: Formal systems that cannot possibly be incomplete except for unknowns and unknowable

On 5/7/2025 1:55 PM, olcott wrote:

On 5/6/2025 11:16 AM, Richard Heathfield wrote:

On 06/05/2025 16:38, Alan Mackenzie wrote:

These aren't particularly difficult things to comprehend.  As I keep
saying, you ought to show a lot more respect for people who are
mathematically educated.

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I concur.
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As someone who is not particularly mathematically educated (I have an A- level in the subject, but that's all), I tend to steer well clear of mathematical debates, although I have occasionally dipped a toe.
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I have *enormous* respect for those who know their tensors from their manifolds and their conjectures from their eigenvalues, even though it's all Greek to me.
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But to understand the Turing proof requires little if any mathematical knowledge. It requires only the capacity for clear thinking.
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Having been on the receiving end of lengthy Usenet diatribes by cranks in my own field, I don't hold out much hope for our current culprits developing either the capacity for clear thought or any measure of respect for expertise any time soon.
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Nor do I believe they are capable of understanding proof by contradiction, which is just about the easiest kind of proof there is. In fact, the most surprising aspect of this whole affair is that

 

(according to Mike) Mr Olcott seems to have (correctly) spotted a minor flaw in the proof published by Dr Linz. How can he get that and not get contradiction? Proof by contradiction is /much/ easier.
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 When THERE IS NO CONTRADICTION then proof by contradiction fails.

False.  The contradiction is that HHH is found to not map the below function after it is assumed that the below function is computable and that HHH computes it:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly