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

On 2025-05-07 18:03:00 +0000, olcott said:

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.

 I concur.
 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.
 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.
 But to understand the Turing proof requires little if any mathematical knowledge. It requires only the capacity for clear thinking.
 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.
 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.
 

 https://www.liarparadox.org/Linz_Proof.pdf
The flaw is on the top of page 3 where
there are two q0 start states in the same sequence
of state transitions.

There is no page 3 there, only unnumbered pages.
No, there is not. Every Turing machine has only one start state and
q0 is that state and no other. Once the Turing machine has started
the start state is like any other state, so a Turing machine may make
a transition to the start state and continue from there.
But going back to the start state is not useful in this context. The
presentation would be clearer if Linz had used a different start state
for each Turing machine or explicitly renamed the start state of H'
when constructing Ĥ.

I encode that in this easier to understand notation.
 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

It is easy to understand that that does not encode anything.
It is also easy to understand that you are trying the deception
known as "fallacy fallacy".
--
Mikko