Yes.

No. Any finite algorithm can be performed on a Turing machine, so if no Turing machine can solve the halting problem, then nor can any finite algorithm.without any proof ?

For your example of coloured Turing machines, I gave an explicit way of doing so (by creating a Turing machine with a symbol for every colour-symbol combination in your coloured Turing machine). You then made the non-standard restriction that I'm not allowed to change the alphabet, but this is irrelevant because there's an easy algorithm to convert an arbitrary Turing machine into a 2-symbol Turing machine.

My point is thatbecause as I argued diagonalization argument does not apply to the set of colored turing machine (considering my new definition)

**there's absolutely no need to apply a diagonal argument to the set of coloured Turing machines**! The unsolvability of the halting problem by any finite algorithm is the logical conclusion of:

1. Turing machines cannot solve the halting problem

2. Anything that can be computed by a finite algorithm can be computed by a Turing machine

There's no diagonal argument involved in proving 2, which is the part that you're seeming to dispute.

There are plenty of such formal systems: the first-order theory of algebraically closed fields of characteristic zero has this property. The caveat is that such systems are necessarily weak in their expressive power; you can't formalise the halting problem in such a system.

Yes, and both approaches are futile. There was a time pre-Godel where mathematicians such as Hilbert pursued this goal -- see https://en.wikipedia.org/wiki/Hilbert%27s_program -- but this is no longer taken seriously because it's been proven impossible.raoofha wrote: ↑June 11th, 2023, 3:33 pmmy approach is similar to James Anderson Transmathematics, he is trying to totalize mathematics, I'm trying to totalizing computation