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I understand that in a sufficiently complicated, consistent formal system, not all statements are true, not all statements are decidable, not all statements' decidability is decidable, 3-decidable, ..., $n$-decidable.

Where a statement is $n$-decidable if its $n-1$-decidability is decidable.

There is no $n$ for which all statements are $n$-decidable.

But is it also true that there are statements that are $n$-undecidable for all $n$?

If so, is there any notion of higher undecidabilities? $\omega$-undecidable?

If not, what prevents Hilbert's program? I don't see how even a boundless $n$ would prevent it (although I am vaguely thinking of something akin to the diagonalization argument).

Zaz
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  • What is "n-undecidable" ? – Peter Mar 11 '22 at 10:39
  • A statement is $n$-decidable if its $n-1$-decidability is decidable – Zaz Mar 11 '22 at 15:59
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    Just to check: a statemement is $0$-decidable if it is decidable; it is $n+1$-decidable if its $n$-decidability is decidable, is that it? In any case, doesn't Rice's theorem imply a positive answer to the title question? – Nagase Mar 11 '22 at 19:58

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