Re: [Haskell-cafe] type metaphysics

On Mon, Feb 02, 2009 at 02:41:36PM -0800, Dan Piponi wrote:

2009/2/2 Luke Palmer :

...

But Nat ~> Bool is computably uncountable, meaning there is no injective (surjective?) function Nat ~> (Nat ~> Bool), by the diagonal argument above.

Given that the Haskell functions Nat -> Bool are computably uncountable, you'd expect that for any Haskell function (Nat -> Bool) -> Nat there'd always be two elements that get mapped to the same value.

So here's a programming challenge: write a total function (expecting total arguments) toSame :: ((Nat -> Bool) -> Nat) -> (Nat -> Bool,Nat -> Bool) that finds a pair that get mapped to the same Nat.

Ie. f a==f b where (a,b) = toSame f

(Warning: sketchy argument ahead.) Let f :: (Nat -> Bool) -> Nat be a total function and let g0 = const True. The application f g0 can only evaluate g0 at finitely many values, so f g0 = f (< k) for any k larger than all these values. So we can write

toSame f = (const True, head [ (< k) | k <- [1..], f (const True) == f (< k) ])

and toSame is total on total inputs. Regards, Reid