Re: [Haskell-cafe] Point-free style

Daniel Fischer wrote:

And could one define

\f g h x y -> f (g x) (h y)

point-free? Any definition can be made point free if you have a complete combinator base at your disposal, e.g., S and K.

Haskell has K (called const), but lacks S. S could be defined as spread f g x = f x (g x) Given that large set of Haskell prelude functions I would not be surprised if spread could already be defined point free in Haskell. :) -- Lennart