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
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