Newbie trying to get through Bird. Could someone provide a clean solution,
with proof (so I can see how these proofs are laid out), to this:
Given:
f :: Integer -> Integerg :: Integer -> (Integer -> Integer)
h :: ...h x y = f (g x y)
Questions:
a. Fill in the type assignment for "h".
b. Which
On Tuesday 18 May 2010 21:49:50, R J wrote:
> Newbie trying to get through Bird. Could someone provide a clean
> solution, with proof (so I can see how these proofs are laid out), to
> this: Given:
> f :: Integer -> Integer
> g :: Integer -> (Integer -> Integer)
> h :: ...
> h x y = f (g x y)
> Q
