Bird problem 1.6.2 is:
If f :: (a, b) -> c, then define a function "swap" such that:
flip (curry f) = curry (f . swap).
I'd very much appreciate if someone could tell me whether there's a rigorous
solution simpler than mine, which is:
Since (.) :: (q -> r) -> (p -> q) -> (p -> r), we have f :: q -> r and swap ::
p -> q. Type unification of f requires q = (a, b) and r = c.
Since f :: (a, b) -> c and curry :: ((l, m) -> n) -> (l -> m -> n),
typeunification requires l = a, b = m, and n = c. Therefore,curry :: ((a, b)
-> c) -> (a -> b -> c), and (curry f) :: a -> b -> c.
Since flip :: (s -> t -> u) -> t -> s -> u, type unification requiress = a, t =
b, and u = c. Therefore, flip :: (a -> b -> c) -> b -> a -> c,and flip (curry
f) :: b -> a -> c.
Therefore, curry (f . swap) :: b -> a -> c, and p :: b -> a. Therefore,swap
:: b -> a -> (a, b), and:
swap :: b -> a -> (a, b)swap x y = (y,
x)
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