[Haskell-cafe] Bird problem 1.6.2 -- is there an easier method?

[R J]

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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