There is no clean proof of that statement because it is false.
(Consider the argument 'undefined'.)
2010/5/23 R J
Given the following definition of "either", from the prelude: either :: (a -> c, b -> c) -> Either a b -> c either (f, g) (Left x) = f x either (f, g) (Right x) = g x what's a clean proof that: h . either (f, g) = either (h . f, g . h)? The only proof I can think of requires the introduction of an anonymous function of z, with case analysis on z (Case 1: z = Left x, Case 2: z = Right y), but the use of anonymous functions and case analysis is ugly, and I'm not sure how to tie up the two cases neatly at the end. For example here's the "Left" case: h . either (f, g) = {definition of "\"} \z -> (h . either (f, g)) z = {definition of "."} \z -> (h (either (f, g) z) = {definition of "either" in case z = Left x} \z -> (h (f x)) = {definition of "."} \z -> (h . f) x = {definition of "."} h . f
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