[Haskell-cafe] Clean proof -- correction
R J
rj248842 at hotmail.com
Sun May 23 12:24:50 EDT 2010
Correction: the theorem is
h . either (f, g) = either (h . f, h . g)
(Thanks to Lennart for pointing out the typo.)
From: rj248842 at hotmail.com
To: haskell-cafe at haskell.org
Subject: Clean proof?
Date: Sun, 23 May 2010 15:41:20 +0000
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 Thanks.
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