Correction: the theorem is
h . either (f, g) = either (h . f, h . g)
(Thanks to Lennart for pointing out the typo.)
From: [email protected]
To: [email protected]
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.
The New Busy is not the too busy. Combine all your e-mail accounts with
Hotmail. Get busy.
_________________________________________________________________
The New Busy is not the old busy. Search, chat and e-mail from your inbox.
http://www.windowslive.com/campaign/thenewbusy?ocid=PID28326::T:WLMTAGL:ON:WL:en-US:WM_HMP:042010_3_______________________________________________
Haskell-Cafe mailing list
[email protected]
http://www.haskell.org/mailman/listinfo/haskell-cafe
