
25 Nov
2010
25 Nov
'10
7:50 a.m.
On 25 November 2010 12:48, Stephen Tetley
Substituting, m x == (r1 -> x)
liftM2 :: (a -> b -> c) -> (r1 -> a) -> (r1 -> b) -> (r1 -> ans)
Apologies, there's an obvious typo, the correct version is: liftM2 :: (a -> b -> ans) -> (r1 -> a) -> (r1 -> b) -> (r1 -> ans)