I'm going to fail introduction to algorithms and be the disgrace of MIT...
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Anonymous2009-12-16 14:52
>>9 \dick -> (.) (.) :: t -> (a1 -> b -> c) -> a1 -> (a -> b) -> a -> c
Since "dick" is unused, it's simpler just to write (.) (.) :: (a1 -> b -> c) -> a1 -> (a -> b) -> a -> c, and it's easier to comprehend if you rename the type variables.
let boobs = (.) (.) :: (b -> c -> d) -> b -> (a -> c) -> a -> d
Lets name the arguments to "boobs" f, b, g and a. It is then likely equivalent to the following definition