Type of flip id

http://www.haskell.org/pipermail/beginners/2011-March/006477.html

The point is that the type of id has to be unified with the type of flip's
(first) argument.

flip :: (a -> b -> c) -> (b -> a -> c)
id :: t -> t

So we have to unify (a -> b -> c) and (t -> t). Fully parenthesized,
a -> b -> c is a -> (b -> c). Now unification yields

t = a
-- id's arg must have the same type as the flip's argument's arg

and

t = (b -> c)
-- id's result must have the same result as flip's argument's result

From that follows a = (b -> c) and *id can be passed to flip only at a more
restricted type than id's most general type, namely at the type
id :: (b -> c) -> (b -> c)*

So,

flip (id :: (b -> c) -> b -> c) :: b -> (b -> c) -> c