PR Stanley:
I know sort of instinctively why it is so but can't think of the formal rationale for it: f g x = g (g x) :: (t -> t) -> (t -> t)
First of all - it is not the definition f g x = ... :: (t-> ... but the type of the function which might be specified: f :: (t->t)->t->t Then, the answer to:
Why not (t -> t) -> t -> (t -> t) to take account of the argument x for g?
is simple. If t is the type of x, then g must be g :: t->t, you're right. So f :: (t->t) -> t -> [the type of the result] But this result is of the type t, it is g(g x), not (t->t), it is as simple as that. Perhaps you didn't recognize that "->" is syntactically a right-associative op, so a->b->c is equivalent to a->(b->c), or (t->t)->t->t equiv. to (t->t)->(t->t) Jerzy Karczmarczuk