My puzzlement came from the fact that : f(in -> out)(in') -> out == f((in -> out), in') -> out and f(in -> out) -> (in' -> out) are similar. But t

My puzzlement came from the fact that : f(in -> out)(in') -> out == f((in -> out), in') -> out and f(in -> out) -> (in' -> out) are similar. But thinking some more about it, I don't see a good reason for which this would be a problem. This simply means that any function defined to take [] elements is also able to take [:x] and return a function taking [x:]... I hadn't thought about this particular thing until now, but this kind of halfway-through computation is interesting.