Notice that, (\x -> x) a reduces to a, so (\a b c -> a b c) x (y-z) z reduces to x (y-z) z. You can therefore simplify your function quite a bit. wierdFunc x y z = if y-z > z then x (y-z) z else (\d e -> d) (y-z) z and you can still apply that lambda abstraction (beta-reduce) wierdFunc x y z = if y-z > z then x (y-z) z else y-z None of these (except, of course, fix and remainder) are recursive. A recursive function is just one that calls itself. For wierdFunc to be recursive, the identifier wierdFunc would have to occur in the right-hand side of it's definition.
Thanks for your help. For some reason I didn't think "x (y - z) z" and "y - z" had the same type. I am still trying to understand how they do. Also, had a feeling the fix function was related to the "Y" combinator; it seems they're the same thing! http://en.wikipedia.org/wiki/Y_combinator thanks again, -andrew