Why do g and g' have different types? g x y = let ?f = \x-> x in ?f x ++ (show (?f y)) g' x y = let f = \x-> x in f x ++ (show (f y)) Is there a way I can use implicit types and let g be as general as g'? -Alex- ______________________________________________________________ S. Alexander Jacobson tel:917-770-6565 http://alexjacobson.com