Hi Folks, Here is an interesting phenomena: let f = (\x y -> x y x) let p = (\x y -> 1) Now let's evaluate f p 1 f p 1 = (\x y -> x y x) p 1 -- by replacing f with its definition = p 1 p -- by substituting x with p and y with 1 = (\x y -> 1) 1 p -- by replacing the first p with its definition = 1 -- the function returns 1 regardless of its arguments However, Haskell will not proceed with that evaluation because it will first determine the type signature of f and judge it to be an infinite type. Conversely, if f is omitted and this expression p 1 p is evaluated then the result 1 is generated. Why does f p 1 (which evaluates to p 1 p) fail whereas p 1 p succeeds? What lesson should I learn from this example? /Roger