rho x = map (((flip mod) x).(\a -> a^2-1)) (rho x) This function will generate an infinite list. And I tested in GHCi, the function type is *Main> :t rho rho :: Integral b => b -> [b] If I define a function like this fun x = ((flip mod) x).(\a -> a^2-1) The type is *Main> :t fun fun :: Integral c => c -> c -> c My question is, how can Haskell deduce the function type to b - [b]? We don't have any [] type data in this function. Thanks!