This works in Haskell just fine, thank you very much! ghci> y f = f (y f) ghci> factorial = y (\f x -> if x == 0 then 1 else x * f (x-1)) ghci> fac

This works in Haskell just fine, thank you very much! ghci> y f = f (y f) ghci> factorial = y (\f x -> if x == 0 then 1 else x * f (x-1)) ghci> factorial 5 120 However, the Y combinator as defined in a non-recursive form is tougher to implement in Haskell [1][2]. It looks a lot better in Racket (and this is coming from someone who doesn't particularly like Racket). (λ (r) (λ (h) (h h)) (λ (g) (r (λ (x) (g g) x)) [1] http://stackoverflow.com/a/5885270/3072788 http://stackoverflow.com/a/5885270/3072788 [2] http://stackoverflow.com/a/13119751/3072788 http://stackoverflow.com/a/13119751/3072788