Alternate strategy for Y1, Y2 * This is (in effect) the strategy used by OCaml. The mutually recursive: let rec f x = A ; A may refer to f or g and g y = B ; B may refer to f or g in C is implemented using regular, non-mutual recursion, like this (`u` is a variable not occurring free in `A`, `B`, or `C`): let rec u g x = (let f = u g in A) in let rec g y = (let f = u g in B) in let f = u g in C or, expanded into the form we've been working with: let u = Y (\u g x. (\f. A) (u g)) in let g = Y ( \g y. (\f. B) (u g)) in let f = u g in C We abstract the Y1 and Y2 combinators from this as follows: let Yu = \ff. Y (\u g. ff ( u g ) g) in let Y2 = \ff gg. Y ( \g. gg (Yu ff g ) g) in let Y1 = \ff gg. (Yu ff) (Y2 ff gg) in let f = Y1 (\f g. A) (\f g. B) in let g = Y2 (\f g. A) (\f g. B) in C * Here's the same strategy extended to three mutually-recursive functions. `f`, `g` and `h`: let v = Y (\v g h x. (\f. A) (v g h)) in let w = Y ( \w h x. (\g. (\f. B) (v g h)) (w h)) in let h = Y ( \h x. (\g. (\f. C) (v g h)) (w h)) in let g = w h in let f = v g h in D Or in Y1of3, Y2of3, Y3of3 form: let Yv = \ff. Y (\v g h. ff ( v g h) g h) in let Yw = \ff gg. Y ( \w h. (\g. gg (Yv ff g h) g h) ( w h)) in let Y3of3 = \ff gg hh. Y ( \h. (\g. hh (Yv ff g h) g h) (Yw ff gg h)) in let Y2of3 = \ff gg hh. Yw ff gg (Y3of3 ff gg hh) in let Y1of3 = \ff gg hh. Yv ff (Y2of3 ff gg hh) (Y3of3 ff gg hh) in let f = Y1of3 (\f g h. A) (\f g h. B) (\f g h. C) in let g = Y2of3 (\f g h. A) (\f g h. B) (\f g h. C) in let h = Y3of3 (\f g h. A) (\f g h. B) (\f g h. C) in D