+* Here's the same strategy extended to three mutually-recursive functions. `f`, `g` and `h`:
+
+ let u = Y (\u g h x. (\f. A) (u g h)) in
+ let w = Y ( \w h x. (\g. (\f. B) (u g h)) (w h)) in
+ let h = Y ( \h x. (\g. (\f. C) (u g h)) (w h)) in
+ let g = w h in
+ let f = u g h in
+ D