How do I define y-combinator without “let rec”?

Question

In almost all examples, a y-combinator in ML-type languages is written like this:

let rec y f x = f (y f) x
let factorial = y (fun f -> function 0 -> 1         


        

Answer 1

Short answer: You can't.

Long answer: The simply typed lambda calculus is strongly normalizing. This means it's not Turing equivalent. The reason for this basically boils down to the fact that a Y combinator must either be primitive or defined recursively (as you've found). It simply cannot be expressed in System F (or simpler typed calculi). There's no way around this (it's been proven, after all). The Y combinator you can implement works exactly the way you want, though.

I would suggest you try scheme if you want a real Church-style Y combinator. Use the applicative version given above, as other versions won't work, unless you explicitly add laziness, or use a lazy Scheme interpreter. (Scheme technically isn't completely untyped, but it's dynamically typed, which is good enough for this.)

See this for the proof of strong normalization: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.127.1794

After thinking some more, I'm pretty sure that adding a primitive Y combinator that behaves exactly the way the letrec defined one does makes System F Turing complete. All you need to do to simulate a Turing machine then is implement the tape as an integer (interpreted in binary) and a shift (to position the head).