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# Re: in defense of types

```Michael Vanier <mvanier@cs.caltech.edu> writes:

> > let z f = (fun x -> f (fun z -> x x z)) (fun x -> f (fun z -> x x z));;
> >
> >   is not typable.  since two 'x' in 'x x z' cannot have the same type.
> >   The key point is distinguish the type of these instances of 'x'.
> >
> > # type 'a t = I of ('a t -> 'a);;
> > type 'a t = I of ('a t -> 'a)
> > # let z f = (function I x -> f (fun z -> x (I x) z))
> >                 (I (function I x -> f (fun z -> x (I x) z)));;
> > val z : (('a -> 'b) -> 'a -> 'b) -> 'a -> 'b = <fun>
> >
> >   voila! you can type it. Now, let's try simple example.
>
> This is a brilliant and non-obvious demonstration; thanks a lot!  You saved
> me hours of scratching my head ;-)
>
> BTW I'm not sure what you mean when you say that Y is not usable in eager
> languages.  You can define Y in scheme, which is eager.  Do you mean in
> eager statically-typed languages?

I don't understand either. And isn't this much simpler and working:

let rec y f e = f (y f) e
let ffact f n = if n = 1 then 1 else n * f (n - 1)
;;
y ffact 5

??

```