On Sat, May 29, 2010 at 9:28 PM, Cory Knapp <cory.m.knapp@gmail.com> wrote:

Hello,

A professor of mine was recently playing around during a lecture with Church booleans (I.e., true = \x y -> x; false = \x y -> y) in Scala and OCaml. I missed what he did, so I reworked it in Haskell and got this:

>type CB a = a -> a -> a

>ct :: CB aC
>ct x y = x

>cf :: CB a
>cf x y = y

>cand :: CB (CB a) -> CB a -> CB a
>cand p q = p q cf

>cor :: CB (CB a) -> CB a -> CB a
>cor p q = p ct q

I found the lack of type symmetry (the fact that the predicate arguments don't have the same time) somewhat disturbing, so I tried to find a way to fix it. I remembered reading about existential types being used for similar type-hackery, so I added quantification to the CB type and got

By the way, I looked on wikipedia and their definitions vary slightly from yours:

cand p q = p q p

cor p q = p p q

I think yours are equivalent though and for the rest of this reply I use the ones from wikipedia.

I think the reason the it doesn't type check with the types you want is because in cand we need to apply p at two different types for the type variable 'a'. In Haskell this requires you to do something different. What you did works (both the CB (CB a) and the rank n type). As does this:

\begin{code}

type CB a = a -> a -> a

ct :: CB a

ct x y = x

cf :: CB a

cf x y = y

cand :: (forall a. CB a) -> CB a -> CB a

cand p q = p q p

\end{code}

And in fact, it still works as we'd hope:

*Main> :t cand ct

cand ct :: CB a -> a -> a -> a

In Church's ë-calc the types are ignored, but in Haskell they matter, and in a type like cand :: CB a -> CB a -> CB a, once the type of 'a' is fixed all uses of p must have the same 'a'. In the type, (forall a1. CB a1) -> CB a -> CB a, then p can be applied at as many instantiations of a1 as we like inside of cand.

I hope that helps,

Jason