Question

So I was playing around with Haskell today, thinking about autogeneration of function definitions given a type.

For example, the definition of the function

twoply :: (a -> b, a -> c) -> a -> (b, c)

is obvious to me given the type (if I rule out use of undefined :: a).

So then I came up with the following:

¢ :: a -> (a ->b) -> b
¢ = flip ($)

Which has the interesting property that

(¢) ¢ ($) :: a -> (a -> b) -> b

Which brings me to my question. Given the relation =::= for “has the same type as”, does the statement x =::= x x ($) uniquely define the type of x? Must x =::= ¢, or does there exist another possible type for x?

I’ve tried to work backward from x =::= x x ($) to deduce x :: a -> (a -> b) -> b, but gotten bogged down.

Answer 1

x =::= x x ($) is also true for x = const, which has the type a -> b -> a. So it does not uniquely identify the type.