Yes.
add :: Num a => a -> a -> a
add = \a -> \b -> a + b
This is also an example of a closure. \b -> a + b
has access to a
through a closure because a is defined outside the scope of that lambda.
Edit
These are called Peano number values
data Nat = Zero | Succ Nat
add Zero = \n -> n
add (Succ a) = \b -> Succ (add a b)
toNat 0 = Zero
toNat n = Succ (toNat (n-1))
fromNat Zero = 0
fromNat (Succ n) = 1 + fromNat n
λ: fromNat $ add (toNat 3) (toNat 4)
7