[SOLVED] Haskell dot (.) operator in de Morgan's law implementation

Haskell dot (.) operator in de Morgan's law implementation

In this question, the author has written an implementation of de Morgan's laws in Haskell. I understand the implementations of notAandnotB, and notAornotB, but I'm struggling to understand the implementation of notAorB which is:

notAorB :: (Either a b -> c) -> (a -> c, b -> c)
notAorB f = (f . Left, f . Right)

Could someone explain how the (f . Left, f . Right) part works? I've seen the . operator used before, but with three arguments, not two.

Thank you in advance.

Solution

Recall that the definition of the . operator is (f . g) x = f (g x), i.e. f . g = \x -> f (g x) (syntactically, it is a binary operator, it is just that Haskell’s syntax sugar permits the latter definition to be restated as the former). So, the definition you have can be rephrased as

notAorB f = ((\x -> f (Left x)), (\y -> f (Right y)))

(this can be done mechanically by Lambdabot on #haskell, tell him to @unpl ‹expr›), or more verbosely

notAorB f = (lt, rt)
  where lt x = f (Left x)
        rt y = f (Right y)

As always, try writing down the types. If (.) :: ∀ s t u. (t -> u) -> (s -> t) -> s -> u, f :: Either a b -> c, Left :: ∀ p q. p -> Either p q, then f . Left or (.) f Left :: a -> c, etc.