Surely that fails for e.g. a?a?a?. I'd imagine you could do some sort of simplification first though to avoid this redundancy.
Oof, my hubris.
import Prelude hiding (null)
import Data.Set (Set, toList, fromList, empty, singleton, isSubsetOf, unions, null)
data Regex = Class [Char] -- character class
| Seq [Regex] -- sequence, ABC
| Choice [Regex] -- choice, A|B|C
| Star Regex -- zero or more, A*
deriving (Show)
-- The language of a regex is either finite or infinite.
-- We only care about the finite case.
data Lang = Finite (Set String) | Infinite deriving (Show, Eq)
zero = Finite empty
one = Finite (singleton "")
isEmpty (Finite s) = null s
isEmpty Infinite = False
cat :: Lang -> Lang -> Lang
cat x y | isEmpty x || isEmpty y = zero
cat (Finite s) (Finite t) = Finite $ fromList [x ++ y | x <- toList s, y <- toList t]
cat _ _ = Infinite
subsingleton :: Lang -> Bool
subsingleton Infinite = False
subsingleton (Finite s) = isSubsetOf s (fromList [""])
eval :: Regex -> Lang
eval (Class chars) = Finite $ fromList [[c] | c <- chars]
eval (Seq rs) = foldr cat one $ map eval rs
eval (Choice rs) | any (== Infinite) langs = Infinite
| otherwise = Finite $ unions [s | Finite s <- langs]
where langs = map eval rs
eval (Star r) | subsingleton (eval r) = one
| otherwise = Infinite