[Haskell-cafe] Re: Subsequence near solved hopefully

Sam Mason writes:

Just to muddy the water a bit. . . What happens if the second string is infinite?

This version should do it: isSubSeq :: (Eq a) => [a] -> [a] -> Bool isSubSeq [] _ = True isSubSeq _ [] = False isSubSeq (x:xs) (y:ys) | x == y = isSubSeq xs ys | otherwise = isSubSeq (x:xs) ys Peter