scanl does not participate in list fusion

nomeata came up with a fusable version of a strict scanl' in a comment on #9345 (closed) that uses the same basic technique as the new foldl. I think we should use something like the following non-strict version to replace the current scanl:

scanl :: (b -> a -> b) -> b -> [a] -> [b]
scanl f a bs = build $ \c n ->
    a `c`
    foldr (\b g x -> let b' = f x b in (b' `c` g b'))
          (const n)
          bs
          a

While we're at it, we should make sure that map g . scanl f a and scanl f a . map g fuse properly.