Challenge 2: The Same Chain, Written With do-Notation — Possible Solution
====================================================================

Main.hs:

findUser :: Int -> Maybe String
findUser 5 = Just "alice@example.com"
findUser _ = Nothing

findOrderByEmail :: String -> Maybe String
findOrderByEmail "alice@example.com" = Just "Order #42"
findOrderByEmail _ = Nothing

findRecentOrderDo :: Int -> Maybe String
findRecentOrderDo userId = do
  email <- findUser userId
  order <- findOrderByEmail email
  return order

main :: IO ()
main = do
  print (findRecentOrderDo 5)
  print (findRecentOrderDo 99)

Output:

    Just "Order #42"
    Nothing

Explanation:
findRecentOrderDo produces IDENTICAL results to Challenge 1's >>=
chain for both inputs, because do-notation is pure syntax sugar over
exactly that same >>= chain -- `email <- findUser userId` desugars to
`findUser userId >>= \email -> ...`, and the whole do block desugars
to the same nested >>= expression Challenge 1 wrote by hand. Nothing
about the underlying behavior changed; only the surface syntax did.

WHY THIS WORKS AS AN ANSWER
------------------------------
This rewrites Challenge 1's exact scenario using do-notation and
confirms both the success and failure cases produce identical
results to the explicit >>= version, directly demonstrating the
chapter's own "do-notation is pure sugar over >>=" claim.