Challenge 1: A Color Type With Derived Eq and Show — Possible Solution
====================================================================

Main.hs:

data Color = Red | Green | Blue
  deriving (Eq, Show)

main :: IO ()
main = do
  print (Red == Green)
  print (Red == Red)
  putStrLn (show Blue)

Output:

    False
    True
    Blue

Explanation:
deriving (Eq, Show) generates both instances automatically from
Color's own structure -- no hand-written code needed. The derived Eq
instance considers two Color values equal only if they're literally
the same constructor, so Red == Green is False and Red == Red is
True. The derived Show instance converts each constructor to its own
name as a String, so show Blue produces "Blue", exactly the
mechanism that has been powering every print call used throughout
this entire course.

WHY THIS WORKS AS AN ANSWER
------------------------------
This uses deriving exactly as the chapter introduces it for a
straightforward sum type, demonstrating both the Eq comparison and
the Show conversion working correctly with zero hand-written instance
code.