Confusing (to me) statement from "Type Classes in Haskell"

Question

I'm reading up on type classes, and started looking at the paper Type Classes in Haskell.

In Section 2.2 - Superclasses, the authors use the following example:

class (Eq a) => Ord a where 
   (<)  :: a -> a -> Bool
   (<=) :: a -> a -> Bool

Then, they proceed to state that "This declares that type a belongs to class Ord if there are operations (<) and (<=) of the appropriate type and if a belongs to class Eq. Thus, if (<) is defined on some type, then (==) must be defined on that type as well."

The second sentence does not make any sense to me; why would a type that defines (<) have to define (==) if it is not declared to be an instance of either Ord or Eq?

Answer 1

Simple answer: When the above class definition is in scope, you simply cannot bind (<) or (<=) except as part of an instance declaration for Ord.

Answer 2

[This is the same answer as Andreas Rossberg's, a bit more elaborated.]

Note that the only way you can overload a function symbol in Haskell is via type classes. So, if I want to define < for some type, say (Int, Int), then the only way to do so is by saying that (Int, Int) is an instance of Ord. But, in order to do that, (Int, Int) must first be an instance of Eq. So, I have to make (Int, Int) an instance of Eq, define its == operation, then make it an instance of Ord and define its < and <= operations.

[You might also see the related question Type classes vs object interfaces where similar issues are mentioned.]