[Chapter 6] Clarify type class signatures

Author: Robin Abbi

text/chapter6.md

@@ -164,6 +164,8 @@ true
 
 The `Ord` type class defines the `compare` function, which can be used to compare two values, for types which support ordering. The comparison operators `<` and `>` along with their non-strict companions `<=` and `>=`, can be defined in terms of `compare`.
 
+_Note_: In the example below, the class signature contains `<=`. This usage of `<=` in this context indicates that Eq is a subclass of Ord and is not intended to represent the use of `<=` as a comparison operator. See the section [Superclasses](#superclasses) below.
+
 ```haskell
 data Ordering = LT | EQ | GT
 
@@ -596,9 +598,9 @@ Note that the `Partial` constraint appears _inside the parentheses_ on the left
 
 Just as we can express relationships between type class instances by making an instance dependent on another instance, we can express relationships between type classes themselves using so-called _superclasses_.
 
-We say that one type class is a superclass of another if every instance of the second class is required to be an instance of the first, and we indicate a superclass relationship in the class definition by using a backwards facing double arrow.
+We say that one type class is a superclass of another if every instance of the second class is required to be an instance of the first, and we indicate a superclass relationship in the class definition by using a backwards facing double arrow ( `<=` ).
 
-We've already seen some examples of superclass relationships: the `Eq` class is a superclass of `Ord`, and the `Semigroup` class is a superclass of `Monoid`. For every type class instance of the `Ord` class, there must be a corresponding `Eq` instance for the same type. This makes sense, since in many cases, when the `compare` function reports that two values are incomparable, we often want to use the `Eq` class to determine if they are in fact equal.
+We've [already seen an example of superclass relationships](#ord): the `Eq` class is a superclass of `Ord`, and the `Semigroup` class is a superclass of `Monoid`. For every type class instance of the `Ord` class, there must be a corresponding `Eq` instance for the same type. This makes sense, since in many cases, when the `compare` function reports that two values are incomparable, we often want to use the `Eq` class to determine if they are in fact equal.
 
 In general, it makes sense to define a superclass relationship when the laws for the subclass mention the members of the superclass. For example, it is reasonable to assume, for any pair of `Ord` and `Eq` instances, that if two values are equal under the `Eq` instance, then the `compare` function should return `EQ`. In other words, `a == b` should be true exactly when `compare a b` evaluates to `EQ`. This relationship on the level of laws justifies the superclass relationship between `Eq` and `Ord`.