class T root pos sel | pos -> root, root -> sel where f :: pos -> sel -> Bool instance T root (Any root) sel
But the same applies to the second functional dependency and the type variable sel. Every instantiation of root determines the instantiation of sel. And that forbids instance T Int (Any Int) Bool and instance T Int (Any Int) Int inside the same scope, doesn't it?
Indeed that is your intent, expressed in the functional dependency. It may help to think of a class declaration as an `interface' and of the set of instances as an `implementation' (of the type class). In the example above, the "class T root pos sel" _declares_ a ternary relation T and specifies some `constraints'. The set of instances of T (in our example, there is only one instance) specifies the triples whose set defines the relation T. In Herbrand interpretation, an unground instance instance C1 x y => C (Foo x) (Bar y) corresponds to a set of instances where the free type variables are substituted by all possible ground types provided the instance constraints (such as C1 x y) hold. In our example, an unground instance |instance T root (Any root) sel| is equivalent to a set of ground instances where |root| and |sel| are replaced with all possible ground types. Including instance T Int (Any Int) Bool instance T Int (Any Int) Int These two instances are in the model for `instance T root (Any root) sel'. A set of instances, an implementation of a type class, must satisfy the interface, that is, constraints imposed by the class declaration, including the functional dependency constraints. In our example, any implementation of T must satisfy root -> sel constraints. The above two instances show there exists a model of T where the functional dependency is violated. That's why both GHC 6.4 and Hugs reject the instance. Again, it is a mystery why GHC 6.6 accepts it.