| instance (Table a c, Show c) => Show a where I would have thought that there is on overlap: the instance in my code above defines how to show a table if the cell is showable;
No, the instance defines how to show values of any type; that type must be an instance of Table. There is no `if' here: instances are selected regardless of the context such as (Table a c, Show c) above. The constraints in the context apply after the selection, not during. Please see ``Choosing a type-class instance based on the context'' http://okmij.org/ftp/Haskell/TypeClass.html#class-based-overloading for the explanation and the solution to essentially the same problem. There are other problems with the instance: | instance (Table a c, Show c) => Show a where For example, there is no information for the type checker to determine the type c. Presumably there could be instances Table [[Int]] Int Table [[Int]] Bool So, when the a in (Table a c) is instantiated to [[Int]] there could be two possibilities for c: Int and Bool. You can argue that the type of the table uniquely determines the type of its cells. You should tell the type checker of that, using functional dependencies or associated types: class Table table where type Cell table :: * toLists :: table -> [[Cell table]]