Bar will extend the methods you implemented in foo (and the ones you did not provide implementations for if any). If you did not provide concrete implementations of the methods from baz in foo, then foo is abstract. If bar doesn't provide concrete implementations of these methods then bar will also be abstract.
posted 13 years ago
Ah, so baz becomes an integral part of foo (or bar) once the interface's abstract methods are made concrete. Thus obviating the need for a subclass (bar) to implement a superclass' interface (baz), as it's already been 'Borg-ed' by the superclass (foo).