Java 8 supports a new use of super:

interface I {
default int f(){return 0;}
}

class X implements I {
public int f() {
return I.super.f();
}
}

This is described in 15.12.1.

However, there is wording in 15.12.1 that says: Let T be the type declaration immediately enclosing the method invocation. It is a compile-time error if I is not a direct superinterface of T, or if there exists some other direct superclass or direct superinterface of T, J, such that J is a subtype of I.

That would seem to require an error on this case, because T has a superclass of J, which is a subtype of I.

interface I {
        default int f(){return 0;}
}
class J implements I { }
class T extends J implements I {
        public int f() {
                return I.super.f();
        }
}