There is no such thing as an "explicitly specialized class". There are classes, class templates, and explicit specializations of class templates (also called "class template explicit specialization"). The first sentence you quote is simply missing "template" at the end. I haven't waded through the rest of it, but likely "template" is missing in a few places.

@jensmaurer I haven't waded through the rest of it

If you have done that, you would find many defects in this whole clause [temp.expl.spec], I believe. I'm implying CWG529

Incidentally, I think it may exist the (member) class explicit specialization, consider the following case

template<class T>
struct A{
  class B{};
};
template<>
struct A<int>::B{ int b;};  // class explicit specialization for the enclosing class template specialization.

In this case, B is a templated class instead of a class template, we can declare the explicit specialization for this templated class. I also think [temp.expl.spec] p6 should uniformly apply to either case(class template explicit specialization or class explicit specialization).

This reading does not make sense: You can also explicitly specialize a member function of a class template, but your reading doesn't provide for that. I think we're simply missing a few "template" words in p6.

template<class T>
struct A{
  class B{};
};
template<>
struct A<int>::B{ int b;};  // class explicit specialization for the enclosing class template specialization.

That's an explicit specialization for A<int>::B. Paragraph 6 would apply if it were a member B of an explicitly specialized A<int>. But in this example, A<int> is not explicitly specialized.

The definition of an explicitly specialized class is unrelated to the definition of a generated specialization.

The above rule in paragraph 6 may cover the case I mentioned if we could say the A<int>::B is an explicitly specialized class. However, What I'm concerning is the above rule should have also covered an explicitly specialized class template that is a member class template or other entity that can be explicitly specialized.

template<class T>
struct D{
  template<class U>
  struct E{};  //#1
};
template<>
template<class U>
struct D<int>::E{
   U member_;
};

Since we explicitly specialized the member class template for class template specialization D< int >, hence the definition of the explicitly specialized member class template is unrelated to that at #1, paragraph 6 should also cover this case. Maybe, the rule could extend to that

The definition of an explicitly specialized entity is unrelated to the definition of that entity or generated specialization from that entity.

Incidentally, template<> struct A<int>::B{/*...*/} is an explicit specialization of the member class B as @JohelEGP said(and I also meant that meaning in the above comment), instead of an explicit specialization of the enclosing class template. This concern comes from the meaning of the wording "explicit specialized", which is given multi-meaning in [temp.expl.spec], which is covered by CWG529.