A virtual member function is used if it is not pure?

Question

C++03 3.2.2 ...An object or non-overloaded function is used if its name appears in a potentially-evaluated expression. A virtual member function is used if it is not pure...

And then later in 3.2.3 we have: Every program shall contain exactly one definition of every non-inline function or object that is used in that program; no diagnostic required. The definition can appear explicitly in the program, it can be found in the standard or a user-defined library, or (when appropriate) it is implicitly defined (see 12.1, 12.4 and 12.8). An inline function shall be defined in every translation unit in which it is used.

Along the lines I am reading: a pure virtual function is not used. The ODR applies only to functions which are used. Doesn't this imply that the following would be legal? I am guessing the answer is no, it doesn't, but then I can't understand why.

//x.h
struct A
{
   virtual void f() = 0;
};

//y.cpp
#include "x.h"
void A::f()
{
}

//z.cpp
#include "x.h"
#include <iostream>
void A::f()
{
   std::cout << "Hello" << std::endl;
}

//main.cpp
#include "x.h"
struct B:A
{
   virtual void f()
   {
      A::f();
   }
};

int main()
{
   A* p = new B;
   p->f();
}

Answer 1

The two clauses are not mutually exclusive. That a virtual function is used if it is not pure, does not mean that the converse holds. If a virtual function is pure it does not mean that it is necessarily not used. It may still be used "if its name appears in a potentially evaluated expression" such as in your example: A::f();.