5.3.4. Theorem 5.3.

(a)      d( , A i ,, A j , )=d( , A j ,, A i , )  for all ij .

(b)      d( , A i ,, A i , )=0

Proof of (a)

By repeated use of theorem 5.2, we have

        d( , A i ,, A j , )=d( , A i + A j ,, A j , )

                        =d( , A i + A j ,, A j ( A i + A j ), )   =d( , A i + A j ,, A i , )

                        =d( , A i + A j A i ,, A i , )   =d( , A j ,, A i , )

                        =d( , A j ,, A i , )                      [axiom 1 used]

Proof of (b)

Using (a), we have

        d( , A i ,, A i , )=d( , A i ,, A i , )=0