(A1) For distinct points x; y 2 X, there is z 2 X, such that d(x; y; z) 6= 0 .
(A2) d(x; y; z) = 0 if two of the triple x; y; z 2 X are equal.
(A3) d(x; y; z) = d(x; z; y) = (symmetry in all three variables),
(A4) d(x; y; z) d(x; y; a) + d(x; a; z) + d(a; y; z),
for all x; y; z; a 2 X,