There is a tiny error in the writeup above by r2001, and
it presents the occasion for a cute little proof.
M1 follows from M2, M3 and M4 as follows:
0 = d(x,x) from M2
d(x,x) < = d(x,y) + d(y,x) from M4
d(x,y) + d(y,x) = d(x,y) + d(x,y) = 2d(x,y) from M3
so...
0 = d(x,x) < = 2d(x,y)
and thus 0 < = d(x,y).
So 'd' need only satisfy M2, M3, and M4 if it is to be a metric.
Of course, only a mathematician would think to take z = y in M4....