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**....