$M(3):=\text{ For all $x,y,z \in X$, $d(x,z) \leq d(x,y)+d(y,z)$}$
$M'(3):= \text { If $x,y,z \in X$ are distinct, then $d(x,z)\leq d(x,y)+d(y,z)$}$. [all the other axioms are identical]
Proof:
To show, $M(3) \implies M'(3)$: $d(x,z) \leq d(x,y)+d(y,z) \quad \forall x, y, z\in X \implies d(x,z) \leq d(x,y)+d(y,z) \quad \forall x \neq y \neq z \neq x \in X $
To show, $M'(3) \implies M(3):$ $d(x,z) \leq d(x,y)+d(y,z) \quad \forall x\neq y\neq z \neq x$. The other axioms remaining identical, we resolve the rest of the cases.
When $x=y=z$, implication holds since $LHS=0$.
When $x=y \neq z$, inequality holds.
When $x =z \neq y $ and when $x \neq y=z$ , the inequality holds. Thereby
$ M(1) \land M(2) \land M'(3) \implies M(1) \land M(2) \land M(3)$ and
$M(3) \implies M'(3) \implies M(1) \land M(2) \land M(3) \implies M(1) \land M(2) \land M'(3)$
So, $M(3)$ can be replaced by the weaker $M'(3)$.
Is this correct? Any suggestions?