Distance in Pseudometric is Non-Negative

Theorem
Let $X$ be a set on which a pseudometric $d: X \times X \to \R$ has been imposed.

Then:
 * $\forall x, y \in X: d \left({x, y}\right) \ge 0$

Proof
By definition of pseudometric, we have that:
 * M1: $d \left({x, x}\right) = 0$
 * M2: $d \left({x, y}\right) = d \left({y, x}\right)$
 * M3: $d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$

Hence: