Null Relation is Antireflexive, Symmetric and Transitive

Theorem
Let $S$ be a set which is not empty.

Let $\mathcal R \subseteq S \times S$ be the null relation.

Then $\mathcal R$ is antireflexive, symmetric and transitive.

If $S = \varnothing$ then Relation on Null Set is Equivalence applies.

Proof
From the definition of null relation, $\mathcal R = \varnothing$.

Antireflexivity
This follows directly from the definition:
 * $\mathcal R = \varnothing \implies \forall x \in S: \left({x, x}\right) \notin \mathcal R$

and so $\mathcal R$ is antireflexive.

Symmetry
It follows vacuously that:
 * $\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \in \mathcal R$

and so $\mathcal R$ is symmetric.

Transitivity
It follows vacuously that:
 * $\left({x, y}\right), \left({y, z}\right) \in \mathcal R \implies \left({x, z}\right) \in \mathcal R$

and so $\mathcal R$ is transitive.