Null Relation is Antireflexive, Symmetric and Transitive

Theorem
Let $S$ be a set which is non-empty.

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

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

If $S = \O$ then Relation on Empty Set is Equivalence applies.

Proof
From the definition of null relation:
 * $\mathcal R = \O$

Antireflexivity
This follows directly from the definition:
 * $\mathcal R = \O \implies \forall x \in S: \tuple {x, x} \notin \mathcal R$

and so $\mathcal R$ is antireflexive.

Symmetry
It follows vacuously that:
 * $\tuple {x, y} \in \mathcal R \implies \tuple {y, x} \in \mathcal R$

and so $\mathcal R$ is symmetric.

Transitivity
It follows vacuously that:
 * $\tuple {x, y}, \tuple {y, z} \in \mathcal R \implies \tuple {x, z} \in \mathcal R$

and so $\mathcal R$ is transitive.