Relation is Antireflexive iff Disjoint from Diagonal Relation

Theorem

Let $\RR \subseteq S \times S$ be a relation on a set $S$.

Then:

$\RR$ is antireflexive

if and only if

$\Delta_S \cap \RR = \O$

where $\Delta_S$ is the diagonal relation.

Proof

Necessary Condition

Let $\RR$ be an antireflexive relation.

Let $\tuple {x, y} \in \Delta_S \cap \RR$.

By definition:

$\tuple {x, y} \in \Delta_S \implies x = y$

Likewise, by definition:

$\tuple {x, y} \in \RR \implies x \ne y$.

Thus:

$\Delta_S \cap \RR = \set {\tuple {x, y}: x = y \land x \ne y}$

and so:

$\Delta_S \cap \RR = \O$

$\Box$

Sufficient Condition

Let $\Delta_S \cap \RR = \O$.

Then by definition:

$\forall \tuple {x, y} \in \RR: \tuple {x, y} \notin \Delta_S$

Thus:

$\not \exists \tuple {x, y} \in \RR: x = y$

Thus by definition, $\RR$ is antireflexive.

$\blacksquare$