Null Relation is Antireflexive, Symmetric and Transitive

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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.

$\Box$


Symmetry

It follows vacuously that:

$\tuple {x, y} \in \mathcal R \implies \tuple {y, x} \in \mathcal R$

and so $\mathcal R$ is symmetric.

$\Box$


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.

$\blacksquare$


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