# Equivalence of Definitions of Strict Ordering

## Theorem

Let $S$ be a set.

Let $\mathcal R$ be a relation on $S$.

The following definitions of the concept of Strict Ordering are equivalent:

### Definition 1

Let $\mathcal R$ be a relation on a set $S$.

Then $\mathcal R$ is a strict ordering (on $S$) if and only if the following two conditions hold:

 $(1)$ $:$ Asymmetry $\displaystyle \forall a, b \in S:$ $\displaystyle a \mathrel {\mathcal R} b$ $\displaystyle \implies$ $\displaystyle \neg \paren {b \mathrel {\mathcal R} a}$ $(2)$ $:$ Transitivity $\displaystyle \forall a, b, c \in S:$ $\displaystyle \paren {a \mathrel {\mathcal R} b} \land \paren {b \mathrel {\mathcal R} c}$ $\displaystyle \implies$ $\displaystyle a \mathrel {\mathcal R} c$

### Definition 2

Let $\mathcal R$ be a relation on a set $S$.

Then $\mathcal R$ is a strict ordering (on $S$) if and only if the following two conditions hold:

 $(1)$ $:$ Antireflexivity $\displaystyle \forall a \in S:$ $\displaystyle \neg \paren {a \mathrel {\mathcal R} a}$ $(2)$ $:$ Transitivity $\displaystyle \forall a, b, c \in S:$ $\displaystyle \paren {a \mathrel {\mathcal R} b} \land \paren {b \mathrel {\mathcal R} c} \implies a \mathrel {\mathcal R} c$

## Proof

Let $\mathcal R$ be transitive.

Then by Transitive Relation is Antireflexive iff Asymmetric it follows directly that:

$(1): \quad$ If $\mathcal R$ is antireflexive then it is asymmetric
$(2): \quad$ If $\mathcal R$ is asymmetric then it is antireflexive.

$\blacksquare$