Equivalence of Definitions of Strict Ordering

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Theorem

Let $S$ be a set.

Let $\RR$ be a relation on $S$.

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


Definition 1

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

Then $\RR$ is a strict ordering (on $S$) if and only if $\RR$ satisfies the strict ordering axioms:

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


Definition 2

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

Then $\RR$ is a strict ordering (on $S$) if and only if $\RR$ satisfies the strict ordering axioms:

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


Proof

Let $\RR$ be transitive.

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

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

$\blacksquare$