Ordering is Directed iff Composite with Inverse is Trivial Ordering

Theorem
Let $\struct {S, \RR}$ be an ordered set.

Then $\RR$ is a directed ordering :
 * $\RR^{-1} \circ \RR = S \times S$

where:
 * $\circ$ denotes composite relation
 * $\RR^{-1}$ denotes inverse relation
 * $S \times S$ denotes the trivial relation, that is, the Cartesian product of $S$ with itself.

Proof
We are given that $\RR$ is an ordering.

Sufficient Condition
Let $\RR$ be a directed ordering on $S$.

Then by definition:
 * $\forall x, y \in S: \exists z \in S: x \mathrel \RR z \land y \mathrel \RR z$

By definition of inverse relation:


 * $\forall x, y \in S: \exists z \in S: z \mathrel {\RR^{-1} } x \land z \mathrel {\RR^{-1} } y$

By definition of composite relation
 * $\RR^{-1} \circ \RR := \set {\tuple {x, y} \in S \times S: \exists z \in S: \tuple {x, z} \in \RR \land \tuple {z, y} \in \RR^{-1} }$

Let $\tuple {x, y} \in S \times S$ be arbitrary.

Then:
 * $\exists z \in \RR: \tuple {x, z} \in \R \land \tuple {y, z} \in \R$

That is:
 * $\exists z \in \RR: \tuple {x, z} \in \R \land \tuple {z, y} \in \R^{-1}$

That is:
 * $\tuple {x, y} \in \RR^{-1} \circ \RR$

As $\tuple {x, y}$ is arbitrary, it follows that:


 * $\RR^{-1} \circ \RR = S \times S$

Necessary Condition
Let $\struct {S, \RR}$ be such that:
 * $\RR^{-1} \circ \RR = S \times S$

Then by definition of composite relation:
 * $\RR^{-1} \circ \RR = S \times S = \set {\tuple {x, y} \in S \times S: \exists z \in S: \tuple {x, z} \in \RR \land \tuple {z, y} \in \RR^{-1} }$

That is:
 * $\forall \tuple {x, y} \in S \times S: \exists z \in S: \tuple {x, z} \in \RR \land \tuple {z, y} \in \RR^{-1}$

That is, by definition of inverse relation:


 * $\forall \tuple {x, y} \in S \times S: \exists z \in S: \tuple {x, z} \in \RR \land \tuple {y, z} \in \RR$

That is, $\RR$ is a directed ordering.