Union of Orderings is not necessarily Ordering

Theorem
Let $A$ be a set.

Let $\RR$ and $\SS$ be orderings on $A$.

Then $\RR \cup \SS$ is not necessarily an ordering on $A$.

Proof
Let $\RR$ be an ordering as asserted.

Let $\SS$ be the dual ordering of $\RR$.

From Dual Ordering is Ordering, $\SS$ is an ordering.

By definition, the dual of $\RR$ is the inverse of $\RR$:
 * $\SS = \RR^{-1}$

From Union of Antisymmetric Relation with Inverse is Antisymmetric iff Diagonal, it is generally not the case that $\RR \cup \RR^{-1}$ is antisymmetric.

Hence, in general, $\RR \cup \RR^{-1}$ is not an ordering.