Dual of Lattice Ordering is Lattice Ordering

Theorem
Let $\preccurlyeq$ be a lattice ordering.

Then its inverse $\preccurlyeq^{-1}$ or $\succcurlyeq$ is also a lattice ordering.

Proof
Let $\left({S, \preccurlyeq}\right)$ be a lattice.

From Inverse of Ordering is Ordering we have that $\succcurlyeq$ is an ordering.

It remains to be shown that for all $x, y \in S$, the ordered set $\left({\left\{{x, y}\right\}, \succcurlyeq}\right)$ admits both a supremum and an infimum.

Let $x, y \in S$.

Then $\left({\left\{{x, y}\right\}, \preccurlyeq}\right)$ admits both a supremum and an infimum.

Let $c = \sup \left({\left\{{x, y}\right\}, \preccurlyeq}\right)$.

Then by definition of supremum:
 * $\forall s \in \left\{{x, y}\right\}: s \preccurlyeq c$
 * $\forall d \in S: c \preccurlyeq d$
 * where $d$ is an upper bound of $\left({\left\{{x, y}\right\}, \preccurlyeq}\right) \subseteq S$.

Hence by definition of inverse relation:
 * $\forall s \in \left\{{x, y}\right\}: c \succcurlyeq s$
 * $\forall d \in S: d \succcurlyeq c$;
 * where $d$ is an upper bound of $\left({\left\{{x, y}\right\}, \preccurlyeq}\right) \subseteq S$.

But by Upper Bound is Lower Bound for Inverse Ordering, $d$ is a lower bound of $\left({\left\{{x, y}\right\}, \succcurlyeq}\right) \subseteq S$.

So by definition of infimum, $c = \inf \left({\left\{{x, y}\right\}, \succcurlyeq}\right)$.

That is, $\left({\left\{{x, y}\right\}, \succcurlyeq}\right)$ admits an infimum.

Using a similar technique it can be shown that:
 * If $c = \inf \left({\left\{{x, y}\right\}, \preccurlyeq}\right)$, then $c = \sup \left({\left\{{x, y}\right\}, \succcurlyeq}\right)$

Hence $\left({\left\{{x, y}\right\}, \succcurlyeq}\right)$ admits both a supremum and an infimum.

That is, $\succcurlyeq$ is a lattice ordering.