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 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.