Dual of Dual Ordering

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

Let $\struct {S, \succeq}$ be its dual.

Then the dual of $\struct {S, \succeq}$ is again $\struct {S, \preceq}$.

Proof
Denote with $\preceq'$ the dual of $\succeq$.

By definition of dual ordering, we thus have for all $a, b \in S$:


 * $a \preceq b$ $b \succeq a$
 * $b \succeq a$ $a \preceq' b$

Hence $a \preceq b$ $a \preceq' b$.

The result follows from Equality of Relations.