# Duality Principle (Order Theory)/Local Duality

This proof is about Local Duality in Order Theory. For other uses, see Duality Principle.

## Theorem

Let $\Sigma$ be a statement about ordered sets (whether in natural or a formal language).

Let $\Sigma^*$ be the dual statement of $\Sigma$.

Let $\left({S, \preceq}\right)$ be an ordered set, and let $\left({S, \succeq}\right)$ be its dual.

Then the following are equivalent:

$(1): \quad \Sigma$ is true for $\left({S, \preceq}\right)$
$(2): \quad \Sigma^*$ is true for $\left({S, \succeq}\right)$

## Proof

### $(1)$ implies $(2)$

By assumption, $\Sigma$ is true for $\left({S, \preceq}\right)$.

By Dual of Dual Ordering, the dual statement $\Sigma^*$ applied to $\left({S, \succeq}\right)$ is the same as $\Sigma$ applied to $\left({S, \preceq}\right)$.

Hence $\Sigma^*$ is true for $\left({S, \succeq}\right)$.

$\Box$

### $(2)$ implies $(1)$

From Dual of Dual Statement (Order Theory), $\left({\Sigma^*}\right)^* = \Sigma$.

From Dual of Dual Ordering, $\left({S, \preceq}\right)$ is the dual of $\left({S, \succeq}\right)$.

The result thus follows from applying the other implication to $\Sigma^*$ and $\left({S, \succeq}\right)$.

$\blacksquare$