Ideals equal Filters in Dual Ordered Set

Theorem
Let $L_1 = \left({S, \preceq_1}\right)$ be an ordered set.

Let $L_2 = \left({S, \preceq_2}\right)$ be a dual ordered set of $L_1$.

Then $\mathit{Ids}\left({L_1}\right) = \mathit{Filt}\left({L_2}\right)$

where
 * $\mathit{Filt}\left({L_2}\right)$ denotes the set of all filters of $L_2$,
 * $\mathit{Ids}\left({L_1}\right)$ denotes the set of all ideals of $L_1$.

Proof
Let $x$ be a set.

By definition of $\mathit{Filt}$:
 * $x \in \mathit{Filt}\left({L_2}\right) \iff x$ is a filter on $L_2$.

By Ideal is Filter in Dual Ordered Set:
 * $x \in \mathit{Filt}\left({L_2}\right) \iff x$ is an ideal in $L_1$.

By definition of $\mathit{Ids}$:
 * $x \in \mathit{Filt}\left({L_2}\right) \iff x \in \mathit{Ids}\left({L_1}\right)$

Hence by definition of set equality:
 * $\mathit{Ids}\left({L_1}\right) = \mathit{Filt}\left({L_2}\right)$