If Ideal and Filter are Disjoint then There Exists Prime Filter Including Filter and Disjoint from Ideal

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a distributive lattice.

Let $I$ be an ideal in $L$.

Let $F$ be a filter on $L$ such that
 * $F \cap I = \varnothing$

Then there exists a prime filter $P$ in $L$: $F \subseteq P$ and $P \cap I = \varnothing$

Proof
By Dual Distributive Lattice is Distributive:
 * $L^{-1}$ is a distributive lattice

where $L^{-1} = \left({S, \succeq}\right)$ denotes the dual of $L$.

By Filter is Ideal in Dual Ordered Set:
 * $I' := F$ as an ideal in $L^{-1}$.

By Ideal is Filter in Dual Ordered Set:
 * $F' := I$ as a filter on $L^{-1}$.

By assumption:
 * $I' \cap F' = \varnothing$

By If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter:
 * there exists prime ideal $P'$ in $L^{-1}$: $I' \subseteq P'$ and $P' \cap F' = \varnothing$

By Prime Filter is Prime Ideal in Dual Lattice:
 * $P := P'$ as a prime filter on $L$.

Thus there exists prime filter $P$ on $L$: $F \subseteq P$ and $P \cap I = \varnothing$