Set of Subsets is Cover iff Set of Complements is Free

Theorem
Let $S$ be a set.

Let $\mathcal C$ be a set of sets.

Then $\mathcal C$ is a cover for $S$ $\left\{{\complement_S \left({X}\right): X \in \mathcal C}\right\}$ is free.

Proof
Let $S$ be a set.

Necessary Condition
Let $\mathcal C$ be a cover for $S$:
 * $S \subseteq \bigcup \mathcal C$

Suppose:
 * $\left\{{\complement_S \left({X}\right): X \in \mathcal C}\right\}$

is not free.

Then there exists an $x \in S$ such that:


 * $\forall X \in \mathcal C: x \in \complement_S \left({X}\right)$

But by the definition of (relative) complement:
 * $\forall X \in \mathcal C: x \notin X$

That is:
 * $x \notin \bigcup \mathcal C$

which contradicts:
 * $S \subseteq \bigcup \mathcal C$

Sufficient Condition
Let:
 * $\left\{{\complement_S \left({X}\right): X \in \mathcal C}\right\}$

be free.

Suppose:
 * $\bigcup \mathcal C$

is not a cover for $S$.

Then there exists some $x \in S$ such that:
 * $\forall X \in \mathcal C: x \notin X$

Hence:
 * $\forall \complement_S \left({X}\right) \in \mathcal C: x \in X$

which contradicts the supposition that:
 * $\left\{{\complement_S \left({X}\right): X \in \mathcal C}\right\}$

is free.