Set of Subsets is Cover iff Set of Complements is Free

Theorem
Definitions 1 and 2 for a cover are equivalent.

Proof
Let $S$ be a set.

Let $\mathcal C$ be a set of subsets for $S$.

Definition 1 Implies Definition 2
Suppose:


 * $\bigcup \mathcal C = S$

and


 * $\{ \complement(X): X \in \mathcal C \}$

is not free.

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


 * $\forall X \in \mathcal C: x \in \complement(X)$

But by the definition of complement:


 * $\forall X \in \mathcal C: x \notin X$

which contradicts:


 * $\bigcup \mathcal C = S$

Definition 2 Implies Definition 1
Suppose:


 * $\{ \complement(X): X \in \mathcal C \}$

is free and:


 * $\bigcup \mathcal C$

is not a cover for $X$.

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


 * $\forall X \in \mathcal C: x \notin X$

Hence:


 * $\forall \complement(X) \in \mathcal C: x \in X$

which contradicts that:


 * $\{ \complement(X): X \in \mathcal C \}$

is free.

$\blacksquare$