Complements of Components of Two-Component Partition form Partition

Theorem
Let $S$ be a set with at least two elements.

Let $A, B \subseteq S$.

Let $\complement_S$ denote the complement relative to $S$.

$A \mid B$ is a partition of $S$ iff $\complement_S \left({A}\right) \mid \complement_S \left({B}\right)$ is a partition of $S$.

Necessary Condition
Let $A \mid B$ be a partition of $S$.

That is, by definition:


 * $(1): \quad A \cap B = \varnothing$
 * $(2): \quad A \cup B = S$
 * $(3): \quad A, B \ne \varnothing$

Thus:

Now suppose WLOG that $\complement_S \left({A}\right) = \varnothing$.

But $B \ne \varnothing$ by hypothesis.

Thus:
 * $A \ne \varnothing \implies \complement_S \left({A}\right) \ne \varnothing$

Mutatis mutandis:
 * $B \ne \varnothing \implies \complement_S \left({B}\right) \ne \varnothing$

Thus all three conditions are satisfied for $\complement_S \left({A}\right) \mid \complement_S \left({B}\right)$ to be a partition of $S$.

Sufficient Condition
Let $\complement_S \left({A}\right) \mid \complement_S \left({B}\right)$ is a partition of $S$.

By the necessary condition, it follows that $\complement_S \left({\complement_S \left({A}\right)}\right) \mid \complement_S \left({\complement_S \left({B}\right)}\right)$ is a partition of $S$.

From Relative Complement of Relative Complement:
 * $\complement_S \left({\complement_S \left({A}\right)}\right) = A$ and $\complement_S \left({\complement_S \left({B}\right)}\right) = B$

and so $A \mid B$ is a partition of $S$.