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 $S$ $\relcomp S A \mid \relcomp S B$ is a partition of $S$.

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

That is, by definition:


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

Thus:

, now suppose that $\relcomp S A = \O$.

Thus by Rule of Transposition:
 * $B \ne \O \implies \relcomp S A \ne \O$

Mutatis mutandis:
 * $A \ne \O \implies \relcomp S B \ne \O$

by hypothesis:
 * $A, B \ne \O$

and so:
 * $\relcomp S A, \relcomp S B \ne \O$

Thus all three conditions are satisfied for $\relcomp S A \mid \relcomp S B$ to be a partition of $S$.

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

By the necessary condition, it follows that $\relcomp S {\relcomp S A} \mid \relcomp S {\relcomp S B}$ is a partition of $S$.

From Relative Complement of Relative Complement:
 * $\relcomp S {\relcomp S A} = A$ and $\relcomp S {\relcomp S B} = B$

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