Closure of Union and Complement imply Closure of Set Difference

Theorem
Let $\RR$ be a system of sets on a universe $\mathbb U$ such that for all $A, B \in \RR$:
 * $(1): \quad A \cup B \in \RR$
 * $(2): \quad \map \complement A \in \RR$

where $\cup$ denotes set union and $\complement$ denotes complement (relative to $\mathbb U$).

Then:
 * $\forall A, B \in \RR: A \setminus B \in \RR$

where $\setminus$ denotes set difference.

Proof
Let $A, B \in \RR$.

As both set union and complement are closed in $\RR$ the result follows.