Closure of Intersection and Symmetric Difference imply Closure of Union

Theorem
Let $\R R$ be a system of sets such that for all $A, B \in \RR$:
 * $(1): \quad A \cap B \in \RR$
 * $(2): \quad A \symdif B \in \RR$

where $\cap$ denotes set intersection and $\symdif$ denotes set symmetric difference.

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

where $\cup$ denotes set union.

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

From Union as Symmetric Difference with Intersection‎:
 * $\paren {A \symdif B} \symdif \paren {A \cap B} = A \cup B$

By hypothesis:
 * $A \cap B \in \RR$

and:
 * $\paren {A \symdif B} \symdif \paren {A \cap B} \in \RR$

and so:
 * $A \cup B \in \RR$