Closure of Intersection and Symmetric Difference imply Closure of Union

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

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

Then:
 * $\forall A, B \in \mathcal R: A \cup B \in \mathcal R$

where $\cup$ denotes set union.

Proof
Let $A, B \in \mathcal R$.

From Union as Symmetric Difference with Intersection‎:
 * $\left({A * B}\right) * \left({A \cap B}\right) = A \cup B$

By hypothesis:
 * $A \cap B \in \mathcal R$

and:
 * $\left({A * B}\right) * \left({A \cap B}\right)$

and so:
 * $A \cup B \in \mathcal R$