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:

$A \cup B \in \RR$

$\blacksquare$