Closure of Intersection and Symmetric Difference imply Closure of Set Difference

Theorem
Let $\RR$ 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:
 * $\forall A, B \in \RR: A \setminus B \in \RR$

where $\setminus$ denotes set difference.

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

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

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

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

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