Closure of Intersection and Symmetric Difference imply Closure of Set Difference

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:
 * $\forall A, B \in \mathcal R: A \setminus B \in \mathcal R$

where $\setminus$ denotes set difference.

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

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

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

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

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