Closure of Intersection and Symmetric Difference imply Closure of Set Difference

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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:

$A \setminus B \in \mathcal R$

$\blacksquare$