Closure of Intersection and Symmetric Difference imply Closure of Union

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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:

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

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:

$A \cup B \in \mathcal R$

$\blacksquare$