Suppose that $A$, $B$, and $C$ are sets.

• $A\triangle B=(A\setminus B)\cup(B\setminus A)$.

• $A\triangle B=A^{c}\triangle B^{c}$, where the superscript $c$ denotes taking complements.

• Note that for any set $A$, the symmetric difference satisfies $A\triangle A=\emptyset$ and $A\triangle\emptyset=A$.

• The symmetric difference operator is commutative since $A\triangle B=(A\setminus B)\cup(B\setminus A)=(B\setminus A)\cup(A\setminus B)=B\triangle A$.

• The symmetric difference operation is associative: $(A\triangle B)\triangle C=A\triangle(B\triangle C)$. This means that we may drop the parentheses without any ambiguity, and we can talk about the symmetric difference of multiple sets.

• Let $A_{1},\ldots,A_{n}$ be sets. The symmetric difference of these sets is written

  $$\substack{n\\ \displaystyle{\triangle}\\ i=1}A_{i}.$$

  In general, an element will be in the symmetric difference of several sets iff it is in an odd number of the sets.

It is worth noting that these properties show that the symmetric difference operation can be used as a group law to define an abelian group on the power set of some fixed set.

Finally, we note that intersection distributes over the symmetric difference operator:

giving us that the power set of a given fixed set can be made into a Boolean ring using symmetric difference as addition, and intersection as multiplication.