I'm studying for my exam and I came up with this little proof, but I'm wary because the professor took a much longer approach. Am I right in saying that a symmetric difference is the same as the difference between a union and an intersection? Thanks in advance.
Suppose $A$ and $B$ are disjoint.
This means $A \cap B = \varnothing$
Since $A \bigtriangleup B$ is the set of all elements that belong to $A$ or $B$ but not to both,
$$A \bigtriangleup B = A \cup B – A \cap B$$
$A \cap B = \varnothing$, therefore $A \bigtriangleup B = A \cup B$.
Best Answer
$A \Delta B=(A-B)\cup(B-A)$ $=(A\cap B^C)\cup(B\cap A^C)$ $=(A\cup B)\cap(A^C\cup B^C)$ $=(A\cup B)\cap(A\cap B)^C=(A\cup B)-(A\cap B)$