I just want to make sure I'm thinking of this correctly.
From what I understand, this is basically saying that:
If the union of everything in set A that's not in set B and everything that's in set B but not A = The union of A and B, then set A and B have no common elements.
Which would be true, because the (A - B) ∪ (B - A) part would exclude shared elements.
Is this correct?
Yes, this is true. Look up symmetric difference. It is defined as $$(A - B) \cup (B - A),$$ although an equivalent definition is $$(A \cup B) - (A \cap B).$$
You know that $$ (A - B) \cup (B - A) = A\cup B $$ and you want to prove that $$ A\cap B = \emptyset. $$ Say that there is an element $x\in A\cap B$. You want to prove that no such exist, so assume that is does. Then Now then $x\in A$ and $x\in B$. So, certainly, $a\in A\cup B = (A - B) \cup (B - A).$ If an element is in the union of two sets, then it is one of the sets (maybe in both). So $x\in A- B$ or $x\in B- A$. But both of these options don't hold. Saying that, for example, $x\in A - B$ is saying that $x\notin B$ which contradicts that $x\in A\cap B$. Hence no such $x$ exists.
