Proving the simple equality $A△B = (A ∪ B) − (A ∩ B)$ I'm trying to prove the equality:
$$A\triangle B = (A ∪ B) − (A ∩ B)$$
It is such a simple equality but I'm not sure how to prove it, I just seem to write the obvious facts.
$x \in A\triangle B ⇔ x \in A$ and $x \in B$ but $x ∉ A∩B ⇔ x \in A ∪ B$ but $x ∉ A∩B$ which is the same as writing $(A ∪ B) - (A ∩ B)$
 A: The symmetric difference $A \triangle B$ is not defined as $x \in A$ AND $x \in B$ but not $x \in A \cap B$. That's a contradiction, since it amounts to saying $x \in A\cap B$ and $x \notin A\cap B$.
The symmetric difference is defined as $x \in A$ and $x \notin B$, or else, $x\in B$ and $x\notin A$: $$A \triangle B = (A-B) \cup (B - A)$$ This in turn, is equivalent to $x \in A$ or $x \in B$, and $x \notin (A \cap B)$. This is indeed equivalent to saying $x \in (A\cup B) - (A \cap B)$.  You can show this by "element chasing".
A: If you define $A\triangle B$ as $(A\setminus B)\cup (B\setminus A)$ (one of severalequivalent definitions), then you can write
$$x\in (A\cup B)\setminus (A\cap B)$$
$$\iff (x\in A \text { or }x \in B)\text{ and (not } (x\in A \text{ and }x\in B))$$
$$\iff (x\in A \text { or }x \in B)\text{ and  }(x\notin A \text{ or }x\notin B)$$
$$\iff ((x\in A \text { or }x \in B)\text{ and  }x\notin A )  \text{ or }((x\in A \text { or }x \in B)\text{ and  }x\notin B)$$
$$\iff x\in B\setminus A\text{ or } x\in A\setminus B\iff x\in A\triangle B.$$
