$(A-B) \cup (B-A) = (A\cup B) - (A \cap B)$ So it is as it seen in the title. Here is why I am stuck:
Let $x \in [(A-B) \cup (B-A)] \Longrightarrow x \in (A-B)$ or $x \in (B-A) $ $\Longrightarrow$ ($x \in A$ and $x \not \in B$) or ($x \not \in A$ and $x \in B$).
Here is where I am stuck. Can I product these parenthesis? Like:
(x in A or x not in A) and ( x in A or x in B) and (x not in B or x not in A) and (x not in B or x in B).
And please, how I can learn write with symbols? Cant find the button. So sorry.

Instead of "multiplying the parentheses" you can see each case separately:

*

*If $x\in A\text{ and }x\not\in B$, then $x\in A\cup B$ (because $A$ is a subset of $A\cup B$ and $x\in A$) and $x\not\in A\cap B$ (because $A\cap B$ is a subset of $B$ and $x\not\in B$), therefore $x\in (A\cup B)-(A\cap B)$.


*If $x\not\in A\text{ and }x\in B$, then $x\in A\cup B$ (because $B$ is a subset of $A\cup B$ and $x\in B$) and $x\not\in A\cap B$ (because $A\cap B$ is a subset of $A$ and $x\not\in A$), therefore $x\in (A\cup B)-(A\cap B)$.
In both cases you conclude that $x\in (A\cup B)-(A\cap B)$, which proves that every element of $(A-B)\cup (B-A)$ is and element of $(A\cup B)-(A\cap B)$.
The only thing left to prove that these two sets are equal is the converse, i.e that every element of $(A\cup B)-(A\cap B)$ is an element of $(A-B)\cup (B-A)$.
