Feedback on my proof that $(A\setminus B)\cup(B\setminus A)=(A\cup B)\setminus (A\cap B)$ I would like to prove that $(A\setminus B)\cup(B\setminus A)=(A\cup B)\setminus (A\cap B)$. Please could you offer some feedback?
Firstly, I will show that:
$(A\setminus B)\cup(B\setminus A)\subseteq(A\cup B)\setminus (A\cap B).\tag{1}$
Assume $x \in (A\setminus B)\cup(B\setminus A)$. Then, by the definition of union, $x 
\in A\setminus B$, or $x \in B\setminus A$, or both. 
(i) but if $x \in A\setminus B$ - then $x \in A$, $x \notin B$ and $x \notin A \cap B$. 
(ii) or if $x \in B\setminus A$, then $x \in B$ and $x \notin A$  and $x \notin B \cap A$. 
Putting the (i) and (ii) together, either $x \in A$ or $x \in B$ but $x\notin A\cap B$.  This proves (1).
$(A\cup B)\setminus (A\cap B)\subseteq(A\setminus B)\cup(B\setminus A)\tag{2}$
Assume $x \in (A \cup B) \setminus (A \cap B)$.Then $x \in A$ or $x \in B$ or both by the definition of union. However we know that  $x \notin A \cap B$. Therefore it must be the case that either $x \in A$ and $x \notin B$ or $x \in B$ and $x \notin A$. This proves (2).
Thanks in advance.
 A: I think you did quite well, and it's the most appropriate approach for you at this point in time:
Just expanding on your work (filling in some gaps which were easy enough to assume):

Assume $x \in (A\setminus B)\cup(B\setminus A)$. Then, by the definition of union, 

(a) $x \in A\setminus B$, or $x \in B\setminus A$.
(a.i) $x \in A\setminus B$, then $x \in A$ and $x \notin B$, so $x \notin A \cap B$. 
or
(a. ii) $x \in B\setminus A$, then $x \in B$ and $x \notin A$, so $x \notin B \cap A$. 
From (a), and (a.i) and (a.ii), we have ($x \in A$ or $x \in B$) and $x\notin (A\cap B)$.  Hence, $x \in (A\cup B)$ and $x \notin (A\cap B)$. That is, $x \in (A\cup B)\setminus(A\cap B)$.
Therefore, $(A\setminus B)\cup(B\setminus A)\subseteq(A\cup B)\setminus (A\cap B).\tag{1}$

Now, Assume $x \in [(A \cup B) \setminus (A \cap B)]$.

Then $x \in (A\cup B)$ and $x\notin (A \cap B)$, by the definition of setminus.
Then $x \in A$ or $x \in B$  by the definition of union. However we also know that  $x \notin (A \cap B)$.  Therefore it must be the case that either $(x \in A$ and $x \notin B)$ or ($x \in B$ and $x \notin A$). That is, either $x \in (A \setminus B)$ or $x \in (B \setminus A)$. So $x \in (A\setminus B) \cup (B \setminus A)$. 
Therefore, $(A\cup B)\setminus (A\cap B)\subseteq(A\setminus B)\cup(B\setminus A)\tag{2}$

Very nicely done, by the way.
A: This is not a feedback, but just a better way to do it. I will use the distributivity of $\cup$ over $\cap$.
$$(A/ B)\cup(B/A)=$$
$$(A\cap B^c)\cup(A^c\cap B)=$$
$$(A\cup A^c)\cap(A\cup B)\cap(B^c\cup A^c)\cap(B^c\cup B)=$$
$$(A\cup B)\cap(B^c\cup A^c)=$$
$$(A\cup B)\cap(A\cap B)^c=$$
$$(A\cup B)/(A\cap B)$$
