Proof of $(A\cup B)-(A\cap B)=(A-B)\cup(B-A)$ I was trying to prove $(A\cup B)-(A\cap B)=(A-B)\cup(B-A)$ and came across issues in translating (pertaining to what I did with $\emptyset$) and got through the proof but was doubting its accuracy so if someone could please affirm or deny its accuracy that would be great. Also, I'm guessing there's much shorter ways of proving this and was thinking of what those ways might be and all I could come up with was set construction. Any ideas on others? 
My proof is as follows:
Suppose $x\in(A\cup B)-(A\cap B)$.
Thus $x\in(A\cup B):x\notin(A\cap B)$.
Hence $x\in A$ or $x\in B:x\notin A$ or $x\notin B$.
This presents us with 4 cases:


*

*$x\in A:x\notin A$

*$x\in A:x\notin B$

*$x\in B:x\notin A$

*$x\in B:x\notin B$


Case 1 and 4 are $\emptyset$ however so we really have $x\in\emptyset$, or $x\in A:x\notin B$, or $x\in B:x\notin A$, or $x\in\emptyset$ which is equivalent to
$x\in\emptyset$, or $x\in(A-B)$, or $x\in(B-A)$, or $x\in\emptyset$.
Which is the same as $x\in\emptyset\cup(A-B)\cup(B-A)\cup\emptyset$
And since the union of the empty set with any set X is the set X itself, we have $x\in(A-B)\cup(B-A)$, which shows that $\forall x[x\in(A\cup B)-(A\cap B)\Rightarrow x\in(A-B)\cup(B-A)]$.
Alternatively, suppose that $x\in(A-B)\cup(B-A)$. Since the union of any set X is the set X itself, we can say that $x\in\emptyset_1\cup(A-B)\cup(B-A)\cup\emptyset_2$ where $\emptyset_1={x\in A:x\notin A}$ and $\emptyset_2={x\in B:x\notin B}$ so $x\in\emptyset_1$, or $x\in(A-B)$, or $x\in(B-A)$, or $x\in\emptyset_2$, thus $x\in [x\in A:x\notin A] $, or $x\in A:x\notin B$, or $x\in B:x\notin A$, or $x\in B:x\notin B$, hence $x\in A:x\notin A$, or $x\in A:x\notin B$, or $x\in B:x\notin A$, or $x\in B:x\notin B$, therefore $x\in A$ or $x\in B$, and $x\notin A$ or $x\notin B$ such that $x\in(A\cup B)$ and $x\notin(A\cap B)$ which means that $x\in(A\cup B)-(A\cap B)$.
The former shows that $\forall x[x\in(A-B)\cup(B-A)\Rightarrow x\in(A\cup B)-(A\cap B)]$
, and thus $(A-B)\cup(B-A)\subset(A\cup B)-(A\cap B)$ by the definition of a subset.
Since, as we showed earlier, $(A\cup B)-(A\cap B)\subset(A-B)\cup(B-A)$ and $(A-B)\cup(B-A)\subset(A\cup B)-(A\cap B)$, by the definition of set equality, $(A\cup B)-(A\cap B)=(A-B)\cup(B-A)$.
Thank you! 
 A: Let $x$ be in $A \cup B$. Then either $x \in A$ and $x \not \in B$, $x \in A$ and $x \in B$, or $x\not \in A$ and $x \in B$. 
Therefore, if $x$ is in $A \cup B$ but not in $A \cap B$, then $x$ is in $A - B$ or $B-A$. Vice versa, if $x$ is in $A - B$ or $B-A$, then $x$ is in $A \cup B$ but not in $A \cap B$.
I'll let you finish the proof (it only involves one sentence).
Note: it's normal to struggle with those kind of proofs at first. Draw diagrams and work hard until you really understand what is going on.
A: 
It means "such that". I know it's not used by everyone but I thought it was relatively conventional. My apologies is that isn't the case. 

It's not quite used this way.   The convention is used in set construction notation to separate the domain of the bound variable and the predicate that constructs the set.   In set construction, we may also use pipes ("|") for this.
For example : $\;2\Bbb N= \{x \in \Bbb N : \exists k\in \Bbb N \; (x=2k)\}\;$, reads as: "the set $2\Bbb N$ is the set of natural numbers, $x$, such that there exists a natural number, $k$ so that $x=2k$."
Thusly, what you appear to be trying to say is:
$$\begin{align} &(A\cup B)-(A\cap B) \; 
\\[1ex] = &  \{ x\in A\cup B: x\notin A\cap B\}
\\[1ex]  & \vdots
\\[1ex] = & \{x\in A:x\notin B\}\cup\{x\in B:x\notin A\}
\\[1ex] = & (A-B)\cup (B-A)
\end{align}$$
Which is okay.

To use my words:

Take an arbitrary element of the union of $A$ and $B$ that is not in their intersection, that is $\;(A\cup B)-(A \cap B)\;$.  This element is in either $A$ or in $B$ but it is not in both $A$ and $B$.   Considering case by case, (1) if it is in $A$, then it is not in $B$, and (2) if it is in $B$, then it is not in $A$.   So it is either in $A$ but not $B$ or it is in $B$ but not $A$.   Which is to say it is in the symmetric difference of the sets. $$(A\cup B)-(A\cap B)\subseteq (A-B)\cup(B-A)$$
Take any arbitrary element of the union of set differences that is $(A-B)\cup(B-A)$.   This element is either in $A$ but not $B$, or it is in $B$ but not $A$.   Hence it is in either $A$ or $B$ but it is not in both. $$(A\cup B)-(A\cap B)\supseteq (A-B)\cup(B-A)$$
Hence the two are demonstrably equivalent, which is what was to be shown. $$(A\cup B)-(A\cap B) \equiv (A-B)\cup(B-A) \quad \Box$$

A: Here's a different understanding:
$$\begin{align*}
(A \cup B) - (A \cap B) &= (A \cup B) \cap (A \cap B)^c \\
&= (A \cup B) \cap (A^c \cup B^c) \quad (1)\\
&= (A \cap B^c) \cup (B \cap A^c) \quad (2) \\
&=(A-B) \cup (B-A)
\end{align*}$$
If you need help understanding how to get from $(1) \to (2)$, let me know.
Hint: It follows from $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.
A: Here is the way to do it by algebra
\begin{align}
(A\cup B)-(A\cap B)
&=(A-A\cap B)\cup(B-A\cap B)
\\
&=(A-B)\cup(B-A)
\end{align}
