Proving properties of set complements I have to prove $(A \cup B) - (A \cap B) = (A - B) \cup (B - A)$, but I am not sure where to start. Can anyone point me in the correct direction?
 A: Using the fact that $A -B=A \cap B^C$, we get:
$$(A \cup B) - (A \cap B)=$$
$$(A \cup B) \cap (A \cap B)^C=$$
$$(A \cup B) \cap (A^C \cup B^C)=$$
$$(A \cap A^C) \cup (A \cap B^C) \cup (B \cap A^C) \cup (B \cap B^C)=$$
$$\emptyset \cup (A \cap B^C) \cup (B \cap A^C) \cup \emptyset=$$
$$(A \cap B^C) \cup (B \cap A^C)=$$
$$(A-B) \cup (B-A)$$
A: Take any $x \in (A \cup B) - (A \cap B)$, then $x\in A\cup B$ and $x\notin A\cap B$ 
so ($x\in A $ or $x\in B$) and ($x\notin A$ or $x\notin B$)
so if $x\in A$ then $x\notin B$ so $x\in A-B$...
A: $\boxed{(A \cup B) - (A \cap B) \subset (A - B) \cup (B - A)}$
$\scriptstyle x\in (A \cup B) - (A \cap B)\implies \left\lbrace\begin{array}l x\in A\;\land \;x\notin A\cap B\iff x\in A-B \\\text{or}\\x\in B\;\land \;x\notin A\cap B \iff x \in B-A \end{array}\right.\implies x\in (A-B)\cup(B-A)$
$\boxed {(A - B) \cup (B - A)\subset (A \cup B) - (A \cap B)} $
$\scriptstyle x\in(A - B) \cup (B - A)\implies\left\lbrace\begin{array}l x\in(A - B) \iff x\in A \;\land \;x\notin B\iff x\in A\;\land \;x\notin A\cap B\\\text{or}\\x\in(B-A)\iff x\in B\; \land\; x\notin A\iff x\in B\;\land\; x\notin A\cap B\end{array}\right.\implies x\in A\cup B-A\cap B$
A: Define
$\tag 1 A \bigtriangleup B = (A \cup B) - (A \cap B)$
and
$\tag 2 A \bigtriangleup^{`} B = (A - B) \cup (B - A)$
Hint: Each of these binary operations have the same general properties. For example, they both satisfy the commutativity law (both formulas are symmetric). After playing around by plugging in stuff and calculating the results, get serious and show that $\bigtriangleup = \bigtriangleup^{`}$.

More Theory and Hints
In what follows the symbol $=^E$ will be used to mean that two expressions are equal, but it remains as an exercise for the student to check the details. Before continuing, they can study [General Properties of Set Complements and Differences][1] and [Set Difference is Right Distributive over Union][2], learning about some 'tools of the trade'.
There are several ways to go about showing that $\bigtriangleup = \bigtriangleup^{`}$, like the direct method used by Bram28. Here we want to use an approach that highlights a general set theoretic concept:
Theorem 1: If $A$ and $B$ are two sets then there exist sets $G$, $H$ and $C$ so that (3) thru (5) hold:
$\tag 3 A = G \cup C$
$\tag 4 B = H \cup C$
$\tag 5 G \cap H = \emptyset$
Proof
Let $C = A \cap B$, $\;G = A - C$ and $H = B - C$. Both (3) and (4) are easily handled. For (5),
$(A - C) \cap (B - C) = (A - (A \cap B)) \, \cap \,  (B - (A \cap B)) =^E (A-B) \cap (B-A) =^E \emptyset$. $\qquad \blacksquare$
Observe that the proof of thereom 1 contains the result that $\bigtriangleup^{`}$ is obtained by taking the union of two disjoint sets.
Proposition 2: $A \bigtriangleup B = G \cup H$.
Proof: Exercise (start with $[ (G \cup C) \cup (H \cup C) ] - [ (G \cup C) \cap (H \cup C) ]$.
Proposition 3: $A \bigtriangleup^{`} B = G \cup H$.
Proof: Exercise.
Proposition 4: $\bigtriangleup = \bigtriangleup^{`}$.
Proof: Follows from proposition 2 and 3.
Finally, theorem 1 shows the existence of the sets $G$, $H$ and $C$, but one wants to wrap it up and understand the following result.
Proposition 5: For the two sets $A$ and $B$, the decomosition specified by (3), (4) and (5), uniquely define the sets $G$, $H$ and $C$.
Proof: A simple exercise.
