Proving that $(A \cup B) \cap \overline{(A \cap B)} = (A \cap \overline{B}) \cup (\overline{A} \cap B) $ I'm trying to teach myself set-theory. I have been unable to prove algebraically that:
$(A \cup B) \cap \overline{(A \cap B)} = (A \cap \overline{B}) \cup (\overline{A} \cap B) $
I know it's elementary but I just want to see an example. (The book doesn't provide one.)
Correct me if I'm wrong...the first step is to use De Morgan's law to expand the left side. i.e. 
$ (A \cup B) \cap  \overline{(A \cap B)} = (A \cup B) \cap \overline{A} \cup \overline{B}$
$ (A \cup B) \cap  \overline{(A \cap B)} = A \cup (B \cap \overline{A}) \cup \overline{B} $, by associative property.
$ (A \cup B) \cap  \overline{(A \cap B)} = (A \cup \overline{A}) \cap (B \cup \overline{B} )$, by commutative property.
after that I don't know what to do. 
I don't know whether I should rewrite these or if different notation would help. i.e. should I expand both sides to use the notations/definition of set-union? i.e. $ A \cup B = \{x \mid x \in A \lor x \in B \}$
 A: $$(A \cup B) \cap \overline{(A \cap B)} = (A \cap \overline{B}) \cup (\overline{A} \cap B)\tag{1}$$

Step (1) $ (A \cup B) \cap  \overline{(A \cap B)} = (A \cup B) \cap \overline{A} \cup \overline{B}$

Yes, correct, but you should keep parentheses: $ (A \cup B) \cap  \overline{(A \cap B)} = (A \cup B) \cap (\overline{A} \cup \overline{B})$

Step (2) Not correct: $ (A \cup B) \cap  \overline{(A \cap B)}$ = $A \cup (B \cap \overline{A}) \cup \overline{B} $, by associative property.

No: the associative property (like the commutative property) applies to a chain of unions, or a chain of intersections, but not a mixed chains of unions and intersections. This is why parentheses in the first step are needed. You need to use distribution: 
$ (A \cup B) \cap  \overline{(A \cap B)}$ = $(A \cup B) \cap (\overline{A} \cup \overline{B}) = [(A\cup B) \cap \overline{A}] \cup [(A \cup B) \cap \overline{B}]$
Can you take it from here? You can use distributivity again...then use that fact that $A \cap \overline{A} = \varnothing$. Likewise for $B\cap \overline{B} = \varnothing$. Simplify.

Another approach is to unpack what proving your equality $(1)$ requires: 
In general, we prove that for two sets $P, Q$, $$P = Q \iff P \subseteq Q \text{ AND}\;\;Q\subseteq P$$
For your equality $(1)$, that means you can prove the equality by proving: 
$$[(A \cup B) \cap \overline{(A \cap B)}] \subseteq [(A \cap \overline{B}) \cup (\overline{A} \cap B)]\tag{2}$$
and by proving $$[(A \cap \overline{B}) \cup (\overline{A} \cap B)] \subseteq [(A \cup B) \cap \overline{(A \cap B)}]\tag{3}$$
Unpack what this means in terms of "chasing elements", for which I'll give you a start: 
$(2)$ If $x \in [(A \cup B)\cap \overline{(A \cap B)}]$ then $x \in (A\cup B)$ AND $x \notin (A \cap B)$, which means $(x \in A$ OR $x \in B)$ AND $(x \notin A$ OR $x \notin B)$...
$(3)$ If $x \in [(A\cap \overline{B})\cup (\overline{A} \cap B)]$, then ...$(x \in A \cap \overline{B})$ OR $(x \in \overline{A} \cap B)$ ....
A: Your first step is almost correct: you dropped the parentheses. Thus, you should have $$\begin{align}(A\cup B)\cap\overline{(A\cap B)}&=(A\cup B)\cap(\overline{A}\cup\overline{B})\\ &= \big[(A\cup B)\cap\overline{A}\big]\cup\big[(A\cup B)\cap\overline{B}\big]\\ &= (B\cap\overline{A})\cup(A\cap\overline{B}).\end{align}$$
As an example, we can take intervals on the real line, say $A=[0,2]$ and $B=[1,3]$. As an exercise, you can try to work it out to verify the identity as an example.
A: Good first effort, but you don't have associativity and commutativity when dealing with the operations $\cup$ and $\cap$, so parentheses are super important.
It looks like you are missing what it means for two sets, call them $X$ and $Y$, to be equal.  The definition is pretty simple: $X = Y$ means that any element $x \in X$ is also in $Y$, and every element $y \in Y$ is also contained in $X$.
Formally, $X = Y$ $\iff$ $X \subseteq Y$ and $Y \subseteq X$.
So you want to break this into two problems, showing that each is a subset of the other.  So start out with "Let $x \in (A \cup B) \cap \overline{(A \cap B)}$..." and go from there.  Let me know if you need more.
A: For future reference, the solution:
prove: $\ (A \cup B) \cap \overline{(A \cap B)} = (\overline{A} \cap B) \cup (\overline{B} \cap A)   $
I will manipulate the left side only:
$\ (A \cup B) \cap (\overline{A} \cup \overline{B}) $, use De Morgan's law
= $\ (( A \cup B) \cap \overline{A} ) \cup ((A \cup B) \cap \overline{B}) $  , distribution
= $\ ( (\overline{A} \cap B) \cup (\overline{A} \cap A)) \cup ( (\overline{B} \cap B) \cup (\overline{B} \cap A)) $ , distribution
remember that $\ \overline{A} \cap A = \emptyset $
= $\ (\overline{A} \cap B) \cup (\overline{B} \cap A) $
A: $x \in A \wedge \not\in B \vee  \not\in A \wedge \in B$ 
So x is in B and not A or in A but not B.
so x is in one of them but not both.
