How to prove that $(A\cup B)-A = B-(A\cap B)$? I'm going through the book Proofs and Fundamentals by Ethan Bloch, and in the chapter of Set Operations there is an exercise (Exercise 3.3.9) that asks you to do the following:

Let $A$ and $B$ be sets. Prove that $(A\cup B)-A = B-(A\cap B)$.

I know that to prove this you just need to show that the right hand side of the equation is a subset of the left side and vice versa. 
I began trying to prove it by choosing an arbitrary element $x$ of $(A\cup B)-A$. 
$x \in (A\cup B)-A$, which is the same as
$x \in A \cup B $ and $x \notin A$, which is the same as
$x \in A $ or  $x \in B $, and $x \notin A$, which I think means
$x \in B $.
I have some intuition on how to proceed but I don't know how to do it formally.
Any corrections and tips and/or complete proofs are welcome.
 A: Your last step should be
$(x\in A \;or\; x\in B) \;and\; (x\notin A)$
Use Distributive law now
$(x\in A \;and\; x\notin A)\;or\;( x\in B \;and\; (x\notin A))$
$\Rightarrow x\in B \;and\; (x\notin A)$
$\Rightarrow x\in B-A$
$\Rightarrow x\in B-(A\cap B)$
This shows $((A\cup B)-A )\subset (B-(A\cap B))$
Now let $x \in (B-(A\cap B)$
$\Rightarrow x\in B \;and \;x\notin (A\cap B)$
$\Rightarrow x\in B \;and\; (x\notin A\; or\; x\notin B)$
$\Rightarrow (x\in B \;and\;x\notin A) \; or\;(x\in B \; and \; x\notin B)$
$\Rightarrow (x\in B \;and\;x\notin A)$
$\Rightarrow x\in B-A$
Also, $B-A \subset (A\cup B)-A$
This completes the proof.
A: Suppose $x \in (A \cup B) - A$. Then $(x \in A \text{ or } x \in B)$ and $x \notin A$. This simplifies to $x \in B$ and $x \notin A$. Since $A \cap B \subset A$ (do you see why this is true?), it follows that $x \notin A \cap B$. Hence $x \in B$ and $x\notin A \cap B$, so $x \in B - (A \cap B)$.
A: A quicker way to notice the solution is the following:
$A \subseteq (A \cup B)$, so $(A \cup B)-A=(A \cup B)- \underbrace{A \cap (A \cup B)}_\text{ distribute}=(A \cup B-A)-A \cap B=B-A \cap B$.

Another idea is the following. We can use your proof, but in the last line you state 
"($x \in A$ or $x \in B$) and $x \notin A$.
But let's use the logically equivalent statement:
($x \in A$ and $ x \notin A$) or ($x \in B$ and $x \notin A$).
The first option is a contradiction, so use the second.
then $x \in B-A \subseteq B-A \cap B$.
But now you only have one direction of the proof.

We show the opposite inclusion:
We want: $B-(A\cap B) \subseteq (A \cup B)-A$.
$x \in B-(A \cap B) \implies (x \in B) \land(x \notin A \cap B) \implies (x \in B) \land (x \notin A \lor x \notin B) \implies$_______
hint: you have two cases now, one of them is a contradiction.
A: Using three properties $A\setminus B=A\cap\bar{B}, B\cap\bar{B}=\emptyset, \overline{A\cup B}=\bar{A}\cap\bar{B}$, where $\bar{B}$ is the complement of $B$, then you have
$$ LHS=A\cup B\setminus A=(A\cup B)\cap\bar{B}=(A\cap\bar{B})\cup (B\cap\bar{B})=A\cap\bar{B} $$
and
$$ RHS=B\setminus A\cap B=A\cap(\overline{A\cap B})=A\cap(\bar{A}\cup\bar{B})=(A\cap\bar{A})\cup(A\cap\bar{B})=A\cap\bar{B} $$
from which you can obtain
$$ A\cup B\setminus A=B\setminus A\cap B.$$
A: To prove that two sets are equal, the most basic tool is extensionality: they are equal iff they have the same elements.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$ In other words, the original statement is equivalent to $$
\tag{0}
\langle \forall x :: x \in (A \cup B) - A \;\equiv\; x \in B - (A \cap B) \rangle
$$
To prove this, let's simplify both sides of this equivalence in turn, by expanding the definitions and then using the laws of logic.  For the left hand side,
$$\calc
    x \in (A \cup B) - A
\op=\hint{expand the definitions of $\;-\;$ and $\;\cup\;$}
    (x \in A \lor x \in B) \;\land\; x \not\in A
\op=\hint{distribute $\;\land\;$ over $\;\lor\;$ -- to bring both occurrences of $\;x \in A\;$ together}
    (x \in A \land x \not\in A) \;\lor\; (x \in B \land x \not\in A)
\op=\hint{simplify}
    \false \;\lor\; (x \in B \land x \not\in A)
\op=\hint{simplify}
    x \in B \land x \not\in A
    \tag{*}
\endcalc$$
Do the same thing with the right hand side of $\ref{0}$, get the same result $\ref{*}$, and conclude that $\ref{0}$ is true.  That concludes the proof of the original statement.
