Help on basic set theory question. Prove or Disprove: For every two sets $A$ and $B$, $(A\cup B)-B=A$. 
I believed it was true, so first I showed that $(A\cup B)-B$ is a subset of $A$. My question is how do I prove that $A$ is a subset of $(A\cup B)-B$?
What I have first is what follows:
Suppose there exists an arbitrary element $x$ in $A$.
If $x$ is in $A$, then $x$ is not in $B$
From here, I'm stuck.
 A: Let $A = \{1, 2 \}$, and $B=\{1, 2 \}$. Then, $(A \cup B)-B= \{1, 2 \} - \{1, 2 \} = \emptyset \neq A.$
A: HINT: You are implicitly assuming that $A\cap B=\varnothing$.
A: One way to approach this is by calculation and simplification using the laws of logic.
This quickly shows that the given statement is even equivalent to $\;A\;$ and $\;B\;$ being disjoint:
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$
$$\calc
(A \cup B) - B \;=\; A
\calcop\equiv{set extensionality, definitions of $\;\cup,-\;$}
\langle \forall x :: (x \in A \lor x \in B) \land x \not\in B \;\equiv\; x \in A \rangle
\calcop\equiv{logic: use $\;x \not\in B\;$ on other side of $\;\land\;$; simplify}
\langle \forall x :: x \in A \land x \not\in B \;\equiv\; x \in A \rangle
\calcop\equiv{logic: $\;P \land Q \equiv P\;$ and $\;\lnot P \lor Q\;$ are both alternative forms of $\;P \Rightarrow Q\;$}
\langle \forall x :: x \not\in A \lor x \not\in B \rangle
\calcop\equiv{logic: DeMorgan; definitions of $\;\cap,\emptyset\;$, and set extensionality}
A \cap B \;=\; \emptyset
\endcalc$$
