Prove that $(A∪B)-B=A$ iff sets $A$ and $B$ are disjoint. I was working on this assignment and was wondering if an assumption I made was correct and if my proofs as a whole were correct. The first part of the question was: 
Use the algebraic method to prove the following set equality: $A∪B-B=A-B$
I've done this part already and use the result of said proof to formulate the start of the next proof.
Let $A$ and $B$ be two sets. Prove that $(A\cup B)-B=A$ iff $A$ and $B$ are disjoint.
Based on the previous proof:
$Let:(A∪B)-B=A-B=A$
$1)\ A-B=A\cap\bar B\text{; by Set Difference}$
$2)\ A\cap\bar B=A\cap(\bar B\cup\emptyset) \text{; by Identity}$
$3)\ A\cap(\bar B\cup\emptyset)=A\cup\overline{(B\cap\bar\emptyset)}\text{; by DeMorgan’s 
Law}$
$4)\ A\cup\overline{(B\cap\bar\emptyset)}=A\cup\overline{(B\cap U)}\text{; by 0/1 Law}$
$5)\ A\cup\overline{(B\cap U)}=A\cup\bar B\text{; by Identity}$
$6)\text{ Then, } A\cap\bar B=A\cup\bar B=A,\text{ which can only be true if }\bar B=A\text{; by Repetition Law}$
$7)\ \bar B=A\text{ implies }B\subseteq\bar A \therefore A\text{ and }B\text{ must be disjoint.}$
I'm wondering if this proof is acceptable (especially the conclusion I draw from points 6 and 7) or if I'm missing out on something (first time taking a course that wants rigorous proofs). Thanks!
 A: You've misused DeMorgan's Law. There is an easier way to proceed. You've already noted that $(A\cup B)- B = A-B$. Thus we want to prove that $A-B = A$ if and only if $A,B$ are disjoint. I'll prove one direction and leave you to prove the other.
I'll prove that if $A-B=A$ then $A$ and $B$ are disjoint. To do so, I'll proceed by contrapositive; that is, I'll prove that if $A$ and $B$ are not disjoint, then $A-B\neq A$. If $A$ and $B$ are not disjoint, then there is $a\in A\cap B$. Thus $a\not\in A-B$. So $A-B\neq A$ since $a\in A$ but $a\not\in A-B$.
The other direction is simple. Don't get hung up on using set operations so heavily. It can make a simple proof rather complicated and difficult to piece together.
A: I'd like to say: "You've misused DeMorgan's Law. There is an easier way to proceed." :)  
Now:  $A∪B-B=A \iff$
$$A-B=A\iff A\cap B^c=A \iff A\subset B^c   $$ $\iff$ $A$ and $B$ are disjoint
A: Since you have your answer, I thought I would show another way to go about it: you can do a truth table.
$$\begin{array}{lllll}
A&B&(A\text{ or }B)&\text{Not }B&(A\text{ or }B)\text{ and }(\text{Not }B)\\
\hline
T&T&T&F&F\\
T&F&T&T&T\\
F&T&T&F&F\\
F&F&F&T&F
\end{array}$$
The first line doesn't matter. We can throw it out because $A$ and $B$ are to not have an element $x$ in common. The top line is basically saying $x$ is in $A$ and $x$ is in $B$ and we don't have that case here.
Other than that $A\equiv(A\text{ or }B)\text{ and }(\text{Not }B)$ are the same thing as long as the $A$ and $B$ are disjoint as we given.
