Prove that $A - B = A $ iff $A \cap B = \emptyset$ I have a lemma below:
$${A}-{B} = {A} \Longleftrightarrow A \cap B = \emptyset.$$
I start with the $\Longrightarrow$:
$$\text{Suppose}\ x \in {A}-{B},\text{then}\  x \in A\ \text{but}\  x \notin B.$$
$${A}-{B} = {A}\ \text{so}\ x \in A\ \text{but}\ x \notin B \equiv x \in A.$$
So the fact that $x \notin B$ does not matter at all. It's getting confused because this lemma is just proved to prove this theorem:
$$P(A-B) = P(A) \Longleftrightarrow A \cap B = \emptyset.$$
With the lemma, I can prove the theorem. Anyone has any advice on how to prove the lemma? Thank you!
 A: I think you're almost there. You have "$x\in A$ and $x\notin B$" $\Longleftrightarrow$ "$x\in A$". In particular, $x\in A \Longrightarrow x\notin B$, meaning that there is no $x$ for which $x\in A$ and $x\in B$, so $A\cap B$ is empty.
Conversely, if $A\cap B$ is empty then for any $x\in A$ we have $x\notin B$, so $x\in A\Longrightarrow x\notin B\Longrightarrow(x\in A\text{ and }x\notin B)$, and clearly $(x\in A\text{ and }x\notin B)\Longrightarrow x\in A$.
A: $A-B$ is defined as $\{a \in A : a \notin B\}$. Then,
$$A-B=A \iff \{a \in A : a \notin B\} =A \iff \{a \in A : a \in B\}= \phi$$
Now, $\{a \in A : a \in B\} = A \cap B$ by definition, so we're done.
A: $\Rightarrow$ We prove the contrapositive statement: $A\cap B\neq\emptyset \Rightarrow A-B\neq A$.
$A\cap B\neq\emptyset\Rightarrow $there exists $x\in A\cap B$ i.e. $x\in A$ and $x\in B$ so $x\in A$ and $x\notin A-B$ thus $A-B\neq A$ (there exists one element, namely $x$, that belongs to one set, $A$, but not to $A-B$).
$\Leftarrow$ We first note that $A-B\subset A$. Now $x\in A\Rightarrow x\notin B$ since $A\cap B =\emptyset$ so it must be $x\in A-B$ thus it is also $A\subset A-B$ which implies that $A-B=A$ and this concludes the proof.
A: $A\cap B'=A\Longrightarrow$ $$A\cap B=(A\cap B')\cap B= A\cap (B'\cap B)=A\cap\emptyset=\emptyset $$
$A\cap B=\emptyset\Longrightarrow$
$$A\cap B'=(A\cap B')\cup(A\cap B)=A\cap(B'\cup B)=A$$
