Proof of $ A \cap B = A \iff A \cup B = B$ I tried to see if this was asked here before but I am pretty sure im the first one - I hope I'm right.
So I am supposed to show"
$$ A \cap B = A \iff A \cup B = B$$ 
So what I get from this obviously I need to prove both the "if" and "only if" but I get stuck really early: 
Assumption (1)
$$ x \in B  $$ 
Premise (2)
$$ A \cup B = B  $$
(1)(2) and def. '= '
$$  x \in A\cup 
B $$ 
def intersection
$$ x \in A \lor x \in B $$ 
but as you can see this isn't really going anywhere since I cannot derive $A$ from there as I have $\lor$. So my guess is that I need to assume $x \in A$. But then this seems very wishy washy. I would really appreciate any pointers or hints on where to start to tackle this problem in a better way. 
EDIT: i did indeed switch up A and B on the equal side, but don't think that makes a difference
 A: Show $A \cap B = B \iff B \subseteq A \iff A \cup B = A.$
A: Expanding the definition of equality of sets, we have the following logical equivalence:
$$(\forall x. x\in A\cap B \Leftrightarrow x \in B) \iff (\forall x.x\in A\cup B\Leftrightarrow x\in A)$$ or completely logicallly:
$$(\forall x. (x\in A\land x\in B) \Leftrightarrow x \in B) \iff (\forall x.(x\in A\lor x\in B)\Leftrightarrow x\in A)$$ 
Proving the $\implies$ direction, we have the assumption $\forall x. (x\in A\land x\in B) \Leftrightarrow x \in B$ and we need to prove $\forall x.(x\in A\lor x\in B)\Leftrightarrow x\in A$. The $\Leftarrow$ case is trivial. For the $\Rightarrow$ case, for an arbitrary $x$, if $x\in A$ then we're done, so assume $x\in B$, then by our first assumption, we know that $x\in A\land x\in B$ from which we immediately derive $x\in A$ completing the $\Rightarrow$ case and the overall $\implies$ case.
The $\impliedby$ case is symmetrical and I leave it as an exercise.
A: If $A\cap B=B$, then $A\cup (A \cap B)=A\cup B$. But $A\cap B\subseteq A$, therefore $A\cup B=A$.
If $A\cup B=A$, then $(A \cup B)\cap B= A\cap B$. But $B\subseteq A\cup B$, therefore $B=A\cap B$.
A: We will prove it step by step and completely below.
Part 1: If $A \cup B = A$ then $A \cap B = B$
Proof:-
To prove $A \cap B = B$, we need to prove $A \cap B \subseteq B$ and $B \subseteq A \cap B$.
Now, let $x \in A \cap B$ be arbitrary.
$$\therefore x \in A \wedge x \in B$$
$$\therefore \forall x \in A \cap B, x \in B$$
$$\therefore A \cap B \subseteq B$$
Now, let $x \in B$ be arbitrary.
$$\because A \cup B = A, \forall x \in A \cup B, x \in A$$
Also, $B \subseteq A \cup B$.
$$\therefore \forall x \in B, x \in A \cup B \implies x \in A$$
Thus, we may conclude that $\forall x \in B, x \in A$.
$$\therefore \forall x \in B, x \in A \cap B$$
$$\therefore B \subseteq A \cap B$$
Thus, from the above two results, we may conclude that $A \cup B = A \implies A \cap B = B$.
Part 2: If $A \cap B = B$ then $A \cup B = A$
Proof:-
Again, we have $\forall x \in A, x \in A \cup B$
$$\therefore A \subseteq A \cup B$$
Now, consider $x \in A \cup B$ to be arbitrary.
$$\therefore x \in A \vee x \in B$$
Case I: If $x \in A$
This is a trivial case and we may say that $x \in A$
Case II: If $x \in B$
Now, given that $B = A \cap B$.
$$\therefore \forall x \in B, x \in A \cap B$$
$$\therefore \forall x \in B, x \in A \wedge x \in B$$
Thus, in both the cases, whether $x \in A$ or $x \in B$, we get that $x \in A$.
$$\therefore \forall x \in A \cup B, x \in A$$
$$\therefore A \cup B \subseteq A$$
From the above two results we may conclude that $A \cap B = B \implies A \cup B = A$.
From the two parts proved above, the proof for iff is done.
So, we have $A \cap B = B \iff A \cup B = A$
A: $A \cap B = B \iff$ $\forall b \in B$ $b\in A \iff$ $ B \subset A$
See that you can understand the implications in both directions.
Lastly, $B \subset A \iff A \cup B = A $ is rather clear. 
