A solid strategy to prove: $A \subset B\Leftrightarrow A=A\cap B \Leftrightarrow B= A\cup B $ To begin i tried to prove that the first statement implies second, but took a while to type that (now let's prove the other 5 implications, which would fullfill the circle....), I think there must be a short way to do this. Any ideas?
 A: 1) Suppose $A \subset B$.  $A \cap B \subset of A$ as intersections are by definition subsets of the intersecting sets.  If $a \in A$ then $a \in B$ as $A \subset B$ $a \in A \cap B$ (because b is in both A and B).  So $ A \subset A \cap B$.  (Because every element of A is in $A \cap B$).  So $A \cap B \subset of A$ and $ A \subset A \cap B$ so $A = A \cap B$.
So 1) => 2.
2) Suppose $A = A \cap B$.  $B \subset A \cup B$ (by definition of union).  Let $b \in A \cup B$.  Then either $b \in A$ or $b \in B$.  If $b \ A = A \cap B$ then $b \in B$.  So all $b \in A \cup B$ is in $B$.  So $A \cup B \subset B$.  So $B = A \cup B$.
So 2) => 3.
3) Suppose $B = A \cup B$.  Suppose $a \in A$.  Then $a \in A \cup B = B$ so $a \in B$.  So every element of $A$ is also an element of $B$.  So $A \subset B$.
So 3) => 1.
We're done.
1 => 2.  And 2=>3=>1 so 1 <=> 2.
3 => 1.  And 1=>2=>3 so 1 <=> 3.
2=>3.  And 3=>1=>2 so 2<=> 3.
A: 
Statement $1$ is $x\in A\implies x\in B$
Statement $2$ is $x\in A= x\in A\land x\in B$
Statement $3$ is $x\in B = x\in A\lor x\in B$

$1\to2:$ $x\in A\implies x\in B \therefore x\in A\implies x\in A \land x\in B$
$2\to1:$ $x\in A=x\in A \land x\in B\therefore x\in B$
$1\to3:$ $x\in A\implies x\in B \therefore x\in A \lor x\in B \implies x\in B \lor x\in B \implies x\in B$
$3\to1:$ $x\in A \lor x\in B=x\in B\therefore x\in A\implies x\in A \lor x\in B\implies x\in B$
