$A \cap(A \cup B) = A$ for all sets $A$ and $B$ 
For all sets $A$ and $B$, $$A \cap(A \cup B) = A.$$

I get that this is true informally, but what would be the notation to formally prove this?
 A: Hint: Frequently, the easiest method for showing some set $X$ and some set $Y$ are equal is this:  first, show that any element $x\in X$ also satisfies $x\in Y$ -- which proves that $X\subseteq Y$.  Then, show that any element $y\in Y$ also satisfies $y\in X$, proving that $Y\subseteq X$.
If $X\subseteq Y$, and $Y\subseteq X$, then necessarily $X=Y$.
So, start off with an element $x$ which is in $A\cap(A\cup B)$, and show that $x\in A$. Then, show that any element $y\in A$ also satisfies $y\in A\cap(A\cup B)$. 
It might help you to remember the following two facts:


*

*Every element of $A\cap X$ is both an element of $A$ and an element of $X$ -- and so in particular is definitely a member of $A$.

*If $x\in X$ and $x\in Y$, then $x\in X\cap Y$.
Edit: Here's a start.
Suppose that $x\in A\cap (A\cup B)$. Then by definition of the intersection, we must have $x\in A$ and $x\in A\cup B$. In particular, $x\in A$.
Now, see if you can show that any element $x\in A$ must also be in $A\cap(A\cup B)$.
A: In general, if $A\subseteq X$ then we have $A\cap X=A\ $ (these are actually equivalent). 
Then apply this for $X:=A\cup B$.
A: $$A\cap(A\cup B)=(A\cap A)\cup(A\cap B)$$ Moving on $A\cup(A\cap B)$ 
If $A \subseteq B$  then $A \cap B=A$ and this equality happens.
