Prove $A\cup (A\cap B) = A$ What I've done so far is stated that 


*

*$(A \cup A) \cap (A \cup B)$  by distribution

*$A\cap (A\cup B)$        by 1, definition of $\cup$

*$A\cap A$              by 2, definition of $\cup$

*$A$                   by 3, definition of $\cap$


I'm not sure if I need to state that $A\cup B = \{x\mid x\in A\lor x\in B\}$ somewhere in there.
I'd really appreciate some help on this one. Thank you!

Your third step is incorrect: it’s not true in general that $A=A\cup B$. 
I would prove the result by element-chasing, i.e., showing that if $x\in A$, then $x\in A\cup(A\cap B)$, which is immediate from the definition of union, and that if $x\in A\cup(A\cap B)$, then $x\in A$, which is also pretty straightforward.
If you’re going to do it algebraically, by manipulating unions and intersections directly, the argument will depend on what rules of manipulation you have available. For instance, there is an absorption rule that says that if $X\subseteq Y$, then $X\cup Y=Y$. If you have that available, you can apply it with $X=A\cap B$ and $Y=A$ to get the result immediately. Or you might have the other absorption rule, that if $X\subseteq Y$, then $X\cap Y=X$; in that case you can apply one of the De Morgan laws to write $A\cup(A\cap B)$ as $(A\cup A)\cap(A\cup B)=A\cap(A\cup B)$ and apply the absorption rule with $X=A$ and $Y=A\cup B$.
