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!

Without context, I would do this without applying rules.  First I would show:
 $$ A \cup (A \cap B) \subset A$$
then I would show $$A \subset A \cup (A \cap B) .$$  These two together imply the relation you're trying to prove.
For the first statement, assume you have $a \in A \cup (A \cap B)$.  Because $a$ is in this union, it is either in $A$ or in the intersection of $A$ and $B$...  It must be in $A$ in either of these cases.
For the second statement, assume you have $a \in A$.  Every element of $A$ is also in $A \cup C$ for any set C so $a \in A\cup (A \cap B)$.  
