Proving that $A \cap (A \cup B) = A$ . Please check solution For homework I need to prove the folloving:
$$
A \cap (A \cup B) = A
$$
I did that in the following manner:
$$
A \cap (A \cup B)\\
x \in A \land (x \in A \lor x \in B)\\
(x \in A\ \land x \in A) \lor (x \in A \land x \in B)\\
x \in A \lor (x \in A \land x \in B)\\
\text{now I have concluded that x belongs to A,}\\
\text{because of the boolean expression, because I get something like this:}\\
x \in A \lor (x \in A \land x \in B) = x \in A\\
\text{when $x \in A$ is true the whole expression will be true}\\
$$
Did I do it right???
thanks
 A: Yes, you are correct. The given equality can be expressed by the two implications:  $x \in A\cap (A \cup B) \implies x \in A$, and the implication $x \in A \implies x \in A \cap(A\cup B)$. Since both implications are satisfied, so is the set equality.
Just be careful in your "equations" (i.e., your use of the "equal sign here"). You wrote: 

$$x \in A \lor (x \in A \land x \in B) \underbrace{=}_{\uparrow ?} x \in A$$

It would be more correct to conclude: $$x \in A \lor (x \in A \land x \in B)\;\underbrace{\iff}_{\text{if and only if}}\; x \in A$$
from which it follows $$A \cap (A \cup B) = A$$
A: Or you can just think that $A\subset A\cup B$ and thus from the definition of $\cap$ you have that $A\cap (A\cup B)=A$
A: In general, for any two sets $\;X,Y\;$ , we always have that
$$X\cap Y=X\iff X\subset Y$$
With the above your exercise is thus trivial...
A: Using fundamental laws of Set Algebra
$$\begin{cases}A \cap (A \cup B) & Given\\
=(A \cap A) \cup (A \cap B) & \text{Distributive Law}\\
=A \cup (A \cap B) & A \cap A = A\\
=A & A \cap B \subset A
\end{cases}$$
