Understanding Set Theory and Proving $A \cap(B\cup A) = A$ I am trying to wrap my head around discrete mathematics in order to help my understanding of self taught programming. I am now trying to understand Set Theory, more specifically proving certain theorems. I understand the basic concept of proofs, but cannot seem to figure out how to approach/complete this one. Any help would be well received and greatly appreciated. 
Prove: 
$$A \cap(B\cup A) = A$$
 A: Let $x$ be in the LHS. Then $x\in A$, $x\in B\cup A$ by definition of intersection, so $x\in A$. This LHS$\subseteq A$.
Let $x\in A$. Then $x\in B\cup A$ by definition of union. Thus $x\in$ LHS by definition of intersection, so $A\subseteq$LHS.
If $X\subseteq Y$ and $Y\subseteq X$ then $X=Y$. This equality follows in your problem.
A: From $x\in A\cap(B\cup A)$ it follows by definition that $x\in A$
If conversely $x\in A$ then also $x\in B\cup A$ and these two facts lead to $x\in A\cap(B\cup A)$
This together proves that $A\cup(B\cap A)=A$
Alternatively if $P\subset Q$ then $P\cap Q=P$. You can apply that for $P=A$ and $Q=B\cup A$.
A: I would approach this as a simplification problem, most easily solved by starting with $\;A \cap (B \cup A)\;$, expanding the definitions and then simplifying using the rules of logic.
So we calculate which $\;x\;$ are in that set:
\begin{align}
& x \in A \cap (B \cup A) \\
\equiv & \qquad \text{"definition of $\;\cap\;$; definition of $\;\cup\;$"} \\
(*) \;\;\; \phantom{\equiv} & x \in A \;\land\; (x \in B \;\lor\; x \in A) \\
\equiv & \qquad \text{"logic: use $\;x \in A\;$ on other side of $\;\land\;$"} \\
& x \in A \;\land\; (x \in B \;\lor\; \text{true}) \\
\equiv & \qquad \text{"logic: simplify"} \\
(**) \;\;\; \phantom{\equiv} & x \in A \\
\end{align}
By set extensionality, this proves the theorem.
In essence, the above proof uses (a generalization of) the absorption law from logic to prove the absorption law for sets.
Another way to go from $(*)$ to $(**)$ is
\begin{align}
& P \land (Q \lor P) \\
\equiv & \qquad \text{"rewrite -- to give both sides of $\;\land\;$ the same shape"} \\
& (P \lor \text{false}) \land (P \lor Q) \\
\equiv & \qquad \text{"extract common disjunct:$\;\lor\;$ distributes over $\;\land\;$"} \\
& P \lor (\text{false} \land Q) \\
\equiv & \qquad \text{"simplify"} \\
& P \\
\end{align}
