I think I have figured out this proof, but was hoping someone could verify it.
\begin{align*}
x \in (A \cap B) \cup (A – B) & \iff x \in A \cap B \land x \in A – B \\
& \iff (x \in A \land x \in B) \lor(x \in A \land x \not \in B) \\
& \iff x \in A \land (x \in B \lor x \not \in B) \\
& \iff x \in A.
\end{align*}
The first line is the definition of union. The second is the definition of interection and set difference. The third uses the rule from propositional logic that $p \wedge (q \lor r) \equiv (p \wedge q) \lor (p \wedge r)$. Finally, $p \lor \neg p$ is a tautology that is always true, and $p \wedge T \equiv p$.
Best Answer
The retranscript from set operations to propositions is immediate:
$$(a\land b)\lor(a\land\lnot b)$$
and this can be rewritten
$$a\land(b\lor\lnot b)=a.$$
You can also use a "membership" table,
$$\begin{array}{|c|c|c|c|c|} A&B&A\cap B&A\setminus B&(A\cap B)\cup(A\setminus B) \\\hline \in&\in&\in&\notin&\in \\\in&\notin&\notin&\in&\in \\\notin&\in&\notin&\notin&\notin \\\notin&\notin&\notin&\notin&\notin \\\hline \end{array}$$