Proving $(A \cap B) \cup (A - B) = A$ 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 $p \wedge T \equiv p$.
 A: 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,
$$\begin{array}{|c|c|c|c|c|}
A&B&A\cap B&A\setminus B&(A\cap B)\cup(A\setminus B)
\\\hline
\in&\in&\in&\notin&\in
\\\in&\notin&\notin&\in&\in
\\\notin&\in&\notin&\notin&\notin
\\\notin&\notin&\notin&\notin&\notin
\\\hline
\end{array}$$
A: You did a fine job, and proved the biconditional.  You also provided good justification: the "third" property can be justified by the "distributive law/rule" of "and" over "or".
In general, distributivity also holds among set union/intersection, so $$A\cap (B\cup C) = (A\cap B) \cup (A\cap C).$$
Also note that one can write $(A\cap B)\cup (A-B)$ as $(A\cap B)\cup (A\cap B^C)$, where $B^C$ is the complement of $B$.
So considere your proof verified enthusiastically.   
A: Yes, your solution is correct. 
Now, let me do it in a less formal way. 
To prove a set identity like $X=Y$, we can prove it in two steps: $X\subset Y$ and $Y \subset X$. 
Now, by definition of "$\subset$", you can easily check:


*

*$A\cap B\subset A$, 

*$A-B\subset A$

*$(A\cap B)\cup (A-B)\subset A$
Secondly, for any element $x\in A$, $x$ either in $B$ or (exclusively) not in $B$. This tells you
$$
A\subset (A\cap B)\cup (A-B)\;.
$$
