Show that $A \cup B = (A$ \ $B ) \cup (A \cap B) \cup (B$ \ $A)$ Let $A, B$ be finite sets. Show that $A \cup B = (A$ \ $B ) \cup (A \cap B) \cup (B$ \ $A)$.
Deduce that $|A| + |B| = |A \cup B| + |A \cap B|$
These are obvious when considering Venn diagrams but I'm at a lose on even how to start to show this algebraically. 
I've tried re-writing $(A$ \ $B ) \cup (A \cap B) \cup (B$ \ $A)$ as $(A \cap \neg B) \cup (A \cap B) \cup (B \cap \neg A)$ but I don't know where to go from there. 
Any help would be appreciated. Thank you. 
 A: On one hand
$$ A\setminus B, A \cap B \subseteq A \subseteq A \cup B, \qquad B \setminus A \subseteq B \subseteq A \cup B $$
hence "$\supseteq$" holds. 
Now let $x \in A \cup B$. Then $x \in A$ or $x \in B$. If $x \in A$, then either $x \in B$ and hence $x \in A \cap B$ or $x \not\in B$ hence $x \in A \setminus B$. If $x \not\in A$, then as $x \in A \cup B$ we must have $x \in B$ and hence $x \in B \setminus A$. So in all cases $x \in (B \setminus A) \cup (A \cap B) \cup (A \setminus B)$. This proves "$\subseteq$".
A: Use distributivity laws. For example:
$$(A\cap \neg B)\cup(A\cap B) = (A\cup A) \cap (A\cup B)\cap(A\cup \neg B) \cap (B\cup \neg B) =\\
=A\cap (A\cup B) \cap (A\cup \neg B)\cap U$$
Now, since $U$ is the universal set, $U\cap X=X$ for any $X$. Also, since $A\subseteq A\cup B$, you know that $A\cap (A\cup B)=A$, so you get $$A\cap(A\cup \neg B)$$
and again, vecause $A\subseteq A\cup\neg B$, you have $A\cap(A\cup\neg B)=A$, so in conclusion:
$$(A\cap \neg B)\cup (A\cap B) = A.$$
A: you can't whisk a proof out of thin air! so what assumption are you starting from? the most reasonable assumption is that if two sets $X$ and $Y$ are mutually disjoint then
$$
|X \cup Y | = |X| + |Y| \tag{1}
$$
see if you can solve the problem using this assumption.
