Given finite sets $A$ and $B$, is $|A \cup B| = |A \cap B| \iff A = B$ true? Given finite sets $A$ and $B$, is
$$
|A \cup B| = |A \cap B| \iff A = B
$$
true? Here is what I tried. For
$$
|A \cup B| = |A \cap B| \implies A = B,
$$
suppose that $|A \cup B| = |A \cap B|$, and from the inclusion-exclusion principle,
\begin{align}
|A \cup B| &= |A| + |B| - |A \cap B| \\
2|A \cup B| &= |A| + |B|.
\end{align}
However, I am not sure how to proceed from here.

For any two sets $\ A\ $ and $\ B,\ A\cup B\ $ can be written as the union of three disjoint sets:
$$A\cup B = (A \cap B)\ \cup\ (A - B)\ \cup (B - A). $$
Therefore, for any two finite sets $\ A\ $ and $\ B,\ $ we have the following:
$$\ |A \cap B| = |A \cup B| \iff |A \cap B| = |(A \cap B)\ \cup\ (A - B)\ \cup (B - A)|$$
$$\iff |A \cap B|= |A \cap B|\ + |A - B|\ + |B - A|\quad\text{(because the three sets are disjoint)} $$
$$ \overset{(*)}{\iff} (A - B) = (B - A) = \emptyset,\quad \text{i.e. there's nothing in A that's not in B and vice versa} $$
$$\iff A=B.$$
Finiteness of sets $\ A\ $ and $\ B\ $ were used for the right implication at $\ (*).$
