= Solution
Every <complement of a set> is taken in the same <universal set> $\mathcal U$. Apply <De Morgan's laws> for <propositional logic> to the membership propositions. For every $x\in\mathcal U$,
$$
\begin{aligned}
x\in\overline{A\cap B}
&\iff\neg(x\in A\land x\in B)\\
&\iff(x\notin A\lor x\notin B)
\iff x\in\overline A\cup\overline B,\\
x\in\overline{A\cup B}
&\iff\neg(x\in A\lor x\in B)\\
&\iff(x\notin A\land x\notin B)
\iff x\in\overline A\cap\overline B.
\end{aligned}
$$
Equality of membership for every element proves the <set identities>
$$
\boxed{\overline{A\cap B}=\overline A\cup\overline B,
\qquad\overline{A\cup B}=\overline A\cap\overline B.}
$$
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