De Morgan's laws

Laws.
- $\overline{A\cup B}=\overline A\cap\overline B$
- $\overline{A\cap B}=\overline A\cup\overline B$

Proof of first. $x\in\overline{A\cup B}\iff x\notin A\cup B\iff x\notin A$ and $x\notin B\iff x\in\overline A\cap\overline B$.

Proof of second. $x\in\overline{A\cap B}\iff$ not ($x\in A$ and $x\in B$) $\iff x\notin A$ or $x\notin B\iff x\in\overline A\cup\overline B$.

Conclusion. Both laws hold.