De Morgan's Laws (Set Theory)

Theorem: Let $$S=\{A_i|i \in I\}$$, where each $$A_i$$ is a set and $$I$$ is some indexing set. Then, $$\overline{\bigcup_{i \in I} A_i}= \bigcap_{i \in I} \overline{A_i}$$ and $$\overline{\bigcap_{i \in I} A_i}= \bigcup_{i \in I} \overline{A_i}$$.

Proof
$$x \in \overline{\bigcup_{i \in I} A_i}$$ iff

$$x\ \not\in\ A_i, \forall i \in I$$ iff

$$x \in \overline{A_i}, \forall i \in I$$ iff

$$x \in \bigcap_{i \in I} \overline{A_i} $$

Therefore, $$\overline{\bigcup_{i \in I} A_i}= \bigcap_{i \in I} \overline{A_i}$$.

Also,

$$x \in \overline{\bigcap_{i \in I} A_i}$$ iff

$$x\ \not\in\ A_i$$, for some $$i \in I$$ iff

$$x \in \overline{A_i}$$, for some $$i \in I$$ iff

$$x \in \bigcup_{i \in I} \overline{A_i}$$

Therefore, $$\overline{\bigcap_{i \in I} A_i}= \bigcup_{i \in I} \overline{A_i}$$.

QED