De Morgan's Laws (Set Theory)

Theorem

 * $$R - \left({S \cap T}\right) = \left({R - S}\right) \cup \left({R - T}\right)$$
 * $$R - \left({S \cup T}\right) = \left({R - S}\right) \cap \left({R - T}\right)$$


 * $$\overline {S \cup T} = \overline S \cap \overline T$$
 * $$\overline {S \cap T} = \overline S \cup \overline T$$

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
So $$R - \left({S \cap T}\right) = \left({R - S}\right) \cup \left({R - T}\right)$$.

So $$R - \left({S \cup T}\right) = \left({R - S}\right) \cap \left({R - T}\right)$$.

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

Also,

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

QED