# De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection/Proof 2

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## Theorem

- $\overline {T_1 \cap T_2} = \overline T_1 \cup \overline T_2$

## Proof

\(\displaystyle \) | \(\) | \(\displaystyle x \in \overline {T_1 \cap T_2}\) | |||||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle x \notin \paren {T_1 \cap T_2}\) | Definition of Set Complement | ||||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle \neg \paren {x \in T_1 \land x \in T_2}\) | Definition of Set Intersection | ||||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle \neg \paren {x \in T_1} \lor \neg \paren {x \in T_2}\) | De Morgan's Laws (Logic): Disjunction of Negations | ||||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle x \in \overline {T_1} \lor x \in \overline {T_2}\) | Definition of Set Complement | ||||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle x \in \overline {T_1} \cup \overline {T_2}\) |

By definition of set equality:

- $\overline {T_1 \cap T_2} = \overline {T_1} \cup \overline {T_2}$

$\blacksquare$

## Sources

- 1977: Gary Chartrand:
*Introductory Graph Theory*... (previous) ... (next): Appendix $\text{A}.1$: Sets and Subsets: Problem Set $\text{A}.1$: $6$

- 2008: Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*... (previous) ... (next): $\S 2$