De Morganin lait

${\displaystyle {\overline {A\cap B}}={\overline {A}}\cup {\overline {B}}}$ jos ja vain jos ${\displaystyle {\overline {A\cap B}}\subseteq {\overline {A}}\cup {\overline {B}}}$ ja ${\displaystyle {\overline {A\cap B}}\supseteq {\overline {A}}\cup {\overline {B}}}$.

mielivaltaiselle ${\displaystyle x}$:lle:

${\displaystyle \subseteq }$:

${\displaystyle x\in {\overline {A\cap B}}}$

${\displaystyle x\notin {A\cap B}}$

${\displaystyle x\notin A}$ tai ${\displaystyle x\notin B}$

${\displaystyle x\in {\overline {A}}}$ tai ${\displaystyle x\in {\overline {B}}}$

${\displaystyle x\in {\overline {A}}\cup {\overline {B}}}$

Joten ${\displaystyle {\overline {A\cap B}}\subseteq {\overline {A}}\cup {\overline {B}}}$

${\displaystyle \supseteq }$:

${\displaystyle x\in {\overline {A}}\cup {\overline {B}}}$

${\displaystyle x\in {\overline {A}}}$ tai ${\displaystyle x\in {\overline {B}}}$

${\displaystyle x\notin A}$ tai ${\displaystyle x\notin B}$

${\displaystyle x\notin {A\cap B}}$

${\displaystyle x\in {\overline {A\cap B}}}$

Joten ${\displaystyle {\overline {A\cap B}}\supseteq {\overline {A}}\cup {\overline {B}}}$

${\displaystyle {\overline {A\cap B}}\subseteq {\overline {A}}\cup {\overline {B}}}$ ja ${\displaystyle {\overline {A\cap B}}\supseteq {\overline {A}}\cup {\overline {B}}}$ joten ${\displaystyle {\overline {A\cap B}}={\overline {A}}\cup {\overline {B}}}$

${\displaystyle {\overline {A\cup B}}={\overline {A}}\cap {\overline {B}}}$ voidaan todistaa käyttämällä samanlaista menetelmää.

• De Morganin lait Matematiikan verkkosanakirja. Matematiikkalehti Solmu. Viitattu 16.7.2020.
• Jan Thompson: Matematiikan käsikirja, s. 59. Tammi, 1991. ISBN 951-31-0471-0.