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