Order, lattices, ordered algebraic structures Proposition

De Morgan's laws in a Boolean algebra

Statement

Let $A$ be a Boolean algebra. Then $\neg(a \vee b) = \neg a \wedge \neg b$ and $\neg(a \wedge b) = \neg a \vee \neg b$ for all $a, b \in A$ .

Remarks

Proof idea: in a Boolean algebra each element has a unique complement. Check that $\neg a \wedge \neg b$ satisfies the two complement equations for $a \vee b$ . In the power set Boolean algebra, the laws become $X \setminus (A \cup B) = (X \setminus A) \cap (X \setminus B)$ and its dual. In the two-element Boolean algebra, they are the laws of logic for not, and, and or.

History

The laws are named after Augustus De Morgan. His book Formal Logic of 1847 states them.

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Boolean algebra definition
De Morgan's laws in a Boolean algebra statement

Tag	`0XYVDWGQX2`
URI	`m2://0XYVDWGQX2`
Kind	statement, proposition
Writing	W0 stub
MSC	06E05
Status	This copy of the site has no build data. Run `m2 status 0XYVDWGQX2`.
Uses	Boolean algebra

De Morgan's laws in a Boolean algebra

Statement

Remarks

History

Reading path

Further reading

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