Absorption Laws (Boolean Algebras)

This proof is about Absorption Laws in the context of Boolean Algebra. For other uses, see Absorption Laws.

Theorem

Then for all $a, b \in S$:

$a = a \vee \paren {a \wedge b}$

$a = a \wedge \paren {a \vee b}$

That is, $\vee$ absorbs $\wedge$, and $\wedge$ absorbs $\vee$.

Let $a, b \in S$.

Then:

$\ds a \vee \paren {a \wedge b}$

$=$

$\ds \paren {a \wedge \top} \vee \paren {a \wedge b}$

$\ds $

$=$

$\ds a \wedge \paren {\top \vee b}$

$\ds $

$=$

$\ds a \wedge \top$

$\ds $

$=$

$\ds a$

as desired.

$\Box$

The result:

$a = a \wedge \paren {a \vee b}$

follows from the Duality Principle.

$\blacksquare$