# Absorption Laws (Boolean Algebras)

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

## Theorem

Let $\struct {S, \vee, \wedge}$ be a Boolean algebra, defined as in Definition 1.

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 $\vee$.

## Proof

Let $a, b \in S$.

Then:

 $\ds a \vee \paren {a \wedge b}$ $=$ $\ds \paren {a \wedge \top} \vee \paren {a \wedge b}$ Boolean Algebra: Axiom $(BA_1 \ 3)$: $\top$ is the identity for $\wedge$ $\ds$ $=$ $\ds a \wedge \paren {\top \vee b}$ Boolean Algebra: Axiom $(BA_1 \ 2)$: $\wedge$ distributes over $\vee$ $\ds$ $=$ $\ds a \wedge \top$ Identities of Boolean Algebra also Zeroes $\ds$ $=$ $\ds a$ Boolean Algebra: Axiom $(BA_1 \ 3)$ $\top$ is the identity for $\wedge$

as desired.

$\Box$

The result:

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

follows from the Duality Principle.

$\blacksquare$