Absorption Laws (Boolean Algebras)

From ProofWiki
Jump to navigation Jump to search

This proof is about the absorption laws as they apply to Boolean algebras. 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:

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

$\blacksquare$


Sources