Absorption Laws (Boolean Algebras)

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This proof is about the absorption laws as they apply to Boolean algebras. For other uses, see Absorption Laws.

Theorem

Let $\left({S, \vee, \wedge}\right)$ be a Boolean algebra, defined as in Definition 1.


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

$a = a \vee \left({a \wedge b}\right)$
$a = a \wedge \left({a \vee b}\right)$

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


Proof

Let $a, b \in S$.


Then:

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

as desired.

$\Box$


The result:

$a = a \wedge \left({a \vee b}\right)$

follows from the Duality Principle.

$\blacksquare$


Sources