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:

$\ds \forall a, b \in S: \, $

$\ds a$

$=$

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

$\ds a$

$=$

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

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

Let $a, b \in S$.

Then:

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

$=$

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

Boolean Algebra Axiom $(\text {BA}_1 3)$: Identity Elements: $\top$ is the identity for $\wedge$

$\ds $

$=$

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

Boolean Algebra Axiom $(\text {BA}_1 2)$: Distributivity: $\wedge$ distributes over $\vee$

$\ds $

$=$

$\ds a \wedge \top$

$\ds $

$=$

$\ds a$

Boolean Algebra Axiom $(\text {BA}_1 3)$: Identity Elements: $\top$ is the identity for $\wedge$

as desired.

$\Box$

The result:

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

follows from the Duality Principle.

$\blacksquare$

1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 1.5$: Lemma $1.1$
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Boolean algebra
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Boolean algebra