Absorption Laws (Boolean Algebras)

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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 $\vee$.


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 $(\text {BA}_1 3)$: 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 $\vee$
\(\ds \) \(=\) \(\ds a \wedge \top\) Identities of Boolean Algebra are also Zeroes
\(\ds \) \(=\) \(\ds a\) 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.

$\blacksquare$


Sources