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 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