Operations of Boolean Algebra are Idempotent
Jump to navigation
Jump to search
Definition
Let $\struct {S, \vee, \wedge}$ be a Boolean algebra.
Then:
- $\forall x \in S: x \wedge x = x = x \vee x$
That is, both $\vee$ and $\wedge$ are idempotent operations.
Proof
Let $x \in S$.
Then:
\(\ds x\) | \(=\) | \(\ds x \vee \bot\) | as $\bot$ is the identity of $\vee$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \vee \paren {x \wedge \neg x}\) | as $x \wedge \neg x = \bot$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \vee x} \wedge \paren {x \vee \neg x}\) | both $\vee$ and $*$ distribute over the other | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \vee x} \wedge \top\) | as $x \vee \neg x = \top$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \vee x\) |
So $x = x \vee x$.
$\Box$
The result $x = x \wedge x$ follows from Duality Principle (Boolean Algebras).
$\blacksquare$
Sources
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 1.5$: Theorem $1.13$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Boolean algebra
- 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): $\S 2$: Exercise $2$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Boolean algebra