Identities of Boolean Algebra are also Zeroes
Jump to navigation
Jump to search
Theorem
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1.
Let the identity for $\vee$ be $\bot$ and the identity for $\wedge$ be $\top$.
Then:
\(\text {(1)}: \quad\) | \(\ds \forall x \in S: \, \) | \(\ds x \vee \top\) | \(=\) | \(\ds \top\) | ||||||||||
\(\text {(2)}: \quad\) | \(\ds \forall x \in S: \, \) | \(\ds x \wedge \bot\) | \(=\) | \(\ds \bot\) |
That is:
- $\bot$ is a zero element for $\wedge$
- $\top$ is a zero element for $\vee$.
Proof
Let $x \in S$.
Then:
\(\ds x \vee \top\) | \(=\) | \(\ds \paren {x \vee \top} \wedge \top\) | Boolean Algebra Axiom $(\text {BA}_1 3)$: Identity Elements: $\top$ is the identity of $\wedge$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \vee \top} \wedge \paren {x \vee \neg x}\) | Boolean Algebra Axiom $(\text {BA}_1 4)$: Complements: $x \vee x' = \top$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \vee \paren {\top \wedge x'}\) | Boolean Algebra Axiom $(\text {BA}_1 2)$: Distributivity: both $\vee$ and $\wedge$ distribute over the other | |||||||||||
\(\ds \) | \(=\) | \(\ds x \vee x'\) | Boolean Algebra Axiom $(\text {BA}_1 3)$: Identity Elements: $\top$ is the identity of $\wedge$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \top\) | Boolean Algebra Axiom $(\text {BA}_1 4)$: Complements: $x \vee x' = \top$ |
So $x \vee \top = \top$.
$\Box$
The result $x \wedge \bot = \bot$ follows from the Duality Principle.
$\blacksquare$
Also known as
These identities can be seen referred to as the Null Laws.
Sources
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 1.5$: Theorem $1.15, \ 1.15 \ \text{(b)}$
- 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