Identities of Boolean Algebra are also Zeroes

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