Identities of Boolean Algebra also Zeroes

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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:

$(1): \quad \forall x \in S: x \vee \top = \top$
$(2): \quad \forall x \in S: x \wedge \bot = \bot$


That is, $\bot$ is a zero element for $\wedge$, and $\top$ 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 $(\text {BA}_1 3)$: $\top$ is the identity of $\wedge$
\(\ds \) \(=\) \(\ds \paren {x \vee \top} \wedge \paren {x \vee \neg x}\) Boolean Algebra: Axiom $(\text {BA}_1 4)$: $x \vee x' = \top$
\(\ds \) \(=\) \(\ds x \vee \paren {\top \wedge x'}\) Boolean Algebra: Axiom $(\text {BA}_1 2)$: both $\vee$ and $\wedge$ distribute over the other
\(\ds \) \(=\) \(\ds x \vee x'\) Boolean Algebra: Axiom $(\text {BA}_1 3)$: $\top$ is the identity of $\wedge$
\(\ds \) \(=\) \(\ds \top\) Boolean Algebra: Axiom $(\text {BA}_1 4)$ $x \vee x' = \top$

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

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