Identities of Boolean Algebra also Zeroes

From ProofWiki
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:

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


That is, $\bot$ is a zero element for $\wedge$, and $\top$ is a zero element for $\vee$.


Proof

Let $x \in S$.

Then:

\(\displaystyle x \vee \top\) \(=\) \(\displaystyle \paren {x \vee \top} \wedge \top\) Boolean Algebra: Axiom $(BA_1 \ 3)$: $\top$ is the identity of $\wedge$
\(\displaystyle \) \(=\) \(\displaystyle \paren {x \vee \top} \wedge \paren {x \vee \neg x}\) Boolean Algebra: Axiom $(BA_1 \ 4)$: $x \vee x' = \top$
\(\displaystyle \) \(=\) \(\displaystyle x \vee \paren {\top \wedge x'}\) Boolean Algebra: Axiom $(BA_1 \ 2)$: both $\vee$ and $\wedge$ distribute over the other
\(\displaystyle \) \(=\) \(\displaystyle x \vee x'\) Boolean Algebra: Axiom $(BA_1 \ 3)$: $\top$ is the identity of $\wedge$
\(\displaystyle \) \(=\) \(\displaystyle \top\) Boolean Algebra: Axiom $(BA_1 \ 4)$ $x \vee x' = \top$

So $x \vee \top = \top$.

$\Box$


The result $x \wedge \bot = \bot$ follows from the Duality Principle.

$\blacksquare$


Sources