Identities of Boolean Algebra 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:

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

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.

$\blacksquare$