Two is Boolean Algebra

Theorem

Let $\mathbf 2$ denote two.

Then $\mathbf 2$ is a Boolean algebra.

Proof

It is useful to first state the Cayley tables for the three logical operations $\lor$, $\land$ and $\neg$:

$\begin{array}{c|cc} \lor & \bot & \top \\ \hline \bot & \bot & \top \\ \top & \top & \top \end{array} \qquad \begin{array}{c|cc} \land & \bot & \top \\ \hline \bot & \bot & \bot \\ \top & \bot & \top \end{array} \qquad \begin{array}{c|cc} & \bot & \top \\ \hline \neg & \top & \bot \end{array}$

Let us now verify the axioms for a Boolean algebra in turn.

$(BA \ 0)$: Closure

It is immediate from the Cayley tables that $S$ is closed under $\lor$, $\land$ and $\neg$.

$\Box$

$(BA \ 1)$: Commutativity

Follows from the Rule of Commutation.

$\Box$

$(BA \ 2)$: Distributivity

Follows from the Rule of Distribution

$\Box$

$(BA \ 3)$: Identities

Follows from Conjunction with Tautology and Disjunction with Contradiction.

$\Box$

$(BA \ 4)$: Complements

Follows from Contradiction is Negation of Tautology and Tautology is Negation of Contradiction.

$\Box$

Having verified all axioms, we conclude $\mathbf 2$ is a Boolean algebra.

$\blacksquare$

Sources

• 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): $\S 2$: Exercise $3$