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 $\neg$.

$(BA \ 1)$: Commutativity
Follows from the Rule of Commutation.

$(BA \ 2)$: Distributivity
Follows from the Rule of Distribution

$(BA \ 3)$: Identities
Follows from Conjunction with Tautology and Disjunction with Contradiction.

$(BA \ 4)$: Complements
Follows from Contradiction Negation of Tautology and Tautology Negation of Contradiction.

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