=== Definition 1 implies Definition 2 === Let $\left({S, \vee, \wedge, \neg}\right)$ be an algebraic system which satisfies the criteria of Definition 1. ;Axiom $(BA_2 \ 0)$: From Axiom $(BA_1 \ 0)$, we have that $\left({S, \vee, \wedge, \neg}\right)$ is closed under $\vee$, $\wedge$ and $\neg$. That is: :$\forall a, b \in S: a \vee b \in S$ :$\forall a, b \in S: a \wedge b \in S$ :$\forall a \in S: \neg a \in S$ ;Axiom $(BA_2 \ 1)$: From Axiom $(BA_1 \ 1)$, we have that $\vee$ and $\wedge$ are commutative on $S$. That is: :$\forall a, b \in S: a \vee b = b \vee a$ :$\forall a, b \in S: a \wedge b = b \wedge a$ ;Axiom $(BA_2 \ 2)$: From Operations of Boolean Algebra are Associative: :$\forall a, b, c \in S: a \vee \left({b \vee c}\right) = \left({a \vee b}\right) \vee c$ :$\forall a, b, c \in S: a \wedge \left({b \wedge c}\right) = \left({a \wedge b}\right) \wedge c$ ;Axiom $(BA_2 \ 3)$: From Absorption Laws (Boolean Algebras): For all $a, b \in S$: :$a = a \vee \left({a \wedge b}\right)$ :$a = a \wedge \left({a \vee b}\right)$ The specific format in which these results are expressed in axiom $(BA_2 \ 3)$ follow from Axiom $(BA_2 \ 1)$: $\vee$ and $\wedge$ are commutative on $S$. ;Axiom $(BA_2 \ 4)$: From Axiom $(BA_1 \ 2)$, we have that both $\vee$ and $\wedge$ distribute over the other. Hence: :$\forall a, b, c \in S: a \wedge \left({b \vee c}\right) = \left({a \wedge b}\right) \vee \left({a \wedge c}\right)$ :$\forall a, b, c \in S: a \vee \left({b \wedge c}\right) = \left({a \vee b}\right) \wedge \left({a \vee c}\right)$ ;Axiom $(BA_2 \ 5)$: We have: {{begin-eqn}} {{eqn | l = \left({a \wedge \neg a}\right) \vee b       | r = \bot \vee b       | c = Boolean Algebra: Axiom $(BA_1 \ 4)$ }} {{eqn | r = b       | c = Boolean Algebra: Axiom $(BA_1 \ 3)$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \left({a \vee \neg a}\right) \wedge b       | r = \top \wedge b       | c = Boolean Algebra: Axiom $(BA_1 \ 4)$ }} {{eqn | r = b       | c = Boolean Algebra: Axiom $(BA_1 \ 3)$ }} {{end-eqn}} All axioms are fulfilled. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\left({S, \vee, \wedge, \neg}\right)$ be an algebraic system which satisfies the criteria of Definition 2. ;Axiom $(BA_1 \ 0)$: From Axiom $(BA_2 \ 0)$, we have that :$\forall a, b \in S: a \vee b \in S, \ a \wedge b \in S, \ \neg a \in S$ That is, $\left({S, \vee, \wedge, \neg}\right)$ is closed under $\vee$, $\wedge$ and $\neg$. ;Axiom $(BA_1 \ 1)$: From Axiom $(BA_2 \ 2)$, we have that: :$\forall a, b \in S: a \vee b = b \vee a, \ a \wedge b = b \wedge a$ That is, $\vee$ and $\wedge$ are commutative on $S$. ;Axiom $(BA_1 \ 2)$: From Axiom $(BA_2 \ 4)$, we have that: :$\forall a, b, c \in S: a \wedge \left({b \vee c}\right) = \left({a \wedge b}\right) \vee \left({a \wedge c}\right)$ :$\forall a, b, c \in S: a \vee \left({b \wedge c}\right) = \left({a \vee b}\right) \wedge \left({a \vee c}\right)$ From Axiom $(BA_1 \ 1)$, we have that $\vee$ and $\wedge$ are commutative on $S$, and so: :$\forall a, b, c \in S: \left({a \vee b}\right) \wedge c = \left({a \wedge c}\right) \vee \left({b \wedge c}\right)$ :$\forall a, b, c \in S: \left({a \wedge b}\right) \vee c = \left({a \vee c}\right) \wedge \left({b \vee c}\right)$ That is, both $\vee$ and $\wedge$ distribute over the other. ;Axiom $(BA_1 \ 4)$: From Meet with Complement is Bottom: :$\exists \bot \in S: \forall a \in S: a \wedge \neg a = \bot$ From Join with Complement is Top: :$\exists \top \in S: \forall a \in S: a \vee \neg a = \top$ These elements $\bot$ and $\top$ are the only elements of $S$ which have these properties. ;Axiom $(BA_1 \ 3)$: We have demonstrated Boolean Algebra: Axiom $(BA_1 \ 4)$ above. Then: {{begin-eqn}} {{eqn | l = \bot \vee b       | r = \left({a \wedge \neg a}\right) \vee b       | c = Boolean Algebra: Axiom $(BA_1 \ 4)$ }} {{eqn | r = b       | c = Boolean Algebra: Axiom $(BA_2 \ 5)$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \top \wedge b       | r = \left({a \vee \neg a}\right) \wedge b       | c = Boolean Algebra: Axiom $(BA_1 \ 4)$ }} {{eqn | r = b       | c = Boolean Algebra: Axiom $(BA_2 \ 5)$ }} {{end-eqn}} From Axiom $(BA_1 \ 1)$, we have that $\vee$ and $\wedge$ are commutative on $S$, and so: :$b \vee \bot = b :$b \wedge \top = b That is, $\bot$ is an identity element for $\vee$, and $\top$ is an identity element for $\wedge$. All axioms are fulfilled. {{qed|lemma}} Hence the result. {{qed}} Category:Boolean Algebras ct7dewa199tpdk5r2hvb615d760yena