Equivalence of Definitions of Boolean Algebra

Definition 1 implies Definition 2
Let $\struct {S, \vee, \wedge, \neg}$ be an algebraic system which satisfies the criteria of Definition 1.


 * Axiom $(\text {BA}_2 \ 0)$:

From Axiom $(\text {BA}_1 \ 0)$, we have that $\struct {S, \vee, \wedge, \neg}$ 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 $(\text {BA}_2 \ 1)$:

From Axiom $(\text {BA}_1 \ 1)$, 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 $(\text {BA}_2 \ 2)$:

From Operations of Boolean Algebra are Associative:


 * $\forall a, b, c \in S: a \vee \paren {b \vee c} = \paren {a \vee b} \vee c$
 * $\forall a, b, c \in S: a \wedge \paren {b \wedge c} = \paren {a \wedge b} \wedge c$


 * Axiom $(\text {BA}_2 \ 3)$:

From Absorption Laws (Boolean Algebras):

For all $a, b \in S$:


 * $a = a \vee \paren {a \wedge b}$
 * $a = a \wedge \paren {a \vee b}$

The specific format in which these results are expressed in axiom $(\text {BA}_2 \ 3)$ follow from Axiom $(\text {BA}_2 \ 1)$: $\vee$ and $\wedge$ are commutative on $S$.


 * Axiom $(\text {BA}_2 \ 4)$:

From Axiom $(\text {BA}_1 \ 2)$, we have that both $\vee$ and $\wedge$ distribute over the other.

Hence:
 * $\forall a, b, c \in S: a \wedge \paren {b \vee c} = \paren {a \wedge b} \vee \paren {a \wedge c}$
 * $\forall a, b, c \in S: a \vee \paren {b \wedge c} = \paren {a \vee b} \wedge \paren {a \vee c}$


 * Axiom $(\text {BA}_2 \ 5)$:

We have:

All axioms are fulfilled.

Definition 2 implies Definition 1
Let $\struct {S, \vee, \wedge, \neg}$ be an algebraic system which satisfies the criteria of Definition 2.


 * Axiom $(\text {BA}_1 \ 0)$:

From Axiom $(\text {BA}_2 \ 0)$, we have that
 * $\forall a, b \in S: a \vee b \in S, \ a \wedge b \in S, \ \neg a \in S$

That is, $\struct {S, \vee, \wedge, \neg}$ is closed under $\vee$, $\wedge$ and $\neg$.


 * Axiom $(\text {BA}_1 \ 1)$:

From Axiom $(\text {BA}_2 \ 2)$, 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 $(\text {BA}_1 \ 2)$:

From Axiom $(\text {BA}_2 \ 4)$, we have that:
 * $\forall a, b, c \in S: a \wedge \paren {b \vee c} = \paren {a \wedge b} \vee \paren {a \wedge c}$
 * $\forall a, b, c \in S: a \vee \paren {b \wedge c} = \paren {a \vee b} \wedge \paren {a \vee c}$

From Axiom $(\text {BA}_1 \ 1)$, we have that $\vee$ and $\wedge$ are commutative on $S$, and so:


 * $\forall a, b, c \in S: \paren {a \vee b} \wedge c = \paren {a \wedge c} \vee \paren {b \wedge c}$
 * $\forall a, b, c \in S: \paren {a \wedge b} \vee c = \paren {a \vee c} \wedge \paren {b \vee c}$

That is, both $\vee$ and $\wedge$ distribute over the other.


 * Axiom $(\text {BA}_1 \ 4)$:

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 $(\text {BA}_1 \ 3)$:

We have demonstrated Boolean Algebra: Axiom $(\text {BA}_1 \ 4)$ above.

Then:

From Axiom $(\text {BA}_1 \ 1)$, 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.

Hence the result.