Equivalence of Definitions of Boolean Algebra

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Theorem

The following definitions of the concept of Boolean Algebra are equivalent:

Definition 1

A Boolean algebra is an algebraic system $\struct {S, \vee, \wedge, \neg}$, where $\vee$ and $\wedge$ are binary, and $\neg$ is a unary operation.

Furthermore, these operations are required to satisfy the following axioms:

\((\text {BA}_1 0)\)   $:$   $S$ is closed under $\vee$, $\wedge$ and $\neg$      
\((\text {BA}_1 1)\)   $:$   Both $\vee$ and $\wedge$ are commutative      
\((\text {BA}_1 2)\)   $:$   Both $\vee$ and $\wedge$ distribute over the other      
\((\text {BA}_1 3)\)   $:$   Both $\vee$ and $\wedge$ have identities $\bot$ and $\top$ respectively      
\((\text {BA}_1 4)\)   $:$   $\forall a \in S: a \vee \neg a = \top, a \wedge \neg a = \bot$      


Definition 2

A Boolean algebra is an algebraic system $\struct {S, \vee, \wedge, \neg}$, where $\vee$ and $\wedge$ are binary, and $\neg$ is a unary operation.

Furthermore, these operations are required to satisfy the following axioms:

\((\text {BA}_2 0)\)   $:$   Closure:      \(\ds \forall a, b \in S:\) \(\ds a \vee b \in S \)      
\(\ds a \wedge b \in S \)      
\(\ds \neg a \in S \)      
\((\text {BA}_2 1)\)   $:$   Commutativity:      \(\ds \forall a, b \in S:\) \(\ds a \vee b = b \vee a \)      
\(\ds a \wedge b = b \wedge a \)      
\((\text {BA}_2 2)\)   $:$   Associativity:      \(\ds \forall a, b, c \in S:\) \(\ds a \vee \paren {b \vee c} = \paren {a \vee b} \vee c \)      
\(\ds a \wedge \paren {b \wedge c} = \paren {a \wedge b} \wedge c \)      
\((\text {BA}_2 3)\)   $:$   Absorption Laws:      \(\ds \forall a, b \in S:\) \(\ds \paren {a \wedge b} \vee b = b \)      
\(\ds \paren {a \vee b} \wedge b = b \)      
\((\text {BA}_2 4)\)   $:$   Distributivity:      \(\ds \forall a, b, c \in S:\) \(\ds a \wedge \paren {b \vee c} = \paren {a \wedge b} \vee \paren {a \wedge c} \)      
\(\ds a \vee \paren {b \wedge c} = \paren {a \vee b} \wedge \paren {a \vee c} \)      
\((\text {BA}_2 5)\)   $:$   Identity Elements:      \(\ds \forall a, b \in S:\) \(\ds \paren {a \wedge \neg a} \vee b = b \)      
\(\ds \paren {a \vee \neg a} \wedge b = b \)      


Proof

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:

\(\ds \paren {a \wedge \neg a} \vee b\) \(=\) \(\ds \bot \vee b\) Boolean Algebra: Axiom $(\text {BA}_1 \ 4)$
\(\ds \) \(=\) \(\ds b\) Boolean Algebra: Axiom $(\text {BA}_1 \ 3)$


\(\ds \paren {a \vee \neg a} \wedge b\) \(=\) \(\ds \top \wedge b\) Boolean Algebra: Axiom $(\text {BA}_1 \ 4)$
\(\ds \) \(=\) \(\ds b\) Boolean Algebra: Axiom $(\text {BA}_1 \ 3)$


All axioms are fulfilled.

$\Box$


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:

\(\ds \bot \vee b\) \(=\) \(\ds \paren {a \wedge \neg a} \vee b\) Boolean Algebra: Axiom $(\text {BA}_1 \ 4)$
\(\ds \) \(=\) \(\ds b\) Boolean Algebra: Axiom $(\text {BA}_2 \ 5)$
\(\ds \top \wedge b\) \(=\) \(\ds \paren {a \vee \neg a} \wedge b\) Boolean Algebra: Axiom $(\text {BA}_1 \ 4)$
\(\ds \) \(=\) \(\ds b\) Boolean Algebra: Axiom $(\text {BA}_2 \ 5)$

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.

$\Box$


Hence the result.

$\blacksquare$

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