Operations of Boolean Algebra are Associative

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Theorem

Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1.


Then:

\(\ds \forall a, b, c \in S: \, \) \(\ds \paren {a \wedge b} \wedge c\) \(=\) \(\ds a \wedge \paren {b \wedge c}\)
\(\ds \forall a, b, c \in S: \, \) \(\ds \paren {a \vee b} \vee c\) \(=\) \(\ds a \vee \paren {b \vee c}\)


That is, both $\vee$ and $\wedge$ are associative operations.


Proof

Let $a, b, c \in S$.

Let:

$x = a \wedge \paren {b \wedge c}$
$y = \paren {a \wedge b} \wedge c$

Then:

\(\ds a \vee x\) \(=\) \(\ds a \vee \paren {a \wedge \paren {b \wedge c} }\)
\(\ds \) \(=\) \(\ds \paren {a \vee a} \wedge \paren {a \vee \paren {b \wedge c} }\) Boolean Algebra: Axiom $(\text {BA}_1 2)$: both $\vee$ and $\wedge$ distribute over the other
\(\ds \) \(=\) \(\ds a \wedge \paren {a \vee \paren {b \wedge c} }\) Operations of Boolean Algebra are Idempotent
\(\ds \) \(=\) \(\ds a\) Absorption Laws (Boolean Algebras)


Similarly:

\(\ds a \vee y\) \(=\) \(\ds a \vee \paren {\paren {a \wedge b} \wedge c}\)
\(\ds \) \(=\) \(\ds \paren {a \vee \paren {a \wedge b} } \wedge \paren {a \vee c}\) Boolean Algebra: Axiom $(\text {BA}_1 2)$: both $\vee$ and $\wedge$ distribute over the other
\(\ds \) \(=\) \(\ds a \wedge \paren {a \vee c}\) Absorption Laws (Boolean Algebras)
\(\ds \) \(=\) \(\ds a\) Absorption Laws (Boolean Algebras)

Thus we see we have $a \vee x = a \vee y$.


Next:

\(\ds \neg a \vee x\) \(=\) \(\ds \neg a \vee \paren {a \wedge \paren {b \wedge c} }\)
\(\ds \) \(=\) \(\ds \paren {\neg a \vee a} \wedge \paren {\neg a \vee \paren {b \wedge c} }\) Boolean Algebra: Axiom $(\text {BA}_1 2)$: both $\vee$ and $\wedge$ distribute over the other
\(\ds \) \(=\) \(\ds \top \wedge \paren {\neg a \vee \paren {b \wedge c} }\) Boolean Algebra: Axiom $(\text {BA}_1 4)$: $\neg a \vee a = \top$
\(\ds \) \(=\) \(\ds \neg a \vee \paren {b \wedge c}\) Boolean Algebra: Axiom $(\text {BA}_1 3)$: $\top$ is the identity of $\wedge$


Similarly:

\(\ds \neg a \vee y\) \(=\) \(\ds \neg a \vee \paren {\paren {a \wedge b} \wedge c}\)
\(\ds \) \(=\) \(\ds \paren {\neg a \vee \paren {a \wedge b} } \wedge \paren {\neg a \vee c}\) Boolean Algebra: Axiom $(\text {BA}_1 2)$: both $\vee$ and $\wedge$ distribute over the other
\(\ds \) \(=\) \(\ds \paren {\paren {\neg a \vee a} \wedge \paren {\neg a \vee b} } \wedge \paren {\neg a \vee c}\) Boolean Algebra: Axiom $(\text {BA}_1 2)$: both $\vee$ and $\wedge$ distribute over the other
\(\ds \) \(=\) \(\ds \paren {\top \wedge \paren {\neg a \vee b} } \wedge \paren {\neg a \vee c}\) Boolean Algebra: Axiom $(\text {BA}_1 4)$: $\neg a \vee a = \top$
\(\ds \) \(=\) \(\ds \paren {\neg a \vee b} \wedge \paren {\neg a \vee c}\) Boolean Algebra: Axiom $(\text {BA}_1 3)$: $\top$ is the identity of $\wedge$
\(\ds \) \(=\) \(\ds \neg a \vee \paren {b \wedge c}\) Boolean Algebra: Axiom $(\text {BA}_1 2)$: both $\vee$ and $\wedge$ distribute over the other

Thus we see we have $\neg a \vee x = \neg a \vee y$.


In conclusion, we have:

$a \vee x = a \vee y$
$\neg a \vee x = \neg a \vee y$

Hence $x = y$ by Cancellation of Join in Boolean Algebra, that is:

$\paren {a \wedge b} \wedge c = a \wedge \paren {b \wedge c}$

$\Box$


The result:

$\paren {a \vee b} \vee c = a \vee \paren {b \vee c}$

follows from the Duality Principle.

$\blacksquare$


Sources

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