Theorem: Operations of Boolean Algebra are Associative

Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1.
Then:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \wedge b} \wedge c
      | r = a \wedge \paren {b \wedge c}
}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \vee b} \vee c
      | r = a \vee \paren {b \vee c}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = a \vee x
      | r = a \vee \paren {a \wedge \paren {b \wedge c} }
      | c = 
}}
{{eqn | r = \paren {a \vee a} \wedge \paren {a \vee \paren {b \wedge c} }
      | c = {{Boolean-algebra-axiom|1|2}}: both $\vee$ and $\wedge$ distribute over the other
}}
{{eqn | r = a \wedge \paren {a \vee \paren {b \wedge c} }
      | c = Operations of Boolean Algebra are Idempotent
}}
{{eqn | r = a
      | c = Absorption Laws (Boolean Algebras)
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = a \vee y
      | r = a \vee \paren {\paren {a \wedge b} \wedge c}
      | c = 
}}
{{eqn | r = \paren {a \vee \paren {a \wedge b} } \wedge \paren {a \vee c}
      | c = {{Boolean-algebra-axiom|1|2}}: both $\vee$ and $\wedge$ distribute over the other
}}
{{eqn | r = a \wedge \paren {a \vee c}
      | c = Absorption Laws (Boolean Algebras)
}}
{{eqn | r = a
      | c = Absorption Laws (Boolean Algebras)
}}
{{end-eqn}}
Thus we see we have $a \vee x = a \vee y$.
Next:
{{begin-eqn}}
{{eqn | l = \neg a \vee x
      | r = \neg a \vee \paren {a \wedge \paren {b \wedge c} }
      | c = 
}}
{{eqn | r = \paren {\neg a \vee a} \wedge \paren {\neg a \vee \paren {b \wedge c} }
      | c = {{Boolean-algebra-axiom|1|2}}: both $\vee$ and $\wedge$ distribute over the other
}}
{{eqn | r = \top \wedge \paren {\neg a \vee \paren {b \wedge c} }
      | c = {{Boolean-algebra-axiom|1|4}}: $\neg a \vee a = \top$
}}
{{eqn | r = \neg a \vee \paren {b \wedge c}
      | c = {{Boolean-algebra-axiom|1|3}}: $\top$ is the identity of $\wedge$
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \neg a \vee y
      | r = \neg a \vee \paren {\paren {a \wedge b} \wedge c}
      | c = 
}}
{{eqn | r = \paren {\neg a \vee \paren {a \wedge b} } \wedge \paren {\neg a \vee c}
      | c = {{Boolean-algebra-axiom|1|2}}: both $\vee$ and $\wedge$ distribute over the other
}}
{{eqn | r = \paren {\paren {\neg a \vee a} \wedge \paren {\neg a \vee b} } \wedge \paren {\neg a \vee c}
      | c = {{Boolean-algebra-axiom|1|2}}: both $\vee$ and $\wedge$ distribute over the other
}}
{{eqn | r = \paren {\top \wedge \paren {\neg a \vee b} } \wedge \paren {\neg a \vee c}
      | c = {{Boolean-algebra-axiom|1|4}}: $\neg a \vee a = \top$
}}
{{eqn | r = \paren {\neg a \vee b} \wedge \paren {\neg a \vee c}
      | c = {{Boolean-algebra-axiom|1|3}}: $\top$ is the identity of $\wedge$
}}
{{eqn | r = \neg a \vee \paren {b \wedge c}
      | c = {{Boolean-algebra-axiom|1|2}}: both $\vee$ and $\wedge$ distribute over the other
}}
{{end-eqn}}
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:
:$\paren {a \wedge b} \wedge c = a \wedge \paren {b \wedge c}$
{{qed|lemma}}
The result:
:$\paren {a \vee b} \vee c = a \vee \paren {b \vee c}$
follows from the Duality Principle.
{{qed}}