Operations of Boolean Algebra are Associative

Theorem
Let $\left({S, \vee, \wedge, \neg}\right)$ be a Boolean algebra, defined as in Definition 1.

Then:


 * $\forall a, b, c \in S: \left({a \wedge b}\right) \wedge c = a \wedge \left({b \wedge c}\right)$
 * $\forall a, b, c \in S: \left({a \vee b}\right) \vee c = a \vee \left({b \vee c}\right)$

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

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

Let:
 * $x = a \wedge \left({b \wedge c}\right)$
 * $y = \left({a \wedge b}\right) \wedge c$

Then:

Similarly:

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

Next:

Similarly:

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, i.e.:


 * $\left({a \wedge b}\right) \wedge c = a \wedge \left({b \wedge c}\right)$

The result:


 * $\left({a \vee b}\right) \vee c = a \vee \left({b \vee c}\right)$

follows from the Duality Principle.