Join is Associative

Theorem
Let $\left({S, \sqcup, \sqcap}\right)$ be a Boolean algebra.

Denote with $\preceq$ the ordering on $S$, and with $'$ the complement operation.

Then for all $a,b,c \in S$:


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

i.e., $\sqcup$ is associative.

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

Applying axiom $(BA \ 2)$ for Boolean algebras twice, we obtain that:


 * $a \sqcup \left({b \sqcup c}\right) \preceq d$ iff $a \preceq d$ and $b \preceq d$ and $c \preceq d$

Similarly, it follows that the right-hand condition is equivalent to:


 * $\left({a \sqcup b}\right) \sqcup c \preceq d$

The result follows from Poset Elements Equal iff Equal Weak Upper Closure.