Meet 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 $a,b,c \in S$:


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

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

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

Applying axiom $(BA \ 3)$ for Boolean algebras twice, we obtain that:


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

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


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

The result follows by Poset Elements Equal iff Equal Weak Lower Closure.