Meet is Commutative

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

Then $\sqcap$ is commutative.

Proof
Denote with $\preceq$ the ordering of the Boolean algebra $S$.

Suppose that for some $a,b,c \in S$, we have:


 * $c \preceq a \sqcap b$

By axiom $(BA \ 3)$ for Boolean algebras, this comes down to:


 * $c \preceq a$ and $c \preceq b$

which is exactly the same condition as for:


 * $c \preceq b \sqcap a$

Therefore, we have established:


 * $c \preceq a \sqcap b \iff c \preceq b \sqcap a$

Since $\preceq$ is reflexive, taking $c = b \sqcap a$ we obtain:


 * $b \sqcap a \preceq a \sqcap b$

and $c = a \sqcap b$ yields the converse.

Since $\preceq$ is also antisymmetric, it follows that:


 * $a \sqcap b = b \sqcap a$

That is, $\sqcap$ is commutative.