Join is Commutative

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

Then $\sqcup$ is commutative.

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

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


 * $a \sqcup b \preceq c$

By axiom $(BA \ 2)$ for Boolean algebras, this comes down to:


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

which is exactly the same condition as for:


 * $b \sqcup a \preceq c$

Therefore, we have established:


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

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


 * $b \sqcup a \preceq a \sqcup b$

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

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


 * $a \sqcup b = b \sqcup a$

That is, $\sqcup$ is commutative.