Join Succeeds Operands

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

Denote with $\preceq$ the ordering on $S$.

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


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

Proof
By axiom $(BA \ 2)$ for Boolean algebras, it follows that:


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

Since $\preceq$ is reflexive, the left-hand side is true.

Hence the result.