Meet Precedes Operands

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

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

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


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

Proof
By axiom $(BA \ 3)$ for Boolean algebras, it follows that:


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

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

Hence the result.