Meet Precedes Operands

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

Let $a, b \in S$ admit a meet $a \wedge b \in S$.

Then:


 * $a \wedge b \preceq a$
 * $a \wedge b \preceq b$

That is, $a \wedge b$ precedes its operands $a$ and $b$.

Proof
By definition of meet:


 * $a \wedge b = \inf \set {a, b}$

where $\inf$ denotes infimum.

Since an infimum is a lower bound:


 * $\inf \set {a, b} \preceq a$
 * $\inf \set {a, b} \preceq b$

as desired.