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 fortiori a lower bound:

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

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

as desired.

$\blacksquare$