Meet Precedes Operands

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Theorem

Let $\left({S, \preceq}\right)$ be an ordered set.

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


Then:

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

i.e., $a \wedge b$ precedes its operands $a$ and $b$.


Proof

By definition of meet:

$a \wedge b = \inf \left\{{a, b}\right\}$

where $\inf$ denotes infimum.


Since an infimum is a fortiori a lower bound:

$\inf \left\{{a, b}\right\} \preceq a$
$\inf \left\{{a, b}\right\} \preceq b$

as desired.

$\blacksquare$