Completely Irreducible implies Meet Irreducible
Jump to navigation
Jump to search
Theorem
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $p \in S$.
Then if $p$ is completely irreducible, then $p$ is meet irreducible.
Proof
Assume that
- $p$ is completely irreducible.
By Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element
- $\exists q \in S: p \prec q \land \left({\forall s \in S: p \prec s \implies q \preceq s}\right) \land p^\succeq = \left\{ {p}\right\} \cup q^\succeq$
where $p^\succeq$ denotes the upper closure of $p$.
Let $a, b \in S$ such that
- $p = a \wedge b$
Aiming for a contradiction, suppose
- $a \ne p$ and $b \ne p$
- $p \preceq b$
By definition of strictly precede:
- $p \prec b$
Then
- $q \preceq b$
By definition of strictly precede:
- $p \preceq q$
- $p \preceq a$
By definition of strictly precede:
- $p \prec a$
Then
- $q \preceq a$
By Meet Semilattice is Ordered Structure:
- $q \preceq p$
By definition of antisymmetry:
- $p = q$
Thus this contradicts $p \prec q$ by definition of strictly precede.
$\blacksquare$
Sources
- Mizar article WAYBEL16:23