Completely Irreducible implies Meet Irreducible

Theorem
Let $L = \left({S, \wedge, \preceq}\right)$ 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 that
 * $a \ne p$ and $b \ne p$

By Meet Precedes Operands:
 * $p \preceq b$

By definition of strictly precede:
 * $p \prec b$

Then
 * $q \preceq b$

By definition of strictly precede:
 * $p \preceq q$

By Meet Precedes Operands:
 * $p \preceq a$

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.