Strictly Precede and Step Condition and not Precede implies Joins are equal

Theorem
Let $\left({S, \vee, \preceq}\right)$ be a join semilattice.

Let $p, q, u \in S$ such that
 * $p \prec q$ and $\left({\forall s \in S: p \prec s \implies q \preceq s}\right)$ and $u \npreceq p$

Then $p \vee u = q \vee u$

Proof
We will prove that
 * $\forall s \in S: p \preceq s \land u \preceq s \implies q \vee u \preceq s$

Let $s \in S$ such that
 * $p \preceq s$ and $u \preceq s$

We have:
 * $p \ne s$

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

By step condition:
 * $q \preceq s$

Thus by Alternative Definition of Join:
 * $q \vee u \preceq s$

By Join Succeeds Operands:
 * $q \preceq q \vee u$ and $u \preceq q \vee u$

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

By definition of transitivity:
 * $p \preceq q \vee u$

Thus by Alternative Definition of Join
 * $p \vee u = q \vee u$