Join Semilattice Ordered Subset Not Always Subsemilattice

Theorem
Let $\struct {S, \circ}$ be a semilattices.

Let $\preceq$ be the ordering on $S$ defined by:
 * $a \preceq b \iff \paren {a \circ b} = b$

Let $T \subseteq S$ be a subset of $S$.

Let $\struct{T, \preceq \restriction_T}$ be a join semilattice.

Let $\vee$ be the binary operation on $S$ defined by:
 * for all $a, b \in S$, $a \vee b$ is the join of $a$ and $b$ with respect to $\preceq$.

Then:
 * $\struct{T, \vee}$ may not be a subsemilattice of $\struct {S, \circ}$.

Proof
Let $S = \set{p, q, r, s}$.

Let $\circ : S \times S \to S$ be defined by:
 * $p \circ q = q \circ p = r$
 * $p \circ r = r \circ p = r$
 * $q \circ r = r \circ q = r$
 * $p \circ s = s \circ p = s$
 * $q \circ s = s \circ q = s$
 * $r \circ s = s \circ r = s$
 * $p \circ p = p$
 * $q \circ q = q$
 * $r \circ r = r$
 * $s \circ s = s$

Then a manual check shows that $\circ$ is:
 * closed
 * associative
 * commutative
 * idempotent

So $\struct{S, \circ}$ is a semilattice with ordering
 * $\preceq = \set{\tuple{p, p}, \tuple{q, q}, \tuple{r, r}, \tuple{s, s}, \tuple{p, r}, \tuple{q,r}, \tuple{p, s}, \tuple{q,s}, \tuple{r, s}}$

Let $T = \set{p, q, s}$.

Then:
 * $\preceq \restriction_T = \set{\tuple{p, p}, \tuple{q, q}, \tuple{s, s}, \tuple{p, s}, \tuple{q,s}}$

Then a manual check shows that all joins exist in $\struct{T, \preceq \restriction_T}$:
 * $p \vee q = q \vee p = s$
 * $p \vee s = s \vee p = s$
 * $q \vee s = s \vee q = s$
 * $p \vee p = p$
 * $q \vee q = q$
 * $s \vee s = s$

So $\struct{T, \preceq \restriction_T}$ is a join semilattice with respect to $\preceq \restriction_T$.

Now:
 * $p \vee q = s \neq r = p * q$

So $\vee$ is not the restiction of $\circ$ to $T$.

That is, $\struct{T, \vee}$ is not a subsemilattice of $\struct {S, \circ}$ by definition.