Equivalence of algebraic and order semilattices

Let $S$ be an algebraic semilattice. (That is, an commutative semigroup in which every element is idempotent.) Recall the natural partial order $\leq$ on $S$ given by $e \leq f$ if and only if $e = ef = fe$. Show that any two elements $e$ and $f$ of $S$ have a greatest lower bound with respect to $\leq$, which is $ef$.

Conversely, suppose $S$ is an order semilattice. (That is, we have a partial order $\leq$ on $S$ such that any two elements have a greatest lower bound.) Show that the operator $\mathsf{glb}$ on $S$ makes $S$ into an algebraic semilattice, whose natural order coincides with $\leq$.

Suppose $S$ is an algebraic semilattice, and let $e,f \in S$. Now $ef = eff$ and $ef = eef = efe$, so that $ef \leq e$ and $ef \leq f$. Now suppose $h \leq e$ and $h \leq f$. Then $h = he$ and $h = hf$, so that $h = hf = hef$, and we have $h \leq ef$. So $ef$ is a greatest lower bound of $e$ and $f$ with respect to the natural order.

Now suppose $(S,\leq)$ is an order semilattice, and denote the greatest lower bound of $e$ and $f$ by $\mathsf{glb}(e,f)$. We claim that $\mathsf{glb}$ is associative. To that end, let $a,b,c \in S$. Note that $\mathsf{glb}(\mathsf{glb}(a,b),c) \leq \mathsf{glb}(a,b) \leq a$, $\mathsf{glb}(\mathsf{glb}(a,b),c) \leq \mathsf{glb}(a,b) \leq b$, and $\mathsf{glb}(\mathsf{glb}(a,b),c) \leq c$. So $\mathsf{glb}(\mathsf{glb}(a,b)) \leq \mathsf{glb}(b,c)$, and we have $\mathsf{glb}(\mathsf{glb}(a,b),c) \leq \mathsf{glb}(a,\mathsf{glb}(b,c))$. Similarly, $\mathsf{glb}(a,\mathsf{glb}(b,c)) \leq a,b,c$, so that $\mathsf{glb}(a,\mathsf{glb}(b,c)) \leq \mathsf{glb}(a,b)$, and we have $\mathsf{glb}(a,\mathsf{glb}(b,c)) \leq \mathsf{glb}(\mathsf{glb}(a,b),c)$. Since $\leq$ is antisymmetric, $\mathsf{glb}(\mathsf{glb}(a,b),c) = \mathsf{glb}(a,\mathsf{glb}(b,c))$. So $S$ is a semigroup under $\mathsf{glb}$. Certainly this operator is commutative and associative, so that $(S,\mathsf{glb})$ is an algebraic semilattice. Now let $\sigma$ denote the natural order on $S$. If $e \sigma f$, we have $e = \mathsf{glb}(e,f)$, so that $e \leq f$. Conversely, if $e \leq f$, then $\mathsf{glb}(e,f) = e$, and thus $e \sigma f$.