every $x,y\in B$, $x\vee y<\infty\Rightarrow x\vee y\in B$. Then for all $x\in B\cap PP^{-1}$, there exist $\mu,\nu\in B\cap P$ such that $x=\mu\nu^{-1}$ and for all $p,q\in P$ with $pq^{-1}=x$ we have $\mu\leq p$ and $\nu\leq q$.

The lemma says that if $x\in PP^{-1}\cap B$, then the minimal elements of must also be contained in $B$. In particular, if $\phi(A\cap P)=B\cap P$, then $\phi(x)\in PP^{-1}$.

Fix $x\in B\cap PP^{-1}$. Say $x=st^{-1}$ with $s,t\in P$. Then $x^{-1}s\in P$ and $x\leq s$. Also $e\leq s$, and so $x\vee e<\infty$. Since $e,x\in B$ we get $x\vee e\in B$. Let $\mu=x\vee e$ and $\nu=x^{-1}(x\vee e)$. Then $\mu\nu^{-1}=x\vee e(x^{-1}(x\vee e))^{-1}=x$.

Fix $p,q\in P$ such that $x=pq^{-1}$. Then $x^{-1}p=(pq^{-1})^{-1}p=q$, and so $x\leq p$. Therefore $\mu=x\vee e\leq p$. Now $\mu^{-1}p\in P$, and then $\nu^{-1}q=\mu^{-1}p\in P$ gives $\nu\leq q$.

We can now prove Theorem \[(G\*,P\*) is left ql\]. Its proof is based on [@Spielberg2012 Theorem 2.11], and our presentation is helped by Emily Irwin’s treatment of [@Spielberg2012 Theorem 2.11] in her University of Otago Honours thesis.

Fix $x\in P^*P^{*-1}$. We shall prove that there exist $\mu,\nu\in P^*$ with $x=\mu\nu^{-1}$ such that whenever $\gamma\delta^{-1}=x$ we have $\mu\leq \gamma$ and $\nu\leq \delta$ (see ).

Choose $\alpha,\beta\in P^*$ such that $x=\alpha\beta^{-1}$. Choose 