the first $2k+2$ terms of $\alpha$ in normal form $$p_0tp_1t\ldots p_{k-1}tr_k$$ where $p_i\in L_A$ for $0\leq i<k$ and $r_k\in P$. There exists $p_k\in L_A$ such that $r_kA=p_kA$ and $p_k\leq r_k$. As above, we can write $r_k=p_ka$ for some $a\in A\cap P$. Then $$\begin{aligned}
      \alpha&=p_0tp_1t\ldots tr_ktq_{k+1}\\
      &=p_0tp_1t\ldots tp_katq_{k+1}\\
      &=p_0tp_1t\ldots tp_kt\phi(a)q_{k+1}.
    \end{aligned}$$ We set $p_{k+1}=\phi(a)q_{k+1}$, which is in $P$ because $\phi(a)$ is. Then $\alpha=p_0tp_1t\ldots tp_ktp_{k+1}$ has form . By induction, every $\alpha$ has normal form . This implies (2).

For (2) $\Rightarrow$ (1) we argue by contradiction: we will assume (2) holds but (1) doesn’t. Let $X$ be a set of coset representatives satisfying (2), and suppose that there exists some coset $gA$ such that $gA\cap P\neq\emptyset$ which has no minimal coset representative. Let $p\in X$ be the coset representative of $gA$. Since $p$ is not minimal, there exists $q\in gA\cap P$ with $p\not\leq q$. Thus $p^{-1}q\not\in P$. Consider $qt\in P^*$ in normal form: $$\label{qt in normal form}
qt=pp^{-1}qt=pt\phi(p^{-1}q).$$ Since $\phi(A\cap P)=B\cap P$ we have $\phi(p^{-1}q)\not \in P$. So is not the normal form , a contradiction. Thus $(2)\Rightarrow (1)$.

\[PP-1-closed is vee closed\] Let $(G,P)$ be a quasi-lattice ordered group and let $B$ be a subgroup of $G$. Suppose that for 