a complete set $X$ of left coset representatives of $G/A$ that contains $$L_A:=\{p\in P: q\in pA\cap P\Rightarrow p\leq q\}.$$ By Lemma \[right lower bound gives unique normal form in P\] we can write $\alpha$ and $\beta$ in unique normal form: $$\alpha=p_0tp_1t\ldots tp_mtr \text{ where $p_i\in L_A$ and $r\in P$};$$ $$\beta=q_0tq_1t\ldots tq_nts \text{ where $q_i\in L_A$ and $s\in P$}.$$ Now $x=\alpha\beta^{-1}$ is equal to $$\alpha\beta^{-1}=p_0tp_1t\ldots tp_mtrs^{-1}t^{-1}q_n^{-1}t^{-1}\ldots t^{-1}q_0^{-1}.$$ First we look for initial cancellations in the middle of $\alpha\beta^{-1}$: if $rs^{-1}\in B$, then we can replace $trs^{-1}t^{-1}$ with $\phi^{-1}(rs^{-1})$. By assumption (3), Lemma \[PP-1-closed is vee closed\] applies and there exist $b_1,b_2\in P\cap B$ such that $rs^{-1}=b_1b_2^{-1}$. Then $$\phi^{-1}(rs^{-1})=\phi^{-1}(b_1b_2^{-1})=\phi^{-1}(b_1)\phi^{-1}(b_2)^{-1}.$$ Since $\phi(A\cap P)=B\cap P$ we have $\phi^{-1}(b_1)\phi^{-1}(b_2)^{-1}\in PP^{-1}$. Then $$\label{alpha beta inverse}
x=\alpha\beta^{-1}=p_0tp_1t\ldots tp_m\phi^{-1}(b_1)\phi^{-1}(b_2)^{-1}q_n^{-1}t^{-1}\ldots t^{-1}q_0^{-1}.$$ We can repeat this process until there are no more cancellations available in the middle, and so we assume this is the case for the expression . This gives the following cases:

1.  there are no $t$ and no more $t^{-1}$,

2.  there are no more $t^{-1}$,

3.  there are no more $t$,

4.  there are $t$ and $t^{-1}$, and then the term with $t$ to the left and $t^{-1}$ to its right is not in $B$.

In 