\in F(B_0,c_0)$ where $B_0= \{b_0 \ | \ b \in B\}$. Among all counterpairs, we choose one with $m=l_b(g)$ minimal. If $l_b(g) = 0$, then both equalities hold in the base group $G^*$ and hence we can only have $g \in F(B_0,c_0)F(B^*)$. But then clearly $g \in F(B,C)F(A,B)$ and we have reached a contradiction, as we wish.

Suppose, then that $l_a(g)>0$. Choosing $\varepsilon_m =-1$, just for definiteness, we obtain $g_mhg_m^{-1} = z \in F(B^*,C^*_+)$ and $g_mh'g_m^{-1} = z' \in F(B^*,C^*_+)$. By the induction hypothesis on $|r|$, we can only have $g_m \in F(B^*,C^*_+)F(B^*) =F(B^*,C^*_+)$. Then $$ghg^{-1}= g_0a^{\varepsilon_1}g_1\ldots a^{\varepsilon_{m-1}}g_{m-1}\overleftarrow {g_m}\overleftarrow h\overleftarrow {g_m}^{-1}g_{m-1}^{-1}a^{-\varepsilon_{m-1}}\ldots g_1^{-1}a^{-\varepsilon_1}g_0^{-1} =k$$ and similarly for $h'$ and $k'$. Since $\overleftarrow h$ and $\overleftarrow {h'}$ are conjugates of $h$ and $h'$, it follows from the minimality of our choice of $m$ that the only conclusion we can have is that $g_0a^{\varepsilon_1}g_1\ldots a^{\varepsilon_{m-1}}g_{m-1}\overleftarrow {g_m}\in F(B,C)F(A,B)$ and from this in turn it follows that $g\in F(B,C)F(A,B)$, which is the required contradiction.

**Case 2.2**To deal with this case we employ techniques similar to those of §5 of [@C] and follow pages 272–273 of [@C] in our notation and terminology. Thus we express $G = \langle X : r=1 \rangle$ as an HNN-extension of 