one of $p_1,p_2$ nontrivial – for otherwise $F(A^*,B^*) \cap F(B^*,C^*)$ would be exceptional, and then, by Proposition 5.2 of [@C], $\tilde v$ would be trivial.

Now the hypotheses of Proposition 5.5 of [@C] are satisfied and therefore each extremal generator appears in just a single syllable of $u^{-1}\tilde v^{-1}v_0\tilde v$. Furthermore $p_2p_0^{-1}p_1q$ is a cyclic rearrangement of $u^{-1}\tilde v^{-1}v_0\tilde v$. This means that $\tilde v$ is *intermediate* (ie omits all four *extremal* generators $\{a_{\kappa}, a_{\lambda}, c_{\mu}, c_{\nu}\}$ – see page 273 of [@C]) and that $p_1 \equiv p_2 \equiv \tilde v$. It follows that ${\overleftarrow {z_{j-1}}} \equiv \tilde v \equiv w_j$ and hence that $$\begin{aligned}
 \tilde v h'_0bh'_1 \ldots bh'_j &=& k'_0bk'_1 \ldots bk'_j \tilde v \\  \tilde vb^{-1}h'_{j+1}b^{\varepsilon_{j+1}} \ldots b{\varepsilon_n}h'_n &=& b^{-1}k'_{j+1}b^{\varepsilon_{j+1}} \ldots b^{\varepsilon_n}k'_n \end{aligned}$$ yielding, of course, $$\begin{aligned}
\tilde v h \tilde v^{-1} &=& k \\
  \tilde vh'_0bh'_1 \ldots bh'_j\tilde v^{-1} &=& k'_0bk'_1 \ldots bk'_j \\ \tilde vb^{-1}h'_{j+1}b^{\varepsilon_{j+1}} \ldots b^{\varepsilon_n}h'_n\tilde v^{-1} &=& b^{-1}k'_{j+1}b^{\varepsilon_{j+1}} \ldots b^{\varepsilon_n}k'_n. \end{aligned}$$ Since $h'$ is the product of $h'_0bh'_1 \ldots bh'_j$ and $b^{-1}h'_{j+1}b^{\varepsilon_{j+1}} \ldots b^{\varepsilon_n}h'_n$ and each has $b$–length less than $h'$, it follows that $h$ commutes with both and therefore with $h'$, which is a contradiction.

In completing the proof of , we 