F(A^*_+,B^*,C^*)$. By Proposition 5.1 of [@C], the basic exceptional relation is either of the form $u=v_0v_2$, with $z_0 \equiv v_2$ or $u=\tilde v^{-1}v_0\tilde v$ with $z_0 \equiv \tilde v$. The former implies that $z_0 = v_0^{-1}u \in F(B,C)F(A,B)$, which we can rule out since $z_0$ is nontrivial of type $(A^*:C^*)$, and so the latter must hold. So we now have a counterpair of the form $$\tilde v h \tilde v^{-1} =k, \quad \tilde v h'\tilde v^{-1} = \tilde vbh'_1b^{\varepsilon_2} \ldots b^{\varepsilon_n}h'_n\tilde v^{-1} = bk'_1b^{\varepsilon_2} \ldots b^{\varepsilon_n}k'_n =k'.$$

$h'$ has uniform signature pattern.

Suppose not; since the initial occurrence of $b$ in $h'$ has exponent $+1$, in the system of equalities yielded by the normal form theorem, we find, for some $j$ (which can be chosen minimal), $z_{j-1} \in U, w_j \in L$ and ${\overleftarrow {z_{j-1}}} h'_j = k'_jw_j$ so that ${\overleftarrow {z_{j-1}}} h'_jw_j^{-1} = k'_j$ defines an element of $F(A^*,B^*,C^*_-) \cap F(B^*,C^*)$ which must be exceptional since $z_{j-1} \neq 1$. Hence we can write $F(A^*,B^*,C^*_-) \cap F(B^*,C^*) = \langle p \rangle * F(B^*,C^*_-) = \langle q \rangle * F(B^*,C^*_-) $ where $p \in F(A^*,B^*,C^*_-)$ and $q \in  F(B^*,C^*)$. Corollary 5.4 of [@C] implies that $p = p_1p_0p_2^{-1}$, with at least 