are no counterpairs satisfying $l_b(g)=0$.

\[Claim 4.1\]There do not exist counterpairs $ghg^{-1}=k, gh'g^{-1}=k'$, with $l_b(g)=0$, such that min$\{l_b(h),l_b(h')\} = 0$.

Suppose not; clearly there is no loss of generality in assuming $l_b(h) =l_b(k) =0$. Among all such counterpairs, we choose one with $l_b(h') = l_b(k')$ minimal. If $l_b(h') = l_b(k')=0$, then $h,h' \in F(A^*,B^*), k, k' \in F(B^*,C^*)$ and we have an immediate contradiction to the overall induction hypothesis.

So we can assume that $l_b(h') = l_b(k')>0$. We can write $$gh'g^{-1} = gh'_0b^{\varepsilon_1}h'_1 \ldots b^{\varepsilon_n}h'_ng^{-1} = k'_0b^{\varepsilon_1}k'_1 \ldots b^{\varepsilon_n}k'_n =k'$$ and there is no loss of generality in assuming that $\varepsilon_1 =1$. Adjusting the equality so that it is standardised form, we obtain $g=k'_0z_0{h'_0}^{-1}$, where $z_0 \in U$ is nontrivial of type $(A^*:C^*)$. Substituting for $g$ and replacing the original $h,k,h'$ and $k'$ by the resulting conjugates, we can rewrite, adjusting our notation, the two equalities in the form $$z_0 h z_0^{-1} = k, \quad z_0h'z_0^{-1} \equiv z_0bh'_1b^{\varepsilon_2} \ldots b^{\varepsilon_n}h'_nz_0^{-1} = bk'_1b^{\varepsilon_2} \ldots b^{\varepsilon_n}k'_n \equiv k'.$$ Since $z_0 \neq 1$ and $k \notin F(B^*,C^*_+)$, because $z_0 h z_0^{-1} = k$ cannot hold in $F(A^*,B^*,C^*_+)$, the equality $h = z_0^{-1} kz_0$ must define an exceptional element of the intersection $F(A^*,B^*) \cap 