show for the case at hand that we cannot have a counterpair when $l_b(h) = l_b(k)= l_b(h')= l_b(k')=0$. We write $g = \tilde gb^{\varepsilon_m}g_m$ as above, and can take $\varepsilon_m =-1$. Then we obtain $\smash{\tilde gb^{-1}g_mhg_m^{-1}b\tilde g^{-1} = k \ , \   \tilde gb^{-1}g_mh'g_m^{-1}b\tilde g^{-1}=k'}$ and hence $g_mhg_m^{-1}= z \in U, \quad g_mh'g_m^{-1}= z' \in U.$ Now these equalities define elements of the intersection $g_mF(A^*,B^*)g_m^{-1} \cap F(A^*_+,B^*,C^*_+)$, which involves Magnus subgroups of a group with a shorter relator. (Note that we cannot have all of $A^*_+,B^*,C^*_+$ empty since clearly $h,h' \neq 1$.) Hence either $h$ and $h'$ commute or $g_m \in F(A^*_+,B^*,C^*_+)F(A^*,B^*)$, say $g_m =z_0h_0$. The former is contradictory and so from the latter we obtain $\smash{z_0h_0hh_0^{-1}z_0^{-1} = z}$ and $\smash{z_0h_0h'h_0^{-1}z_0^{-1} = z'}$ whence $\smash{h_0hh_0^{-1} \in U, h_0h'h_0^{-1} \in U}$. Let us write $\smash{x= h_0hh_0^{-1}}$ and $\smash{x'= h_0h'h_0^{-1}}$. Then we have $\smash{\tilde gb^{-1}z_0xz_0^{-1}b\tilde g^{-1}}{=} k$ and $\smash{s \tilde gb^{-1}z_0x'z_0^{-1}b\tilde g^{-1}{=}k'}$. This yields $\tilde g{\overleftarrow {z_0}}\overleftarrow x\smash{{\overleftarrow {z_0}}}^{\!-1}\tilde g^{-1} {=} k$ and $\tilde gb^{-1}{\overleftarrow {z_0}}\overleftarrow {x\smash{'}\strut}\smash{{\overleftarrow {z_0}}}^{\!-1}b\tilde g^{-1}=k'.$ By the minimality of $l_b(g)$, we deduce that conjugates of $h$ and $\smash{h'}$ commute (either $k$ and $k'$ or $\overleftarrow x$ and $\smash{\overleftarrow {x\smash{'}\strut}}$) or $\smash{\tilde g{\overleftarrow {z_0}}} \in F(B,C)F(A,B)$. In the latter 