k.$$ Now two subcases arise depending on whether or not (the detailed expression for) $ghg^{-1}$ is or is not reduced. If the latter occurs, then either $\zeta_1=1$ and $g_mh_0 =z \in U$ or $\zeta_l=-1$ and $h_lg_m^{-1} =z \in U$. It suffices to assume the first occurs. Substituting for $g_m$ we obtain $$\tilde gb^{-1}zh_0^{-1}hh_0z^{-1}b\tilde g^{-1} = k,
\tilde gb^{-1}zh_0^{-1}h'h_0z^{-1}b\tilde g^{-1}  = k'$$ and hence $$\tilde g\overleftarrow z(b^{-1}h_0^{-1}hh_0b){\overleftarrow z}^{-1}b\tilde g^{-1} = k, \tilde g\overleftarrow z(b^{-1}h_0^{-1}h'h_0b){\overleftarrow z}^{-1}\tilde g^{-1} = k'.$$ By the minimality of $l_b(g)$, we deduce that conjugates of $h$ and $h'$ commute, which of course is a contradiction, or that $\tilde g\overleftarrow z \in F(B,C)F(A,B)$. However, if the latter holds then $g = \tilde gb^{-1}g_m = \tilde gb^{-1}zh_0^{-1} = \tilde g\overleftarrow z b^{-1}h_0^{-1} \in F(B,C)F(A,B)$ and here too we have the necessary contradiction.

To complete the proof of the Reduction we have to see what happens when our expression for $ghg^{-1}$ is reduced. Then of course $l_b(k) >0$, say $k = k_0b^{\xi_1}k_1 \ldots b^{\xi_n}k_n$ (where $n= 2m+l$) and $h= g^{-1}kg$, with $g^{-1}kg$ not reduced. But now we can argue exactly as we did when $ghg^{-1}$ was not reduced.

To complete the inductive step in the subsidiary induction, it remains only to 