$${\overleftarrow {\tilde v}} h_1 = k_1\tilde v.$$ This does not hold freely and so (c) applies, giving $${\overleftarrow {\tilde v}} \equiv \tilde v$$ which is impossible. More generally, pick the least $i$ such that ${\overleftarrow {z_{i-1}}} h_i = k_iz_i$, does not hold freely, with the appropriate interpretation for $i=1$ or $i=n$. Then $${\overleftarrow {\tilde v}} \equiv z_1, \quad{\overleftarrow {z_1}} \equiv z_2,\quad \ldots \quad{\overleftarrow {z_{i-2}}} \equiv z_{i-1}$$ and so $L(\overleftarrow{z_{i-1}}) = L(\tilde v)$. Again (c) applies giving ${\overleftarrow {z_{i-1}}} \equiv \tilde v$ and we have another impossible situation. This is the contradiction we require to conclude the proof of .

Unfortunately this is the point at which the argument becomes even more complicated.

Maintaining our notation $ghg^{-1} =k, gh'g^{-1}=k'$ for counterpairs, we shall write $\rho_b(h), \rho_b(h')$ for the number of times that $b$ changes sign in reduced expressions for $h, h'$. We essentially argue by induction on $\rho_b(h)+\rho_b(h')$.

\[Claim 4.2\] There do not exist counterpairs $ghg^{-1}=k, gh'g^{-1}=k'$ satisfying $l_b(g)=0$ and min$\{l_b(h),l_b(h')\} > 0$ such that $\rho_b(h) +\rho_b(h')= 0$.

Suppose not; then, without loss of generality, there exists a counterpair $$\begin{aligned}
ghg^{-1} &= gh_0bh_1\ldots h_{m-1}bh_mg^{-1} = k_0bk_1 \ldots  k_{m-1}bk_m  = k \\ gh'g^{-1} &= gh'_0bh'_1\ldots h'_{n-1}bh'_ng^{-1} = k'_0bk'_1 \ldots  k'_{n-1}bk'_n  = k.\tag*{\hbox{and}}\end{aligned}$$ 