can thus assume that $h'= bh'_1b\ldots bh'_n$ and $k'= bk'_1b\ldots bk'_n$. However it should be noted that, unlike in the proof of the Subclaim, we do not have any information about $F(A^*,B^*,C^*_{-}) \cap F(B^*,C^*)$ and so we do not at present know that $\tilde v$ is intermediate.

We obtain the usual system of equalities $${\overleftarrow {\tilde v}} h'_1 = k'_1z_1, \ldots , {\overleftarrow {z_{n-1}}} h'_n = k'_n\tilde v.$$ If all these equalities hold freely, then we obtain $$h'_i=k'_i, i=1,2, \ldots,n \quad \hbox{and}
\quad {\overleftarrow {\tilde v}} \equiv z_1, {\overleftarrow {z_1}} \equiv z_2,\ldots , {\overleftarrow {z_{n-1}}} \equiv \tilde v$$ which is clearly impossible. It follows, therefore that some equality does not hold freely and we have to analyse the sequence $${\overleftarrow {\tilde v}} h'_1 = k'_1z_1, {\overleftarrow {z_1}} h'_2 = k'_2z_2, \ldots {\overleftarrow {z_{n-1}}} h'_n = k'_n\tilde v.$$ To complete the argument for Case 4.1 we require two further results which are similar in nature to Proposition 6.2 of [@C]. These are stated and proved in and will also be used below.

Since one of the inequalities does not hold freely, it follows from below that $\tilde v$ must be intermediate and hence below can be applied. If $n=1$ we have 