F(Y,C)F(X,Y) = F(B,C)F(A,B)$ and our counterpair assumption rules this out and we have the required contradiction.

This completes the logical structure of the proof of but of course it remains to deal with Case 2.2.

Case 2.2: the inductive step when $l_b(g) >0$ {#sec3}
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As noted above, Case 2.2 is dealt with by a subsidiary induction on $l_b(g)$. In this section we deal with the inductive step of this subsidiary induction and hence reduce Case 2.2 to the initial step of the subsidiary induction when $l_b(g) =0$.

Our standpoint here is, therefore, that we have an overall inductive hypothesis which asserts that the theorem holds for relators of shorter length and, arguing by contradiction, we are assuming that there exist counterpairs $ghg^{-1}=k$ and $gh'g^{-1}=k'$. For the purposes of the subsidiary induction we know there is a counterpair where $l_b(g)$ is minimal but strictly positive.

In any such counterpair $l_b(h) = l_b(k)= l_b(h')= l_b(k')=0$.

Suppose not; without loss of generality, we may assume that $l_b(h)>0$. Let us write $g = \tilde gb^{\varepsilon_m}g_m$ where $m=l_b(g)$ and $l_b(\tilde g)=m-1$. Still without loss of generality we also assume that $\varepsilon_m = -1$. Then we can write $$ghg^{-1}= \tilde gb^{-1}g_mh_0b^{\zeta_1}h_1 \ldots b^{\zeta_l}h_lg_m^{-1}b\tilde g^{-1} = 