comparatively straightforward and we deal with it in . Then in we tackle the core of the argument, namely the case when $l_b(g)=0$.

**Case 2.3**As described below, the standard method for dealing with the case when no generator has exponent sum zero in $r$ reduces this case to Case 2.2.

We have at least three generators $a,b,c$ where $A=\{a\}, b\in B, C=\{c\}$; suppose that $r$ has exponent sum $\alpha \neq 0$ in $a$ and exponent sum $\beta \neq 0$ in $b$. Introduce new generators $x,y$ setting $b=y^\alpha$ and $a=xy^{-\beta}$ so that we have embedded $G$ in the amalgamated free product $\hat G = \langle G*F(y) \ |\ b=y^\alpha \rangle$ and then replaced $a$ by $x=ay^{\beta}$. The resulting relator $\hat r \equiv r(xy^{-\beta},y^{\alpha}, \ldots , c)$ has exponent sum zero in $y$ and our equalities become $g\hat h(X,Y)g^{-1}=\hat k(Y,C)$ and $g\hat h'(X,Y)g^{-1}=\hat k'(Y,C)$ where $X=\{x\}, Y=\{y,B'\}, C=\{c\}$ and $B = \{b,B'\}$.

Now Case 2.2 applies and gives the conclusion that either $gF(X,Y)g^{-1}\cap F(Y,C)$ is cyclic or $g \in F(Y,C)F(X,Y)$. Clearly $F(A,B) \subseteq F(X,Y)$ and $F(B,C) \subseteq F(Y,C)$, so that $gF(A,B)g^{-1}\cap F(B,C) \subseteq gF(X,Y)g^{-1}\cap F(Y,C)$. By our counterpair assumption, $gF(X,Y)g^{-1}\cap F(Y,C)$ cannot be cyclic and therefore $g \in F(Y,C)F(X,Y)$. But $G \cap 