A, t^n x \rangle \le G$. Suppose $[G: \langle A, t^n x \rangle] < \infty$, then $[F: F \cap \langle A, t^n x \rangle] < \infty$. Set $K = F \cap \langle A, t^n x \rangle = \cup_k(i_x \phi^n)^{-k}(A)$. As $[F: K] < \infty$, K is finitely generated and there exists a $k_0 \ge 1$ such that $K = (i_x \phi^n)^{-k_0}(A)$. The statements $[F:K] < \infty$, $K = (i_x \phi^n)^{-k_0}(A)$, and $A$ is a proper free factor of $F$ imply $F = (i_x \phi^n)^{-k_0}(A)$, which contradicts the minimal assumption on $\phi$. Therefore, $[G: \langle A, t^n x \rangle] = \infty$ as needed. This concludes the reverse direction.

For the forward direction, suppose $\phi$ is irreducible and atoroidal and let $H \le G$ be a finitely generated noncyclic group with $\chi(H) = 0$. We need to show $[G: H] < \infty$. Let $p:G=F*_\phi \to \mathbb Z$ be the natural fibration. Note that $\ker p = \cup_i t^i F t^{-i}$. Then $\left.p\right|_H$ is not trivial since $\ker p$ is locally free yet $H$ is finitely generated but not free: it is not cyclic and $\chi(H) = 0$. Also $\left.\ker p\right|_H = H \cap \ker p$ is not trivial as $H \not \cong 