I am currently studying the book of Graham Higman and Elizabeth Scott, The Existentially Closed Groups, London Mathematical Society Monographs New Series, Clarendon Press Oxford, 1988. In the Section 2, page 14, it gives a proof of the following theorem Let $ G $ be any group, and let $ B_1 , ... , B_k $ be any subgroups of $ G $. Let $ L $ be the repeated HNN-extension $$ \langle G, s_1 , ... , s_k \vert (s_i)^{-1} b s_i = b , \forall b \in B_i, i=1,...,k \rangle $$ of $G$. Then the subgroup $K \subseteq L $ generated by $G$ and $t = (s_1 \cdots s_k)^2$ is an HNN-extension of $G$ with $t$ centralizing $ \cap_{1 \leq i \leq k} B = B$. So $$ K = \langle G, t \vert t^{-1}bt = b , \forall b \in B \rangle . $$
Here, in the proof, we define an HNN-extension $ X = \langle G, d \vert d^{-1}bd = b \forall b \in B \rangle $, then we define a homomorphism $ \theta: X \to K $ by $ d \mapsto t $ and $ g \mapsto g $ for all $ g \in G $, and clearly this homomorphism is an epimorphism. In order to show that $ \theta $ is injective, we consider an arbitrary element of $ X $, say $ w= g_0 d^{\varepsilon_1} g_1 \cdots g_{n-1} d^{\varepsilon_n} g_n $ where $ \varepsilon_i = \pm 1 $ and if $ \varepsilon_i = -\varepsilon_{i+1} $, then $ g_i \notin B $ (since elements of $ X $ can be written in this way uniquely as a normal form), then if $ (w)\theta = 1 $, we want to show $ w = 1 $. Then we assume that $ n \geq 1 $ to lead to a contradiction. My problem starts here. The authors say that for $ 1 \leq i \leq n $, if $ \varepsilon_i = -\varepsilon_{i+1} = 1 $, then $ g_i \notin B $, then we can choose $ i_1 \leq k $ such that
$$ g_i \in B_k \cap B_{k-1} \cap \cdots \cap B_{i_1 +1} , g_i \notin B_{i_1} .$$
I dont get this part of the proof. Could not it be that $ g_i \in G - (B_1 \cup \cdots \cup B_k) $? Is there something that prevents this?