changing direction in $c$. Let $\gamma\subset D$ be a path from $a$ to $b$. Composing these paths we obtain a simple cycle $\gamma"$ through $c$ consisting of subpaths $\gamma_1'\subseteq \gamma_1$ of $\gamma'\subseteq\gamma$, respectively, from some $a'$ to $b'$. In particular $\gamma_1$ is not a flag. We may assume inductively that $\gamma_1'\cap D=\{a',b'\}$. If $a'\in V_j$, then possibly after exchanging the roles of $a',b'$, there is some $x\in\gamma\cap (V_{j-1}\cup V_j)$ closest to $b'$. Then $a',x$ are connected by an $E_j$-path entirely contained in $D$ and $(x,\ldots b')$ is a flag. Since the $E_j$-path changes direction in every vertex, we also find an $E_{j-1}$-path inside $D$ to the next vertex of the flag $(x,\ldots b')$. Since $\gamma_1'$ doesn’t meet $D$ except in $a',b'$, we see that ${\gamma'_1}^\smallfrown (b',\ldots, x)$ is a reduced path. Now Lemma \[l:nicepath\] implies $c\in D$, a contradiction. Hence $c\notin\operatorname{acl}(ab)$.

Note that the proof shows in fact the following:

\[r:flagcl\]A vertex $x\neq a,b$ is in $\operatorname{acl}(ab)$ if and only if there is a reduced path from $a$ to $b$ that changes direction in $x$. Hence $\operatorname{acl}(ab)=\{a,b\}$ if and only if $a,b$ is a flag or $a$ and $b$ are not connected. In fact, we have $\operatorname{dcl}(ab)=\operatorname{acl}(ab)$.

\[l:treeacl\] If $g\in\operatorname{acl}(A)$, 