-  $Y^{*} = \{y\}$ , that is,  $Y^{*}$  is a single vertex. From Lemma A.8,  $W^{*}$  is a district in  $\mathcal{H}[W^{*}]$ , and  $\text{Anc}_{\mathcal{H}[W^{*}]}(y) = W^{*}$ . By definition of hedge, the vertices  $W^{*}$  form a hedge for  $Q[y]$  in  $\mathcal{H}$ . Note that  $y \in Y^{mcip}$ , and from Lemma A.2 it is a district of  $\mathcal{H}[Y^{mcip}]$ . As a result,  $Q[Y^{mcip}]$  is not identifiable in  $\mathcal{H}$ .
-  $|Y^{*}| \geq 2$ . Let  $y_{i}$  and  $y_{j}$  be the first and the last vertices of  $Y^{*}$  in the topological order. Define  $Y^{ij*} = \{y_k^{ij}|y_k \in Y^*\} \cup \{y_{kl}^{ij}|y_k, y_l \in Y^*\}$ .  $Y^{ij*}$  are the vertices in  $V_{bot}^{\mathcal{H}}$  with superscript  $ij$  corresponding to the vertices in  $Y^{*}$ . Note that  $y_{i}^{ij}, y_{j}^{ij} \in Y^{ij*}$ , since  $y_{i}, y_{j} \in Y^{*}$ . Since  $y_{j}^{ij} \in Y^{mcip}$  and from Lemma A.2  $y_{j}^{ij}$  is a district in  $\mathcal{H}[Y^{mcip}]$ , it suffices to show that there is a hedge formed for  $y_{j}^{ij}$  in  $\mathcal{H}$ . We show that the vertices  $W = W^{*} \cup Y^{ij*}$  form a hedge for  $y_{j}^{ij}$  in  $\mathcal{H}$ . From Lemma A.8,  $\text{Anc}_{\mathcal{H}[W^{*}]}(Y^{*}) = W^{*}$ , that is, all of the vertices in  $W^{*}$  are ancestors of  $Y^{*}$  in  $\mathcal{H}[W^{*}]$ , and therefore in  $\mathcal{H}[W]$ . Also, the vertices  $y_{kl}^{ij}$  in  $Y^{ij*}$  have a direct edge to their corresponding vertex in  $W^{*}$ , i.e.,  $y_{kl}^{b}$ , and therefore are ancestors of  $Y^{*}$  in  $\mathcal{H}[W]$  as well. Further, each vertex in  $Y^{*}$  such as  $y_{k}$  is a parent of  $y_{k}^{ij}$ , which is in turn a parent of  $y_{i}^{ij}$  (or is  $y_{i}^{ij}$  itself if  $k = i$ ). Finally,  $y_{i}^{ij}$  has a directed edge to  $y_{j}^{ij}$  by construction. As a result, all of the vertices  $W$  have a direct path to  $y_{j}^{ij}$  in  $\mathcal{H}[W]$ . That is,  $\text{Anc}_{\mathcal{H}[W]}(y_{j}^{ij}) = W$ . It now remains to show that  $W$  is a district in  $\mathcal{H}[W]$ . From Lemma A.8,  $W^{*}$  is a district in  $\mathcal{H}[W^{*}]$ . As a result, the vertices  $W^{*}$  are connected through bidirected edges in  $\mathcal{H}[W]$ . There is a bidirected edge between  $y_{j}$  and  $y_{i}^{ij}$  by construction of  $\mathcal{H}$ . It suffices to show that for any  $v \in Y^{ij*}$ , there exists a bidirected path between  $v$  and  $y_{i}^{ij}$  in  $\mathcal{H}[W]$ . A vertex  $y_{kl}^{ij} \in Y^{ij*}$  (with double subscript, which are due to the bidirected edges among  $Y^{*}$ ) has bidirected edges to  $y_{k}^{ij}$  and  $y_{l}^{ij}$ , which are both in  $Y^{ij*}$  by definition. Now take an arbitrary vertex  $y_{k}^{ij} \in Y^{ij*}$  (with single subscript, due to vertices in  $Y^{*}$ ). We know  $y_{k} \in Y^{*}$ , as  $y_{k}^{ij} \in Y^{ij*}$ , by definition of  $Y^{ij*}.Y^{*}$  is a district in  $G[Y^{*}]$ . That is, there exists a bidirected path from  $y_{k}$  to  $y_{i}$  in  $G[Y^{*}]$ . From Lemma A.8 by taking  $W = Y^{*}$ , there is a bidirected path  $p$  from  $y_{k}$  to  $y_{i}$  in  $H[Y^{*}] \cup\{y_{lm}|y_l,y_m\in Y^*\}$ . By construction of  $\mathcal{H}$ , if we replace each vertex  $v$  on  $p$  by  $v^{ij}$ , we achieve a bidirected path  $p'$  with vertices in  $Y^{ij*}$  from  $y_{k}^{ij'}$  to  $y_{i}^{ij'}$ , which completes the proof.