SlowGuess/NeurIPS_2023_OCR huggingface dataset:
in  $G$  still exist in  $G'$ . Therefore,  $D(G'(I))$  can only have more edges, and any potential additional edges in  $D(G'(I))$  must be between  $e$  and some observed node  $X_i$ . There are two cases. (a) Suppose  $X_i \in \mathrm{ch}_X(H_i)$ , then because  $e$  is always in the same clique as  $\mathrm{ch}_X(H_i)$ , the edge  $X_i - e$  must already exist in  $D(G'(I))$ . (b) Suppose  $X_i \in \mathrm{ch}_X(H_j)$  where  $H_i \neq H_j$ . Then  $X_i$  and  $e$  are  $d$ -connected because there is an active path between  $H_i$  and  $H_j$  under intervention target  $I$ . By Lemma D.13, there exists a maximal clique  $C \in \Omega_G^{(I)}$  where  $\mathrm{ch}_X(H_i) \cup \mathrm{ch}_X(H_j) \subseteq C$ . However, we also know that  $\mathrm{ch}_X(H_i) \subseteq \mathrm{ch}_X(H_j) \cup \{e\} \subseteq X' \subseteq C$ . Therefore, the edge  $X_i - e$  must already exist in  $D(G^{(I)})$ . Then,  $\{\Omega_G^{(I)}\}_{I \in \mathcal{I}}$  is the same as  $\{\Omega_{G'}^{(I)}\}_{I \in \mathcal{I}}$ . And by Lemma D.10,  $\{\mathcal{T}_X(G^{(I)})\}_{I \in \mathcal{I}}$  is the same as  $\{\mathcal{T}_X(G'^{(I)})\}_{I \in \mathcal{I}}$ . By Lemma D.17, this is not possible because  $G$  is maximal.
D.1 Replaceable and imaginary subsets
Somewhat unintuitively, it is possible that a maximal valid subset can be contained in another maximal valid subset. The reason is that  $X''$  in Definition 3.2 is not allowed to depend on the clique  $C$ .
Example 7. Consider  $G$  defined in Figure 8. When  $I = \{H_1\}$  or  $I = \emptyset$ , there is one maximal clique  $C = \{X_1, X_2, X_3\} \in \Omega_G^{(I)}$ . When  $I = \{H_2\}$ , then there are two maximal cliques in  $\Omega_G^{(I)}$ :  $\{X_1, X_2\}$  and  $\{X_2, X_3\}$ . Thus,
$$
\Omega = \{\{X _ {1}, X _ {2} \}, \{X _ {2}, X _ {3} \}, \{X _ {1}, X _ {2}, X _ {3} \} \},
$$
and the valid subsets are  $\{X_1\}$ ,  $\{X_2\}$ ,  $\{X_3\}$ ,  $\{X_1, X_2\}$ , and  $\{X_2, X_3\}$ . It is clear that  $\{X_1, X_2\}$  is a maximal valid subset by the definition. But it turns out that  $\{X_2\}$  is also maximal: Even though there are valid sets between  $\{X_2\}$  and each clique in  $\Omega$ , there is not a single valid subset with this property.
Proof.  $(\Rightarrow)$  Suppose there exists a non-replaceable maximal valid subset  $X'$  such that there is no hidden variable  $H_i$  with  $\mathrm{ch}_X(H_i) = X'$ . Then  $X'$  is imaginary, which is a contradiction. Note that  $X'$  cannot be a proper set of any  $\mathrm{ch}_X(H_i)$  either, because it is non-replaceable and all  $\mathrm{ch}_X(H_i)$  are maximal valid subsets by Proposition 4.1.
$(\Leftarrow)$  On the other hand, there exists a hidden variable  $H_{i}$  such that  $\mathrm{ch}_X(H_i) = X'$ . Then it must be non-replaceable. Suppose it is replaceable, then, by definition, there exists another maximal valid subset  $X''$  such that  $X' \subsetneq X''$ . Then by the subset condition (Lemma C.1),  $X''$  must be imaginary. Suppose  $X''$  is replaceable, then there must exist another non-replaceable imaginary subset that contains both  $X'$  and  $X''$ , which is also a contradiction.
There might exist a hidden variable  $H_{i}$  such that  $\mathrm{ch}_X(H_i)$  is replaceable (Example 8).
Example 8 (Non-imaginary set can be replaceable). Consider Figure 9 below. When the intervention target  $I$  is  $\emptyset$ ,  $\{H_4\}$  or  $\{H_1\}$ , then there are two maximal cliques:  $\{X_1, X_2, X_3, X_5\}$  and  $\{X_2, X_3, X_4, X_5\}$ . When the intervention target  $I$  is  $\{H_2\}$ , then there are three maximal cliques:  $\{X_1, X_5\}$ ,  $\{X_4\}$  and  $\{X_2, X_3, X_5\}$ . When the intervention target is  $\{H_3\}$ , then there are three maximal cliques:  $\{X_2, X_4\}$ ,  $\{X_3, X_5\}$  and  $\{X_1, X_2, X_5\}$ . Both  $\{X_2\}$  and  $\{X_2, X_5\}$  are maximal valid subsets. But  $\{X_2, X_5\}$  is imaginary. One might be tempted to add another edge  $H_2 \to X_5$  which would create a clique of  $\{X_2, X_4, X_5\}$  when intervening on  $H_3$ .
Figure 9: In this example,  $\{X_2\}$  is a maximal valid subset and is contained in another maximal valid subset  $\{X_2, X_5\}$  even though  $\{X_2\}$  is the child set of  $H_2$ .
D.2 Single source node
It can be shown that imaginary subsets do not exist if there is only one single hidden source node, i.e. a hidden variable such that  $\mathrm{pa}(H_i) = \emptyset$ . Note that this still allows for arbitrarily many hidden variables  $|H| > 1$ .
For any two observed variables  $X_{a}$  and  $X_{b}$ . Let  $P^{(a,b)}$  be the set of all active paths between  $X_{a}$  and  $X_{b}$  in  $G^{(\emptyset)}$ .
Definition D.1 (Dominated path). An active path  $P$  between  $X_{a}$  and  $X_{b}$  is called a  $(X_{a}, X_{b})$ -dominated path if either of the followings is true:
(1)  $P$  is a common parent path (i.e.,  $P$  is  $X_{a} \gets H_{k} \to X_{b}$  for some hidden variable  $H_{k}$ ) and every directed path to  $\mathrm{so}(P)$  has at least one node  $H_{i} \neq H_{k}$  such that such that  $\{X_{a}, X_{b}\} \subseteq \mathrm{ch}_{X}(H_{i})$ .
(2)  $P$  is not a common parent path and (a) every directed path to  $\mathrm{so}(P)$  has at least one node  $H_{i}$  such that  $\{X_a,X_b\} \subseteq \mathrm{ch}_X(H_i)$  or (b) there exists a hidden node  $H_{j}$  in  $P$  such that  $X_{a}\gets H_{j}\to X_{b}$  is a non-  $(X_{a},X_{b})$  -dominated path (see (1)).
Remark D.1. We consider the empty path from  $H_{i}$  to  $H_{i}$  as a directed path.
Remark D.2. To slightly abuse notation, when  $\mathrm{so}(P)$  only has one element, we also use  $\mathrm{so}(P)$  to mean that element.
Appendix D.2.1 gives some lemmas to characterize dominated paths.
Proof. Suppose there is an imaginary subset  $X'$ . Then it must have at least three elements in it. If  $|X'| = 1$ , then it must be a subset of  $\operatorname{ch}_X(H_i)$  for some  $H_i$ . If  $|X'| = 2$ , then these two nodes do not share the same parent. Thus, there must exist at least two active paths between them in  $G^{(\emptyset)}$  because of Lemma E.4. By definition, these two active paths are non-dominated. But because there is only one single source node in the latent DAG, by Proposition D.5, this is not possible.
For every pair of nodes in  $X'$ , there could only be one non-dominated path between them because the necessary conditions in Proposition D.5 cannot be satisfied when there is only one single source node and Lemma D.1 shows that there is at least one non-dominated path.
By Proposition D.6,  $X^{\prime}$  cannot be imaginary.
D.2.1 Key lemmas and propositions for proving Theorem 4.4
The following two lemmas characterize dominated paths.
Lemma D.1. If  $X_{a}$  and  $X_{b}$  are  $d$ -connected in  $G^{(\emptyset)}$ , then  $P^{(a,b)}$  has at least one non- $(X_{a},X_{b})$ -dominated path.
Proof. Because  $X_{a}$  and  $X_{b}$  are  $d$ -connected in  $G^{(\emptyset)}$ , by definition,  $P^{(a,b)}$  is not empty.
Now let's consider two cases:
(a) There exists at least one hidden node that is a common parent of  $X_{a}$  and  $X_{b}$ . Without loss of generality, let  $H_{1}$  be a common parent of  $X_{a}$  and  $X_{b}$ . Because there is no circle and there is only a finite number of hidden variables, there exists a hidden variable  $H_{2}$  that could be  $H_{1}$  or an ancestor of  $H_{1}$  such that  $\{X_{a}, X_{b}\} \subseteq \mathrm{ch}_{X}(H_{2})$  and  $H_{2}$  does not have an ancestor such that both  $X_{a}$  and  $X_{b}$  are its children. By definition,  $X_{a} \gets H_{2} \to X_{b}$  is non- $(X_{a}, X_{b})$ -dominated.
(b) There is no hidden node that is a common parent of  $X_{a}$  and  $X_{b}$ . Then every path in  $P^{(a,b)}$  is non- $(X_{a}, X_{b})$ -dominated. Because  $P^{(a,b)}$  is not empty,  $P^{(a,b)}$  has at least one non- $(X_{a}, X_{b})$ -dominated path.
Lemma D.2. If  $\{X_a, X_b\}$  is a valid subset and  $P^{a,b}$  has only one non- $(X_a, X_b)$ -dominated path, then the non- $(X_a, X_b)$ -dominated path must be a common parent path.
Proof. Let's consider two cases:
(a) There is no hidden node that is a common parent of  $X_{a}$  and  $X_{b}$ . Because  $\{X_{a}, X_{b}\}$  is a valid subset, there must exist at least two active paths between  $X_{a}$  and  $X_{b}$  in  $G^{\emptyset}$ . Otherwise, by Lemma E.4,  $X_{a}$  and  $X_{b}$  must be  $d$ -separated for an intervention target. By definition, any active paths between  $X_{a}$  and  $X_{b}$  are non- $(X_{a}, X_{b})$ -dominated paths. So, this case is not possible.
(b) There exists at least one hidden node that is a common parent of  $X_{a}$  and  $X_{b}$ . Without loss of generality, let  $H_{1}$  be a common parent of  $X_{a}$  and  $X_{b}$ . If the active path  $X_{a} \gets H_{1} \to X_{b}$  is  $(X_{a}, X_{b})$ -dominated, we can find an ancestor of  $H_{1}$  called  $H_{2}$  such that  $\{X_{a}, X_{b}\} \subseteq \mathrm{ch}_{X}(H_{2})$ . Because there is only a finite number of hidden variables and there is no circle, we can eventually find a common parent  $H'$  of  $X_{a}$  and  $X_{b}$  such that  $X_{a} \gets H' \to X_{b}$  is a non- $(X_{a}, X_{b})$ -dominated path. Therefore at least one of the non- $(X_{a}, X_{b})$ -dominated paths is a common parent path. And if there is only one non- $(X_{a}, X_{b})$ -dominated path, it must be a common parent path.
Lemma D.3. For any two nodes  $X_{a}$  and  $X_{b}$  where  $P^{a,b}$  has only one non- $(X_{a}, X_{b})$ -dominated path  $P_{*}$ , then there does not exist a directed path from  $H_{1}$  to  $H_{2}$  with either  $H_{1} \to X_{a}$ ,  $H_{2} \to X_{b}$  or  $H_{2} \to X_{a}$ ,  $H_{1} \to X_{b}$  such that  $\mathrm{so}(P) \in \mathrm{de}(H_{2})$ .
Proof. By Lemma D.2,  $P_*$  is a common parent path. Let  $H_*$  be the common parent. Now here are two cases:
A.  $H_{1} = H_{2}$ . In this case,  $P_{*}$  would not be a non-dominated path by definition. So this is not possible.
B.  $H_{1} \neq H_{2}$ . Because this path is dominated, then either (a)  $H_{*}$  must be on this path or (b) there exists a common parent of  $X_{a}, X_{b}$ :  $H'$  such that  $H' \in \overline{\mathrm{an}}(H_{1})$ . Case (a) is not possible because  $H_{*} \in \mathrm{de}(H_{2})$ . Case (b) is also not possible by Lemma D.4.
Lemma D.4. For any two nodes  $X_{a}$  and  $X_{b}$  where  $P^{a,b}$  has only one non- $(X_{a},X_{b})$ -dominated path  $P_{*}$ , for any hidden node  $H'$  that is a common parent of  $X_{a},X_{b}$  and not the same as  $\mathrm{so}(P_{*})$ ,  $\mathrm{so}(P_{*})$  is an ancestor of  $H'$ .
Proof. Suppose  $\mathrm{so}(P_{*})$  is not an ancestor of  $H'$ . Then because  $X_{a} \gets H' \gets X_{b}$  is a  $(X_{a}, X_{b})$ -dominated path, and there is only a finite number of hidden variables, we can eventually find a common parent  $H''$  of  $X_{a}$  and  $X_{b}$  such that  $X_{a} \gets H'' \to X_{b}$  is a non-  $(X_{a}, X_{b})$ -dominated path. But  $\mathrm{so}(P_{*})$  is not an ancestor of  $H'$ . So it cannot be  $H''$ . This is a contradiction to the fact that  $P^{a,b}$  has only one non-  $(X_{a}, X_{b})$ -dominated path.
Proposition D.5. If  $\{X_a, X_b\} \subseteq X'$  where  $X'$  is an imaginary subset and  $P^{(a,b)}$  has at least two non- $(X_a, X_b)$ -dominated paths, then the following two conditions hold:
(a) There exists a pair of non- $(X_{a}, X_{b})$ -dominated paths  $(P_{1}, P_{2})$  such that given any intervention target  $I \in \mathcal{I}$ , at least one of  $P_{1}$  and  $P_{2}$  is active in  $G^{(I)}$
(b) For any pair of non- $(X_{a}, X_{b})$ -dominated paths  $(P_{1}, P_{2})$  that satisfies (a), there does not exist a single hidden node  $H_{\#}$  that is shared by all directed path to  $\mathrm{so}(P_{1})$  and  $\mathrm{so}(P_{2})$  in  $G^{(\emptyset)}$ .
Proof. First, let's prove the first condition. We consider two conditions:
(a) There is no hidden node that is a common parent of  $X_{a}$  and  $X_{b}$ . Because  $\{X_{a}, X_{b}\}$  is a valid subset, there must exist at least two active paths between  $X_{a}$  and  $X_{b}$  in  $G^{(\emptyset)}$ . Otherwise, by Lemma E.4,  $X_{a}$  and  $X_{b}$  must be  $d$ -separated for an intervention target. By definition, any active paths between  $X_{a}$  and  $X_{b}$  are non-  $(X_{a}, X_{b})$ -dominated paths.
(b) There exists at least one hidden node that is a common parent of  $X_{a}$  and  $X_{b}$ . Without loss of generality, let  $H_{1}$  be a common parent of  $X_{a}$  and  $X_{b}$ . If the active path  $X_{a} \gets H_{1} \to X_{b}$  is  $(X_{a}, X_{b})$ -dominated, we can find an ancestor of  $H_{1}$  called  $H_{2}$  such that  $\{X_{a}, X_{b}\} \subseteq \mathrm{ch}_{X}(H_{2})$ . Because there is only a finite number of hidden variables and there is no circle, we can eventually find a common parent  $H'$  of  $X_{a}$  and  $X_{b}$  such that  $X_{a} \gets H' \to X_{b}$  is a non- $(X_{a}, X_{b})$ -dominated path. Therefore at least one of the non- $(X_{a}, X_{b})$ -dominated paths is a common parent path. Because  $P^{(a,b)}$  has at least two non- $(X_{a}, X_{b})$ -dominated paths, and given any intervention target, the common parent path will not be destroyed, the first condition is true.
Let's denote  $P_{1}$  to be a non- $(X_{a}, X_{b})$ -dominated active path with hidden variables  $H_{a,1}$  and  $H_{b,1}$  such that  $X_{a} \gets H_{a,1}$  and  $X_{b} \gets H_{b,1}$  are in  $P_{1}$ . Let  $P_{2}$  to be an a non- $(X_{a}, X_{b})$ -dominated active path with hidden variables  $H_{a,2}$  and  $H_{b,2}$  such that  $X_{a} \gets H_{a,2}$  and  $X_{b} \gets H_{b,2}$  are in  $P_{2}$ .
Now let's prove the second condition. Suppose, on the contrary, there exists such a hidden node  $H_{\#}$ . Because  $P_{1}$  and  $P_{2}$  are non- $(X_{a}, X_{b})$ -dominated, there must exist one directed path to  $P_{1}$  and one directed path to  $P_{2}$  such that these two directed paths do not have any hidden node  $H_{i}$  with  $\{X_{a}, X_{b}\} \subseteq \mathrm{ch}_{X}(H_{i})$ , except for maybe  $\mathrm{so}(P_{1})$  and  $\mathrm{so}(P_{2})$ . On the other hand, these two directed paths share at least one common node  $H_{\#}$ . Let  $H_{*}$  be the common nodes with the lowest ranking in topological ordering. Then  $H_{*}$  has the property that, under any intervention target  $I$ , one of the two directed paths  $P_{1,*}$  and  $P_{2,*}$ , where  $P_{1,*}$  is the directed path from  $H_{*}$  to  $\mathrm{so}(P_{1})$  and  $P_{2,*}$  is the directed path from  $H_{*}$  to  $\mathrm{so}(P_{2})$ , still exists. If this is not true, then there is one intervention target  $I_{*}$  such that intervening on this node would break  $P_{1,*}$  and  $P_{2,*}$ . Then  $I_{*}$  would be a common node on  $P_{1,*}$  and  $P_{2,*}$  other than  $H_{*}$ . This violates the assumption  $H_{*}$  has the lowest ranking in the topological ordering. Note that  $\mathrm{so}(P_{1})$  or  $\mathrm{so}(P_{2})$  only has one element in it by Lemma E.3.
We now have four paths  $P_{1}$ ,  $P_{2}$ ,  $P_{1,*}$  and  $P_{2,*}$ . Under any intervention target  $I$ , one of  $P_{1}$ ,  $P_{2}$  is active and one of  $P_{1,*}$  and  $P_{2,*}$  is still active. Let's consider the following three cases:
A.  $P_{1,*}$  has no shared hidden node with  $P_{2}$  and  $P_{2,*}$  has no shared hidden node with  $P_{1}$ .
First of all,  $H_{*}$  cannot be a common parent of  $X_{a}$  and  $X_{b}$ . Suppose  $H_{*}$  is a common parent of  $X_{a}$  and  $X_{b}$ . Then by our construction, it is only possible if  $H_{*} = \mathrm{so}(P_{1})$  and  $H_{*} = \mathrm{so}(P_{2})$ . Because  $P_{1}$  and  $P_{2}$  are both non-dominated paths, they must both be common parent paths in this case. This would imply they are the same path, which is a contradiction.
Note that by Lemma E.3, an active path between hidden nodes  $H_{1}$  and  $H_{2}$  can only be a common ancestor path or a directed path. Either way, there is only one source node of the path and every other node in the path is a descendant of that source node. Therefore,  $P_{1,*}$  cannot have a shared hidden node with  $P_{1}$  except for  $\mathrm{so}(P_1)$ , otherwise there will be a directed circle. Similarly,  $P_{2,*}$  cannot have a shared hidden node with  $P_{2}$  except for  $\mathrm{so}(P_2)$ . For any intervention target  $I$ , there are three cases:
(a) None of  $P_{1}, P_{2}, P_{1,*}$  and  $P_{2,*}$  are affected by  $I$ . Then all of  $H_{a,1}, H_{a,2}, H_{b,1}, H_{b,2}$  are descendants of  $H_{*}$ .
(b) One of  $P_{1,*}$  and  $P_{2,*}$  is destroyed. Without loss of generality, suppose  $P_{2,*}$  no longer exists. The intervention target  $I$  must be a part of the path  $P_{2,*}$  and it cannot be  $H_{*}$ . Because  $P_{2,*}$  has no shared hidden node with  $P_{1}$ , intervening on  $I$  means that  $P_{1}$  is still intact. Therefore,  $H_{a,1}, H_{b,1}$  are descendents of  $H_{*}$ .
(c) One of  $P_{1}$  and  $P_{2}$  is destroyed. Without loss of generality, suppose  $P_{2}$  no longer exists. The intervention target  $I$  must be a part of the path  $P_{2}$  and it cannot be so  $(P_{2})$ . Therefore,  $P_{1,*}$  is still intact.  $H_{a,1}, H_{b,1}$  are descendants of  $H_{*}$ .
Finally, in all three cases, at least one pair of  $(H_{a,1}, H_{b,1})$  and  $(H_{a,2}, H_{b,2})$  is a set of descendants of  $H_*$ . Now consider graph  $G' = (V, E')$  with  $E' = E \cup \{H_* \to X_a\} \cup \{H_* \to X_b\}$ . Note that  $E'$  is not the same as  $E$  because at least one of  $\{H_* \to X_a\}$  and  $\{H_* \to X_b\}$  is missing. Otherwise,  $H_*$  would be a common parent of  $X_a$  and  $X_b$ . By Lemma D.15,  $G'$  violates the maximality.
B. Without loss of generality,  $P_{1,*}$  has shared hidden nodes with  $P_{2}$  and  $P_{1}$  is not a common parent path.
Note that in this case,  $P_{2,*}$  cannot have a shared hidden node with  $P_{1}$  except for one special case. Suppose the opposite is true. Let the shared node between  $P_{1,*}$  and  $P_{2}$  be  $H_{1}$ . Let the shared node between  $P_{2,*}$  and  $P_{1}$  be  $H_{2}$ . There must be a circle between  $H_{1}$  and  $H_{2}$  if  $H_{1} \neq H_{2}$  which is a contradiction. If  $H_{1} = H_{2}$ , then  $H_{1} = H_{2} = \mathrm{so}(P_{1}) = \mathrm{so}(P_{2})$ . In this case, there is always at least one pair of  $(H_{a,1}, H_{b,1})$  and  $(H_{a,2}, H_{b,2})$  that is a set of descendants of  $\mathrm{so}(P_{1})$  and  $\mathrm{so}(P_{2})$  under any intervention target. In this case, adding  $\{X_{a}, X_{b}\}$  to  $\mathrm{ch}_{X}(\mathrm{so}(P_{1}))$  would not change  $\Omega_{G}^{(I)}$  by the proof of Lemma D.15. By Lemma D.17, this is not possible. So we will ignore this special case.
Now let's consider the following cases under any intervention target  $I$  for when  $P_{2,*}$  does not have a shared hidden node with  $P_1$ :
(a) None of  $P_{1}, P_{2}, P_{1,*}$  and  $P_{2,*}$  are affected by  $I$ . Then  $H_{a,1}, H_{b,1}$  are descendants of  $\mathrm{so}(P_1)$ .
(b)  $P_{1,*}$  is destroyed. Then  $P_{1}$  must be intact because the intervention target is in  $P_{1,*}$ .  $H_{a,1}, H_{b,1}$  are descendants of  $\mathrm{so}(P_1)$ .
(c)  $P_{2,*}$  is destroyed. Because  $P_{2,*}$  does not have a shared hidden node with  $P_1, H_{a,1}, H_{b,1}$  are descendants of  $\mathrm{so}(P_1)$ .
(d)  $P_{2}$  is destroyed. Then  $P_{1}$  must be intact. Then  $H_{a,1}, H_{b,1}$  are descendents of  $\mathrm{so}(P_1)$ .
(e)  $P_{1}$  is destroyed. Then  $P_{2}$  must be intact.  $P_{1,*}$  must also be intact because the intervention target is in  $P_{1}$  and it is not  $\mathrm{so}(P_1)$ . Because  $P_{2,*}$  cannot have a shared hidden node with  $P_{1}, P_{2,*}$  is also active. Therefore,  $H_{a,2}, H_{b,2}$  are connected to  $\mathrm{so}(P_1)$  via the common ancestor  $H_{*}$ .
Let's consider graph  $G' = (V, E')$  with  $E' = E \cup \{\mathrm{so}(P_1) \to X_a\} \cup \{\mathrm{so}(P_1) \to X_b\}$ . Note that  $E'$  is not the same as  $E$  because at least one of  $\{\mathrm{so}(P_1) \to X_a\}$  and  $\{\mathrm{so}(P_1) \to X_b\}$  is missing because  $P_1$  is not a common parent path and  $P_1$  is also non- $(X_a, X_b)$ -dominated.
For any intervention target  $I$  that induces  $(a), (b), (c), (d)$ , adding  $\{X_a, X_b\}$  to  $\mathrm{ch}_X(\mathrm{so}(P_1))$  would not change  $\Omega_G^{(I)}$  by the proof of Lemma D.15.
For any intervention target  $I$  that induces  $(e)$ ,  $D(G^{\prime I})$  does not have more edges than  $D(G^{I})$ . If the opposite is true, then the new edge in  $D(G^{\prime I})$  must involve the added node  $X_{a}$  or  $X_{b}$ . Without loss of generality, suppose the new edge is between  $X_{a}$  and  $X_{k}$ , it must be that a parent of  $X_{k}$  is  $d$ -connected to  $\mathrm{so}(P_1)$ . Suppose that the parent of  $X_{k}$  is  $H_{k}$ . By Lemma E.3, there are 2 cases. If  $\mathrm{so}(P_1)$  has a directed path to  $H_{k}$ , then  $H_{k}$  is connected to  $X_{a}$  and  $X_{b}$  via the common ancestor  $H_{*}$ . If  $H_{k}$  has a directed path to  $\mathrm{so}(P_1)$  or  $H_{k}$  and  $\mathrm{so}(P_1)$  share a common ancestor, by our assumption,  $H_{k}$  and  $\{X_{a}, X_{b}\}$  are connected by common ancestor  $H_{\#}$ . Therefore,  $X_{k}$  and  $X_{a}$  are  $d$ -connected in  $D(G^{I})$ .
So adding  $\{X_{a},X_{b}\}$  to  $\mathrm{ch}_X(\mathrm{so}(P_1))$  would not change  $\Omega_G^{(I)}$ .
Finally, by Lemma D.17, this is not possible.
C. Without loss of generality,  $P_{1,*}$  has shared hidden nodes with  $P_{2}$  and  $P_{1}$  is a common parent path.
Let  $H!$  be the shared node between  $P_{1,*}$  and  $P_{2}$  that has the lowest topological ranking. Therefore, for any intervention target  $I$ , at least one of  $P_{2}$  and the directed path from  $H!$  to  $\mathrm{so}(P_1)$  is intact. Otherwise,  $H!$  is not the shared node between  $P_{1,*}$  and  $P_{2}$  that has the lowest topological ranking.
First, let's consider two scenarios.
(1) Both  $P_{1}$  and  $P_{2}$  are common parent paths. First of all,  $\mathrm{so}(P_1)$  is not the same as  $\mathrm{so}(P_2)$ . Otherwise,  $P_{1}$  and  $P_{2}$  would be the same path. On the other hand, if  $\mathrm{so}(P_2)$  appears in  $P_{1,*}$  because there is the only possible node that can be shared and is not  $\mathrm{so}(P_1)$ , it would violate how  $P_{1,*}$  is chosen. So this scenario is not possible.
(2)  $P_{1}$  is a common parent path and  $P_{2}$  is not. First,  $H_{!}$  is not the same as  $\mathrm{so}(P_1)$  because  $H_{!}$  is also on  $P_{2}$  which is a non-  $(X_{a},X_{b})$  -dominated path. By definition, this is not possible. On the other hand,  $H_{!}$  is not a common parent of  $X_{a}$  and  $X_{b}$  because it would violate how  $P_{1,*}$  is chosen. We'll study this scenario next.
Similar to the previous case,  $P_{2,*}$  cannot have a shared hidden node with  $P_1$ . Now let's consider the following cases under any intervention target  $I$ :
(a) None of  $P_{1}, P_{2}, P_{1,*}$  and  $P_{2,*}$  are affected by  $I$ . Then  $H_{a,1}, H_{b,1}$  are descendants of  $H_{1}$ .
(b)  $P_{2,*}$  is destroyed. Because  $P_{2,*}$  does not have a shared hidden node with  $P_1, H_{a,1}, H_{b,1}$  are descendants of  $H_!$ .
(c)  $P_{2}$  is destroyed. Then the directed path from  $H_{!}$  to  $\mathrm{so}(P_1)$  is still intact. Then  $H_{a,1}, H_{b,1}$  are descendants of  $H_{!}$ .
(d)  $P_{1,*}$  is destroyed but the directed path from  $H!$  to  $\mathrm{so}(P_1)$  is still intact. Then  $P_{1}$  must be intact because the intervention target is in  $P_{1,*}$ .  $H_{a,1}, H_{b,1}$  are descendants of  $H!$ .
(e)  $P_{1}$  is destroyed or the directed path from  $H_{!}$  to  $\mathrm{so}(P_1)$  is destroyed. In either case,  $P_{2}$  must be intact. The directed path from  $H_{*}$  to  $H_{!}$  must also be intact. Because  $P_{2,*}$  cannot have a shared hidden node with  $P_{1}$  or  $P_{1,*}$ ,  $P_{2,*}$  is also active. Therefore,  $H_{a,2}, H_{b,2}$  are connected to  $H_{!}$  via the common ancestor  $H_{*}$ .
Now consider graph  $G' = (V, E')$  with  $E' = E \cup \{H_! \to X_a\} \cup \{H_! \to X_b\}$ . Note that  $E'$  is not the same as  $E$  because at least one of  $\{H_! \to X_a\}$  and  $\{H_! \to X_b\}$  is missing. By the same argument as fore,  $G'$  also violates the maximality assumption by Lemma D.17.