SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Proof.  $X_{a}$  and  $X_{b}$  do not share the same parent. Therefore, there must exist at least two hidden nodes  $H_{a,1}$  and  $H_{b,1}$  such that  $X_{a} \in H_{a,1}$  and  $X_{b} \in H_{b,1}$ . However, if there are only two such hidden nodes, Lemma E.4 suggests there exists an intervention target  $I$  that makes  $X_{a}$  and  $X_{b}$ $d$ -separated and thus  $X_{a}$  and  $X_{b}$ $d$ -separated. This is a violation of the fact that  $\{X_{a}, X_{b}\}$  is a valid subset.
Lemma D.15. For a hidden variable  $H_{i}$ , there does not exist an observed node  $X_{i}$  such that  $X_{i} \notin \mathrm{ch}_{X}(H_{i})$  but under any intervention target  $I \in \mathcal{I}$ ,  $X_{i}$  is a descendant of  $H_{i}$ .
Proof. This violates maximality.
Consider  $G' = (V, E')$  with  $E' = E \cup \{H_i \to X_i\}$ . For any intervention target  $I$ ,  $D(G'(I))$  does not have more edges than  $D(G^I)$ . If the opposite is true, then the new edge in  $D(G'(I))$  must involve  $X_i$ . Suppose the new edge is between  $X_i$  and  $X_j$ , it must be that a parent of  $X_j$  is  $d$ -connected to  $H_i$ . However, because  $X_i$  is a descendant of  $H_i$ , by Lemma E.3,  $X_i$  must be  $d$ -connected to  $X_j$  in  $D(G^I)$ . Therefore, 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, the existence of  $G'$  violates the maximality condition.
Lemma D.16.  $\{\mathrm{ch}_X(H_i)\}_{i = 1}^m$  is a complete cover.
Proof. By Assumption 1(a) and Assumption 1(b), two observed variables  $X_{i}$  and  $X_{j}$  are dependent if and only if  $X_{i}$  and  $X_{j}$  are  $d$ -connected. Suppose  $\{\mathrm{ch}_X(H_i)\}_{i=1}^m$  is not a complete cover. Then for some intervention target  $I \in \mathcal{I}$ , there exists one clique  $C$  in  $D(G^{(I)})$  that is not shattered and that clique contains one edge that is not covered by other shattered cliques in  $D(G^{(I)})$ . Suppose that edge is between  $X_{1}$  and  $X_{2}$ , then there exist  $H_{1}$  and  $H_{2}$  with  $X_{1} \in \mathrm{ch}_{X}(H_{1})$  and  $X_{2} \in \mathrm{ch}_{X}(H_{2})$  and  $H_{1}$  is  $d$ -connected to  $H_{2}$ . By Lemma D.13, this is not possible.
Lemma D.17. If  $G = (X \cup H, E)$  is a maximal measurement model, then there does not exist a measurement model  $G' = (X \cup H, E')$  with  $E' = E \cup \{H_i \to X_i\}$  for some hidden variable  $H_i$  and observed variable  $X_i$  such that for any intervention target  $I \in \mathcal{I}$ , where  $\mathcal{I}$  is the complete family of intervention targets,
$$
\mathcal {T} _ {X} (G ^ {(I)}) = \mathcal {T} _ {X} (G ^ {\prime (I)})
$$
Proof. By the definition of the maximal measurement model, we just need to show that  $G'$  also satisfies Assumption 1. If  $G'$  satisfies Assumption 1, then the existence of  $G'$  would mean that  $G$  is not a maximal measurement model. Before delving into the graphical assumptions, one first notices that because for any intervention target  $I \in \mathcal{I}$ ,  $\mathcal{T}_X(G^{(I)}) = \mathcal{T}_X(G^{\prime (I)})$ , then  $X_{i}$  must always be  $d$ -connected to  $\mathrm{ch}_X(H_i)$  in both  $G^{(I)}$  and  $G^{\prime (I)}$  under any intervention target.
For Assumption 1(a), if a distribution  $P$  is Markov to  $G$ , then it must be Markov to  $G'$ . This is because  $G'$  has one more edge than  $G$ , any active path in  $G$  would remain active in  $G'$  as well. Thus  $\mathcal{T}(G') \subseteq \mathcal{T}(G) \subseteq \mathcal{T}(P)$ . A similar argument can be made for the interventional case.
For Assumption 1(b), we know that for any intervention target  $I$ ,  $X_{i} \perp \mathbb{1}_{P^{(I)}}X_{j} \Rightarrow \mathrm{d - sep}_{G^{(I)}}(\{X_{i}\} \{X_{j}\} | \emptyset)$ . Because  $\mathcal{T}_X(G^{(I)}) = \mathcal{T}_X(G^{\prime (I)})$ , Assumption 1(b) is true for  $G^{\prime}$  as well.
For Assumption 1(c), because  $\mathcal{T}_X(G^{(I)}) = \mathcal{T}_X(G^{\prime (I)})$  for any intervention target and  $G^{\prime},G$  only differs in biparite graph. If Assumption 1(c) is satisfied by  $G$ , it must be satisfied by  $G^{\prime}$  as well.
Therefore,  $G'$  satisfies Assumption 1 and  $G$  is not maximal, which is a contradiction.
E Learning the skeleton of latent structures
Once we have learned the bipartite graph  $G_{B}$ , the next step is to learn the DAG  $G_{H}$  over the latent variables  $H$ . This turns out to be straightforward: Assumption 1(c) suggests that two hidden variables  $H_{i}$  and  $H_{j}$  are  $d$ -separated if and only if  $\mathrm{ch}_X(H_i)$  and  $\mathrm{ch}_X(H_j)$  are in different cliques (Lemma E.2). Therefore, the idea is to learn causal graphs using unconditional  $d$ -separations of latent variables under interventions.
Define  $\mathcal{M}_H$  as follows:
$$
\mathcal {M} _ {H} (G) := \left(\left\{\langle A, B \rangle : \mathrm {d - s e p} _ {G} (A B \mid \emptyset) \text {f o r d i s j o n t s u b s e t s} A, B \subseteq H \right\}\right).
$$
Note that  $\mathcal{M}_H(G) \subseteq \mathcal{T}_H(G)$  because  $\mathcal{M}_H(G)$  only stores unconditional  $d$ -separations. Therefore, to learn the latent DAG, we would like to answer the following question:
$$
G i v e n \{\mathcal {M} _ {H} (G ^ {(I)})) \} _ {I \in \mathcal {I}}, c a n w e r e c o r e c e R G _ {H}?
$$
Remark E.1. This is harder than the fully observational case where we have access to all conditional  $d$ -separations.
The answer to the previous question is affirmative as shown by the next theorem (see Appendix E.2).
Theorem E.1. Under Assumptions 1-2 and suppose the bipartite graph  $G_B$  is correct, the skeleton of  $G_H$  is identifiable.
E.1 Identifying unconditional  $d$ -separations of the latents
Given the bipartite graph  $G_B$  and  $\{\mathcal{T}_X(P^{(I)})\}_{I \in \mathcal{I}}$ , we can easily construct  $\{\mathcal{M}_H(G^{(I)}))\}_{I \in \mathcal{I}}$  by the following lemma.
Lemma E.2. Given intervention target  $I$ , two hidden variables  $H_{i}$  and  $H_{j}$  are  $d$ -separated in  $G^{(I)}$  if and only if there does not exist a clique  $C \in \Omega_P^{(I)}$  such that  $\mathrm{ch}_X(H_i) \cup \mathrm{ch}_X(H_j) \subseteq C$ .
Proof. By Assumption 1(a) and Assumption 1(b), two observed variables  $X_{i}$  and  $X_{j}$  are dependent if and only if  $X_{i}$  and  $X_{j}$  are  $d$ -connected.
If  $H_{i}$  and  $H_{j}$  are  $d$ -separated in  $G^{(I)}$ , then, by Assumption 1(c), there exists  $X_{i}, X_{j} \in X$  where  $X_{i} \in \operatorname{ch}_{X}(H_{i}), X_{j} \in \operatorname{ch}_{X}(H_{j})$ , such that  $X_{i}$  and  $X_{j}$  are independent. Therefore,  $\operatorname{ch}_X(H_i) \cup \operatorname{ch}_X(H_j)$  is not a clique.
On the other hand, if there exist a clique  $C \in \Omega_P^{(I)}$  such that  $\mathrm{ch}_X(H_i) \cup \mathrm{ch}_X(H_j) \subseteq C$ . This means that  $\mathrm{ch}_X(H_i) \cup \mathrm{ch}_X(H_j)$  is a clique. Then  $H_i$  and  $H_j$  must be  $d$ -connected, which is a contradiction.
Now we have Algorithm 1 to find the skeleton.
Algorithm 1 Learning Skeleton of  $G_{H}$    
Input:  $\{\mathcal{M}_H(G^{(I)}))\}_{I\in \mathcal{I}}$ $E\gets \emptyset$  // Set of unoriented edges   
/\* Step 0 \*/   
if  $\{\mathcal{M}_H(G^{(I)})\}_{I\in \mathcal{I}}$  only has one distinct element then Output:E   
\*/Step 1: Remove any pair of hidden variables that appears twice \*/   
for  $\mathcal{M}_H(G^{(I)}))\in \{\mathcal{M}_H(G^{(I)})\}_{I\in \mathcal{I}}$  do for  $\langle H_1,H_2\rangle \in \mathcal{M}_H(G^{(I)})$  do If  $\langle H_1,H_2\rangle$  appears twice with at least two different intervention targets, delete  $\langle H_1,H_2\rangle$  from  $\{\mathcal{M}_H(G^{(I)}))\}_{I\in \mathcal{I}}$    
\*/Step 2:Add unoriented edges \*/   
for  $\mathcal{M}_H(G^{(I)})\in \{\mathcal{M}_H(G^{(I)})\}_{I\in \mathcal{I}}$  do for  $\langle H_1,H_2\rangle \in \mathcal{M}_H(G^{(I)})$  do E E U {H1H2}   
Output:E
Proof. The proof relies on the correctness of Algorithm 1 which we show here.
First, suppose  $\{\mathcal{M}_H(G^{(I)}))\}_{I\in \mathcal{I}}$  only has one distinct element. Then there is no edge between latent variables in  $G_{H}$ . Suppose, on the contrary, there is an edge  $H_{1}\rightarrow H_{2}$  between two hidden variables  $H_{1}$  and  $H_{2}$  in  $G_{H}$ , then by Lemma E.2,  $\mathrm{ch}_X(H_1)\cup \mathrm{ch}_X(H_2)$  must be a clique in  $D(G^{(\emptyset)})$ . But by Lemma E.3, when the intervention target  $I$  is  $H_{2}$ ,  $H_{1}$  and  $H_{2}$  would be  $d$ -separated, and by Lemma E.2,  $\mathrm{ch}_X(H_1)\cup \mathrm{ch}_X(H_2)$  must not be a clique in  $D(G^{(I)})$ . Therefore, there must be at least two distinct elements in  $\{\mathcal{M}_H(G^{(I)}))\}_{I\in \mathcal{I}}$ .
Now, let's consider the case after step 0. Let's denote the unoriented edge set returned by Algorithm 1 as  $E_{1}$  and the true unoriented edge set of  $G_{H}$  as  $E_{2}$ .
A.  $E_2 \subseteq E_1$ . By Lemma E.3,  $\{\mathcal{M}_H(G^{(I)}))\}_{I \in \mathcal{I}}$  has all pairs of hidden variables. And we removed all the pair that does not have an edge in  $E_2$  by Lemma E.5.
B.  $E_{1} \subseteq E_{2}$ . Suppose  $\exists \langle H_{1}, H_{2} \rangle \in E_{1}$  such that  $\langle H_{1}, H_{2} \rangle \notin E_{2}$ . Note that  $\langle H_{1}, H_{2} \rangle$  only appears with one intervention target and by Lemma E.5, this is not possible.
E.3 Useful Lemmas
Lemma E.3. Given a DAG  $G = (V, E)$ . If two nodes  $A, B \in V$  are  $d$ -connected by the empty set, then at least one of the following must be true
A. They share a common ancestor (common ancestor path)
B. There are directed paths between the two nodes and all the directed paths must have the same direction (directed path)
Proof. Because  $A, B$  are d-connected, there must exist active paths between the two nodes. There are four possibilities of active path
A.  $A \rightarrow \ldots \rightarrow B$ . This could be a direct path if every edge points to the same direction. If one of the intermediate edges points in the opposite direction. Then we would have a collider which makes the path inactive.
B.  $A\gets \dots \leftarrow B$  . The same as the first case.
C.  $A \to \ldots \leftarrow B$ . This is not possible because there must exist a collider on the path.
D.  $A \leftarrow \ldots \rightarrow B$ .  $A$  and  $B$  must share a common ancestor in this path unless there is a collider which is not possible.
Obviously, we cannot have two directed paths pointing in opposite directions because that would create a circle.
Lemma E.4. For any two hidden variables  $H_{1}$  and  $H_{2}$ , there exists at least one intervention target  $I \in \mathcal{I}$ , such that  $\langle H_1, H_2 \rangle \in \mathcal{M}_H(G^{(I)})$ .
Proof. If  $H_{1}$  and  $H_{2}$  are  $d$ -separated in  $G^{(\emptyset)}$ , then we can just choose  $I = \emptyset$ .
If  $H_{1}$  and  $H_{2}$  are d-connected in  $G^{(\emptyset)}$  and assume, without loss of generality, the directed paths between  $H_{1}$  and  $H_{2}$ , if exist, point to  $H_{2}$ , then based on Lemma E.3, if we intervene on  $H_{2}$ , all the active paths between  $H_{1}$  and  $H_{2}$  would disappear.
Lemma E.5. If  $\{\mathcal{M}_H(G^{(I)})\}_{I\in \mathcal{I}}$  has at least two distinct elements, then for any two hidden variables  $H_{1}$  and  $H_{2}$ , there is no direct edge between  $H_{1}$  and  $H_{2}$  if and only if there exists at least two intervention targets  $I_{1}, I_{2} \in \mathcal{I}$ , such that  $\langle H_1, H_2 \rangle \in \mathcal{M}_H(G^{(I_1)})$  and  $\langle H_1, H_2 \rangle \in \mathcal{M}_H(G^{(I_2)})$ .
Proof. If there is no direct edge between  $H_{1}$  and  $H_{2}$  and there exists only one intervention target  $I_{1} \in \mathcal{I}$ , such that  $\langle H_{1}, H_{2} \rangle \in \mathcal{M}_{H}(G^{(I_{1})})$ , then  $I_{1} \neq \emptyset$ . There is because if  $I_{1} = \emptyset$  and  $\langle H_{1}, H_{2} \rangle \in \mathcal{M}_{H}(G^{(\emptyset)})$ . Then there are no active paths between  $H_{1}$  and  $H_{2}$  and deleting edges via interventions would not create active paths. Because  $\{\mathcal{M}_{H}(G^{(I)})\}_{I \in \mathcal{I}}$  has at least two distinct elements, there must exist another intervention target such that  $\langle H_{1}, H_{2} \rangle$  appears. By the proof of Lemma E.6, one of  $H_{1}$  and  $H_{2}$  is the intervention target  $I_{1}$ . Without loss of generality, suppose  $I_{1} = \{H_{1}\}$ , we can add  $H_{1} \gets H_{2}$  to the graph and it would violate the maximality condition. Thus, if there is no direct edge between  $H_{1}$  and  $H_{2}$ , then there exists at least two intervention targets  $I_{1}, I_{2} \in \mathcal{I}$ , such that  $\langle H_{1}, H_{2} \rangle \in \mathcal{M}_{H}(G^{(I_{1})})$  and  $\langle H_{1}, H_{2} \rangle \in \mathcal{M}_{H}(G^{(I_{2})})$ .
If there exists at least two intervention targets  $I_1, I_2 \in \mathcal{I}$ , such that  $\langle H_1, H_2 \rangle \in \mathcal{M}_H(G^{(I_1)})$  and  $\langle H_1, H_2 \rangle \in \mathcal{M}_H(G^{(I_2)})$ , then suppose there is a direct edge between  $H_1$  and  $H_2$ . Without loss of generality, let's assume  $H_1 \to H_2$ . Then two variables would only be  $d$ -separated when intervening on  $H_2$  which is a contradiction.
Lemma E.6. In Algorithm 1, after step 1, for any intervention target  $I$  such that  $\mathcal{M}_H(G^{(I)}) \neq \mathcal{M}_H(G^{(\emptyset)})$ , then there exist a hidden variable  $H_1$  such that  $H_1 \in \langle H_i, H_j \rangle$  for all  $\langle H_i, H_j \rangle \in \mathcal{M}_H(G^{(I)})$  and  $H_1$  is the intervention target  $I$ .
Proof. We just need to show that if  $\langle H_i, H_j \rangle \in \mathcal{M}_H(G^{(I)})$  and  $\langle H_i, H_j \rangle \notin \mathcal{M}_H(G^{(I')})$  for  $I' \neq I$ , then one of  $H_i$  or  $H_j$  must be the intervention target  $I$ . This is because, after step 1, every pair of hidden variables only appear once.
Suppose the opposite is true, then neither  $H_{i}$  nor  $H_{j}$  is being intervened on. So there is no direct edge between  $H_{i}$  and  $H_{j}$  since otherwise  $H_{i}$  and  $H_{j}$  would be  $d$ -connected. On the other hand,  $\langle H_{i}, H_{j} \rangle \notin \mathcal{M}_{H}(G^{(I')})$  for  $I' \neq I$ , implies that  $H_{i}$  and  $H_{j}$  are  $d$ -connected in the observational distribution because  $\mathcal{M}_{H}(G^{(I)}) \neq \mathcal{M}_{H}(G^{(\emptyset)})$ . Then by Lemma E.3, we can find another intervention target  $H_{i}$  or  $H_{j}$  such that  $\langle H_{i}, H_{j} \rangle$  would appear again which would be a contradiction.
F Limitations of edge orientations
Unlike known interventions, with only access to CI information, edge orientation might not always be possible even when there are no latent variables as shown in Example 1. But it is also possible as demonstrated by Example 13. This raises the question of which edges provably cannot be oriented.
In this section, we attempt to answer this question by studying equivalence relations between DAGs under unknown interventions. Such equivalence relations are finer partitions of the Markov equivalence class. The results are purely graphical and can be studied in their own right. If assuming Markov and faithfulness, it also shows the limitations of edge orientations with access to CI information only. In addition, one could use conditional invariances to improve the identifiability of edge orientations. But it might require additional assumptions like direct  $\mathcal{L}$ -faithfulness[58] which we do not make in this paper.
Example 13. Consider the CPDAG  $G$  shown in Figure 14. Let  $G_{1}$  be a consistent extension of  $G$  such that  $X_{1} \to X_{2}$  and  $I_{1} = X_{2}$ . Then  $\{(X_{1}, X_{2}, \emptyset), (X_{3}, X_{2}, \emptyset)\} \subseteq \mathcal{T}(G_{1}^{(I_{1})})$ .  $G_{2}$  is another consistent extension of  $G$  where  $X_{2} \to X_{1}$  but there does not exist an intervention target that can induce  $\mathcal{T}(G_{1}^{(I_{1})})$ . Therefore, if we observe  $\mathcal{T}(G_{1}^{(I_{1})})$ , then we know that  $X_{1} \to X_{2}$ .
Figure 14: CPDAG of three variables with no latent variables.
In this section, we consider the general case of an arbitrary, fully observed DAG (i.e.  $H = \emptyset$ ) instead of the measurement model. Given the skeleton of the DAG, all the compelled edges in the MEC can already be identified without access to interventional distributions. Therefore, without loss of generality, we assume the CPDAG is given and study if unknown hard interventions on single nodes can help orient reversible edges. Furthermore, because isolated edges are covered, they must be reversible in the MEC as well (cf. Lemma G.1), there is no loss of generality in focusing on reversible edges.
To state the problem formally:
Given a CPDAG  $G$ , we want to orient its reversible edges given a list  $\{\mathcal{T}(G_{*}^{(I)})\}_{I \in \mathcal{I}}$  for a family of intervention targets  $\mathcal{I}$  and  $G_{*}$ , which is the true consistent extension of  $G$ . In other words, is the tuple  $\langle G_{*}, \mathcal{I} \rangle$  the unique one that can induce  $\{\mathcal{T}(G_{*}^{(I)})\}_{I \in \mathcal{I}}$ ?
Let's start with some definitions. Recall the definition of covered edges[12]:
In particular, an isolated edge is a special covered edge.
Remark F.1. Isolated edges are trivially covered.
F.1 Isolated equivalence class
Chickering [12] show that two Markov equivalent DAGs can be transformed into one another by a sequence of covered edge reversals. The isolated equivalence class is also defined in a transformational fashion. We restate the definition below.
Definition F.1. (Isolated equivalence class) Two DAGS  $G_{1}$  and  $G_{2}$  are isolated equivalent, denoted  $G_{1} \sim_{E} G_{2}$ , if there exists a sequence of isolated edge reversals.
The next two theorems show that it is impossible to distinguish DAGs in an IEC by looking at  $d$ -separations only.
Theorem F.2. Let  $G_{1}$  and  $G_{2}$  be two consistent extensions of CPDAG  $G = (V, E)$ . Suppose
1.  $G_{1}$  and  $G_{2}$  only differs in one isolated edge
2. All the interventional targets are single nodes and the intervention is hard.
Suppose the tuple  $(G_{1},\mathcal{I}_{1})$  induces  $\{\mathcal{T}(G_1^{(I)})\}_{I\in \mathcal{I}_1}$
Then there exists a family of interventional targets  $\mathcal{I}_2$  such that the tuple  $(G_2, \mathcal{I}_2)$  also induces  $\{\mathcal{T}(G_1^{(I)})\}_{I \in \mathcal{I}_1}$ .
Proof. Because  $G_{1}$  and  $G_{2}$  are in the same IEC, then, by definition, there exists a sequence of isolated edge reversals. Then we can prove by applying Theorem F.2 repeatedly.
F.1.1 Proof of Theorem F.2
Definition F.3. (Augmented Active Subgraph) Let  $G = (V, E)$  be a DAG and let path  $P: V_1 \to \ldots \to V_k$  be an active path given  $C$  where  $C \subseteq V$ . Suppose  $H_i$  is a collider, then there must exist at least a node  $V_d$  in  $\overline{\mathrm{de}}(V_i)$  where  $V_d \in C$ . We define an active descendant path of  $V_i$  to be a directed path from  $V_i$  to  $V_d$ . An augmented active subgraph of  $P$  is a subgraph including the active path  $P$  and all the active descendant paths of colliders in the original active path.
Remark F.2. Our definition also considers the special case where  $V_{i} = V_{d}$  and the directed path is just  $V_{i}\rightarrow V_{i}$ .
Lemma F.4. Let  $G = (V, E)$  be a DAG. Let  $I$  be a single node hard intervention target, then,
$$
\mathcal {T} (G) \subseteq \mathcal {T} (G ^ {I})
$$
Proof. Doing a hard intervention on any node  $V_{i} \in V$  removes all incoming edges to  $H_{i}$ .
Suppose the statement is not true, then there exists  $(A,B,C)\in \mathcal{T}(G)$  such that  $(A,B,C)\notin \mathcal{T}(G^{I})$ . This means that intervening on  $V_{i}$  creates an active path between  $A,B$  given  $C$ . But deleting edges would not create a new path. The only way an active path can emerge is if a previously blocked path becomes unblocked after an intervention.
A path is blocked if either:
1. any non-collider is being conditioned on.
2. no descendants of collider (including itself) is being conditioned on
For the first condition, we know that deleting edges would not make non-colliders colliders. On the other hand, deleting edges would also not make non-descendants descendants (intervening on the collider itself would destroy the path). Therefore, hard intervening on a single node would not create new active paths.
Lemma F.5. Let  $G$  be a DAG. Suppose there is an active path  $P$  in  $G$ , and the hard intervention target  $I$  is not on the augmented active subgraph of the active path. Then this path stays active.
Proof. Because the intervened node is not on the path nor on the directed path going out from one of the colliders on the active path. Deleting its incoming edges would not change these paths.  $\square$
Lemma F.6. Let  $G_{1}$  and  $G_{2}$  be two DAGs that share the same augmented active subgraph of active path  $P$  given conditioning set  $C$ . If there is a single node hard intervention on one of the nodes of that augmented active subgraph in both  $G_{1}$  and  $G_{2}$ , then either the path is active in both  $G_{1}$  and  $G_{2}$  or it is not active in either  $G_{1}$  or  $G_{2}$ .
Proof. Intervening on any node on the active path would destroy (if the intervened node has incoming edges) or sustain (if the intervened node has no incoming edges) the path in both  $G_{1}$  and  $G_{2}$ .
Suppose we intervene on a descendant of a collider on the active path that is not the collider itself. Any active descendant path that has the intervened node on it would get destroyed. The path would stay active if there exists an alternative active descendant path that does not involve the intervened node, which is shared between  $G_{1}$  and  $G_{2}$  because they share the same augmented active subgraph. Note that intervening on a node would not create a new descendant path.
Proof of Theorem F.2. For notation, let's suppose the isolated edge is  $V_{i} \to V_{j}$  in  $G_{1}$ .
To prove this, we just need to show that for any  $I_{i}\in \mathcal{I}_{1}$ , there exists  $I_{j}$  such that
$$
\mathcal {T} \left(G _ {1} ^ {\left(I _ {i}\right)}\right) = \mathcal {T} \left(G _ {2} ^ {\left(I _ {j}\right)}\right)
$$
In particular, we will show that we can choose  $I_{j} = \{I_{i}\}$  when  $I_{i} \neq \{V_{i}\}$  and  $I_{j} = \{V_{j}\}$  when  $I_{i} = \{V_{i}\}$ .
By the definition of Markov equivalence, we know that
$$
\mathcal {T} (G _ {1}) = \mathcal {T} (G _ {2})
$$
First of all, note that an augmented active subgraph given conditioning set  $C$  in  $G_{1}$  must exist and be active given conditioning set  $C$  in  $G_{2}$  and vice versa. We know that two DAGs in the same Markov equivalence class share the same skeleton and v-structures. Therefore, for any augmented active subgraph given conditioning set  $C$  in  $G_{1}$ , there is a subgraph with the same skeleton in  $G_{2}$ . If the edges of the augmented active subgraph in  $G_{1}$  have the same orientations in  $G_{2}$ , then it is active in  $G_{2}$ . If some edges of the augmented active subgraph in  $G_{1}$  have different orientations than that of  $G_{2}$ . Then the subgraph can only differ in one isolated edge. Because the endpoints of isolated edge cannot have other parents, the isolated edge cannot be in the descendant paths. The isolated edge can be in the active path itself, with edges going out from endpoints of isolated edge. When that edge is reversed, the path is still active.
To show that  $\mathcal{T}(G_1^{(I_i)}) = \mathcal{T}(G_2^{(I_j)})$ , we first need to show that for any  $(A,B,C) \in \mathcal{T}(G_1^{(I_i)}), (A,B,C) \in \mathcal{T}(G_2^{(I_j)})$ . By Lemma F.4,  $\mathcal{T}(G_1) \subseteq \mathcal{T}(G_1^{(I_i)})$  and  $\mathcal{T}(G_2) \subseteq \mathcal{T}(G_2^{(I_j)})$ . We can only consider tuple  $(A,B,C)$  where  $(A,B,C) \notin \mathcal{T}(G_1)$ .
If  $(A,B,C) \notin \mathcal{T}(G_1)$ , but  $(A,B,C) \in \mathcal{T}(G_1^{(I_i)})$ , then all the active paths between  $A$  and  $B$  given  $C$  must be blocked or deleted after intervention on  $I_i$ . Let's consider the following two cases: