SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Therefore, the variables  $a$  and  $t$  could be such that (69) is different from (70). We note that similar steps can be used to show that (66) is not equivalent to (10). In conclusion, Assumption 2 is crucial for the formulae in (9) and (10) to hold.
First, we state the following d-separation criterion used to prove Lemma 2(b) and Lemma 2(d). See Appendix F.1 for a proof.
Lemma 3. Suppose Assumptions 1 to 3 hold. Then,  $t \perp_{d} \mathbf{b} | \mathbf{z}^{(i)}$  in  $\mathcal{G}_{\underline{t}}$ .
Now, we prove each part of Lemma 2 one-by-one.
Proof of Lemma 2(a) In  $\mathcal{G}_{\mathbf{b}}^{t}$ , all edges going into  $\mathbf{b}$  are removed. Under Assumption 3, this implies that all edges going out of  $t$  are removed. Now, consider any path  $\mathcal{P}(F_t,y)$  between  $F_{t}$  and  $y$  in  $\mathcal{G}_{\mathbf{b}}^{t}$ . This path takes one of the following two forms:  $(a) F_{t} \longrightarrow t \longleftrightarrow \dots y$  or  $(b) F_{t} \longrightarrow t \longleftrightarrow \dots y$ . In either case, there is a collider at  $t$  in  $\mathcal{P}(F_t,y)$ . This collider is blocked when  $\mathbf{z}$  and  $\mathbf{b}$  are conditioned on because  $t \notin \mathbf{z}$ ,  $t \notin \mathbf{b}$ , and  $t$  does not have any descendants in  $\mathcal{G}_{\mathbf{b}}^{t}$ . Therefore,  $y \perp_{d} F_{t} | \mathbf{z}, \mathbf{b}$  in  $\mathcal{G}_{\mathbf{b}}^{t}$ . Similarly, the collider is blocked when  $\mathbf{z}^{(i)}$  and  $\mathbf{b}$  are conditioned on because  $t \notin \mathbf{z}^{(i)}$ ,  $t \notin \mathbf{b}$ , and  $t$  does not have any descendants in  $\mathcal{G}_{\mathbf{b}}^{t}$ . Therefore,  $y \perp_{d} F_{t} | \mathbf{z}^{(i)}$ ,  $\mathbf{b}$  in  $\mathcal{G}_{\mathbf{b}}^{t}$ .
Proof of Lemma 2(b) We prove this by contradiction. Assume there exists at least one unblocked path between  $t$  and some  $b \in \mathbf{b}$  in  $\mathcal{G}_{\underline{t}}$ . Let  $\mathcal{P}(t, b)$  denote any such unblocked path.
Suppose we condition on  $\mathbf{z}^{(i)}$ . From Lemma 3,  $\mathcal{P}(t,b)$  is blocked in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{z}^{(i)}$  is conditioned on. Let  $\nu$  be any node at which  $\mathcal{P}(t,b)$  is blocked in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{z}^{(i)}$  is conditioned on. We must have that  $\nu \in \mathcal{P}(t,b) \setminus \{t,b\}$  and  $\nu \in \mathbf{z}^{(i)}$ . Then, the path  $\mathcal{P}(t,\nu) \subset \mathcal{P}(t,b)$  is unblocked in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{z}^{(i)}$  is unconditioned on. However, this contradicts  $t \perp_{d} \mathbf{z}^{(i)}$  in  $\mathcal{G}_{\underline{t}}$  (which follows from (8)(i) and Fact 1).
Proof of Lemma 2(c) We prove this by contradiction. Assume there exists at least one unblocked path between  $t$  and some  $z^{(i)} \in \mathbf{z}^{(i)}$  in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{b}$  is conditioned on. Let  $\mathcal{P}(t, z^{(i)})$  denote any such unblocked path.
Suppose, we uncondition on  $\mathbf{b}$ . From (8)(i) and Fact 1, we have  $t \perp_{d} \mathbf{z}^{(i)}$  in  $\mathcal{G}_{\underline{t}}$ . Therefore,  $\mathcal{P}(t, z^{(i)})$  is blocked in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{b}$  is unconditioned on. Now, we create a set  $\mathbf{v}$  consisting of all the nodes at which  $\mathcal{P}(t, z^{(i)})$  is blocked in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{b}$  is unconditioned on. Define the set  $\mathbf{v}$  such that for any  $\nu \in \mathbf{v}$ , the following are true:  $(a)\nu \in \mathcal{P}(t, z^{(i)}) \setminus \{t, z^{(i)}\}$ ,  $(b)\mathcal{P}(t, z^{(i)})$  contains a collider at  $\nu$  in  $\mathcal{G}_{\underline{t}}$ , and  $(c)$  there exists an unblocked descendant path from  $\nu$  to some  $b \in \mathbf{b}$  in  $\mathcal{G}_{\underline{t}}$ .
Now, we must have  $\mathbf{v} \neq \emptyset$ , since  $\mathcal{P}(t,z^{(i)})$  is blocked in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{b}$  is unconditioned on. Let  $\nu_{c} \in \mathbf{v}$  be that node which is closest to  $t$  in the path  $\mathcal{P}(t,z^{(i)})$ , and let  $\mathcal{P}(\nu_c,b)$  be an unblocked descendant path from  $\nu$  to some  $b \in \mathbf{b}$  in  $\mathcal{G}_{\underline{t}}$  (there must be one from the definition of the set  $\mathbf{v}$ ). Consider the path  $\mathcal{P}(t,b)$  obtained after concatenating  $\mathcal{P}(t,\nu_c) \subset \mathcal{P}(t,z^{(i)})$  and  $\mathcal{P}(\nu_c,b)$ . By the definition of  $\mathbf{v}$  and the choice of  $\nu_{c}, \mathcal{P}(t,b)$  is unblocked in  $\mathcal{G}_{\underline{t}}$  since  $(a)\mathcal{P}(t,\nu_c)$  is unblocked in  $\mathcal{G}_{\underline{t}}, (b)\mathcal{P}(\nu_c,b)$  is unblocked in  $\mathcal{G}_{\underline{t}}$ , and  $(c)$  there is no collider at  $\nu_{c}$  in  $\mathcal{P}(t,b)$ . However, this contradicts  $t \perp d_{\mathbf{b}}$  in  $\mathcal{G}_{\underline{t}}$  (which follows from Lemma 2(b)).
Proof of Lemma 2(d) We prove this by contradiction. Assume there exists at least one unblocked path between  $t$  and  $F_{\mathbf{b}}$  in  $\mathcal{G}^{\mathbf{b}}$  when  $\mathbf{z}^{(i)}$  is conditioned on. Let  $\mathcal{P}(t, F_{\mathbf{b}})$  denote the shortest of these unblocked path. By definition of  $\mathcal{G}^{\mathbf{b}}$ , this path has to be of the form:  $t \cdots, b \gets F_{\mathbf{b}}$  for some  $b \in \mathbf{b}$ . Now, we have the following three cases:
(i)  $\mathcal{P}(t,F_{\mathbf{b}})$  contains  $t\longrightarrow b$  : In this case, because a path is a sequence of distinct nodes,  $\mathcal{P}(t,F_{\mathbf{b}})$  has to be  $t\longrightarrow b\gets F_{\mathbf{b}}$  . By assumption,  $\mathcal{P}(t,F_{\mathbf{b}})$  is unblocked when  $\mathbf{z}^{(i)}$  is conditioned on. Since there is a collider at  $b$  in  $\mathcal{P}(t,F_{\mathbf{b}})$  , there exists at least one unblocked descendant path from  $b$  to  $\mathbf{z}^{(i)}$  when  $\mathbf{z}^{(i)}$  is conditioned on. Let  $\mathcal{P}(b,z^{(i)})$  denote the shortest of these paths from  $b$  to some  $z^{(i)}\in \mathbf{z}^{(i)}$  in  $\mathcal{G}^{\mathbf{b}}$  . We note that this path also exists in  $\mathcal{G}$  and is of the form  $b\longrightarrow \dots \longrightarrow z^{(i)}$
Suppose we uncondition on  $\mathbf{z}^{(i)}$ . Consider the path  $\mathcal{P}(t, z^{(i)}) \supset \mathcal{P}(b, z^{(i)})$  between  $t$  and  $z^{(i)}$  of the form  $t \longrightarrow b \longrightarrow \dots \longrightarrow z^{(i)}$  in  $\mathcal{G}$ . This path remains unblocked even when  $\mathbf{z}^{(i)}$  is unconditioned on as it does not have any colliders. This contradicts  $z^{(i)} \perp_{d} t$  (which follows from (8)).
(ii)  $\mathcal{P}(t,F_{\mathbf{b}})$  contains  $t\longrightarrow b_1$  for some  $b_{1}\in \mathbf{b}$  such that  $b_{1}\neq b$  : In this case, the path  $\mathcal{P}(t,F_{\mathbf{b}})$  has to be of the form  $t\longrightarrow b_1\dots b\gets F_{\mathbf{b}}$  . Therefore, there exists at least one collider on the path  $\mathcal{P}(t,F_{\mathbf{b}})$  . Let  $\nu \in \mathcal{P}(t,F_{\mathbf{b}})\setminus \{t,F_{\mathbf{b}}\}$  be the collider on the path  $\mathcal{P}(t,F_{\mathbf{b}})$  that is closest to  $b_{1}$  . Consider the path  $\mathcal{P}(t,\nu)\subset \mathcal{P}(t,F_{\mathbf{b}})$  . We note that this path also exists in  $\mathcal{G}$  and is of the form  $t\longrightarrow b_1\longrightarrow \dots \longrightarrow \nu$
By assumption,  $\mathcal{P}(t,F_{\mathbf{b}})$  is unblocked when  $\mathbf{z}^{(i)}$  is conditioned on. Since there is a collider at  $\nu$  in  $\mathcal{P}(t,F_{\mathbf{b}})$ , there exists at least one unblocked descendant path from  $\nu$  to  $\mathbf{z}^{(i)}$  when  $\mathbf{z}^{(i)}$  is conditioned on. Let  $\mathcal{P}(\nu ,z^{(i)})$  denote the shortest of these paths from  $\nu$  to some  $z^{(i)}\in \mathbf{z}^{(i)}$  in  $\mathcal{G}^{\mathbf{b}}$ . We note that this path also exists in  $\mathcal{G}$  and is of the form  $\nu \longrightarrow \dots \longrightarrow z^{(i)}$ .
Suppose we uncondition on  $\mathbf{z}^{(i)}$ . Consider the path  $\mathcal{P}(t, z^{(i)})$  between  $t$  and  $z^{(i)}$  in  $\mathcal{G}$  obtained after concatenating  $\mathcal{P}(t, \nu) \subset \mathcal{P}(t, F_{\mathbf{b}})$  and  $\mathcal{P}(\nu, z^{(i)})$ . This path, of the form  $t \longrightarrow b_1 \longrightarrow \dots \longrightarrow \nu \longrightarrow \dots \longrightarrow z^{(i)}$ , remains unblocked even when  $\mathbf{z}^{(i)}$  is unconditioned on as it does not have any colliders. This contradicts  $z^{(i)} \perp_{d} t$  (which follows from (8)).
(ii)  $\mathcal{P}(t,F_{\mathbf{b}})$  does not contain  $t\longrightarrow b_1$  for every  $b_{1}\in \mathbf{b}$ : By assumption,  $\mathcal{P}(t,F_{\mathbf{b}})$  is unblocked in  $\mathcal{G}^{\mathbf{b}}$  when  $\mathbf{z}^{(i)}$  is conditioned on. Therefore, if  $\mathcal{P}(t,F_{\mathbf{b}})$  does not contain the edge  $t\longrightarrow b_{1}$  for any  $b_{1}\in \mathbf{b}$ , there exists a path  $\mathcal{P}(t,b)$  between  $t$  to  $b$  in  $\mathcal{G}$  that is unblocked when  $\mathbf{z}^{(i)}$  is conditioned on, and takes one of the following two forms: (a)  $t\longleftarrow \dots b$  or (b)  $t\leftrightarrow \dots b$ . Then, it is easy to see that the path  $\mathcal{P}(t,b)$  also remains unblocked in  $\mathcal{G}_{\underline{t}}$  while  $\mathbf{z}^{(i)}$  is conditioned on. However, this contradicts  $t\perp \perp_d\mathbf{b}|\mathbf{z}^{(i)}$  in  $\mathcal{G}_{\underline{t}}$  (which follows from Lemma 3).
Proof of Lemma 2(e) We prove this by contradiction. Assume there exists at least one unblocked path between  $y$  and some  $b \in \mathbf{b}$  in  $\mathcal{G}_{\mathbf{b}}$  when  $t$  and  $\mathbf{z}^{(i)}$  are conditioned on. Let  $\mathcal{P}(b, y)$  denote the shortest of these unblocked path. Therefore, no  $b_1 \in \mathbf{b}$ , such that  $b_1 \neq b$ , is on the path  $\mathcal{P}(b, y)$ , i.e.,  $b_1 \notin \mathcal{P}(b, y)$ . Further,  $\mathcal{P}(b, y)$  takes one of the following two forms because all the edges going out of  $\mathbf{b}$  are removed in  $\mathcal{G}_{\mathbf{b}}$ : (a)  $b \leftarrow \dots y$  or (b)  $b \leftrightarrow \dots y$ .
Suppose we condition on  $\mathbf{z}^{(o)}$  (while  $t$  and  $\mathbf{z}^{(i)}$  are still conditioned on). From (7) and Fact 1, we have  $y \perp_{d} b|t, \mathbf{z}$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$ . Therefore, the path  $\mathcal{P}(b, y)$  is blocked in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when  $\mathbf{z}^{(o)}$  is conditioned on (while  $t$  and  $\mathbf{z}^{(i)}$  are still conditioned on). Let  $\nu$  be any node at which  $\bar{\mathcal{P}}(b, y)$  is blocked in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when  $\mathbf{z}^{(o)}$  is conditioned on (while  $t$  and  $\mathbf{z}^{(i)}$  are still conditioned on). We must have that  $\nu \in \mathcal{P}(b, y) \setminus \{y, b\}$  and  $\nu \in \mathbf{z}^{(o)}$ . Suppose we uncondition on  $\mathbf{z}^{(o)}$  (while  $t$  and  $\mathbf{z}^{(i)}$  are still conditioned on). Then, the path  $\mathcal{P}(b, \nu) \subset \mathcal{P}(b, y)$  is unblocked in  $\mathcal{G}_{\underline{\mathbf{b}}}$ .
We consider the following two scenarios depending on whether or not  $\mathcal{P}(b,\nu)$  contains  $t$ . In both scenarios, we show that there is an unblocked path between  $t$  and  $\nu$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when we condition on  $\mathbf{b}$  (while  $t$  and  $\mathbf{z}^{(i)}$  are still conditioned on).
(i)  $\mathcal{P}(b, \nu)$  contains  $t$ : Consider the path  $\mathcal{P}(t, \nu) \subset \mathcal{P}(b, \nu)$  which is unblocked in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when  $t$  and  $\mathbf{z}^{(i)}$  are conditioned on. Further, by the choice of  $\mathcal{P}(b, y)$ , no  $b_1 \in \mathbf{b}$  is on the path
$\mathcal{P}(t,\nu)$ . Therefore, the path  $\mathcal{P}(t,\nu)$  in  $\mathcal{G}_{\mathbf{b}}$  remains unblocked when we condition on  $\mathbf{b}$  (while  $t$  and  $\mathbf{z}^{(i)}$  are still conditioned on).
(ii)  $\mathcal{P}(b,\nu)$  does not contain  $t$ : Consider the path  $\mathcal{P}(t,\nu)\supset \mathcal{P}(b,\nu)$  (by including the extra edge  $t\to b$ ) which takes one of the following two forms:  $(a)t\longrightarrow b\longleftarrow \dots \nu$  or  $(b)$ $t\longrightarrow b\leftrightarrow \dots \nu$ . Further, by the choice of  $\mathcal{P}(b,y)$ , no  $b_{1}\in \mathbf{b}$ $(b_{1}\neq b)$  is on the path  $\mathcal{P}(t,\nu)$ . Suppose we condition on  $\mathbf{b}$  (while  $t$  and  $\mathbf{z}^{(i)}$  are still conditioned on). Then, the path  $\mathcal{P}(t,\nu)$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$  is unblocked because  $(a)$  the collider at  $b$  is unblocked when  $\mathbf{b}$  is conditioned on and  $(b)$  the path  $\mathcal{P}(b,\nu)$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$  remains unblocked when  $\mathbf{b}$  is conditioned on (while  $t$  and  $\mathbf{z}^{(i)}$  are still conditioned on).
Now, suppose we uncondition on  $t$  (while  $\mathbf{b}$  and  $\mathbf{z}^{(i)}$  are still conditioned on). We have the following two scenarios depending on whether or not  $\mathcal{P}(t,\nu)$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$  remains unblocked. In both scenarios, we show that there is an unblocked path between  $t$  and  $\nu$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when we uncondition on  $t$  (while  $\mathbf{b}$  and  $\mathbf{z}^{(i)}$  are still conditioned on).
1. If  $\mathcal{P}(t,\nu)$  remains unblocked: In this case,  $\mathcal{P}(t,\nu)$  in  $\mathcal{G}_{\mathbf{b}}$  is an unblocked path between  $t$  and  $\nu$  when  $\mathbf{z}^{(i)}$  and  $\mathbf{b}$  are conditioned on, as desired.
2. If  $\mathcal{P}(t,\nu)$  does not remain unblocked: In this case, it is the unconditioning on  $t$  (while  $\mathbf{b}$  and  $\mathbf{z}^{(i)}$  are still conditioned on) that blocks  $\mathcal{P}(t,\nu)$ . Now, we create a set  $\mathbf{w}$  consisting of all the nodes at which  $\mathcal{P}(t,\nu)$  is blocked in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when  $t$  is unconditioned on (while  $\mathbf{b}$  and  $\mathbf{z}^{(i)}$  are still conditioned on). Define the set  $\mathbf{w}$  such that for any  $w\in \mathbf{w}$ , the following are true:  $(a) w\in \mathcal{P}(t,\nu)\setminus \{t,\nu \}$ ,  $(b)\mathcal{P}(t,\nu)$  contains a collider at  $w$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$ , and  $(c)$  there exists an unblocked descendant path from  $w$  to  $t$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$ .
Now, we must have  $\mathbf{w} \neq \emptyset$ , since  $\mathcal{P}(t, \nu)$  is blocked in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when  $t$  is unconditioned on (while  $\mathbf{b}$  and  $\mathbf{z}^{(i)}$  are still conditioned on). Let  $w_{c} \in \mathbf{w}$  be that node which is closest to  $\nu$  in the path  $\mathcal{P}(t, \nu)$ , and let  $\mathcal{P}(w_{c}, t)$  be an unblocked descendant path from  $w_{c}$  to  $t$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$  (there must be one from the definition of the set  $\mathbf{w}$ ). Consider the path  $\mathcal{P}'(\nu, t)$  obtained after concatenating  $\mathcal{P}(\nu, w_{c}) \subset \mathcal{P}(t, \nu)$  and  $\mathcal{P}(w_{c}, t)$ . By the definition of  $\mathbf{w}$  and the choice of  $w_{c}$ ,  $\mathcal{P}'(\nu, t)$  is unblocked in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when  $t$  is unconditioned on (while  $\mathbf{b}$  and  $\mathbf{z}^{(i)}$  are still conditioned on) since (a)  $\mathcal{P}(\nu, w_{c})$  is unblocked, (b)  $\mathcal{P}(w_{c}, t)$  is unblocked, and (c) there is no collider at  $w_{c}$  in  $\mathcal{P}'(\nu, t)$ . Therefore, we have an unblocked path between  $t$  and  $\nu$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when  $\mathbf{z}^{(i)}$  and  $\mathbf{b}$  are conditioned on, as desired.
To conclude the proof, we note that the existence of an unblocked path between  $t$  and  $\nu \in \mathbf{z}^{(o)}$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$  when  $\mathbf{z}^{(i)}$  and  $\mathbf{b}$  are conditioned on contradicts  $\mathbf{z}^{(o)} \perp_{d} t | \mathbf{b}, \mathbf{z}^{(i)}$  in  $\mathcal{G}_{\underline{\mathbf{b}}}$  (which follows from (8) and Fact 1).
F.1 Proof of Lemma 3
First, we claim  $t \perp_{d} \mathbf{b} | \mathbf{z}$  in  $\mathcal{G}_{\underline{t}}$ . We assume this claim and proceed to prove the statement in the Lemma by contradiction. Assume there exists at least one unblocked path between  $t$  and some  $b \in \mathbf{b}$  in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{z}^{(i)}$  is conditioned on. Let  $\mathcal{P}(t, b)$  denote the shortest of these unblocked path. Therefore, no  $\bar{b}_1 \in \mathbf{b}$  such that  $b_1 \neq b$  is not on the path  $\mathcal{P}(t, b)$ , i.e.,  $b_1 \notin \mathcal{P}(t, b)$ .
Suppose we condition on  $\mathbf{z}^{(o)}$  (while  $\mathbf{z}^{(i)}$  is still conditioned on). From the claim,  $\mathcal{P}(t,b)$  is blocked in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{z}^{(o)}$  is conditioned on (while  $\mathbf{z}^{(i)}$  is still conditioned on). Let  $\nu$  be any node at which  $\mathcal{P}(t,b)$  is blocked in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{z}^{(o)}$  is conditioned on (while  $\mathbf{z}^{(i)}$  is still conditioned on). We must have that  $\nu \in \mathcal{P}(t,b) \setminus \{t,b\}$  and  $\nu \in \mathbf{z}^{(o)}$ . Then, the path  $\mathcal{P}(t,\nu) \subset \mathcal{P}(t,b)$  is unblocked in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{z}^{(o)}$  is unconditioned on (while  $\mathbf{z}^{(i)}$  is still conditioned on). Further, no  $b \in \mathbf{b}$  is on the path  $\mathcal{P}(t,\nu)$ . As a result, the path  $\mathcal{P}(t,\nu)$  remains unblocked when  $\mathbf{b}$  is conditioned on (while  $\mathbf{z}^{(i)}$  is still conditioned on). However, this contradicts  $t \perp_{d} \mathbf{z}^{(o)}|b, \mathbf{z}^{(i)}$  in  $\mathcal{G}_{\underline{t}}$  (which follows from (8)(ii) and Fact 1).
Proof of Claim -  $t \perp_{d} \mathbf{b} | \mathbf{z}$  in  $\mathcal{G}_{\underline{t}}$ : It remains to prove the claim  $t \perp_{d} \mathbf{b} | \mathbf{z}$  in  $\mathcal{G}_{\underline{t}}$ . We prove this by contradiction. Assume there exists at least one unblocked path between  $t$  and some  $b \in \mathbf{b}$  in  $\mathcal{G}_{\underline{t}}$  when  $\mathbf{z}$  is conditioned on. Let  $\mathcal{P}(t, b)$  denote any such unblocked path. This path takes one of the following two forms: (a)  $t \longleftarrow \dots b$  or (b)  $t \leftrightarrow \dots b$  because all edges going out of  $t$  are removed in  $\mathcal{G}_{\underline{t}}$ .
Suppose we condition on  $t$  (while  $\mathbf{z}$  is still conditioned on). The path  $\mathcal{P}(t, b)$  remains unblocked because  $t \notin \mathcal{P}(t, b)$  (a path is a sequence of distinct nodes). Then, the path  $\mathcal{P}(y, b) \supset \mathcal{P}(t, b)$  of the form  $(a) y \leftrightarrow t \leftarrow \dots b$  or  $(b) y \leftrightarrow t \leftrightarrow t \leftrightarrow \dots b$  is unblocked because the additional conditioning on  $t$  (while  $\mathbf{z}$  is still conditioned on) unblocks the collider at  $t$ . However, this contradicts  $\mathbf{b} \perp_{d} y | t, \mathbf{z}$  in  $\mathcal{G}_t$  (which follows from (7) and Fact 1).
G Experimental Results
In this section, we provide additional experimental results. First, we provide more details regarding the numerical example in Section 3.1. Next, we demonstrate the applicability of our method on a class of graphs slightly different from the one in Section 4.1. Then, we provide the 6 random graphs from Section 4.2 as well as ATE estimation results on specific choices of SMCMs including the one in Figure 2. Finally, we provide histograms analogous to Figure 6 for the second choice of  $\mathbf{b}$  on German credit dataset as well as details about our analysis with Adult dataset.
G.1 Numerical example in Section 3.1
The observed variables for this example also follow the structural equation model in (11). Also, to generate the true ATE, we intervene on the generation model in (11) by setting  $t = 0$  and  $t = 1$ .
G.2 Applicability to a class of random graphs
As in Section 4.1, we create a class of random SMCMs, sample 100 SMCMs from this class, and check if (7) and (8) hold by checking for corresponding d-separations in the SMCMs. The class of random graphs considered here is analogous to the class of random graphs considered in Section 4.1 expect for the choice of  $t$ . Here, we choose any variable that is ancestor of  $y$  but not its parent as  $t$ . This is in contrast to Section 4.1 where we choose any variable that is ancestor of  $y$  but not its parent or grandparent as  $t$ . We compare the success rate of the same two approaches: (i) exhaustive search for  $\mathbf{z}$  satisfying (7) and (8) and (ii) search for a  $\mathbf{z}$  of size at-most 5 satisfying (7) and (8). We provide the number of successes of these approaches as a tuple in Table 2 for various  $p$ ,  $d$ , and  $q$ . As before, we see that the two approaches have comparable performances and the IDP algorithm gives 0 successes across various  $p$ ,  $d$ , and  $q$  even though it is supplied with the true PAG. Also, as expected the number of successes for this class of graphs is much lower than the class considered in Section 4.1.
Table 2: Number of successes out of 100 random graphs for our methods shown as a tuple. The first method searches a  $\mathbf{z}$  exhaustively and the second method searches a  $\mathbf{z}$  with size at-most 5.
p=10	p=15
d=2	d=3	d=4	d=2	d=3	d=4
q=0.0	(6,6)	(3,3)	(1,1)	(11,11)	(2,2)	(1,1)
q=0.5	(3,3)	(0,0)	(0,0)	(5,5)	(2,2)	(1,1)
q=1.0	(1,1)	(0,0)	(0,0)	(1,1)	(0,0)	(0,0)
G.3 ATE estimation
We also conduct ATE estimation experiments on four specific SMCMs. The first SMCM is the graph  $\mathcal{G}^{toy}$  in Figure 2. The remaining graphs, named  $\mathcal{G}_i^{toy}$ ,  $i \in \{1,2,3\}$ , are shown in Figure 8, and are obtained by adding additional edges and modifying  $\mathcal{G}^{toy}$ . These SMCMs are designed in a way such that there exists  $\mathbf{z} = (\mathbf{z}^{(i)}, \mathbf{z}^{(o)})$  satisfying the conditional independence statements in Theorem 3.2.
We follow a data generation procedure similar to the one in Section 4.2. In contrast, we show the performance of our approach for a fixed  $n$  but different thresholds of p-value  $p_v$ . We average the ATE error over 50 runs where in each run we set  $n = 50000$ . As we see in Figure 9, both the ATE estimates returned by Algorithm 1 are far superior compared to the naive front-door adjustment using b.
Figure 8: The causal graphs used to further validate our theoretical results. These are obtained by adding additional edges (shown in red) to  $\mathcal{G}^{toy}$  in Figure 2. We denote these graphs (from left to right) by  $\mathcal{G}_1^{toy}$ ,  $\mathcal{G}_2^{toy}$ , and  $\mathcal{G}_3^{toy}$ , respectively.
Figure 9: Performance of Algorithm 1 for different p-value thresholds  $p_v$  on  $\mathcal{G}^{toy}$  in Figure 2 on top left, on  $\mathcal{G}_1^{toy}$  from Figure 8 on top right, on  $\mathcal{G}_2^{toy}$  in Figure 8 on bottom left, and on  $\mathcal{G}_3^{toy}$  from Figure 8 on bottom right
In Figure 11, we provide the 6 random SMCMs used in Section 4.2. As mentioned in Section 4.1, we choose the last variable in the causal ordering as  $y$  and a variable that is ancestor of  $y$  but not its parent or grandparent as  $t$ . We also show the corresponding  $\mathbf{z} = (\mathbf{z}^{(i)},\mathbf{z}^{(o)})$  satisfying (7) and (8).
G.4 German Credit dataset
As in Section 4.3, we assess the conditional independence associated with the selected  $\mathbf{z}$  for the choice of  $\mathbf{b} = \{\# \text{ of people financially dependent on the applicant, applicant's savings}\}$ , Algorithm 1 results in  $\mathbf{z}^{(i)} = \{\text{purpose for which the credit was needed, applicant's checking account status with the bank}\}$  via 100 random bootstraps. We show the corresponding p-values for these bootstraps in a histogram in Figure 10 below. As expected, we observe the p-values to be spread out.