SlowGuess/NeurIPS_2023_OCR huggingface dataset:
For any  $\mathbf{v} \in \mathcal{W}$ , let  $\mathcal{G}_{\overline{\mathbf{v}}}$  be the graph obtained by removing the edges going into  $\mathbf{v}$  in  $\mathcal{G}$ , and let  $\mathcal{G}_{\underline{\mathbf{v}}}$  be the graph obtained by removing the edges going out of  $\mathbf{v}$  in  $\mathcal{G}$ .
Theorem C.1 (Rules of do-calculus, Pearl [1995]). For any disjoint subsets  $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4 \subseteq \mathcal{W}$ , we have the following rules.
Rule 1:  $\mathbb{P}(\mathbf{v}_1|do(\mathbf{v}_2),\mathbf{v}_3,\mathbf{v}_4) = \mathbb{P}(\mathbf{v}_1|do(\mathbf{v}_2),\mathbf{v}_3)$  if  $\mathbf{v}_1$ $\perp_{d}\mathbf{v}_{4}|\mathbf{v}_{2},\mathbf{v}_{3}$  in  $\mathcal{G}_{\overline{\mathbf{v}}_2}$
Rule 2:  $\mathbb{P}(\mathbf{v}_1|do(\mathbf{v}_2),\mathbf{v}_3,do(\mathbf{v}_4)) = \mathbb{P}(\mathbf{v}_1|do(\mathbf{v}_2),\mathbf{v}_3,\mathbf{v}_4)$  if  $\mathbf{v}_1$ $\perp_{d}\mathbf{v}_4|\mathbf{v}_2,\mathbf{v}_3$  in  $\mathcal{G}_{\overline{\mathbf{v}}_2,\underline{\mathbf{v}}_4}$
Rule 3:  $\mathbb{P}(\mathbf{v}_1|do(\mathbf{v}_2),\mathbf{v}_3,do(\mathbf{v}_4)) = \mathbb{P}(\mathbf{v}_1|do(\mathbf{v}_2),\mathbf{v}_3)$  if  $\mathbf{v}_1$ $\perp_{d}\mathbf{v}_{4}|\mathbf{v}_{2},\mathbf{v}_{3}$  in  $\mathcal{G}_{\overline{\mathbf{v}}_2,\overline{\mathbf{v}_4(\mathbf{v}_3)}}$
where  $\mathbf{v}_4(\mathbf{v}_3)$  is the set of nodes in  $\mathbf{v}_4$  that are not ancestors of any node in  $\mathbf{v}_3$  in  $\mathcal{G}_{\overline{\mathbf{v}}_2}$ . Pearl [1995] also gave an alternative criterion for Rule 3.
Rule 3a:  $\mathbb{P}(\mathbf{v}_1|do(\mathbf{v}_2),\mathbf{v}_3,do(\mathbf{v}_4)) = \mathbb{P}(\mathbf{v}_1|do(\mathbf{v}_2),\mathbf{v}_3)$  if  $\mathbf{v}_1$ $\perp_{d}F_{\mathbf{v}_4}|\mathbf{v}_2,\mathbf{v}_3$  in  $\mathcal{G}_{\mathbf{v}_2}^{\mathbf{v}_4}$
where  $\mathcal{G}^{\mathbf{v}_4}$  is the graph obtained from  $\mathcal{G}$  after adding (a) a node  $F_{\mathbf{v}_4}$  and (b) edges from  $F_{\mathbf{v}_4}$  to every node in  $\mathbf{v}_4$ .
Also, throughout our proofs, we use the following fact.
Fact 1. Consider any  $\mathcal{G}'$  obtained by removing any edge(s) from  $\mathcal{G}$ . For any sets of variables  $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \subseteq \mathcal{W}$ , if  $\mathbf{v}_1$  and  $\mathbf{v}_2$  are  $d$ -separated by  $\mathbf{v}_3$  in  $\mathcal{G}$  than  $\mathbf{v}_1$  and  $\mathbf{v}_2$  are  $d$ -separated by  $\mathbf{v}_3$  in  $\mathcal{G}'$ .
D Causal Identifiability
In this section, we derive the causal effect for the SMCM in Figure 3(top), i.e., (6), as well as prove Theorem 3.1 one by one.
D.1 Proof of (6)
First, using the law of total probability, we have
$$
\mathbb {P} (y | d o (t = t)) = \sum_ {z _ {1}, z _ {2}} \mathbb {P} (y | d o (t = t), z _ {1} = z _ {1}, z _ {2} = z _ {2}) \mathbb {P} \left(z _ {1} = z _ {1}, z _ {2} = z _ {2} \mid d o (t = t)\right). \tag {12}
$$
Now, we show that the two terms in RHS of (12) can be simplified as follows
$$
\mathbb {P} (y \mid d o (t = t), z _ {1} = z _ {1}, z _ {2} = z _ {2}) = \sum_ {t ^ {\prime}} \mathbb {P} (y \mid z _ {1} = z _ {1}, z _ {2} = z _ {2}, t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime} \mid z _ {1} = z _ {1}). \tag {13}
$$
$$
\mathbb {P} \left(z _ {1} = z _ {1}, z _ {2} = z _ {2} \mid d o (t = t)\right) = \sum_ {\boldsymbol {b}} \mathbb {P} \left(z _ {2} = z _ {2} \mid \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} \left(\boldsymbol {b} = \boldsymbol {b} \mid t = t, z _ {1} = z _ {1}\right) \mathbb {P} \left(z _ {1} = z _ {1}\right), \tag {14}
$$
Combining (12) to (14) results in (6).
Proof of (13): We have
$$
\mathbb {P} (y \mid d o (t = t), z _ {1} = z _ {1}, z _ {2} = z _ {2}) \tag {15}
$$
$$
\stackrel {(a)} {=} \mathbb {P} (y = y | d o (t = t), z _ {1} = z _ {1}, z _ {2} = z _ {2}, \mathbf {b} = \mathbf {b}) \tag {16}
$$
$$
\stackrel {(b)} {=} \mathbb {P} (y = y | d o (t = t), z _ {1} = z _ {1}, z _ {2} = z _ {2}, d o (\boldsymbol {b} = \boldsymbol {b})) \tag {17}
$$
$$
\stackrel {(c)} {=} \mathbb {P} (y = y | z _ {1} = z _ {1}, z _ {2} = z _ {2}, d o (\boldsymbol {b} = \boldsymbol {b})) \tag {18}
$$
$$
\stackrel {(d)} {=} \sum_ {t ^ {\prime}} \mathbb {P} (y = y | z _ {1} = z _ {1}, z _ {2} = z _ {2}, d o (\boldsymbol {b} = \boldsymbol {b}), t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime} | z _ {1} = z _ {1}, z _ {2} = z _ {2}, d o (\boldsymbol {b} = \boldsymbol {b})) \tag {19}
$$
$$
\stackrel {(e)} {=} \sum_ {t ^ {\prime}} \mathbb {P} (y = y | z _ {1} = z _ {1}, z _ {2} = z _ {2}, t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime} | z _ {1} = z _ {1}, z _ {2} = z _ {2}, d o (\boldsymbol {b} = \boldsymbol {b})) \tag {20}
$$
$$
\stackrel {(f)} {=} \sum_ {t ^ {\prime}} \mathbb {P} (y = y | z _ {1} = z _ {1}, z _ {2} = z _ {2}, t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime} | z _ {1} = z _ {1}, d o (\boldsymbol {b} = \boldsymbol {b})) \tag {21}
$$
$$
\stackrel {(g)} {=} \sum_ {t ^ {\prime}} \mathbb {P} (y = y | z _ {1} = z _ {1}, z _ {2} = z _ {2}, t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime} | z _ {1} = z _ {1}), \tag {22}
$$
where  $(a)$  and  $(f)$  follow from Rule 1,  $(b)$  follows from Rule 2,  $(c)$ ,  $(e)$ , and  $(g)$  follow from Rule 3a, and  $(d)$  follows from the law of total probability.
Proof of (14): From the law of total probability, we have
$$
\mathbb {P} \left(z _ {1} = z _ {1}, z _ {2} = z _ {2} \mid d o (t = t)\right) \tag {23}
$$
$$
= \sum_ {\boldsymbol {b}} \mathbb {P} \left(z _ {1} = z _ {1}, z _ {2} = z _ {2} \mid d o (t = t), \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} (\boldsymbol {b} = \boldsymbol {b} \mid d o (t = t)) \tag {24}
$$
$$
\stackrel {(a)} {=} \sum_ {\boldsymbol {b}} \mathbb {P} \left(z _ {2} = z _ {2} \mid d o (t = t), \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} \left(z _ {1} = z _ {1} \mid d o (t = t), \boldsymbol {b} = \boldsymbol {b}, z _ {2} = z _ {2}\right) \mathbb {P} \left(\boldsymbol {b} = \boldsymbol {b} \mid d o (t = t)\right) \tag {25}
$$
$$
\stackrel {(b)} {=} \sum_ {\boldsymbol {b}} \mathbb {P} \left(\boldsymbol {z} _ {2} = z _ {2} \mid \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} \left(\boldsymbol {z} _ {1} = z _ {1} \mid d o (t = t), \boldsymbol {b} = \boldsymbol {b}, \boldsymbol {z} _ {2} = z _ {2}\right) \mathbb {P} \left(\boldsymbol {b} = \boldsymbol {b} \mid d o (t = t)\right) \tag {26}
$$
$$
\stackrel {(c)} {=} \sum_ {\boldsymbol {b}} \mathbb {P} \left(\boldsymbol {z} _ {2} = z _ {2} \mid \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} \left(z _ {1} = z _ {1} \mid d o (t = t), \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} (\boldsymbol {b} = \boldsymbol {b} | d o (t = t)) \tag {27}
$$
$$
\stackrel {(d)} {=} \sum_ {\boldsymbol {b}} \mathbb {P} \left(\boldsymbol {z} _ {2} = z _ {2} \mid \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} \left(\boldsymbol {z} _ {1} = z _ {1}, \boldsymbol {b} = \boldsymbol {b} \mid d o (t = t)\right) \tag {28}
$$
$$
\stackrel {(e)} {=} \sum_ {\boldsymbol {b}} \mathbb {P} \left(z _ {2} = z _ {2} \mid \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} \left(z _ {1} = z _ {1} \mid d o (t = t)\right) \mathbb {P} (\boldsymbol {b} = \boldsymbol {b} \mid d o (t = t), z _ {1} = z _ {1}) \tag {29}
$$
$$
\stackrel {(f)} {=} \sum_ {\boldsymbol {b}} \mathbb {P} \left(\boldsymbol {z} _ {2} = z _ {2} \mid \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} \left(\boldsymbol {z} _ {1} = z _ {1}\right) \mathbb {P} \left(\boldsymbol {b} = \boldsymbol {b} \mid d o (\boldsymbol {t} = t), \boldsymbol {z} _ {1} = z _ {1}\right) \tag {30}
$$
$$
\stackrel {(g)} {=} \sum_ {\boldsymbol {b}} \mathbb {P} \left(\boldsymbol {z} _ {2} = z _ {2} \mid \boldsymbol {b} = \boldsymbol {b}\right) \mathbb {P} \left(z _ {1} = z _ {1}\right) \mathbb {P} \left(\boldsymbol {b} = \boldsymbol {b} \mid t = t, z _ {1} = z _ {1}\right) \tag {31}
$$
where  $(a)$ ,  $(d)$ , and  $(e)$  follow from the definition of conditional probability,  $(b)$  and  $(f)$  follows from Rule 3a,  $(c)$  follows from Rule 1, and  $(g)$  follows from Rule 2.
D.2 Proof of Theorem 3.1
Let  $\mathrm{An}(y)$  denote the union of  $y$  and the set of ancestors of  $y$ , and let  $\mathcal{G}^{\mathrm{An}(y)}$  denote the subgraph of  $\mathcal{G}$  composed only of nodes in  $\mathrm{An}(y)$ . First, we show that if  $\mathbf{b} \perp_{d} y|t$ ,  $\mathbf{z}$  holds for some  $\mathbf{z}$ , then there is no bi-directed path between  $t$  to  $\mathbf{b}$  in  $\mathcal{G}^{\mathrm{An}(y)}$ .
Lemma 1. Suppose Assumptions 1 to 3 hold. Suppose there exists a set  $\mathbf{z} \subseteq \mathcal{V} \setminus \{t, \mathbf{b}, y\}$  such that  $\mathbf{b} \perp_{d} y|t, \mathbf{z}$ . Then, there is no bi-directed path between  $t$  and  $\mathbf{b}$  in  $\mathcal{G}^{\mathrm{An}(y)}$ .
Given this claim, Theorem 3.1 follows from Tian and Pearl [2002, Theorem 4]. It remains to prove Lemma 1.
Proof of Lemma 1. We prove this result by contradiction. First, from Assumptions 1 and 3,  $t \in \mathrm{An}(y)$  and  $\mathbf{b}_0 \in \mathrm{An}(y)$  for some  $\mathbf{b}_0 \subset \mathbf{b}$ . Assume there exists a bi-directed path between  $t$  and some  $b \in \mathbf{b}_0$  in  $\mathcal{G}^{\mathrm{An}(y)}$ . Let  $\mathcal{P}(t,b)$  denote the shortest of these paths. This path is of the form  $t \leftrightarrow v_1 \leftrightarrow \dots \leftrightarrow v_r \leftrightarrow b$  for some  $r \geq 0$  where  $v_q \in \mathcal{G}^{\mathrm{An}(y)}$  for every  $q \in [r]$ . We have the following two cases depending on the value of  $r$ .
(i)  $r = 0$ : In this case, consider the path  $\mathcal{P}(y,b)\supset \mathcal{P}(t,b)$  in  $\mathcal{G}$  of the form:  $y\leftrightarrow t\leftrightarrow b$  in  $\mathcal{G}$  (such a path exists because of Assumption 2). The path  $\mathcal{P}(y,b)$  is unblocked when  $t$  and  $\mathbf{z}$  are conditioned on contradicting  $\mathbf{b}\perp_{d}y|t,\mathbf{z}$
(ii)  $r \geq 1$ : In this case, consider the path  $\mathcal{P}(y,b) \supset \mathcal{P}(t,b)$  in  $\mathcal{G}$  of the form:  $y \leftrightarrow t \leftrightarrow v_1 \leftrightarrow \dots \leftrightarrow v_r \leftrightarrow b$  (such a path exists because of Assumption 2). We have the following two scenarios depending on whether the path  $\mathcal{P}(y,b)$  is unblocked or blocked when  $t$  and  $\mathbf{z}$  are conditioned on. Suppose we condition on  $t$  and  $\mathbf{z}$ .
(a) The path  $\mathcal{P}(y,b)$  is unblocked: In this case, by assumption,  $\mathbf{b} \perp_{d} y|t, \mathbf{z}$  is contradicted.
(b) The path  $\mathcal{P}(y, b)$  is blocked: We create a set  $\mathbf{w}$  such that for any  $w \in \mathbf{w}$  the following are true:  $(a) w = v_q$  for some  $q \in [r]$ ,  $(b) w \notin \mathbf{z}$ ,  $(b)$  there is no descendant path  $\mathcal{P}(w, z)$  between  $w$  and some  $z \in \mathbf{z}$ , and  $(c)$  there is no descendant path  $\mathcal{P}(w, t)$  between  $w$  and  $t$ .
In this scenario,  $\mathbf{w} \neq \emptyset$  because  $\mathcal{P}(y, b)$  is blocked. Let  $w_c \in \mathbf{w}$  be that node which is closest to  $b$  in the path  $\mathcal{P}(y, b)$ . By the choice of  $w_c$ , the path  $\mathcal{P}(w_c, b) \subset \mathcal{P}(y, b)$  is unblocked (when  $t$  and  $\mathbf{z}$  are conditioned on). Furthermore, by the definition of  $\mathbf{w}$ ,  $(a) w_c \in \mathcal{G}^{\mathrm{An}(y)}$  (because  $w_c = v_q$  for some  $q \in [r]$ ) and  $(b)$  there exists a descendant path  $\mathcal{P}(w_c, y)$  between  $w_c$  and  $y$  such that  $t \notin \mathcal{P}(w_c, y)$  as well as  $z \notin \mathcal{P}(w_c, y)$  for every  $z \in \mathbf{z}$ . Therefore, the path  $\mathcal{P}(w_c, y)$  is unblocked (when  $t$  and  $\mathbf{z}$  are conditioned on).
Consider the path  $\mathcal{P}'(y, b)$  obtained after concatenating  $\mathcal{P}(w_c, y)$  and  $\mathcal{P}(w_c, b)$  at  $w_c$ . This path is unblocked (when  $t$  and  $\mathbf{z}$  are conditioned on) because:  $(a) \mathcal{P}(w_c, b)$  is unblocked,  $(b) \mathcal{P}(w_c, y)$  is unblocked, and  $(c)$  there is no collider at  $w_c$  in this path (because  $\mathcal{P}(w_c, y)$  is a descendant path to  $y$ ). However, this contradicts  $\mathbf{b} \perp_{d} y|t, \mathbf{z}$ .
E A generalized front-door condition
In this section, we prove Theorem 3.2. We begin by stating a few d-separation statements used in this proof. See Appendix F for a proof.
Lemma 2. Suppose Assumptions 1 to 3 and  $d$ -separation criteria in Theorem 3.2, i.e., (7) and (8), hold. Then,
(a)  $y \perp_{d} F_{t} | \mathbf{z}, \mathbf{b} \text{ in } \mathcal{G}_{\mathbf{b}}^{t}$  and  $y \perp_{d} F_{t} | \mathbf{z}^{(i)}, \mathbf{b} \text{ in } \mathcal{G}_{\mathbf{b}}^{t}$ ,
(b)  $t\perp \perp_{d}\mathbf{b}$  in  $\mathcal{G}_{\mathfrak{t}}$
(c)  $t \perp_{d} \mathbf{z}^{(i)}|_{\mathbf{b}}$  in  $\mathcal{G}_{t}$ ,
(d)  $t \perp_{d} F_{\mathbf{b}} | \mathbf{z}^{(i)}$  in  $\mathcal{G}^{\mathbf{b}}$ , and
(e)  $y\perp \perp_{d}\mathbf{b}|t,\mathbf{z}^{(i)}$  in  $\mathcal{G}_{\mathbf{b}}$
Now, we proceed with the proof in two parts. In the first part, we prove (9), and in the second part, we prove (10).
E.1 Proof of (9)
$$
\mathbb {P} (y = y \mid d o (t = t)) = \sum_ {\boldsymbol {z}} \mathbb {P} (y = y \mid d o (t = t), \boldsymbol {z} = \boldsymbol {z}) \mathbb {P} (\boldsymbol {z} = \boldsymbol {z} \mid d o (t = t)). \tag {32}
$$
Now, we show that the two terms in RHS of (32) can be simplified as follows
$$
\mathbb {P} (y = y \mid d o (t = t), \mathbf {z} = \mathbf {z}) = \sum_ {t ^ {\prime}} \mathbb {P} (y = y \mid \mathbf {z} = \mathbf {z}, t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime}). \tag {33}
$$
$$
\mathbb {P} (\boldsymbol {z} = \boldsymbol {z} | d o (t = t)) = \mathbb {P} (\boldsymbol {z} = \boldsymbol {z} | t = t), \tag {34}
$$
Combining (33) and (34) completes the proof of (9).
Proof of (33): We have
$$
\mathbb {P} (y = y \mid d o (t = t), \mathbf {z} = \mathbf {z}) \stackrel {(a)} {=} \mathbb {P} (y = y \mid d o (t = t), \mathbf {z} = \mathbf {z}, \mathbf {b} = \mathbf {b}) \tag {35}
$$
$$
\stackrel {(b)} {=} \mathbb {P} (y = y | d o (t = t), \mathbf {z} = \mathbf {z}, d o (\mathbf {b} = \mathbf {b})) \tag {36}
$$
$$
\stackrel {(c)} {=} \mathbb {P} (y = y | \mathbf {z} = \mathbf {z}, d o (\mathbf {b} = \mathbf {b})) \tag {37}
$$
$$
\stackrel {(d)} {=} \sum_ {t ^ {\prime}} \mathbb {P} (y = y | \mathbf {z} = z, d o (\mathbf {b} = \mathbf {b}), t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime} | \mathbf {z} = z, d o (\mathbf {b} = \mathbf {b})), \tag {38}
$$
where  $(a)$  follows from Rule 1, (7), and Fact 1,  $(b)$  follows from Rule 2, (7), and Fact 1, and  $(c)$  follows from Rule 3a and Lemma 2(a), and  $(d)$  follows from the law of total probability.
Now, we simplify the first term in (38) as follows:
$$
\mathbb {P} (y = y \mid \mathbf {z} = \mathbf {z}, d o (\mathbf {b} = \mathbf {b}), t = t ^ {\prime}) \stackrel {(a)} {=} \mathbb {P} (y = y \mid \mathbf {z} = \mathbf {z}, \mathbf {b} = \mathbf {b}, t = t ^ {\prime}) \tag {39}
$$
$$
\stackrel {(7)} {=} \mathbb {P} (y = y | \mathbf {z} = z, t = t ^ {\prime}), \tag {40}
$$
where  $(a)$  follows from Rule 2, (7), and Fact 1. Likewise, we simplify the second term in (38) as follows:
$$
\begin{array}{l} \mathbb {P} (t = t ^ {\prime} \mid d o (\mathbf {b} = \mathbf {b}), \mathbf {z} = \mathbf {z}) \stackrel {(a)} {=} \mathbb {P} (t = t ^ {\prime} \mid d o (\mathbf {b} = \mathbf {b}), \mathbf {z} ^ {(i)} = \mathbf {z} ^ {(i)}) \stackrel {(b)} {=} \mathbb {P} (t = t ^ {\prime} \mid \mathbf {z} ^ {(i)} = \mathbf {z} ^ {(i)}) (41) \\ \stackrel {(c)} {=} \mathbb {P} (t = t ^ {\prime}), (42) \\ \end{array}
$$
where  $(a)$  follows from Rule 1, (8), and Fact 1,  $(b)$  follows from Rule 3a and Lemma 2(d), and  $(c)$  follows (8).
Putting together (38), (40), and (42) results in (33).
Proof of (34): From the law of total probability, we have
$$
\mathbb {P} (\mathbf {z} = \mathbf {z} | d o (t = t)) = \sum_ {\mathbf {b}} \mathbb {P} (\mathbf {z} = \mathbf {z} | d o (t = t), \mathbf {b} = \mathbf {b}) \mathbb {P} (\mathbf {b} = \mathbf {b} | d o (t = t)). \tag {43}
$$
Now, we simplify the first term in (43) as follows:
$$
\mathbb {P} (\mathbf {z} = \mathbf {z} | d o (t = t), \mathbf {b} = \mathbf {b}) \tag {44}
$$
$$
\stackrel {(a)} {=} \mathbb {P} (\mathbf {z} ^ {(i)} = \mathbf {z} ^ {(i)} | d o (t = t), \mathbf {b} = \mathbf {b}) \cdot \mathbb {P} (\mathbf {z} ^ {(o)} = \mathbf {z} ^ {(o)} | d o (t = t), \mathbf {b} = \mathbf {b}, \mathbf {z} ^ {(i)} = \mathbf {z} ^ {(i)}) \tag {45}
$$
$$
\stackrel {(b)} {=} \mathbb {P} \left(\boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)} \mid d o (t = t), \boldsymbol {b} = \boldsymbol {b}\right) \cdot \mathbb {P} \left(\boldsymbol {z} ^ {(o)} = \boldsymbol {z} ^ {(o)} \mid t = t, \boldsymbol {b} = \boldsymbol {b}, \boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)}\right) \tag {46}
$$
$$
\stackrel {(c)} {=} \mathbb {P} \left(\boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)} \mid t = t, \boldsymbol {b} = \boldsymbol {b}\right) \cdot \mathbb {P} \left(\boldsymbol {z} ^ {(o)} = \boldsymbol {z} ^ {(o)} \mid t = t, \boldsymbol {b} = \boldsymbol {b}, \boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)}\right) \tag {47}
$$
$$
\stackrel {(d)} {=} \mathbb {P} (\mathbf {z} = \mathbf {z} | t = t, \mathbf {b} = \mathbf {b}), \tag {48}
$$
where  $(a)$  and  $(d)$  follow from the definition of conditional probability,  $(b)$  follows from Rule 2, (8), and Fact 1, and  $(c)$  follows from Rule 2 and Lemma 2(c). Likewise, we simplify the second term in (43) as follows:
$$
\mathbb {P} (\boldsymbol {b} = \boldsymbol {b} | d o (t = t)) \stackrel {(a)} {=} \mathbb {P} (\boldsymbol {b} = \boldsymbol {b} | t = t), \tag {49}
$$
where  $(a)$  follows from Rule 2 and Lemma 2(b).
Putting together (43), (48), and (49), results in (34) as follows:
$$
\mathbb {P} (\boldsymbol {z} = \boldsymbol {z} | d o (t = t)) = \sum_ {\boldsymbol {b}} \mathbb {P} (\boldsymbol {z} = \boldsymbol {z} | t = t, \boldsymbol {b} = \boldsymbol {b}) \mathbb {P} (\boldsymbol {b} = \boldsymbol {b} | t = t) \stackrel {(a)} {=} \mathbb {P} (\boldsymbol {z} = \boldsymbol {z} | t = t), \tag {50}
$$
where  $(a)$  follows from the law of total probability.
E.2 Proof of (10)
$$
\mathbb {P} (y = y \mid d o (t = t)) = \sum_ {\boldsymbol {b}} \mathbb {P} (y = y \mid d o (t = t), \boldsymbol {b} = \boldsymbol {b}) \mathbb {P} (\boldsymbol {b} = \boldsymbol {b} \mid d o (t = t)). \tag {51}
$$
Now, we show that the first term in RHS of (51) can be simplified as follows
$$
\begin{array}{l} \mathbb {P} (y = y \mid d o (t = t), \mathbf {b} = \mathbf {b}) (52) \\ = \sum_ {\boldsymbol {z} ^ {(i)}} \left(\sum_ {t ^ {\prime}} \mathbb {P} (y = y | \boldsymbol {s} = \boldsymbol {s}, t = t ^ {\prime}) \mathbb {P} \left(t = t ^ {\prime}\right)\right) \mathbb {P} \left(\boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)} | \boldsymbol {b} = \boldsymbol {b}, t = t\right). (53) \\ \end{array}
$$
where  $\mathbf{s} \triangleq (\mathbf{b}, \mathbf{z}^{(i)})$ . Using (49) and (53) in (51), completes the proof of (10) as follows:
$$
\mathbb {P} (y = y \mid d o (t = t)) \tag {54}
$$
$$
= \sum_ {\boldsymbol {s}} \left(\sum_ {t ^ {\prime}} \mathbb {P} (y = y | \boldsymbol {s} = \boldsymbol {s}, t = t ^ {\prime}) \mathbb {P} \left(t = t ^ {\prime}\right)\right) \mathbb {P} \left(\boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)} \mid \boldsymbol {b} = \boldsymbol {b}, t = t\right) \mathbb {P} (\boldsymbol {b} = \boldsymbol {b} | t = t) \tag {55}
$$
$$
\stackrel {(a)} {=} \sum_ {\mathbf {s}} \left(\sum_ {t ^ {\prime}} \mathbb {P} (y = y | \mathbf {s} = \mathbf {s}, t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime})\right) \mathbb {P} (\mathbf {s} = \mathbf {s} | t = t), \tag {56}
$$
where  $(a)$  follows from the definition of conditional probability.
Proof of (53): From the law of total probability, we have
$$
\begin{array}{l} \mathbb {P} (y = y \mid d o (t = t), \mathbf {b} = \mathbf {b}) (57) \\ = \sum_ {\boldsymbol {z} ^ {(i)}} \mathbb {P} (y = y | \boldsymbol {b} = \boldsymbol {b}, \boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)}, d o (t = t)) \mathbb {P} (\boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)} | \boldsymbol {b} = \boldsymbol {b}, d o (t = t)). (58) \\ \end{array}
$$
Now, we simplify the first term in (58) as follows:
$$
\mathbb {P} (y = y \mid \mathbf {b} = \mathbf {b}, \mathbf {z} ^ {(i)} = \mathbf {z} ^ {(i)}, d o (t = t)) \tag {59}
$$
$$
\stackrel {(a)} {=} \mathbb {P} (y = y | d o (\boldsymbol {b} = \boldsymbol {b}), \boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)}, d o (t = t)) \tag {60}
$$
$$
\stackrel {(b)} {=} \mathbb {P} (y = y | d o (\boldsymbol {b} = \boldsymbol {b}), \boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)}) \tag {61}
$$
$$
\stackrel {(c)} {=} \sum_ {t ^ {\prime}} \mathbb {P} (y = y | d o (\mathbf {b} = \mathbf {b}), \mathbf {z} ^ {(i)} = \mathbf {z} ^ {(i)}, t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime} | d o (\mathbf {b} = \mathbf {b}), \mathbf {z} ^ {(i)} = \mathbf {z} ^ {(i)}), \tag {62}
$$
where  $(a)$  follows from Rule 2, Lemma 2(e), and Fact 1,  $(b)$  follows from Rule 3a and Lemma 2(a), and  $(c)$  follows from the law of total probability. We further simplify the first term in (62) as follows:
$$
\mathbb {P} (y = y \mid d o (\boldsymbol {b} = \boldsymbol {b}), \boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)}, t = t ^ {\prime}) \stackrel {(a)} {=} \mathbb {P} (y = y \mid \boldsymbol {b} = \boldsymbol {b}, \boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)}, t = t ^ {\prime}), \tag {63}
$$
where  $(a)$  follows from Rule 2 and Lemma 2(e). Using (63) and (42) in (62), we have
$$
\mathbb {P} (y = y \mid \mathbf {b} = \mathbf {b}, \mathbf {z} ^ {(i)} = \mathbf {z} ^ {(i)}, d o (t = t)) = \sum_ {t ^ {\prime}} \mathbb {P} (y = y \mid \mathbf {b} = \mathbf {b}, \mathbf {z} ^ {(i)} = \mathbf {z} ^ {(i)}, t = t ^ {\prime}) \mathbb {P} (t = t ^ {\prime}). \tag {64}
$$
Now, we simplify the second term in (58) as follows:
$$
\mathbb {P} \left(\boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)} \mid \boldsymbol {b} = \boldsymbol {b}, d o (t = t)\right) \stackrel {(a)} {=} \mathbb {P} \left(\boldsymbol {z} ^ {(i)} = \boldsymbol {z} ^ {(i)} \mid \boldsymbol {b} = \boldsymbol {b}, t = t\right), \tag {65}
$$
where  $(a)$  follows from Rule 2 and Lemma 2(c).
Putting together (58), (64), and (65) results in (53).
E.3 Necessity of Assumption 2
In this section, we provide an example to signify the importance of Assumption 2 to Theorem 3.2. Consider the semi-Markovian causal model in Figure 7 where Assumptions 1 and 3 hold but Assumption 2 does not hold.
Figure 7: An SMCM signifying the importance of Assumption 2
While  $\mathbf{z} = (\mathbf{z}^{(i)},\mathbf{z}^{(o)})$  satisfies (7) and (8) where  $\mathbf{z}^{(i)} = \emptyset$ , the causal effect is not equal to the formulae in (9) or (10). To see this, we note that the set  $\{a\}$  is a back-door set in Figure 7 implying
$$
\mathbb {P} (y | d o (t = t)) = \sum_ {a} \mathbb {P} (y | a, t) \mathbb {P} (a). \tag {66}
$$
Now, we simplify the right hand side of (66) to show explicitly that it is not equivalent to (9). From the law of total probability, we have
$$
\mathbb {P} (y | a, t) = \sum_ {\mathbf {z}} \mathbb {P} (y | \mathbf {z}, a, t) \mathbb {P} (\mathbf {z} | a, t) \tag {67}
$$
$$
\stackrel {(a)} {=} \sum_ {\mathbf {z}} \left(\sum_ {t ^ {\prime}} \mathbb {P} (y | \mathbf {z}, a, t) \mathbb {P} \left(t ^ {\prime}\right)\right) \mathbb {P} (\mathbf {z} | a, t) \stackrel {(b)} {=} \sum_ {\mathbf {z}} \left(\sum_ {t ^ {\prime}} \mathbb {P} (y | \mathbf {z}, t ^ {\prime}) \mathbb {P} \left(t ^ {\prime}\right)\right) \mathbb {P} (\mathbf {z} | a, t), \tag {68}
$$
where  $(a)$  follows because  $\sum_{t'} \mathbb{P}(t') = 1$  and  $(b)$  follows  $y$  is independent of every other variable conditioned on  $\mathbf{z}$ . Plugging (68) in (66), we have
$$
\mathbb {P} (y | d o (t = t)) = \sum_ {\mathbf {z}} \left(\sum_ {t ^ {\prime}} \mathbb {P} (y | \mathbf {z}, t ^ {\prime}) \mathbb {P} \left(t ^ {\prime}\right)\right) \left(\sum_ {a} \mathbb {P} (\mathbf {z} | a, t) \mathbb {P} (a)\right). \tag {69}
$$
Lastly, using the law of total probability, (9) can be rewritten as
$$
\mathbb {P} (y | d o (t = t)) = \sum_ {\mathbf {z}} \left(\sum_ {t ^ {\prime}} \mathbb {P} (y | \mathbf {z}, t ^ {\prime}) \mathbb {P} \left(t ^ {\prime}\right)\right) \left(\sum_ {a} \mathbb {P} (\mathbf {z} | a, t) \mathbb {P} (a | t)\right). \tag {70}
$$