which is not possible if those elements are parentless. Now, observe that the only distinction between the graphs $G_{\text{do}}(X s_Z)$ for $s_Z \subsetneq Z$ and $G_{\text{do}}(XZ)$ is that the former has a strictly larger set of edges (the incoming edges to nodes in $Z \setminus s_Z$ are present in the former and not in the latter). Since d-connection cannot be lost by adding edges, $(X \not\perp^d Y | s_Z)_{G_{\text{do}}(XZ)}$ implies $(X \not\perp^d Y | s_Z)_{G_{\text{do}}(X s_Z)}$. We then use the fact that $\text{Clus}_3$ implies $X \not\ni Y | \text{do}(s_Z)$ for all $s_Z \subsetneq Z$, and consequently that $(X \not\perp Y | s_Z)_{G_{\text{do}}(XZ)}$ for all $s_Z \subsetneq Z$ (Lemma 4.1), which when taken together with the d-connection $(X \not\perp^d Y | s_Z)_{G_{\text{do}}(X s_Z)}$ indicates that the causal model must be fine-tuned. $\square$

**Lemma 4.14** For any disjoint sets X, Y and Z of observed nodes, $X \models Y | \text{do}(Z)$ is Irred$_3 \Rightarrow$ each $e_Z \in Z$ is a cause of at least one element $e_Y \in Y$.

**Proof:** $X \models Y | \text{do}(Z)$ satisfies Irred$_3$ implies in particular that for each $e_Z \in Z$, either $e_Z \models Y | \text{do}(XZ \setminus e_Z)$ or $e_Z \models Y | \text{do}(Z \setminus e_Z)$. As $e_Z$ is a singleton, each of these implies by Lemma 4.13 that $e_Z$ is a cause of at least one $e_Y \in Y$, which establishes the claim. $\square$

**Lemma 4.18** For any disjoint sets X, Y and Z of observed nodes, $X \models Y | \text{do}(Z)$ is Irred$_2$ implies that both $(s_Y \not\perp^d \tilde{s}_Y | Z)_{G_{\text{do}(Z)}}$ and $(X \not\perp^d s_Y | Z \tilde{s}_Y)_{G_{\text{do}(X(\tilde{s}_Y)Z)}}$ for any partition of Y into subsets $s_Y, \tilde{s}_Y$. Here, $X(W) := X \setminus (X \cap \text{anc}(W))$ for $\text{anc}(W)$ denoting the set of all ancestors of W in the original graph $G$.

**Proof:** If $X \models Y | \text{do}(Z)$ is Irred$_2$, then $X \models s_Y | \text{do}(Z)$, $\tilde{s}_Y$ must hold and either $(s_Y \not\perp \tilde{s}_Y | XZ)_{G_{\text{do}(XZ)}}$ or $(s_Y \not\perp \tilde{s}_Y | Z)_{G_{\text{do}(Z)}}$ must hold, for each partition of $Y = s_Y \tilde{s}_Y$. Using the d-separation property, this implies that either $(s_Y \not\perp^d \tilde{s}_Y | XZ)_{G_{\text{do}(XZ)}}$ or $(s_Y \not\perp^d \tilde{s}_Y | Z)_{G_{\text{do}(Z)}}$ must hold. Noting that in $G_{\text{do}(XZ)}$, $X$ only consists of parentless nodes, it cannot act as a collider and therefore removing it from the conditioning set cannot remove d-connection. Moreover, in going from $G_{\text{do}(XZ)}$ to $G_{\text{do}(XZ)}$ we would introduce additional edges, which cannot remove d-connection. Thus in both cases we have $(s_Y \not\perp^d \tilde{s}_Y | Z)_{G_{\text{do}(Z)}}$. The first condition, i.e. $X \models s_Y | \text{do}(Z)$, $\tilde{s}_Y$ further yields by Lemma E.5, $(X \not\perp^d s_Y | Z \tilde{s}_Y)_{G_{\text{do}(X(\tilde{s}_Y)Z)}}$. $\square$

## G.2 Proofs for Section 5

**Lemma 5.4** For $d \ge 2$, $d+1$-Minkowski space-time is a conical poset.

**Proof:** In $d+1$-dim. Minkowski space-time $\mathcal{T}$, the boundary of the light cone of a space-time point $(\vec{x}_0, t_0) \in \mathcal{T}$ is given as follows, where $\vec{x}_0 = (x_0^1, x_0^2, \dots, x_0^d)$ are the $d$-dimensional spatial co-ordinates of the point (in some chosen co-ordinate system).

$$c^2(t - t_0)^2 = \sum_{1<i\le d} (x^i - x_0^i)^2. \qquad (55)$$

Observe that for each time slice (fixed value of $t$), the above equation corresponds to a $d$-dimensional sphere with radius $c^2(t - t_0)^2$. Whenever $d \ge 2$, for a fixed $t$, given any finite portion of this spherical boundary, we can determine the radius of the sphere and therefore its center, which is associated with the spatial co-ordinate $\vec{x}_0$. If we are considering the future light cone, this is non-empty only for $t > t_0$. Given a non-empty time slice of the future light cone of a space-time point $(\vec{x}_0, t_0) \in \mathcal{T}$, we can consider light rays emanating from each point on the boundary of the sphere and extending towards the negative $t$ direction, these light rays would necessarily intersect at a unique time $t_0$ which allows us to fully determine the original point $(\vec{x}_0, t_0)$, starting from any portion of the future light cone boundary on a given non-empty time slice.

Now let $L$ be a finite subset of points in $d+1$-dim. Minkowski space-time and consider the joint future $f(L) = \bigcap_{x \in L} J^+(x)$ of all the points in $L$. The boundary of this joint future is compiled from pieces of the light cone boundaries of points in $L$. In particular, the boundary of $f(L)$ associated with a non-empty time slice is compiled from spherical pieces which are portions of light cone boundaries of points in $L$, associated with that time slice (for $d = 2$, these are circular arcs). We can show that each $x \in \text{span}(L)$ contributes a unique, finite portion to the boundary of the joint future region $f(L)$ at every non-empty time slice. For this, recall that by definition, if $x \in \text{span}(L)$, then $x \in L' \subseteq L$ such that $f(L') = f(L)$ and there is no $L'' \subsetneq L'$ satisfying $f(L'') = f(L)$. If we assume that $x$ doesn't contribute a finite piece to the boundary of $f(L)$, then this yields a contradiction as it would mean that for $L'' := L' \setminus x$, we have
$f(L'') = f(L)$. To show uniqueness, suppose $x \in \text{span}(L)$ contributes a finite portion to the boundary of $f(L)$ at a time slice $t$. No other $x' \in \mathcal{T}$ can contribute the same portion of the boundary of $f(L)$, due to the argument presented in the first paragraph of this proof: each such portion allows us to uniquely reconstruct the point $x$ as it is also a portion of the future light cone of $x$.

Thus, the distinct spherical portions that comprise the boundary of $f(L)$ can each be mapped back uniquely to a distinct spanning element of $L$. Since $L$ is arbitrary here, this implies that $f(L)$ is invertible on all sets $L$ with $L = \text{span}(L)$, which proves the conicality of Minkowski space-times with 2 or more spatial dimensions. $\square$

**Lemma 5.12** Let $\mathcal{T}$ be a conical poset and $L \subset \mathcal{T}$ be finite. Let $\mathcal{X}$ be a set of ORVs on this poset. Then, the knowledge of $\bar{\mathcal{F}}_s(\mathcal{X})$ implies the locations $O(\mathcal{X}_i)$ for all its spanning elements $\mathcal{X}_i \in \text{span}(\mathcal{X})$.

**Proof:** Recalling that $O(\mathcal{X}) = \{O(\mathcal{X}_i)|\mathcal{X}_i \in \mathcal{X}\}$ and $f(L) := \bigcap_{x \in L} J^+(x)$, then by definition, for any $\mathcal{X} \subset \mathcal{S}$,

$$ \bar{\mathcal{F}}_s(\mathcal{X}) = \bigcap_{\mathcal{X}_i \in \mathcal{X}} J^+(O(\mathcal{X}_i)) = f(O(\mathcal{X})) = f(O(\text{span}(\mathcal{X}))). \quad (56) $$

If the poset is conical, by Definition 5.3, $f(\text{span}(O(\mathcal{X})))$ is injective. Hence, it remains to show that $\text{span}(O(\mathcal{X})) \stackrel{!}{=} O(\text{span}(\mathcal{X}))$. To this purpose, we differentiate between the case that the embedding is degenerate or not. If the embedding is non-degenerate, we have a bijection between $\mathcal{X}$ and $L_\mathcal{X}$, which immediately yields $\text{span}(O(\mathcal{X})) = O(\text{span}(\mathcal{X}))$. Otherwise, there are multiple ORVs $u_\mathcal{X} \in \mathcal{X}$ sharing the same location. In this case, it is easy to see that either all of them or none of them belong to $\text{span}(\mathcal{X})$.<fnref n="18" /> Therefore, we can group every set of ORVs embedded degenerately at the same location, into a single ORV embedded at the same location without changing the span. Thus we have reduced a degenerate problem to a non-degenerate one such that $\text{span}(O(\mathcal{X})) = O(\text{span}(\mathcal{X}))$ holds in the former if and only if it holds in the latter. Since we have established that this always holds in the non-degenerate case, this proves the claim. $\square$

As outlined in the main text, our primary results rely on showing that conicality of a poset is equivalent to another poset property called location symmetry. We proceed by motivating this property. Suppose we have a family $\{S_i\}_{i \in I}$ where each element is a set of locations $S_i \subset \mathcal{T}$ all of which share the same span, let us denote this by $K$. Then we can write $S_i = KL_i$ (where $KL_i$ is short for the union of the sets), making clear that the locations in $K$ must be shared among the sets and here the different $L_i$ may also share locations. Plugging $S_i$ for $L_i$ and $K$ for $L_j$ into Equation (16), which holds in conical space-times, we obtain that the joint futures of these sets must be identical in such space-times,

$$ f(K) = f(S_i) \subseteq f(S_i \setminus K) = f(L_i), \forall i \in I \quad (57) $$

where $f(L) := \bigcap_{x \in L} J^+(x)$. Here, the subset relation follows from the fact that removing sets from an intersection always yields a (potentially improper) superset of the original intersection. As this holds for all $i \in I$, we can replace $f(L_i)$ with $\bigcap_{i \in I} f(L_i)$. This yields that in conical space-times where $\text{span}(KL_i) = K$ for all $i \in I$ (or equivalently $\bigcap_{i \in I} \text{span}(KL_i) = K$),

$$ f(KL_i) = f(KL_k) \quad \forall i, k \in I \quad \implies \quad f(K) \subseteq \bigcap_{i \in I} f(L_i) \quad (58) $$

More generally, we can consider sets $S_i = KL_i$ where $K$ is not the common span of the sets. Then $\bigcap_{i \in I} \text{span}(KL_i) \neq K$. This implies that there must be further shared locations in $\text{span}(KL_i) \setminus K$ for each $i$, that are commonly shared between all the $L_i$'s, and using conicality, it can be shown that for all $i$, $\text{span}(KL_i) \setminus K$ must be identical in this case (as we do in the proofs below). Based on this motivation, we have the following definition of location symmetry which captures the two alternative conditions discussed above.

<footnote n="18">¹⁸If at least one $u_\mathcal{X} \ni e_\mathcal{X} \in \text{span}(\mathcal{X})$, for each allowed choice of $s_\mathcal{X}$ (as according to the definition), $|s_\mathcal{X} \cap u_\mathcal{X}| = 1$. However, due to considering the union of such sets we then regain the entirety of $u_\mathcal{X} \subset \bigcup s_\mathcal{X}$.</footnote>
**Definition G.1 (Location Symmetry)** Let $\mathcal{T}$ be a poset and $I$ be an index set. Let $K, L_i \subset \mathcal{T}$ discrete with $i \in I$. Then $\mathcal{T}$ satisfies location symmetry if

$$ \begin{gathered} f(K) \cap f(L_i) = f(K) \cap f(L_k) \quad \forall i, k \in I \\ \implies f(K) \subseteq \bigcap_{i \in I} f(L_i) \quad \lor \quad \exists! M \subset \mathcal{T} : \operatorname{span}(KL_i) \setminus K = M \neq \emptyset \quad \forall i \in I \end{gathered} \tag{59} $$

where

$$ f(L) := \bigcap_{x \in L} J^+(x) $$

If $K$ or $L_i$ is empty, we understand their intersection as being equal to the full poset $\mathcal{T}$.

Alternatively, we can also phrase location symmetry in terms of a set of ORVs $S$ on this poset.

**Definition G.2 (Location Symmetry for ORVs)** Let $\mathcal{T}$ be a poset and $I$ be an index set. Let $\mathcal{X}, \mathcal{Y}_i \subset S$ with $i \in I$, not necessarily disjoint or non-empty. Then $\mathcal{T}$ satisfies location symmetry

$$ \begin{gathered} \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}_i) = \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}_k) \quad \forall i, k \in I \\ \implies \bar{\mathcal{F}}_s(\mathcal{X}) \subseteq \bar{\mathcal{F}}_s(\bigcup_{i \in I} \mathcal{Y}_i) \quad \lor \quad \exists! M \subset \mathcal{T} : \mathcal{O}(\operatorname{span}(\mathcal{X}\mathcal{Y}_k)) \setminus \mathcal{O}(\mathcal{X}) = M \neq \emptyset \quad \forall k \in I \end{gathered} \tag{60} $$

Especially in the respective second alternative, these definitions takes care to play well with the respective sets / their locations potentially not being disjoint, as this will be of importance for considering degenerate embeddings of ORVs.

**Lemma G.3** $\mathcal{T}$ is a conical poset if and only if it satisfies location symmetry.

**Proof:** For this proof, we will refer to ORVs directly, as that is more compact and matches the notation we will use for the proofs going forward. All notation matches Definition G.2.

“⇒”: Here we assume location symmetry and prove conicality. We only require the former for the special case where $\mathcal{X} = \emptyset$ while $\mathcal{Y}_1$ and $\mathcal{Y}_2$ are any disjoint sets of ORVs with $\bar{\mathcal{F}}_s(\mathcal{Y}_1) = \bar{\mathcal{F}}_s(\mathcal{Y}_2)$. Then, applying Definition 5.9 to empty sets of ORVs, we recover $\bar{\mathcal{F}}_s(\mathcal{X}) = \mathcal{T}$. Moreover, the pre-condition $\bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}_1) = \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}_2)$ for applying location symmetry is satisfied as it is equivalent to $\bar{\mathcal{F}}_s(\mathcal{Y}_1) = \bar{\mathcal{F}}_s(\mathcal{Y}_2)$. Therefore, location symmetry implies that

$$ \mathcal{T} = \bar{\mathcal{F}}_s(\mathcal{Y}_1\mathcal{Y}_2) \lor \mathcal{O}(\operatorname{span}(\mathcal{X}\mathcal{Y}_1)) \setminus \mathcal{O}(\mathcal{X}) = \mathcal{O}(\operatorname{span}(\mathcal{X}\mathcal{Y}_2)) \setminus \mathcal{O}(\mathcal{X}). \tag{61} $$

Here, the second alternative immediately reduces to $\mathcal{O}(\operatorname{span}(\mathcal{Y}_1)) = \mathcal{O}(\operatorname{span}(\mathcal{Y}_2))$, due to $\mathcal{X}$ being empty. The first is an equality as $\bar{\mathcal{F}}_s(\mathcal{Y}_1\mathcal{Y}_2) \subseteq \mathcal{T}$ trivially holds. Moreover, notice that the first case $\mathcal{T} = \bar{\mathcal{F}}_s(\mathcal{Y}_1\mathcal{Y}_2)$ is equivalent to $\mathcal{T} = \bar{\mathcal{F}}_s(\mathcal{Y}_1) = \bar{\mathcal{F}}_s(\mathcal{Y}_2)$, and this can only be satisfied if both sets of ORVs are embedded into the minimal element of the poset (which may not exist). If the minimal element exists, we therefore have $\mathcal{O}(\mathcal{Y}_1) = \mathcal{O}(\mathcal{Y}_2)$, which in particular implies $\mathcal{O}(\operatorname{span}(\mathcal{Y}_1)) = \mathcal{O}(\operatorname{span}(\mathcal{Y}_2))$ In other words, for both alternatives of location symmetry we get that for any disjoint sets of ORVs $\mathcal{Y}_1$ and $\mathcal{Y}_2$, $\bar{\mathcal{F}}(\mathcal{Y}_1) = \bar{\mathcal{F}}(\mathcal{Y}_2) \implies \mathcal{O}(\operatorname{span}(\mathcal{Y}_1)) = \mathcal{O}(\operatorname{span}(\mathcal{Y}_2))$, reproducing the definition of conicality (Equation (16)).

“$\Leftarrow$”: Let $\mathcal{X}, \mathcal{Y}_i \subset S$ with $i \in I$ such that

$$ \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}_i) = \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}_k) \quad \forall i, k \in I. \tag{62} $$

We now show that in any conical space-time, this implies one of the two conditions in the second line of Equation (60), thus proving location symmetry.

Applying the property Equation (16) of conical space-times to Equation (62), we see that for any choice of $i \in I$, we can deduce the value of $\mathcal{O}(\mathcal{A})$ for all $\mathcal{A} \in \operatorname{span}(\mathcal{X}\mathcal{Y}_i)$ for ORVs embedded in any conical poset. This implies for the set of respective locations to be identical.

$$ \mathcal{O}(\operatorname{span}(\mathcal{X}\mathcal{Y}_i)) = \mathcal{O}(\operatorname{span}(\mathcal{X}\mathcal{Y}_k)) \quad \forall i, k \in I. \tag{63} $$

We now consider two cases and show that these yield respectively the two alternatives for location symmetry, thus concluding the proof.
1. Case 1 ($\exists i : O(\mathcal{X}) = O(\operatorname{span}(\mathcal{X}\mathcal{Y}_i))$) : In this case, we can deduce that the same holds for all $\mathcal{Y}_k$ with $k \in I$, due to Equation (62). Hence,

$$ \bar{\mathcal{F}}_s(\mathcal{X}) = \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}_i) \qquad (64) $$

$$ \Rightarrow \bar{\mathcal{F}}_s(\mathcal{X}) \subseteq \bar{\mathcal{F}}_s(\mathcal{Y}_i) \quad \forall i \in I \qquad (65) $$

$$ \iff \bar{\mathcal{F}}_s(\mathcal{X}) \subseteq \bigcap_{i \in I} \bar{\mathcal{F}}_s(\mathcal{Y}_i) = \bar{\mathcal{F}}_s\left(\bigcup_{i \in I} \mathcal{Y}_i\right), \qquad (66) $$

yielding the first alternative of Equation (60).

2. Case 2 ($\forall i : O(\mathcal{X}) \neq O(\operatorname{span}(\mathcal{X}\mathcal{Y}_i))$) : This implies that $O(\operatorname{span}(\mathcal{X}\mathcal{Y}_i)) \setminus O(\mathcal{X}) \neq \emptyset$. We have already established in Equation (63) that in conical space-times where Equation (62) is satisfied, $O(\operatorname{span}(\mathcal{X}\mathcal{Y}_k))$ must be identical for all $i \in I$, this implies the same holds for $O(\operatorname{span}(\mathcal{X}\mathcal{Y}_i)) \setminus O(\mathcal{X})$, and yields the second alternative of Equation (60).

□

We conclude by providing a weaker, but significantly shorter version of the second alternative that will be useful for clarity in some later proofs.

**Lemma G.4** Let $M \subseteq \mathcal{T}$ non-empty. Let $I, \mathcal{X}, \mathcal{Y}_k$ as in Definition G.2. Then

$$ O(\operatorname{span}(\mathcal{X}\mathcal{Y}_k)) \setminus O(\operatorname{span}(\mathcal{X})) = M \quad \forall k \in I \quad \Rightarrow \quad \bigcap_k O(\mathcal{Y}_k) \neq \emptyset. \qquad (67) $$

**Proof:** First, we see that $\operatorname{span}(\mathcal{X}\mathcal{Y}_i) \setminus \mathcal{X} \subseteq \mathcal{Y}_i$. As this carries over to their respective locations, we obtain

$$ \emptyset \neq M = O(\operatorname{span}(\mathcal{X}\mathcal{Y}_k)) \setminus O(\operatorname{span}(\mathcal{X})) \quad \forall k \in I \qquad (68) $$

$$ \iff \emptyset \neq M = O(\operatorname{span}(\mathcal{X}\mathcal{Y}_k)) \cap O(\mathcal{Y}_k) \setminus O(\operatorname{span}(\mathcal{X})) \quad \forall k \in I \qquad (69) $$

$$ \Rightarrow \emptyset \neq M \subseteq O(\operatorname{span}(\mathcal{X}\mathcal{Y}_k)) \cap O(\mathcal{Y}_k) \quad \forall k \in I \qquad (70) $$

$$ \Rightarrow \emptyset \neq M \subseteq \bigcap_k O(\operatorname{span}(\mathcal{X}\mathcal{Y}_k)) \cap O(\mathcal{Y}_k) \qquad (71) $$

$$ \Rightarrow \emptyset \neq \bigcap_k O(\mathcal{Y}_k), \qquad (72) $$

where the last implication holds as intersecting larger sets always yields a (potentially improper) superset.

□

## G.3 Proofs for Section 6

To prove Theorem 6.3, we require some auxiliary lemmas, which we first prove before stating and proving the main theorem.

**Lemma G.5** Let $A, B \subset S$. Then

$$ \bar{\mathcal{F}}_s(A) \cap \bar{\mathcal{F}}_s(B \setminus e_B) \subseteq \bar{\mathcal{F}}(e_B) \quad \forall e_B \in B \quad \Rightarrow \quad \bar{\mathcal{F}}_s(AB) = \bar{\mathcal{F}}_s(AB \setminus e_B) \quad \forall e_B \in B. \qquad (73) $$

**Proof:** Let $e_B^i := e_B$. We can transform

$$ \bar{\mathcal{F}}_s(A) \cap \bar{\mathcal{F}}_s(B \setminus e_B^i) \subseteq \bar{\mathcal{F}}(e_B^i) \qquad (74) $$

$$ \Rightarrow \bar{\mathcal{F}}_s(A) \cap \bar{\mathcal{F}}_s(B \setminus e_B^i) \subseteq \bar{\mathcal{F}}_s(A) \cap \bar{\mathcal{F}}(e_B^i) \qquad \forall e_B^i \in B \qquad (75) $$

$$ \Rightarrow \bar{\mathcal{F}}_s(A) \cap \bar{\mathcal{F}}_s(B \setminus e_B^i) \subseteq \bar{\mathcal{F}}_s(A) \cap \bar{\mathcal{F}}(e_B^i) \cap \bar{\mathcal{F}}_s(B \setminus \{e_B^i, e_B^j\}) \qquad \forall e_B^i, e_B^j \in B \qquad (76) $$

$$ \iff \bar{\mathcal{F}}_s(A) \cap \bar{\mathcal{F}}_s(B \setminus e_B^i) \subseteq \bar{\mathcal{F}}_s(A) \cap \bar{\mathcal{F}}_s(B \setminus e_B^j) \qquad \forall e_B^i, e_B^j \in B. \qquad (77) $$
Combining these for all $e_B$, we get

$$ \bar{\mathcal{F}}_s(\mathcal{A}) \cap \bar{\mathcal{F}}_s(\mathcal{B} \setminus e_B^i) = \bar{\mathcal{F}}_s(\mathcal{A}) \cap \bar{\mathcal{F}}_s(\mathcal{B} \setminus e_B^j) \quad \forall e_B^i, e_B^j \in \mathcal{B} \quad (78) $$

$$ \implies \bar{\mathcal{F}}_s(\mathcal{AB}) = \bar{\mathcal{F}}_s(\mathcal{AB} \setminus e_B) \quad \forall e_B \in \mathcal{B}. \quad (79) $$

This is precisely the claim.

□

**Lemma G.6** Let A be a set of unconditional affects relations and $A' \subseteq A$ consist of all affects relations in A that are Irred$_3$. Then for any non-degenerate embedding $\mathcal{E}$ into an arbitrary space-time $\mathcal{T}$ satisfying compat we have

$$ (X \models Y | \mathrm{do}(Z)) \in A' \implies \bar{\mathcal{F}}_s(\mathcal{YX}) \cap \bar{\mathcal{F}}_s(\mathcal{Z} \setminus e_Z) \subseteq \bar{\mathcal{F}}(e_Z) \quad \forall e_Z \in \mathcal{Z}. \quad (80) $$

**Proof:** As $X \models Y$ | do($Z$) satisfies Irred$_3$, we have for $\tilde{Z} := Z \setminus e_Z$

$$ \forall e_Z \in Z: \quad e_Z \models Y | \mathrm{do}(\tilde{Z}) \lor e_Z \models Y | \mathrm{do}(\tilde{Z}X), \quad (81) $$

either of which is Irred$_1$, if it holds. Therefore, by compat, we get

$$ \bar{\mathcal{F}}_s(\mathcal{Y}) \cap \bar{\mathcal{F}}_s(\tilde{Z}) \subseteq \bar{\mathcal{F}}(e_Z) \lor \bar{\mathcal{F}}_s(\mathcal{Y}) \cap \bar{\mathcal{F}}_s(\mathcal{X}\tilde{Z}) \subseteq \bar{\mathcal{F}}(e_Z) \quad (82) $$

$$ \implies \bar{\mathcal{F}}_s(\mathcal{YX}) \cap \bar{\mathcal{F}}_s(\tilde{Z}) \subseteq \bar{\mathcal{F}}(e_Z) \quad (83) $$

for any $e_Z$.

□

With these three auxiliary lemmas in place, we can continue to prove the theorem.

**Lemma G.7** Let A be a set of unconditional affects relations and $A' \subseteq A$ consist of all affects relations in A that are Irred$_3$. Then for any non-degenerate embedding $\mathcal{E}$ into a conical space-time $\mathcal{T}$ satisfying compat, we have

$$ (X \models Y | \mathrm{do}(Z)) \in A' \implies \bar{\mathcal{F}}_s(\mathcal{Y}) \cap \bar{\mathcal{F}}_s(\mathcal{X}) \subseteq \bar{\mathcal{F}}_s(\mathcal{Z}). \quad (84) $$

**Proof:** By conjoining Lemma G.6 with Lemma G.5 for $\mathcal{A} = \mathcal{YX}$ and $\mathcal{B} = \mathcal{Z}$, for any affects relation $X \models Y$ | do($Z$) irreducible in the third argument, we get

$$ \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}\mathcal{Z}) = \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}\mathcal{Z} \setminus e_Z) \quad \forall e_Z \in \mathcal{Z}. \quad (85) $$

Due to conicality, Lemma G.3 implies that $\mathcal{T}$ satisfies location symmetry. By using this property on the chain of equalities (setting $\mathcal{X}$ in Equation (60) to $\mathcal{X}\mathcal{Y}$, and $\mathcal{Y}_i$ to $\mathcal{Z}_i = \mathcal{Z} \setminus e_Z^i$) as well as the transformation of Lemma G.4, we arrive at

$$ \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Y}) \subseteq \bar{\mathcal{F}}_s(\mathcal{Z}) \lor \bigcap_i \mathrm{O}(\mathcal{Z}_i) \neq \emptyset, \quad (86) $$

where $\mathcal{Z}_i := \mathcal{Z} \setminus e_i$.

For the second alternative, beware that the different $\mathcal{Z}_i$ are not disjoint. Nonetheless, $\bigcap_i \mathcal{Z}_i = \emptyset$. Therefore, multiple ORVs must share the same location. As this corresponds to a degenerate embedding, we arrive at the claim.

□

**Theorem 6.3** Let A be a set of unconditional affects relations and $A' \subseteq A$ consist of all affects relations in A that are both Irred$_1$ and Irred$_3$. Then for any non-degenerate embedding $\mathcal{E}$ into a conical space-time $\mathcal{T}$ satisfying compat, we have

$$ (X \models Y | \mathrm{do}(Z)) \in A' \implies \bar{\mathcal{F}}_s(\mathcal{Y}) \subseteq \bar{\mathcal{F}}_s(\mathcal{X}) \cap \bar{\mathcal{F}}_s(\mathcal{Z}). \quad (22) $$
**Proof:** Due to Lemma G.7, compat implies the following for an Irred₁ and Irred₃ affects relation

$$ (X \models Y | \mathrm{do}(Z)) \in A' \implies \bar{\mathcal{F}}_s(\mathcal{Y}) \cap \bar{\mathcal{F}}_s(\mathcal{X}) \subseteq \bar{\mathcal{F}}_s(\mathcal{Z}). \qquad (87) $$

Conjoining this with compat for this affects relation (which implies $\bar{\mathcal{F}}_s(\mathcal{Y}) \cap \bar{\mathcal{F}}_s(\mathcal{Z}) \subseteq \bar{\mathcal{F}}_s(\mathcal{X})$) and using elementary set theory yields

$$ \bar{\mathcal{F}}_s(\mathcal{Y}) \cap \bar{\mathcal{F}}_s(\mathcal{Z}) \subseteq \bar{\mathcal{F}}_s(\mathcal{X}) \cap \bar{\mathcal{F}}_s(\mathcal{Y}) \wedge \bar{\mathcal{F}}_s(\mathcal{Y}) \cap \bar{\mathcal{F}}_s(\mathcal{X}) \subseteq \bar{\mathcal{F}}_s(\mathcal{Z}) \cap \bar{\mathcal{F}}_s(\mathcal{Y}) \qquad (88) $$

$$ \implies \bar{\mathcal{F}}_s(\mathcal{Y}) \cap \bar{\mathcal{F}}_s(\mathcal{X}) = \bar{\mathcal{F}}_s(\mathcal{Y}) \cap \bar{\mathcal{F}}_s(\mathcal{Z}) \qquad (89) $$

Due to conicality, Lemma G.3 implies that $\mathcal{T}$ satisfies location symmetry. Using location symmetry, we arrive at

$$ \bar{\mathcal{F}}_s(\mathcal{Y}) \subseteq \bar{\mathcal{F}}_s(\mathcal{X}\mathcal{Z}) \vee \mathcal{O}(\mathcal{X}) \cap \mathcal{O}(\mathcal{Z}) \neq \emptyset. \qquad (90) $$

As the second option would imply the embedding to be degenerate, the claim, corresponding to the first option, follows. $\square$

**Theorem 6.4** Let $\mathcal{A}$ be a set of unconditional affects relations arising from a causal model not yielding any affects relations with $Clus_3$ and let $A' \subseteq \mathcal{A}$ consist of all affects relations in $\mathcal{A}$ that are both Irred₁ and Irred₃. Then compatibility of $\mathcal{A}$ with an embedding $\mathcal{E}$ in any space-time $\mathcal{T}$ implies Equation (22).

**Proof:** Every affects relation $X \models Y|\mathrm{do}(Z)$ in $A'$ is Irred₁ and Irred₃. By Lemma D.6, due to the absence of affects relations with $Clus_3$, there exists $e_X^1 \in X$ such that $e_X^1 \models Y$. Therefore, for any compatible embedding $\mathcal{E}$, we have $\bar{\mathcal{F}}_s(\mathcal{Y}) \subseteq \bar{\mathcal{F}}(e_X^1)$. However, due to $X \models Y|\mathrm{do}(Z)$ being Irred₁, we also have $X \setminus e_X^1 \models Y | \mathrm{do}(Ze_X^1)$, which again by Lemma D.6 implies that there is $e_X^2 \in X \setminus e_X^1$ such that $e_X^2 \models Y$. Repeating this argument recursively, we find that any compatible embedding $\mathcal{E}$ of such affects relations in a space-time will satisfy $\bar{\mathcal{F}}_s(\mathcal{Y}) \subseteq \bar{\mathcal{F}}(e_X) \forall e_X \in \mathcal{X}$, or in short, $\bar{\mathcal{F}}_s(\mathcal{Y}) \subseteq \bar{\mathcal{F}}_s(\mathcal{X})$.

Next we use the fact that $X \models Y|\mathrm{do}(Z)$ is Irred₃, which gives $s_Z \models Y | \mathrm{do}(\tilde{s}_Z X)$ or $s_Z \models Y | \mathrm{do}(\tilde{s}_Z)$ for all choices of $s_Z \subseteq Z$. Applying Lemma D.6, both these alternatives yield $\exists e_Z^1 \in Z \ \mathrm{s.t.} \ e_Z^1 \models Y$ and $\bar{\mathcal{F}}_s(\mathcal{Y}) \subseteq \bar{\mathcal{F}}(e_Z^1)$. Moreover, $X \models Y|\mathrm{do}(Z)$ being Irred₃ implies in particular that $Z \setminus e_Z^1 \models Y | \mathrm{do}(e_Z^1 X)$ or $Z \setminus e_Z^1 \models Y | \mathrm{do}(e_Z^1)$, both of which imply by Lemma D.6 that $\exists e_Z^2 \in Z \setminus e_Z^1 \ \mathrm{s.t.} \ e_Z^2 \models Y$. Repeating this procedure recursively, we obtain that for any compatible embedding $\mathcal{E}$ of these affects relations, we have $\bar{\mathcal{F}}_s(\mathcal{Y}) \subseteq \bar{\mathcal{F}}(e_Z) \forall e_Z \in \mathcal{Z}$, or in short, $\bar{\mathcal{F}}_s(\mathcal{Y}) \subseteq \bar{\mathcal{F}}_s(\mathcal{Z})$.

Combining both results, we recover the claim. $\square$

## G.4 Proofs for Appendix E

**Lemma E.4** For a causal model over a set $S$ of RVs, where $X, Y, Z, W \subset S$ disjoint, $X \not\models Y|\mathrm{do}(Z)$, $W \implies (X \perp Y|ZW)_{\mathcal{G}_{\mathrm{do}}(XZ)}$.

**Proof:** Suppose by contradiction that $X \not\models Y|\{\mathrm{do}(Z), W\}$ (which amounts to $P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW) = P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|ZW)$) and $(X \not\perp Y|ZW)_{\mathcal{G}_{\mathrm{do}(XZ)}}$. The latter is equivalent to saying that there exist values $z$ of $Z$, $w$ of $W$ and distinct values $x$ and $x'$ of $X$ such that $P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|X = x, Z = z, W = w) \neq P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|X = x', Z = z, W = w)$. However, this implies that $P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW)$ does have a non-trivial dependence on $X$ and contradicts $P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW) = P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|ZW)$ (which would imply that it is equal to an $X$-independent quantity). This proves the result. $\square$

**Lemma E.13** If $X \models Y|\mathrm{do}(Z)$, $W$ is a Red₄ affects relation, then there exists $\tilde{s}_W \subseteq W$ such that $X \models Y|\mathrm{do}(Z)$, $\tilde{s}_W$ holds.

**Proof:** Suppose $X \models Y|\mathrm{do}(Z)$, $W$ holds and is a Red₄ affects relation. Then writing out this affects relation along with the two conditions implied by the reducibility (while recalling that $s_W \cup \tilde{s}_W = W$), we have

$$ \begin{align} P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW) & \neq P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|ZW) \nonumber \\ P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Ys_W|XZ\tilde{s}_W) &= P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZ\tilde{s}_W)P_{\mathcal{G}_{\mathrm{do}(XZ)}}(s_W|XZ\tilde{s}_W) \tag{91} \\ P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Ys_W|Z\tilde{s}_W) &= P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|Z\tilde{s}_W)P_{\mathcal{G}_{\mathrm{do}(XZ)}}(s_W|Z\tilde{s}_W). \nonumber \end{align} $$
The last two equalities are equivalent to the following.

$$ 
\begin{align} 
P_{\mathcal{G}_{\text{do}(XZ)}}(Y|XZW) &= P_{\mathcal{G}_{\text{do}(XZ)}}(Y|XZ\tilde{s}_W) \tag{92} \\ 
P_{\mathcal{G}_{\text{do}(XZ)}}(Y|ZW) &= P_{\mathcal{G}_{\text{do}(XZ)}}(Y|Z\tilde{s}_W) \nonumber 
\end{align} 
$$

Plugging these back into the first expression of Equation (91) we obtain $P_{\mathcal{G}_{\text{do}(XZ)}}(Y|XZ\tilde{s}_W) \neq P_{\mathcal{G}_{\text{do}(Z)}}(Y|Z\tilde{s}_W)$, which is equivalent to $X \models Y|\text{do}(Z), \tilde{s}_W$. $\square$

**Theorem E.18** *Clustering implies irreducibility* For any affects relation $X \models Y|\text{do}(Z), W$, $\text{Clus}_i \Rightarrow \text{Irred}_i$ for all $i \in \{1, 2, 3, 4\}$.

**Proof:** For $i \in \{1, 2, 3\}$, the proofs are entirely analogous to that of Theorem 4.11 and will not be repeated. We carry out the proof for $i = 4$ below.

Again, start by assuming that $X \models Y|\text{do}(Z), W$ holds and satisfies $\text{Clus}_4$ and $\text{Red}_4$. From Definition E.17 and Definition E.9 this implies the following conditions (see also Equation (92)).

$$ 
\begin{align} 
P_{\mathcal{G}_{\text{do}(XZ)}}(Y|XZ\tilde{s}_W) &= P_{\mathcal{G}_{\text{do}(Z)}}(Y|Z\tilde{s}_W), \quad \forall \tilde{s}_W \subsetneq W, \nonumber \\ 
P_{\mathcal{G}_{\text{do}(XZ)}}(Y|XZW) &= P_{\mathcal{G}_{\text{do}(XZ)}}(Y|XZ\tilde{s}_W) \quad \exists \tilde{s}_W \subseteq W, \tag{93} \\ 
P_{\mathcal{G}_{\text{do}(XZ)}}(Y|ZW) &= P_{\mathcal{G}_{\text{do}(XZ)}}(Y|Z\tilde{s}_W) \quad \exists \tilde{s}_W \supseteq W. \nonumber 
\end{align} 
$$

These imply that $P_{\mathcal{G}_{\text{do}(XZ)}}(Y|XZW) = P_{\mathcal{G}_{\text{do}(Z)}}(Y|ZW)$, which is equivalent to $X \nvDash Y|\text{do}(Z), W$. This contradicts our initial assumption that $X \models Y|\text{do}(Z), W$ and establishes the result. $\square$

**Theorem E.19** *Clustering implies fine-tuning (conditional version)* Any affects relation $X \models Y|\text{do}(Z), W$ that satisfies $\text{Clus}_i$ for $i \in \{1, 2, 3, 4\}$ and generated in a causal graph $\mathcal{G}$ where $X$ is not a cause of $W$, necessarily arises from a fine-tuned causal model on $\mathcal{G}$.

**Proof:** The proofs for $i \in \{1, 2, 3\}$ are entirely analogous as those of Theorem 4.12 whenever we are given that $X$ is not a cause of $W$. This is because the main step of the proof is to infer from $X \models Y|\text{do}(Z)$ that $(X \nvDash^d Y|Z)_{\mathcal{G}_{\text{do}(XZ)}}$, and the analogous inference holds for conditional affects relations when $X$ is not a cause of $W$, due to Lemma E.5 i.e., $X \models Y|\{\text{do}(Z), W\}$ implies $(X \nvDash^d Y|ZW)_{\mathcal{G}_{\text{do}(XZ)}}$ whenever $X$ is not a cause of $W$. We therefore do not repeat the proof.

The proof for $i = 4$ is also analogous, but we repeat it for completeness. Suppose that $X \models Y|\text{do}(Z), W$ holds and is a $\text{Clus}_4$ affects relation. This implies in particular that for all $e_W \in W$, $X \nvDash Y|\text{do}(Z), e_W$. Writing this out, we have

$$ 
\begin{align} 
P_{\mathcal{G}_{\text{do}(XZ)}}(Y|XZW) &\neq P_{\mathcal{G}_{\text{do}(Z)}}(Y|ZW) \tag{94} \\ 
P_{\mathcal{G}_{\text{do}(XZ)}}(Y|XZe_W) &= P_{\mathcal{G}_{\text{do}(Z)}}(Y|Ze_W), \quad \forall e_W \in W. \nonumber 
\end{align} 
$$

Further, using Lemma E.4, the non-affects relations $X \nvDash Y|\text{do}(Z), e_W$ imply the conditional independences $(X \perp Y|Z, e_W)_{\mathcal{G}_{\text{do}(XZ)}}$. Now consider the corresponding d-separation for each $e_W \in W$, we can either have $(X \perp^d Y|Z, e_W)_{\mathcal{G}_{\text{do}(XZ)}}$ or $(X \nvDash^d Y|Z, e_W)_{\mathcal{G}_{\text{do}(XZ)}}$. In the latter case we could have a d-connection with a corresponding conditional independence, which would make the model fine-tuned. So we consider the case where $(X \perp^d Y|Z, e_W)_{\mathcal{G}_{\text{do}(XZ)}}$ holds for all $e_W \in W$ (for all other cases the above argument establishes the result about fine-tuning). However, by assumption that $X$ is not a cause of $W$, we have $X(W) = W$ and we know from Lemma E.5 that $(X \perp^d Y|Z, e_W)_{\mathcal{G}_{\text{do}(XZ)}}$ implies $X \nvDash Y|\{\text{do}(Z), W\}$, which contradicts our initial assumption and established the result. $\square$

**Corollary E.20** For any unconditional affects relation $X \models Y|\text{do}(Z)$ which is $\text{Clus}_2$ and has $|Y| \ge 3$, $X \models s_Y|\text{do}(Z), \tilde{s}_Y$ is $\text{Clus}_4$ for each partition $s_Y \tilde{s}_Y = Y$ with $|\tilde{s}_Y| \ge 2$.

**Proof:** Any $X \models Y|\text{do}(Z)$ which is $\text{Clus}_2$, by Lemma E.2.3 is equivalent to $X \models s_Y|\text{do}(Z), \tilde{s}_Y$ for each partition $s_Y \tilde{s}_Y = Y$.

If for the case of $\tilde{s}_Y \ge 2$ – which is only possible for $Y \ge 2$ – this affects relation were not $\text{Clus}_4$, there would exist $X \models s_Y|\text{do}(Z), \tilde{s}'_Y$ with $\tilde{s}'_Y \subsetneq \tilde{s}_Y$, implying $X \models s_Y \tilde{s}'_Y|\text{do}(Z)$ by Lemma E.2.1. This is in contradiction to the original affects relation being $\text{Clus}_2$ and proves the claim. $\square$