A Characterization of Lewisian Causal Models

Abstract

An important component in the interventionist account of causal explanation is an interpretation of counterfactual conditionals as statements about consequences of hypothetical interventions. The interpretation receives a formal treatment in the framework of functional causal models. In Judea Pearl’s influential formulation, functional causal models are assumed to satisfy a “unique-solution” property; this class of Pearlian causal models includes the ones called recursive. Joseph Halpern showed that every recursive causal model is Lewisian, in the sense that from the causal model one can construct a possible worlds model in David Lewis’s well-known semantics that satisfies the exact same formulas in a certain language. Moreover, he demonstrated that some Pearlian (non-recursive) models are not Lewisian in this sense. This raises the question regarding the exact contour of Lewisian causal models. In this paper, we provide a characterization of the class of Lewisian causal models and a complete axiomatization with respect to this class. Our results have philosophically interesting consequences, two of which are especially worth noting. First, the class of Stalnakerian causal models, a subclass of Lewisian causal models, is precisely the class of Pearlian models that do not contain any cycle of counterfactual dependence (in a sense of counterfactual dependence akin to Lewis’s famous relation between distinct events). Second, a more natural class of causal models is actually a superclass of Lewisian causal models, the logic of which respects only weak centering rather than centering.

Appendix: Proofs of Theorems 1 and 2

Proof of Theorem 1

The “only if” direction is established by Lemmas 1–4. Here we focus on the “if” direction. Suppose that a causal model T over signature S is Solution-Determinate and Solution-Transitive in Cycles, and all its submodels are Solutionful and Solution-Conservative. We first construct a possible worlds model $M=\langle \varOmega , R,\pi \rangle$ over S and prove it is Lewisian. Let $\varOmega$ be the set of all possible value assignments to $\textbf{U}\cup \textbf{V}$. The definition of $\pi$ is obvious.

To define R, we use the following notations. Since T is Solution-Determinate and every submodel of T is Solutionful, we use $\textbf{s}(\textbf{u})$ to denote the unique solution to T relative to a context $\textbf{u}$. Let $\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})$ denote one solution $\textbf{v}$ to $T_{\textbf{X}=\textbf{x}}$ relative to context $\textbf{u}$. Let $\varOmega _{{\textbf{s}(\textbf{u})}}$ be the set of worlds of the form ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}$ where $\textbf{x}$ is the value configuration of $\textbf{X}$ and $\textbf{v}$ is one of the solutions of $T_{\textbf{X}=\textbf{x}}$ relative to context $\textbf{u}$.¹ For a world ${\textbf{s}(\textbf{u})}$, we define $\preceq ^0_{{\textbf{s}(\textbf{u})}}$ over $\varOmega _{{\textbf{s}(\textbf{u})}}$ as follows²: ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ iff ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ assigns $\textbf{x}$ to $\textbf{X}$. Let $\preceq ^1_{{\textbf{s}(\textbf{u})}}$ be the transitive closure of $\preceq ^0_{{\textbf{s}(\textbf{u})}}$. Then we inductively define $\preceq _{{\textbf{s}(\textbf{u})}}$ as below. If $\preceq ^i_{{\textbf{s}(\textbf{u})}}$ is not yet strongly-connected, let $w_{ia}$ and $w_{ib}$ be two incomparable worlds, and let $A_i=\{w\in \varOmega _{{\textbf{s}(\textbf{u})}}\mid w\preceq ^i_{{\textbf{s}(\textbf{u})}}w_{ia}\}$ and $B_i=\{w\in \varOmega _{{\textbf{s}(\textbf{u})}}\mid w_{ib}\preceq ^i_{{\textbf{s}(\textbf{u})}}w\}$. Then define $\preceq ^{i+1}_{{\textbf{s}(\textbf{u})}}:=\preceq ^i_{{\textbf{s}(\textbf{u})}}\cup (A_i\times B_i)$. Let $\preceq _{{\textbf{s}(\textbf{u})}}$ be the first in this process that is strongly-connected.

We assert that R associates with each ${\textbf{s}(\textbf{u})}$ a total preorder $\preceq _{{\textbf{s}(\textbf{u})}}$ over $\varOmega _{{\textbf{s}(\textbf{u})}}$ such that ${\textbf{s}(\textbf{u})}\in \varOmega _{{\textbf{s}(\textbf{u})}}$ and ${\textbf{s}(\textbf{u})}\prec _{{\textbf{s}(\textbf{u})}} v$ for every $v\not ={\textbf{s}(\textbf{u})}\in \varOmega _{{\textbf{s}(\textbf{u})}}$. $\preceq _{{\textbf{s}(\textbf{u})}}$ is strongly-connected by its construction. We now show that $\preceq ^i_{{\textbf{s}(\textbf{u})}}$ is transitive by induction. The base case $\preceq ^1_{{\textbf{s}(\textbf{u})}}$ is obvious. Assume that $\preceq ^k_{{\textbf{s}(\textbf{u})}}$ is transitive, we show that so is $\preceq ^{k+1}_{{\textbf{s}(\textbf{u})}}$. Given any ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}, {\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}, {\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})}\in \varOmega _{{\textbf{s}(\textbf{u})}}$, suppose ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq ^{k+1}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ and ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\preceq ^{k+1}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})}$. There are four cases to consider:

• Case 1: ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ and ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})}$. By the inductive hypothesis, ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})}$. So ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq ^{k+1}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})}$.

• Case 2: ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ and $({\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}, {\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})})\in A_{k}\times B_{k}$. There are two $\preceq ^{k}_{{\textbf{s}(\textbf{u})}}$-incomparable worlds $w_{ka}$ and $w_{kb}$ such that ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}w_{ka}$ and $w_{kb}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})}$, thus ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}w_{ka}$. It follows that $({\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})},{\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})})\in A_{k}\times B_{k}$, and so ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq ^{k+1}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})}$.

• Case 3: $({\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})},{\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})})\in A_{k}\times B_{k}$ and ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ $\preceq ^{k}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})}$. Similar to Case 2.

• Case 4: $({\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})},{\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})})\in A_{k}\times B_{k}$ and $({\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}, {\textbf{s}(\textbf{u},\textbf{z}\mid \mathbf {v''})})\in A_{k}\times B_{k}$. Then $w_{kb}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ and ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}w_{ka}$, and so $w_{kb}\preceq ^{k}_{{\textbf{s}(\textbf{u})}}w_{ka}$, which contradicts the supposition that $w_{ka}$ and $w_{kb}$ are $\preceq ^{k}_{{\textbf{s}(\textbf{u})}}$-incomparable.

Therefore, $\preceq ^{k+1}_{{\textbf{s}(\textbf{u})}}$ is also transitive. By construction, so is $\preceq _{{\textbf{s}(\textbf{u})}}$.

Next, we prove the other Lewisian constraints on R. Obviously, ${\textbf{s}(\textbf{u})}\in \varOmega _{{\textbf{s}(\textbf{u})}}$, and for every $v\in \varOmega _{{\textbf{s}(\textbf{u})}}$, ${\textbf{s}(\textbf{u})}\preceq ^0_{{\textbf{s}(\textbf{u})}} v$, so ${\textbf{s}(\textbf{u})}\preceq _{{\textbf{s}(\textbf{u})}} v$. Suppose that there is some ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}$ such that ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\not ={\textbf{s}(\textbf{u})}$ and ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq _{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u})}$. That means either there exist ${\textbf{s}(\textbf{u},\mathbf {x_1}\mid \mathbf {v_1})}$,..., ${\textbf{s}(\textbf{u},\mathbf {x_m}\mid \mathbf {v_m})}$ such that ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\mathbf {x_1}\mid \mathbf {v_1})}\preceq ^0_{{\textbf{s}(\textbf{u})}}...\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\mathbf {x_m}\mid \mathbf {v_m})}\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u})}$ or there are two $\preceq ^0_{{\textbf{s}(\textbf{u})}}$-incomparable worlds in the chain from ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}$ to ${\textbf{s}(\textbf{u})}$. For the former case, as ${\textbf{s}(\textbf{u})}$ assigns $\mathbf {x_m}$ to $\mathbf {X_m}$ and T is Solution-Conservative, $\mathbf {v_m}$ is the solution to T relative to $\textbf{u}$ due to Solution-Determinateness. Thus ${\textbf{s}(\textbf{u},\mathbf {x_m}\mid \mathbf {v_m})}={\textbf{s}(\textbf{u})}$. Repeat the same argument, we would have ${\textbf{s}(\textbf{u},\textbf{x}\mid \textbf{v})}={\textbf{s}(\textbf{u})}$, a contradiction. For the latter case, for any world ${\textbf{s}(\textbf{u},\mathbf {x_k}\mid \mathbf {v_k})}$ such that ${\textbf{s}(\textbf{u},\mathbf {x_k}\mid \mathbf {v_k})}\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u})}$, ${\textbf{s}(\textbf{u},\mathbf {x_k}\mid \mathbf {v_k})}={\textbf{s}(\textbf{u})}$. Assume that the pair of $\preceq ^0_{{\textbf{s}(\textbf{u})}}$-incomparable worlds that are closest to ${\textbf{s}(\textbf{u})}$ is $s\preceq _{{\textbf{s}(\textbf{u})}}t$ with $t={\textbf{s}(\textbf{u})}$, then s and t would be $\preceq ^0_{{\textbf{s}(\textbf{u})}}$-comparable, also a contradiction.

So the constructed possible worlds model is Lewisian as desired. What remains to be shown is that $T,(\textbf{u},\textbf{v})\models \psi$ iff $M,\mu (\textbf{u},\textbf{v})\models \psi$ for every $\psi \in \mathcal {L}(S)$, where $\mu$ assigns ${\textbf{s}(\textbf{u})}(\in \varOmega )$ to $(\textbf{u},\textbf{v})(\in Sol(T))$. We prove this claim by induction on the structure of $\psi$.

At first we show that for any $\mathcal {L}(S)$ formula $\beta$ that does not contain ‘$\mathrel {\mathop \Box }\mathrel {}\rightarrow$’, $T_{\textbf{X}=\textbf{x}},(\mathbf {u'},\mathbf {v'})\models \beta$ iff $M, {\textbf{s}(\mathbf {u'},\textbf{x}\mid \mathbf {v'})}\models \beta$. For the case of $Y=y$, suppose $T_{\textbf{X}=\textbf{x}},(\mathbf {u'},\mathbf {v'})\models Y=y$, then $\mathbf {v'}$ assigns y to Y. Therefore, we have $M, {\textbf{s}(\mathbf {u'},\textbf{x}\mid \mathbf {v'})}\models Y=y$. Conversely, if $M, {\textbf{s}(\mathbf {u'},\textbf{x}\mid \mathbf {v'})}\models Y=y$, then $Y=y$ is consistent with $\mathbf {v'}$. Thus, $T_{\textbf{X}=\textbf{x}},(\mathbf {u'},\mathbf {v'})\models Y=y$. The Boolean cases are routine.

From the above result, we have $T,(\textbf{u},\textbf{v})\models Y=y$ iff $M,\mu (\textbf{u},\textbf{v})\models Y=y$. We then show that $T,(\textbf{u},\textbf{v})\models \textbf{X}=\textbf{x}\mathrel {\mathop \Box }\mathrel {}\rightarrow \beta$ iff $M,{\textbf{s}(\textbf{u})}\models \textbf{X}=\textbf{x}\mathrel {\mathop \Box }\mathrel {}\rightarrow \beta$. The other cases are straightforward by inductive hypothesis.

From left to right, if $T,(\textbf{u},\textbf{v})\models \textbf{X}=\textbf{x}\mathrel {\mathop \Box }\mathrel {}\rightarrow \beta$, then for all $(\textbf{u},\mathbf {v^i})\in Sol(T_{\textbf{X}=\textbf{x}})$, $T_{\textbf{X}=\textbf{x}},(\textbf{u},\mathbf {v^i})\models \beta$. As $\beta$ is a formula that does not contain ‘$\mathrel {\mathop \Box }\mathrel {}\rightarrow$’, we have $M, {\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}\models \beta$ for $(\textbf{u},\mathbf {v^i})$. Since ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}\in \varOmega _{{\textbf{s}(\textbf{u})}}$ and ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}\preceq ^0_{{\textbf{s}(\textbf{u})}}v$ for any $v\in \varOmega _{{\textbf{s}(\textbf{u})}}$ such that v assigns $\textbf{x}$ to $\textbf{X}$, then ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}\preceq _{{\textbf{s}(\textbf{u})}}v$, so ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}\in C_M({\textbf{s}(\textbf{u})},\textbf{X}=\textbf{x})$. Suppose that there is a world ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\in \varOmega _{{\textbf{s}(\textbf{u})}}$ such that ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\not ={\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$ where ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$ is the corresponding world of $(\textbf{u},\mathbf {v^i})\in Sol(T_{\textbf{X}=\textbf{x}})$, and ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\in C_M({\textbf{s}(\textbf{u})},\textbf{X}=\textbf{x})$. Then ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ assigns $\textbf{x}$ to $\textbf{X}$ and ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\preceq _{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$. As ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ according to the definition of $\preceq ^0_{{\textbf{s}(\textbf{u})}}$, they cannot be $\preceq ^1_{{\textbf{s}(\textbf{u})}}$-incomparable and then cannot be $\preceq ^j_{{\textbf{s}(\textbf{u})}}$-incomparable. So, based on the definition of $\preceq _{{\textbf{s}(\textbf{u})}}$, it can only be the case that ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\preceq ^1_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$ from the fact that ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\preceq _{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$. Thus there exist ${\textbf{s}(\textbf{u},\mathbf {z_1}\mid \mathbf {v_1})}$,..., ${\textbf{s}(\textbf{u},\mathbf {z_k}\mid \mathbf {v_k})}$ such that ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\mathbf {z_1}\mid \mathbf {v_1})}\preceq ^0_{{\textbf{s}(\textbf{u})}}...\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\mathbf {z_k}\mid \mathbf {v_k})}\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$. Together with the fact that ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}\preceq ^0_{{\textbf{s}(\textbf{u})}}{\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$, that means $T_{\textbf{X}=\textbf{x}}$ relative to $\textbf{u}$ has a solution $\mathbf {v^i}$ that is consistent with $\mathbf {Z_k}=\mathbf {z_k}$,..., $T_{\textbf{Y}=\textbf{y}}$ relative to $\textbf{u}$ has a solution that is consistent with $\textbf{X}=\textbf{x}$. Since T is Solution-Transitive in Cycles, $T_{\textbf{X}=\textbf{x}}$ has a solution relative to $\textbf{u}$ that is consistent with $\textbf{Y}=\textbf{y}$. Suppose that the solution is $\mathbf {v^k}$, ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^k})}$ assigns $\textbf{y}$ to $\textbf{Y}$. As ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}$ assigns $\textbf{x}$ to $\textbf{X}$, the intervention $\textbf{X}=\textbf{x}$ and $\textbf{Y}=\textbf{y}$ do not contradict with each other. We have the world ${\textbf{s}(\textbf{u},\textbf{x},\textbf{y}\mid \mathbf {v'})}$ ($={\textbf{s}(\textbf{u},\textbf{y},\textbf{x}\mid \mathbf {v'})}$). For every submodel of T is Solution-Conservative, $\mathbf {v'}$ is one of the solutions to $T_{\textbf{X}=\textbf{x}}$ relative to $\textbf{u}$, but ${\textbf{s}(\textbf{u},\textbf{y}\mid \mathbf {v'})}\not ={\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$, contradiction. Therefore $C_M({\textbf{s}(\textbf{u})},\textbf{X}=\textbf{x})$ only has the worlds of the form ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$. Then we have that for all $v\in C_M({\textbf{s}(\textbf{u})},\textbf{X}=\textbf{x})$, $M,v\models \beta$, so $M,{\textbf{s}(\textbf{u})}\models \textbf{X}=\textbf{x}\mathrel {\mathop \Box }\mathrel {}\rightarrow \beta$.

From right to left, suppose $M,{\textbf{s}(\textbf{u})}\models \textbf{X}=\textbf{x}\mathrel {\mathop \Box }\mathrel {}\rightarrow \beta$. As we have proved that the elements in $C_M({\textbf{s}(\textbf{u})},\textbf{X}=\textbf{x})$ are the worlds of the form ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$, then for each ${\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}$, $M, {\textbf{s}(\textbf{u},\textbf{x}\mid \mathbf {v^i})}\models \beta$. It follows that $T_{\textbf{X}=\textbf{x}}, (\textbf{u},\mathbf {v^i})\models \beta$. Then for all $(\textbf{u},\mathbf {v^i})\in Sol(T_{\textbf{X}=\textbf{x}})$, $T_{\textbf{X}=\textbf{x}}, (\textbf{u},\mathbf {v^i})\models \beta$. Hence, $T,(\textbf{u},\textbf{v})\models \textbf{X}=\textbf{x}\mathrel {\mathop \Box }\mathrel {}\rightarrow \beta$.

Proof of Theorem 2

The soundness of $AX_\text {L}(S)$ is easy to verify and omitted to save space. To prove completeness, we follow the canonical model approach. That is, for an $AX_\text {L}(S)$-consistent formula $\varphi$, we construct a Lewisian causal model from a maximally $AX_\text {L}(S)$-consistent set containing $\varphi$ and prove $\varphi$ is satisfied in that model.

Given an $AX_\text {L}(S)$-consistent formula $\varphi$, we can extend it into a maximally $AX_\text {L}(S)$-consistent set C. According to the formulas in C, we define structural equations for the canonical model as follows: for any endogenous variable X, $f_X(\textbf{u},\textbf{y})=x$ iff $\textbf{Y}=\textbf{y}\diamond \!\!\rightarrow X=x\in C$ (well-defined by L1 and L2). As L3 is in C, we can determine a value configuration $\textbf{v}^c$ for $\textbf{V}$ which is not relative to any context. Then the canonical model is denoted as $T^c,(\textbf{u},\textbf{v}^c)$ for every context $\textbf{u}$.

Before we prove that $\varphi$ is true in $T^c,(\textbf{u},\textbf{v}^c)$, we shall show that $T^c$ is a solutionful causal model, that is, $T^c$ has at least one solution given every context $\textbf{u}$. Since L4 and $\textbf{V}=\mathbf {v^c}$ are in C, $\textbf{V}_{V_1}=\mathbf {v^c}\diamond \!\!\rightarrow V_1=v_1\wedge ...\wedge \textbf{V}_{V_n}=\mathbf {v^c}\diamond \!\!\rightarrow V_n=v_n\in C$. That means $f_{V_i}(\textbf{u},\mathbf {v^c}_{V_i})=v_i$, where $\mathbf {v^c}_{V_i}$ and $v_i$ are the respective values of $\textbf{V}_{V_i}$ and $V_i$ in $\mathbf {v^c}$ for any context $\textbf{u}$. Hence, given any context $\textbf{u}$, $\mathbf {v^c}$ can solve all the functions. That is to say, $T^c$ has one solution $\mathbf {v^c}$ relative to every context, and hence is a solutionful causal model.

Now we can prove that $\psi \in C$ iff $T^c,(\textbf{u},\mathbf {v^c})\models \psi$ for every context $\textbf{u}$ and every $\psi \in \mathcal {L}(S)$ by induction on the structure of $\psi$. If $\psi$ is $X=x$, suppose that $X=x\in C$, then $\textbf{V}=\mathbf {v^c}$ is consistent with $X=x$ and thus $T^c,(\textbf{u},\mathbf {v^c})\models X=x$. The other direction is similar. To prove the case of $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \beta$, we follow the strategy to reduce it into simpler formulas by applying some axioms and rules. Basically, due to L0 and RE, $\beta$ can be written as a disjunctive normal form. Then thanks to L5, $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \beta$ can be separated into several formulas $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \beta _i$ where $\beta _i$ is a conjunction of formulas of the form $Y=y$ or its negation. According to L3, we have $X\not =x\Leftrightarrow \bigvee _{x'\in \mathcal {R}(X){\setminus }\{x\}}X=x'\in C$. After applying the rule RE with L5 repeatedly, we can delete the negations in $\beta _i$ and reduce $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \beta$ into formulas of the form $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \textbf{Y}=\textbf{y}$. According to L6, to prove the case of $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \textbf{Y}=\textbf{y}$, it suffices to show that $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \textbf{W}=\textbf{w}\in C$ iff $T^c,(\textbf{u},\mathbf {v^c})\models \textbf{X}=\textbf{x}\diamond \!\!\rightarrow \textbf{W}=\textbf{w}$ in which $\textbf{W}=\textbf{V}{\setminus }\textbf{X}$.

We establish the above clause by induction on $|\textbf{V}{\setminus }\textbf{X}|$. When $|\textbf{V}{\setminus }\textbf{X}|=1$, suppose $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow W=w\in C$, we have $f_W(\textbf{u},\textbf{x})=w$ and thus there is a solution $(\textbf{x},w)$ to $T^c_{\textbf{X}=\textbf{x}}$ relative to $\textbf{u}$. Hence $T^c,(\textbf{u},\mathbf {v^c})\models \textbf{X}=\textbf{x}\diamond \!\!\rightarrow W=w$. The other direction is similar. Assume that the above clause holds for $|\textbf{V}{\setminus }\textbf{X}|<k$, we now show the case of $|\textbf{V}{\setminus }\textbf{X}|=k$. As $k\ge 2$, we can write $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \textbf{W}=\textbf{w}$ as $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow (W_1=w_1\wedge W_2=w_2\wedge \mathbf {W_3}=\mathbf {w_3})$. Suppose that $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow (W_1=w_1\wedge W_2=w_2\wedge \mathbf {W_3}=\mathbf {w_3})\in C$, then $(\textbf{X}=\textbf{x}\wedge W_1=w_1)\diamond \!\!\rightarrow ( W_2=w_2\wedge \mathbf {W_3}=\mathbf {w_3})\in C$ due to L7. According to the inductive hypothesis, $T^c,(\textbf{u},\mathbf {v^c})\models (\textbf{X}=\textbf{x}\wedge W_1=w_1)\diamond \!\!\rightarrow ( W_2=w_2\wedge \mathbf {W_3}=\mathbf {w_3})$. Similarly, $T^c,(\textbf{u},\mathbf {v^c})\models (\textbf{X}=\textbf{x}\wedge W_2=w_2)\diamond \!\!\rightarrow ( W_1=w_1\wedge \mathbf {W_3}=\mathbf {w_3})$. Since L8 holds in any causal model, we have $T^c,(\textbf{u},\mathbf {v^c})\models \textbf{X}=\textbf{x}\diamond \!\!\rightarrow ( W_1=w_1\wedge W_2=w_2\wedge \mathbf {W_3}=\mathbf {w_3})$. Conversely, if $T^c,(\textbf{u},\mathbf {v^c})\models \textbf{X}=\textbf{x}\diamond \!\!\rightarrow ( W_1=w_1\wedge W_2=w_2\wedge \mathbf {W_3}=\mathbf {w_3})$, then $T^c,(\textbf{u},\mathbf {v^c})\models (\textbf{X}=\textbf{x}\wedge W_1=w_1)\diamond \!\!\rightarrow ( W_2=w_2\wedge \mathbf {W_3}=\mathbf {w_3})$ as L7 holds in any causal model. According to the inductive hypothesis, $(\textbf{X}=\textbf{x}\wedge W_1=w_1)\diamond \!\!\rightarrow ( W_2=w_2\wedge \mathbf {W_3}=\mathbf {w_3})\in C$. Similarly, $(\textbf{X}=\textbf{x}\wedge W_2=w_2)\diamond \!\!\rightarrow ( W_1=w_1\wedge \mathbf {W_3}=\mathbf {w_3})\in C$. Since L8 is in C, $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow (W_1=w_1\wedge W_2=w_2\wedge \mathbf {W_3}=\mathbf {w_3})\in C$.

It is then easy to show that $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \beta \in C$ iff $T^c,(\textbf{u},\mathbf {v^c})\models \textbf{X}=\textbf{x}\diamond \!\!\rightarrow \beta$, using the aforementioned strategy of reducing $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \beta$ via $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \beta ^{dnf}$, where $\beta ^{dnf}$ is a disjunctive normal form of $\beta$.

The further cases of the inductive step concern Boolean combinations of formulas of the form $X=x$ and formulas of the form $\textbf{X}=\textbf{x}\diamond \!\!\rightarrow \beta$. These cases are very straightforward. Therefore, for every $\psi \in \mathcal {L}(S)$, $\psi \in C$ iff $T^c,(\textbf{u},\mathbf {v^c})\models \psi$ for every context $\textbf{u}$.

Finally, we need to show that $T^c\in \mathcal {T}_\text {L}(S)$. This is trivial given what has been shown. Since L1, L9, L10 and L11 are in C, they hold relative to $T^c,(\textbf{u},\mathbf {v^c})$. Hence, by Lemmas 5–8, $T^c$ is solution-determinate and solution-transitive in cycles, and every submodel of $T^c$ is solutionful and solution-conservative. That is, $T^c$ is indeed a Lewisian causal model.