An event algebra for causal counterfactuals

Abstract

“If the tower is any taller than 320 ms, it may collapse,” Eiffel thinks out loud. Although understanding this counterfactual poses no trouble, the most successful interventionist semantics struggle to model it because the antecedent can come about in infinitely many ways. My aim is to provide a semantics that will make modeling such counterfactuals easy for philosophers, computer scientists, and cognitive scientists who work on causation and causal reasoning. I first propose three desiderata that will guide my theory: it should be general, yet conservative, yet useful. Next, I develop a formalization of events in the form of an algebra. I identify an event with all the ways in which it can be brought about and provide rules for determining the referent of an arbitrary event description. I apply this algebra to counterfactuals expressed using underdeterministic causal models-models that encode non-probabilistic causal indeterminacies. Specifically, I develop semaphore interventions, which represent how the target system may be modified from without in a coordinated fashion. This, in turn, allows me to bring about any event within a single model. Finally, I explain the advantages of this semantics over other interventionist competitors.

Appendix

1.1 The syntax and semantics of model statements

Here, I present the syntax and semantics of event names and model statements—the statements that can be true or false in a model—in one place.

First, the grammar:

		production rule
brick sentence	$B \rightarrow \!$	an expression over some variables from $\vec {\mathcal {V}}$;
		the exact syntax depends on the ranges
event name	$E \rightarrow \!$	B,$(\lnot E)$, $(E \wedge E)$, $(E \vee E)$, $(E \rightarrow E)$, $(E \equiv E)$,
modal sentence	$M \rightarrow \!$	$\mathop {\Box }E$, $\mathop {\Diamond }E$,
counterfactual	$C \rightarrow \!$	, , , ,
model statement	$S \rightarrow \!$	(C), $(\lnot S)$, $(S \wedge S)$, $(S \vee S)$.

In all cases used throughout, brick sentences were arithmetic equations or inequalities. The grammar allows for model statements like , but not for $\mathop {\Box }\mathop {\Diamond }\varphi$ or .

Let $\mathfrak {M}$ be the (pre-intervention) model of evaluation, $\mathfrak {M}^{\top }$ be the set of solutions to $\mathfrak {M}$, $\psi$ be a name over $\vec {X}$ ($\vec {X}\subseteq \vec {\mathcal {V}}$), $\varphi$ be any name of a consistent event (i.e., $\llbracket \varphi \rrbracket \ne \emptyset$) over some subset of $\vec {\mathcal {V}}$, $\vec {\sigma }$ be any assignment over all its nodes, ,²⁹ and $S_1$, $S_2$ be any model statements.³⁰ Below is the full semantics; although it doesn’t need to be expressed in terms of the event algebra, it can, and I provide the alternative rules too:

(32)

(33)

(34)

(35)

(36)

$$\begin{aligned}&\mathfrak {M}\!\models \lnot S_1{} & {} \textit{iff}\,\ \mathfrak {M}\!\nvDash S_1 \text {, i.e., it's not the case that } \mathfrak {M}\!\models S_1 {,}{} & {} \end{aligned}$$

(37)

$$\begin{aligned}&\mathfrak {M}\!\models S_1 \wedge S_2\!\!{} & {} \textit{iff}\,\ \mathfrak {M}\!\models S_1 \text { and }\mathfrak {M}\!\models S_2{,}{} & {} \end{aligned}$$

(38)

$$\begin{aligned}&\mathfrak {M}\!\models S_1 \vee S_2\!\!{} & {} \textit{iff}\,\ \mathfrak {M}\!\models S_1 \text { or }\mathfrak {M}\!\models S_2.{} & {} \end{aligned}$$

(39)

1.2 An event has at most one canonical form

Theorem

A non-empty set of assignments over at least one variable has a single canonical form, which is the junction of the maximum number of assignment sets (juncts) over non-overlapping variables:

(40)

where: $S_{\vec {X}}$ is over variables $\vec {X}$, $\vec {X}\ne \emptyset$, assignments from $J_i$ don’t share variables with assignments from $J_j$ ($i\ne j$), and $J_i \ne \{\langle \rangle \}$.

Proof

The proof will be by induction on the maximum number of juncts the event decomposes into. First, notice that if a set of assignments is a junction of other sets of assignments, then the projection of the former onto some variables is a junction of the projections of the latter:

(41)

where $\vec {X}=\vec {Y}\cup \vec {Z}$, $\vec {P}\subseteq \vec {X}$, and $U_{\vec {X}}[\vec {P}]$ is a projection of $U_{\vec {X}}$ onto $\vec {P}$ (read $V_{\vec {Y}}[\vec {P}\cap \vec {Y}]$ and $W_{\vec {Z}}[\vec {P}\cap \vec {Z}]$ analogously).³¹ This property follows from (1), the definition of junction, because if an assignment is a spatial concatenation of two other assignments, then a projection of the former is a spatial concatenation of the corresponding projections of the latter.

Now, for the inductive part of the proof.

Base step The event denoted by any sentence over one variable is already in the canonical form. (The only non-empty set of assignments over no variables is $\llbracket \top \rrbracket = \{\langle \rangle \}$; as a convention, assume that this is its canonical form.)

Inductive step Assume property (40) holds for sentences over fewer than k variables. Take $S_{\vec {X}}$, a set of assignments over k variables. If $S_{\vec {X}}$ doesn’t decompose any further, it trivially satisfies the property, as the set is already in the canonical form.

Otherwise, decompose $S_{\vec {X}}$ as

(42)

where , $\vec {Y}\ne \emptyset \ne \vec {Z}$, and $L_{\vec {Z}}$ doesn’t decompose any further. As $K_{\vec {Y}}$ is now over fewer than k variables, in virtue of the inductive assumption, for some n and conjuncts $\left\{ J_i\right\} _{i=1}^n$. Therefore, the canonical form for the target set is:

(43)

What’s left is to prove that this decomposition is unique. Decompose $S_{\vec {X}}$ into any two juncts:

(44)

where , and $\vec {P}\ne \emptyset \ne \vec {Q}$. Either $\vec {Z}\subseteq \vec {Q}$ or $\vec {Z}\subseteq \vec {P}$, because otherwise, per (41),$L_{\vec {Z}}$ could be decomposed further into two juncts: one over some variables from $\vec {Q}$ and the other over some variables from $\vec {P}$. For focus, assume $\vec {Z}\subseteq \vec {Q}$. This means that , where $\vec {R}= \vec {Q}\backslash \vec {Z}$. Therefore,

(45)

In virtue of the inductive assumption, has a unique canonical form, and it’s the same as the one for $K_{\vec {Y}}$, because . That in turn means that (43) still gives the canonical form of $S_{\vec {X}}$. Therefore, $S_{\vec {X}}$ has a unique canonical form. This completes the inductive proof. $\square$