Non-strict Interventionism: The Case Of Right-Nested Counterfactuals

Abstract

The paper focuses on a recent challenge brought forward against the interventionist approach to the meaning of counterfactual conditionals. According to this objection, interventionism cannot account for the interpretation of right-nested counterfactuals, the problem being its strict interventionism. We will report on the results of an empirical study supporting the objection. Furthermore, we will extend the well-known logic of intervention with a new operator expressing an alternative notion of intervention that does away with strict interventionism (and thus can account for some critical examples). This new notion of intervention operates on the valuation of the variables in a causal model, and not on their functional dependencies.

Proofs

Proof of Theorem 1

Soundness follows from the validity of axioms and rules. Completeness relies on Halpern (2000), which shows that MP and the axioms D0-D11 (Table 3) are sound and complete with respect to recursive causal models for a slightly different language ${\mathcal {L}}^+$. Here is a sketch of the argument.

Table 3 Axiom system for ${\mathcal {L}}^+$ w.r.t. recursive causal models over the given signature (from Halpern (2000))

First, formulas in the language ${\mathcal {L}}^+$ are Boolean combinations of expressions of the form $[\overrightarrow{X}{=}\overrightarrow{x}]\gamma$, with $\overrightarrow{X} \subseteq {\mathcal {V}}$, $x_i \in {\mathcal {R}}(X_i)$ and $\gamma$ a Boolean combination of atoms of the form $Y(\overrightarrow{u}){=}y$, with $Y \in {\mathcal {V}}$, $y \in {\mathcal {R}}(Y)$ and $\overrightarrow{u}$ a vector of values for all variables in ${\mathcal {U}}$. An atom $Y(\overrightarrow{u}){=}y$ is true in a model when setting variables in ${\mathcal {U}}$ to $\overrightarrow{u}$ gives Y the value y. As mentioned, MP and axioms D0-D11 are sound and complete for recursive causal models (Theorem 3.3 in Halpern (2000)).²⁴ Then, note that the referred paper also mentions not only that D10 implies D2 and D9, but also that, within recursive models, D2 and D10 are equivalent. Thus, MP together with axioms D0-D8 and D11 are sound and complete for ${\mathcal {L}}^+$ within recursive models.

Now, for our purposes, observe first that axioms D0, MP, D4, D6 and D8 in the table above are identical to our axioms A0, MP, A4, A6 and A8. Moreover, axioms D1 and D2 are derivable in our system by using A8 over, respectively, A1 and A2. Then, axioms D3 and D5 are also derivable in our system, this time via propositional reasoning and $\langle \overrightarrow{X}{=}\overrightarrow{x}\rangle \phi \leftrightarrow [\overrightarrow{X}{=}\overrightarrow{x}]\phi$, which is derivable from A7.1 and propositional reasoning.²⁵ Note also how D7 can be derived from A7.1, A7.2 and propositional reasoning.²⁶ Finally, axiom D11 is derivable in our system too (an instance of repetitive applications of A7.2). Thus, MP and axioms A0-A8 are sound and complete within recursive models for the fragment of ${\mathcal {L}}_{[\,]}$ that corresponds to ${\mathcal {L}}^+$.

To complete the argument note that, besides allowing explicit interventions on exogenous variables variables (with axioms A4 and A$_{\mathcal {U}}$ describing the effects), the only significant way in which ${\mathcal {L}}_{[\,]}$ extends ${\mathcal {L}}^+$ is by imposing no restrictions on a counterfactual’s consequent. The extension means, effectively, that nested counterfactuals are allowed. In order to deal with them, our axiom system has axioms A7.1, A7.2 and A9: the first and the second ‘push’ intervention operators inside the counterfactual’s consequent (commuting with negations and distributing over conjunctions), and the third collapses two sequential interventions into a single one. Thus, with the help of axiom A$_{[]}$ for the basic cases, every formula in ${\mathcal {L}}_{[\,]}$ can be translated into a semantically equivalent one in ${\mathcal {L}}^+$, for which MP and A0-A8 are sound and complete.

Proof of Proposition 1

Let $\langle {\mathcal {S}},{\mathcal {A}} \rangle$ be a causal model. Let $\overrightarrow{X}{=}\overrightarrow{x}$ be a counterfactual antecedent, with (i) $\overrightarrow{X_d}$ the variables in $\overrightarrow{X}$ whose ${\mathcal {A}}$-value differs from the intended $\overrightarrow{x}$, and (ii) $\overrightarrow{Z}$ the endogenous variables not occurring in $\overrightarrow{X}$ that are not causally dependent on variables in $\overrightarrow{X_d}$, with $\overrightarrow{z}$ their ${\mathcal {A}}$-values. By construction, $\overrightarrow{X}$ and $\overrightarrow{Z}$ are disjoint. Now, use ${\mathcal {A}}_1$ to abbreviate ${\mathcal {A}}^{\overrightarrow{X}{=}\overrightarrow{x}}$ (Definition 8), so ${\mathcal {A}}_1$ assigns the respective $x_i$ to each variable $X_i$ in $\overrightarrow{X}$ and, for variables not in $\overrightarrow{X}$, follows ${\mathcal {A}}$ for those not causally dependent of any variable in $\overrightarrow{X_d}$, using ${\mathcal {S}}$ and the values already in ${\mathcal {A}}_1$ to calculate those causally dependent of some variable in $\overrightarrow{X_d}$. Finally, let ${\mathcal {A}}_2$ be the (unique) valuation that is identical with ${\mathcal {A}}$ with respect to exogenous variables not in $\overrightarrow{X}$, assigns to exogenous variables in $\overrightarrow{X}$ their respective value in $\overrightarrow{x}$, and complies with the structural functions in ${\mathcal {S}}_{(\overrightarrow{X}{=}\overrightarrow{x},\overrightarrow{Z}{=}\overrightarrow{z})}$. Take any $Y \in {\mathcal {U}}\cup {\mathcal {V}}$; it will be shown that ${\mathcal {A}}_1(Y) = {\mathcal {A}}_2(Y)$.

First, for exogenous variables. (i) If $Y \in {\mathcal {U}}$ is not in $\overrightarrow{X}$, then both ${\mathcal {A}}_1$ and ${\mathcal {A}}_2$ agree with ${\mathcal {A}}$, the first because Y is not causally dependent on any other variable, and the second by definition. (ii) If $Y \in {\mathcal {U}}$ is in $\overrightarrow{X}$, both ${\mathcal {A}}_1$ and ${\mathcal {A}}_2$ assign to it the value indicated by $\overrightarrow{x}$.

Now, for endogenous variables. (i) If $Y \in {\mathcal {V}}$ is in $\overrightarrow{X}$, then ${\mathcal {A}}_1$ gives it the value indicated by $\overrightarrow{x}$ (by de finition), and so does ${\mathcal {A}}_2$ (by complying with ${\mathcal {S}}_{(\overrightarrow{X}{=}\overrightarrow{x},\overrightarrow{Z}{=}\overrightarrow{z})}$). (ii) If $Y \in {\mathcal {V}}$ is in $\overrightarrow{Z}$ then, ${\mathcal {A}}_1$ uses the value in ${\mathcal {A}}$ (by definition), and so does ${\mathcal {A}}_2$ (via the values $\overrightarrow{z}$, taken from ${\mathcal {A}}$). (iii)Finally, suppose Y is neither in $\overrightarrow{X}$ nor in $\overrightarrow{Z}$; then, the structural functions used by ${\mathcal {A}}_1$ and ${\mathcal {A}}_2$ to calculate Y’s value are the same: from ${\mathcal {S}}$ for the first, and from ${\mathcal {S}}_{(\overrightarrow{X}{=}\overrightarrow{x},\overrightarrow{Z}{=}\overrightarrow{z})}$ for the second. On its own, this does not guarantee that Y’s value under both $\langle {\mathcal {S}}, {\mathcal {A}}_1 \rangle$ and $\langle {\mathcal {S}}_{(\overrightarrow{X}{=}\overrightarrow{x},\overrightarrow{Z}{=}\overrightarrow{z})}, {\mathcal {A}}_2 \rangle$ is the same: there is a unique function $F_Y$, and yet the values of its parameters (all other variables) might be different. But, according to the previous items, the only variables in which ${\mathcal {A}}_1$ and ${\mathcal {A}}_2$ might differ are precisely the endogenous variables in neither $\overrightarrow{X}$ nor in $\overrightarrow{Z}$. Then, relying on the recursiveness of the model, one can use an inductive argument to show that, when the process that assigns values to variables calculates the value of such a Y, the values of all its parents will be the same in both ${\mathcal {A}}_1$ and ${\mathcal {A}}_2$. Indeed, the step $\#_0$ in the process assigns values to the variables in the set $S_{0} := \{ Y \in {\mathcal {V}}\setminus (\overrightarrow{X} \cup \overrightarrow{Z}) \mid \text {the parents of } Y \text { are in } {\mathcal {U}}\cup \overrightarrow{X} \cup \overrightarrow{Z} \}$. The valuations ${\mathcal {A}}_1$ and ${\mathcal {A}}_2$ coincide in the values of the variables in ${\mathcal {U}}\cup \overrightarrow{X} \cup \overrightarrow{Z}$, so they will coincide in the values of variables in $S_0$. Crucially, the model is recursive, so $S_{0} \ne \emptyset$. Then, each step $\#_{k+1}$ assigns values to the variables in the set $S_{k+1} := \{ Y \in {\mathcal {V}}\setminus (\overrightarrow{X} \cup \overrightarrow{Z}) \mid \text {the parents of } Y \text { are in } {\mathcal {U}}\cup \overrightarrow{X} \cup \overrightarrow{Z} \cup \bigcup _{0 \le i \le k} S_i \}$. Now, ${\mathcal {A}}_1$ and ${\mathcal {A}}_2$ coincide in the values of the variables in ${\mathcal {U}}\cup \overrightarrow{X} \cup \overrightarrow{Z} \cup \bigcup _{0 \le i \le k} S_i$, so they will coincide in the values of variables in $S_{k+1}$. But, again, the model is recursive, so $S_{k+1} \ne \emptyset$. Thus, eventually the values of all variables in ${\mathcal {V}}\setminus (\overrightarrow{X} \cup \overrightarrow{Z})$ will be calculated, and the values will be the same in both ${\mathcal {A}}_1$ and ${\mathcal {A}}_2$.

Proof of Theorem 2

The proof follows the reduction axioms strategy frequently used in dynamic epistemic logicBaltag et al. (1998), van Ditmarsch et al. (2008), van Benthem (2011). In our case, the strategy relies on a sound and complete axiom system for the -less fragment of . For the remaining formulas, those involving , the strategy uses ‘reduction axioms’: valid formulas and validity-preserving rules indicating how to translate a formula with occurrences of into a provably equivalent one without them. Soundness follows from the validity and validity-preserving properties of the new axioms and rules (so a formula and its translation are semantically equivalent); completeness follows from the completeness of the axiom system for the -less fragment, as the recursion axioms define a recursive validity-preserving translation from the full into the latter. The reader is referred to Wang and Cao (2013) and (van Ditmarsch et al. 2008, Chapter 7) for a detailed explanation of this technique.

For the underlying system, MP and axioms A0-A9, A$_{\mathcal {U}}$ and A$_{[]}$ constitute a sound and complete axiomatization for ${\mathcal {L}}_{[\,]}$ over causal models (Theorem 1). But, the fragment in belonging to ${\mathcal {L}}_{[\,]}$ is not affected by the change of models (in particular, because strict interventions force the valuation to agree with the structural functions, thus always producing a causal model). Thus, as argued in Footnote 19, they also constitute a sound and complete axiom system for ${\mathcal {L}}_{[\,]}$ over general causal models.

For dealing with , axioms A10-A13 define the recursive translation that takes any formula in and returns a logically equivalent one without . Here are arguments for their validity.

• Axiom A10 is the basic case for the translation, as it eliminates by showing how the valuation that results from a non-strict intervention is equivalent to a valuation that results from a strict intervention. Its validity follows from Proposition 1.

• Axioms A11 and A12 indicate how to deal with negations (commute and $\lnot$) and conjunctions (distribute over $\wedge$). In particular, the validity of the former comes from the fact that a non-strict intervention is deterministic.

• Axiom A13 eliminates a non-strict intervention that precedes a strict one. For its validity, take a causal model $\langle {\mathcal {S}},{\mathcal {A}} \rangle$, and note the following.

  • if and only if $\langle {\mathcal {S}},{\mathcal {A}}^{\overrightarrow{X}{=}\overrightarrow{x}} \rangle \models [\overrightarrow{Z}{=}\overrightarrow{z}]\phi$, that is, if and only if $\langle {\mathcal {S}}_{\overrightarrow{Z} {=}\overrightarrow{z}}, ({\mathcal {A}}^{\overrightarrow{X}{=}\overrightarrow{x}})^{{\mathcal {S}}_{\overrightarrow{Z} {=}\overrightarrow{z}}} \rangle \models \phi$.

  • $\langle {\mathcal {S}},{\mathcal {A}} \rangle \models [\overrightarrow{X'}{=}\overrightarrow{x'}][\overrightarrow{Z}{=}\overrightarrow{z}]\phi$ with $\overrightarrow{X'}{=}\overrightarrow{x'}$ the subassignment of $\overrightarrow{X}{=}\overrightarrow{x}$ for $\overrightarrow{X'}=\overrightarrow{X}\setminus {\mathcal {V}}$ if and only if $\langle {\mathcal {S}},{\mathcal {A}} \rangle \models [\overrightarrow{X''}{=}\overrightarrow{x''}, \overrightarrow{Z}{=}\overrightarrow{z}]\phi$ with $\overrightarrow{X''}{=}\overrightarrow{x''}$ the subassignment of $\overrightarrow{X'}{=}\overrightarrow{x'}$ for $\overrightarrow{X''}=\overrightarrow{X'}\setminus \overrightarrow{Z}$ (Axiom A9), that is, if and only if $\langle {\mathcal {S}}_{\overrightarrow{X''}{=}\overrightarrow{x''}, \overrightarrow{Z}{=}\overrightarrow{z}}, {\mathcal {A}}^{{\mathcal {S}}_{\overrightarrow{X''}{=}\overrightarrow{x''}, \overrightarrow{Z}{=}\overrightarrow{z}}} \rangle \models \phi$.

  Thus, it is enough to show both

  $$\begin{aligned} {\mathcal {S}}_{\overrightarrow{Z} {=}\overrightarrow{z}} = {\mathcal {S}}_{\overrightarrow{X''}{=}\overrightarrow{x''}, \overrightarrow{Z}{=}\overrightarrow{z}} \qquad \text {and}\qquad ({\mathcal {A}}^{\overrightarrow{X}{=}\overrightarrow{x}})^{{\mathcal {S}}_{\overrightarrow{Z} {=}\overrightarrow{z}}} = {\mathcal {A}}^{{\mathcal {S}}_{\overrightarrow{X''}{=}\overrightarrow{x''}, \overrightarrow{Z}{=}\overrightarrow{z}}}. \end{aligned}$$

  with $\overrightarrow{X''} {=}\overrightarrow{x''}$ the subassignment of $\overrightarrow{X}{=}\overrightarrow{x}$ for $\overrightarrow{X''}=\overrightarrow{X}\setminus ({\mathcal {V}}\cup \overrightarrow{Z})$ (thus, $\overrightarrow{X''}$ contains only exogenous variables). The first part is straightforward, as $\overrightarrow{X''}$ contains no endogenous variables. For the second part, it will be shown that both valuations agree in the values of every variable. Here are the cases for each exogenous variable $U \in {\mathcal {U}}$:

  • if $U \in \overrightarrow{Z}$, then $U \notin \overrightarrow{X''}$ and both valuations return the appropriate value in $\overrightarrow{z}$;

  • if $U \notin \overrightarrow{Z}$ but $U \in \overrightarrow{X}$, then $U \in \overrightarrow{X''}$ and both valuations return the appropriate value in $\overrightarrow{x}$;

  • if $U \notin \overrightarrow{Z}$ and $U \notin \overrightarrow{X}$, both valuations return the value ${\mathcal {A}}(U)$;

  Then, here are the cases for each endogenous variable $V \in {\mathcal {V}}$ (thus, $V \notin \overrightarrow{X''}$):

  • if $V \in \overrightarrow{Z}$, then both valuations return the appropriate value in $\overrightarrow{z}$.

  • if $V \notin \overrightarrow{Z}$, then both valuations return the unique solution for the respective (but identical) structural equations under the (identical) values for variables in ${\mathcal {U}}\cup \overrightarrow{Z}$.