Interventionist counterfactuals

Abstract

A number of recent authors (Galles and Pearl, Found Sci 3 (1):151–182, 1998; Hiddleston, Noûs 39 (4):232–257, 2005; Halpern, J Artif Intell Res 12:317–337, 2000) advocate a causal modeling semantics for counterfactuals. But the precise logical significance of the causal modeling semantics remains murky. Particularly important, yet particularly under-explored, is its relationship to the similarity-based semantics for counterfactuals developed by Lewis (Counterfactuals. Harvard University Press, 1973b). The causal modeling semantics is both an account of the truth conditions of counterfactuals, and an account of which inferences involving counterfactuals are valid. As an account of truth conditions, it is incomplete. While Lewis’s similarity semantics lets us evaluate counterfactuals with arbitrarily complex antecedents and consequents, the causal modeling semantics makes it hard to ascertain the truth conditions of all but a highly restricted class of counterfactuals. I explain how to extend the causal modeling language to encompass a wider range of sentences, and provide a sound and complete axiomatization for the extended language. Extending the truth conditions for counterfactuals has serious consequences concerning valid inference. The extended language is unlike any logic of Lewis’s: modus ponens is invalid, and classical logical equivalents cannot be freely substituted in the antecedents of conditionals.

Appendix

1.1 Disjunctive Normal Form

Every Boolean sentence is exactly equivalent to a sentence in disjunctive normal form; i.e., a disjunction of conjunctions of atomic sentences. This disjunctive normal form can be derived in two steps.

The first step is to generate a sentence whose subsentences are all positive—i.e., all atomic sentences, negated atomic sentences, conjunctions, or disjunctions. Given a sentence ϕ _n with non-positive subsentences, one can generate a sentence ϕ _n+1 by the following procedure. First, find ϕ _n ’s longest non-positive subsentence ψ. (If two or more of these subsentences are tied for longest, let ψ be leftmost.)

• If ψ is of the form $\neg \neg \theta$, then ϕ _n+1 is the sentence that results from replacing ψ with θ in ϕ _n . (By $(\neg)$, this preserves exact equivalence.)

• If ψ is of the form $\theta_1 \supset \theta_2$, then ϕ _n+1 is the sentence that results from replacing ψ with $\neg \theta_1 \vee \theta_2$ in ϕ _n . (By $(\supset^{+})$, this preserves exact equivalence.)

• If ψ is of the form $\neg(\theta_1 \supset \theta_2)$, then ϕ _n+1 is the sentence that results from replacing ψ with $\theta_1 \wedge \neg \theta_2$ in ϕ _n . (By $(\supset^{-})$, this preserves exact equivalence.)

• If ψ is of the form $\neg(\theta_1 \wedge \theta_2)$, then ϕ _n+1 is the sentence that results from replacing ψ with $\neg \theta_1 \vee \neg \theta_2$ in ϕ _n . (By (∧⁻), this preserves exact equivalence.)

• If ψ is of the form $\neg(\theta_1 \vee \theta_2)$, then ϕ _n+1 is the sentence that results from replacing ψ with $\neg \theta_1 \wedge \neg \theta_2$ in ϕ _n . (By (∨⁻), this preserves exact equivalence.)

Since this procedure replaces each non-positive subsentence of ϕ with a positive sentence, and generates either no new non-positive sentences, or one positive sentence which is shorter than the one replaced, it is guaranteed to terminate in a sentence with only positive subsentences. Since each step in the procedure preserves exact equivalence, the positive sentence derived at the end is exactly equivalent to the original sentence.

The next step is to “push the conjunctions inward”. Begin with a sentence ϕ₀, all of whose subsentences are positive. Given a sentence ϕ _n , all of whose subsentences are positive, but not necessarily in disjunctive normal form, one can generate a sentence ϕ _n+1 by the following procedure. First, find ϕ _n ’s longest subsentence ψ that is a conjunction of disjunctions $(\theta^{1}_{1} \vee \ldots \vee \theta^{1}_{f(1)})\vee \ldots \vee (\theta^{n}_{1} \vee \ldots \vee \theta^{n}_{f(n)}).$ Consider all the conjunctions of θ ⁱ _j s that have at least one conjunct of the form θ ^k _g(k) for each $k \in \{1\ldots n\}$ (where g(k) is an arbitrary number between 1 and f(k) inclusive). Let ψ′ be the disjunction of all these conjunctions. Let ϕ _n+1 be the sentence that results from replacing ψ with the disjunction of all sentences of the form ψ′ in ϕ _n . By (∧⁺) and (∨⁺), replacing ϕ _n with ϕ _n+1 preserves exact equivalence.

Each step of the second procedure eliminates the longest disjunction inside the scope of a conjunction, and without introducing any new disjunction inside the scope of a conjunction. Each step keeps the sentence positive. After a finite number of steps, the length of the longest disjunction inside the scope of a conjunction must equal 1, at which point ϕ = ϕ _n+1. At this stage, the sentence will be entirely positive, and have no disjunctions inside the scope of conjunctions: it will be in disjunctive normal form.

1.2 Soundness

The validity of (R1) and (CM1) is ensured by the definitions of the Boolean connectives.

(R2)’s valdity is ensured by the selection semantics for $\,\square{\kern-4.1pt}\rightarrow\,:$ Suppose $(\psi_1 \wedge \ldots \wedge \psi_n) \supset \theta$ is true in every model, and $v_M((\phi \,\square{\kern-4.1pt}\rightarrow\, \psi_1) \wedge \ldots \wedge (\phi \,\square{\kern-4.1pt}\rightarrow\, \psi_n)) = 1$. It remains to show that $v_M(\phi \,\square{\kern-4.1pt}\rightarrow\, \theta) = 1$. The semantic rule for ∧ ensures that for each of $\psi_i \in \{\psi_1 \ldots \psi_n\}, v_M (\psi_i) = 1$. By the semantic rule for $\,\square{\kern-4.1pt}\rightarrow\,$, for every $\psi_i \in \{\psi_1 \ldots \psi_n\}$ and $M' \in f(\phi, M), v_{M'}(\psi_i) = 1$. By hypothesis, $v_{M'}((\psi_1 \wedge \ldots \wedge \psi_n) \supset \theta) = 1$ for every $M' \in f(\phi, M)$. By tautological reasoning ${v_{M}^{\prime}}(\theta)$ v _M ′(θ) = 1 for every $M' \in f(\phi, M)$, and so $v_M(\phi \,\square{\kern-4.1pt}\rightarrow\, \theta) = 1$.

The validity of (CM2) and (CM3) is ensured by the behaviour of variables. Every variable takes on some value or other (hence (CM2)), and no variable takes on two values at once (hence (CM3)).

(CM4) is valid because every sentence of the form A is made true by a single state, and each state corresponds to exactly one intervention. So f(A, M) contains exactly one model M′. For every sentence ϕ, either ${v_{M}^{\prime}}(\theta)$ = 1, in which case $v_M({\bf A} \,\square{\kern-4.1pt}\rightarrow\, \phi) = 1$, or ${v_{M}^{\prime}}(\theta)$ (ϕ) = 0, in which case $v_M({\bf A} \,\square{\kern-4.1pt}\rightarrow\, \neg \phi) = 1$. In either case $v_M(({\bf A} \,\square{\kern-4.1pt}\rightarrow\, \phi) \vee ({\bf A} \,\square{\kern-4.1pt}\rightarrow\, \neg \phi)) = 1$.

(CM5) is valid because every sentence of the form A is made true by some state or other, and each state corresponds to an intervention. Therefore, there is at least one submodel in f(A, M). Since $\bot$ is true in no model, $\bot$ is not true in every submodel in f(A, M). Hence, $v_M(\neg({\bf A}\,\square{\kern-4.1pt}\rightarrow\, \bot)) =1$.

(CM6) is valid because the order of interventions on a causal model does not affect the end result of those interventions, provided one doesn’t intervene twice on the same variable. To generate M _A ∧ V = v from M _A , one intervenes on M _A to set V = v. Suppose v _{M_A} (V = v) = 0. Then by the semantic rules for material conditional and counterfactual, $v_M({\bf A} \,\square{\kern-4.1pt}\rightarrow\, (V = v \supset B)) = 1.$ By tautological reasoning $v_M((({\bf A} \wedge V = v) \,\square{\kern-4.1pt}\rightarrow\, B) \supset ({\bf A} \,\square{\kern-4.1pt}\rightarrow\, (V = v \supset B))) = 1.$ On the other hand, suppose that in M _A , V = v. Intervening to change V’s structural equation will not affect the values of V’s nondescendants, for their values do not depend on V’s value. Nor (by hypothesis) will it affect V’s value. Nor will it affect the values of V’s descendants, since in a recursive model, their values are determined by the value of V and V’s nondescendants. Ergo, v _{M_A} (V′ = v′) = v _{M_A} ∧ V = v(V′ = v′) for all $V' \in \mathcal{V}.$ Hence, where B is any Boolean combination of atomic sentences, v _{M_A} (B) = v _{M_A} ∧ V = v(B), and so once again, $v_M((({\bf A} \wedge V = v) \,\square{\kern-4.1pt}\rightarrow\, B) \supset ({\bf A} \,\square{\kern-4.1pt}\rightarrow\, (V = v \supset B))) = 1.$

(CM7) is valid because any intervention that makes A true also generates a model in which A is true. This holds where A is an atomic sentence—intervening to set V = v generates a model in which V = v—or a negated atomic sentence—intervening to set V = v′ generates a model in which V ≠ v, so long as v′ ≠ v. The conditions on truthmaking ensure that if it holds for ϕ and ψ, it also holds for $\phi \vee \psi, \neg( \phi \vee \psi), \phi \wedge \psi, \neg ( \phi \wedge \psi), \phi \supset \psi$ and $\neg(\phi \supset \psi).$

(CM8) is valid for the class of recursive models. To show this, we must appeal to a notion of counterfactual dependence among variables. Say that V′ counterfactually depends on V iff for some variables $V_1\ldots V_m$, all distinct from V and V′, some values $v_1\ldots v_m$ of $V_1 \ldots V_m$, some values v and v ^* of V, and some value v′ of V′, the intervention setting $V_1= v_1, \ldots V_m = v_m$, and V = v generates a submodel in which V′ = v′, while the intervention setting $V_1= v_1, \ldots V_{m} = v_{m}$, and V = v ^* generates a submodel in which V′ ≠ v′.

We can show that V′ stands in the ancestral of counterfactual dependence to V iff V is an ancestor of V′. First, the left-to-right direction: If V′ counterfactually depends on V, then the structural equations ensure that V′’s value depends non-trivially on V’s value. Hence, V is an ancestor of V′. Since the relation of being an ancestor is transitive, it follows that if V′ stands in the ancestral of counterfactual dependence to V, then V is an ancestor of V′. Next, the right-to-left direction: if V ₁ is an ancestor of V _n , there is a sequence of variables $V_1, \ldots V_n$ such that V ₁ is the parent of V ₂, and $\ldots V_{n-1}$ is the parent of V _n . Since each variable counterfactually depends on each of its parents, it follows that V′ stands in the ancestral of counterfactual dependence to V.

Furthermore, it follows from the definition of the selection function and the semantic rule for the counterfactual that V′ counterfactually depends on V iff there is a sentence of the form d(V, V′) such that v _M (d(V, V′)) = 1. As a corollary, if there exist sentences $d_1(V_1, V_2)\ldots d_{n-2}(V_{n-2}, V_{n-1})$ such that $v_M( d_1(V_1, V_2) \wedge \ldots \wedge d_{n-1}(V_{n-1}, V_{n})) =1$, then V _n stands in the ancestral of counterfactual dependence to V ₁, and so V ₁ is an ancestor of V _n .

In order to show the validity of (CM8), we can assume that $v_M( d_1(V_1, V_2) \wedge \ldots \wedge d_{n-1}(V_{n-1}, V_{n})) = 1$, and show that there is no sentence of the form d _n (V _n , V ₁) such that v _M (d _n (V _n , V ₁)) = 1. By our supposition, in M, V _n stands in the ancestral of the counterfactual dependence relation to V ₁. Therefore, V ₁ is an ancestor of V _n . On the assumption that M is recursive, V _n-1 is not an ancestor or V ₁, and so cannot counterfactually depend on V _n-1. Therefore, no sentence of the form d _n (V _n-1, V ₁) can be true in M; QED.

(CM9) is sound according to the recursive conditions for truthmaking. $v_M((A \vee B) \,\square{\kern-4.1pt}\rightarrow\, \phi) = 1$ iff v _M ′(ϕ) = 1 for every submodel M′ generated by an intervention on M corresponding to a state that makes A ∨ B true. But according to condition $(\vee^{+}), s\, \vDash\, A \vee B$ iff $s\, \vDash\, A$ or $s\, \vDash\, B$ or $s = t \sqcup r$ where $t \,\vDash\, A$ and $r\, \vDash\, B$. In other words, rephrasing the third disjunct in terms of condition $(\wedge^{+}), s \,\vDash\, A \vee B$ iff $s \,\vDash\, A$ or $s\, \vDash\, B$ or $s \vdash A \wedge B$. So $v_M((A \vee B) \,\square{\kern-4.1pt}\rightarrow\, \phi) = 1$ iff v _M ′(ϕ) = 1 for every submodel M′ generated by an intervention on M corresponding to a state that makes A true, and v _M ′(ϕ) = 1 for every submodel M′ generated by an intervention on M corresponding to a state that makes B true, and v _M ′(ϕ) = 1 for every submodel M′ generated by an intervention on M corresponding to a state that makes A ∧ B true. In other words, $v_M((A \vee B) \,\square{\kern-4.1pt}\rightarrow\, \phi) = 1$ iff $v_M(A \,\square{\kern-4.1pt}\rightarrow\, \phi) = 1$, and $v_M(B \,\square{\kern-4.1pt}\rightarrow\, \phi) = 1$, and $v_M((A \wedge B) \,\square{\kern-4.1pt}\rightarrow\, phi) = 1$. The semantic rules for Boolean connectives ensure that $v_M(((A \vee B) \,\square{\kern-4.1pt}\rightarrow\, \phi) \equiv ((A \,\square{\kern-4.1pt}\rightarrow\, \phi) \wedge (B \,\square{\kern-4.1pt}\rightarrow\, \phi) \wedge ((A \wedge B) \,\square{\kern-4.1pt}\rightarrow\, \phi))) = 1$.

(CM10) is valid because any exact equivalent A′ of A is made true by the same states as A; hence f(A, M) = f(A′, M).

(CM11) is valid because where $\mathcal{V}_{\bf A}$ is the set of variables mentioned in ${\bf A}, \mathcal{V}_{\bf B}$ is the set of variables mentioned in B and B′, and $\mathcal{V}_{\bf C}$ is the set of variables mentioned in C, the procedure of performing the interventions sanctioned by each of A, B, and C has the same results as the procedure by performing the interventions sanctioned by A and B′, then performing the interventions sanctioned by B (thereby reversing the interventions sanctioned by B′) and performing the interventions sanctioned by C.

1.3 Completeness

A proof system is complete over a class of models iff for every consistent set of sentences $\Upgamma$, there is a model in which every sentence in $\Upgamma$ is true. Suppose, then, that we have a consistent set of sentences $\Upgamma_0$. If we can specify a procedure for constructing a model in which every member of $\Upgamma_0$ is true, then we will have proved completeness. We begin by using Lindenbaum’s Lemma to construct a maximal consistent set $\Upgamma$ such that $\Upgamma_0 \subseteq \Upgamma$. We can use $\Upgamma$ to build a causal model $M = \langle\mathcal{V},\mathcal{S}, \mathcal{A}\rangle$, by the following procedure.

Let $\mathcal{V}$ contain one variable V _i for each variable V _i for which the language contains a name.

To define $\mathcal{S}$, we must show that for every $V_i \in \mathcal{V}$ and partial assignment function $\mathcal{A}^{-}$, where $\mathcal{A}^{-}(V_j) \in \mathcal{R}(V_j)$ for the $V_j \in \mathcal{V} - V_i, \Upgamma$ contains exactly one sentence of the form $\wedge_{V_j \in \mathcal{V} -V_{i}} (V_{j} = \mathcal{A}^{-}(V_{j})) \,\square{\kern-4.1pt}\rightarrow\, V_i = v_i$. To show this, we must prove both an existence claim and a uniqueness claim.

The existence claim: for every $V_i \in \mathcal{V}$ and partial assignment function $\mathcal{A}^{-}, \Upgamma$ contains at least one sentence of the form $\wedge_{j \in \mathcal{V} -V_i} (V_j = \mathcal{A}^{-}(V_{j})) \,\square{\kern-4.1pt}\rightarrow\, V_i = v_i$. This can be shown by assuming otherwise, and proving a contradiction. Since $\Upgamma$ is maximal, if the existence claim is false, then for every $v_i \in \mathcal{R}(V_i)$,

$$ \neg (\wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, V_i = v_i \in \Upgamma) $$

Since $\Upgamma$ contains every instance of every axiom schema, and (CM4) is an axiom, for each $v_i \in \mathcal{R}(V_i)$,

$$ (\wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, V_i = v_i) \vee (\wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, V_i \neq v_i) \in \Upgamma $$

Since $\Upgamma$ is closed under tautological consequence, for each $v_i \in \mathcal{R}(V_i)$,

$$ \wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, V_i \neq v_i \in \Upgamma $$

By (R2) and the closure of $\Upgamma$ under logical consequence,

$$ \wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, \wedge_{v \in {\mathcal{R}}(V_i)} V_i \neq v\in \Upgamma $$

By (CM1), (CM2), and (R1),

$$ \wedge_{v \in {\mathcal{R}}(V_i)} V_i \neq v \vdash \bot $$

and so by (R2), and the closure of $\Upgamma$ under logical consequence,

$$ \wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, \bot \in \Upgamma $$

But since every instance of (CM5) is an axiom,

$$ \neg (\wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, \bot) \in \Upgamma $$

We have derived a contradiction, thus completing the existence proof.

Now, the uniqueness claim: for every $V_i \in \mathcal{V}$ and partial assignment function $\mathcal{A}^{-}, \Upgamma$ contains at most one sentence of the form $\wedge_{j \in \mathcal{V} -V_i} (V_j = \mathcal{A}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, V_i = v_i.$ Again, this can be shown by assuming otherwise and proving a contradiction. We will suppose the uniqueness claim is false, i.e., that $\Upgamma$ contains sentences of the form

$$ \wedge_{j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, V_i = v_i $$

and

$$ \wedge_{j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, V_i = v'_i $$

where v _i  ≠ v′ _i . And we will prove a contradition.

By (R2) and the closure of $\Upgamma$ under entailment, we can infer that

$$ \wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, (V_i = v_i \wedge V_i = v'_i) \in \Upgamma $$

But by (CM3)

$$ V_i = v_i \wedge V_i = v'_i \vdash \bot $$

and so by (R2),

$$ \wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, \bot \in \Upgamma $$

But since every instance of (CM5) is an axiom,

$$ \neg (\wedge_{V_j \in {\mathcal{V}} -V_i} (V_j = {\mathcal{A}}^{-}(V_j)) \,\square{\kern-4.1pt}\rightarrow\, \bot) \in \Upgamma $$

We have derived a contradiction, thus completing the uniqueness proof.

Now we can define $\mathcal{S}$. Since there are many possible partial assignment functions for each set of variables, let us number each possible $\mathcal{A}^{-}$ with a subscript. First, let us write, for each V _i , an equation of the form

$$ f_{i} = v_{k} \begin{array}{l} \hbox{if} (V_{j} = {\mathcal{A}}^-_k(j)) \hbox{for each} V_{j} \in {\mathcal{V}} - \{V_{i}\}\\ \hbox{where} \wedge_{j \in {\mathcal{V}} -V_{i}} (V_{j} = {\mathcal{A}}^{-}_k(j)) \,\square{\kern-4.1pt}\rightarrow\, V_{i} = v_{k} \in \Upgamma \\ \end{array} $$

By our existence and uniqueness claim, there is exactly one such v _k for every i and k. Now, let us replace each of these equations with an equation that has the same solutions in all cases, but does not mention any redundant variables. (If all variables are redundant, delete the equation altogether.) The set of equations generated this way is $\mathcal{S}.\,\mathcal{S}$ will contain at most one structural equation for each variable.

The last step in defining M is to define $\mathcal{A}$. (CM2) and (CM3) ensure that for each $V \in \mathcal{V}, \Upgamma$ contains exactly one sentence of the form V = v. Let $\mathcal{A}(V_i)$ be the v _i such that $V_i = v_i \in \Upgamma$.

Having defined M, we must show that it is recursive. (CM8) ensures that for any two variables V and V′, if V′ stands in the ancestral of counterfactual dependence to V, then V does not counterfactually depend on V′. I have shown in the soundness portion of the appendix that the ancestral of counterfactual dependence is the ancestor relation. It follows that for any two variables V and V′, if V′ is an ancestor of V, then V is not counterfactually dependent on V′. Furthermore, since all variables counterfactually depend on their parents, we can infer that for any two variables V and V′, if V′ is an ancestor of V, then V is not parent of V′. This entails that no variable is its own descendant. For if a variable V were its own descendant, it would have a parent V′ that was also V’s descendant. Thus, no variable in M can be its own descendant, and so M is recursive.

The last step of the completeness proof is to show that if $\phi \in \Upgamma$, then v _M (ϕ) = 1. This can be proved in two steps. First, we prove that every sentence is equivalent to a Boolean compound of atomic sentences and simple conditionals of the form ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, B$. Next, we show that for any such Boolean compound, $\phi \in \Upgamma \,\, \hbox{iff} \,\, v_{M}(\phi) = 1$.

I will prove that every sentence is equivalent to a Boolean compound of atomic sentences and simple conditionals of the form ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, B$ by providing a translation procedure.

The first part of the translation procedure eliminates all complex antecedents. Begin with a sentence ϕ₀. Let ψ _n be ϕ _n ’s shortest subsentence of the form $A \,\square{\kern-4.1pt}\rightarrow\, \theta$. By (CM10) and the disjunctive normal form theorem, θ is equivalent to a sentence of the form ${\bf A}_1 \vee \ldots \vee {\bf A}_n \,\square{\kern-4.1pt}\rightarrow\, \psi$. By (CM9), ψ _n is equivalent to a conjunction of the form $({\bf A}_1 \,\square{\kern-4.1pt}\rightarrow\, \psi) \wedge \ldots \wedge ({\bf A}_n \,\square{\kern-4.1pt}\rightarrow\, \psi)$. Let ϕ _n+1 be the result of replacing Replace ψ with this equivalent sentence. Repeated applications of (R2) ensure ϕ _n+1 is equivalent to ϕ _n . Each step eliminates a counterfactual with a complex antecedent without setting up a new one in its place, and so the procedure will terminate when there are no remaining subsentences with complex antecedents.

The second part of the translation procedure is to eliminate counterfactual consequents. Begin with a sentence ϕ₀, all of whose conditional subsentences have antecedents of the form A. Let ψ _n be ϕ _n ’s shortest subsentence of the form ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, \theta$ where θ is not of the form B. Let $(\theta_1^1 \wedge \ldots \wedge \theta_1^{f(1)}) \vee \ldots \vee (\theta_n^1 \wedge \ldots \wedge \theta_n^{f(n)})$ be the (classical) disjunctive normal form of θ. By (R2), ψ′ is equivalent to ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, ((\theta_1^1 \wedge \ldots \wedge \theta_1^{f(1)}) \vee \ldots \vee (\theta_n^1 \wedge \ldots \wedge \theta_n^{f(n)})),$ and by (CM4) and (R2) together, ϕ′ is equivalent to $(({\bf A} \,\square{\kern-4.1pt}\rightarrow\, \theta_1^1) \wedge \ldots \wedge ({\bf A} \,\square{\kern-4.1pt}\rightarrow\, \theta_1^{f(1)})) \vee \ldots \vee (({\bf A} \,\square{\kern-4.1pt}\rightarrow\, \theta_n^1) \wedge \ldots \wedge ({\bf A} \,\square{\kern-4.1pt}\rightarrow\, \wedge \theta_n^{f(n)})).$ Each θ ⁱ _j is either atomic or of the form ${\bf B} \,\square{\kern-4.1pt}\rightarrow\, C.$ If θ ^j _i is of the form ${\bf B}^j_i \,\square{\kern-4.1pt}\rightarrow\, C^j_i,$ then (CM11) ensures that ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, \theta^j_i$ is equivalent to ${\bf A}^j_i \wedge {\bf B}^j_i \,\square{\kern-4.1pt}\rightarrow\, C^j_i,$ where A ^j _i the conjunction that results from removing any conjuncts from A that share variables with B ^j _i . Finally, let A ^j _i  = A and C ^j _i  = θ ^j _i if θ ^j _i is atomic. Thus, ψ _n is equivalent to $(({\bf A}_1^1 \,\square{\kern-4.1pt}\rightarrow\, C_{1}^{1}) \wedge \ldots \wedge ({\bf A}^{f(1)}_1 \,\square{\kern-4.1pt}\rightarrow\, C_{1}^{f(1)})) \vee \ldots \vee (({\bf A}^{1}_{n} \,\square{\kern-4.1pt}\rightarrow\, C_n^1) \wedge \ldots \wedge ({\bf A}^{f(n)}_{n} \,\square{\kern-4.1pt}\rightarrow\, \wedge C_n^{f(n)}))$ (call this sentence ψ′ _n . Let ϕ _n+1 be the sentence that results from replacing ψ _n with ψ′ _n in ϕ _n . Repeated applications of (R2) ensure that ϕ _n+1 is equivalent to ϕ _n . Each step eliminates a counterfactual consequent without setting up a new one in its place, and so the procedure will terminate when there are no remaining subsentences with counterfactual consequents.

All that remains now is to show that if ϕ is a Boolean compound of atomic sentences and sentences of the form ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, B$, then v _M (ϕ) = 1. In the base case, where ϕ is an atomic sentence, this follows straightforwardly from the definition of $\mathcal{A}$. Suppose it holds for all sentences shorter than ϕ. Then we can show that it holds for ϕ as well.

• Suppose ϕ is of the form V ≠ v and $\phi \in \Upgamma.$ We have already shown that if $\phi \in \Upgamma,$ there is some v′ ≠ v such that $V = v' \in \Upgamma$. By the definition of $\mathcal{A}, \mathcal{A}(V) = v'$. By the semantic rule for atomic sentences v _M (V = v) = 0, and so by the semantic rule for negation, v _M (ϕ) = 1.

• Where ϕ is of the form $\psi_1 \vee \psi_2, \neg (\psi_1 \vee \psi_2), \psi_1 \wedge \psi_2, \neg(\psi_1 \wedge \psi_2), \psi_1 \supset \psi_2,$ or $\neg( \psi_1 \supset \psi_2),$ we can use the usual definitions of truth functional connectives to prove on the basis of the inductive hypothesis that $\phi \in \Upgamma$.

• Suppose ϕ is of the form ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, V =v$ and $\phi \in \Upgamma.$ We can prove by induction on the number of variables in A (written ‘|A|’) that v _M (ϕ) = 1. (CM7) and (CM5) ensure that this hypothesis holds for the case of $|{\bf A}| = |\mathcal{V}|;$ the definition of the selection function ensures that it holds where $|{\bf A}| = |\mathcal{V}| -1.$ Suppose it holds for $|{\bf A}| > |\mathcal{V}| - n.$ We can show that it holds for $|{\bf A}| = |\mathcal{V}| - n.$ For any two distinct variables V and V′ not mentioned in A, there will be a v and a v′ such that $\Upgamma$ contains sentences of the form $({\bf A} \wedge V' = v') \,\square{\kern-4.1pt}\rightarrow\, V = v$ and $({\bf A} \wedge V = v) \,\square{\kern-4.1pt}\rightarrow\, V' = v'$.

• (We know there must be such sentences by (CM4), (CM2), (R2), and (CM8). (CM4), (CM2), and (R2) ensure that $\Upgamma$ contains sentences of the form $({\bf A} \wedge V' = v') \,\square{\kern-4.1pt}\rightarrow\, V = v_1$ and $({\bf A} \wedge V = v) \,\square{\kern-4.1pt}\rightarrow\, V' = v'_1$ for each $v \in \mathcal{R}(V)$ and $v' \in \mathcal{R}(V').$ Since M is recursive, one of the variables is a nondescendant of the other. Let us assume without loss of generality that V is a nondescendant of V′. There must be a v such that $({\bf A} \wedge V' =v'_i) \,\square{\kern-4.1pt}\rightarrow\, V = v \in \Upgamma$ for every $v'_i \in\mathcal{R}(V').$ We can simply choose for our v′ the $v \in \mathcal{R}(V)$ such that $({\bf A} \wedge V= v) \,\square{\kern-4.1pt}\rightarrow\, V' = v'.$)

• By the inductive hypothesis, $v_M(({\bf A} \wedge V' = v') \,\square{\kern-4.1pt}\rightarrow\, V = v) = v_M(({\bf A} \wedge V = v) \,\square{\kern-4.1pt}\rightarrow\, V' = v') = 1.$ Since (we’ve assumed without loss of generality) V is a nondescendant of $V', ({\bf A} \wedge V' = v'_i) \,\square{\kern-4.1pt}\rightarrow\, V = v \in \Upgamma$ for every $v'_i \in \mathcal{R}(V')$. By (CM6), ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, (V' = v'_i \supset V = v) \in \Upgamma$ for every $v'_i \in \mathcal{R}(V')$. By (R2) and (CM2), ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, V = v \in \Upgamma.$

• We know that $v_M({\bf A} \,\square{\kern-4.1pt}\rightarrow\, V = v) = 1,$ since ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, V =v$ follows validly from other sentences that are true in M, and all our axioms and rules are sound. (R2), (CM2), and (CM5) ensure that $\Upgamma$ contains no sentence of the form ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, V =v^*$ for v ^* ≠ v.

• Thus, for arbitrary V and $v \in \mathcal{R}(V),$ if $|{\bf A}| \leq |\mathcal{V} -(n+1),$ if ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, V =v \in \Upgamma,$ then $v_M(A \,\square{\kern-4.1pt}\rightarrow\, V =v) = 1.$

• Suppose ϕ is of the form ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, B.$ (CM4) ensures that for every atomic subsentence V _i  = v _i of B, either ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, V_i = v_i \in \Upgamma$ or ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, V_i \neq v_i \in \Upgamma.$ (R2) and (CM5) entail that the conjunction of the $V_i = v_i \in \Upgamma$ and the $V_i \neq v_i \in \Upgamma$ is compatible with B; therefore this conjunction entails B. By our inductive hypothesis, $v_M({\bf A} \,\square{\kern-4.1pt}\rightarrow\, V_i = v_i) = 1$ for all V _i such that ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, V_i = v_i \in \Upgamma,$ and $v_M({\bf A} \,\square{\kern-4.1pt}\rightarrow\, V_i \neq v_i) = 1$ for all V _i such that ${\bf A} \,\square{\kern-4.1pt}\rightarrow\, V_i \neq v_i \in \Upgamma.$ By the selection semantics for the counterfactual, $v_M({\bf A} \,\square{\kern-4.1pt}\rightarrow\, B) = 1$.