Optimal and Causal Counterfactual Worlds

Let $L$ denote the language of Peano arithmetic. A (counterfactual) world $W$ is any subset of $L$ . These worlds need not be consistent. Let $W$ denote the set of all worlds. The actual world $W_{N} \in W$ is the world consisting of all sentences that are true about $N$ .

Consider the function $C : L \to W$ which sends the sentence $ϕ$ to the world we get by “correctly” counterfactually assuming $ϕ$ . The function $C$ is not formally defined, because we do not yet have a satisfactory theory of logical counterfactuals.

Hopefully we all agree that $ϕ \in C (ϕ)$ and $ϕ \in W_{N} \Rightarrow C (ϕ) = W_{N}$ .

Given an (infinite) directed acyclic graph $G$ , and a map $v$ from sentences to vertices of $G$ , we say that $C$ is consistent with $G$ and $v$ if $C (ϕ) = C (ψ)$ for all $v (ϕ) = v (ψ)$ , and whenever $W_{N}$ and $C (ϕ)$ disagree on a sentence $ψ$ there must exist some causal chain $ψ_{1}, \dots ψ_{n}$ such that:

$v (ψ_{1}) = v (ϕ)$ ,
$ψ_{n} = ψ$ ,
$W_{N}$ and $C (ϕ)$ disagree on every $ψ_{i}$ , and
$v (ψ_{i})$ is a parent of $v (ψ_{i + 1})$ .

These conditions give a kind of causal structure such that changes from $W_{N}$ and $C (ϕ)$ must propagate through the graph $G$ .

Given a function $f : W \to R$ , we say that $C$ optimizes $f$ if for all $ϕ \in W$ and $W \neq C (ϕ)$ we have $f (C (ϕ)) > f (W)$ .

Many approaches to logical counterfactuals can be described either as choosing the optimal world (under some function) in which $ϕ$ is true or observing the causal consequences of setting $ϕ$ to be true. The purpose of this post is to prove that these frameworks are actually equivalent, and to provide a strategy for possibly showing that no attempt at logical counterfactuals which could be described within either framework could ever be what we mean by “correct” logical counterfactuals.

A nontrivial cycle in $C$ is a list of sentences $ϕ_{1}, \dots, ϕ_{n}$ , such that $ϕ_{i} \in C (ϕ_{i + 1})$ , $ϕ_{n} \in C (ϕ_{1})$ , and the worlds $C (ϕ_{i})$ are not all the same for all $i$ .

Given a partial order $≻$ , we say that $C$ optimizes $≻$ if for all $ϕ \in W$ and $W \neq C (ϕ)$ we have $C (ϕ) ≻ W$ .

Our main result is that the following are equivalent:

$C$ optimizes $f$ for some function $f$ .
$C$ is consistent with $G$ and $v$ for some DAG $G$ and map $v$ .
$C$ has no nontrivial cycles.
$C$ optimizes $≻$ for some partial order $≻$ .

Proof:

1 $\Rightarrow$ 2: Construct the graph $G$ with a vertex for every world in the image of $C$ . The map $v$ sends $ϕ$ to the vertex associated with $C (ϕ)$ . Insert an edge to $W_{N}$ from every other vertex. Insert an edge from the vertex associated with $W_{1} \neq W_{N}$ to the vertex associated with $W_{2} \neq W_{N}$ whenever $f (W_{1}) < f (W_{2})$ . Clearly $C (ϕ) = C (ψ)$ for all $v (ϕ) = v (ψ)$ .

Assume $W_{N}$ and $C (ϕ)$ disagree on a sentence $ψ$ . If $ψ \in W_{N}$ then $v (ϕ) \to v (ψ)$ , since $v (ψ)$ is the vertex associated with $W_{N}$ which is a child of every vertex. Therefore you get a length 1 path from $v (ϕ)$ to $v (ψ)$ . Otherwise, $ψ \notin W_{N}$ , so $ψ \in C (ϕ)$ . In this case, note that since $ψ$ is in both $C (ϕ)$ and $C (ψ)$ , and these worlds are not the same, it must be that $f (C (ψ)) > f (C (ϕ))$ . Again, this means that you get a length 1 path from $v (ϕ)$ to $v (ψ)$ . Therefore $C$ is consistent with $G$ and $v$ .

2 $\Rightarrow$ 3: Consider a nontrivial cycle $ϕ_{1}, \dots, ϕ_{n}$ . If any of these sentences were in $W_{N}$ , then they would all be true in $W_{N}$ , since $C (ϕ) = W_{N}$ whenever $ϕ \in W_{N}$ . This would contradict the fact that $ϕ_{1}, \dots, ϕ_{n}$ is a nontrivial cycle.

Otherwise, since $ϕ_{i} \in C (ϕ_{i + 1})$ , there must be a path from $v (ϕ_{i + 1})$ to $v (ϕ_{i})$ in $G$ , since $C (ϕ_{i + 1})$ and $W_{N}$ differ on $ϕ_{i}$ . Concatenating these paths together would give a cycle in $G$ unless the $v (ϕ_{i})$ are all the same vertex. However, if all of the $v (ϕ_{i})$ are the same vertex, then all of the $C (ϕ_{i})$ would be the same world, which would contradict the fact that $ϕ_{1}, \dots, ϕ_{n}$ is a nontrivial cycle.

3 $\Rightarrow$ 4: Consider the partial order on the image of $C$ constructed by saying that $W_{1} ≻ W_{2}$ if $W_{1} = C (ϕ)$ for some $ϕ \in W_{2}$ , and taking the transitive closure of these rules. If this were not a partial order, it would have to be because we created a cycle of worlds $W_{1}, \dots, W_{n}$ , such that each $W_{i} = C (ϕ_{i})$ with $ϕ_{i} \in W_{i + 1}$ and $ϕ_{n} \in W_{1} .$ This is would be a nontrivial cycle.

Extend this partial order to all of $W$ by saying that if $W_{1}$ is in the image of $C$ and $W_{2}$ is not, then $W_{1} ≻ W_{2}$ . Note that this $C$ optimizes this partial order by definition.

4 $\Rightarrow$ 1: Let $C$ optimize the partial order $≻$ . Consider the restriction of $≻$ to the image of $C$ . This is a partial order on a countable set. Order the worlds in this partial order $W_{1}, W_{2}, \dots .$ Embed the partial order into $R$ by repeatedly defining $f (W_{n})$ such that:

$f (W_{n}) > 0$ ,
$f (W_{n}) > f (W_{i})$ for all $i < n$ and $W_{n} ≻ W_{i}$ , and
$f (W_{n}) < f (W_{i})$ for all $i < n$ and $W_{i} ≻ W_{n}$ .

Define $f (W) = 0$ if $W$ is not in the image of $C$ . Clearly this function $f$ is constructed such that C optimizes $f$ .

$□$

This result is useful not just for establishing the equivalence of (a certain type of) optimization and acyclic causal networks, but also for providing a strategy for showing that “correct” logical counterfactuals cannot arise as an optimization process or through an acyclic causal network. To show this, one only has to exhibit a single nontrivial cycle. In the simplest case, this can be done by exhibiting a pair of sentences $ϕ$ and $ψ$ such that $ϕ$ and $ψ$ are clearly counterfactual consequences of each other, but do not correspond to identical counterfactual worlds.

Do the “correct” logical counterfactuals exhibit a nontrivial cycle?