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 $\varphi$ to the world we get by "correctly" counterfactually assuming $\varphi$. The function $C$ is not formally defined, because we do not yet have a satisfactory theory of logical counterfactuals.

Hopefully we all agree that $\varphi \in C\left(\varphi \right)$ and $\varphi \in W_N\Rightarrow C\left(\varphi \right)=W_N$.

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 $C\left(\varphi \right)=C\left(\psi \right)$ for all $v\left(\varphi \right)=v\left(\psi \right)$, and whenever $W_N$ and $C\left(\varphi \right)$ disagree on a sentence $\psi$ there must exist some causal chain ${\psi }_1,\dots {\psi }_n$ such that:

1. $v\left({\psi }_1\right)=v\left(\varphi \right)$,

2. ${\psi }_n=\psi$,

3. $W_N$ and $C\left(\varphi \right)$ disagree on every ${\psi }_i$, and

4. $v\left({\psi }_i\right)$ is a parent of $v\left({\psi }_{i+1}\right)$.

These conditions give a kind of causal structure such that changes from $W_N$ and $C\left(\varphi \right)$ must propagate through the graph $G$.

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

Many approaches to logical counterfactuals can be described either as choosing the optimal world (under some function) in which $\varphi$ is true or observing the causal consequences of setting $\varphi$ 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 ${\varphi }_1,\dots ,{\varphi }_n$, such that ${\varphi }_i\in C\left({\varphi }_{i+1}\right)$, ${\varphi }_n\in C\left({\varphi }_1\right)$, and the worlds $C\left({\varphi }_i\right)$ are not all the same for all $i$.

Given a partial order $\succ$ , we say that $C$ optimizes $\succ$ if for all $\varphi \in W$ and $W\ne C\left(\varphi \right)$ we have $C\left(\varphi \right)\succ W$.

Our main result is that the following are equivalent:

1. $C$ optimizes $f$ for some function $f$.

2. $C$ is consistent with $G$ and $v$ for some DAG $G$ and map $v$.

3. $C$ has no nontrivial cycles.

4. $C$ optimizes $\succ$ for some partial order $\succ$.

Proof:

1 $\Rightarrow$ 2: Construct the graph $G$ with a vertex for every world in the image of $C$. The map $v$ sends $\varphi$ to the vertex associated with $C\left(\varphi \right)$. Insert an edge to $W_N$ from every other vertex. Insert an edge from the vertex associated with $W_1\ne W_N$ to the vertex associated with $W_2\ne W_N$ whenever $f\left(W_1\right)<f\left(W_2\right)$. Clearly $C\left(\varphi \right)=C\left(\psi \right)$ for all $v\left(\varphi \right)=v\left(\psi \right)$.

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

2 $\Rightarrow$ 3: Consider a nontrivial cycle ${\varphi }_1,\dots ,{\varphi }_n$. If any of these sentences were in $W_N$, then they would all be true in $W_N$, since $C\left(\varphi \right)=W_N$ whenever $\varphi \in W_N$. This would contradict the fact that ${\varphi }_1,\dots ,{\varphi }_n$ is a nontrivial cycle.

Otherwise, since ${\varphi }_i\in C\left({\varphi }_{i+1}\right)$, there must be a path from $v\left({\varphi }_{i+1}\right)$ to $v\left({\varphi }_i\right)$ in $G$, since $C\left({\varphi }_{i+1}\right)$ and $W_N$ differ on ${\varphi }_i$. Concatenating these paths together would give a cycle in $G$ unless the $v\left({\varphi }_i\right)$ are all the same vertex. However, if all of the $v\left({\varphi }_i\right)$ are the same vertex, then all of the $C\left({\varphi }_i\right)$ would be the same world, which would contradict the fact that ${\varphi }_1,\dots ,{\varphi }_n$ is a nontrivial cycle.

3 $\Rightarrow$ 4: Consider the partial order on the image of $C$ constructed by saying that $W_1\succ W_2$ if $W_1=C\left(\varphi \right)$ for some $\varphi \in W_2$, 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 $W_i=C\left({\varphi }_i\right)$ with ${\varphi }_i\in W_{i+1}$ and ${\varphi }_n\in W_1.$ 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 $W_1\succ W_2$. Note that this $C$ optimizes this partial order by definition.

4 $\Rightarrow$ 1: Let $C$ optimize the partial order $\succ$. Consider the restriction of $\succ$ 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 $f\left(W_n\right)$ such that:

1. $f\left(W_n\right)>0$,

2. $f\left(W_n\right)>f\left(W_i\right)$ for all $i<n$ and $W_n\succ W_i$, and

3. $f\left(W_n\right)<f\left(W_i\right)$ for all $i<n$ and $W_i\succ W_n$.

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

$\square$

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 $\varphi$ and $\psi$ such that $\varphi$ and $\psi$ are clearly counterfactual consequences of each other, but do not correspond to identical counterfactual worlds.

Do the "correct" logical counterfactuals exhibit a nontrivial cycle?