# 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 $\mathcal{W}$ denote the set of all worlds. The actual world $W_\mathbb{N}\in\mathcal{W}$ is the world consisting of all sentences that are true about $\mathbb{N}$. 

Consider the function $C:L\rightarrow\mathcal{W}$ which sends the sentence $\phi$ to the world we get by "correctly" counterfactually assuming $\phi$. The function $C$ is not formally defined, because we do not yet have a satisfactory theory of logical counterfactuals.

Hopefully we all agree that $\phi\in C(\phi)$ and
$\phi\in W_\mathbb{N}\Rightarrow C(\phi)=W_\mathbb{N}$.

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 $C(\phi)=C(\psi)$ for all $v(\phi)=v(\psi)$, and whenever $W_{\mathbb{N}}$ and $C(\phi)$ disagree on a sentence $\psi$ there must exist some causal chain $\psi_1,\ldots \psi_n$ such that:

1) $v(\psi_1)=v(\phi)$,

2) $\psi_n=\psi$,

3) $W_{\mathbb{N}}$ and $C(\phi)$ disagree on every $\psi_i$, and

4) $v(\psi_i)$ is a parent of $v(\psi_{i+1})$.

These conditions give a kind of causal structure such that changes from $W_{\mathbb{N}}$ and $C(\phi)$ must propagate through the graph $G$.

Given a function $f:\mathcal{W}\rightarrow\mathbb{R}$, we say that $C$ optimizes $f$ if for all $\phi\in W$ and $W\neq C(\phi)$ we have $f(C(\phi))>f(W)$.

Many approaches to logical counterfactuals can be described either as choosing the optimal world (under some function) in which $\phi$ is true or observing the causal consequences of setting $\phi$ 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 $\phi_1,\ldots,\phi_n$, such that $\phi_{i}\in C(\phi_{i+1})$, $\phi_n\in C(\phi_1)$, and the worlds $C(\phi_i)$ are not all the same for all $i$.

Given a partial order $\succ$ , we say that $C$ optimizes $\succ$ if for all $\phi\in W$ and $W\neq C(\phi)$ we have $C(\phi)\succ W$.

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 $\phi$ to the vertex associated with $C(\phi)$. Insert an edge to $W_{\mathbb{N}}$ from every other vertex. Insert an edge from the vertex associated with $W_1\neq W_{\mathbb{N}}$ to the vertex associated with $W_2\neq W_{\mathbb{N}}$ whenever $f(W_1)<f(W_2)$. Clearly $C(\phi)=C(\psi)$ for all $v(\phi)=v(\psi)$. 

Assume $W_{\mathbb{N}}$ and $C(\phi)$ disagree on a sentence $\psi$. If $\psi\in W_{\mathbb{N}}$ then $v(\phi)\rightarrow v(\psi)$, since $v(\psi)$ is the vertex associated with $W_\mathbb{N}$ which is a child of every vertex. Therefore you get a length 1 path from $v(\phi)$ to $v(\psi)$. Otherwise, $\psi\notin W_{\mathbb{N}}$, so $\psi\in C(\phi)$. In this case, note that since $\psi$ is in both $C(\phi)$ and $C(\psi)$, and these worlds are not the same, it must be that $f(C(\psi))>f(C(\phi))$. Again, this means that you get a length 1 path from $v(\phi)$ to $v(\psi)$. Therefore $C$ is consistent with $G$ and $v$.

2 $\Rightarrow$ 3:
Consider a nontrivial cycle $\phi_1,\ldots,\phi_n$. If any of these sentences were in $W_\mathbb{N}$, then they would all be true in $W_\mathbb{N}$, since $C(\phi)=W_\mathbb{N}$ whenever $\phi\in W_\mathbb{N}$. This would contradict the fact that $\phi_1,\ldots,\phi_n$ is a nontrivial cycle.

Otherwise, since $\phi_i\in C(\phi_{i+1})$, there must be a path from $v(\phi_{i+1})$ to $v(\phi_i)$ in $G$, since $C(\phi_{i+1})$ and $W_\mathbb{N}$ differ on $\phi_i$. Concatenating these paths together would give a cycle in $G$ unless the $v(\phi_i)$ are all the same vertex. However, if all of the $v(\phi_i)$ are the same vertex, then all of the $C(\phi_i)$ would be the same world, which would contradict the fact that $\phi_1,\ldots,\phi_n$ 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 $W_1=C(\phi)$ for some $\phi\in W_2$, 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 $W_1,\ldots,W_n$, such that each $W_i=C(\phi_i)$ with $\phi_i\in W_{i+1}$ and $\phi_n\in W_{1}.$ This is would be a nontrivial cycle.

Extend this partial order to all of $\mathcal{W}$ 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 $W_1,W_2,\ldots.$ Embed the partial order into $\mathbb{R}$ by repeatedly defining $f(W_n)$ such that:

1) $f(W_n)>0$, 

2) $f(W_n)>f(W_i)$ for all $i<n$ and $W_n\succ W_i$, and

3) $f(W_n)<f(W_i)$ for all $i<n$ and $W_i\succ W_n$.

Define $f(W)=0$ if $W$ is not in the image of $C$. Clearly this function $f$ is constructed such that C optimizes $f$. 

$\square$

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 $\phi$ and $\psi$ such that $\phi$ 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?