A New Halpern-Pearl Definition of Actual Causality by Appealing to the Default World

Abstract

Halpern and Hitchcock appealed to the normality of witness worlds to solve the problem of isomorphism in the Halpern-Pearl definition of actual causality. This paper first proposes a new isomorphism counterexample, called “bogus permission,” to show that their approach is unsuccessful. Then, to solve the problem of isomorphism, I propose a new improvement over the Halpern-Pearl definition by introducing default worlds. Finally, I demonstrate that my new definition can resolve more potential counterexamples than similar approaches in the current literature, including the Lewisian causal dependence, Menziesian causal dependence, and modified version of the Halpern-Pearl definition. Some other advantages of my definition are also discussed.

Appendix 1

Proofs of theorems require a property of interventionist counterfactuals that have been proved by Galles and Pearl (1997).

Composition. If $\left( {M,\vec{u}} \right) \vDash \left[ {\vec{X} = \vec{x}} \right]\vec{W} = \vec{w}$, then: $\left( {M,\vec{u}} \right) \vDash \left[ {\vec{X} = \vec{x},\vec{W} = \vec{w}} \right]\vec{Y} = \vec{y}$, if and only if, $\left( {M,\vec{u}} \right) \vDash \left[ {\vec{X} = \vec{x}} \right]\vec{Y} = \vec{y}$.

Theorem 1

Let $\overrightarrow{X}=\overrightarrow{x}$ be $X=x$, $\varphi$ be $Y=y$, and they are Boolean variables, the following conditions hold:

• (1) If they have Menziesian causal dependence, they satisfy AC2^def (a).

• (2) If they satisfy AC2^def and for any $x^{\prime}$, $\left( {M,\vec{u}^{d} } \right) \vDash \left[ {X = x^{\prime}} \right]\vec{W} = \vec{w}$ where $\overrightarrow{W}=\overrightarrow{w}$ is the contingencies, then they have Menziesian causal dependence.

Proof:

(1) Let $\overrightarrow{W}$ be $\varnothing$. AC2^def (a) trivially holds.

(2) Assuming that $X = x$ and $Y = y$ satisfy AC2^def, and $\left( {M,\vec{u}^{d} } \right) \vDash \left[ {X = x} \right]\vec{W} = \vec{w}$, we at least have $\left( {M,\vec{u}^{d} } \right) \vDash \left[ {X = x,\vec{W} = \vec{w}} \right]Y = y$ and $\left( {M,\vec{u}^{d} } \right) \vDash \left[ {X = x} \right]\vec{W} = \vec{w}$. By composition, $\left( {M ,\vec{u}^{d} } \right) \vDash \left[ {\vec{X} = \vec{x}} \right]Y = y$. Since $X = x$ and $Y = y$ satisfy AC2^def, we have there is a $X = x^{\prime}$ such that $\left( {M ,\vec{u}^{d} } \right) \vDash \left[ {X = x^{\prime},\vec{W} = \vec{w}} \right]\neg \left( {Y = y} \right)$. Therefore, $\left( {M_{X = x^{\prime}} ,\vec{u}^{d} } \right) \vDash \left[ {\vec{W} = \vec{w}} \right]\neg \left( {Y = y} \right)$. Since $\left( {M_{X = x^{\prime}} ,\vec{u}^{d} } \right) \vDash \vec{W} = \vec{w}$, again by the composition, $\left( {M ,\vec{u}^{d} } \right) \vDash \left[ {X = x^{\prime}} \right]Y = y^{\prime}$.

Theorem 2

Let $\overrightarrow{X}=\overrightarrow{x}$ be $X=x$, and $\varphi$ be $Y=y$, and they are Boolean variables, if AC2^def holds when taking $\left( {M,\vec{u}} \right) \vDash \vec{W} = \vec{w}$ as contingencies and $\left( {M,\vec{u}} \right) \vDash X = x$, then for the same $\vec{W} = \vec{w}$, they satisfy AC2^mod (b), and hence AC2^org (b).

Proof:

Assuming that $X = x$ and $Y = y$ satisfy AC2^def when taking $\left( {M,\vec{u}} \right) \vDash \vec{W} = \vec{w}$ as the contingency, for all subsets $\vec{Z}^{\prime}$ of $\vec{Z}$ and $\vec{z}^{*}$ such that $\left( {M,\vec{u}} \right) \vDash \vec{Z}^{\prime} = \vec{z}^{*}$, we have $\left( {M_{{\vec{W} = \vec{w}}} ,\vec{u}^{d} } \right) \vDash \left[ {X = x,\vec{Z}^{\prime} = \vec{z}^{*} } \right]Y = y$. Let $\vec{Z}^{\prime}$ be $\vec{Z}$, since $\left( {M,\vec{u}} \right) \vDash \vec{W} = \vec{w}$, we have $\left( {M_{{\vec{Z} = \vec{z}^{*} ,\vec{W} = \vec{w}}} ,\vec{u}^{d} } \right) = \left( {M_{{\vec{W} = \vec{w}}} ,\vec{u}} \right) = \left( {M,\vec{u}} \right)$. Since $\left( {M_{{\vec{Z} = \vec{z}^{*} ,\vec{W} = \vec{w}}} ,\vec{u}^{d} } \right) \vDash \left[ {X = x} \right]Y = y$, $\left( {M,\vec{u}} \right) \vDash \left[ {X = x} \right]Y = y$. Because $\left( {M,\vec{u}} \right) \vDash X = x$, we have $\left( {M,\vec{u}} \right) \vDash Y = y$. s$\left( {M,\vec{u}} \right) \vDash \vec{W} = \vec{w} \wedge X = x \wedge \vec{Z}^{\prime} = \vec{z}^{*}$ for all subsets $\vec{Z}^{\prime}$ of $\vec{Z}$ and $\vec{z}^{*}$ such that $\left( {M,\vec{u}} \right) \vDash \vec{Z}^{\prime} = \vec{z}^{*}$. Therefore, for the same $\vec{W} = \vec{w}$ and $\vec{Z}^{\prime} = \vec{z}^{*}$, $\left( {M_{{\vec{W} = \vec{w}}} ,\vec{u}} \right) \vDash \left[ {X = x,\vec{Z}^{\prime} = \vec{z}^{*} } \right]Y = y$.

proof from AC2^mod (b) to AC2^org(b) is trivial.