Action Type Deontic Logic

Abstract

A new deontic logic, Action Type Deontic Logic, is presented. To motivate this logic, a number of benchmark cases are shown, representing inferences a deontic logic should validate. Some of the benchmark cases are singled out for further comments and some formal approaches to deontic reasoning are evaluated with respect to the benchmark cases. After that follows an informal introduction to the ideas behind the formal semantics, focussing on the distinction between action types and action tokens. Then the syntax and semantics of Action Type Deontic Logic is presented and it is shown to meet the benchmarks. Finally, possibilities for further research are indicated. In the appendix, decidability of the satisfiability of formulas is proved via a technique known from monadic First Order Logic.

Appendix

I now state the definitions and lemmas needed to prove theorem 2 stating that Action Type Deontic Logic is decidable.

Definition 9

Let $M=\langle G, V, \theta \rangle$ and $M'=\langle G', V', \theta '\rangle$ be models. Define a strong homomorphism from $M$ to $M'$ to be a function $h$ from $dom(M)$ to $dom(M')$, such that.

• 1. For any $p_i\in P$, $\theta (p_i)=\theta '(p_i)$.

• 2. For any $T_i\in BT$, $\alpha \in G$ ($\alpha \in V(T_i)$ iff $h(\alpha )\in V'(T_i)$).

For the following lemma to go through we need a surjective strong homomorphism. Let $M$ and $M'$ be models and $h$ a surjective, strong homomorphism from $M$ to $M'$.

Lemma 1

• 1. For any $T\in ACT$, $\alpha \in G$ ( $\alpha \in V(T)$ iff $h(\alpha )\in V'(T)$).

• 2. For any $\phi \in WFF$, $M\vDash \phi$ iff $M'\vDash \phi$

Part 2 of the above lemma shows that satisfiability is invariant under surjective, strong homomorphisms.

Proof

• 1. We define the complexity of an action term as the number of action type terms in the action type term. The proof is by induction on the complexity of action terms.

  • 1. Induction basis. Let $T_i\in BT$, $\alpha \in G$. We have $\alpha \in V(T_i)$ iff $h(\alpha )\in V'(T_i)$ by definition of surjective, strong homomorphism.

  • 2. Suppose the lemma holds for any action type term of complexity lower than $n$. Now suppose $T$ is an action type term with complexity $n$. The proof divides into several cases.

    • (a) $T$ is of the form $-S$. Left to right. If $\alpha \in V(-S)$, then $\alpha \in G\setminus V(S)$, and so $\alpha \not \in V(S)$. Then $h(\alpha )\not \in V'(S)$ (by induction hypothesis), so $h(\alpha )\in V'(-S)$. Right to left. If $h(\alpha )\in V'(-S)$, then $h(\alpha )\in G'\setminus V'(S)$, and so $h(\alpha )\not \in V'(S)$. Then $\alpha \not \in V(S)$ (by induction hypothesis), so $\alpha \in V(-S)$.

    • (b) $T$ is of the form $(S\cap R)$. Left to right. If $\alpha \in V(S\cap R)$, then $\alpha \in V(S)$ and $\alpha \in V(R)$. Hence, by induction hypothesis, $h(\alpha )\in V'(S)$ and $h(\alpha )\in V'(R)$. Hence $h(\alpha )\in V'(S\cap R)$. Right to left. If $h(\alpha )\in V'(S\cap R)$, then $h(\alpha )\in V'(S)$ and $h(\alpha )\in V'(R)$. Hence, by induction hypothesis, $\alpha \in V(S)$ and $\alpha \in V(R)$. Hence $\alpha \in V(S\cap R)$.

    • (c) $T$ is of the form $(S\uplus R)$. Left to right If $\alpha \in V(S\uplus R)$, then 1) $\alpha \in V(S)$ or 2) $\alpha \in V(R)$. Assume 1. By induction hypothesis $h(\alpha )\in V'(S)$. Since $V(R)\ne \emptyset$ by definition of $\uplus$, there is $\beta \in V(R)$. By induction hypothesis $h(\beta )\in V'(R)$. Hence $h(\alpha )\in V'(S\uplus R)$. The other case is similar. Right to left. If $h(\alpha )\in V(S\uplus R)$, then 1) $h(\alpha )\in V'(S)$ or 2) $h(\alpha )\in V'(R)$. Assume 1. By induction hypothesis $\alpha \in V(S)$. $V'(R)\ne \emptyset$ by definition of $\uplus$. Let $\delta \in V'(R)$. Since $h$ is surjective, there is $\beta \in G$ such that $h(\beta )=\delta$. By induction hypothesis $\beta \in V(R)$. Hence, $\alpha \in V(S\uplus R)$. The other case is similar.

• 2. We define the complexity of a formula as the number of propositional connectives in the formula. The proof is by induction on the complexity of formulas.

  • 1. There are two basic cases, according to whether $\phi$ is a propositional variable $p_i\in P$ or $\phi$ is of the form $\mathbf{may}[T]$ for some $T\in ACT$. $M\vDash p_i$ iff$M'\vDash p_i$ is obvious from definition of surjective, strong homomorphism. Assume $M\vDash \mathbf{may}[T]$. Then there is $\alpha \in V(T)$. By part 1 of this lemma, $h(\alpha )\in V'(T)$. Hence $M'\vDash \mathbf{may}[T]$. Assume $M'\vDash \mathbf{may}[T]$. Then there is $\beta \in V'(T)$. Since $h$ is surjective, there is $\delta \in G$, such that $h(\delta )=\beta$. Hence by part 1 of this lemma, $\delta \in V(T)$. Therefore, $M\vDash \mathbf{may}[T]$.

  • 2. Suppose the lemma holds for all formulas of complexity lower than $n$. Now, let $\phi$ be a formula of complexity $n$. We consider two cases.

    • (a) Suppose $\phi$ is of the form $\lnot \psi$. Left to right. If $M\vDash \lnot \psi$, then $M\nvDash \psi$ and hence $M'\nvDash \psi$ (by induction hypothesis) and so $M'\vDash \lnot \psi$, i.e. $M'\vDash \phi$. Right to left. If $M'\vDash \lnot \psi$, then $M'\nvDash \psi$ and hence $M\nvDash \psi$ (by induction hypothesis) and so $M\vDash \lnot \psi$.

    • (b) Suppose $\phi$ is of the form $\psi \wedge \chi$. Left to right. If $M\vDash \psi \wedge \chi$, then $M\vDash \psi$ and $M\vDash \chi$. Hence $M'\vDash \psi$ and $M'\vDash \chi$ (by induction hypothesis). So, $M'\vDash \psi \wedge \chi$ i.e. $M'\vDash \phi$. Right to left. If $M'\vDash \psi \wedge \chi$, then $M'\vDash \psi$ and $M'\vDash \chi$. Hence $M\vDash \psi$ and $M\vDash \chi$ (by induction hypothesis). So, $M\vDash \psi \wedge \chi$.$\square$

Now we show decidability of satisfiability with a technique used for showing this for monadic First Order Logic, see e.g. Jeffrey et al. (1997). We prove the following lemma.

Lemma 2

If a formula $\phi$ is satisfiable, then it is satisfiable in a model with a domain of size at most $2^n$, where $n$ is the number of basic action type terms occurring in $\phi$.

Proof

Let $\phi$ be a formula and $M=\langle G, V, \theta \rangle$ a model such that $M\vDash \phi$. Let $S=\{s_1,\ldots ,s_n\}$ be the basic action types occurring in $\phi$, and $Q=\{q_1,\ldots ,q_m \}$ be the propositional variables occurring in $\phi$. We first derive the model $M'=\langle G', V', \theta '\rangle$ from $M$ as follows.

• 1. $dom(M')=dom(M).$

• 2. For any $T_i\in BT$, if $T_i\in S$, then $V'(T_i)=V(T_i)$. Otherwise $V'(T_i)=\emptyset$.

• 3. For any $p_i\in P$, if $p_i\in Q$, then $\theta '(p_i)=T(p_i)$. Otherwise $\theta '(p_i)=f$.

In a sense, $M'$ is $M$ stripped of all information not pertaining to $\phi$. Since the truth of $\phi$ only depends on its subformulas, an easy inductive argument shows that $M'\vDash \phi$.

Let the signature of an $\alpha \in dom(M')$ be a sequence $(j_1, \ldots , j_n)$ of $0's$ and $1's$, such that, for each $s_k\in S$

• 1. $j_i=1$, if $\alpha \in V(s_k).$

• 2. $j_i=0$, if $\alpha \not \in V(s_k)$.$\square$

There are at most $2^n$ signatures. Call $\alpha$ and $\beta$ similar, if they have the same signature. Obviously, similarity is an equivalence relation. For an element $\alpha \in G'$, we use the notation $|\alpha |$ for the equivalence class $C$ s.t. $\alpha \in C$. Now we define $N= \langle F, V^N, \theta ^N \rangle$ as follows. The domain of $N$ consists of exactly one element from each of the equivalence classes induced by the similarity relation. Hence $dom(N)$ is of size at most $2^n$. Further, we define:

Definition 10

• 1. For each $T_i\in BT, \beta \in dom(N)$ ( $\beta \in V^N(T_i)$ iff $\beta \in V'(T_i)$).

• 2. For each $p_i\in P, \theta ^N(p_i)=\theta '(p_i)$.

We now define a function $h$, from $dom(M')$ to $dom(N)$ as follows. For each $\alpha \in G', h(\alpha )= \delta$, where $\delta \in dom(N)$, and $|\alpha |=|\delta |$. Clearly, there is exactly one such $\delta$, hence $h$ is a surjective function. Further, by Definition 10 $h$ is a strong homomorphism. Hence by Lemma 1, $N\vDash \phi$.

Theorem 2 follows directly from Lemma 2.