An axiomatic characterization of temporalised belief revision in the law

Abstract

This paper presents a belief revision operator that considers time intervals for modelling norm change in the law. This approach relates techniques from belief revision formalisms and time intervals with temporalised rules for legal systems. Our goal is to formalise a temporalised belief base and corresponding timed derivation, together with a proper revision operator. This operator may remove rules when needed or adapt intervals of time when contradictory norms are added in the system. For the operator, both constructive definition and an axiomatic characterisation by representation theorems are given.

Appendix

Theorem 1 An operator $\otimes$ is a prioritised legal revision for ${\mathbb {K}}$ if and only if it satisfies the postulates of (TBR-1) Success, (TBR-2) Inclusion, (TBR-3) Consistency, (TBR-4) Uniformity, and (TBR-5) Safe Retainment.

Proof

Proof has two parts. First, we start from the satisfaction of postulates to the construction as a legal revision operator. Second, we prove that an operator is a legal revision if previous postulates are satisfied.

$\underline{\Leftarrow )~\hbox {Postulates~to~construction}:}$

Let $*$ be an operator that satisfies Success, Inclusion, Consistency, Uniformity and Safe Retainment. We have to show that $*$ is a legal revision operator.

• (1) Let $\sigma ^{c}$ be a function such that for every temporalised base ${\mathbb {K}}$ and for every temporalised sentence $\alpha ^{J}$ holds $\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})) = {\mathbb {K}} \setminus {\mathbb {K}} * \alpha ^{J}$.

$\square$

We first show that $\sigma ^{c}$ is an incision function. To do this we show that the conditions in Definition 9 are satisfied by $\sigma ^{c}$; that is:

• $\sigma ^{c}$ is a well-defined function: if $\lnot \alpha ^{J}$ and $\lnot \beta ^{J}$ are such that $\varPi (\lnot \alpha ^{J},{\mathbb {K}}) = \varPi (\lnot \beta ^{J},{\mathbb {K}})$ then $\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})) = \sigma ^{c}(\varPi (\lnot \beta ^{J},{\mathbb {K}}))$.

  Let $\lnot \alpha ^{J}$ and $\lnot \beta ^{J}$ be two temporalised sentences such that $\varPi (\lnot \alpha ^{J},{\mathbb {K}}) = \varPi (\lnot \beta ^{J},{\mathbb {K}})$. We need to show that $\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})) = \sigma ^{c}(\varPi (\lnot \beta ^{J},{\mathbb {K}}))$. By Definition 6 and Definition 5, for all subset ${\mathbb {K}} '$ of ${\mathbb {K}}$, $\lnot \alpha ^{J} \in Cn^t({\mathbb {K}} ')$ if and only if $\lnot \beta ^{J} \in Cn^t({\mathbb {K}} ')$. Then $\lnot \alpha ^{J} \cup {\mathbb {K}} '$ is temporally inconsistent if and only if $\lnot \beta ^{J} \cup {\mathbb {K}} '$ is temporally inconsistent. Thus, by uniformity, ${\mathbb {K}} \cap ({\mathbb {K}} *\alpha ^{J}) = {\mathbb {K}} \cap ({\mathbb {K}} *\beta ^{J})$. Then, ${\mathbb {K}} \setminus ({\mathbb {K}} *\alpha ^{J}) = {\mathbb {K}} \setminus ({\mathbb {K}} *\beta ^{J})$. Therefore, by (1), $\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})) = \sigma ^{c}(\varPi (\lnot \beta ^{J},{\mathbb {K}}))$.

• $\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})) \subseteq \bigcup (\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$.

  Let $\beta ^{P} \in \sigma (\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$. By (1), we have that $\beta ^{P} \in {\mathbb {K}} \setminus {\mathbb {K}} *\alpha ^{J}$. Then, it holds that $\beta ^{P} \not \in {\mathbb {K}} *\alpha ^{J}$, and from ${\textit{Safe Retainment}}$ we have that $\beta ^{P}$ is not a safe element, otherwise it will be part of the revision. Since $\beta ^{P}$ is not a safe element then it holds that $\beta ^{P}$ is a sentence that can produce effects in favour of a possible temporal contradiction with $\alpha ^{J}$ where J and P are overlapped time interval. Then, $\beta ^{P}$ is in a minimal subset (under set inclusion) ${\mathbb {H}}$ of ${\mathbb {K}}$ such that ${\mathbb {H}} \cup \alpha ^{J}$ is temporally inconsistent. By Definition 6, if ${\mathbb {H}}$ is a minimal subset such that ${\mathbb {H}} \cup \alpha ^{J}$ is temporally inconsistent then ${\mathbb {H}} \in \varPi (\lnot \alpha ^{J},{\mathbb {K}})$, and therefore $\beta ^{P} \in \bigcup (\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$. Since this holds for any arbitrary $\beta ^{P} \in \sigma (\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$ we have that $\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})) \subseteq \bigcup (\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$.

• If ${\mathbb {H}} \in \varPi (\alpha ^{J},{\mathbb {K}})$, ${\mathbb {H}} \cap \sigma ^{c}(\varPi (\alpha ^{J},{\mathbb {K}})) \ne \emptyset$.

  Let ${\mathbb {H}} \in \varPi (\alpha ^{J},{\mathbb {K}})$. We need to show that ${\mathbb {H}} \cap \sigma ^{c}(\varPi (\alpha ^{J},{\mathbb {K}})) \ne \emptyset$. We should prove that, there exists $\beta ^{P} \in {\mathbb {H}}$ such that $\beta ^{P} \in \sigma ^{c}(\varPi (\alpha ^{J},{\mathbb {K}}))$. Suppose $\lnot \alpha ^{J}$ is consistent. Since we have assumed that ${\mathbb {K}}$ is temporally consistent, by consistency, ${\mathbb {K}} *\lnot \alpha ^{J}$ is temporally consistent. Since ${\mathbb {H}}$ is inconsistent with $\lnot \alpha ^{J}$ then ${\mathbb {H}} \not \subseteq {\mathbb {K}} *\lnot \alpha ^{J}$ by success. This means that there is some $\beta ^{P} \in {\mathbb {H}}$ and $\beta ^{P} \not \in {\mathbb {K}} *\lnot \alpha ^{J}$. Since ${\mathbb {H}} \subseteq {\mathbb {K}}$ it follows that $\beta ^{P} \in {\mathbb {K}} \setminus {\mathbb {K}} *\lnot \alpha ^{J}$; i.e., by our definition of $\sigma ^{c}$, $\beta ^{P} \in \sigma ^{c}(\varPi (\alpha ^{J},{\mathbb {K}}))$. Therefore, ${\mathbb {H}} \cap \sigma ^{c}(\varPi (\alpha ^{J},{\mathbb {K}})) \ne \emptyset$.

• $\beta ^{P} \in \sigma ^{c}(\varPi (\alpha ^{J},{\mathbb {K}}))$ then for some ${\mathbb {H}} \in \varPi (\alpha ^{J},{\mathbb {K}})$ such that $\beta ^{P} \in {\mathbb {H}}$ it holds that $\beta = \alpha$ or $\beta ^{P} = \delta ^{Q} \rightarrow \alpha ^{P} {\text { and }} \delta \in {\mathbb {L}}$.

  Let $\beta ^{P} \in \sigma ^{c}(\varPi (\alpha ^{J},{\mathbb {K}}))$. Then $\beta ^{P} \in {\mathbb {K}} \setminus ({\mathbb {K}} *\lnot \alpha ^{J})$, hence $\beta ^{P} \not \in {\mathbb {K}} *\lnot \alpha ^{J}$. Therefore, by Safe Retainment we have that $\beta ^{P}$ is not a safe element. Since $\beta ^{P}$ is not a safe element then it holds that $\beta ^{P}$ is a sentence that can produce effects in favour of a possible temporally contradiction with $\alpha ^{J}$ where J and P are overlapped time interval. Then, $\beta ^{P}$ is in a minimal subset (under set inclusion) ${\mathbb {H}}$ of ${\mathbb {K}}$ such that ${\mathbb {H}} \cup \alpha ^{J}$ is temporally inconsistent. By Definition 6, if ${\mathbb {H}}$ is a minimal subset such that ${\mathbb {H}} \cup \alpha ^{J}$ is temporally inconsistent then ${\mathbb {H}} \in \varPi (\alpha ^{J},{\mathbb {K}})$. Then, since $\beta ^{P}$ is a sentence that can produce effects in favour of a possible temporally contradiction with $\alpha ^{J}$, $\beta = \alpha$ or $\beta ^{P} = \delta ^{Q} \rightarrow \alpha ^{P} {\text { and }} \delta \in {\mathbb {L}}$.

Once we have proven that $\sigma ^{c}$ is a proper incision function, to finalise the proof we must show that ${\mathbb {K}} *\alpha ^{J} = {\mathbb {K}} \otimes \alpha ^{J}$.

($\subseteq$):

Let $\beta ^{[t_i]} \in {\mathbb {K}} *\alpha ^{J}$.

It follows by inclusion that $\beta ^{[t_i]} \in {\mathbb {K}} \cup \{\alpha ^{J}\}$.

Then, $\beta ^{[t_i]} \in {\mathbb {K}}$.

It follows from $\beta ^{[t_i]} \in {\mathbb {K}} *\alpha ^{J}$ and $\beta ^{[t_i]} \in {\mathbb {K}}$ that $\beta ^{[t_i]} \not \in {\mathbb {K}} \setminus {\mathbb {K}} *\alpha ^{J}$.

Thus, by (1), $\beta ^{[t_i]} \not \in \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$. Hence, $\beta ^{[t_i]} \in {\mathbb {K}} \otimes \alpha ^{J}$.

($\supseteq$):

Let $\beta ^{[t_i]} \in {\mathbb {K}} \otimes \alpha ^{J}$.

By definition, $\beta ^{[t_i]} \in ({\mathbb {K}} \setminus \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))) \cup out(\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})),J) \cup \ \{\alpha ^{J}\}$.

From Remark 4, $out(\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})),J) \subseteq {\mathbb {K}}$ and then, $\beta ^{[t_i]} \in {\mathbb {K}}$ and $\beta ^{[t_i]} \not \in \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$.

Thus, by (1), $\beta ^{[t_i]} \not \in {\mathbb {K}} \setminus {\mathbb {K}} *\alpha ^{J}$.

Hence, $\beta ^{[t_i]} \in {\mathbb {K}} *\alpha ^{J}$.

The second part of the demonstration follows.

$\Rightarrow )$Construction to postulates: Let $\sigma ^{c}$ be a consequence incision function and $\otimes$ its associated operator and ${\mathbb {K}}$ a knowledge base. Then, for all $\alpha ^{J}$:

${\mathbb {K}} \otimes \alpha ^{J}$$=$$({\mathbb {K}} \setminus S) \cup out(S,J) \cup \ \{\alpha ^{J}\}$ where $S = \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$.

We prove that the postulates hold for the given construction, as follows.

• Success$\alpha ^{J} \in {\mathbb {K}} \otimes \alpha ^{J}$.

  Straightforward by Definition 11.

• Inclusion If $\beta ^{[t_i]} \in {\mathbb {K}} \otimes \alpha ^{J}$ then $\beta ^{[t_i]} \in {\mathbb {K}} \cup \{\alpha ^{J}\}$.

  Let $\beta ^{[t_i]} \in {\mathbb {K}} \otimes \alpha ^{J}$. From Definition 11 we have that ${\mathbb {K}} \otimes \alpha ^{J}$$=$$({\mathbb {K}} \setminus S) \cup out(S,J) \cup \ \{\alpha ^{J}\}$ where $S = \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$. Following Remark 4, for all $\beta ^{P} \in S$ there exists $\beta ^{Q} \in out(S,J)$ such that $Q \subseteq P$. Then, $out(S,J) \subseteq {\mathbb {K}}$. Therefore, $\beta ^{[t_i]} \in {\mathbb {K}} \cup \{\alpha ^{J}\}$.

• Consistence if $\alpha ^{J}$ is consistent then ${\mathbb {K}} \otimes \alpha ^{J}$ is temporally consistent.

  Suppose $\alpha$ is consistent. By Definition 8, $\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$ returns a set of sentences which they are selected from every subset of ${\mathbb {K}}$ temporally inconsistent with $\alpha ^{J}$. Then, since ${\mathbb {K}}$ is temporally consistent (Remark 1), ${\mathbb {K}} \setminus \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$ is temporally consistent. From Definition 10, if there exists $\lnot \alpha ^{Q}$ or $\beta ^{P} \rightarrow \alpha ^{Q}$ in $out(\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})),J)$ then the intervals J and Q are not overlapped. Therefore, ${\mathbb {K}} \setminus \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})) \cup out(\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})),J)$ is temporally consistent. Then, following Definition 11, ${\mathbb {K}} \otimes \alpha ^{J}$ is temporally consistent.

• Uniformity if for all ${\mathbb {K}} ' \subseteq {\mathbb {K}}$, $\{\alpha ^{J}\} \cup {\mathbb {K}} '$ is temporally inconsistent if and only if $\{\beta ^{J}\} \cup {\mathbb {K}} '$ is temporally inconsistent then ${\mathbb {K}} \cap ({\mathbb {K}} \otimes \alpha ^{J}) = {\mathbb {K}} \cap ({\mathbb {K}} \otimes \beta ^{J})$.

  Let $\alpha$ and $\beta$ be consistent sentences and J a time interval. Suppose that for all subset ${\mathbb {K}}$ ’ of ${\mathbb {K}}$, $\{\alpha ^{J}\} \cup {\mathbb {K}} '$ is temporally inconsistent if and only if $\{\beta ^{J}\} \cup {\mathbb {K}} '$ is temporally inconsistent. Then $\varPi (\alpha ^{J},{\mathbb {K}}) = \varPi (\beta ^{J},{\mathbb {K}})$ and since $\sigma ^{c}$ is a well defined function then $\sigma ^{c}(\varPi (\alpha ^{J},{\mathbb {K}})) = \sigma ^{c}(\varPi (\beta ^{J},{\mathbb {K}}))$. In the same way $out(\sigma ^{c}(\varPi (\alpha ^{J},{\mathbb {K}})),J) = out(\sigma ^{c}(\varPi (\beta ^{J},{\mathbb {K}})),J)$. Therefore, ${\mathbb {K}} \cap ({\mathbb {K}} \otimes \alpha ^{J}) = {\mathbb {K}} \cap ({\mathbb {K}} \otimes \beta ^{J})$.

• Safe retaiment$\beta ^{P} \in {\mathbb {K}} \otimes \alpha ^{J}$ if and only if $\beta ^{P}$ is a safe element with respect to $\alpha ^{J}$ in ${\mathbb {K}}$.

  • Proof that if $\beta ^{P} \in {\mathbb {K}} \otimes \alpha ^{J}$ then $\beta ^{P}$ is a safe element with respect to $\alpha ^{J}$ in ${\mathbb {K}}$.

    Let $\beta ^{P} \in {\mathbb {K}} \otimes \alpha ^{J}$ then by Definition 11 we have two alternatives:

    • $\beta ^{P} \not \in \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$. We can identify two different cases: either $\beta ^{P}$ does not belong to any minimal proof, or it does. Let us consider the two cases separately.

      If $\beta ^{P}\not \in X$ for every $X \in \varPi (\lnot \alpha ^{{\mathbb {K}}},J)$ then $\lnot \alpha ^{J}$ does not belong to any minimal set (under set inclusion) B of ${\mathbb {K}}$ such that $\lnot \alpha ^{Q} \in Cn^t(B)$ with $J \top Q$ and then it is a safe element with respect to $\alpha ^{J}$ in ${\mathbb {K}}$.

      Now consider the case where $\beta ^{P} \in X$ for every $X \in \varPi (\lnot \alpha ^{{\mathbb {K}}},J)$. Since $\beta ^{P} \not \in \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$ then by Definition 9 and Definition 8, $\beta \ne \alpha$ and $\beta \ne \delta \rightarrow \alpha$. Then $\beta ^{P}$ is not a sentence that can produce effects in favour of a possible temporally contradiction with $\alpha ^{J}$. Hence, $\beta ^{P}$ is a safe element with respect to $\alpha ^{J}$ in ${\mathbb {K}}$.

    • $\beta ^{P} \in \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$. Since $\beta ^{P} \in {\mathbb {K}} \otimes \alpha ^{J}$ then, by Definition 11, $\beta ^{P} \in out(\sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}})),J)$. Then, by Definition 10, the time intervals P and J are not overlapped. Therefore, $\beta ^{P}$ is a safe element with respect to $\alpha ^{J}$ in ${\mathbb {K}}$.

  • Proof that if $\beta ^{P}$ is a safe element with respect to $\alpha ^{J}$ in K then $\beta ^{P} \in {\mathbb {K}} \otimes \alpha ^{J}$.

    Let $\beta ^{P} \in {\mathbb {K}}$ be a safe element with respect to the revision of ${\mathbb {K}}$ by $\alpha ^{J}$. Then, $\beta ^{P}$ is not a sentence that can produce effects in favour of a possible temporally contradiction with $\alpha ^{J}$ where J and P are overlapped time interval. Then, $\beta ^{P}$ does not belong to any minimal subset under set inclusion X of ${\mathbb {K}}$ such that $\lnot \alpha ^{J} \in Cn^t(X)$ with $J \top P$. Thus, by Definition 6, $\beta ^{P} \not \in \bigcup (\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$. Following Definition 9, $\beta ^{P} \not \in \sigma ^{c}(\varPi (\lnot \alpha ^{J},{\mathbb {K}}))$. Therefore, by Definition 11, $\beta ^{P} \in {\mathbb {K}} \otimes \alpha ^{J}$.