Norm-system revision: theory and application

Abstract

This paper generalises classical revision theory of the AGM brand to sets of norms. This is achieved substituting input/output logic for classical logic and tracking the changes. Operations of derogation and amendment—analogues of contraction and revision—are defined and characterised, and the precise relationship between contraction and derogation, on the one hand, and derogation and amendment on the other, is established. It is argued that the notion of derogation, in particular, is a very important analytical tool, and that even core deontic concepts such as that of permission resists a satisfactory analysis without it. By way of illustration the last section of the paper analyses the much debated concept of positive permission, of which there turns out to be more than one kind.

Appendix

1.1 Proof of theorem 8

Proof

We need to show that any operation ‘–’ satisfying the listed properties coincides with a partial meet derogation operation. That is, we need to show the existence of a selection function δ such that

$$ out(G) - (a, b) = \bigcap\delta (out(G)\perp(a, b)) $$

We construct the selection function δ, using a familiar technique, as follows:

$$ \delta(out(G)\perp(a, b))= \left\{ \begin{array}{ll} \{out(G)\} \quad \hbox{ if }out(G)\perp(a, b) = \emptyset,\hbox{ otherwise } \\ \{F\in out(G)\perp(a, b):out(G)-(a, b)\subseteq F\} \end{array}\right. $$

We need to show, first of all, that δ is well-defined and that it is a selection function: Well-definedness. Starting with well-definedness we need to show that $\delta(out(G)\perp(a, b)) = \delta(out(G)\perp(c, d))$ whenever $out(G)\perp(a, b) = out(G)\perp(c, d)$. So assume that $out(G)\perp(a, b) = out(G)\perp (c, d)$.

In the limiting case that $out(G)\perp(a, b) = \emptyset = out(G)\perp(c, d)$, we have $\delta(out(G)\perp(a, b)) = \{out(G)\} = \delta(out(G)\perp(c, d))$, by the first case of the definition of δ, so we may assume that $out(G)\perp(a, b) \neq \emptyset \neq out(G)\perp(c, d)$.

Proceeding on that assumption, we we turn to the case where either (a, b) or (c, d), say wlog. (a, b), is not in out(G). Then $out(G)\perp (a, b) = \{out(G)\}$, whence $out(G)\perp (c, d) = \{out(G)\}$ as well. Hence, it suffices to show that $\delta(out(G)\perp (g, h))\neq \emptyset$ for any (g, h) whenever $out(G)\perp (g, h)\neq \emptyset$. The proof splits into two cases:

• 1. Suppose $(g, h)\notin out(G)$, then $out(G)\subseteq out(G)-(g, h)$ by D-Vacuity. Hence $out(G)\in \delta(out(G)\perp(g, h))$ by the second case of the definition of δ, and we are done.

• 2. Suppose on the contrary that (g, h) ∈ out(G). Since we are assuming that $out(G)\perp (g, h)$ is non-empty it follows that $\nvdash h$ by the properties of out-entailment. By D-Success, therefore, it follows that $(g, h)\notin out(G)-(g, h)$, whence, by D-inclusion and lemma 8, out(G) − (g, h) can be expanded to a subset F of out(G) such that maximally $(g, h)\notin out(F)$. It follows that $F\in out(G)\perp(g, h)$ and since $out(G)-(g, h)\subseteq F$ we also have $F\in \delta(out(G)\perp(g, h))$. Hence $\delta(out(G)\perp(a, b))= \delta(out(G)\perp(c, d))$ as desired.

This completes the limiting cases. $\square$

Now, for the principal case where $out(G)\perp(a, b) = out(G)\perp (c, d) \neq \emptyset$ and (a, b), (c, d) ∈ out(G), note that $F\in \delta(out(G)\perp(a, b))$ implies $out(G)-(a, b)\subseteq F$ by the second case of the definition of δ. Since $out(G)\perp(a, b) = out(G)\perp(c, d)$ and (a, b), (c, d) ∈ out(G) it follows, by lemma 13 that out((a, b)) = out((c, d)), whence out(G) − (a, b) = out(G) − (c, d) by D-Extensionality. Hence $out(G)-(c, d) \subseteq F$, so $F\in \delta(out(G)\perp(c, d))$ by the definition of δ. Therefore $\delta(out(G)\perp(a, b))\subseteq \delta(out(G)\perp(c, d))$. The other direction is similar so δ is well-defined.

δ is a selection function. To prove that δ is a selection function in the sense of definition 3, we need to show that $\emptyset \subset \delta(out(G)\perp(a, b))\subseteq out(G)\perp(a, b)$ whenever $out(G)\perp(a, b)\neq \emptyset$, and that $\delta(out(G)\perp(a, b)) = \{out(G)\}$ otherwise. The is immediate from the first case of the definition of δ. For $\emptyset\subseteq \delta(out(G)\perp(a, b))$ we have already shown that this holds whenever $out(G)\perp (g, h)\neq \emptyset$. The remaining case where $out(G)\perp (g, h) = \emptyset$ is again immediate from the first case of the definition of δ. It only remains to show therefore, that $\delta(out(G)\perp(a, b))\subseteq out(G)\perp(a, b)$, which is immediate from the second case of the definition of δ.

For${\bf out(G) - (a, b)} = \varvec{\bigcap\delta }{\bf (out(G)}\varvec{\perp} ({\bf a, b})).$ Finally, we need to show that $out(G) - (a, b) = \bigcap\delta (out(G)\perp(a, b))$. There are two cases to consider:

• (a) Suppose $out(G)\perp(a, b) = \emptyset$: Then $\bigcap\delta(out(G)\perp(a, b)) = out(G)$, by the definition of δ, whence $out(G)-(a, b) \subseteq out(G) = \bigcap\delta(out(G)\perp(a, b))$ by D-Inclusion. For the converse inclusion, note that $out(G)\perp(a, b) = \emptyset$ implies $\vdash b$. Hence out(G) − (a, b) = out(G) by D-Failure, and therefore $\bigcap\delta (out(G)\perp(a, b)) = out(G) \subseteq out(G)-(a, b)$ as desired.

• (b) Suppose $out(G)\perp(a, b) \neq \emptyset$: Since $F\in \delta (out(G)\perp(a, b)) \; iff\; F\in out(G)\perp(a, b)$ and $out(G)-(a, b)\subseteq F$, the inclusion $out(G)-(a, b) \subseteq \bigcap\delta(out(G)\perp(a, b))$ follows immediately from the second case of the definition of δ. For the converse inclusion suppose $(c, d)\in out(G)\setminus out(G)-(a, b)$. We need to find an $F\in \delta (out(G)\perp(a, b))$ such that $(c, d)\notin F$. Since $(c, d)\in out(G)\setminus out(G)-(a, b)$, we have $a\vdash c$ by Input Dependence, whence $(c, d) \in out((out(G)- (a, b))\cup (c, b))$, by Local Recovery. From lemma 4 it follows that (c, b→d) ∈ out(G) − (a, b). By SI and D-Closure, therefore, we have (a, b→d) ∈ out(G) − (a, b), whence $(a, \neg b\rightarrow d)\notin out(G)-(a, b)$ by another application of D-Closure together with AND.Therefore $(a, b\vee d)\notin out(G)-(a, b)$ by WO, whence out(G) − (a, b) can be extended to a set $F\in out(G)\perp (a, b\vee d)$ by lemma 8. Since $F\in out(G)\perp (a, b \vee d)$ we have $(a, b\vee d)\notin F$, whence $(a, d)\notin F$ by WO and $(c, d)\notin F$ by SI. Moreover, since $(a, b\vee d)\notin F$ we also have $(a, b)\notin F$, again by WO, so $F\in out(G)\perp (a, b)$ by lemma 10. Taking stock we have $F\in out(G)\perp (a, b)$ and $(c, d)\notin F$, so the proof is complete.

This completes the proof.

1.2 Proof of theorem 12

Proof: It suffices to find a partial meet derogation operation which yields $\dotplus$ via the Levi identity. Let δ be defined as follows:

$$ \delta(out(G)\perp (a, \neg b))= \left\{ \begin{array}{ll} \{out(G)\} \qquad\qquad \hbox{ if } out(G)\perp (a, \neg b) \hbox{is empty, otherwise }\\ \{F\in G\perp (a, \neg b): (out(G)\dotplus (a, b))\cap out(G)\subseteq F\} \end{array}\right. $$

We need to check that δ is a selection function and that it is well-defined:

Well-definedness. We need to show that $\delta(out(G)\perp(a, b)) = \delta(out(G)\perp(c, d))$ whenever $out(G)\perp(a, b) = out(G)\perp(c, d)$. All the limiting cases are similar to theorem 8. For the principal case where $out(G)\perp (a, b)\neq \emptyset$ and (a, b), (c, d) ∈ out(G) we reason as follows: Since $out(G)\perp(a, b) = out(G)\perp(c, d)$, we have out((a, b)) = out((c, d)) by lemma 13. Moreover, since $out(G)\perp(a, b)\neq \emptyset$ we also have $\nvdash b$, whence a≡c by lemma 1. By A-Extensionality it follows that $out(G)\dotplus (a, b) = out(G)\dotplus (c, d)$, so $\delta(out(G)\perp(a, b)) = \delta(out(G)\perp(c, d))$ as desired.

$\varvec{\delta}$is a selection function. To show that δ is a selection function, it suffices to show that $\delta(out(G)\perp(a, \neg b))\neq \emptyset$, whenever $out(G)\perp(a, \neg b)\neq \emptyset$, since, as is easy to check, all other cases are similar to theorem 8. In other words, we need to show, on the assumption that $\delta(out(G)\perp(a, \neg b))\neq \emptyset$, that there is an $F\in out(G)\perp (a, \neg b)$ such that $(out(G)\dotplus(a, b))\cap out(G)\subseteq F$. Clearly $(out(G)\dotplus(a, b))\cap out(G)\subseteq out(G)$, so it suffices to show that $(a, \neg b)\notin out(G)\dotplus(a, b)$, because then $(a, \neg b)\notin (out(G)\dotplus(a, b))\cap out(G)$ so this set can be extended to an $F\in out(G)\perp (a, \neg b)$ by lemma 8. Suppose for reductio ad absurdum that $(a, \neg b)\in out(G)\dotplus (a, b)$. Then since $(a, b)\in out(G)\dotplus(a, b)$ by A-Success it follows that $(a, f)\in out(G)\dotplus(a, b)$ by AND. By A-Consistency therefore $b\vdash f$ whence $\vdash\neg b$. But then $out(G)\perp(a, \neg b) = \emptyset$ contrary to assumption.

Finally we verify the identity

$$ out(G)\dotplus (a, b) = out(\bigcap\delta (out(G)\perp(a, \neg b))\cup (a, b)) $$

As a mnemonic device put;

$$ out(G)\dotplus_\delta (a, b) : = out(\bigcap\delta (out(G)\perp(a, \neg b))\cup (a, b)) $$

We thus need to prove that $out(G)\dotplus_\delta (a, b) = out(G)\dotplus (a, b)$. We split the proof into two mutually exclusive and jointly exhaustive cases:

• (a) Suppose b is inconsistent: Then we have $\vdash \neg b$, so $\bigcap\delta(out(G)\perp(a, \neg b)) = out(G)$ whence $out(G)\dotplus_\delta (a, b) = out(G\cup (a, b))$ by the definition of $\dotplus_\delta$. It suffices to show, therefore, that $out(G\cup (a, b)) = out(G)\dotplus (a, b)$. The right-in-left is is just A-Inclusion. For the converse we need only show that $out(G) \subseteq out(G)\dotplus(a, b)$, since we then have $out(G)\cup (a, b) \subseteq out(G)\dotplus (a, b)$ by A-Success, whence $out(out(G)\cup (a, b)) \subseteq out(out(G)\dotplus (a, b))$, by monotony for out, so $out(G\cup (a, b)) \subseteq out(G)\dotplus (a, b)$ by the closure properties of the out operator together with A-Closure. Now, by general set-theory $out(G)\subseteq out(G)\dotplus (a, b)$ whenever $out(G)\setminus out(G)\dotplus (a, b) = \emptyset$, so it suffices to show the latter. Suppose for reduction that $(c, d)\in out(G)\setminus out(G)\dotplus(a, b)$. Then, by A-Relevance there is an $F\subseteq out(G)$ such that $(a, \neg b)\notin out(F)$. However b is inconsistent so $\vdash \neg b$ contradicting $(a, \neg b)\notin out(F)$. Therefore $out(G)\setminus out(G)\dotplus (a, b) = \emptyset$, which completes the case.

• (b) In the principal case where b is consistent we argue for each direction of the desired identity separately:

  • [$\Rightarrow$]: We wish to prove that $out(G)\dotplus (a, b)\subseteq out(G)\dotplus_\delta(a, b)$. By the definition of δ we have $out(G)\cap (out(G)\dotplus (a, b))\subseteq \bigcap\delta (out(G)\perp (a, \neg b))$, from which it follows that $(out(G)\cap (out(G)\dotplus (a, b)))\cup (a, b) \subseteq \bigcap\delta (out(G)\perp (a, \neg b))\cup (a, b)$. By monotony for out, therefore, we have $out((out(G)\cap (out(G)\dotplus (a, b)))\cup (a, b)) \subseteq out(\bigcap\delta (out(G)\perp (a, \neg b))\cup (a, b))$, which, by the definition of $out(G)\dotplus_\delta (a, b)$ implies $out((out(G)\cap (out(G)\dotplus (a, b)))\cup (a, b)) \subseteq out(G)\dotplus_\delta (a, b)$. By lemma 18 it thus follows that $out(G)\dotplus (a, b)\subseteq out(G)\dotplus_\delta (a, b)$, as desired.

  • [$\Leftarrow$]: We wish to show that $out(G)\dotplus_\delta (a, b) \subseteq out(G)\dotplus (a, b)$, that is, we want to show that $out(\bigcap\delta (out(G)\perp(a, \neg b))\cup (a, b))\subseteq out(G)\dotplus (a, b)$. By monotony for out and A-Closure it suffices to demonstrate that $\bigcap\delta (out(G)\perp (a, \neg b))\cup (a, b)\subseteq out(G)\dotplus (a, b)$. Now, $(a, b)\in out(G)\dotplus (a, b)$, by A-Success, so we need only show that $\bigcap\delta (out(G)\perp(a, \neg b))\subseteq out(G)\dotplus (a, b)$. The argument proceeds by contraposition: Suppose $(c, d)\notin out(G)\dotplus (a, b)$. In the limiting case where $(c, d)\notin out(G)$ we have $(c, d)\notin \bigcap\delta (out(G)\perp(a, \neg b)) \subseteq out(G)$ so we are done. So, suppose (c, d) ∈ out(G). Then $(c, d)\in out(G)\setminus out(G)\dotplus (a, b)$. By A-Relevance there is an F such that

    • 1. $out(G)\cap (out(G)\dotplus (a, b))\subseteq F\subseteq out(G)$

    • 2. $(a, \neg b)\notin out(F)$, and

    • 3. $(a, \neg b)\in out(F\cup (c, d))$.

    By condition 1 and 2 and lemma 8 it follows that F can be extended to an $F'\in out(G)\perp(a, \neg b)$. By condition 3 it follows that $(c, d)\notin F'$, and by 1 again it follows that $out(G)\cap (out(G)\dotplus(a, b))\subseteq F'$. Hence, $F'\in \delta(G\perp (a, \neg b))$, whence $(c, d)\notin \bigcap\delta(out(G)\perp(a, \neg b)) = out(G)\dotplus_\delta(a, b)$ as desired.

This completes the proof.