Contraction in Interrogative Belief Revision

Abstract

In the paper “On the role of the research agenda in epistemic change”, Olsson and Westlund have suggested that the notion of epistemic state employed in the standard framework of belief revision (Alchourrón et al. 1985; Gärdenfors 1988) should be extended to include a representation of the agent’s research agenda (Olsson and Westlund 2006). The resulting framework will here be referred to as interrogative belief revision. In this paper, I attempt to deal with the problem of how research agendas should change in contraction, a problem largely left open by Olsson and Westlund. Two desiderata of an appropriate solution are suggested: one is a principle of continuity, stating that changes in the research agenda should somehow reflect that certain long term research interests are kept fixed. The other desideratum, which is based on part of Olsson and Westlund’s motivation for adding research agendas to the epistemic states, is that we should be able to account for how contraction may serve to open up new, fruitful hypotheses for investigation. In order to achieve these desiderata, I base my solution on a revised version of Olsson and Westlund’s notion of epistemic state.

Appendix

1.1 Proofs of Some Observations

Proof of observation 2.1: in order to show that DF(Q) is a T₁-question and T₁-equivalent to Q, it is sufficient to note that for each $i \in \left\{ {1, \ldots ,n} \right\}$, we have $T_{1} \cap X_{i} = T_{1} \cap Q/X_{i}$. To see this, since Q is a T₁-question, for each $i,j \in \left\{ {1, \ldots ,n} \right\}$ such that i ≠ j we have $T_{1} \cap X_{\text{i}} \cap X_{\text{j}} = \varnothing$, hence $T_{1} \cap X_{\text{i}} \subseteq - X_{\text{j}}$. From this follows that $T_{ 1} \cap \, X_{\text{i}} \, \subseteq \, \cap \bigcap \{ - X_{\text{j}} | i \ne j, j \in \, \left\{ { 1, \ldots , n} \right\}\}$, hence $T_{ 1} \cap X_{\text{i}} = T_{ 1} \cap X_{\text{i}} cap \bigcap \{ - X_{\text{j}} | i \ne j,j \in \left\{ { 1, \ldots ,n} \right\}\} = T_{ 1} \cap Q /X_{\text{i}}$. DF(Q) is clearly exclusive w.r.t T₂ since its members are pairwise disjoint. Finally, suppose that DF(Q) was exhaustive w.r.t T₂. Then we would have $T_{ 2} \subseteq \bigcup {\text{DF}}(Q)$. A moments reflection shows that, for each $i,j \in \, \left\{ {1, \ldots ,n} \right\}$ such that i ≠ j, we have $\bigcup \,{\text{DF}}(Q) \cap \, X_{\text{i}} \cap \, X_{\text{j}} = \, \varnothing$, and so by elementary set theory we have $T_{2} \cap X_{\text{i}} \cap X_{\text{j}} = \varnothing$. But this means that Q is exclusive w.r.t T₂, contradiction.□

Proof of observation 2.2: in order to show that NEX(Q) is a T₁-question and T₁-equivalent to Q, it is sufficient to note that for each $i \in \left\{ {1, \ldots ,n} \right\}$, we have $T_{1} \cap X_{\text{i}} = T_{1} \cap Q//X_{\text{i}}$. To see this, since Q is a T₁-question, for each $i,j \in \left\{ {1, \ldots ,n} \right\}$ such that i ≠ j we have $T_{1} \cap X_{\text{i}} \cap X_{\text{j}} = \varnothing$, hence $T_{1} \cap X_{\text{i}} \subseteq T_{1} \cap - X_{\text{j}}$. From this follows that $T_{ 1} \cap X_{\text{i}} \subseteq T_{ 1} \cap \, \bigcap \{ - X_{\text{j}} | i \ne j, j \in \left\{ { 1, \ldots ,n} \right\}\}$. Conversely, from the exhaustiveness of Q w.r.t T₁ one can easily show that $T_{ 1} \cap \, \bigcap \left\{ { - X_{\text{j}} | i \ne j, j \in \{ 1, \ldots ,n\} } \right\} \subseteq T_{ 1} \cap X_{\text{i}}$, hence $T_{ 1} \cap X_{\text{i}} = T_{ 1} \cap \bigcap \left\{ { - X_{\text{j}} | i \ne j, j \in {\text{\{ 1}}, \ldots ,n\} } \right\} = T_{ 1} \cap Q //X_{\text{i}}$ as desired. Finally, suppose that NEX(Q) were exclusive w.r.t T₂. Then we would have $T_{2} \cap Q //X_{1} \cap \ldots \cap Q //X_{\text{n}} = \varnothing$. But a moments reflection shows that $Q //X_{1} \cap \ldots \cap Q //X_{n} = - X_{1} \cap \ldots \cap - X_{\text{n}} = - (X_{1} \cup \ldots \cup X_{\text{n}} )$, so it follows that $T_{2} \cap - (X_{1} \cup \ldots \cup X_{n} ) = \varnothing {,}\, {\text{i}} . {\text{e }}T_{2} \subseteq X_{1} \cup \ldots \cup X_{n}$. This violates our assumption that Q is not exhaustive w.r.t T₂.□

Proof of observation 2.3: left to right is easy. For right to left: suppose all members of Q = {X₁,…,X_n} are pairwise disjoint. Then for each member X_i, we have $X_{i} \subseteq - X_{j}$ for each j ≠ i, and so it follows that $X_{\text{i}} \subseteq - X_{1} \cap \ldots \cap - X_{{{\text{i}} - 1}} \cap - X_{{{\text{i}} + 1}} \cap \ldots \cap - X_{\text{n}}$, hence $X_{\text{i}} \cap - X_{1} \cap \ldots \cap - X_{{{\text{i}} - 1}} \cap - X_{{{\text{i}} + 1}} \cap \ldots \cap - X_{\text{n}} = X_{\text{i}}$. From this observation it clearly follows that DF(Q) = Q.□

Proof of observation 3.6: clearly COM(Q,T) is a finite set of propositions. That the exclusiveness precondition holds w.r.t T is clear since the members of COM(Q,T) are pairwise disjoint, and the consistency precondition holds as an immediate consequence of the definition of COM(Q,T). The exhaustiveness condition follows easily since the union of all combinations of Q equals W, and from this also follows that if T is non-empty, COM(Q,T) is non-empty. As we have already noticed that the members of COM(Q,T) are pairwise disjoint, it follows by observation 2.3 that it is in disjoint form.□

Proof of observation 5.2: (i) → (ii) is trivial. (ii) → (iii): let M be a T-topic for Q. Clearly, the members of COM(M,T) are disjoint, and so by observation 2.3, COM(M,T) = Q is in disjoint form. (iii) → (i): suppose Q is in disjoint form, i.e. Q = DF(Q). But since Q is a T-question, by observation 3.7 we have COM(Q,T) = DF(Q). Hence Q = COM(Q,T), i.e. Q is a topic for itself.□

Proof of observation 5.8: the first part of the observation is trivial to prove. The second part is proved by applying observation 5.2.□