The Irreducibility of Iterated to Single Revision

Abstract

After a number of decades of research into the dynamics of rational belief, the belief revision theory community remains split on the appropriate handling of sequences of changes in view, the issue of so-called iterated revision. It has long been suggested that the matter is at least partly settled by facts pertaining to the results of various single revisions of one’s initial state of belief. Recent work has pushed this thesis further, offering various strong principles that ultimately result in a wholesale reduction of iterated to one-shot revision. The present paper offers grounds to hold that these principles should be significantly weakened and that the reductionist thesis should ultimately be rejected. Furthermore, the considerations provided suggest a close connection between the logic of iterated belief change and the logic of evidential relevance.

Appendix:

In the proofs that follow, we shall be appealing to the following principles, which are not defined in the main body of the paper. They are to be read as holding for all doxastic states Ψ and all sentences A, B, C:

• (AGM2) A ∈ [Ψ ∗ A]

• (AGM3) If B ∈ [Ψ ∗ A], then B ∈ Cn ([Ψ]∪{A})

• (AGM4) If ¬A ∉ [Ψ] and B ∈ Cn([Ψ] ∪ {A}), then B ∈ [Ψ ∗ A]

• (AGM5) If A is consistent , then [Ψ ∗ A] is also consistent

• (AGM7) [Ψ ∗ A ∧ B]⊆ Cn ([Ψ ∗ A] ∪ {B})

• (AGM8) If ¬B ∉ [Ψ ∗ A], then Cn([Ψ ∗ A] ∪ {B}) ⊆ [Ψ ∗ A ∧ B]

• (DP1) If C ∈ Cn (A), then B ∈ [Ψ ∗ A] iff B ∈ [(Ψ ∗ C) ∗ A]

• (DP2) If ¬C ∈ Cn (A), then B ∈ [Ψ ∗ A] iff B ∈ [(Ψ ∗ C) ∗ A]

• (DP3) If B ∈ [Ψ ∗ A], then B ∈ [(Ψ ∗ B) ∗ A]

• (DP4) If¬B ∉ [Ψ ∗ A],then¬B ∉ [(Ψ ∗ B) ∗ A]

Observation 1

An agent’s single-shot revision dispositions with respect to any truth-functional combination C of sentences A and B are fully determined by the restriction of the preference relation to the members of the sets of maximal A ∧ B-, A ∧ ¬B-, ¬A ∧ B- and ¬A ∧ ¬B-worlds.

Proof

Let ≽ be a complete weak ordering of the set W of possible worlds and ≻ its strict part. Let [[ψ]] denote {w ∈ W:w⊧ψ} and, where S ⊆ W, max(S) denote {x ∈ S: For all y ∈ S, x≽y}. Since C is a truth-functional combination of A and B, the set [[C]] will be equal to the union of one or more of the cells of the partition $\mathcal {P}=\{[{\kern -2.3pt}[ A\wedge B]{\kern -2.3pt}], [{\kern -2.3pt}[ A\wedge \neg B]{\kern -2.3pt}], [{\kern -2.3pt}[ \neg A\wedge B]{\kern -2.3pt}], [{\kern -2.3pt}[ \neg A\wedge \neg B]{\kern -2.3pt}]\}$ of W. We now show that x ∈ max([[C]]) iff $x\in \max (\bigcup \{\max (S):S\in \mathcal {P} \text { and } S\subseteq [{\kern -2.3pt}[ C]{\kern -2.3pt}]\})$, the nature of this last set being determined, as required, by the restriction of the preference relation to the members of the sets of maximal A ∧ B-, A ∧ ¬B-, ¬A ∧ B- and ¬A ∧ ¬B-worlds.

Regarding the left-to-right direction: Assume that x ∈ max([[C]]). Now assume for reductio that $x\notin \max (\bigcup \{\max (S):S\in \mathcal {P} \text { and } S\subseteq [{\kern -2.3pt}[ C]{\kern -2.3pt}]\})$. Now either (i) $x\in \bigcup \{\max (S):S\in \mathcal {P} \text { and } S\subseteq [{\kern -2.3pt}[ C]{\kern -2.3pt}]\}$ or (ii) $x\notin \bigcup \{\max (S):S\in \mathcal {P} \text { and } S\subseteq [{\kern -2.3pt}[ C]{\kern -2.3pt}]\}$. Assume (i). Then there exists a y in $\max (\bigcup \{\max (S):S\in \mathcal {P} \text { and } S\subseteq [{\kern -2.3pt}[ C]{\kern -2.3pt}]\})$, such that y ≻ x. Since $\max (\bigcup \{\max (S):S\in \mathcal {P} \text { and } S\subseteq [{\kern -2.3pt}[ C]{\kern -2.3pt}]\})\subseteq [{\kern -2.3pt}[ C ]{\kern -2.3pt}]$, we also have y ∈ [[C]], contradicting our initial assumption. So assume (ii). Since $\mathcal {P}$ partitions [[C]], there exists an $S\in \mathcal {P}$ such that x ∈ S. Given (ii), we know that x∉ max(S). So there exists a y ∈ max(S) such that y ≻ x. Since S ⊆ [[C]], we also have y ∈ [[C]], again contradicting our initial assumption.

Regarding the right-to-left direction: Assume that $x\in \max (\bigcup \{\max (S):S\in \mathcal {P}$ and S ⊆ [[C]]}). Assume for reductio that x∉ max([[C]]) and hence that there exists a y ∈ [[C]] such that y ≻ x. Since $\mathcal {P}$ partitions [[C]], there exists an $S\in \mathcal {P}$ such that y ∈ S and a z ∈ max(S) such that z≽y. But since $x\in \max (\bigcup \{\max (S):S\in \mathcal {P} \text { and } S\subseteq [{\kern -2.3pt}[ C]{\kern -2.3pt}]\})$, we have x≽z and hence, by transitivity of ≽, x≽y. Contradiction. □

Observation 2

In the presence of the AGM postulates, the DP postulates are jointly equivalent to DP ^′.

Proof

Regarding the left-to-right direction: We consider three cases:

• (1) Suppose A ∈ [(Ψ ∗ A) ∗ B]. Then, by AGM7 and AGM8, it follows that [(Ψ ∗ A) ∗ B] = [(Ψ ∗ A) ∗ A ∧ B]. But from DP1, we know that [(Ψ ∗ A) ∗ A ∧ B] = [Ψ ∗ A ∧ B]. Hence [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B], as required.

• (2) Suppose A, ¬A ∉ [(Ψ ∗ A) ∗ B]. By AGM8, this gives [(Ψ ∗ A) ∗ B]⊆[(Ψ ∗ A)∗¬A ∧ B]∩[(Ψ ∗ A) ∗ A ∧ B] while the converse inclusion to this also holds by AGM7. Hence [(Ψ ∗ A) ∗ B] = [(Ψ ∗ A)∗¬A ∧ B]∩[(Ψ ∗ A) ∗ A ∧ B]. Applying DP1 and DP2 to the right-hand side yields [(Ψ ∗ A) ∗ B] = [Ψ∗¬A ∧ B]∩[Ψ ∗ A ∧ B]. We now split into two cases: (i) ¬A ∈ [Ψ ∗ B] and (ii) ¬A ∉ [Ψ ∗ B]. Assume (i). It follows that [Ψ∗¬A ∧ B] = [Ψ ∗ B] and we recover the desired conclusion. Assume (ii). then we also have A ∉ [Ψ ∗ B] from DP3 and the fact that A ∉ [(Ψ ∗ A) ∗ B]. Hence, by AGM8 and AGM7, we have [Ψ ∗ B] = [Ψ∗¬A ∧ B]∩[Ψ ∗ A ∧ B]. Hence [(Ψ ∗ A) ∗ B] = [Ψ ∗ B]. But since [Ψ ∗ B]⊆[Ψ ∗ A ∧ B], we have [(Ψ ∗ A) ∗ B] = [Ψ ∗ B]∩[Ψ ∗ A ∧ B], as required.

• (3) Suppose ¬A ∈ [(Ψ ∗ A) ∗ B]. Then, by AGM7 and AGM8, [(Ψ ∗ A) ∗ B] = [(Ψ ∗ A)∗¬A ∧ B]. By DP2, we have [(Ψ ∗ A)∗¬A ∧ B] = [Ψ∗¬A ∧ B] and so [(Ψ ∗ A) ∗ B] = [Ψ∗¬A ∧ B]. From ¬A ∈ [(Ψ ∗ A) ∗ B] and DP4 we know that ¬A ∈ [Ψ ∗ B]. Hence, by AGM7 and AGM8, we have [Ψ ∗ B] = [Ψ∗¬A ∧ B] and so [(Ψ ∗ A) ∗ B] = [Ψ ∗ B], as required.

Regarding the right-to-left direction:

• (1) Regarding DP1: If A ∈ Cn(B), then we must be in Case (i) of DP ^′ and so [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B]. Since A ∧ B and B are logically equivalent, we know that [Ψ ∗ A ∧ B] = [Ψ ∗ B] and so [(Ψ ∗ A) ∗ B] = [Ψ ∗ B].

• (2) Regarding DP2: If ¬A ∈ Cn(B), then we must be in Case (iii) of DP ^′, which immediately gives us [(Ψ ∗ A) ∗ B] = [Ψ ∗ B].

• (3) Regarding DP3: We prove the contrapositive. Assume A ∉ [(Ψ ∗ A) ∗ B]. If ¬A ∈ [(Ψ ∗ A) ∗ B], then [(Ψ ∗ A) ∗ B] = [Ψ ∗ B] from Clause (iii) of DP ^′ and so A ∉ [Ψ ∗ B], as required. If ¬A ∉ [(Ψ ∗ A) ∗ B], then [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B]∩[Ψ ∗ B] by Clause (ii) of DP ^′. We know that A ∈ [Ψ ∗ A ∧ B], so if A ∉ [(Ψ ∗ A) ∗ B], then we must have A ∉ [Ψ ∗ B], again as required.

• (4) Regarding DP4: If ¬A ∈ [(Ψ ∗ A) ∗ B], then we must be in Case (iii) of DP ^′, so [(Ψ ∗ A) ∗ B] = [Ψ ∗ B] and hence ¬A ∈ [Ψ ∗ B].

□

Observation 3

AGM2, AGM4 and DP3 jointly entail that A ∈ [(Ψ ∗ A) ∗ B] if A ∈ [Ψ ∗ B] or ¬B ∉ [Ψ ∗ A].

Proof

Assume that ¬B ∉ [Ψ ∗ A]. By AGM2, A ∈ [Ψ ∗ A]. It then follows by AGM4 that A ∈ [(Ψ ∗ A) ∗ B]. Assume that A ∈ [Ψ ∗ B]. It then follows by DP3 that A ∈ [(Ψ ∗ A) ∗ B]. □

Observation 4

AGM2 entails that ¬A ∈ [(Ψ ∗ A) ∗ B] if ¬A ∈ Cn(B) and $\neg B\notin \text {Cn}(\varnothing )$.

Proof

Trivial: Assume that ¬A ∈ Cn(B) and that $\neg B\notin \text {Cn}(\varnothing )$. From the latter, by AGM2, we have B ∈ [(Ψ ∗ A) ∗ B]. By deductive closure of belief sets, and the fact that ¬A ∈ Cn(B), it then follows from this that ¬A ∈ [(Ψ ∗ A) ∗ B]. □

Observation 5

The following two statements are equivalent:

• (1) If two doxastic states are 1-equivalent, then they are 2-equivalent

• (2) If two doxastic states are 1-equivalent, then they are equivalent

Proof

Clearly (2) entails (1), from the definitions of equivalence and k-equivalence. It remains to show that (1) entails (2). So suppose that (a) holds and that Ψ and Ψ^′ are 1-equivalent. We will show by induction on k that Ψ and Ψ^′ are k-equivalent for all k. The base case, k = 1, holds by assumption. Regarding the inductive step, assume that Ψ and Ψ^′ are k-equivalent. We need to show that they are are (k+1)-equivalent, i.e. that for any (k+1)-tuple 〈A ₁, A ₂,…,A _k , A _k+1〉, [((((Ψ∗A ₁)∗A ₂)∗…)∗A _k )∗A _k+1] = [((((Ψ^′ ∗ A ₁)∗A ₂)∗…)∗A _k )∗A _k+1]. Since Ψ and Ψ^′ are k-equivalent, we know that [(((Ψ∗A ₁)∗A ₂)∗…)∗A _k−1] and [(((Ψ^′ ∗ A ₁)∗A ₂)∗…)∗A _k−1] are 1-equivalent. Hence, by (1), they are also 2-equivalent and so [((((Ψ∗A ₁)∗A ₂)∗…)∗A _k )∗A _k+1] = [((((Ψ^′ ∗ A ₁)∗A ₂)∗…)∗A _k )∗A _k+1], as required. □

Observation 6

In the presence of the AGM and DP postulates, NR, RR and LR are respectively equivalent to

$$\begin{array}{@{}rcl@{}} &&\text{(NR}^{\prime})\qquad[(\Psi * A) * B] =\left\{\begin{array}{ll} [\Psi * A\wedge B], \quad if \neg B\notin[\Psi*A] \\[0.25em] [\Psi*B], \quad\quad\,\,\,\,otherwise \end{array}\right.\\ &&\text{(NR}^{\prime})\qquad [(\Psi * A) * B] =\left\{\begin{array}{ll} [\Psi * A\wedge B], \quad if\,\neg A\notin[\Psi *B] \text{ or } \neg B\!\notin[\Psi * A] \\[0.25em] [\Psi*B], \quad\quad\,\,\,\, otherwise \end{array}\right.\\ &&\text{(LR}^{\prime})\qquad[(\Psi * A) * B] =\left\{\begin{array}{ll} [\Psi * A\wedge B],\quad if~K*A\wedge B~is~consistent\\[0.25em] [\Psi*B], \quad\quad\,\,\,~otherwise \end{array}\right. \end{array} $$

Proof

Regarding the equivalence between NR and NR ^′:

• From NR to NR ^′: Assume ¬B ∉ [Ψ ∗ A]. It follows by NR that A ∈ [(Ψ ∗ A) ∗ B]. By DP ^′, we then recover [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B], as required. Assume ¬B ∈ [Ψ ∗ A]. Now either (i) ¬A ∉ [Ψ ∗ B], or (ii) ¬A ∈ [Ψ ∗ B]. Assume (i). Then, by NR, ¬A ∈ [(Ψ ∗ A) ∗ B] and hence, by DP ^′, [(Ψ ∗ A) ∗ B] = [Ψ ∗ B], as required. Assume (ii). On the one hand, it follows by AGM8 that Cn([Ψ ∗ B] ∪ {A}) ⊆ [Ψ ∗ A ∧ B] and hence that (iii) [Ψ ∗ B]⊆[Ψ ∗ A ∧ B]. On the other hand, it it follows by NR that A, ¬A ∉ [(Ψ ∗ A) ∗ B] and hence, by DP ^′, that (iv) [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B]∩[Ψ ∗ B]. By (iii) and (iv), we have [(Ψ ∗ A) ∗ B] = [Ψ ∗ B], as required.

• From NR ^′ to NR: Assume that A ∈ [Ψ ∗ B] or ¬B ∉ [Ψ ∗ A]. From the latter, by NR ^′, we recover [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B] and hence, by AGM2 and closure of belief sets, A ∈ [(Ψ ∗ A) ∗ B], as required. Assume that ¬B ∈ [Ψ ∗ A]. From this, by NR ^′, we recover [(Ψ ∗ A) ∗ B] = [Ψ ∗ B]. Assume further that A, ¬A ∉ [Ψ ∗ B]. It follows that A, ¬A ∉ [(Ψ ∗ A) ∗ B], as required. Finally, alternatively, assume that ¬A ∉ [Ψ ∗ B]. It follows that ¬A ∈ [(Ψ ∗ A) ∗ B], again as required.

Regarding the equivalence between RR and RR ^′:

• From RR to RR ^′: Assume that either ¬A ∉ [Ψ ∗ B] or ¬B ∉ [Ψ ∗ A]. By RR, we have A ∈ [(Ψ ∗ A) ∗ B] and hence, by DP ^′, [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B], as required. Assume instead that ¬A ∈ [Ψ ∗ B] and ¬B ∈ [Ψ ∗ A]. By RR, we have ¬A ∈ [(Ψ ∗ A) ∗ B] and hence, by DP ^′, [(Ψ ∗ A) ∗ B] = [Ψ ∗ B], as required.

• From RR ^′ to RR: Assume that either ¬A ∉ [Ψ ∗ B] or ¬B ∉ [Ψ ∗ A]. By RR ^′ it follows that [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B] and hence, by AGM2 and closure of belief sets, that A ∈ [(Ψ ∗ A) ∗ B], as required. So assume instead that ¬A ∈ [Ψ ∗ B] and ¬B ∈ [Ψ ∗ A]. It follows by RR ^′ that [(Ψ ∗ A) ∗ B] = [Ψ ∗ B] and hence, since ¬A ∈ [Ψ ∗ B], that ¬A ∈ [(Ψ ∗ A) ∗ B], as required.

Regarding the equivalence between LR and LR ^′:

• From LR to LR ^′: Assume that [Ψ ∗ A ∧ B] is consistent. By AGM2, it follows that ¬B∉Cn(A) and hence, by LR, that A ∈ [(Ψ ∗ A) ∗ B]. By DP ^′, we then recover the required result that [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B]. Assume instead that [Ψ ∗ A ∧ B] is inconsistent. By AGM5, it follows that ¬B ∈ Cn(A). By LR, we therefore have ¬A ∈ [(Ψ ∗ A) ∗ B] and hence, by DP ^′, [(Ψ ∗ A) ∗ B] = [Ψ ∗ B], as required.

• From LR ^′ to LR: Assume that ¬B∉Cn(A). It follows, by AGM5, that [Ψ ∗ A ∧ B] is consistent. and hence, by LR ^′, that [(Ψ ∗ A) ∗ B] = [Ψ ∗ A ∧ B]. By AGM2 and closure of belief sets, we then recover A ∈ [(Ψ ∗ A) ∗ B], as required. Assume instead that ¬B ∈ Cn(A). Since, by AGM2, we have B ∈ [(Ψ ∗ A) ∗ B], it then follows by closure of belief sets that ¬A ∈ [(Ψ ∗ A) ∗ B], as required.

□

Observation 7

RED3, IR and the negation of RED2 are jointly inconsistent.

Proof

Assume the negation of RED2, i.e. that there exist sentences A and B and doxastic states Ψ and Ψ^′, such that [Ψ ∗ C] = [Ψ^′ ∗ C], for any truth-functional combination C of A and B, but [(Ψ ∗ A) ∗ B]≠[(Ψ^′ ∗ A) ∗ B]. By IR, there then exists a Ψ^″ such that [Ψ^″ ∗ C] = [Ψ^′ ∗ C], for any C, and [(Ψ^″ ∗ A) ∗ B] = [(Ψ ∗ A) ∗ B]. Since, by assumption, [(Ψ ∗ A) ∗ B]≠[(Ψ^′ ∗ A) ∗ B], we therefore have [(Ψ^″ ∗ A) ∗ B]≠[(Ψ^′ ∗ A) ∗ B]. But this contradicts RED3, which would require, since [Ψ^″ ∗ C] = [Ψ^′ ∗ C], for any C, that [(Ψ^″ ∗ A) ∗ B] = [(Ψ^′ ∗ A) ∗ B]. □