Abstract
Belief revision theories standardly endorse a principle of intensionality to the effect that ideal doxastic agents do not discriminate between pieces of information that are equivalent within classical logic. I argue that this principle should be rejected. Its failure, on my view, does not require failures of logical omniscience on the part of the agent, but results from a view of the update as mighty: as encoding what the agent learns might be the case, as well as what must be. The view is motivated by consideration of a puzzle case, obtained by transposing into the context of belief revision a kind of scenario that Kit Fine has used to argue against intensionalism about counterfactuals. Employing the framework of truthmaker semantics, I go on to develop a novel account of belief revision, based on a conception of the update as mighty, which validates natural hyperintensional counterparts of the usual AGM postulates.
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Notes
The main reason for countenancing the impossible \(\blacksquare \), in the present context, is to avoid technical inconveniences such as having to allow for empty propositions.
2 There are several other, narrower relations of entailment that may be defined within the truthmaker framework. For present purposes, however, we may confine attention to this one.
Modulo some irrelevant differences resulting from the subtly different treatment of inconsistent updates using \(\blacksquare \).
References
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Acknowledgements
For helpful comments on earlier drafts of this paper and/or discussion of the ideas here developed, I would like to thank Singa Behrens, Alexander Dinges, Martin Glazier, Stephan Leuenberger, Stefan Roski, Pierre Saint-Germier, Daniel Rothschild, Josh Schechter, Benjamin Schnieder, Moritz Schulz, Peter Verdée, Bruno Whittle, Steve Yablo. Special thanks are due to Kit Fine for numerous extended exchanges about these topics. My work on this paper was done as part of the Emmy Noether project Relevance, funded by the Deutsche Forschungsgemeinschaft (Grant number KR 4516/2-1); I am very grateful for that support.
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Appendices
Appendix A: Doxastic States in a Truthmaker Framework
The basic structure of truthmaker semantics is that of a state-space, which is a special kind of a partially ordered set. Recall that a partial order on a set S is a two-place relation \(\sqsubseteq \) that is reflexive—\(s \sqsubseteq s\) for all s ∈ S—, transitive—\(s \sqsubseteq t\) and \(t \sqsubseteq u\) implies \(s \sqsubseteq u\) for all s,t,u ∈ S—, and anti-symmetric—\(s \sqsubseteq t\) and \(t \sqsubseteq s\) implies s = t for all s,t ∈ S. Call a partial order \(\sqsubseteq \) on a set S complete iff for every subset T of S, a least upper bound exists, i.e. there is an element s ∈ S such that \(t \sqsubseteq s\) for all t ∈ T, and \(s \sqsubseteq u\) whenever \(t \sqsubseteq u\) for all t ∈ S. By designating a certain subset of the states as the set of possible or consistent states, we obtain a modalized state-space.
Definition 1
A modalized state-space is a triple \(\big (S, S^{\lozenge }, \sqsubseteq \big )\) such that
-
1.
S is a non-empty set,
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2.
\(\sqsubseteq \) is a complete partial order on S, and
-
3.
\(S^{\lozenge }\) is a non-empty subset of S such that \(s \in S^{\lozenge }\) whenever \(s \sqsubseteq t\) and \(t \in S^{\lozenge }\).
Informally, the members of S are the states, \(\sqsubseteq \) is the parthood-relation, and \(S^{\lozenge }\) is the set of possible, or consistent states. For \(T \subseteq S\), we write \(\bigsqcup T\) for the least upper bound of T, which we also call the fusion of the members of T, and often write as t1 ⊔ t2 ⊔… when T = {t1,t2,…}. We call a (possible) world any possible state that contains every state it is compatible with, and we denote their set by Sw. A modalized state-space is called a W-space iff every possible state is part of a possible world, and it is called topsy if it contains only one impossible state. This will then be the one state, written \(\blacksquare \), which contains every state. Throughout this section and the next, we shall be working within some fixed, topsy W-space \(\mathcal {S} ~\text {=}~ \big (S, S^{\lozenge }, \sqsubseteq \big )\).
A proposition P is any non-empty subset of S. A proposition is consistent iff one of its members is, and propositions P and Q are compatible iff some member of P is compatible with some member of Q. The conjunction P ∧ Q of propositions P and Q is {p ⊔ q : p ∈ P and q ∈ Q}. Note that this is non-empty even if P and Q are incompatible, in which case P ∧ Q is \(\{\blacksquare \}\).Footnote 1 The disjunction P ∨ Q is P ∪ Q. A proposition P is said to (looselyFootnote 2) entail (| =) a proposition Q iff every world containing a truthmaker of P contains a truthmaker of Q .
We now turn to the task of defining the class of permissible doxastic states. We do this using a notion of a coherent pair of a plausibility ordering and a transition relation.
Definition 2
A plausibility ordering is a two-place relation ≤ on \(S^{w} \cup \{\blacksquare \}\) satisfying the following conditions, where g(X) :={w ∈ X : w ≤ v for all v ∈ X} for all \(X \subseteq S^{w} \cup \{\blacksquare \}\):
These are exactly the conditions imposed under the possible worlds approach, except for the added clause dealing with \(\blacksquare \). Given any plausibility ordering, we often use B to refer to g(Sw), since this is the set of worlds at which the agent’s beliefs are true.
Definition 3
A transition relation is a three-place relation → on S subject to the following conditions:
Definition 4
Let ≤ and → be a pair of plausibility ordering and transition relation. The operations of wayward revision∘ and (filtered) revision ∗ induced by ≤ and → are defined as follows, for P a non-empty subset of S:
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\(B \circ P~ \text {:=}~ \{t \in S~\text {:} ~p \rightarrow _{b} t\) for some p ∈ P and b ∈ B}
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B ∗ P := g(B ∘ P)
Definition 5
A pair of a plausibility ordering ≤ and a transition relation → is coherent iff, whenever b ∈ B:
Theorem 1
Let ∗ be the revision function induced by some coherent pair of plausibility ordering and transition relation. Then
Proof
Note that \(B * P \subseteq S^{w} \cup \{\blacksquare \}\) for all non-empty P. So to establish that any revision entails some proposition Q, we need to show that every truthmaker of the revision contains a truthmaker of Q as part.
(R-Success): If s ∈ B ∗ P, then \(p \rightarrow _{b} s\) for some p ∈ P and b ∈ B, and by (T-Success), \(s \sqsupseteq p\).
(R-Vacuity): Suppose B is compatible with P, and let b ∈ B be compatible with p ∈ P. Since b is a world, it follows that \(b \sqsupseteq p\), so by (T-Vacuity) \(p \rightarrow _{b} b\), and hence b ∈ B ∘ P. Since b ∈ g(Sw), it follows that b ≤ v for all v ∈ B ∘ P and hence b ∈ B ∗ P. But then v ≤ b for all v ∈ B ∗ P, so \(B * P \subseteq B\) and hence B ∗ P | =B.
(R-Inclusion): Note that by (PT-Existence), B ∗ P is non-empty. Suppose s ∈ B ∧ P, and let b ∈ B and p ∈ P be such that s=b ⊔ p. If \(s~ \text {=} ~\blacksquare \), s contains every state, and hence some verifier of B ∗ P. If s is consistent, then b is compatible with p, so since b ∈ Sw, \(b \sqsupseteq p\), and thus s=b. By (T-Vacuity), \(p \rightarrow _{b} b\), b ∈ B ∘ P. Since b ∈ B = g(Sw), b ≤ w for all w ∈ B ∘ P, hence s = b ∈ B ∗ P.
(R-Consistency): Suppose s ∈ P is consistent and let b ∈ B. By (T-Consistency), v ∈ Sw for some v with \(s \rightarrow _{b} v\), so B ∘ P has a consistent member. By (P-Inconsistency), it follows that g(B ∘ P) has some (indeed, only) consistent members.
(R-Superexpansion(∧)): Note first that by (PT-Existence), B ∗(P ∧ Q) is non-empty. Now suppose s ∈ (B ∗ P) ∧ Q. Then s = t ⊔ q for some t ∈ B ∗ P and q ∈ Q. If \(s ~\text {=}~ \blacksquare \), s contains every state, and so some verifier of B ∗(P ∧ Q). If s is consistent, then t is a possible world, and hence contains q as part. Since t ∈ B ∗ P, \(p \rightarrow _{b} t\) for some p ∈ P and b ∈ B. By (T-Incorporation), also \(p \sqcup q \rightarrow _{b} t\), and hence t ∈ B ∘(P ∧ Q). Now consider any v ∈ B ∘(P ∧ Q). Let \(p^{\prime } \in P\) and \(q^{\prime } \in Q\) be such that \(v \in B \circ \{p^{\prime } \sqcup q^{\prime }\}\). Then by (PT-Link) there is some u ≤ v with u ∈ B ∘{p} and hence u ∈ B ∘ P. Since t ∈ B ∗ P, it follows that t ≤ u and hence by (P-Transitivity) t ≤ v. So t ∈ g(B ∘(P ∧ Q)) = B ∗(P ∧ Q), as desired.
(R-Subexpansion(∨)): Suppose B ∗(P ∨ Q) is compatible with P. Suppose s ∈ B ∗ P. As before, the case in which s is inconsistent is easy. Suppose instead s = w ∈ Sw and \(p \rightarrow _{b} w\) with p ∈ P, and b ∈ B. We wish to show that w ∈(B ∗(P ∨ Q)) ∧ P. Since \(w \sqsupseteq p\) and p ∈ P, it suffices to show that w ∈ B ∗(P ∨ Q). Since w ∈ B ∗ P, w ∈ B ∘ P and hence w ∈ B ∘(P ∨ Q). It remains to show that w ≤ v for all v ∈ B ∘(P ∨ Q). Now note that since B ∗(P ∨ Q) is compatible with P, there is some \(w^{\prime } \in g\big (B \circ \big (P \vee Q\big )\big )\) with \(w^{\prime } \leq v\) for all v ∈ B ∘(P ∨ Q), and \(w^{\prime } \sqsupseteq p^{\prime }\) for some \(p^{\prime } \in P\). It then suffices to show that \(w \leq w^{\prime }\). Now either \(w^{\prime } \in B \circ P\) or \(w^{\prime } \in B \circ Q\). If \(w^{\prime } \in B \circ P\), then \(w \leq w^{\prime }\) is immediate from w ∈ g(B ∘ P). So suppose \(w^{\prime } \in B \circ Q\), so \(q \rightarrow _{b} w^{\prime }\) for some q ∈ Q. Since \(w^{\prime } \sqsupseteq p^{\prime }\), by (T-Incorporation), \(p^{\prime } \sqcup q \rightarrow _{b} w^{\prime }\). Then by (PT-Link), \(u \in B \circ \{p^{\prime }\}\), so u ∈ B ∘ P, and hence w ≤ u. By (P-Transitivity), \(w \leq w^{\prime }\), as desired. □
(R-Success), (R-Vacuity), (R-Inclusion), and (R-Consistency) are the obvious counterparts in our (semantic) setting to the (syntactically formulated) AGM postulates of Success, Vacuity, Inclusion and Consistency. The postulate of Closure serves mainly to ensure intensionality with respect to belief states, which is guaranteed under our account by the identification of belief states with the set of possible worlds at which they are true. The Intensionality postulate, of course, does not hold. Within our semantic setting, the only valid version of this principle is the triviality that B ∗ P = B ∗ Q if P = Q. Under a syntactic formulation of the theory, though, we would have the non-trivial principle that K ∗ α = K ∗ β if α and β are exactly equivalent, i.e. have the same exact truthmakers.Footnote 3
Moreover, as expected, all Finean rules from Section ?? except for the intensionalist rule of Substitution are valid under the obvious interpretation of ⇒.
Theorem 2
Let ∗ be the revision function induced by some coherent pair of plausibility ordering and transition relation. For any propositions P,Q, let P ⇒ Q hold iff B ∗ P | =Q. Then
Proof
(R-Entailment) and (R-Conjunction) are immediate from the definition of ⇒ and the fact that the | =-consequences of a proposition are closed under conjunction. (R-Disjunction) is immediate from the observation that \(g\big (X \cup Y\big ) \subseteq g\big (X\big ) \cup g\big (Y\big )\).
(R-Transitivity): Assume B ∗ P | =Q and B ∗(P ∧ Q) | =R. If P is inconsistent, P ⇒ R follows immediately given (Success). So suppose P is consistent. Then \(B * P \subseteq S^{w}\). So let w ∈ B ∗ P, and suppose \(p \rightarrow _{b} w\) with p ∈ P and b ∈ B. We need to show that \(w \sqsupseteq r\) for some r ∈ R. Since B ∗ P | =Q, we have \(w \sqsupseteq q\) for some q ∈ Q. Then p ⊔ q is consistent. By (T-Incorporation), \(p \sqcup q \rightarrow _{b} w\), so w ∈ B ∘(P ∧ Q). Now let v ∈ B ∘(P ∧ Q), and let \(p^{\prime } \in P\) and \(q^{\prime } \in Q\) be such that \(v \in B \circ \{p^{\prime } \sqcup q^{\prime }\}\). By (PT-Link), \(u \in B \circ \{p^{\prime }\}\) and hence u ∈ B ∘ P for some u ≤ v. But since w ∈ B ∗ P, w ≤ u and hence w ≤ v. So w ∈ B ∗(P ∧ Q). Since B ∗(P ∧ Q) | =R, \(w \sqsupseteq r\) for some r ∈ R, as desired. □
Appendix : B: Expansion and Collapse
We now show that the validity of Subexpansion(∧) or Superexpansion(∨) would collapse our account into intensional AGM. Some additional notation will be helpful. For any proposition Q, let Qw be the set of Q-worlds, i.e. \(\{w \in S^{w}: w \sqsupseteq q\) for some q ∈ Q}. The AGM revision of B by P is then simply g(Pw). Now consider the following constraint on the connection between plausibility orderings and transition relations:
It turns out that whenever (PT-Expansion) is satisfied, the resulting revision function is an AGM revision function.Footnote 4
Proposition 1
Let ≤ and → be a coherent pair of plausibility ordering and transition relation that satisfy (PT-Expansion). Then B ∗ P = g(Pw) whenever P is consistent.
Proof
Let P be consistent. Then \(B * P ~\text {=}~ B * \big (P \cap S^{\lozenge }\big )\) and \( g\big (P^{w}\big ) ~\text {=}~ g\big (\big (P \cap S^{\lozenge }\big )^{w}\big )\), so we may assume without loss of generality that \(P \subseteq S^{\lozenge }\).
Suppose first that w ∈ g(Pw). Let p ∈ P be such that \(w \sqsupseteq p\). Then w ∈ g({p}w), so by (PT-Expansion), w ∈ B ∘{p} and hence w ∈ B ∘ P. Suppose v ∈ B ∘ P. Then v ∈ Pw, so since w ∈ g(Pw), w ≤ v. It follows that w ∈ B ∗ P.
Suppose now that w ∈ B ∗ P, so w ∈ g(B ∘ P). Then w ∈ Pw. Now let v ∈ Pw. Pick a world u ∈ g(Pw), so u ≤ v, and let \(p^{\prime } \in P\) be such that \(u \sqsupseteq p^{\prime }\) and hence u ∈ g({p}w). By (PT-Expansion), u ∈ B ∘{p} and hence u ∈ B ∘ P. Since w ∈ g(B ∘ P), it follows that w ≤ u and hence w ≤ v. So w ∈ g(Pw), as desired. □
Any violation of (PT-Expansion), however, yields a violation of both Subexpansion(∧) and Superexpansion(∨).
Proposition 2
Let ≤ and → be a coherent pair of plausibility ordering and transition relation. Let w ∈ Sw, \(p \sqsubseteq w\), and w ≤ v for all v ∈ B ∘{p}, but w∉B ∘{p}. Then both Subexpansion(∧) and Superexpansion(∨) are invalid.
Proof
For Subexpansion(∧), set Q = {p,w} and P = {p}. It suffices to show that (i) w∉B ∗ P, (ii) B ∗ P is compatible with Q, and (iii) w ∈ B ∗(P ∧ Q). But (i) is immediate from the assumption that w∉B ∘{p}. For (ii), note that since p is consistent, by (T-Consistency) and (T-Completeness), for some world u, u ∈ B ∗ P, and by (T-Success), \(u \sqsupseteq p\), so B ∗ P is compatible with Q. For (iii), note that B ∘(P ∧ Q) = B ∘{p,w} = (B ∘{p})∪(B ∘{w}). Since w is a world, by (T-Success) and (T-Consistency), \(w \rightarrow _{b} u\) implies u = w for all b ∈ B, so B ∘{w} = {w}. It follows that w ∈ B ∘(P ∧ Q). Moreover, for all v ∈ B ∘(P ∧ Q), either v = w or v ∈ B ∘{p}. By assumption, either way we have w ≤ v, and hence w ≤ v for all v ∈ B ∘(P ∧ Q), so w ∈ B ∗(P ∧ Q).
For Superexpansion(∨), set Q = {w} and P = {p}. As before, w∉B ∗ P. So it suffices to show that w ∈ B ∗(P ∨ Q), since then by \(w \sqsupseteq p\) also w ∈(B ∗(P ∨ Q)) ∧ P. But P ∨ Q = {p,w}, and we already showed above that w ∈ B ∗{w,p}. □
It is natural to wonder if the collapse may be avoided by invalidating Subexpansion(∨) and Superexpansion(∧) instead of Subexpansion(∧) and Superexpansion(∨). But this is not so. Indeed, (PT-Expansion) implies (PT-Link), the principle required to establish Subexpansion(∨) or Superexpansion(∧). For assume \(s \sqcup t \rightarrow _{b} w\). If s ⊔ t is inconsistent, \(w ~\text {=}~ \blacksquare \) and w ≤ v for any v ∈ Sw. If s ⊔ t is consistent, w is a world containing s ⊔ t and hence w ∈{s}w. By (PT-Expansion) \(g\big (\{s\}^{w}\big ) \subseteq B \circ \{s\}\). By (P-Limit), g({s}w) is non-empty, so let v ∈ g({s}w). Then v ≤ w and v ∈ B ∘{s} as required.
Appendix : C: A Space for Dominos
Finally, we construct a model of a doxastic state that satisfies the assumptions of the domino case and whose revision function is obtained from a coherent pair of plausibility ordering and transition relation. Let f1,f2,… be a countable infinity of sentence letters, and let L be the corresponding set of literals, i.e. the set including exactly the sentence letters fn as well as their negations, which we write as \(\overline {f_{n}}\). Say that a subset s of L is consistent iff for all n, at most one of fn and \(\overline {f_{n}}\) is a member of s. Let \(S^{\lozenge } ~\text {=}~ \{s \subseteq L: s\) is consistent} and \(S ~\text {=}~ S^{\lozenge } \cup \{L\}\). It is straightforward to show that \(\mathcal {S} ~\text {=}~ \big (S, S^{\lozenge }, \sqsubseteq \big )\), with \(\sqsubseteq \) interpreted as the subset-relation, is a topsy W-space, the set of worlds Sw being the set of the maximal consistent subsets of L.
We now define a plausibility ordering ≤ on the worlds. First, let \(b ~\text {=}~ \{\overline {f_{n}}: n \in N\}\). Say that a world w is regular if for some n, \(w ~\text {=}~ \{\overline {f_{m}}: m < n\} \cup \{f_{m}: m \geq n\}\), and irregular if not regular and distinct from b. Then for w,v ∈ Sw, we let w ≤ v iff (a) w = b, or (b) w is regular and v≠b, or (c) w is irregular and v is irregular or identical to \(\blacksquare \), or (d) \(w ~\text {=}~ v ~\text {=}~ \blacksquare \). It is readily verified that ≤ satisfies the conditions of (P-Connectedness), (P-Transitivity), (P-Limit), and (P-Inconsistency).
Next, we define a transition relation →. It will be sufficient to specify revisions of b by any state s. Indeed, for each state s we shall always specify a unique revision of b by s. If s is \(\blacksquare \), we let \(s \rightarrow _{b} t\) iff \(t ~\text {=}~ \blacksquare \). If s is consistent, then for each n, we consider the largest m ≤ n, if any, for which either fm or \(\overline {f_{m}}\) is a member of s, and we include fn in our output state if s contains fm, and \(\overline {f_{n}}\) otherwise. More precisely, say that s is n-positive iff (a) fm ∈ s for some m ≤ n, and (b) fm ∈ s for m the greatest number ≤ n for which either fm ∈ s or \(\overline {f_{m}} \in s\). If s is not n-positive, then it is n-negative. Then let ϕ(s,n) = fn if s is n-positive and \(\overline {f_{n}}\) otherwise, and for consistent s, let \(s \rightarrow _{b} t\) iff t = {ϕ(s,n) : n ∈ N}.
Proposition 3
≤ and → are a coherent pair of plausibility ordering and transition relation on \(\mathcal {S}\).
Proof
We skip the straightforward proof that ≤ is a plausibility ordering.
(T-Success): if \(s ~\text {=}~ \blacksquare \), then \(s \rightarrow _{b} t\) implies \(t ~\text {=}~ \blacksquare \) and hence \(t \sqsupseteq s\). If s is consistent, then \(s \rightarrow _{b} t\) implies t = {ϕ(s,n) : n ∈ N}. Suppose fn ∈ s. Then s is n-positive and hence fn ∈ t. Suppose \(\overline {f_{n}} \in s\). Since s is consistent, fn∉s. So s is not n-positive, and hence \(\overline {f_{n}} \in t\). So \(s \sqsubseteq t\), as required.
(T-Completeness): \(s \rightarrow _{b} t\) implies that either \(t ~\text {=}~ \blacksquare \) or t = {ϕ(s,n) : n ∈ N}. By construction, for each n, {ϕ(s,n) : n ∈ N} has either fn or \(\overline {f_{n}}\) as a member, so {ϕ(s,n) : n ∈ N}∈ Sw.
(T-Consistency): \(s \rightarrow _{b} t\) implies t = {ϕ(s,n):n ∈ N} given that s is consistent. By construction of {ϕ(s,n):n ∈ N}, it follows that for all n, at most one of fn and \(\overline {f_{n}}\) are members of {ϕ(s,n):n ∈ N}, so t is consistent.
(T-Vacuity): If \(s \sqsubseteq b\), then s contains no fn as member. By construction, then neither does {ϕ(s,n) : n ∈ N}, so {ϕ(s,n) : n ∈ N} = b.
(T-Incorporation): Assume \(s \rightarrow _{b} t\) and \(r \sqsubseteq t\). We need to show that \(s \sqcup r \rightarrow _{b} t\). Suppose first that s is inconsistent. Then s ⊔ r = s, and \(s \sqcup r \rightarrow _{b} t\) follows immediately. Suppose then that s is consistent, so t = {ϕ(s,n):n ∈ N}. By definition of →, \(s \sqcup r \rightarrow _{b} \{\phi \big (s \sqcup r, n\big ): n \in N\}\), so it suffices to show that for all n, s is n-positive iff s ⊔ r is.
Suppose first that s ⊔ r is n-positive. So for m the greatest number ≤ n such that s ⊔ r contains either fm or \(\overline {f_{m}}\), we have fm ∈ s ⊔ r. Since \(s \sqcup r \sqsubseteq t ~\text {=}~ \{\phi \big (s, n\big )\text {:} n \in N\}\), it follows that ϕ(s,m) is fm, so s is m-positive. But since m is the greatest number ≤ n for which s ⊔ r contains either fm or \(\overline {f_{m}}\), it follows that there can be no number k between m and n for which s contains \(\overline {f_{k}}\), and hence it follows that s is n-positive also.
Suppose now that s ⊔ r is n-negative. Then either (a) there is no m ≤ n with fm ∈ s ⊔ r, or (b) we have \(\overline {f_{m}} \in s \sqcup r\) for m the greatest number ≤ n for which either fm ∈ s ⊔ r or \(\overline {f_{m}} \in s \sqcup r\). If (a), then there is no m ≤ n with fm ∈ s, so s is n-negative. If (b), then since \(s \sqcup r \sqsubseteq t ~\text {=}~ \{\phi \big (s, n\big )\text {:} n \in N\}\), it follows that ϕ(s,m) is \(\overline {f_{m}}\), so s is m-negative. But since m is the greatest number ≤ n for which s ⊔ r contains either fm or \(\overline {f_{m}}\), it follows that there can be no number k between m and n for which s contains fk, and hence it follows that s is n-negative also.
(PT-Existence): note that g(Sw) = {b}, and \(s \rightarrow _{b} \{\phi \big (s, n\big )~\text {:} ~n \in N\}\) for all \(s \in S^{\lozenge }\), and \(s \rightarrow _{b} \blacksquare \) for \(s ~\text {=}~ \blacksquare \).
(PT-Link): Suppose v ∈ B ∘{s ⊔ r}, so \(s \sqcup t \rightarrow _{b} v\). We need to show that \(s \rightarrow _{b} u\) for some u ≤ v. To that end, we establish
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1.
if \(s \sqcup t \rightarrow _{b} b\), then \(s \rightarrow _{b} b\)
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2.
if \(s \sqcup t \rightarrow _{b} w\) for some regular w, then \(s \rightarrow _{b} v\) for some regular v
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3.
If \(s \sqcup t \rightarrow _{b} w\) for some irregular, consistent w, then \(s \rightarrow _{b} v\) for some consistent v
For (1), note that if \(s \sqcup t \rightarrow _{b} b\), then \(s \sqcup t \sqsubseteq b\) by (T-Success), so \(s \sqsubseteq b\), so \(s \rightarrow _{b} b\) by (T-Vacuity). For (2), we prove the contrapositive. Suppose that \(s \rightarrow _{b} v\) and v = {ϕ(s,n):n ∈ N} is irregular. By definition of irregularity there are m < k with both fm and \(\overline {f_{k}}\) members of v. By definition of ϕ and the fact that \(b ~\text {=}~ \{\overline {f_{n}}\text {:} n \in N\}\) it follows that for some \(m^{\prime } < k^{\prime }\), both \(f_{m^{\prime }}\) and \(\overline {f_{k^{\prime }}}\) are members of s, and hence of s ⊔ t. By (T-Success), both \(f_{m^{\prime }}\) and \(\overline {f_{k^{\prime }}}\) are members of w, which rules out w being regular. For (3), note that if \(s \sqcup t \rightarrow _{b} w\) with w consistent, then s ⊔ t is consistent, hence so is s, and so by consistency so is v with \(s \rightarrow _{b} v\). □
We now show that our doxastic state satisfies the assumptions of the domino case under their obvious interpretation. Since some of these assumptions concern negated propositions, we will move to a bilateral conception of propositions as a pair of set of truthmakers and a set of falsitymakers. We shall take the revision of a belief state by a bilateral proposition to be simply the revision by the set of truthmakers, so our overall account of revision is not changed.
More precisely, we call a bilateral propositionP any pair of unilateral propositions. The first (second) coordinate of P is denoted by P+ (P−) and comprises the truthmakers (falsitymakers) of P. Let P ∧Q = (P+ ∧Q+,P−∨Q−), P ∨Q = (P+ ∨Q+,P−∧Q−), and ¬P = (P−,P+). P is said to be exhaustive iff every w ∈ Sw contains either a member of P+ or a member of P− as a part, and it is said to be exclusive iff no w ∈ Sw contains both a member of P+ and a member of P− as a part. Both properties can be shown to be preserved under the boolean operations, and it can also be shown that the logic of loose entailment over exclusive and exhaustive propositions is classical (cf. [10, pp. 665ff]).
Now let \(\mathbf {F}_{n} ~\text {=}~ \big (\{f_{n}\}, \{\overline {f_{n}}\}\big )\) and B = g(Sw). Note that Fn is always exclusive and exhaustive. Let P ⇒Q hold iff B ∗P+ | =Q+, and let ⇒Q hold iff B | =Q+. Then
Proposition 4
The belief state and revision function induced by ≤ and → satisfy the assumptions of the domino case:
Proof
(B) is immediate from the facts that B = {b} and the definition \(b ~\text {=}~ \{\overline {f_{n}}: n \in N\}\).
For (D.+) and (D. −), note that fn is m-positive iff m ≥ n, so \(f_{n} \rightarrow _{b} t\) iff \(t ~\text {=}~ \{\overline {f_{m}}\text {:} m < n\} \cup \{f_{m}\text {:} m \geq n\}\). So \(B * \mathbf {F}_{n}^{+} ~\text {=}~ \{\{\overline {f_{m}}~\text {:} ~m < n\} \cup \{f_{m}~\text {:} ~m \geq n\}\}\). So the sole truthmaker of \(B * \mathbf {F}_{n}^{+}\) contains a truthmaker of Fm whenever m ≥ n, as required for (D.+), and a truthmaker of ¬Fm whenever m < N, as required for (D. −). □
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Krämer, S. Mighty Belief Revision. J Philos Logic 51, 1175–1213 (2022). https://doi.org/10.1007/s10992-022-09663-7
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DOI: https://doi.org/10.1007/s10992-022-09663-7