Segerberg on the Paradoxes of Introspective Belief Change

Abstract

The aim of the chapter is to provide a critical assessment of Krister Segerberg’s solution to the problems of introspective belief change. We present three alternative ways in which the paradoxes may be avoided. The first is a solution due to Lindström and Rabinowicz, using a two-dimensional semantics for DDL. The second is found in a logic for belief change suggested by Bonanno, in which the operator for belief is replaced by a class of operators for belief, each supplied with a temporal index. The third solution consists in a logic for belief change due to van Benthem, founded on the method of Dynamic Epistemic Logic in which the dynamics is modelled by operations on entire models, rather than on some structure within the models. We argue that, while there are some differences between these approaches, there is a strong structural similarity between them, and they avoid the paradoxes of DDL in essentially the same way. Furthermore, this way of avoiding the paradoxes is both different from and, we think, more natural than Segerberg’s own solution.

Appendix: Proofs of Main Results

1.1 Proof of Theorem 1

The proof is based on constructing models for $S_{2D}$ out of models for $S_{Temp}$, in the following manner:

Definition 7

Given a temporal belief model $\mathfrak {A}= \langle W,\{B_n\}_{n\in \omega }, \{I_n\}_{n\in \omega }, V \rangle$, we define the two-dimensional revision model

$$\begin{aligned} \mathfrak {A}_{2D} =\langle W^*, \{B_u\}_{u\in W}, \{R^*_u\}_{u\in W},V^*\rangle \end{aligned}$$

as follows. We set

$$ \begin{aligned} W^*= \{(u,n) : u\in W \; \& \; n\in \omega \} \end{aligned}$$

For all $(u,n),(v,m),(w,k) \in W^*$, we set $(u,n) B_{(v,m)} (w,k)$ iff $u B_n w$ and $k = m$. We set $(u,n) R^*_{(v,m)}(X) (w,k)$ iff

• $u = w$,

• $k = n+1$ and

• $Z = I_k(u)$, where $Z = \{t\in W : (t,m)\in X\}$

Finally, we set $(u,n)\in V^*(p)$ iff $u \in V(p)$.

The construction is sound by the following proposition:

Proposition 2

$\mathfrak {A}_{2D}$ is a two-dimensional revision model, for any temporal belief model $\mathfrak {A}$.

Proof

We need to check that, for each $X\subseteq W^*$, if $(u,m) R^*_{(v,n)}(X) (w,k)$ then

• 1. $B_{(v,n)}(w,k)\subseteq X$

• 2. if $X\ne \emptyset$ then $B_{(v,n)}(w,k) \ne \emptyset$

• 3. if $B_{(v,n)}(u,m) \cap X \ne \emptyset$ then $B_{(v,n)}(w,k) = B_{(v,n)}(u,m)\cap X$

So suppose $(u,m) R^*_{(v,n)}(X) (w,k)$. Then $u = w$, $k = m+1$ and

$$\begin{aligned} I_{m+1}(u) = \{t \in W : (t,n)\in X\} \end{aligned}$$

Now, since

$$\begin{aligned} B_{m+1}(u) \subseteq I_{m+1}(u) \end{aligned}$$

item (1) follows easily by definition of the relation $B_{(v,n)}$: for if

$(u,m+1) B_{v,n} (w^\prime ,k^\prime )$, then $k^\prime = n$ and $u B_{m+1} w^\prime$, so $w^\prime \in I_{m+1}(u)$, so $(w^\prime ,n) = (w^\prime , k^\prime )\in X$.

For (2), note that $X \ne \emptyset$ implies $I_{m+1}(u)\ne \emptyset$, so $B_{m+1}(u) \ne \emptyset$. Pick $w^\prime$ such that $u B_{m+1} w^\prime$. Then $(u,m+1) B_{v,n} (w^\prime ,n)$ so $B_{(v,n)}(u,m+1)\ne \emptyset$.

Lastly, for (3), suppose $B_{(v,n)}(u,m) \cap X \ne \emptyset$. Let $(w^\prime ,k^\prime ) \in B_{(v,n)}(u,m) \cap X \ne \emptyset$; then $k^\prime = n$ and $u B_{m} w^\prime$. Since $(w^\prime ,n)\in X$, $w^\prime \in I_{m+1}(u)$. So

$$\begin{aligned} B_m (u) \cap I_{m+1} (u) \ne \emptyset \end{aligned}$$

and hence

$$\begin{aligned} B_{m+1}(u) = B_m (u) \cap I_{m+1} (u) \end{aligned}$$

This means that

$$\begin{aligned} B_{(v,n)} (u,m+1) = B_{(v,n)} (u,m) \cap X \end{aligned}$$

To see this, suppose $(u,m+1) B_{(v,n)} (s,i)$. Then $i = n$, and $u B_{m+1} s$. But then $u B_m s$ and $s \in I_{m+1}(u)$. So $(u,m) B_{(v,n)} (s,n)$ and $(s,n) \in X$.

Conversely, suppose $(u,m+1) B_{(v,n)} (s,i)$ and $(s,i) \in X$. By definition of $B_{v,n}$, $i = n$. So $(s,n) \in X$ and therefore $s \in I_{m+1}(u)$. Furthermore, $u B_{m+1} s$. So $s \in B_m (u) \cap I_{m+1} (u)$, hence $s \in B_{m+1}(u)$. By definition this means that $(u,m+1) B_{(v,n)} (s,n)$, i.e. $(u,m+1)B_{(v,n)}(s,i)$ as required.

We now define a mapping $G$ from pointed temporal belief models to pointed two-dimensional revision models by setting

$$\begin{aligned} G(\mathfrak {A},u) =_{df.} (\mathfrak {A}_{2D},(u,0),(u,0)) \end{aligned}$$

for each pointed temporal revision model $(\mathfrak {A},u)$. We then have the following result, which gives the key to the soundness result for $F$:

Lemma 1

For any pointed temporal model $(\mathfrak {A},u)$ and any $\mathcal {L}_{2D}$-formula $\alpha$, we have

$$\begin{aligned} (\mathfrak {A},u)\vDash F(\alpha ) \Longleftrightarrow G(\mathfrak {A},u)\vDash \alpha \end{aligned}$$

Proof

We show, for any formula $\alpha$, that for each world $u$ in the universe of $\mathfrak {A}$, we have both

$$\begin{aligned} (1)\quad (\mathfrak {A},u)\vDash \tau _{n,m}(\alpha ) \Longrightarrow \forall v\in W: (\mathfrak {A}_{2D},(v,n),(u,m))\vDash \alpha \end{aligned}$$

and

$$\begin{aligned} (2)\quad (\mathfrak {A},u)\nvDash \tau _{n,m}(\alpha ) \Longrightarrow \forall v\in W: (\mathfrak {A}_{2D},(v,n),(u,m))\nvDash \alpha \end{aligned}$$

for all $v\in W$. From $(1)$ and $(2)$ together it follows that

$$\begin{aligned} (\mathfrak {A},u)\vDash \tau _{0,0}(\alpha ) \Longleftrightarrow (\mathfrak {A}_{2D},(u,0),(u,0))\vDash \alpha \end{aligned}$$

i.e.

$$\begin{aligned} (\mathfrak {A},u)\vDash F(\alpha ) \Longleftrightarrow G(\mathfrak {A},u)\vDash \alpha \end{aligned}$$

as desired.

The proof goes by induction on the length of $\alpha$. For propositional variables, both clauses are immediate, and the steps for Boolean connectives are easy.

Step for $B$: Suppose $(\mathfrak {A},u)\vDash \tau _{n,m}(B\alpha )$, i.e. $(\mathfrak {A},u)\vDash B_m \tau _{n,n}(\alpha )$. Let $v\in W$ and let $(w,k)$ be such that $(u,m) B_{v,n} (w,k)$. Then by definition $u B_m w$ and $k = n$, so we must have $(\mathfrak {A},w)\vDash \tau _{n,n} (\alpha )$ and by clause $(1)$ of the IH we get $(\mathfrak {A}_{2D},(v,n),(w,n))\vDash \alpha$. So we must have $(\mathfrak {A}_{2D},(v,n),(u,m))\vDash B\alpha$. This shows that

$$\begin{aligned} (1)\quad (\mathfrak {A},u)\vDash \tau _{n,m}(B \alpha ) \Longrightarrow \forall v\in W: (\mathfrak {A}_{2D},(v,n),(u,m))\vDash B \alpha \end{aligned}$$

Suppose that $(\mathfrak {A},u)\nvDash \tau _{n,m}(B\alpha )$, i.e. $(\mathfrak {A},u)\nvDash B_m \tau _{n,n}(\alpha )$. Then there exists $w\in W$ such that $u B_m w$ and $(\mathfrak {A},v)\nvDash \tau _{n,n}\alpha$. Let $v\in W$; then we have $(u,m) B_{(v,n)} (w,n)$ and by clause $(2)$ of IH we have $(\mathfrak {A}_{2D},(v,n),(w,n))\nvDash \alpha$, hence $(\mathfrak {A}_{2D},(v,n),(u,m))\nvDash B \alpha$. We have shown that

$$\begin{aligned} (2)\quad (\mathfrak {A},u)\nvDash \tau _{n,m}(B \alpha ) \Longrightarrow \forall v\in W: (\mathfrak {A}_{2D},(v,n),(u,m))\nvDash B \alpha \end{aligned}$$

as required.

Step for $*$: Suppose $(\mathfrak {A},u)\vDash \tau _{n,m}([*\alpha ]\beta )$, i.e.

$$\begin{aligned} (\mathfrak {A},u)\vDash I_{m+1}\tau _{n,n}(\alpha )\rightarrow \tau _{n,m+1}(\beta ) \end{aligned}$$

We note that by the IH we have, for each $v\in W$,

$$\begin{aligned} (\ddagger )\quad \left\| \tau _{n,n}(\alpha )\right\| _{\mathfrak {A}} = \{t \in W : (t,n) \in \left\| \alpha \right\| _{(v,n)}\} \end{aligned}$$

Suppose for $v\in W$ that $(u,m) R^*_{(v,n)}(\left\| \alpha \right\| _{(v,n)})(w,k)$. Then $k = m+1$. Furthermore, by definition and by $(\ddagger )$ we get

$$\begin{aligned} I_{m+1}(u) = \{t \in W : (t,n) \in \left\| \alpha \right\| _{(v,n)}\} = \left\| \tau _{n,n}(\alpha )\right\| _{\mathfrak {A}} \end{aligned}$$

So $(\mathfrak {A},u)\vDash I_{m+1}(\tau _{n,n}(\alpha ))$. Thus, we get $(\mathfrak {A},u)\vDash \tau _{n,m+1}(\beta )$. By clause $(1)$ of the IH, this gives $(\mathfrak {A}_{2D},(v,n),(w,m+1))\vDash \beta$, i.e. $(\mathfrak {A}_{2D},(v,n),(w,k))\vDash \beta$. So $(\mathfrak {A}_{2D},(v,n),(u,m))\vDash [*\alpha ]\beta$. We have thus shown

$$\begin{aligned} (1)\quad (\mathfrak {A},u)\vDash \tau _{n,m}([*\alpha ]\beta ) \Longrightarrow \forall v\in W: (\mathfrak {A}_{2D},(v,n),(u,m))\vDash [*\alpha ]\beta \end{aligned}$$

Suppose $(\mathfrak {A},u)\nvDash \tau _{n,m}([*\alpha ]\beta )$, i.e. $(\mathfrak {A},u)\vDash I_{m+1}\tau _{n,n}(\alpha )$ but $(\mathfrak {A},u)\nvDash \tau _{n,m+1}(\beta )$. Pick $v\in W$. Using $(\ddagger )$ we obtain

$$\begin{aligned} I_{m+1}(u) = \left\| \tau _{n,n}(\alpha )\right\| _{\mathfrak {A}} = \{t \in W : (t,n) \in \left\| \alpha \right\| _{(v,n)}\} \end{aligned}$$

From this we can conclude that $(u,m) R^*_{(v,n)}(\left\| \alpha \right\| _{(v,n)})(u,m+1)$. Furthermore, by clause $(2)$ of the IH we have $(\mathfrak {A}_{2D},(v,n),(u,m+1)\nvDash \beta )$, so $(\mathfrak {A}_{2D},(v,n),(u,m) \nvDash [*\alpha ]\beta )$. We have thus shown

$$\begin{aligned} (2)\quad (\mathfrak {A},u)\nvDash \tau _{n,m}([*\alpha ]\beta ) \Longrightarrow \forall v\in W: (\mathfrak {A}_{2D},(v,n),(u,m))\nvDash [*\alpha ]\beta \end{aligned}$$

as required.

Step for $\dagger$: Given that the IH holds for $\alpha$, suppose first that we have $(\mathfrak {A},u)\vDash \tau _{n,m}(\dagger \;\alpha )$, i.e. $(\mathfrak {A},u)\vDash \tau _{m,m}(\alpha )$. Then we have by clause $(1)$ of IH: for all $v\in W$, $(\mathfrak {A}_{2D},(v,m),(u,m))\vDash \alpha$. In particular, $(\mathfrak {A}_{2D},(u,m),(u,m))\vDash \alpha$. This means that, for all $v\in W$, $(\mathfrak {A}_{2D},(v,n),(u,m))\vDash \dagger \;\alpha$. We have established:

$$\begin{aligned} (1)\quad (\mathfrak {A},u)\vDash \tau _{n,m}(\dagger \;\alpha ) \Longrightarrow \forall v\in W: (\mathfrak {A}_{2D},(v,n),(u,m))\vDash \dagger \;\alpha \end{aligned}$$

On the other hand, suppose $(\mathfrak {A},u)\nvDash \tau _{n,m}\dagger \,\alpha$, i.e. $(\mathfrak {A},u)\nvDash \tau _{m,m}(\alpha )$. Then we have by clause $(2)$ of IH: for all $v\in W$, $(\mathfrak {A}_{2D},(v,m),(u,m))\nvDash \alpha$. In particular, $(\mathfrak {A}_{2D},(u,m),(u,m))\nvDash \alpha$. This means that, for all $v\in W$, $(\mathfrak {A}_{2D},(v,n),(u,m))\nvDash \dagger \;\alpha$. We have established:

$$\begin{aligned} (2)\quad (\mathfrak {A},u)\nvDash \tau _{n,m}(\dagger \;\alpha ) \Longrightarrow \forall v\in W: (\mathfrak {A}_{2D},(v,n),(u,m))\nvDash \dagger \;\alpha \end{aligned}$$

This ends the proof.

We now prove Theorem 1 as follows: suppose $F(\Gamma )\nvDash _{Temp} F(\alpha )$. Then there is a pointed temporal belief model $(\mathfrak {A},u)$ such that $(\mathfrak {A},u)\vDash F(\Gamma )$ but $(\mathfrak {A},u)\nvDash F(\alpha )$. By the previous theorem, $G(\mathfrak {A},u)\vDash \Gamma$ but $G(\mathfrak {A},u)\nvDash \alpha$. Hence $\Gamma \nvDash _{2D} \alpha$. This ends the proof of the theorem.

1.2 Proof of Theorem 2

We use the same strategy as in the previous section:

Definition 8

Given a two-dimensional model $\mathfrak {A}$ and a world $v$ in the universe of $\mathfrak {A}$, we define the conditional belief model

$$\begin{aligned} \mathfrak {A}_{DEL}[v] = \langle W^*, \{\sigma _u\}_{u\in W^*}, V^*\rangle \end{aligned}$$

as follows: we set $W^*= W$ and $V^*= V$. For each $u\in W$ and $X\subseteq W$, we set

$$\begin{aligned} \sigma _u(X) = \{w \in W : \exists p \in W [ u R^*_v(X) p \text { and } p B_v w ] \} \end{aligned}$$

It is easily checked that $\mathfrak {A}_{DEL}[v]$ is a conditional belief model. We define the mapping $G$ from pointed two-dimensional revision models to pointed conditional belief models by setting $G(\mathfrak {A},v,u) = (\mathfrak {A}_{DEL}[v],u)$ for a pointed two-dimensional revision model $(\mathfrak {A},v,u)$. We have the following result:

Lemma 2

For any pointed two-dimensional model $(\mathfrak {A},u,v)$ and any static ${\mathcal {L}}_{DEL}$-formula $\alpha$ we have

$$\begin{aligned} (\mathfrak {A},u,v)\vDash \tau (\alpha ) \Longleftrightarrow G(\mathfrak {A},u,v)\vDash \alpha \end{aligned}$$

Proof

By induction over the length of static formulas we show that, for all $v\in W$ we have

$$\begin{aligned} (\mathfrak {A},u,v)\vDash \tau (\alpha ) \Longleftrightarrow (\mathfrak {A}_{DEL}[u],v)\vDash \alpha \end{aligned}$$

The steps for atomic formulas and Boolean connectives are trivial.

Step for $A$: suppose $(\mathfrak {A},u,v)\vDash \tau (A\alpha )$, i.e. $(\mathfrak {A},u,v)\vDash [*\lnot \tau (\alpha )]B\!\perp$. By seriality of $R^*_u(\left\| \lnot \tau (\alpha )\right\| ^{\mathfrak {A}}_u)$ there must be some $w$ such that $v R^*_u(\left\| \lnot \tau (\alpha )\right\| ^{\mathfrak {A}}_u) w$. Furthermore, clearly we have $B_u(w) = \emptyset$, and this means that ${\left\| \lnot \tau (\alpha )\right\| ^{\mathfrak {A}}_u} = \emptyset$. Hence $\left\| \tau (\alpha )\right\| ^\mathfrak {A}_u = W = W^*$. By the IH, $\left\| \alpha \right\| _{\mathfrak {A}_{DEL}[u]} = W^*$, and so we have $(\mathfrak {A}_{DEL}[u],v)\vDash A\alpha$ as required.

Conversely, suppose $(\mathfrak {A},u,v)\nvDash \tau (A\alpha )$, i.e. $(\mathfrak {A},u,v)\nvDash [*\lnot \tau (\alpha )]B\!\!\perp$. Then there is some $w$ such that $v R^*_u (\left\| \lnot \tau (\alpha )\right\| ^\mathfrak {A}_u) w$ and $B_u(w)\ne \emptyset$. Hence there is some $s$ such that $w B_u s$. By the definition of a two-dimensional model, $s\in \left\| \lnot \tau (\alpha )\right\|$ i.e. $(\mathfrak {A},u,s)\vDash \lnot \tau (\alpha )$. Hence $(\mathfrak {A},u,s)\nvDash \tau (\alpha )$, and by the IH $(\mathfrak {A}_{DEL}[u],s)\nvDash \alpha$. Hence $(\mathfrak {A}_{DEL}[u],v)\nvDash A \alpha$ as required.

Step for $B$: suppose $(\mathfrak {A},u,v) \vDash \tau (B(\alpha \mid \beta ))$, i.e.

$$\begin{aligned} (\mathfrak {A},u,v)\vDash [*\tau (\beta )] B \tau (\alpha ) \end{aligned}$$

Suppose $w \in \sigma _v(\left\| \beta \right\| _{\mathfrak {A}_{DEL}[u]})$. By the IH this means that $w \in \sigma _v (\left\| \tau (\beta )\right\| ^{\mathfrak {A}}_u)$, so there is some $s$ such that $v R^*_u(\left\| \alpha \right\| ^\mathfrak {A}_u) s$ and $s B_u w$. Since $(\mathfrak {A},u,v)\vDash [*\tau (\beta )] B \tau (\alpha )$ we have $(\mathfrak {A},u,s)\vDash B\tau (\alpha )$ so $(\mathfrak {A},u,w)\vDash \tau (\alpha )$. By IH we get $(\mathfrak {A}_{DEL}[u],w)\vDash \alpha$. We have thus shown that $(\mathfrak {A},u,v)\vDash B(\alpha \mid \beta )$ as required.

Conversely, suppose that $(\mathfrak {A},u,v)\nvDash \tau (B(\alpha \mid \beta ))$, i.e.

$$\begin{aligned} (\mathfrak {A},u,v)\nvDash [*\tau (\beta )] B \tau (\alpha ) \end{aligned}$$

Then there is some $s$ such that $v R^*_u (\left\| \tau (\beta )\right\| ^\mathfrak {A}_u) s$ and $(\mathfrak {A},u,s)\nvDash B\tau (\alpha )$. This means that for some $w$ we have $s B_u w$ and $(\mathfrak {A},u,w)\nvDash \tau (\alpha )$. By the IH we have $v R^*_u (\left\| \beta \right\| _{\mathfrak {A}_{DEL}[u]}) s$, and thus we have $w\in \sigma _v(\left\| \beta \right\| _{\mathfrak {A}_{DEL}[u]})$. Furthermore, by the IH again, we have $(\mathfrak {A}_{DEL}[u],w)\nvDash \alpha$. Thus $(\mathfrak {A}_{DEL}[u],v)\nvDash B(\alpha \mid \beta )$ as required.

Using the fundamental property of the translation $\rho$ used in the construction of $F$, this lemma immediately entails:

Corollary 2

For any pointed two-dimensional model $(\mathfrak {A},u,v)$ and any ${\mathcal {L}}_{DEL}$-formula $\alpha$ we have

$$\begin{aligned} (\mathfrak {A},u,v)\vDash F(\alpha ) \Longleftrightarrow G(\mathfrak {A},u,v)\vDash \alpha \end{aligned}$$

From this result, we can prove Theorem 2 just like we proved Theorem 1.