Judgment aggregation and minimal change: a model of consensus formation by belief revision

Abstract

When a group of agents attempts to reach an agreement on certain issues, it is usually desirable that the resulting consensus be as close as possible to the original judgments of the individuals. However, when these judgments are logically connected to further beliefs, the notion of closeness should also take into account to what extent the individuals would have to revise their entire belief set to reach an agreement. In this work, we present a model for generation of agreement with respect to a given agenda which allows individual epistemic entrenchment to influence the value of the consensus. While the postulates for the transformation function and their construction resemble those of AGM belief revision, the notion of an agenda is adapted from the theory of judgment aggregation. This allows our model to connect both frameworks.

Proofs

1.1 Section 3

Proposition 1

Let $\varPhi$ and $\varPsi$ be agendas and $\mathbf {K}$ and $\mathbf {H}$ belief profiles.

• (a) If $\varPhi \le \varPsi$, then $\mu _\varPhi \vdash \mu _\varPsi$.

• (b) If $\varPhi \le \varPsi$, then $\mathbf {K}\Box \varPhi$ implies $\mathbf {K}\Box \varPsi$.

• (c) If $\varPhi \le \varPsi$, then $\mathbf {K}\Diamond \varPhi$ implies $\mathbf {K}\Diamond \varPsi$.

• (d) If $\mathbf {H}\subseteq \mathbf {K}$, then $\mathbf {K}\Diamond \varPhi$ implies $\mathbf {H}\Diamond \varPhi$.

Proof

• (a) Let $\varPhi = \{\varphi _1, \ldots , \varphi _n\}$ and $\varPsi = \{\psi _1, \ldots , \psi _m\}$ such that $\varPhi \le \varPsi$. For each $\varphi _i\in \varPhi$, there exists $\psi _j\in \varPsi$ such that $\varphi _i \vdash \psi _j$, so $\varphi _i \vdash \psi _1 \vee \cdots \vee \psi _m$ for all $1 \le i \le n$; hence, $\varphi _1 \vee \cdots \vee \varphi _n \vdash \psi _1 \vee \cdots \vee \psi _m$.

• (b) Assume $\varPhi \le \varPsi$ and $\mathbf {K}\Box \varPhi$, let $\psi \in \varPsi$, and let i, j be agents. Then by (a), $\mu _\varPhi \vdash \mu _\varPsi$, so $\mu _\varPsi \in K_i$. And then by refinement, $\psi \in K_i$ if and only if $\bigvee \{\varphi \in \varPhi \ |\ \varphi \vdash \psi \} \in K_i$, which by unanimity on $\varPhi$ holds if and only if $\bigvee \{\varphi \in \varPhi \ |\ \varphi \vdash \psi \} \in K_j$, which is equivalent to $\psi \in K_j$. Similarly, $\lnot \psi \in K_i$ if and only if $\bigwedge \{\lnot \varphi \in \varPhi \ |\ \varphi \vdash \psi \} \in K_i$ if and only if $\bigwedge \{\lnot \varphi \in \varPhi \ |\ \varphi \vdash \psi \} \in K_j$ if and only if $\lnot \psi \in K_j$.

• (c) Assume $\mathbf {K}\Diamond \varPhi$, then there exists a $\varphi \in \varPhi$ such that $\lnot \varphi \not \in K_i$ for all agents i. By refinement, there exists a $\psi \in \varPsi$ such that $\varphi \vdash \psi$, therefore, $\lnot \psi \not \in K_i$ for all i, and $\mathbf {K}\Diamond \varPsi$.

• (d) Assume $\mathbf {K}\Diamond \varPhi$, then there exists a $\varphi \in \varPhi$ such that $\lnot \varphi \not \in K_i$ for all agents i. By $\mathbf {H}\subseteq \mathbf {K}$, $\lnot \varphi \not \in H_i$ for all i, so $\mathbf {H}\Diamond \varPhi$. $\square$

Lemma 1

Let $\oplus$ be a belief set transformation function satisfying $(\oplus 1)$ through $(\oplus 3)$. Then for all belief profiles $\mathbf {K}$ and all agendas $\varPhi$,

$$\begin{aligned} \left( \mathbf {K}+ \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) \right) \subseteq (\mathbf {K}\oplus \varPhi ). \end{aligned}$$

Proof

Let $\varphi \in \varPhi$ such that for some agent i, $\lnot \varphi \in K_i$. Then by ($\oplus$3), $\lnot \varphi \in (\mathbf {K}\oplus \varPhi )_i$, and by ($\oplus$2), $\lnot \varphi \in (\mathbf {K}\oplus \varPhi )_j$ for any agent j. In addition, by ($\oplus$2), $\mu _\varPhi = \bigvee \varPhi \in (\mathbf {K}\oplus \varPhi )_i$ for all i. Therefore, for all agents i,

$$\begin{aligned} \bigwedge \left\{ \lnot \varphi \ |\ \varphi \in \varPhi \text { and } \lnot \varphi \in K_i \text { for some } i \right\} \wedge \bigvee \varPhi \in (\mathbf {K}\oplus \varPhi )_i. \end{aligned}$$

By modus tollendo ponens, it follows that

$$\begin{aligned} \bigvee \left\{ \varphi \in \varPhi \ |\ \lnot \varphi \not \in K_i \text { for all } i \right\} = \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) \in (\mathbf {K}\oplus \varPhi )_i. \end{aligned}$$

By ($\oplus$3), $K_i \subseteq (\mathbf {K}\oplus \varPhi )_i$, so by ($\oplus$1), $\mathrm {Cn}\left( K_i \cup \left\{ \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) \right\} \right) \subseteq (\mathbf {K}\oplus \varPhi )_i$. $\square$

Theorem 1

A belief profile transformation function $\oplus$ satisfies ($\oplus$1) through ($\oplus$4) if and only if

$$\begin{aligned} (\mathbf {K}\oplus \varPhi ) = \left( \mathbf {K}+ \bigvee \mathrm {Cs}(\varPhi ,\mathbf {K}) \right) \end{aligned}$$

(2)

for all belief profiles $\mathbf {K}$ and all agendas $\varPhi$, where $\mathrm {Cs}(\varPhi , \mathbf {K})$ is the consistency set as introduced in Definition 5 and $\bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) = \bot$ for $\mathrm {Cs}(\varPhi , \mathbf {K}) = \emptyset$.

Proof

(“if” part) ($\oplus$1) and ($\oplus$3) are obvious.

($\oplus$2):

If $\mathrm {Cs}(\varPhi , \mathbf {K}) = \emptyset$, then $(\mathbf {K}\oplus \varPhi ) = \mathbf {K}_\bot$ and $\mathbf {K}\Box \varPhi$ holds trivially, therefore, assume $\mathrm {Cs}(\varPhi , \mathbf {K}) \ne \emptyset$. It then follows immediately that $\mu _\varPhi = \bigvee \varPhi \in (\mathbf {K}\oplus \varPhi )_i$ for all i.

Let $\varphi \in \varPhi$. If $\varphi \in \mathrm {Cs}(\varPhi , \mathbf {K})$, then $\lnot \varphi \not \in (\mathbf {K}\oplus \varPhi )_i$ for all i because $\bigvee \mathrm {Cs}(\varPhi , \mathbf {K})$ is consistent with all belief sets in $\mathbf {K}$. If furthermore $|\mathrm {Cs}(\varPhi , \mathbf {K})| = 1$, then $\varphi \in K_i$ for all i, otherwise $\varphi \not \in K_i$.

If $\varphi \not \in \mathrm {Cs}(\varPhi , \mathbf {K})$, then $\lnot \varphi \in (\mathbf {K}\oplus \varPhi )_i$ for all i because all agenda formulas are pairwise inconsistent, and $\varphi \not \in (\mathbf {K}\oplus \varPhi )_i$ for all i.

($\oplus$4):

Follows from Lemma 1.

$\square$

Proof

(“only if” part) Let $\oplus$ satisfy ($\oplus$1) through ($\oplus$4). By Lemma 1, $\left( \mathbf {K}+ \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) \right) \subseteq (\mathbf {K}\oplus \varPhi )$. By the “if” part of this theorem, $\left( \mathbf {K}+ \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) \right)$ satisfies ($\oplus$1) through ($\oplus$3), so by ($\oplus$4), $(\mathbf {K}\oplus \varPhi ) \subseteq \left( \mathbf {K}+ \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) \right)$. $\square$

Proposition 2

Let $\mathbf {K}$ be a belief profile and $\varPhi$ an agenda.

• (a) $(\mathbf {K}\oplus \varPhi ) = \mathbf {K}$ if and only if $\mathbf {K}\Box \varPhi$.

• (b) $(\mathbf {K}\oplus \varPhi ) \ne \mathbf {K}_\bot$ if and only if $\mathbf {K}\Diamond \varPhi$.

• (c) For all $\mathbf {H}\subseteq \mathbf {K}$, $(\mathbf {H}\oplus \varPhi ) \subseteq (\mathbf {K}\oplus \varPhi )$.

• (d) For all $\varPsi \le \varPhi$, $(\mathbf {K}\oplus \varPhi ) \subseteq (\mathbf {K}\oplus \varPsi )$.

Proof

• (a) Assume $\mathbf {K}\Box \varPhi$ and let i be an agent. Define $\mathrm {Cs}(\varPhi , \mathbf {K}, i) = \left\{ \varphi \in \varPhi \ \big |\ \lnot \varphi \not \in K_i \right\}$, then $\bigvee \mathrm {Cs}(\varPhi , \mathbf {K}, i) \in K_i$. Because $\mathbf {K}\Box \varPhi$, $\mathrm {Cs}(\varPhi , \mathbf {K}, i) = \mathrm {Cs}(\varPhi , \mathbf {K})$ as defined in (1). Therefore, for all i, $\bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) \in K_i$, so $(\mathbf {K}\oplus \varPhi )_i = K_i$. The converse theorem follows from ($\oplus$2).

• (b) Assume not $\mathbf {K}\Diamond \varPhi$, then $\mathrm {Cs}(\varPhi , \mathbf {K})$ is empty and $\bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) = \bot$, so $(\mathbf {K}\oplus \varPhi ) = \mathbf {K}_\bot$. Now assume $(\mathbf {K}\oplus \varPhi ) = \mathbf {K}_\bot$. By (1), $K_i \cup \left\{ \bigvee \mathrm {Cs}(\varPhi , \mathbf {K})\right\}$ can only be inconsistent for any i if either $\mathrm {Cs}(\varPhi , \mathbf {K})$ is empty, in which case $\mathbf {K}\Diamond \varPhi$ does not hold, or if $K_i = K_\bot$, again implying that $\mathrm {Cs}(\varPhi , \mathbf {K})$ is empty.

• (c) Let $\mathbf {H}\subseteq \mathbf {K}$, then $\mathrm {Cs}(\varPhi , \mathbf {K}) \subseteq \mathrm {Cs}(\varPhi , \mathbf {H})$, so $\mathrm {Cn}\left( \left\{ \bigvee \mathrm {Cs}(\varPhi , \mathbf {H})\right\} \right) \subseteq \mathrm {Cn}\left( \left\{ \bigvee \mathrm {Cs}(\varPhi , \mathbf {K})\right\} \right)$. By monotonicity of $\mathrm {Cn}$, $\mathrm {Cn}\left( H_i \cup \left\{ \bigvee \mathrm {Cs}(\varPhi , \mathbf {H})\right\} \right) \subseteq \mathrm {Cn}\left( K_i \cup \left\{ \bigvee \mathrm {Cs}(\varPhi , \mathbf {K})\right\} \right)$ for all i.

• (d) Let $\varPsi \le \varPhi$ and $\psi \in \varPsi$ such that $\lnot \psi \not \in K_i$ for all agents i. Then $\lnot \varphi \not \in K_i$ for the $\varphi \in \varPhi$ such that $\psi \vdash \varphi$ and all i, so

  $$\begin{aligned} \mathrm {Cs}(\varPsi , \mathbf {K}) \subseteq \varSigma _{\varPhi } = \left\{ \psi \in \varPsi \ \big |\ \lnot \varphi \not \in K_i \text { for the } \varphi \in \varPhi \text { such that } \psi \vdash \varphi \text { and all } i \right\} . \end{aligned}$$

  Therefore, $\bigvee \mathrm {Cs}(\varPsi , \mathbf {K}) \vdash \bigvee \varSigma _{\varPhi } \equiv \bigvee \mathrm {Cs}(\varPhi , \mathbf {K})$, so $\mathrm {Cn}\left( K_i \cup \left\{ \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}) \right\} \right) \subseteq \mathrm {Cn}\left( K_i \cup \left\{ \bigvee \mathrm {Cs}(\varPsi , \mathbf {K}) \right\} \right)$ for all i. $\square$

Lemma 2

Let $\mathbf {K}$ be a belief profile, $\varPhi$ an agenda, i an agent and $\mathbf {K}' \in \mathbf {K}\triangle {\varPhi }$. If $A \in K_i$ and $A \not \in K'_i$, then there exists a $\varphi \in \varPhi$ such that (a) $\lnot \varphi \in K_i$, $\lnot \varphi \not \in K'_j$ for all agents j, and $A \rightarrow \lnot \varphi \in K'_i$, and (b) $\lnot \varphi ' \in K'_i$ for all $\varphi ' \in \varPhi$ such that $\lnot \varphi ' \in K_i$ and $\varphi ' \ne \varphi$.

Proof

(a) follows directly from the maximality condition on the elements of $\mathbf {K}\triangle \varPhi$. For (b), let $\varphi ' \in \varPhi$ such that $\lnot \varphi ' \in K_i$ and $\varphi ' \ne \varphi$, and assume for reductio that $\lnot \varphi ' \not \in K'_i$. Then by (a), there exists a $\varphi '' \in \varPhi$ such that $\lnot \varphi '' \not \in K'_i$ and $\lnot \varphi ' \rightarrow \lnot \varphi '' \in K'_i$. From the definition of an agenda and logical closure of $K_i'$ we know that $\lnot (\varphi ' \wedge \varphi '') \in K'_i$ and thereby $\varphi ' \rightarrow \lnot \varphi '' \in K'_i$, so $\lnot \varphi '' \in K'_i$, contradicting $\lnot \varphi '' \not \in K'_i$. $\square$

Theorem 2

A belief profile transformation function satisfies ($\ominus$1) through ($\ominus$6) if and only if it is a partial meet agreement contraction function.

Proof

(“if” part)

($\ominus$1):

By definition, $\mathbf {K}\triangle \varPhi$ contains only belief profiles, and the intersection of belief sets is a belief set, therefore, $\mathbf {K}\ominus \varPhi$ is a belief profile.

($\ominus$2):

By definition, $\mathbf {K}\triangle \varPhi$ contains only subprofiles of $\mathbf {K}$; therefore, $\mathbf {K}\ominus \varPhi$ is also a subprofile of $\mathbf {K}$.

($\ominus$3):

If $\mathbf {K}\Diamond \varPhi$, then $\mathbf {K}\triangle \varPhi = \{\mathbf {K}\}$; hence, $\mathbf {K}\ominus \varPhi = \mathbf {K}$.

($\ominus$4):

If $\mathbf {K}' \in \mathbf {K}\triangle \varPhi$, then $\mathbf {K}' \Diamond \varPhi$, and this property holds for any subprofile of $\mathbf {K}'$; therefore, $(\mathbf {K}\ominus \varPhi ) \Diamond \varPhi$.

($\ominus$5):

Let i be an agent, $A \in K_i$, $\mathbf {K}' \in \mathbf {K}\triangle \varPhi$ and $\sigma = \lnot \bigvee \mathrm {Cs}_\ominus (\varPhi , \mathbf {K}, i)$ with $\mathrm {Cs}_\ominus (\varPhi , \mathbf {K}, i)$ as defined in (3). If $A \in K'_i$, there is nothing more to show, so assume $A \not \in K'_i$. By weakening, $\sigma \rightarrow A \in K_i$. Assume for reductio that $\sigma \rightarrow A \not \in K'_i$. Then by Lemma 2, there exists a $\varphi \in \varPhi$ such that $\lnot \varphi \in K_i$, $\lnot \varphi \not \in K'_i$ and $(\sigma \rightarrow A) \rightarrow \lnot \varphi \in K'_i$. By closure, $(\sigma \rightarrow A) \rightarrow \sigma \in K'_i$, and $((\sigma \rightarrow A) \rightarrow \sigma ) \rightarrow \sigma \in K'_i$ because it is a tautology, so $\sigma \in K'_i$. By the definition of $\sigma$, this implies $\lnot \varphi \in K'_i$, which contradicts the result above. Hence, $\sigma \rightarrow A \in K'_i$, so $\sigma \rightarrow A \in \bigcap \left\{ K'_i \ \big |\ \mathbf {K}' \in S(\mathbf {K}\triangle \varPhi )\right\}$ for any selection function S, and, therefore, $A \in \mathrm {Cn}\left( (\mathbf {K}' \ominus \varPhi )_i \cup \left\{ \lnot \bigvee \mathrm {Cs}_\ominus (\varPhi , \mathbf {K}, i) \right\} \right)$.

($\ominus$6):

Let $\varPhi \equiv \varPsi$, then by Proposition 1 (c), $\mathbf {K}' \Diamond \varPhi$ if and only if $\mathbf {K}' \Diamond \varPsi$ for any belief profile $\mathbf {K}'$, so $\mathbf {K}\triangle \varPhi = \mathbf {K}\triangle \varPsi$ and, therefore, $\mathbf {K}\ominus \varPhi = \mathbf {K}\ominus \varPsi$.$\square$

Proof

(“only if” part) Let $\ominus$ satisfy ($\ominus$1) through ($\ominus$6). Define S as

$$\begin{aligned} S(\mathbf {K}\triangle \varPhi ) = \left\{ \mathbf {K}' \in \mathbf {K}\triangle \varPhi \ \big |\ (\mathbf {K}\ominus \varPhi ) \subseteq \mathbf {K}' \right\} . \end{aligned}$$

By ($\ominus$2) and ($\ominus$4), $S(\mathbf {K}\triangle \varPhi )$ is non-empty. Let i be an agent. It follows from the definition of S that $(\mathbf {K}\ominus \varPhi )_i \subseteq \bigcap \left\{ K'_i \ \big |\ \mathbf {K}' \in S(\mathbf {K}\triangle \varPhi )\right\}$, so we have to show that $\bigcap \left\{ K'_i \ \big |\ \mathbf {K}' \in S(\mathbf {K}\triangle \varPhi )\right\} \subseteq (\mathbf {K}\ominus \varPhi )_i$. Let $A \not \in (\mathbf {K}\ominus \varPhi )_i$, we then show that $A \not \in \bigcap \left\{ K'_i \ \big |\ \mathbf {K}' \in S(\mathbf {K}\triangle \varPhi )\right\}$. If $A \not \in K_i$, this holds trivially, so assume $A \in K_i$. Let $\sigma = \lnot \bigvee \mathrm {Cs}_\ominus (\varPhi , \mathbf {K}, i)$ with $\mathrm {Cs}_\ominus (\varPhi , \mathbf {K}, i)$ as defined in (3). Then by ($\ominus$5), $\sigma \rightarrow A \in (\mathbf {K}\ominus \varPhi )_i$, so with $A \not \in (\mathbf {K}\ominus \varPhi )_i$, it follows that $\sigma \vee A \not \in (\mathbf {K}\ominus \varPhi )_i$. We define a belief profile $\mathbf {K}'$ where $K'_i$ is a maximal belief set such that $(\mathbf {K}\ominus \varPhi )_i \subseteq K'_i \subseteq K_i$ and $\sigma \vee A \not \in K'_i$. By (3), there exists a $\varphi \in \varPhi$ such that $\lnot \varphi \not \in K'_i$ and $\lnot \varphi \not \in (\mathbf {K}\ominus \varPhi )_j$ for all j. For all agents $j \ne i$, set $K'_j$ to a maximal belief set such that $(\mathbf {K}\ominus \varPhi )_j \subseteq K'_j \subseteq K_j$ and $\lnot \varphi \not \in K'_j$. Therefore, $\mathbf {K}' \Diamond \varPhi$ and because it is maximal for all agents, $\mathbf {K}' \in \mathbf {K}\triangle \varPhi$, and since it is a superprofile of $\mathbf {K}\ominus \varPhi$, $\mathbf {K}\in S(\mathbf {K}\triangle \varPhi )$. Finally, because $A \not \in K'_i$, $A \not \in \bigcap \left\{ K''_i \ \big |\ \mathbf {K}'' \in S(\mathbf {K}\triangle \varPhi )\right\}$. $\square$

Theorem 3

Let $\ominus$ be a transformation function satisfying $(\ominus 1)$ through $(\ominus 4)$ and $(\ominus 6)$. Then the function $\circledast$ defined for all belief profiles $\mathbf {K}$ and all agendas $\varPhi$ as

$$\begin{aligned} (\mathbf {K}\circledast \varPhi ) = ((\mathbf {K}\ominus \varPhi ) \oplus \varPhi ) \end{aligned}$$

(6)

satisfies $(\circledast 1)$ through $(\circledast 6)$. If, in addition, $\ominus$ satisfies $(\ominus 5)$, then $\circledast$ satisfies $(\circledast 7)$.

Proof

($\circledast$1) and ($\circledast$2) follow from ($\oplus$1) and ($\oplus$2), respectively. ($\circledast$3) follows from ($\ominus$2) and Proposition 2 (c). ($\circledast$4) follows from ($\ominus$3). ($\circledast$5) follows from ($\ominus$4) and Proposition 2 (b). ($\circledast$6) follows from ($\ominus$6). For ($\circledast$7), observe that $\mathrm {Cs}(\varPhi , \mathbf {K}\circledast \varPhi ) = \mathrm {Cs}\left( (\varPhi , \mathbf {K}\ominus {\varPhi }) + \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}\ominus \varPhi ) \right) = \mathrm {Cs}(\varPhi , \mathbf {K}\ominus \varPhi )$; therefore, $\mathrm {Cs}_\ominus (\varPhi , \mathbf {K}, i) = \mathrm {Cs}_\circledast (\varPhi , \mathbf {K}, i)$ for all i. $\square$

Theorem 4

Let $\circledast$ be a transformation function satisfying $(\circledast 1)$ through $(\circledast 6)$. Then the function $\ominus$ defined for all belief profiles $\mathbf {K}$ and all agendas $\varPhi$ as

$$\begin{aligned} (\mathbf {K}\ominus \varPhi ) = \mathbf {K}\cap (\mathbf {K}\circledast \varPhi ) \end{aligned}$$

(7)

satisfies ($\ominus$1) through ($\ominus$4), ($\ominus$5’), and ($\ominus$6). If, in addition, $\circledast$ satisfies ($\circledast$7), then $\ominus$ satisfies ($\ominus$5).

Proof

($\ominus$1) follows from ($\circledast$1). ($\ominus$2) is obvious. For ($\ominus$3), assume $\mathbf {K}\Diamond \varPhi$; then by ($\circledast$4) and ($\oplus$2), $\mathbf {K}\subseteq (\mathbf {K}\circledast \varPhi )$; therefore, $\mathbf {K}\subseteq (\mathbf {K}\ominus \varPhi )$. ($\ominus$4) follows from ($\circledast$2) and ($\circledast$5). For ($\ominus$5), the same observation as in the proof of ($\circledast$7) in Theorem 3 applies. For ($\ominus$5’), let i be an agent, $A \in K_i$, and $B = \bigvee \mathrm {Cs}(\varPhi , \mathbf {K})$. By ($\circledast$2), $B \in (\mathbf {K}\circledast \varPhi )_i$ and, therefore, $\lnot B \rightarrow A \in (\mathbf {K}\circledast \varPhi )_i$. Since also $\lnot B \rightarrow A \in K_i$, $\lnot B \rightarrow A \in (\mathbf {K}\cap (\mathbf {K}\circledast \varPhi ))_i$ and, therefore, $A \in (\mathbf {K}\cap (\mathbf {K}\circledast \varPhi ))_i + \lnot B$. ($\ominus$6) follows from ($\circledast$6). $\square$

Theorem 5

• (a) Let $\ominus$ be a transformation function satisfying ($\ominus$1) through ($\ominus$4), ($\ominus$5’), and ($\ominus$6). Then for all belief profiles $\mathbf {K}$ and all agendas $\varPhi$, $(\mathbf {K}\ominus \varPhi ) = \mathbf {K}\cap ((\mathbf {K}\ominus \varPhi ) \oplus \varPhi )$.

• (b) Let $\circledast$ be a transformation function satisfying ($\circledast$1) through ($\circledast$6). Then for all belief profiles $\mathbf {K}$ and all agendas $\varPhi$, $(\mathbf {K}\circledast \varPhi ) = ((\mathbf {K}\cap (\mathbf {K}\circledast \varPhi )) \oplus \varPhi )$.

Proof

• (a) Let $\ominus$, $\mathbf {K}$, and $\varPhi$ be defined as stated. It follows from ($\ominus$2) that $(\mathbf {K}\ominus \varPhi ) \subseteq \mathbf {K}$, and from ($\oplus$3) that $(\mathbf {K}\ominus \varPhi ) \subseteq ((\mathbf {K}\ominus \varPhi ) \oplus \varPhi )$, so $(\mathbf {K}\ominus \varPhi ) \subseteq \mathbf {K}\cap ((\mathbf {K}\ominus \varPhi ) \oplus \varPhi )$. For the converse inclusion, let i be an agent and A a formula such that $A \in K_i$ and $A \in ((\mathbf {K}\ominus \varPhi ) \oplus \varPhi )_i = (\mathbf {K}\ominus \varPhi )_i + B$ where $B = \bigvee \mathrm {Cs}(\varPhi , (\mathbf {K}\ominus \varPhi ))$, so $B \rightarrow A \in (\mathbf {K}\ominus \varPhi )_i$. We now show that $A \in (\mathbf {K}\ominus \varPhi )_i$. From ($\ominus$5’) it follows that $A \in (\mathbf {K}\ominus \varPhi )_i + \lnot B$. Therefore, both $B \rightarrow A \in (\mathbf {K}\ominus \varPhi )_i$ and $\lnot B \rightarrow A \in (\mathbf {K}\ominus \varPhi )_i$, so $A \in (\mathbf {K}\ominus \varPhi )_i$.

• (b) Let $\circledast$, $\mathbf {K}$, and $\varPhi$ be defined as stated. Since $(\mathbf {K}\circledast \varPhi ) \subseteq \mathbf {K}\cap (\mathbf {K}\circledast \varPhi )$, it follows from Proposition 2 (c) and ($\oplus$3) that $(\mathbf {K}\circledast \varPhi ) \subseteq ((\mathbf {K}\cap (\mathbf {K}\circledast \varPhi )) \oplus \varPhi )$. For the converse inclusion, let i be an agent and $A \in ((\mathbf {K}\cap (\mathbf {K}\circledast \varPhi )) \oplus \varPhi )_i = (K_i \cap (\mathbf {K}\circledast \varPhi )_i) + \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}\cap (\mathbf {K}\circledast \varPhi ))$. With $B = \bigvee \mathrm {Cs}(\varPhi , \mathbf {K}\cap (\mathbf {K}\circledast \varPhi ))$, it follows that $B \rightarrow A \in (\mathbf {K}\circledast \varPhi )_i$. We now show that $A \in (\mathbf {K}\circledast \varPhi )_i$ by proving that $B\in (\mathbf {K}\circledast \varPhi )_i$. From ($\circledast$2) it follows that $\{\varphi \in \varPhi \ \big |\ \lnot \varphi \not \in (\mathbf {K}\circledast \varPhi )_i\} = \mathrm {Cs}(\varPhi , (\mathbf {K}\circledast \varPhi ))$, so $\bigvee \mathrm {Cs}(\varPhi , (\mathbf {K}\circledast \varPhi )) \in (\mathbf {K}\circledast \varPhi )_i$. And because $\mathrm {Cs}(\varPhi , (\mathbf {K}\circledast \varPhi )) \subseteq \mathrm {Cs}(\varPhi , \mathbf {K}\cap (\mathbf {K}\circledast \varPhi ))$, $B \in (\mathbf {K}\circledast \varPhi )_i$. $\square$

1.2 Section 4

Proposition 3

Let $\mathbf {K}$ be a belief profile and $\varPhi$ an agenda. Then $\mathbf {K}\Box \varPhi$ if and only if there exists a set $\varPi$ of interpretation profiles such that $\pi \Box \varPhi$ for all $\pi \in \varPi$ and $[K_i] = \left\{ \pi _i \ \big | \ \pi \in \varPi \right\}$ for all i.

Proof

(“if” part) Let $\varPi$ be a set of interpretation profiles such that $\pi \Box \varPhi$ for all $\pi \in \varPi$ and $[K_i] = \left\{ \pi _i \ \big | \ \pi \in \varPi \right\}$ for all i. Let i, j be agents and $\varphi \in \varPhi$. Then $\varphi \in K_i$ if and only if $\pi _i \models \varphi$ for all $\pi \in \varPi$, which by unanimity of all $\pi \in \varPi$ holds if and only if $\pi _j \models \varphi$ for all $\pi \in \varPi$, and this holds just in case $\varphi \in K_j$. Similarly, it follows that $\lnot \varphi \in K_i$ if and only if $\lnot \varphi \in K_j$. And for all $\pi \in \varPi$, $\pi _i \models \mu _\varPhi$ by $\pi \Box \varPhi$, so $\mathbf {K}\Box \varPhi$.   $\square$

Proof

(“only if” part) Assume $\mathbf {K}\Box \varPhi$. Define

$$\begin{aligned} \varPi = \left\{ \pi \in [\mathbf {K}] \ \big |\ \pi \Box \varPhi \right\} . \end{aligned}$$

Let i be an agent and $\omega _i \in [K_i]$. Then $\omega _i\models \mu _\varPhi$, and there exists a $\varphi \in \varPhi$ such that $\omega _i \models \varphi$. Therefore, $\lnot \varphi \not \in K_i$ and by $\mathbf {K}\Box \varPhi$, $\lnot \varphi \not \in K_j$ for any agent j. Then for each j, there exists a model $\omega _j \in [K_j]$ such that $\omega _j \models \varphi$ as well as $\omega _j \models \mu _\varPhi$. Let $\varphi ' \in \varPhi$ such that $\varphi ' \ne \varphi$, then because $\vdash \lnot (\varphi \wedge \varphi ')$, $\omega _i \not \models \varphi '$ and $\omega _j \not \models \varphi '$. Hence, $\omega _i$ and $\omega _j$ are unanimous on all $\varphi \in \varPhi$. Then the profile $\pi = (\omega _k)$ with $\omega _k = \omega _i$ for $k = i$ and $\omega _k = \omega _j$ for all other k is unanimous on $\varPhi$ and is a model for $\mu$; hence, $\pi \in \varPi$ and $\omega _i \in \left\{ \pi _i \ \big | \ \pi \in \varPi \right\}$, and finally $[K_i] = \left\{ \pi _i \ \big | \ \pi \in \varPi \right\}$. $\square$

Lemma 3

For any belief profile $\mathbf {K}$ and any agenda $\varPhi$,

$$\begin{aligned}{}[\mathbf {K}\oplus \varPhi ] = \left\{ \pi \in [\mathbf {K}] \ \big |\ \pi \Box \varPhi \right\} . \end{aligned}$$

(23)

Proof

We prove that the transformation function defined by (23) satisfies the postulates ($\oplus$1) through ($\oplus$4). ($\oplus$1) follows directly from (23). ($\oplus$2) follows from Proposition 3 (“if”). By (23), $[\mathbf {K}\oplus \varPhi ] \subseteq [\mathbf {K}]$; therefore, ($\oplus$3) holds. To prove ($\oplus$4), let $\mathbf {K}'$ be any other profile that satisfies ($\oplus$1) through ($\oplus$3). By ($\oplus$2) and Proposition 3 (“only if”), $\pi \Box \varPhi$ for all $\pi \in [\mathbf {K}']$. By ($\oplus$2), $[\mathbf {K}'] \subseteq [\mathbf {K}]$. Therefore, $[\mathbf {K}'] \subseteq [\mathbf {K}\oplus \varPhi ]$. Finally, by Theorem 1 (“only if”), $\oplus$ is identical to the expansion agreement function. $\square$

Theorem 6

Let $\preceq$ be an interpretation profile order. Then the belief set transformation function $\circledast$ with

$$\begin{aligned} \left[ \mathbf {K}\circledast \varPhi \right] = \min \left( \varPi _{\varPhi }, \preceq _\mathbf {K}\right) \end{aligned}$$

(10)

where

$$\begin{aligned} \varPi _{\varPhi } = \left\{ \pi \in \varOmega ^n \ \big |\ \pi \Box \varPhi \right\} \end{aligned}$$

(11)

satisfies conditions $(\circledast 1)$ through $(\circledast 6)$.

Proof

Let $\circledast$ be a transformation function based on an interpretation profile order as defined in (10).

($\circledast$1):

Follows directly from (10).

($\circledast$2):

By (11), all $\omega \in \varPi _{\varPhi }$ and hence all $\omega \in \left[ \mathbf {K}\circledast \varPhi \right]$ satisfy $\omega \Box \varPhi$, so by Proposition 3 (“if” part), $(\mathbf {K}\circledast \varPhi ) \Box \varPhi$.

($\circledast$3):

It follows from (8) and (23) that $[\mathbf {K}\oplus \varPhi ] \subseteq [\mathbf {K}\circledast \varPhi ]$.

($\circledast$4):

Let $\mathbf {K}\Diamond \varPhi$. Then there exists a $\varphi \in \varPhi$ such that for all agents i, $\lnot \varphi \not \in K_i$, so for each i, there exists an $\omega \in [K_i]$ such that $\omega \models \varphi$. Define the profile $\pi = (\omega _i)$. Then $\pi \in [\mathbf {K}]$ and $\pi \Box \varPhi$, so $\varPi _{\varPhi } \cap [\mathbf {K}]$ is non-empty. With (8) and (23) it follows that $[\mathbf {K}\circledast \varPhi ] \subseteq [\mathbf {K}\oplus \varPhi ]$.

($\circledast$5):

For any $\varPhi$, there exists an interpretation profile $\pi$ such that $\pi \Box \varPhi$, so $\varPi _{\varPhi }$ is non-empty. Because $\preceq _\mathbf {K}$ is total and $\varPi _{\varPhi }$ is finite due to finiteness of the set of propositional symbols P, $\min \left( \varPi _{\varPhi }, \preceq _\mathbf {K}\right)$ is also non-empty and, therefore, $(\mathbf {K}\circledast \varPhi ) \ne \mathbf {K}_\bot$.

($\circledast$6):

Let $\varPhi \equiv \varPsi$. For any $\pi \in \varOmega ^n$, $\pi \Box \varPhi$ if and only if $\pi \Box \varPsi$, so $\varPi _{\varPhi } = \varPi _{\varPsi }$ and $(\mathbf {K}\circledast \varPhi ) = (\mathbf {K}\circledast \varPsi )$.$\square$

1.3 Section 5.1

Proposition 4

Let K be a belief set and A a contingent formula. Define the belief profile $\mathbf {K}$ with $n= 1$ as $K_1 = K$.

• (a)
  $$\begin{aligned} K +A = (\mathbf {K}\oplus \{A\})_1 \end{aligned}$$

  (17)

• (b) Let $\ominus$ be an agreement contraction function satisfying $(\ominus 1)$ through $(\ominus 6)$. Then the function $-$ defined as

  $$\begin{aligned} K -A = (\mathbf {K}\ominus \{\lnot A\})_1 \end{aligned}$$

  (18)

  satisfies $(-1)$ through $(-6)$.

• (c) Let $\circledast$ be an agreement contraction function satisfying $(\circledast 1)$ through $(\circledast 6)$. Then the function $*$ defined as

  $$\begin{aligned} K *A = (\mathbf {K}\circledast \{A\})_1 \end{aligned}$$

  (19)

  satisfies $(*1)$ through $(*6)$.

Proof

• (a) If $\lnot A \in K$, then $\mathrm {Cs}(\{A\}, \mathbf {K}) = \emptyset$, so $K +A = K_\bot = K_1 +\bigvee \mathrm {Cs}(\{A\}, \mathbf {K}) = (\mathbf {K}\oplus \{A\})_1$. If $\lnot A \not \in K$, then $\mathrm {Cs}(\{A\}, \mathbf {K}) = \{A\}$, so $K +A = K_1 +\bigvee \mathrm {Cs}(\{A\}, \mathbf {K}) = (\mathbf {K}\oplus \{A\})_1$.

• (b) Conditions ($-$1), ($-$2), and ($-$6) follow immediately from ($\ominus$1), ($\ominus$2), and ($\ominus$6), respectively.

  ($-$3):

  Assume $A \not \in K$. Then $\mathbf {K}\Diamond \{\lnot A\}$, and, therefore, by ($\ominus$3), $(\mathbf {K}\ominus \{\lnot A\}) = \mathbf {K}$, so $K -A = K$.

  ($-$4):

  By ($\ominus$4), $\lnot \lnot A \not \in (\mathbf {K}\ominus \{\lnot A\})_1$, so $A \not \in K -A$.

  ($-$5):

  If $A \not \in K$, then $K -A = K$ and ($-$5) holds trivially. If $A \in K$, then $\mathrm {Cs}_\ominus (\{\lnot A\}, \mathbf {K}, 1) = \{\lnot A\}$, so by ($\ominus$5), $K_1 \subseteq (\mathbf {K}\ominus \{\lnot A\})_1 +\lnot \bigvee \{\lnot A\}$; therefore, $K \subseteq (K -A) +A$.

• (c) Conditions ($*$1) through ($*$6) follow immediately from the respective conditions ($\circledast$1) through ($\circledast$6).$\square$

1.4 Section 5.2

Proposition 5

For any JA function F satisfying universal domain, collective rationality, and unanimity preservation, and any JA agenda X, there exists an agreement revision function $\circledast _F$ satisfying $(\circledast 1)$ through $(\circledast 6)$ such that F is the associated JA function of $\circledast _F$ with respect to X for profiles of complete judgment sets.

Proof

Let $\circledast$ be a agreement revision function satisfying ($\circledast$1) through ($\circledast$6) (for instance a distance-based function as defined in Sect. 4.3), F any JA function satisfying the mentioned conditions, and X any JA agenda. For any belief profile $\mathbf {K}$ and any agreement agenda $\varPhi$, define $\circledast _F$ as follows: if $\varPhi = \varPhi (X)$ and $p_X(\mathbf {K})$ is complete with respect to X, then $(\mathbf {K}\circledast _F \varPhi ) = (\mathbf {K}\circledast \{\bigwedge F(p_X(\mathbf {K}))\})$, and $(\mathbf {K}\circledast _F \varPhi ) = (\mathbf {K}\circledast \varPhi )$ otherwise. By collective consistency of F, $\bigwedge F(p_X(\mathbf {K})))$ is consistent, and by universal domain, it is defined for all $\mathbf {K}$, so the domain of $\circledast _F$ is the set of all belief profiles. In the following, assume $\varPhi = \varPhi (X)$ and $p_X(\mathbf {K})$ is complete; otherwise $\circledast _F$ inherits the respective properties from $\circledast$. That $\circledast _F$ satisfies ($\circledast$1), ($\circledast$5), and ($\circledast$6) follows directly from the corresponding properties of $\circledast$. ($\circledast$2) follows from the fact that F is collectively complete and, therefore, there exists a $\varphi \in \varPhi (X)$ such that $\varphi \equiv \bigwedge F(p_X(\mathbf {K}))$. For ($\circledast$3) and ($\circledast$4), assume $\mathbf {K}\Diamond \varPhi (X)$, then by completeness of $p_X(\mathbf {K})$ and unanimity preservation of F, $\mathrm {Cs}(\varPhi (X), \mathbf {K}) = \bigwedge F(p_X(\mathbf {K}))$ and, therefore, $(\mathbf {K}\circledast \bigwedge F(p_X(\mathbf {K}))) = (\mathbf {K}\oplus \varPhi (X))$. The associated JA function $F'$ of $\circledast _F$ is

$$\begin{aligned} F'(\mathbf {J})&= \bigcap _i p_{X} \left( \mathrm {\mathbf {Cn}}(\mathbf {J}) \circledast \bigwedge F(p_X(\mathrm {\mathbf {Cn}}(\mathbf {J})))\right) _i \\&= \bigcap _i p_{X} \left( \mathrm {\mathbf {Cn}}(\mathbf {J}) \circledast \bigwedge F(\mathbf {J})\right) _i \end{aligned}$$

for all judgment profiles $\mathbf {J}$. Let $x \in F'(\mathbf {J})$, then $x \in X$ and $x \in \left( \mathrm {\mathbf {Cn}}(\mathbf {J}) \circledast \bigwedge F(\mathbf {J})\right) _i$ for all i. By ($\circledast$2), x must be consistent with $\bigwedge F(\mathbf {J})$, which because of completeness of F is only possible if $x \in F(\mathbf {J})$. Now let $x \in F(\mathbf {J})$, then by ($\circledast$2), $x \in \left( \mathrm {\mathbf {Cn}}(\mathbf {J}) \circledast \bigwedge F(\mathbf {J})\right) _i$ for all i. Since $x \in X$ by the definition of a JA function, $x \in F'(\mathbf {J})$. Therefore, $F(\mathbf {J}) = F'(\mathbf {J})$. $\square$

Proposition 6

There exist belief profiles $\mathbf {K}, \mathbf {K}'$, an agenda $\varPhi$ and an agreement revision function $\circledast$ such that for any JA function F and any belief revision function profile $(*)$, either

$$\begin{aligned} (\mathbf {K}\circledast \varPhi )\ne & {} (\mathbf {K}(*)\bigwedge F(p_{X(\varPhi )}(\mathbf {K}))) \text{ or }\\ (\mathbf {K}' \circledast \varPhi )\ne & {} (\mathbf {K}' (*)\bigwedge F(p_{X(\varPhi )}(\mathbf {K}'))). \end{aligned}$$

Proof

Define $\mathbf {K}= (\mathrm {Cn}(p \wedge q), \mathrm {Cn}(\lnot p \wedge q))$, $\mathbf {K}' = (\mathrm {Cn}(p \wedge \lnot q), \mathrm {Cn}(\lnot p \wedge q))$, $\varPhi = \{p, \lnot p\}$, and $\circledast$ such that $(\mathbf {K}\circledast \varPhi ) = (\mathrm {Cn}(p \wedge q), \mathrm {Cn}(p \wedge \lnot q))$ and $(\mathbf {K}' \circledast \varPhi ) = (\mathrm {Cn}(p \wedge \lnot q), \mathrm {Cn}(p \wedge q))$. Let F be a JA function and $(*)$ a belief revision function profile. Assume for reductio that both identities hold. For brevity of notation, let $G(\mathbf {K}) = \bigwedge F(p_{X(\varPhi )}(\mathbf {K}))$ and for $\mathbf {K}'$ accordingly. By success of $(*)$, $G(\mathbf {K}) = G(\mathbf {K}') = p$. But because $\mathbf {K}_2 = \mathbf {K}'_2$, $(\mathbf {K}(*)G(\mathbf {K}))_2 = (\mathbf {K}' (*)G(\mathbf {K}'))_2$, but by assumption, $(\mathbf {K}\circledast \varPhi )_2 \ne (\mathbf {K}' \circledast \varPhi )_2$, contradiction. $\square$