Judgment aggregation in nonmonotonic logic

Abstract

Judgment aggregation studies how to aggregate individual judgments on logically correlated propositions into collective judgments. Different logics can be used in judgment aggregation, for which Dietrich and Mongin have proposed a generalized model based on general logics. Despite its generality, however, all nonmonotonic logics are excluded from this model. This paper argues for using nonmonotonic logic in judgment aggregation. Then it generalizes Dietrich and Mongin’s model to incorporate a large class of nonmonotonic logics. This generalization broadens the theoretical boundaries of judgment aggregation by proving that, even if these nonmonotonic logics are employed, certain typical impossibility results still hold.

Appendix: Proof of the representation of \(\mathcal {NL}\)

Given a choice model $\mathfrak {M}=(M,\models ,f)$, we write $X\models \varphi$ if for all $m\in X$, $m\models \varphi$. For any $X\subseteq M$, we denote by $\overline{X}$ the set of all $\varphi \in \mathcal {L}$ such that $X\models \varphi$. The following Galois connection is easily verified.

Lemma 9

For any $\varGamma ,\varSigma \subseteq \mathcal {L}$ and $X,Y\subseteq M$,

• (1) $\varGamma \subseteq \varSigma \Longrightarrow \widehat{\varSigma }\subseteq \widehat{\varGamma }$, $X\subseteq Y\Longrightarrow \overline{Y}\subseteq \overline{X}$

• (2) $\varGamma \subseteq \overline{\widehat{\varGamma }}$, $X\subseteq \widehat{\overline{X}}$

• (3) $\widehat{\varGamma }=\widehat{\overline{\widehat{\varGamma }}}$, $\overline{X}=\overline{\widehat{\overline{X}}}$

• (4) $\widehat{\varGamma \cup \varSigma }=\widehat{\varGamma }\cap \widehat{\varSigma }$, $\overline{X\cup Y}=\overline{X}\cap \overline{Y}$

1.1 Proof of Theorem 3

Proof

For NTR, suppose $\mid \!\approx _{\mathfrak {M}}\varphi$ and $\mid \!\approx _{\mathfrak {M}}\lnot \varphi$. Since $\mathfrak {M}$ is classical for negation, there is no m in $\mathfrak {M}$ such that $m\models \varphi$ and $m\models \lnot \varphi$. It follows that $f(\widehat{\emptyset })=f(M)=\emptyset$. Since $M\ne \emptyset$, this contradicts Ntr.

For INC, suppose $\varphi \in \varGamma$. By Lemma 9, we have $\widehat{\varGamma }\subseteq \widehat{\varphi }$. By Inc, $f(\widehat{\varGamma })\subseteq \widehat{\varGamma }\subseteq \widehat{\varphi }$.

For CM, suppose $\varGamma \mid \!\approx _{\mathfrak {M}}\varSigma$ and $\varGamma \mid \!\approx _{\mathfrak {M}}\varphi$. By the former, $f(\widehat{\varGamma })\subseteq \widehat{\varSigma }$. By Inc, $f(\widehat{\varGamma })\subseteq \widehat{\varGamma }$. Hence, $f(\widehat{\varGamma })\subseteq \widehat{\varGamma }\cap \widehat{\varSigma }\subseteq \widehat{\varGamma }$. By the latter $f(\widehat{\varGamma })\subseteq \widehat{\varphi }$. By Lemma 9 and cum, it follows that $f(\widehat{\varGamma \cup \varSigma })=f(\widehat{\varGamma }\cap \widehat{\varSigma })=f(\widehat{\varGamma })\subseteq \widehat{\varphi }$.

For CC, suppose $\varGamma \mid \!\approx _{\mathfrak {M}}\varSigma$ and $\varGamma \cup \varSigma \mid \!\approx _{\mathfrak {M}}\varphi$. By the former, $f(\widehat{\varGamma })\subseteq \widehat{\varSigma }$. Hence, $f(\widehat{\varGamma })\subseteq \widehat{\varGamma }\cap \widehat{\varSigma }\subseteq \widehat{\varGamma }$. By the latter, $f(\widehat{\varGamma \cup \varSigma })\subseteq \widehat{\varphi }$. By cum and Lemma 9, it follows that $f(\widehat{\varGamma })=f(\widehat{\varGamma }\cap \widehat{\varSigma })=f(\widehat{\varGamma \cup \varSigma })\subseteq \widehat{\varphi }$.

For PBC, suppose $\varGamma ,\lnot \varphi \mid \!\approx _{\mathfrak {M}}\psi$ and $\varGamma ,\lnot \varphi \mid \!\approx \lnot \psi$ for some $\varphi \in \mathcal {L}$. Then $f(\widehat{\varGamma \cup \{\lnot \varphi \}})\subseteq \widehat{\psi }\cap \widehat{\lnot \psi }$. Since $\mathfrak {M}$ is classical for negation, $\widehat{\psi }\cap \widehat{\lnot \psi }=\emptyset$. Hence, $f(\widehat{\varGamma \cup \{\lnot \varphi \}})=\emptyset$. By Ntr, we have $\widehat{\varGamma }\cap \widehat{\lnot \varphi }=\widehat{\varGamma \cup \{\lnot \varphi \}}=\emptyset$. By Inc and that $\mathfrak {M}$ is classical for negation, $f(\widehat{\varGamma })\subseteq \widehat{\varGamma }\subseteq M-\widehat{\lnot \varphi }=\widehat{\varphi }$.

RTA is proved analogously to PBC.

For NonP, suppose $\varGamma \mid \!\approx _{\mathfrak {M}}\psi$ and $\varGamma \mid \!\approx _{\mathfrak {M}}\lnot \psi$ for some $\varphi \in \mathcal {L}$. Then $f(\widehat{\varGamma })\subseteq \widehat{\psi }\cap \widehat{\lnot \psi }$. Since $\mathfrak {M}$ is classical for negation, $\widehat{\psi }\cap \widehat{\lnot \psi }=\emptyset$. Hence, $f(\widehat{\varGamma })=\emptyset$. It follows that $f(\widehat{\varGamma })\subseteq \widehat{\varphi }$ for all $\varphi \in \mathcal {L}$. $\square$

1.2 Proof of Theorem 4

Given a logic $\mid \!\sim$, let $C_{\mid \!\sim }$ be the consequence operator of it, i.e. $C_{\mid \!\sim }(\varGamma )=\{\varphi \mid \varGamma \mid \!\sim \varphi \}$. The following is a well known result in nonmonotonic logic (see e.g. Makinson 1989). We recall it here for later use.

Lemma 10

If $\mid \!\sim$ satisfies INC, CM and CC, then for all $\varGamma \subseteq \mathcal {L}$,

$$\begin{aligned} \text{ IDE } \quad C_{\mid \!\sim }(\varGamma )=C_{\mid \!\sim }C_{\mid \!\sim }(\varGamma ). \end{aligned}$$

The following lemma shows that RTA and weak compactness are sufficient for proving Lindenbaum’s lemma.

Lemma 11

Let $\mid \!\sim$ be a logic satisfying RTA, and weak compactness. Then every consistent set in $\mid \!\sim$ can be extended to a maximal consistent set in it.

Proof

Let $\varGamma$ be any consistent set in $\mid \!\sim$. Let $\varphi _{0},\ldots ,\varphi _{n},\ldots$ enumerate all formulas in $\mathcal {L}$. Define $\varGamma _{n}$ as follows:

$$\begin{aligned} \begin{aligned}\varGamma _{0}&=\varGamma \\ \varGamma _{n+1}&={\left\{ \begin{array}{ll} \varGamma _{n}\cup \{\varphi _{n}\} &{} \text{ if } \varGamma _{n}\cup \{\varphi _{n}\} \text{ is } \text{ consistent } \text{ in } \mid \!\sim \\ \varGamma _{n}\cup \{\lnot \varphi _{n}\} &{} \text{ if } \varGamma _{n}\cup \{\lnot \varphi _{n}\} \text{ is } \text{ consistent } \text{ in } \mid \!\sim \end{array}\right. } \end{aligned} \end{aligned}$$

Let $\varGamma ^{+}=\bigcup _{n\in \mathbb {N}}\varGamma _{n}$. By RTA, for any consistent $\varGamma$ and any $\varphi$, either $\varGamma \cup \{\varphi \}$ is consistent, or $\varGamma \cup \{\lnot \varphi \}$ is consistent. Hence, every $\varGamma _{n}$ is consistent. By weak compactness, it follows that $\varGamma ^{+}$ is consistent. It can be easily verified that $\varGamma ^{+}$ is also maximal. $\square$

Given any logic $\mid \!\sim$, let $MCS^{\mid \!\sim }$ be the set of all maximal consistent sets in $\mid \!\sim$. The following result is well known in classical logic. It can be shown that it also holds for all logics that satisfy INC and RTA.

Lemma 12

Let $\mid \!\sim$ be any logic satisfying INC and RTA. If $m\in MCS^{\mid \!\sim }$ then for any $\varphi \in \mathcal {L}$, $\lnot \varphi \in m$ iff $\varphi \notin m$.

Proof

For the direction from left to right, suppose $\lnot \varphi \in m$ but also $\varphi \in m$. Then by INC, $m\mid \!\sim$, contradicting that m is consistent. For the other direction, suppose $\varphi \notin m$ but also $\lnot \varphi \notin m$. Then by RTA, either $m\cup \{\varphi \}$ or $m\cup \{\lnot \varphi \}$ is consistent, contradicting that m is maximal. $\square$

Lemma 13

Let $\mid \!\sim$ be a logic in $\mathcal {NL}^{*}$. For any $\varGamma \subseteq \mathcal {L}$,

$$\begin{aligned} C_{\mid \!\sim }(\varGamma )=\bigcap \{m\in MCS^{\mid \!\sim }\mid C_{\mid \!\sim }(\varGamma )\subseteq m\}. \end{aligned}$$

Proof

The direction $C_{\mid \!\sim }(\varGamma )\subseteq \bigcap \{m\in MCS^{\mid \!\sim }\mid C_{\mid \!\sim }(\varGamma )\subseteq m\}$ is obvious. For the other direction, suppose $\varphi \notin C_{\mid \!\sim }(\varGamma )$. We prove that $\varphi \notin \bigcap \{m\in MCS^{\mid \!\sim }\mid C_{\mid \!\sim }(\varGamma )\subseteq m\}$. By IDE, $\varphi \notin C_{\mid \!\sim }C_{\mid \!\sim }(\varGamma )$, i.e. $C_{\mid \!\sim }(\varGamma )\mid \!\not \sim \varphi$. By PBC, $C_{\mid \!\sim }(\varGamma )\cup \{\lnot \varphi \}$ is consistent. By Lemma 11, there exists $m\in MCS^{\mid \!\sim }$ such that $C_{\mid \!\sim }(\varGamma )\cup \{\lnot \varphi \}\subseteq m$. It follows that $C_{\mid \!\sim }(\varGamma )\subseteq m$ and $\varphi \notin m$. Hence, $\varphi \notin \bigcap \{m\in MCS^{\mid \!\sim }\mid C_{\mid \!\sim }(\varGamma )\subseteq m\}$, as desired.$\square$

Definition 13

Given a logic $\mid \!\sim$ in $\mathcal {NL}^{*}$, the canonical choice model $\mathfrak {M}^{c}=(M^{c},\models ^{c},f^{c})$ for $\mid \!\sim$ is defined as follows:

• $M^{c}=MCS^{\mid \!\sim }$;

• $m\models ^{c}\varphi$ iff $\varphi \in m$;

• $f^{c}(X)={\left\{ \begin{array}{ll} \widehat{C_{\mid \!\sim }(\varGamma )} &{} \text{ if } \text{ there } \text{ exists } \varGamma \text{ such } \text{ that } \widehat{C_{\mid \!\sim }(\varGamma )}\subseteq X\subseteq \widehat{\varGamma }\\ X &{} \text{ otherwise } \end{array}\right. }$.

By NTR, we have $M^{c}\ne \emptyset$. The following lemma shows that $f^{c}$ is well-defined.

Lemma 14

Let $\mid \!\sim$ be a logic in $\mathcal {NL}^{*}$ and $\mathfrak {M}^{c}=(M^{c},\models ^{c},f^{c})$ be the canonical model for $\mid \!\sim$. For any $\varGamma ,\varSigma \subseteq \mathcal {L}$, if $\widehat{C_{\mid \!\sim }(\varGamma )}\subseteq X\subseteq \widehat{\varGamma }$ and $\widehat{C_{\mid \!\sim }(\varSigma )}\subseteq X\subseteq \widehat{\varSigma }$, then $C_{\mid \!\sim }(\varGamma )=C_{\mid \!\sim }(\varSigma )$.

Proof

We prove that $C_{\mid \!\sim }(\varGamma )=C_{\mid \!\sim }(\varSigma )=C{}_{\mid \!\sim }(\varGamma \cup \varSigma )$. Since $\widehat{C_{\mid \!\sim }(\varGamma )}\subseteq X\subseteq \widehat{\varGamma }\cap \widehat{\varSigma }=\widehat{\varGamma \cup \varSigma }$, we have $\varGamma \cup \varSigma \subseteq \overline{\widehat{\varGamma \cup \varSigma }}\subseteq \overline{X}\subseteq \overline{\widehat{C_{\mid \!\sim }(\varGamma )}}=\bigcap \{m\in MCS^{\mid \!\sim }\mid C_{\mid \!\sim }(\varGamma )\subseteq m\}=C_{\mid \!\sim }(\varGamma )$, by Lemmas 9 and  13. Hence, $\varSigma \subseteq C_{\mid \!\sim }(\varGamma )$. Then by CM and CC, it follows that $C_{\mid \!\sim }(\varGamma )=C{}_{\mid \!\sim }(\varGamma \cup \varSigma )$. Similarly, we have $C_{\mid \!\sim }(\varSigma )=C_{\mid \!\sim }(\varGamma \cup \varSigma )$.¹¹ $\square$

The following lemma says that the canonical choice model for any $\mid \!\sim$ in $\mathcal {NL}^{*}$ is non-trivial, cumulative, and classical for negation.

Lemma 15

Let $\mid \!\sim$ be a logic in $\mathcal {NL}^{*}$ and $\mathfrak {M}^{c}=(M^{c},\models ^{c},f^{c})$ the canonical choice model for $\mid \!\sim$. Then $\mathfrak {M}^{c}$ is a non-trivial and cumulative choice model that is classical for negation.

Proof

For Ntr, suppose $f^{c}(X)=\emptyset$. If $f^{c}(X)=X$, then we are done. If $f^{c}(X)=\widehat{C_{\mid \!\sim }(\varGamma )}=\emptyset$, then $C_{\mid \!\sim }(\varGamma )$ is inconsistent. Then by INC, CM, CC, and NonP, $\varGamma$ is also inconsistent. Hence, $X\subseteq \widehat{\varGamma }=\emptyset$.

Inc holds by the definition of $f^{c}$.

For Cum, suppose $f^{c}(X)\subseteq Y\subseteq X$. If $f^{c}(X)=X$, then $X\subseteq Y\subseteq X$ and hence $f^{c}(X)=f^{c}(Y)$. If $f^{c}(X)=\widehat{C_{\mid \!\sim }(\varGamma )}$, then $\widehat{C_{\mid \!\sim }(\varGamma )}\subseteq Y\subseteq X\subseteq \widehat{\varGamma }$. Hence, $f^{c}(Y)=\widehat{C_{\mid \!\sim }(\varGamma )}=f^{c}(X)$.

Finally, that $\mathfrak {M}^{c}$ is classical for negation follows from Lemma 12. $\square$

Now we are ready to prove Theorem 3. By Lemma 15, it suffices to prove that $\mid \!\sim \,=\,\mid \!\approx _{\mathfrak {M}^{c}}$, i.e. for any $\varGamma \subseteq \mathcal {L}$, $C_{\mid \!\sim }(\varGamma )=\overline{f^{c}(\widehat{\varGamma })}$. Note that for any $\varGamma \subseteq \mathcal {L}$, $f^{c}(\widehat{\varGamma })=\widehat{C_{\mid \!\sim }(\varGamma )}$. Hence, $\overline{f^{c}(\widehat{\varGamma })}=\overline{\widehat{C_{\mid \!\sim }(\varGamma )}}=\bigcap \{m\in MCS^{\mid \!\sim }\mid C_{\mid \!\sim }(\varGamma )\subseteq m\}=C_{\mid \!\sim }(\varGamma )$, by Lemma 13.