Rational evaluation in belief revision

Abstract

We introduce a new operator, called rational evaluation, in belief change. The operator evaluates new information according to the agent’s core beliefs, and then exports the plausible part of the new information. It belongs to the decision module in belief change. We characterize rational evaluation by axiomatic postulates and propose two functional constructions for it, based on the well-known constructions of kernel sets and remainder sets, respectively. The main results of the paper are two representation theorems with respect to the two functional constructions. We also compare and connect the evaluation operator with some related operators in belief revision, including contraction, set contraction, and prioritized multiple revision.

Appendix: Proofs

1.1 Proof of Proposition 1

Proof

• (1) Let $A\vdash \varphi$ and $\varphi \in D$. Suppose $\varphi \notin D\triangleleft A$. From E-Retainment, we know that there exists $C\subseteq D$ such that $C\cup A\nvdash \bot$ and $C\cup A\cup \{\varphi \}\vdash \bot$. Hence, $A\nvdash \varphi$. Contradiction.

• (2) By reductio ad absurdum, the verification is easy and left to the reader.

• (3) Let $\varphi \in D\backslash D\triangleleft A$ and $C=D\triangleleft A$. It is obvious that $C\cup \{\varphi \}\nsubseteq D\triangleleft A$. By E-Inclusion, we have $C=D\triangleleft A\subseteq D$. Hence, $D\triangleleft A\subseteq C\cup \{\varphi \}\subseteq D$. From $C\cup \{\varphi \}\nsubseteq D\triangleleft A$ and E-Maximality, we have $A\cup C\cup \{\varphi \}\vdash \bot$. Since E-Compatibility holds, $A\cup C=A\cup (D\triangleleft A)\nvdash \bot$. Therefore, E-Relevance holds.

• (4) Let $A\vdash \varphi \leftrightarrow \psi$ and $\{\varphi ,\psi \}\subseteq D$. Without loss of generality, we only prove that Left $\Longrightarrow$ Right. Let ‘Left’ hold, i.e., $\varphi \in D\triangleleft A$. From $\varphi \in D\triangleleft A$ and $A\vdash \varphi \leftrightarrow \psi$, we have $\psi \in Cn(A\cup (D\triangleleft A))$. Since $\psi \in D$ and E-Closure holds, $\psi \in D\triangleleft A$, i.e., ‘Right’ holds.

• (5) Let $A\cup D\nvdash \bot$. By reductio ad absurdum, suppose $D\triangleleft A\ne D$. Since E-Inclusion holds, $D\triangleleft A\subseteq D$. Thus, $D\nsubseteq D\triangleleft A$. Hence, there exists $\tau \in D\backslash D\triangleleft A$. From E-Retainment, we conclude that there exists $C\subseteq D$ such that $C\cup A\nvdash \bot$ and $C\cup A\cup \{\tau \}\vdash \bot$. Since $\tau \in D\backslash D\triangleleft A$ and $C\subseteq D$, $C\cup A\cup \{\tau \}\subseteq A\cup D$. Therefore, $A\cup D\vdash \bot$. Contradiction.

• (6) By reductio ad absurdum, suppose $Cn(A\cup (D\triangleleft A))\cap D\nsubseteq D\triangleleft A$. Then there exists $\tau \in D\backslash D\triangleleft A$ such that $A\cup (D\triangleleft A)\vdash \tau$. From $\tau \in D\backslash D\triangleleft A$ and E-Relevance, we conclude that there exists $C\subseteq D$ such that $D\triangleleft A\subseteq C$, $C\cup A\nvdash \bot$, and $C\cup A\cup \{\tau \}\vdash \bot$. Since $D\triangleleft A\subseteq C$, $C\cup A\nvdash \bot$, and $C\cup A\cup \{\tau \}\vdash \bot$, we have $A\cup (D\triangleleft A)\nvdash \tau$. Contradiction.$\square$

1.2 Proof of Lemma 1

Proof

Let $Z\subseteq D$ and $Z\cup A\vdash \bot$. By reductio ad absurdum, suppose that

$$\begin{aligned} (*)\ \forall Y\subseteq Z, Y\notin D\triangle _{\bot }A. \end{aligned}$$

From compactness, $Z\cup A\vdash \bot$ and $A\nvdash \bot$, we know that there exists a finite set $X\subseteq Z$ such that $X\cup A\vdash \bot$ and X is not empty. By ($*$), we have $X\notin D\triangle _{\bot }A$. From $X\subseteq D$, $X\cup A\vdash \bot$ and Definition 2, it follows that there is at least one subset $X_{0}$ of X such that $|X_{0}|=|X|-1$ and $X_{0}\cup A\vdash \bot$. Otherwise, we have $X\in D\triangle _{\bot }A$, which leads to a contradiction. From ($*$), we have $X_{0}\notin D\triangle _{\bot }A$. Since $X_{0}\subseteq D$ and $X_{0}\cup A\vdash \bot$, we have, similarly, the following conclusion: if $|X_{0}|\ge 1$, then there is at least one subset $X_{1}$ of $X_{0}$ such that $|X_{1}|=|X_{0}|-1$ and $X_{1}\cup A\vdash \bot$. So on and so forth. In finite steps it will arrive at the subset $\emptyset$ of X. Since $\emptyset$ is the only proper subset of all subsets of X whose cardinalities are 1, $\emptyset \cup A\vdash \bot$. Contradiction. Therefore, there exists $Y\subseteq Z$ such that $Y\in D\triangle _{\bot }A$. $\square$

1.3 Proof of Proposition 2

Proof

• (1) ($\Longleftarrow$): Let $D\cup A$ be consistent. Then all subsets of D are consistent with A. So all subsets of D do not satisfy Definition 2(1). It means that $D\triangle _{\bot }A=\emptyset$. ($\Longrightarrow$): Let $D\triangle _{\bot }A=\emptyset$. Suppose that $D\cup A$ is inconsistent. From $D\subseteq D$, $D\cup A\vdash \bot$ and Lemma 1, it follows that there exists $Y\subseteq D$ such that $Y\in D\triangle _{\bot }A$. Hence, $D\triangle _{\bot }A\ne \emptyset$. Contradiction.

• (2) Since $\emptyset \cup A\nvdash \bot$, $\emptyset$ does not satisfy Definition 2(1). Hence, $\emptyset \notin D\triangle _{\bot }A$.

• (3) It follows directly from Definitions 1 and 2.

• (4) Obvious.

• (5) Let $X\in D\triangle _{\bot }A$. By reductio ad absurdum, suppose that $X\cap Cn(A)\ne \emptyset$. Then there exists $\tau \in X\cap Cn(A)$. Thus, $X\backslash \{\tau \}\subset X$. By Definition 2(2), we have $X\backslash \{\tau \}\cup A\nvdash \bot$. Since $\tau \in X\cap Cn(A)$, $Cn(X\backslash \{\tau \}\cup A)=Cn(X\cup A)$. From $X\in D\triangle _{\bot }A$, it follows that $X\cup A\vdash \bot$. Hence, $X\backslash \{\tau \}\cup A\vdash \bot$. Contradiction. $\square$

1.4 Proof of Lemma 2

Proof

We need to prove that $\sigma (D\triangle _{\bot }A)$ satisfies all three conditions of Definition 3 for all $D\triangle _{\bot }A$.

• (1) It is obvious from the definition of $\sigma$.

• (2) Let $X\in D\triangle _{\bot }A$. By Proposition 2(2), we have $\emptyset \notin D\triangle _{\bot }A$. So $X\ne \emptyset$. From the definition of $\sigma$, we have $X\subseteq \bigcup (D\triangle _{\bot }A)=\sigma (D\triangle _{\bot }A)$. Hence, $\sigma (D\triangle _{\bot }A)\cap X=X\ne \emptyset$.

• (3) Let $\tau \in \sigma (D\triangle _{\bot }A)$. By reductio ad absurdum, suppose that $A\cup D\backslash \sigma (D\triangle _{\bot }A)\vdash \tau$. From $\tau \in \sigma (D\triangle _{\bot }A)$, there exists $X\in D\triangle _{\bot }A$ such that $\tau \in X$. Since $A\cup D\backslash \sigma (D\triangle _{\bot }A)\vdash \tau$, $Cn(X\cup A)\subseteq Cn(A\cup D\backslash \sigma (D\triangle _{\bot }A)\cup X\backslash \{\tau \})$. Since $X\in D\triangle _{\bot }A$, we have $X\cup A\vdash \bot$. So $A\cup D\backslash \sigma (D\triangle _{\bot }A)\cup X\backslash \{\tau \}\vdash \bot$. Hence, from $D\backslash \sigma (D\triangle _{\bot }A)\cup X\backslash \{\tau \}\subseteq D$ and Lemma 1, it follows that there exists $Y\subseteq D\backslash \sigma (D\triangle _{\bot }A)\cup X\backslash \{\tau \}$ such that $Y\in D\triangle _{\bot }A$. By the definition of $\sigma$, we have $Y\subseteq \sigma (D\triangle _{\bot }A)$. So $Y\cap (D\backslash \sigma (D\triangle _{\bot }A))=\emptyset$. Since $Y\subseteq D\backslash \sigma (D\triangle _{\bot }A)\cup X\backslash \{\tau \}$, we have $Y\subseteq X\backslash \{\tau \}$. So $\tau \notin Y$. Since $\tau \in X$, we have $Y\subset X$. From $X\in D\triangle _{\bot }A$ and Definition 2(2), it follows that $Y\cup A\nvdash \bot$. This result contradicts $Y\in D\triangle _{\bot }A$. Therefore, $A\cup D\backslash \sigma (D\triangle _{\bot }A)\nvdash \tau$.$\square$

1.5 Proof of Theorem 1

Proof

Construction $\Longrightarrow$ Postulates: Let $\triangleleft$ be a kernel evaluation. We need to prove that it satisfies the following postulates.

E-Inclusion::

From Definition 4, it is obvious that $D\triangleleft A=D\backslash \sigma (D\triangle _{\bot }A)\subseteq D$.

E-Compatibility::

By reductio ad absurdum, suppose that $A\cup (D\triangleleft A)\vdash \bot$, i.e., $A\cup (D\backslash \sigma (D\triangle _{\bot }A))\vdash \bot$. From $D\backslash \sigma (D\triangle _{\bot }A)\subseteq D$ and Lemma 1, there exists $Y\subseteq D\backslash \sigma (D\triangle _{\bot }A)$ such that $Y\in D\triangle _{\bot }A$. But by Definition 3, we have $\sigma (D\triangle _{\bot }A)\cap Y\ne \emptyset$. Hence, $Y\nsubseteq D\backslash \sigma (D\triangle _{\bot }A)$. Contradiction.

E-Closure::

Let $\tau \in Cn(A\cup (D\triangleleft A))\cap D$, i.e., $\tau \in Cn(A\cup (D\backslash \sigma (D\triangle _{\bot }A)))\cap D$. So $A\cup (D\backslash \sigma (D\triangle _{\bot }A))\vdash \tau$. From Definition 3(3), we have $\tau \notin \sigma (D\triangle _{\bot }A)$. Since $\tau \in D$, $\tau \in D\backslash \sigma (D\triangle _{\bot }A)=D\triangleleft A$. Hence, $Cn(A\cup (D\triangleleft A))\cap D\subseteq D\triangleleft A$.

E-BiUniformity::

Let $D_{1}$ and $D_{2}$ be two sets that have the same minimal subsets inconsistent with the two consistent sets $A_{1}$ and $A_{2}$ respectively, namely $D_{1}\bigtriangleup _{\bot }A_{1}$ = $D_{2}\bigtriangleup _{\bot }A_{2}$. By Definition 3, we have $\sigma (D_{1}\bigtriangleup _{\bot }A_{1})=\sigma (D_{2}\bigtriangleup _{\bot }A_{2})$. From Definitions 3 and 4, we have $D_{1}\backslash D_{1}\triangleleft A_{1}=D_{1}\backslash (D_{1}\backslash \sigma (D_{1}\bigtriangleup _{\bot }A_{1}))=\sigma (D_{1}\bigtriangleup _{\bot }A_{1})=\sigma (D_{2}\bigtriangleup _{\bot }A_{2})=D_{2}\backslash (D_{2}\backslash \sigma (D_{2}\bigtriangleup _{\bot }A_{2}))=D_{2}\backslash D_{2}\triangleleft A_{2}$.

E-Retainment::

Let $\tau \in D\backslash D\triangleleft A$, i.e., $\tau \in D\backslash (D\backslash \sigma (D\bigtriangleup _{\bot }A))$. Then $\tau \in \sigma (D\triangle _{\bot }A)$. Hence, there exists $X\in D\triangle _{\bot }A$ such that $\tau \in X$. Let $C=X\backslash \{\tau \}$. From $C\subset X\in D\triangle _{\bot }A$ and Definition 2 we have $C\cup A\nvdash \bot$. But $C\cup A\cup \{\tau \}=X\cup A\vdash \bot$. Thus, there exists $C\subseteq D$ such that $C\cup A\nvdash \bot$ and $C\cup A\cup \{\tau \}\vdash \bot$.

Postulates $\Longrightarrow$ Construction: Let $\triangleleft$ satisfy the above postulates. We need to prove that it is a kernel evaluation.

Define $\sigma$ such that $\sigma (D\triangle _{\bot }A)=D\backslash D\triangleleft A$ for all subsets D and consistent subsets A of $\mathcal {L}$. We need to prove the following propositions, namely (a), (b) and (c).

• (a) $\sigma$ is well defined. That is to prove that: if $D_{1}\bigtriangleup _{\bot }A_{1}=D_{2}\bigtriangleup _{\bot }A_{2}$, then $\sigma (D_{1}\bigtriangleup _{\bot }A_{1})=\sigma (D_{2}\bigtriangleup _{\bot }A_{2})$. Let $D_{1}\bigtriangleup _{\bot }A_{1}=D_{2}\bigtriangleup _{\bot }A_{2}$. From E-BiUniformity, we have $D_{1}\backslash D_{1}\triangleleft A_{1}=D_{2}\backslash D_{2}\triangleleft A_{2}$, i.e., $\sigma (D_{1}\bigtriangleup _{\bot }A_{1})=\sigma (D_{2}\bigtriangleup _{\bot }A_{2})$.

• (b) $\sigma$ is an evaluative incision function. Condition (1): Let $\tau \in \sigma (D\triangle _{\bot }A)$, i.e., $\tau \in D\backslash D\triangleleft A$. From E-Retainment, we know that there exists $C\subseteq D$ such that $C\cup A\nvdash \bot$ and $C\cup A\cup \{\tau \}\vdash \bot$. From $C\cup \{\tau \}\subseteq D$, $C\cup A\cup \{\tau \}\vdash \bot$, and Lemma 1, it follows that there exists $Y\subseteq C\cup \{\tau \}$ such that $Y\in D\triangle _{\bot }A$. Hence, $Y\cup A\vdash \bot$. Suppose $\tau \notin Y$. Then by $Y\subseteq C$, we have $C\cup A\vdash \bot$. Contradiction. Hence, $\tau \in Y$. Therefore, from $\tau \in Y\in D\triangle _{\bot }A$, we have $\tau \in \bigcup (D\triangle _{\bot }A)$. Condition (2): Let $X\in D\triangle _{\bot }A$. Then $X\cup A\vdash \bot$. Suppose $\sigma (D\triangle _{\bot }A)\cap X=\emptyset$, i.e., $(D\backslash D\triangleleft A)\cap X=\emptyset$. Since $X\subseteq D$. Then $X\subseteq D\triangleleft A$. And from $X\cup A\vdash \bot$, we have $D\triangleleft A\cup A\vdash \bot$. This result contradicts E-Compatibility. Condition (3): Let $\tau \in \sigma (D\triangle _{\bot }A)$, i.e., $\tau \in D\backslash D\triangleleft A$. By $\tau \notin D\triangleleft A$ and E-Closure, we have $\tau \notin Cn(A\cup (D\triangleleft A))\cap D$. Since $\tau \in D$, we have $A\cup (D\triangleleft A)\nvdash \tau$. From E-Inclusion and the above definition of $\sigma$, we have $D\triangleleft A=D\backslash (D\backslash D\triangleleft A)=D\backslash \sigma (D\triangle _{\bot }A)$. Hence, $A\cup D\backslash \sigma (D\triangle _{\bot }A)\nvdash \tau$.

• (c) $D\triangleleft A$ = $D\backslash \sigma (D\triangle _{\bot }A)$, for all subsets D and consistent subsets A of $\mathcal {L}$. From the above definition of $\sigma$, we have $D\backslash \sigma (D\triangle _{\bot }A)$ = $D\backslash (D\backslash D\triangleleft A)$. And by E-Inclusion, i.e., $D\triangleleft A\subseteq D$, we have $D\backslash \sigma (D\triangle _{\bot }A)=D\triangleleft A$. $\square$

1.6 Proof of Proposition 3

Proof

Let $\forall \tau \in \sigma (D\bigtriangleup _{\bot }A)(A\cup D\backslash \sigma (D\triangle _{\bot }A)\cup \{\tau \}\vdash \bot )$. By reductio ad absurdum, suppose that $\triangleleft _{\sigma }$ does not satisfy E-Maximality, i.e., there exists C such that $D\triangleleft _{\sigma }A\subseteq C\subseteq D$, $A\cup C\nvdash \bot$ and $C\nsubseteq D\triangleleft _{\sigma }A$. By $C\nsubseteq D\triangleleft _{\sigma }A$, there exists $\varphi \in C\backslash D\triangleleft _{\sigma }A$. Since $\varphi \in C\subseteq D$ and $\varphi \notin D\triangleleft _{\sigma }A=D\backslash \sigma (D\triangle _{\bot }A)$, we have $\varphi \in \sigma (D\triangle _{\bot }A)$. Since $\forall \tau \in \sigma (D\bigtriangleup _{\bot }A)(A\cup D\backslash \sigma (D\triangle _{\bot }A)\cup \{\tau \}\vdash \bot )$, we have $A\cup D\backslash \sigma (D\triangle _{\bot }A)\cup \{\varphi \}\vdash \bot$. Since $D\backslash \sigma (D\triangle _{\bot }A)=D\triangleleft _{\sigma }A\subseteq C$ and $\varphi \in C$, we have $D\backslash \sigma (D\triangle _{\bot }A)\cup \{\varphi \}\subseteq C$. Hence, $A\cup C\vdash \bot$. Contradiction. $\square$

1.7 Proof of Lemma 3

Proof

Let $Y\subseteq D$ and $Y\cup A\nvdash \bot$. Then $Y\backslash A\cup A\subseteq D\backslash A\cup A$ and $Cn(Y\backslash A\cup A)\cap \{\bot \}=\emptyset$. From the upper bound property of remainder sets (Alchourrón and Makinson 1981), there exists X such that $Y\backslash A\cup A\subseteq X\in (D\backslash A\cup A)\bigtriangledown \{\bot \}$. Since $X\subseteq D\backslash A\cup A$, we have $X\backslash A\subseteq D\backslash A$. Since $X\in (D\backslash A\cup A)\bigtriangledown \{\bot \}$, we have $X\nvdash \bot$. Since $A\subseteq X$, we have $X=X\backslash A\cup A$. Hence, $X\backslash A\cup A\nvdash \bot$.

Now we prove that $Y\subseteq (X\backslash A)\cup (D\cap A)\in D\bigtriangledown _{\top }A$.

By reductio ad absurdum, suppose that $(X\backslash A)\cup (D\cap A)\notin D\bigtriangledown _{\top }A$. From $X\backslash A\subseteq D\backslash A$ and $X\backslash A\cup A\nvdash \bot$, we have $(X\backslash A)\cup (D\cap A)\subseteq D$ and $(X\backslash A)\cup (D\cap A)\cup A=X\backslash A\cup A\nvdash \bot$. And from $(X\backslash A)\cup (D\cap A)\notin D\bigtriangledown _{\top }A$ and Definition 6, there exists Z such that $(X\backslash A)\cup (D\cap A)\subset Z\subseteq D$ and $Z\cup A\nvdash \bot$. It follows that there exists $\psi \in Z$ such that $\psi \notin (X\backslash A)\cup (D\cap A)$. Hence, $\psi \notin X\backslash A$ and $\psi \notin D\cap A$. Since $\psi \in Z\subseteq D$, we have $\psi \notin A$. Hence, $\psi \notin X\backslash A\cup A$. Since $(X\backslash A)\cup (D\cap A)\subset Z$, we have $X\backslash A\cup A\subseteq Z\cup A$. Since $\psi \in Z\cup A$ and $\psi \notin X\backslash A\cup A$, we have $X\backslash A\cup A\subset Z\cup A$. Hence, $X\subset Z\cup A\subseteq D\backslash A\cup A$. Since $X\in (D\backslash A\cup A)\bigtriangledown \{\bot \}$, we have $Z\cup A\vdash \bot$. Contradiction. Therefore, $(X\backslash A)\cup (D\cap A)\in D\bigtriangledown _{\top }A$.

Let $\varphi \in Y$. If $\varphi \in A$, then from $Y\subseteq D$, we have $\varphi \in D\cap A\subseteq (X\backslash A)\cup (D\cap A)$. If $\varphi \notin A$, then by $Y\subseteq Y\backslash A\cup A\subseteq X$, we have $\varphi \in X\backslash A\subseteq (X\backslash A)\cup (D\cap A)$. Hence, $\varphi \in (X\backslash A)\cup (D\cap A)$ in both cases. Therefore, $Y\subseteq (X\backslash A)\cup (D\cap A)$. $\square$

1.8 Proof of Proposition 4

Proof

• (1) From $\emptyset \subseteq D$, $\emptyset \cup A\nvdash \bot$ and Lemma 3, there exists X such that $\emptyset \subseteq X\in D\bigtriangledown _{\top }A$. Hence, $D\bigtriangledown _{\top }A\ne \emptyset$.

• (2) The equivalence between the former two is obvious. Here we only prove that $\emptyset \in D\bigtriangledown _{\top }A$ iff $D\ne \emptyset \Longrightarrow \forall \varphi \in D(A\vdash \lnot \varphi )$. Left $\Longrightarrow$ Right: Let $\emptyset \in D\bigtriangledown _{\top }A$ and $D\ne \emptyset$. Choose $\varphi \in D$ arbitrarily. Since $\emptyset \subset \{\varphi \}\subseteq D$ and $\emptyset \in D\bigtriangledown _{\top }A$, we have $\{\varphi \}\cup A\vdash \bot$. Hence, $A\vdash \lnot \varphi$. Right $\Longrightarrow$ Left: Let ‘Right’ hold. Suppose that $\emptyset \notin D\bigtriangledown _{\top }A$. From $\emptyset \subseteq D$, $\emptyset \cup A\nvdash \bot$, and Definition 6, it follows that there exists X such that $\emptyset \subset X\subseteq D$ and $X\cup A\nvdash \bot$. Since $\emptyset \subset X\subseteq D$, we have $D\ne \emptyset$. By ‘Right’, we have $\forall \varphi \in D(A\vdash \lnot \varphi )$. Choose $\tau \in X$ arbitrarily. Then $A\vdash \lnot \tau$. Hence, $\{\tau \}\cup A\vdash \bot$. Thus, $X\cup A\vdash \bot$. Contradiction.

• (3) Let $A=\emptyset$. $D\bigtriangledown _{\top }A=D\bigtriangledown \{\bot \}$ follows directly from Definitions 5 and 6.

• (4) The equivalences of these three conditions are obvious.

• (5) Let $X\in D\bigtriangledown _{\top }A$. By reductio ad absurdum, suppose that $Cn(A)\cap D\nsubseteq X$. Then there exists $\tau \in Cn(A)\cap D$ such that $\tau \notin X$. So $X\subset X\cup \{\tau \}\subseteq D$. By $X\in D\bigtriangledown _{\top }A$ and Definition 6, we have $X\cup \{\tau \}\cup A\vdash \bot$. Since $\tau \in Cn(A)$, we have $Cn(X\cup \{\tau \}\cup A)=Cn(X\cup A)$. Hence, $X\cup A\vdash \bot$. But by $X\in D\bigtriangledown _{\top }A$, we have $X\cup A\nvdash \bot$. Contradiction. $\square$

1.9 Proof of Lemma 4

Proof

Left $\Longrightarrow$ Right: Let $D_{1}\bigtriangledown _{\top }A_{1}=D_{2}\bigtriangledown _{\top }A_{2}$. We need to prove that ‘Right’ hold.

• (1) By reductio ad absurdum, suppose that $\exists \psi \in D_{1}\backslash (D_{1}\cap D_{2})(\{\psi \}\cup A_{1}\nvdash \bot )$. From $\{\psi \}\subseteq D_{1}$ and Lemma 3, there exists X such that $\{\psi \}\subseteq X\in D_{1}\bigtriangledown _{\top }A_{1}$. Hence, $\{\psi \}\subseteq X\in D_{2}\bigtriangledown _{\top }A_{2}$. It follows that $\psi \in D_{2}$. So $\psi \notin D_{1}\backslash (D_{1}\cap D_{2})$. Contradiction.

• (2) Similar to (1) and omitted here.

• (3) By reductio ad absurdum, suppose that $\exists X\subseteq D_{1}\cap D_{2}(X\cup A_{1}\nvdash \bot \nLeftrightarrow X\cup A_{2}\nvdash \bot )$. Without loss of generality, suppose that $X\cup A_{1}\nvdash \bot$ and $X\cup A_{2}\vdash \bot$. From Lemma 3, there exists Y such that $X\subseteq Y\in D_{1}\bigtriangledown _{\top }A_{1}$. Hence, $X\subseteq Y\in D_{2}\bigtriangledown _{\top }A_{2}$. It follows that $Y\cup A_{2}\nvdash \bot$. Hence, $X\cup A_{2}\nvdash \bot$. Contradiction.

Left $\Longleftarrow$ Right: Let ‘Right’ hold. We need to prove that ‘Left’ hold. Without loss of generality, we only prove that $D_{1}\bigtriangledown _{\top }A_{1}\subseteq D_{2}\bigtriangledown _{\top }A_{2}$. Let $Z\in D_{1}\bigtriangledown _{\top }A_{1}$. Then $Z\cup A_{1}\nvdash \bot$ and $Z\subseteq D_{1}$. Then by condition (1), we have $Z\cap (D_{1}\backslash (D_{1}\cap D_{2}))=\emptyset$; otherwise $Z\cup A_{1}\vdash \bot$, which leads to a contradiction. From $Z\subseteq D_{1}$ and $Z\cap (D_{1}\backslash (D_{1}\cap D_{2}))=\emptyset$, it follows that $Z\subseteq D_{1}\cap D_{2}$. Then by condition (3) and $Z\cup A_{1}\nvdash \bot$, we have $Z\cup A_{2}\nvdash \bot$. Suppose that $Z\notin D_{2}\bigtriangledown _{\top }A_{2}$. Then from $Z\subseteq D_{2}$, $Z\cup A_{2}\nvdash \bot$, and Definition 6, it follows that there exists Y such that $Z\subset Y\subseteq D_{2}$ and $Y\cup A_{2}\nvdash \bot$. Then by condition (2) we have $Y\cap (D_{2}\backslash (D_{1}\cap D_{2}))=\emptyset$; otherwise $Y\cup A_{2}\vdash \bot$, which leads to a contradiction. From $Y\subseteq D_{2}$ and $Y\cap (D_{2}\backslash (D_{1}\cap D_{2}))=\emptyset$, we have $Y\subseteq D_{1}\cap D_{2}$. Then from $Y\subseteq D_{1}\cap D_{2}$, $Y\cup A_{2}\nvdash \bot$, and condition (3), we have $Y\cup A_{1}\nvdash \bot$. By $Z\subset Y\subseteq D_{1}$ and $Y\cup A_{1}\nvdash \bot$, we have $Z\notin D_{1}\bigtriangledown _{\top }A_{1}$. Contradiction. $\square$

1.10 Proof of Theorem 2

Proof

Construction $\Longrightarrow$ Postulates: Let $\triangleleft$ be a partial meet evaluation. We need to prove that it satisfies the following postulates.

E-Inclusion::

From Definitions 6, 7, and  8, it is obvious that $D\triangleleft A=\bigcap \gamma (D\bigtriangledown _{\top }A)\subseteq D$.

E-Compatibility::

By Definition 7(2), we have $\gamma (D\bigtriangledown _{\top }A)$ $\ne \emptyset$. Since for all $X\in \gamma (D\bigtriangledown _{\top }A)$, we have $X\cup A\nvdash \bot$ and $\bigcap \gamma (D\bigtriangledown _{\top }A)\subseteq X$. Then $A\cup \bigcap \gamma (D\bigtriangledown _{\top }A)\nvdash \bot$, i.e., $A\cup (D\triangleleft A)\nvdash \bot$.

E-BiIrrelevance::

Let $A_{1},A_{2},D_{1},D_{2}\subseteq \mathcal {L}$ and $A_{1},A_{2}$ be both consistent. And let

$$\begin{aligned}&\forall \varphi \in D_{1}\backslash (D_{1}\cap D_{2})(\{\varphi \}\cup A_{1}\vdash \bot ), \\&\forall \varphi \in D_{2}\backslash (D_{1}\cap D_{2})(\{\varphi \}\cup A_{2}\vdash \bot ),\ \mathrm{and} \\&\forall X\subseteq D_{1}\cap D_{2}(X\cup A_{1}\nvdash \bot \Longleftrightarrow X\cup A_{2}\nvdash \bot ). \end{aligned}$$

By Lemma 4, we have $D_{1}\bigtriangledown _{\top }A_{1}=D_{2}\bigtriangledown _{\top }A_{2}$. Then by Definition 8, we have $D_{1}\triangleleft A_{1}=\bigcap \gamma (D_{1}\bigtriangledown _{\top }A_{1})=\bigcap \gamma (D_{2}\bigtriangledown _{\top }A_{2})=D_{2}\triangleleft A_{2}$, i.e., $D_{1}\triangleleft A_{1}=D_{2}\triangleleft A_{2}$.

E-Relevance::

Let $\varphi \in D\backslash D\triangleleft A$, i.e., $\varphi \in D\backslash \bigcap \gamma (D\bigtriangledown _{\top }A)$. From $\varphi \notin \bigcap \gamma (D\bigtriangledown _{\top }A)$ and $\gamma (D\bigtriangledown _{\top }A)\ne \emptyset$, it follows that there exists $X\in \gamma (D\bigtriangledown _{\top }A)$ such that $\varphi \notin X$. By $X\in \gamma (D\bigtriangledown _{\top }A)$, we have $X\subseteq D$, $X\cup A\nvdash \bot$, and $D\triangleleft A=\bigcap \gamma (D\bigtriangledown _{\top }A)\subseteq X$. Since $X\subseteq D$, $\varphi \in D$ and $\varphi \notin X$, we have $X\subset X\cup \{\varphi \}\subseteq D$. Since $X\in \gamma (D\bigtriangledown _{\top }A)$, we have $X\cup \{\varphi \}\cup A\vdash \bot$. Hence, there exists $X\subseteq D$ such that $D\triangleleft A\subseteq X$, $X\cup A\nvdash \bot$, and $X\cup A\cup \{\varphi \}\vdash \bot$.

Postulates $\Longrightarrow$ Construction: Let $\triangleleft$ satisfy the above postulates. We need to prove that it is a partial meet evaluation.

Define a function $\gamma$ such that for all subsets D and consistent subsets A of $\mathcal {L}$, $\gamma (D\bigtriangledown _{\top }A)=\{X\in D\bigtriangledown _{\top }A|D\triangleleft A\subseteq X\}$. And define the operator generated by $\gamma$ such that $D\triangleleft |A=\bigcap \gamma (D\bigtriangledown _{\top }A)$, for all subsets D and consistent subsets A of $\mathcal {L}$. We need to prove the following propositions, namely (a), (b), and (c).

• (a) $\gamma$ is well defined. That is to prove that: for all $D_{1},D_{2}\subseteq \mathcal {L}$ and consistent subsets $A_{1},A_{2}\subseteq \mathcal {L}$, if $D_{1}\bigtriangledown _{\top }A_{1}=D_{2}\bigtriangledown _{\top }A_{2}$, then $\gamma (D_{1}\bigtriangledown _{\top }A_{1})=\gamma (D_{2}\bigtriangledown _{\top }A_{2})$. Let $D_{1}\bigtriangledown _{\top }A_{1}=D_{2}\bigtriangledown _{\top }A_{2}$. By Lemma 4, we have

  $$\begin{aligned}&\forall \varphi \in D_{1}\backslash (D_{1}\cap D_{2})(\{\varphi \}\cup A_{1}\vdash \bot ), \\&\forall \varphi \in D_{2}\backslash (D_{1}\cap D_{2})(\{\varphi \}\cup A_{2}\vdash \bot ),\ \mathrm{and} \\&\forall X\subseteq D_{1}\cap D_{2}(X\cup A_{1}\nvdash \bot \Longleftrightarrow X\cup A_{2}\nvdash \bot ). \end{aligned}$$

  And from E-BiIrrelevance, we have $D_{1}\triangleleft A_{1}=D_{2}\triangleleft A_{2}$. With $D_{1}\bigtriangledown _{\top }A_{1}=D_{2}\bigtriangledown _{\top }A_{2}$ and the above definition of $\gamma$, we have $\gamma (D_{1}\bigtriangledown _{\top }A_{1})=\gamma (D_{2}\bigtriangledown _{\top }A_{2})$.

• (b) $\gamma$ is an evaluative selection function. Condition (1): From the above definition of $\gamma$, it is obvious that $\gamma (D\bigtriangledown _{\top }A)=\{X\in D\bigtriangledown _{\top }A|D\triangleleft A\subseteq X\}\subseteq D\bigtriangledown _{\top }A$. Condition (2): By E-Inclusion and E-Compatibility, we have $D\triangleleft A\subseteq D$ and $D\triangleleft A\cup A\nvdash \bot$. From $D\triangleleft A\subseteq D$, $D\triangleleft A\cup A\nvdash \bot$, and Lemma 3, it follows that there exists Y such that $D\triangleleft A\subseteq Y\in D\bigtriangledown _{\top }A$. By the above definition of $\gamma$, we have $Y\in \gamma (D\bigtriangledown _{\top }A)$. Hence, $\gamma (D\bigtriangledown _{\top }A)\ne \emptyset$.

• (c) For all subsets D and consistent subsets A of $\mathcal {L}$, $D\triangleleft A=D\triangleleft |A$.

$\subseteq$ direction::

From Condition (2) in (b) we know that $\gamma (D\bigtriangledown _{\top }A)\ne \emptyset$. By the above definition of $\gamma$, we have $X\supseteq D\triangleleft A$ for every $X\in \gamma (D\bigtriangledown _{\top }A)$. Hence, $D\triangleleft A\subseteq \bigcap \gamma (D\bigtriangledown _{\top }A)=D\triangleleft |A$.

$\supseteq$ direction::

By reductio ad absurdum, suppose that $D\triangleleft A\nsupseteq D\triangleleft |A$. Then there exists $\varphi \in D\triangleleft |A=\bigcap \gamma (D\bigtriangledown _{\top }A)$ such that $\varphi \notin D\triangleleft A$. Since $\varphi \in \bigcap \gamma (D\bigtriangledown _{\top }A)$, we have $\forall X\in \gamma (D\bigtriangledown _{\top }A)(\varphi \in X)$. Hence, $\varphi \in D$. From $\varphi \in D$, $\varphi \notin D\triangleleft A$, and E-Relevance, it follows that there exists $C\subseteq D$ such that $D\triangleleft A\subseteq C$, $C\cup A\nvdash \bot$ and $C\cup A\cup \{\varphi \}\vdash \bot$. And from $C\subseteq D$, $C\cup A\nvdash \bot$, and Lemma 3, it follows that there exists $Y\supseteq C$ such that $Y\in D\bigtriangledown _{\top }A$. From $D\triangleleft A\subseteq C\subseteq Y\in D\bigtriangledown _{\top }A$ and the above definition of $\gamma$, we have $Y\in \gamma (D\bigtriangledown _{\top }A)$. Since $C\subseteq Y$, $Y\cup A\nvdash \bot$, and $C\cup A\cup \{\varphi \}\vdash \bot$, we have $\varphi \notin Y$. Hence, $\varphi \notin Y\in \gamma (D\bigtriangledown _{\top }A)$. It contradicts $\forall X\in \gamma (D\bigtriangledown _{\top }A)(\varphi \in X)$.$\square$

1.11 Proof of Proposition 5

Proof

Let $\triangleleft$ be a maximal evaluation. By reductio ad absurdum, suppose that there exists C such that $D\triangleleft A\subseteq C\subseteq D$, $A\cup C\nvdash \bot$ and $C\nsubseteq D\triangleleft A$. From Definition 9, it follows that there exists a maximal evaluative selection function $\gamma _{m}$ such that: for all subsets D and consistent subsets A of $\mathcal {L}$, $D\triangleleft A=D\triangleleft _{\gamma _{m}}A=\bigcap \gamma _{m}(D\bigtriangledown _{\top }A)$. Then there exists $X\in D\bigtriangledown _{\top }A$ such that $D\triangleleft A=X$. By $C\nsubseteq D\triangleleft A$, it follows that there exists $\tau \in C\backslash D\triangleleft A$. Since $D\triangleleft A\subseteq C\subseteq D$, we have $D\triangleleft A\subset C\subseteq D$, i.e., $X\subset C\subseteq D$. From $X\in D\bigtriangledown _{\top }A$, $X\subset C\subseteq D$, and Definition 6, we have $C\cup A\vdash \bot$. Contradiction. $\square$

1.12 Proof of Proposition 6

Proof

Let $\otimes$ be a prioritized multiple partial meet revision for D. We prove that $\triangleleft _{\otimes }$ satisfies the following postulates.

E-Inclusion::

By Definition 10, it is obvious that $D\triangleleft _{\otimes }A=D\otimes A\cap D\subseteq D$.

E-Compatibility::

That is to prove that $A\cup (D\otimes A\cap D)\nvdash \bot$. From $A\nvdash \bot$ and Weak Success of $\otimes$, we have $A\subseteq D\otimes A$. Hence, $A\cup (D\otimes A\cap D)\subseteq D\otimes A$. From $A\nvdash \bot$ and Consistency of $\otimes$, we have $D\otimes A\nvdash \bot$. Hence, $A\cup (D\otimes A\cap D)\nvdash \bot$.

E-Closure::

That is to prove that $Cn(A\cup (D\triangleleft _{\otimes }A))\cap D\subseteq D\triangleleft _{\otimes }A$. By reductio ad absurdum, suppose that $Cn(A\cup (D\triangleleft _{\otimes }A))\cap D\nsubseteq D\triangleleft _{\otimes }A$. Then there exists $\tau \in D\backslash D\triangleleft _{\otimes }A$ such that $A\cup (D\triangleleft _{\otimes }A)\vdash \tau$. Since $\tau \in D\backslash D\triangleleft _{\otimes }A=D\backslash (D\otimes A\cap D)$, we have $\tau \in D\backslash D\otimes A$. From Relevance of $\otimes$, it follows that there exists $C\subseteq D\cup A$ such that $D\otimes A\subseteq C$ and $C\cup A\nvdash \bot$ but $C\cup A\cup \{\tau \}\vdash \bot$. Since $A\subseteq D\otimes A$ and $A\cup (D\triangleleft _{\otimes }A)\vdash \tau$, we have $D\otimes A\vdash \tau$. Since $D\otimes A\subseteq C$, we have $C\vdash \tau$. Hence, $Cn(C\cup A)=Cn(C\cup A\cup \{\tau \})$. Then $C\cup A\cup \{\tau \}\nvdash \bot$. Contradiction.

E-Retainment::

Let $\tau \in D\backslash D\triangleleft _{\otimes }A$. Then $\tau \in D\backslash D\otimes A$. From Relevance of $\otimes$, it follows that there exists $C\subseteq D\cup A$ such that $D\otimes A\subseteq C$ and $C\cup A\nvdash \bot$ but $C\cup A\cup \{\tau \}\vdash \bot$. Let $C'=C\backslash A$. Since $C\subseteq D\cup A$, we have $C'\subseteq D$. From $C\cup A\nvdash \bot$ and $C\cup A\cup \{\tau \}\vdash \bot$, we have $C'\cup A\nvdash \bot$ and $C'\cup A\cup \{\tau \}\vdash \bot$. Hence, there exists $C'\subseteq D$ such that $C'\cup A\nvdash \bot$ and $C'\cup A\cup \{\tau \}\vdash \bot$.

E-Relevance::

Let $\tau \in D\backslash D\triangleleft _{\otimes }A$. Then $\tau \in D\backslash D\otimes A$. From Relevance of $\otimes$, it follows that there exists $C\subseteq D\cup A$ such that $D\otimes A\subseteq C$ and $C\cup A\nvdash \bot$ but $C\cup A\cup \{\tau \}\vdash \bot$. Let $C'=C\cap D$. Since $D\otimes A\subseteq C$, we have $D\triangleleft _{\otimes }A=D\otimes A\cap D\subseteq C\cap D=C'$, i.e., $D\triangleleft _{\otimes }A\subseteq C'$. Since $C\cup A\nvdash \bot$, we have $C'\cup A=(C\cap D)\cup A\nvdash \bot$. From $C\subseteq D\cup A$, we have $C\backslash A\subseteq D$. Hence, $C\backslash A\subseteq C\cap D=C'$. Then $C\cup A\cup \{\tau \}=C\backslash A\cup A\cup \{\tau \}\subseteq C'\cup A\cup \{\tau \}$. And from $C\cup A\cup \{\tau \}\vdash \bot$, it follows that $C'\cup A\cup \{\tau \}\vdash \bot$. Hence, there exists $C'\subseteq D$ such that $D\triangleleft _{\otimes }A\subseteq C'$, $C'\cup A\nvdash \bot$ and $C'\cup A\cup \{\tau \}\vdash \bot$.$\square$