Rational metabolic revision based on core beliefs

Abstract

When an agent can not recognize, immediately, the implausible part of new information received, she will usually first expand her belief state by the new information, and then she may encounter some belief conflicts, and find the implausible information based on her criteria to consolidate her belief state. This process indicates a new kind of non-prioritized multiple revision, called metabolic revision. I give some axiomatic postulates for metabolic revision and propose two functional constructions for it, namely kernel metabolic revision and partial meet metabolic revision, with respect to which the representation theorems are proved. I also compare metabolic revision with some related works in the literature, including semi-revision, merging with integrity constraints, evaluation, and evaluative multiple revision.

Appendix: Proofs

Proof of Proposition 1

• (1) Let $\diamond$ satisfy M-ExRetainment and $\tau \in A$. From $(B,A)\in \mathcal {B}$, we have $A\subseteq B\subseteq B\cup D$. So $\tau \in B\cup D$. By reductio ad absurdum, suppose that $\tau \notin B\diamond _{A}D$. So $\tau \in (B\cup D)\backslash B\diamond _{A}D$. By M-ExRetainment, we know that there exists $C\subseteq B\cup D$ such that $C\cup A\nvdash \bot$ but $C\cup \{\tau \}\cup A\vdash \bot$. Since $C\cup A\nvdash \bot$ and $C\cup \{\tau \}\cup A\vdash \bot$, we have $A\nvdash \tau$. Contradiction. Hence, $A\subseteq B\diamond _{A}D$.

• (2) It is obvious.

• (3) Let $\diamond$ satisfy M-ExRetainment and $B\cup D\nvdash \bot$. By reductio ad absurdum, suppose that $B\cup D\nsubseteq B\diamond _{A}D$. So there exists $\psi \in (B\cup D)\backslash B\diamond _{A}D$. From M-ExRetainment, we know that there exists $C\subseteq B\cup D$ such that $C\cup A\nvdash \bot$ and $C\cup \{\psi \}\cup A\vdash \bot$. Since $C\cup \{\psi \}\cup A\subseteq B\cup D$, we have $B\cup D\vdash \bot$. Contradiction. Hence, M-Vacuity holds.

• (4) Let $\diamond$ satisfy M-ExRelevance. By reductio ad absurdum, suppose that $Cn(B\diamond _{A}D)\cap (B\cup D)\nsubseteq B\diamond _{A}D$. So there exists $\psi \in (B\cup D)\backslash B\diamond _{A}D$ such that $B\diamond _{A}D\vdash \psi$. From $\psi \in (B\cup D)\backslash B\diamond _{A}D$ and M-ExRelevance, we know that there exists $C\subseteq B\cup D$ such that $B\diamond _{A}D\subseteq C$ and $C\cup A\nvdash \bot$, but $C\cup \{\psi \}\cup A\vdash \bot$. Since $B\diamond _{A}D\subseteq C$ and $B\diamond _{A}D\vdash \psi$, we have $C\vdash \psi$. So $Cn(C\cup A)=Cn(C\cup \{\psi \}\cup A)$. So $C\cup A\vdash \bot$. Contradiction. Hence, M-RelClosure holds.

• (5) The proof is easy and left to the reader.

• (6) Let $\diamond$ satisfy M-CoreInvariance, M-Inclusion, and M-ExRetainment. By Definition 1, we need to prove that $A\diamond _{B}D\nvdash \bot$, $A\diamond _{B}D\subseteq B\diamond _{A}D$, and $Cn(A\diamond _{B}D)\cap B\diamond _{A}D\subseteq A\diamond _{B}D$. From $(B,A)\in \mathcal {B}$, we have $A\nvdash \bot$. Since $\diamond$ satisfies M-CoreInvariance, $A\diamond _{B}D=Cn(A)\cap (B\cup D)\nvdash \bot$. Suppose that $A\diamond _{B}D\nsubseteq B\diamond _{A}D$. So there exists $\psi \in A\diamond _{B}D$ such that $\psi \notin B\diamond _{A}D$. So $\psi \in Cn(A)\cap (B\cup D)$. So $\psi \in (B\cup D)\backslash B\diamond _{A}D$. From M-ExRetainment, we know that there exists $C\subseteq B\cup D$ such that $C\cup A\nvdash \bot$ and $C\cup \{\psi \}\cup A\vdash \bot$. Since $C\cup \{\psi \}\cup A\vdash \bot$ and $\psi \in Cn(A)$, we have $C\cup A\vdash \bot$. Contradiction. Hence, $A\diamond _{B}D\subseteq B\diamond _{A}D$. Now suppose that $Cn(A\diamond _{B}D)\cap B\diamond _{A}D\nsubseteq A\diamond _{B}D$. So there exists $\varphi \in Cn(A\diamond _{B}D)\cap B\diamond _{A}D$ such that $\varphi \notin A\diamond _{B}D$. Since $\varphi \in Cn(A\diamond _{B}D)$, $Cn(A)\cap (B\cup D)\vdash \varphi$. So $A\vdash \varphi$. Since $\varphi \notin Cn(A)\cap (B\cup D)$, $\varphi \notin B\cup D$. From M-Inclusion, we have $\varphi \notin B\diamond _{A}D$. Contradiction. Thus, $Cn(A\diamond _{B}D)\cap B\diamond _{A}D\subseteq A\diamond _{B}D$. Therefore, $\diamond$ also satisfies M-BelState. $\square$

Proof of Proposition 3

Here we need to prove that: for all $D\subseteq \mathcal {L}$, $\theta ((B\cup D)\bigtriangleup _{\bot }A)$ satisfies Definition 3(1), (2), and (3).

Clause (1)::

By the definition of $\theta$, it is obvious that $\theta ((B\cup D)\bigtriangleup _{\bot }A)\subseteq \bigcup ((B\cup D)\bigtriangleup _{\bot }A)$.

Clause (2)::

Let $X\in (B\cup D)\bigtriangleup _{\bot }A$. So $X\cup A\vdash \bot$. From $(B,A)\in \mathcal {B}$, we have $A\nvdash \bot$. Since $X\cup A\vdash \bot$ and $A\nvdash \bot$, we have $X\ne \emptyset$. And from $X\in (B\cup D)\bigtriangleup _{\bot }A$ we know that $X\subseteq \bigcup ((B\cup D)\bigtriangleup _{\bot }A)=\theta ((B\cup D)\bigtriangleup _{\bot }A)$. So $\theta ((B\cup D)\bigtriangleup _{\bot }A)\cap X=X\ne \emptyset$. Hence, clause (2) holds.

Clause (3)::

Let $\tau \in \theta ((B\cup D)\bigtriangleup _{\bot }A)$, namely $\tau \in \bigcup ((B\cup D)\bigtriangleup _{\bot }A)$. By reductio ad absurdum, suppose that $(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)\vdash \tau$. From compactness, we know that there exists a finite set $Y\subseteq (B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)$ such that $Y\vdash \tau$. Since $\tau \in \bigcup ((B\cup D)\bigtriangleup _{\bot }A)$, we know that there exists $X\in (B\cup D)\bigtriangleup _{\bot }A$ such that $\tau \in X$. From $X\in (B\cup D)\bigtriangleup _{\bot }A$, we have $X\cup A\vdash \bot$. Hence, from $Y\vdash \tau$, $\tau \in X$, and $X\cup A\vdash \bot$, we have $X\backslash \{\tau \}\cup Y\cup A\vdash \bot$. Since $Y\subseteq B\cup D$ and $X\subseteq B\cup D$, we have $X\backslash \{\tau \}\cup Y\subseteq B\cup D$. From $X\backslash \{\tau \}\cup Y\subseteq B\cup D$, $X\backslash \{\tau \}\cup Y\cup A\vdash \bot$, and lower bound property (Lemma 1), we know that there exists $Z\subseteq X\backslash \{\tau \}\cup Y$ such that $Z\in (B\cup D)\bigtriangleup _{\bot }A$. From $Z\in (B\cup D)\bigtriangleup _{\bot }A$ and the definition of $\theta$, we have $Z\subseteq \bigcup ((B\cup D)\bigtriangleup _{\bot }A)=\theta ((B\cup D)\bigtriangleup _{\bot }A)$. Suppose that $Z\cap Y=\emptyset$. From $Z\subseteq X\backslash \{\tau \}\cup Y$, we have $Z\subseteq X\backslash \{\tau \}$. By $Z\in (B\cup D)\bigtriangleup _{\bot }A$, we have $Z\cup A\vdash \bot$. And since $Z\subseteq X\backslash \{\tau \}$, we have $X\backslash \{\tau \}\cup A\vdash \bot$. But from $\tau \in X$ we have $X\backslash \{\tau \}\subset X$, and by $X\in (B\cup D)\bigtriangleup _{\bot }A$ we know that $X\backslash \{\tau \}\cup A\nvdash \bot$. Contradiction. Hence, $Z\cap Y\ne \emptyset$. So there exists $\psi \in Z\cap Y$. Since $\psi \in Z\cap Y$ and $Z\subseteq \theta ((B\cup D)\bigtriangleup _{\bot }A)$, we have $\psi \in \theta ((B\cup D)\bigtriangleup _{\bot }A)$. But from $\psi \in Z\cap Y$ and $Y\subseteq (B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)$, we know that $\psi \in (B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)$. So $\psi \notin \theta ((B\cup D)\bigtriangleup _{\bot }A)$. Contradiction. Hence, $(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)\nvdash \tau$, and clause (3) holds. $\square$

Proof of Theorem 1

Construction $\Longrightarrow$ Postulates: Let $\diamond$ be a KMR operator for (B, A). We need to prove that it satisfies the following postulates.

M-CoreInvariance::

From Definition 4, we have $A\diamond _{B}D=A'=Cn(A)\cap B'$. So we need only to prove $Cn(A)\cap B'=Cn(A)\cap (B\cup D)$. $Cn(A)\cap B'\subseteq Cn(A)\cap (B\cup D)$ is obvious. Suppose that $Cn(A)\cap B'\nsupseteq Cn(A)\cap (B\cup D)$. So there exists $\tau \in Cn(A)\cap (B\cup D)$ such that $\tau \notin B'=(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)$. So $\tau \in \theta ((B\cup D)\bigtriangleup _{\bot }A)$. And from Definition 3, we know that there exists $X\in (B\cup D)\bigtriangleup _{\bot }A$ such that $\tau \in X$. So $X\backslash \{\tau \}\subset X$. Since $X\in (B\cup D)\bigtriangleup _{\bot }A$, we have $X\cup A\vdash \bot$ and $X\backslash \{\tau \}\cup A\nvdash \bot$. But from $\tau \in Cn(A)$ we have $Cn(X\backslash \{\tau \}\cup A)=Cn(X\cup A)$. Contradiction. Hence, $Cn(A)\cap B'\supseteq Cn(A)\cap (B\cup D)$.

M-Inclusion::

From Definition 4, it is obvious that $B\diamond _{A}D=(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)\subseteq B\cup D$.

M-Consistency::

By reductio ad absurdum, suppose that $B\diamond _{A}D\vdash \bot$. So $B\diamond _{A}D\cup A\vdash \bot$. From $B\diamond _{A}D\subseteq B\cup D$ and lower bound property, we know that there exists $Y\subseteq B\diamond _{A}D$ such that $Y\in (B\cup D)\bigtriangleup _{\bot }A$. Since $Y\subseteq B\diamond _{A}D$, we have $Y\subseteq (B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)$. So $Y\cap \theta ((B\cup D)\bigtriangleup _{\bot }A)=\emptyset$. But from $Y\in (B\cup D)\bigtriangleup _{\bot }A$ and Definition 3(2), we have $\theta ((B\cup D)\bigtriangleup _{\bot }A)\cap Y\ne \emptyset$. Contradiction.

M-RelClosure::

By reductio ad absurdum, suppose that $Cn(B\diamond _{A}D)\cap (B\cup D)\nsubseteq B\diamond _{A}D$. So there exists $\psi \in (B\cup D)\backslash B\diamond _{A}D$ such that $B\diamond _{A}D\vdash \psi$. Since $\psi \in (B\cup D)\backslash B\diamond _{A}D=(B\cup D)\backslash ((B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A))$, we have $\psi \in \theta ((B\cup D)\bigtriangleup _{\bot }A)$. From Definition 3(3), we have $(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)\nvdash \psi$. However, from $B\diamond _{A}D\vdash \psi$, we have $(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)\vdash \psi$. Contradiction.

M-Uniformity::

Let $B\cup D_{1}$ and $B\cup D_{2}$ have the same minimal subsets inconsistent with A, namely $(B\cup D_{1})\bigtriangleup _{\bot }A=(B\cup D_{2})\bigtriangleup _{\bot }A$. So $\theta ((B\cup D_{1})\bigtriangleup _{\bot }A)=\theta ((B\cup D_{2})\bigtriangleup _{\bot }A)$. Thus, $(B\cup D_{1})\backslash B\diamond _{A}D_{1}=(B\cup D_{1})\backslash ((B\cup D_{1})\backslash \theta ((B\cup D_{1})\bigtriangleup _{\bot }A))=\theta ((B\cup D_{1})\bigtriangleup _{\bot }A)=\theta ((B\cup D_{2})\bigtriangleup _{\bot }A)=(B\cup D_{2})\backslash ((B\cup D_{2})\backslash \theta ((B\cup D_{2})\bigtriangleup _{\bot }A))=(B\cup D_{2})\backslash B\diamond _{A}D_{2}$. So $(B\cup D_{1})\backslash B\diamond _{A}D_{1}=(B\cup D_{2})\backslash B\diamond _{A}D_{2}$. Hence, M-Uniformity holds.

M-ExRetainment::

Let $\psi \in (B\cup D)\backslash B\diamond _{A}D$. So $\psi \in (B\cup D)\backslash ((B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A))=\theta ((B\cup D)\bigtriangleup _{\bot }A)$. Thus, there exists $Y\in (B\cup D)\bigtriangleup _{\bot }A$ such that $\psi \in Y$. So $Y\cup A\vdash \bot$. Let $C=Y\backslash \{\psi \}$. So $C\subset Y\subseteq B\cup D$. From $Y\in (B\cup D)\bigtriangleup _{\bot }A$ and $C\subset Y$, we have $C\cup A\nvdash \bot$ and $C\cup \{\psi \}\cup A=Y\cup A\vdash \bot$. Hence, there exists $C\subseteq B\cup D$ such that $C\cup A\nvdash \bot$ and $C\cup \{\psi \}\cup A\vdash \bot$. So M-ExRetainment holds.

Postulates $\Longrightarrow$ Construction: Let $\diamond$ satisfy the above postulates. We need to prove that it is a KMR operator for (B, A).

Define a map $\theta$ for (B, A) such that: for all $D\subseteq \mathcal {L}$, $\theta ((B\cup D)\bigtriangleup _{\bot }A)=(B\cup D)\backslash B\diamond _{A}D$. And define an operator $\circ$ for (B, A) such that: for all $D\subseteq \mathcal {L}$, $(B,A)\circ D=(B',A')$, in which $B'=(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)$, $A'=Cn(A)\cap B'$. We need to prove: (a) $\theta$ is well defined; (b) $\theta$ is a metabolic incision function for (B, A); (c) for all $D\subseteq \mathcal {L}$, $(B,A)\diamond D = (B,A)\circ D$.

• (a) $\theta$ is well defined. That is to prove: for all $D_{1},D_{2}\subseteq \mathcal {L}$, if $(B\cup D_{1})\bigtriangleup _{\bot }A=(B\cup D_{2})\bigtriangleup _{\bot }A$, then $\theta ((B\cup D_{1})\bigtriangleup _{\bot }A)=\theta ((B\cup D_{2})\bigtriangleup _{\bot }A)$. Let $(B\cup D_{1})\bigtriangleup _{\bot }A=(B\cup D_{2})\bigtriangleup _{\bot }A$. From M-Uniformity, we have $(B\cup D_{1})\backslash B\diamond _{A}D_{1}=(B\cup D_{2})\backslash B\diamond _{A}D_{2}$, namely $\theta ((B\cup D_{1})\bigtriangleup _{\bot }A)=\theta ((B\cup D_{2})\bigtriangleup _{\bot }A)$.

• (b) $\theta$ is a metabolic incision function for (B, A).

  Clause (1)::

  Let $\tau \in \theta ((B\cup D)\bigtriangleup _{\bot }A)$, namely $\tau \in (B\cup D)\backslash B\diamond _{A}D$. From M-ExRetainment, we know that there exists $C\subseteq B\cup D$ such that $C\cup A\nvdash \bot$ and $C\cup \{\tau \}\cup A\vdash \bot$. From $C\cup \{\tau \}\subseteq B\cup D$, $C\cup \{\tau \}\cup A\vdash \bot$, and lower bound property, we know that there exists $Y\subseteq C\cup \{\tau \}$ such that $Y\in (B\cup D)\bigtriangleup _{\bot }A$. So $Y\cup A\vdash \bot$. From $Y\subseteq C\cup \{\tau \}$, $C\cup A\nvdash \bot$, and $Y\cup A\vdash \bot$, we have $\tau \in Y$. Since $\tau \in Y\in (B\cup D)\bigtriangleup _{\bot }A$, we have $\tau \in \bigcup ((B\cup D)\bigtriangleup _{\bot }A)$. So clause (1) holds.

  Clause (2)::

  Let $X\in (B\cup D)\bigtriangleup _{\bot }A$. By reductio ad absurdum, suppose that $\theta ((B\cup D)\bigtriangleup _{\bot }A)\cap X=\emptyset$, namely $((B\cup D)\backslash B\diamond _{A}D)\cap X=\emptyset$. From $X\subseteq B\cup D$ we have $X\subseteq B\diamond _{A}D$. Since $X\cup A\vdash \bot$, $B\diamond _{A}D\cup A\vdash \bot$. From M-ExRetainment and Proposition 1(1), we know that M-Protection, namely $A\subseteq B\diamond _{A}D$, holds. So $B\diamond _{A}D\vdash \bot$. It contradicts M-Consistency. Hence, $\theta ((B\cup D)\bigtriangleup _{\bot }A)\cap X\ne \emptyset$, and clause (2) holds.

  Clause (3)::

  Let $\tau \in \theta ((B\cup D)\bigtriangleup _{\bot }A)$, namely $\tau \in (B\cup D)\backslash B\diamond _{A}D$. From M-RelClosure, we have $B\diamond _{A}D\nvdash \tau$. And from M-Inclusion, we have $(B\cup D)\backslash ((B\cup D)\backslash B\diamond _{A}D)\nvdash \tau$, namely $(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)\nvdash \tau$. So clause (3) holds.

• (c) for all $D\subseteq \mathcal {L}$, $(B,A)\diamond D = (B,A)\circ D$.

  That is to prove: $B\diamond _{A}D=B'$, $A\diamond _{B}D=A'$, and $(B,A)\circ D\in \mathcal {B}$.

  $B\diamond _{A}D=B'$::

  From the definition of $\circ$, we know that $B'=(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)=(B\cup D)\backslash ((B\cup D)\backslash B\diamond _{A}D)$. From M-Inclusion, we have $B\diamond _{A}D\subseteq B\cup D$. So $B'=(B\cup D)\backslash ((B\cup D)\backslash B\diamond _{A}D)=B\diamond _{A}D$.

  $A\diamond _{B}D=A'$::

  From M-CoreInvariance and the definition of $\circ$, we need to prove that $Cn(A)\cap (B\cup D)=Cn(A)\cap B'$. $\supseteq$ direction is obvious. Suppose that $\subseteq$ direction does not hold, namely $Cn(A)\cap (B\cup D)\nsubseteq Cn(A)\cap B'$. So there exists $\tau \in Cn(A)\cap (B\cup D)$ such that $\tau \notin B'=(B\cup D)\backslash \theta ((B\cup D)\bigtriangleup _{\bot }A)$. Hence, $\tau \in \theta ((B\cup D)\bigtriangleup _{\bot }A)=(B\cup D)\backslash B\diamond _{A}D$. From M-ExRetainment, we know that there exists $C\subseteq B\cup D$ such that $C\cup A\nvdash \bot$ and $C\cup \{\tau \}\cup A\vdash \bot$. So $A\nvdash \tau$. It contradicts $\tau \in Cn(A)$. Hence, $\subseteq$ direction holds too.

  $(B,A)\circ D\in \mathcal {B}$::

  From M-CoreInvariance, M-Inclusion, M-ExRetainment, and Proposition 1(6), we know that $\diamond$ satisfies M-BelState. Since $B\diamond _{A}D=B'$ and $A\diamond _{B}D=A'$, we have $(B,A)\circ D\in \mathcal {B}$.

$\square$

Proof of Lemma 3

Left $\Longrightarrow$ Right: Let ‘Left’ hold. By reductio ad absurdum, suppose that $\exists \varphi \in (M_{1}\cup M_{2})\backslash (M_{1}\cap M_{2})$ such that $\{\varphi \}\cup A\nvdash \bot$. Since $\varphi \in (M_{1}\cup M_{2})\backslash (M_{1}\cap M_{2})$, we derive contradictions from the two following aspects. (a) $\varphi \in M_{1}\backslash M_{2}$: From $\{\varphi \}\subseteq M_{1}$, $\{\varphi \}\cup A\nvdash \bot$, and upper bound property, we know that there exists X such that $\{\varphi \}\subseteq X\in M_{1}\bigtriangledown _{\top }A$. From ‘Left’, we have $\{\varphi \}\subseteq X\in M_{2}\bigtriangledown _{\top }A$. So $\varphi \in M_{2}$. Contradiction. (b) $\varphi \in M_{2}\backslash M_{1}$: By symmetry, it is easy to derive a contradiction. Hence, ‘Right’ holds.

Right $\Longrightarrow$ Left: Let ‘Right’ hold. We need to prove that ‘Left’ holds. $\subseteq$ direction: Let $X\in M_{1}\bigtriangledown _{\top }A$. So $X\subseteq M_{1}$ and $X\cup A\nvdash \bot$. Suppose that $X\nsubseteq M_{1}\cap M_{2}$. Then $X\nsubseteq M_{2}$. So there exists $\phi \in X\backslash M_{2}$. Thus, $\phi \in (M_{1}\cup M_{2})\backslash (M_{1}\cap M_{2})$. From ‘Right’, we have $\{\phi \}\cup A\vdash \bot$. So $X\cup A\vdash \bot$. Contradiction. Hence, $X\subseteq M_{1}\cap M_{2}\subseteq M_{2}$. Now we prove that $X\in M_{2}\bigtriangledown _{\top }A$. By reductio ad absurdum, suppose that $X\notin M_{2}\bigtriangledown _{\top }A$. Since $X\subseteq M_{2}$ and $X\cup A\nvdash \bot$, we know that there exists $Y\supset X$ such that $Y\subseteq M_{2}$ and $Y\cup A\nvdash \bot$. Similarly, suppose that $Y\nsubseteq M_{1}\cap M_{2}$. Then we have $Y\nsubseteq M_{1}$. So there exists $\phi \in Y\backslash M_{1}$. Thus, $\phi \in (M_{1}\cup M_{2})\backslash (M_{1}\cap M_{2})$. By ‘Right’, we have $\{\phi \}\cup A\vdash \bot$. So $Y\cup A\vdash \bot$. Contradiction. Thus, $Y\subseteq M_{1}\cap M_{2}\subseteq M_{1}$. From $X\subset Y\subseteq M_{1}$ and $Y\cup A\nvdash \bot$, we have $X\notin M_{1}\bigtriangledown _{\top }A$. Contradiction. So $X\in M_{2}\bigtriangledown _{\top }A$. Hence, $M_{1}\bigtriangledown _{\top }A\subseteq M_{2}\bigtriangledown _{\top }A$. $\supseteq$ direction: By symmetry, it is easy to prove. Therefore, $M_{1}\bigtriangledown _{\top }A=M_{2}\bigtriangledown _{\top }A$. $\square$

Proof of Proposition 4

By reductio ad absurdum, suppose that there exists $\tau \in (B\cup D)\backslash \bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ such that $\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\vdash \tau$. From Definition 6(2), we have $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\ne \emptyset$. And since $\tau \notin \bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$, we know that there exists $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ such that $\tau \notin X$. From $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ and $\tau \in B\cup D$, we have $X\cup A\nvdash \bot$ and $X\cup \{\tau \}\subseteq B\cup D$. Since $\tau \notin X$, $X\subset X\cup \{\tau \}$. Since $\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\vdash \tau$, we know that $\forall Y\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)(Y\vdash \tau )$. And from $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$, we have $X\vdash \tau$. Since $X\vdash \tau$, $Cn(X)=Cn(X\cup \{\tau \})$. From $X\cup A\nvdash \bot$, we have $X\cup \{\tau \}\cup A\nvdash \bot$. Hence, there exists $Z=X\cup \{\tau \}$ such that $X\subset Z\subseteq B\cup D$ and $Z\cup A\nvdash \bot$. From Definition 5, we have $X\notin (B\cup D)\bigtriangledown _{\top }A$. It contradicts $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$.

Proof of Theorem 2

Construction $\Longrightarrow$ Postulates: Let $\diamond$ be a PMMR operator for (B, A). We need to prove that it satisfies the following postulates.

M-CoreInvariance::

From Definition 7, we have $A\diamond _{B}D=A'=Cn(A)\cap B'$. So we need only to prove $Cn(A)\cap B'=Cn(A)\cap (B\cup D)$. $Cn(A)\cap B'\subseteq Cn(A)\cap (B\cup D)$ is obvious. Suppose that $Cn(A)\cap B'\nsupseteq Cn(A)\cap (B\cup D)$. So there exists $\tau \in Cn(A)\cap (B\cup D)$ such that $\tau \notin B'=\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$. From Definition 6, we have $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\ne \emptyset$. Since $\tau \notin \bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$, we know that there exists $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ such that $\tau \notin X$. From $\tau \in B\cup D$ and $\tau \notin X$, we have $X\subset X\cup \{\tau \}\subseteq B\cup D$. And since $X\in (B\cup D)\bigtriangledown _{\top }A$, we have $X\cup A\nvdash \bot$ and $X\cup \{\tau \}\cup A\vdash \bot$. However, since $\tau \in Cn(A)$, $Cn(X\cup A)=Cn(X\cup \{\tau \}\cup A)$. Contradiction. Hence, $Cn(A)\cap B'\supseteq Cn(A)\cap (B\cup D)$.

M-Inclusion::

From Definition 7, we have $B\diamond _{A}D=\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\subseteq B\cup D$.

M-Consistency::

From Definition 6(2), we have $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\ne \emptyset$. Choose $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ arbitrarily. Since $\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\subseteq X$ and $X\cup A\nvdash \bot$, we have $\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\cup A\nvdash \bot$, namely $B\diamond _{A}D\cup A\nvdash \bot$. Hence, $B\diamond _{A}D\nvdash \bot$.

M-ExIrrelevance::

Let $A\cup \{\varphi \}\vdash \bot$ for all $\varphi \in ((B\cup D_{1})\cup (B\cup D_{2}))\backslash ((B\cup D_{1})\cap (B\cup D_{2}))$. By equivalence property above, we have $(B\cup D_{1})\bigtriangledown _{\top }A=(B\cup D_{2})\bigtriangledown _{\top }A$. So $\varUpsilon ((B\cup D_{1})\bigtriangledown _{\top }A)=\varUpsilon ((B\cup D_{2})\bigtriangledown _{\top }A)$. From Definition 7, we know that $B\diamond _{A}D_{1}=\bigcap \varUpsilon ((B\cup D_{1})\bigtriangledown _{\top }A)=\bigcap \varUpsilon ((B\cup D_{2})\bigtriangledown _{\top }A)=B\diamond _{A}D_{2}$. Hence, M-ExIrrelevance holds.

M-ExRelevance::

Let $\psi \in (B\cup D)\backslash B\diamond _{A}D$. So $\psi \notin B\diamond _{A}D=\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$. From Definition 6(2), we know that $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\ne \emptyset$. So there exists $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ such that $\psi \notin X$. So $X\subset X\cup \{\psi \}$. Since $\psi \in B\cup D$, we have $X\cup \{\psi \}\subseteq B\cup D$. Let $C=X$. It is obvious that $C\subseteq B\cup D$. From $X\in (B\cup D)\bigtriangledown _{\top }A$, we derive that $C\cup A\nvdash \bot$. Since $X\subset X\cup \{\psi \}\subseteq B\cup D$ and $X\in (B\cup D)\bigtriangledown _{\top }A$, we have $C\cup \{\psi \}\cup A\vdash \bot$. From $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$, we derive $B\diamond _{A}D=\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\subseteq X=C$, namely $B\diamond _{A}D\subseteq C$. Hence, there exists $C\subseteq B\cup D$ such that $B\diamond _{A}D\subseteq C$, $C\cup A\nvdash \bot$, and $C\cup \{\psi \}\cup A\vdash \bot$. Thus, M-ExRelevance holds.

Postulates $\Longrightarrow$ Construction: Let $\diamond$ satisfy the above postulates. We need to prove that it is a PMMR operator for (B, A).

Define a map $\varUpsilon$ for (B, A) such that: for all $D\subseteq \mathcal {L}$, $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)=\{X\in (B\cup D)\bigtriangledown _{\top }A|B\diamond _{A}D\subseteq X\}$. Define an operator $\circ$ for (B, A) such that: for all $D\subseteq \mathcal {L}$, $(B,A)\circ D=(B',A')$, in which $B'=\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$, and $A'=Cn(A)\cap B'$. We need to prove that: (a) $\varUpsilon$ is well defined; (b) $\varUpsilon$ is a metabolic selection function for (B, A); (c) for all $D\subseteq \mathcal {L}$, $(B,A)\diamond D= (B,A)\circ D$.

• (a) $\varUpsilon$ is well defined.

  That is to prove that: if $(B\cup D_{1})\bigtriangledown _{\top }A=(B\cup D_{2})\bigtriangledown _{\top }A$, then $\varUpsilon ((B\cup D_{1})\bigtriangledown _{\top }A)=\varUpsilon ((B\cup D_{2})\bigtriangledown _{\top }A)$. Let $(B\cup D_{1})\bigtriangledown _{\top }A=(B\cup D_{2})\bigtriangledown _{\top }A$. From equivalence property, we know that $A\cup \{\varphi \}\vdash \bot$ for all $\varphi \in ((B\cup D_{1})\cup (B\cup D_{2}))\backslash ((B\cup D_{1})\cap (B\cup D_{2}))$. And from M-ExIrrelevance, we have $B\diamond _{A}D_{1}=B\diamond _{A}D_{2}$. Since $(B\cup D_{1})\bigtriangledown _{\top }A=(B\cup D_{2})\bigtriangledown _{\top }A$ and $B\diamond _{A}D_{1}=B\diamond _{A}D_{2}$, we know that $\varUpsilon ((B\cup D_{1})\bigtriangledown _{\top }A)=\{X\in (B\cup D_{1})\bigtriangledown _{\top }A|B\diamond _{A}D_{1}\subseteq X\}=\{X\in (B\cup D_{2})\bigtriangledown _{\top }A|B\diamond _{A}D_{1}\subseteq X\}$ $=\{X\in (B\cup D_{2})\bigtriangledown _{\top }A|B\diamond _{A}D_{2}\subseteq X\}=\varUpsilon ((B\cup D_{2})\bigtriangledown _{\top }A)$. Hence, $\varUpsilon ((B\cup D_{1})\bigtriangledown _{\top }A)=\varUpsilon ((B\cup D_{2})\bigtriangledown _{\top }A)$.

• (b) $\varUpsilon$ is a metabolic selection function for (B, A).

  Clause (1)::

  From the definition of $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$, obviously we have $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\subseteq (B\cup D)\bigtriangledown _{\top }A$.

  Clause (2)::

  From M-ExRelevance and Proposition 1, we know that M-Protection, namely $A\subseteq B\diamond _{A}D$, holds. From M-Consistency, we have $B\diamond _{A}D\cup A=B\diamond _{A}D\nvdash \bot$. From M-Inclusion, we have $B\diamond _{A}D\subseteq B\cup D$. Then by upper bound property, we know that there exists X such that $B\diamond _{A}D\subseteq X\in (B\cup D)\bigtriangledown _{\top }A$. Hence, from the definition of $\varUpsilon$, we have $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\ne \emptyset$.

• (c) for all $D\subseteq \mathcal {L}$, $(B,A)\diamond D= (B,A)\circ D$.

  That is to prove that $B\diamond _{A}D=B'$, $A\diamond _{B}D=A'$, and $(B,A)\circ D\in \mathcal {B}$.

  $B\diamond _{A}D=B'$::

  From clause (2) in (b), we know that $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\ne \emptyset$. Since $\forall Y\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)(B\diamond _{A}D\subseteq Y)$, we have $B\diamond _{A}D\subseteq \bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)=B'$. Suppose that $B\diamond _{A}D\nsupseteq B'$. So there exists $\tau \in B'$ such that $\tau \notin B\diamond _{A}D$, namely there exists $\tau \in \bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ such that $\tau \notin B\diamond _{A}D$. So $\tau \in (B\cup D)\backslash B\diamond _{A}D$. From M-ExRelevance, we know that there exists $C\subseteq B\cup D$ such that $B\diamond _{A}D\subseteq C$, $C\cup A\nvdash \bot$, and $C\cup \{\tau \}\cup A\vdash \bot$. From $C\subseteq B\cup D$, $C\cup A\nvdash \bot$, and upper bound property, we know that there exists X such that $C\subseteq X\in (B\cup D)\bigtriangledown _{\top }A$. Since $B\diamond _{A}D\subseteq C$, we have $B\diamond _{A}D\subseteq X$. Thus, $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$. From $X\in (B\cup D)\bigtriangledown _{\top }A$, we have $X\cup A\nvdash \bot$. Since $X\cup A\nvdash \bot$, $C\subseteq X$, and $C\cup \{\tau \}\cup A\vdash \bot$, we have $\tau \notin X$. So there exists $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ such that $\tau \notin X$ (denoted by (*)). However, from $\tau \in \bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ and $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\ne \emptyset$, we know that $\forall Y\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)(\tau \in Y)$. It contradicts (*). Hence, $B\diamond _{A}D\supseteq B'$.

  $A\diamond _{B}D=A'$::

  By M-CoreInvariance and the definition of $\circ$, that is to prove that $Cn(A)\cap (B\cup D)=Cn(A)\cap B'$. $\supseteq$ direction is obvious. Suppose that $\subseteq$ direction does not hold, namely $Cn(A)\cap (B\cup D)\nsubseteq Cn(A)\cap B'$. So there exists $\tau \in Cn(A)\cap (B\cup D)$ such that $\tau \notin B'=\bigcap \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$. From $\varUpsilon ((B\cup D)\bigtriangledown _{\top }A)\ne \emptyset$ (proved in clause (2) of (b)), we know that there exists $X\in \varUpsilon ((B\cup D)\bigtriangledown _{\top }A)$ such that $\tau \notin X$. Since $\tau \notin X$ and $\tau \in B\cup D$, we have $X\subset X\cup \{\tau \}\subseteq B\cup D$. And from $X\in (B\cup D)\bigtriangledown _{\top }A$, we have $X\cup A\nvdash \bot$ and $X\cup \{\tau \}\cup A\vdash \bot$. But from $\tau \in Cn(A)$, we derive that $Cn(X\cup A)=Cn(X\cup \{\tau \}\cup A)$. Contradiction. Hence, $Cn(A)\cap (B\cup D)\subseteq Cn(A)\cap B'$.

  $(B,A)\circ D\in \mathcal {B}$::

  From M-CoreInvariance, M-Inclusion, M-ExRelevance, and Proposition 1, we know that $\diamond$ satisfies M-BelState. Since $B\diamond _{A}D=B'$ and $A\diamond _{B}D=A'$, we have $(B,A)\circ D\in \mathcal {B}$. $\square$