Residual Contraction

Abstract

In this paper, we propose and axiomatically characterize residual contractions, a new kind of contraction operators for belief bases. We establish that the class of partial meet contractions is a strict subclass of the class of residual contractions. We identify an extra condition that may be added to the definition of residual contractions, which is such that the class of residual contractions that satisfy it coincides with the class of partial meet contractions. We investigate the interrelations in the sense of (strict) inclusion among the class of residual contractions and other classes of well known contraction operators for belief bases.

Appendix: Proofs

Lemma 1

(Hansson 1999) If $A\subseteq B\subseteq Cn(A)$, then $Cn(A)=Cn(B)$.

Lemma 2

(Hansson 1981, Observation 2.2) Let A be a set of sentences and $\alpha$ a sentence. Then, $A\perp \alpha =\emptyset$ if and only if $\vdash \alpha$ (provided that the consequence operation Cn is compact).

Lemma 3

(Hansson 1999, Observation 1.39) Let A be a set of sentences and $\alpha$ and $\beta$ be sentences. Then the two following conditions are equivalent:

• 1. $A\perp \alpha = A\perp \beta$;

• 2. For all subsets D of A: $D \vdash \alpha$ if and only if $D \vdash \beta$.

Lemma 4

Let A be a belief base. $A\perp \alpha =\emptyset$ if and only if .

Proof

(Right to left) Follows trivially by Observation 8.⁶

(Left to right) Let $A\perp \alpha =\emptyset$. Hence, by Lemma 2, $\vdash \alpha$, from which it follows that . $\square$

Proof

(Proof of Observation 8) Let $X\in A\perp \alpha$. Hence $X\subseteq A$ and $X\not \vdash \alpha$. Let $X'\subseteq A$ be such that $Cn(X)\subset Cn(X')$. We intend to prove that $X' \vdash \alpha$. From $Cn(X)\subset Cn(X')$ it follows that there exists $\delta \in X' {\setminus } X$. Let $Y=Cn(X')\cap A$. It holds that $X\subset X\cup \{\delta \}\subseteq Y\subseteq A$. Thus $X\subset Y\subseteq A$. Therefore $Y\vdash \alpha$ (since $X\in A\perp \alpha$). It holds that $Y\subseteq Cn(X')$. Hence $X'\vdash \alpha$. $\square$

Proof

(Proof of Observation 9) Let . Hence $Y\subseteq A$ and $Y\not \vdash \alpha$. By the upper bound property it follows that there exists X such that $Y\subseteq X\in A\perp \alpha$. From $Y\subseteq X$ it follows, by monotony of Cn, that $Cn(Y)\subseteq Cn(X)$. It holds that $X\not \vdash \alpha$ and $X\subseteq A$ (since $X\in A\perp \alpha$). Hence $Cn(Y)\subset Cn(X)$ does not hold since . Therefore $Cn(Y)=Cn(X)$. $\square$

Proof

(Proof of Observation 10) Let $X\in A\perp \alpha$ and $Y\subseteq X$ be such that $Cn(X)=Cn(Y)$. From $X\in A\perp \alpha$ it holds that $X\not \vdash \alpha$ and $X\subseteq A$. Hence $Y\not \vdash \alpha$ and $Y\subseteq A$. It remains to show that Y satisfies condition (3) of Definition 10. Let $Z\subseteq A$ be such that $Cn(Y)\subset Cn(Z)$. Hence $Cn(X)\subset Cn(Z)$. Therefore there exists $\delta \in Z{\setminus } X$. Let $W=Cn(Z)\cap A$. Hence $X\subset W$. Therefore $W\vdash \alpha$ (since $X\in A\perp \alpha$). Thus $Cn(Z)\vdash \alpha$. Thus, by iteration of Cn, $Z\vdash \alpha$. $\square$

Proof

(Proof of Observation 11) (2) implies (1): Let . It follows immediately from Definition 10 and (2) that .

(1) implies (2): Suppose that (2) does not hold. Assume, without loss of generality, that there is some subset X of A such that $X\not \vdash \alpha$ and $X \vdash \beta$. By the residuum upper bound property it follows that there is some set $X'$ such that . On the other hand, from $X\subseteq X'$ it follows that $X'\vdash \beta$. Hence . This contradicts (1). $\square$

Proof

(Proof of Observation 12) Let and Y be such that $X\subseteq Y\subseteq Cn(X) \cap A$. Hence $Y\subseteq A$ and $X\subseteq Y\subseteq Cn(X)$. It follows from Lemma 1, that $Cn(X)=Cn(Y)$. Therefore, $Y \not \vdash \alpha$ and for any subset Z of A if $Cn(Y)\subset Cn(Z)$, then $Z\vdash \alpha$. Therefore . $\square$

Proof

(Proof of Theorem 1) (Construction to postulates)

Let A be a belief base and $\div$ be a residual contraction operator on A. Hence there is a selection function $\gamma$ for A such that for all sentences $\alpha$:

• 1. if $A\vdash \alpha$, then and

• 2. if $A\not \vdash \alpha$, then $A\div \alpha =A$.

We will show that $\div$ satisfies success, inclusion, uniformity, vacuity, failure and weak relevance.

Success Let $\not \vdash \alpha$. Hence, according to Lemmas 2 and 4, . Therefore, by definition of a selection function, is a non-empty subset of . If $A\not \vdash \alpha$, then $A\div \alpha =A$, therefore $A\div \alpha \not \vdash \alpha$. If $A\vdash \alpha$, then , therefore $A\div \alpha \not \vdash \alpha$ (since it holds, for any element X of , that $X\not \vdash \alpha$).

Inclusion Let $\alpha \in \mathcal {L}$. It follows trivially by the definition of $\div$, if $A\not \vdash \alpha$. Assume now that $A\vdash \alpha$. We will consider two cases:

Case 1$\vdash \alpha$. Hence, according to Lemmas 2 and 4, . Therefore, . Hence $A\div \alpha =A$.

Case 2$\not \vdash \alpha$. Hence, according to Lemmas 2 and 4, . Thus is a non-empty subset of . Any element of is a subset of A, hence any element of is also a subset of A. Therefore, .

Uniformity Let $\alpha , \beta$ be two sentences such that it holds for all subsets $A'$ of A that $\alpha \in Cn(A')$ if and only if $\beta \in Cn(A')$. It follows from Observation 11 that . Thus, . We will consider two cases:

Case 1$A\not \vdash \alpha$. Hence, by hypothesis, $A\not \vdash \beta$. Therefore, $A\div \alpha =A=A\div \beta$.

Case 2$A\vdash \alpha$. Hence, by hypothesis, $A \vdash \beta$. Therefore, .

Vacuity Follows trivially by the definition of $\div$.

Failure Let $\vdash \alpha$. Hence, according to Lemmas 2 and 4, . Therefore . Thus .

Weak Relevance Let $\beta$ be such that $\beta \in A{\setminus } A\div \alpha$. Therefore, $A\div \alpha \not = A$. Hence it holds that $A\vdash \alpha$ and consequently that . It also holds that . From $\beta \not \in A\div \alpha$, it follows that there exists such that $\beta \not \in X$. Therefore $A\div \alpha \subseteq X$. On the other hand, $X\subseteq A$ (since ) and, since $\beta \not \in X$, it follows that $X\subseteq A{\setminus } \{\beta \}$. Furthermore, from it follows that $X\not \vdash \alpha$ and for any subset Y of A such that $Cn(X)\subset Cn(Y)$ it holds that $Y\vdash \alpha$.

(Postulates to construction)

Let $\div$ be an operator on A that satisfies success, inclusion, uniformity, vacuity, failure and weak relevance. Let $\gamma$ be such that:

• (i) $\gamma (\emptyset )=\{A\}$;

• (ii) If , then .

We need to show that: (1) $\gamma$ is a (well-defined) function; (2) $\gamma$ is a selection function; (3) for all $\alpha$

(3.1) if $A\vdash \alpha$, then and

(3.2) if $A\not \vdash \alpha$, then $A\div \alpha =A$.

(1) We must prove that for all $\alpha , \beta$ if then .

Suppose that . It follows trivially if . Assume now that . By Observation 11 it follows that for all subsets B of A: $B\vdash \alpha$ if and only if $B\vdash \beta$. Hence, uniformity yields that $A\div \alpha =A\div \beta$. Hence, by definition of $\gamma$, it holds that .

(2) By definition $\gamma (\emptyset )=\{A\}$. Hence, in order to prove that $\gamma$ is a selection function it is sufficient to show that if , then . That follows trivially by the definition of $\gamma$. It remains to prove that . Since it must be the case that $\not \vdash \alpha$. Thus by success, $A\div \alpha \not \vdash \alpha$. It follows by inclusion that $A\div \alpha \subseteq A$. By the residuum upper bound property there is some X such that . Therefore, according to the definition of . Thus .

(3) Let $\alpha$ be an arbitrary sentence. We will prove that (3.1) and (3.2) hold.

(3.1) $A \vdash \alpha$. We will consider two cases:

Case 1$\vdash \alpha$. Thus . By definition of $\gamma$ it follows that . Hence . On the other hand, by failure$A\div \alpha =A$. Therefore .

Case 2$\not \vdash \alpha$. Thus . Therefore, as shown in (2), . It follows from the definition of $\gamma$ that $A\div \alpha$ is a subset of every element of . Thus . It remains to show that . Let $\beta \not \in A\div \alpha$. We will show that . This is obvious if $\beta \not \in A$. Assume now that $\beta \in A$. Hence $\beta \in A{\setminus } A\div \alpha$. It follows by weak relevance that there exists $X\subseteq A{\setminus } \{\beta \}$ such that $A\div \alpha \subseteq X$ and $X\not \vdash \alpha$ but for all $Y\subseteq A$ such that $Cn(X)\subset Cn(Y)$ it holds that $Y\vdash \alpha$. Hence . Therefore, by definition of $\gamma$, it holds that . From $\beta \not \in X$ it follows that .

(3.2) $A\not \vdash \alpha$. Then by vacuity and inclusion it follows that $A\div \alpha =A$. $\square$

Proof

(Proof of Observation 13) Let A be a belief base and $\div$ an operator on A that satisfies relevance. Let $\beta \in A{\setminus } A\div \alpha$. By relevance there exists X such that $A\div \alpha \subseteq X\subseteq A$, $X\not \vdash \alpha$ but $X\cup \{\beta \}\vdash \alpha$. By the upper bound property there exists $X\subseteq X'$ such that $X'\in A\perp \alpha$. It must hold that $\beta \not \in X'$ (otherwise it would follow that $X'\vdash \alpha$). On the other hand, by Observation 8, . Therefore $X'\subseteq A{\setminus } \{\beta \}$, $A\div \alpha \subseteq X'$, $X'\not \vdash \alpha$ and for any $Y\subseteq A$ such that $Cn(X')\subset Cn(Y)$ it holds that $Y\vdash \alpha$. $\square$

Proof

(Proof of Observation 14) Let $X\in A\perp \alpha$. By Observation 8 it follows that . Assume now, by reductio ad absurdum, that there exists some such that $X\subset Y$. From it follows that $Y\subseteq A$ and $Y\not \vdash \alpha$. Hence, from $Y\not \vdash \alpha$ and $X\subset Y\subseteq A$, it follows (by condition (3) of Definition 2) that $X\not \in A\perp \alpha$. Contradiction. Hence .

Let . Hence and there exists no $Y'$ in such that $X'\subset Y'$. It follows from that $X'\subseteq A$ and $X'\not \vdash \alpha$. Let $Z\subseteq A$ be such that $X'\subset Z$. From $X'\in \Gamma$ it follows that . Assume by reductio ad absurdum that $Z\not \vdash \alpha$. Hence, by condition (3) of Definition 10, there exists $Z'\subseteq A$ such that $Cn(Z)\subset Cn(Z')$ and $Z'\not \vdash \alpha$. From $X'\subset Z$ it follows that $Cn(X')\subseteq Cn(Z)\subset Cn(Z')$ and $Z'\not \vdash \alpha$. Hence . Contradiction. Therefore $Z\vdash \alpha$, and we can conclude that $X'\in A\perp \alpha$. Therefore . $\square$

Proof

(Proof of Theorem 2) Let A be a belief base. We start by showing that every maximal residual contraction on A is a partial meet contraction on A.

Let be a maximal selection function and let $\div _{\gamma _{m}}$ be the maximal residual contraction based on $\gamma _{m}$. Now let $\gamma :\{A\perp \epsilon : \epsilon \in \mathcal {L}\}\longrightarrow \mathcal {P}(\mathcal {P}(A))$ be such that (for all $\alpha \in \mathcal {L}$):

We will show that: (1) $\gamma$ is a (well-defined) function; (2) $\gamma$ is a selection function; (3) For all $\alpha$, $A\div _{\gamma _m}\alpha =A-_{\gamma }\alpha$, where $-_{\gamma }$ is the partial meet contraction based on $\gamma$.

(1) Let $A\perp \alpha =A\perp \beta$. Combining Lemma 3 and Observation 11 we can conclude that . Hence it follows from the definition of $\gamma _m$ and from the fact that $\gamma _m$ is a function, that $\gamma (A\perp \alpha )=\gamma (A\perp \beta )$.

(2) Since if $\vdash \alpha$, then , it follows from the definition of $\gamma$ that $\gamma (\emptyset )=\gamma _{m}(\emptyset )=\{A\}$ (since $\gamma _m$ is a selection function). It remains to show that if $A\perp \alpha \not =\emptyset$, then $\emptyset \not =\gamma (A\perp \alpha )\subseteq A\perp \alpha$. Assume that $A\perp \alpha \not =\emptyset$, then and, according to Lemma 4, . Since $\gamma _m$ is a selection function, it holds that . Furthermore, since $\gamma _m$ is maximal, if then there is no such that $X\subset Y$. Therefore it follows by Observation 14 that . Therefore $\emptyset \not =\gamma (A\perp \alpha )\subseteq A\perp \alpha$.

(3) If $A\not \vdash \alpha$, then $A\div _{\gamma _{m}}\alpha =A=A-_\gamma \alpha$. Now assume that $A\vdash \alpha$. Then .

Now we show that every partial meet contraction on A is a maximal residual contraction on A.

Let $\gamma :\{A\perp \epsilon : \epsilon \in \mathcal {L}\}\longrightarrow \mathcal {P}(\mathcal {P}(A))$ be a selection function and let $\div _{\gamma }$ be the partial meet contraction based on $\gamma$. Now let be such that (for all $\alpha \in \mathcal {L}$):

We will show that: (1) $\gamma _m$ is a (well-defined) function; (2) $\gamma _m$ is a maximal selection function; (3) For all $\alpha$, $A\div _{\gamma }\alpha =A-_{\gamma _m}\alpha$, where $-_{\gamma _m}$ is the maximal residual contraction based on $\gamma _m$.

(1) Let . Combining Lemma 3 and Observation 11 we can conclude that $A\perp \alpha =A\perp \beta$. Hence it follows from the definition of $\gamma _m$ and from the fact that $\gamma$ is a function, that .

(2) Since if $\vdash \alpha$, then , it follows, from the definition of $\gamma _m$, that $\gamma (\emptyset )=\gamma _{m}(\emptyset )=\{A\}$ (since $\gamma$ is a selection function). It remains to show that if , then:

(a) ;

(b) If , then there is no such that $X\subset Y$.

Assume that . Then, according to Lemma 4, $A\perp \alpha \not =\emptyset$. Therefore, since $\gamma$ is a selection function, it holds that $\emptyset \not =\gamma (A\perp \alpha )\subseteq A\perp \alpha$. By Observation 8. Hence, since , we can conclude that (a) holds.

Next we prove (b). Let . Since it holds that $X\in A\perp \alpha$. Therefore, it follows from Observation 14 that there is no such that $X\subset Y$. Thus (b) holds.

(3) If $A\not \vdash \alpha$, then $A\div _{\gamma } \alpha =A=A-_{\gamma _{m}} \alpha$. Now assume that $A\vdash \alpha$. Then . $\square$