belief revision in non-classical logics

8/12/2019 Belief Revision in Non-Classical Logics

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Contents

1 Introduction 5

1.1 Epistemic States and Belief Sets . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Logical Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Expansion, Contraction and Revision 8

2.1 Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.3 Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.4 Motivation Behind Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.5 Interrelations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Constructing Contraction and Revision 14

3.1 Epistemic Entrenchment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2 Partial Meet Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3 System of Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.4 Implementation of Belief Revision . . . . . . . . . . . . . . . . . . . . . . . 20

3.4.1 Finite Partial Entrenchment Ranking . . . . . . . . . . . . . . . . . 21

3.4.2 A PROLOG Implementation . . . . . . . . . . . . . . . . . . . . . . 22

4 Three-Valued Logic and Belief Revision 24

4.1 The Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.2 A Sequent Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.2.1 Truth of a Sequent . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.2.2 Provability of a Sequent . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.2.3 Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.3 Automated Proof Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.3.1 Semantic Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.3.2 Tertiary Decision Diagrams . . . . . . . . . . . . . . . . . . . . . . . 28

4.4 Contraction and Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.5 Interrelations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.6 Epistemic Entrenchment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.7 Partial Meet Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.8 System of Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4.9 Interpretation and Applicability . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.10 Generalization ton-valued Logics . . . . . . . . . . . . . . . . . . . . . . . . 35

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CONTENTS 3

5 Rough Belief Revision 375.1 Rough Truth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.1.1 The semantics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.2 Rough belief change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395.3 Partial Meet Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.4 Rough Consistency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.5 Maximal Consistent Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.6 Revision from System of Spheres . . . . . . . . . . . . . . . . . . . . . . . . 455.7 Epistemic Entrenchment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465.8 Interrelations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

6 Conclusions 526.1 Scope for Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53


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Acknowledgements

I would like to take this opportunity to thank my guide Dr. Mohua Banerjee for her supportand guidance throughout this project and her patience at some trying times. I would alsolike to thank her for providing the initial motivation, which led me to take up this topicas a reading project and later as my M.Sc. project. I also thank all my friends and wingmates for being such good company that I could always refresh my efforts at tackling the

challenges posed by this project.Pankaj Kumar Singh

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Chapter 1

Introduction

Belief revision models the change in beliefs of a person when some new information is

provided or an old belief has to be retracted. This kind of study helps in building intelligentagents like robots, who have to manage beliefs about the world in order to achieve theirgoals. We all know that beliefs can sometimes be wrong, so intelligent agents should be ableto revise beliefs when new correct information is acquired. When giving up a belief one hasto decide which of its consequences to retain. Thus a fundamental issue in belief revision ishow to decide, which information has to be retracted in order to avoid contradiction. Otherissue is that logical consequences alone do not tell which beliefs to give up. What do wechoose, when contradiction can be avoided by retracting one of the two given beliefs?

The problem of belief revision is very well studied in classical background, where anystatement is either true or false. There are well-defined functions which model the beliefchange and give out the changes explicitly. But gradually it is realized that belief change is

not that naive. In real world settings an agent is bombarded with all sorts of information,that cannot be rejected or accepted decisively. So there have to be some beliefs which areweakly held or they state vague truth. In any case we have to incorporate vagueness, whichcan be achieved by weakening the background classical logic.

In this thesis we work on two methods of changing the background logic. First is thatwe insert more truth values into classical logic. More truth values increase the varietyof beliefs and hence increase vagueness notions. Other way is to decrease the credibilityof existing beliefs by including incomplete information as beliefs. With these motivationswe investigate 3-valued logic in which there is an extra truth value and rough logic whereincomplete information is incorporated into the formal system.

After investigating the 3-valued logic we realized that change is only with respect to

interpretation of new beliefs, which more or less satisfy the same rules or postulates of beliefchange. We expect similar results with general n-valued logics. But for the rough logic werealize that rules of belief change do not fit well and require modifications. Currently weare working on explicit definition of belief change functions as in the classical case and alsothinking of right postulates of belief change with respect to rough logic. In the remainingchapter we present a formal treatment of preliminaries for belief change.

1.1 Epistemic States and Belief Sets

Epistemic states are the central entities in the belief change theory as they represent anactual or a possible cognitive state of some individual at a given point of time. Our notionof epistemic states is idealization of psychological concepts like consistency of the state

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6 CHAPTER 1. INTRODUCTION

and closeness under logical sequences. Since a persons beliefs tend to be in equilibriumstate as regulated by above two conditions, its interesting to see how they alter themselveswhen disturbed by contradictory notions. Before entering into dynamics of these changes

we introduce propositional model for epistemic state called as belief sets. In this modelepistemic states are sets of propositions, expressed by sentences from some given objectlanguage. The intended interpretation of such a model is that the set consists of thosepropositions that the person accepts in the present state. The central criterion of change isconsistency and logical closure.

1.2 Logical Preliminaries

We already know that belief set is a propositional model of an epistemic states where aproposition or sentence in a belief set means that the individual believes it to be true or

regards it as certain. This model is linguistic and presupposes a language L. One canassume that L consists of expressions formed by sentential connectives : Negation : ,Conjunction : , Disjunction : , Implication :. The symbols and stand for thesentential constants falsity and truth. Formally our epistemic states are based on setof all propositional well formed formulae (wff). Propositional wffs can be described byfollowing inductive definition:

(i) Any propositional letterp which can be directly assigned truth values {T, F}is a wff.

(ii) IfA andB are wffs then A, A B, A B and A B are wffs.

We use L to denote set of all wffs. For defining concepts in belief change we use classical

logic and present the equivalent notions whenever talking in other logics. After realizingour language we have to define logical consequence. In classical background we define anentailment relation between subsets ofL and L. This relation is syntactic inference rela-tion. Similar to there is a binary relation|= based on semantics, if |= then wheneverall propositions in are T, is T. We know the (completeness and soundness):

iff|=

Other deduction properties like:

compactness-

If , where L, then1...n s.t. {1,...,n}

weakening-

If and then

deduction theorem-

If {} then

modus ponens-

If and then


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1.2. LOGICAL PRELIMINARIES 7

, are known to us. With language L and inference relation we can define logical closureCn() of a subset as

Cn() ={ L: }

We define theories ofL as subsets satisfying

Cn() =

From here we can assume that belief sets (generally denoted byK) are theories ofL. Thelargest belief set, denoted as K is the set of all beliefs or propositions. The falsityconstant entails everything and hence Cn() = K. Similarly truth constant isentailed by every proposition, meaning is in every belief set. In addition any belief setcontains alltheorems (wffs s.t. or ). We call subsets ofL consistentif -

B L s.t. and

and inconsistent otherwise. Note that by the definition any set containing isinconsistent. Following result is very useful in classical logic-

If {} is inconsistent then

Notion of maximal consistent sets is also useful, a subset M ofL is maximal consistentif it is consistent and / M implies M {} is inconsistent. Every consistent set has amaximal consistent extension.


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2.3. CONTRACTION 9

(K1) K Ais a belief set(K2) A K A(K3) K A K+ A

(K4) If A /K then K+ A K A(K5) K A= K if and only if A(K6) If A B thenK A= K B(K7) K (A B) (K A) + B(K8) If B /K Athen (K A) + B K (A B)

Above postulates dont explicitly tell which beliefs to be retained and which to be removed.They only circumscribe the set of rational solutions and dont lead to a unique solution asin case of expansion.

2.3 Contraction

The change in epistemic state is called as contraction when some beliefAhas to be retractedwithout adding extra information. Contraction as the name suggests requires removal ofall those beliefs which can imply the currently held beliefA. A contraction function mustsatisfy following postulates-

(K1) K Ais a belief set(K2) K A K(K3) If A /K thenK A= K(K4) If Athen A /K A(K5) K(K A) + A

(K6) If A B thenK A= K B(K7) (K A) (K B) K (A B)(K8) If A /K (A B) then K (A B) K A

Similar to revision, the postulates are not sufficient to produce a unique contraction func-tion and a notion of preference among the beliefs is necessary. Like for retractingA Bfrom C n{A B}, removing either A or B is sufficient. Thus a sense of preference is mustto make a choice among the two possibilities.

2.4 Motivation Behind Postulates

Postulates for revision and contraction, as already said only regulate the desired propertiesof these functions and dont lead to explicit definitions. K1 6 are elementary require-ments that connect K, A and K A and are called basic set of postulates. K7 and K8are concerned with minimal change. To ensure minimal change in K when revised withA, expansion(+) ofK A with another formula B and revision ofKdirectly with A Bmust be same provided K A+B does not become K. Following alternative postulatesfor revision look more closer to K7 andK8:

K7 K A K B K A BK8 IfA /K A B thenK A B K A

Lemma 2.4.1 (i) K7 is equivalent to K7 and(ii) K8 is equivalent to K8.


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10 CHAPTER 2. EXPANSION, CONTRACTION AND REVISION

Proof (i) Suppose K7 is true. Let K A K B, usingK A= K (A B) AK A B + Awe getA K A B. SimilarlyB K A B. But that is sameas A B K A B. Since A B K A B, K A B.

LetK7 holds. If K A B thenB K A B. SinceB K A B and B (B ),B K A B. K7 now givesB K (A B) (A B) =K A. Hence K A + B.

(ii) Suppose K8 holds. Taking A =A B and B =A in K8 directly gives K A B+ A K A, where K A B K A B+ A.

Suppose K8 holds. LetB /K A= K (A B) A, hence A B (A B) /K (A B) A. TakingA =A B andB =A in K8 we getK A K A B. ThusK A + B K A B+ B = K A B.

Following also follow from K 7, 8 and are useful sometimes:

(i) K A B = K A or K A B = K B or K A B = K A K B(ii) K A= K B iff B K A and A K B

Proof (ii) only if follows directly. So for if part, suppose B K Aand A K B. IfAor B is a theorem then K A= K B =K. So suppose A and B. Thus K Aand K B can not be K and hence B /K A and A /K B. But then using K

8,K A= K A + B = K A B = K B+ A= K B.

Similar to revision postulatesK16 are basic. IfA KthenK14 giveKA+AK. Thus withK5 (recovery), (K A) + A= K whenever A K.

Lemma 2.4.2 K7 is equivalent to

K

7

K A Cn({A}) K A B

ProofLet K7 holds. Suppose KACn({A}). IfA /Kthen KA= KAB = Kand hence there is nothing to prove. So we assumeA Kand show that K (A B),so that K A K (A B) K A (A B) =K A B. Using recovery weget (A B) A K (A B), but thenA (A B) A implies A K (A B)and Cn({A}) gives K (A B).

Let K7 holds. Let K A K B. Since K A and A C n{A}, we getA K A Cn{A} K A B. SimilarlyB K A B. Thus (A B) (A B) K A B. Using recovery we also have (A B) K A B, andhence K A B.

Now we prove a result regarding equality of contraction functions.

Theorem 2.4.1 If B A KA and A B KB then KA = KB. IfA, B K thenK A= K B impliesB A K A andA B K B.

ProofWe assumeB A K Aand A B K B and prove thatK A= K B =KAB. IfAorB KABthenAB KABwhich impliesKA= KB = K( A B). Thus we consider the case when A, B / K A B, hence by K8 we haveKAB KA, KB. So we only need to prove KA, KB KAB. Let KA.Using recovery and assumption we have (A B) KA and (A B) KB .Thus using K7 we get (A B) K (A B). Using recovery again on KA Bwith K we have (A B ) KA B . Since A and B are equivalent inK A B we have A K A B. Now we prove thatA A is also in


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2.5. INTERRELATIONS 11

K A B, so that we get K A B. Now ifA /K then A B /K which impliesK= KAB KA, KB, that is to sayKA= K B = K. So we takeA K. Us-ing recovery we get (A B) A A K(A B). ThusA K (A B), K A.

K7 now gives A K (A B ) A = K A B. Hence K A K (A B).Similarly K B K (A B) and we have proved K A= K B = K A B.For the second part, taking A, B Kand recovery gives thatA B K A= K B.

Using the above result we prove following factoring condition for contractions.

K A B = K A or K A B = K B or K A B = K A K B

Proof Taking A = A B and B = A in previous lemma, we have, ifA (A B) K A B and A B A K A then K A B =K A. But since A B Aand A (A B) A B, our condition modifies to: if A B KA B then

K A B = K A. Now, if eitherA B K A B or B A K A B thenKA B = KA or KA B = KB . When both the conditions fail, that isA B / K A B and B A / K A B then because B (A B) we haveB /K A B andA /K A B. K8 then gives K A B = K A K B.

Observation 2.4.1 If A K then A K (A B). We have used this result severaltimes in above proofs and will use it again in next chapter.

Observation 2.4.2 IfA K A B orB A K A B thenK A B = K B.A K A B gives B A KA B which means B (A B) KA B.(A B) B being a theorem is in KB and hence using previous theorem we have

K A B = K B.

2.5 Interrelations

Changes in epistemic states modelled as expansion, revision and contraction are interrelated.Revision of a belief set K with respect to a sentence A can be seen as removal of anyincompatible information A and then addition ofA along with all logical consequences.This relationship is called Levi-identity and is written as-

K A= (K A) + A (2.2)

Let us call the revision function aboveR(). Following theorem tells that functionR()defined through Levi - identity is indeed a revision function.

Theorem 2.5.1 If a contraction function satisfies (K1) to (K6) thenR()satisfies(K1) to (K6). Furthermore if also satisfies (K7) and (K8) then R() satisfies(K7) and (K8) respectively.

Proof

(K1) K A= (K A) +A is closed under logical consequences by 2.1 and hence is abelief set.

(K2) A K A as (K A) {A} A (dilution or weakening).


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12 CHAPTER 2. EXPANSION, CONTRACTION AND REVISION

(K3) From (K2), K A K or (K A) {A} K {A} . Using weakeningCn((K A) {A}) Cn(K {A}). 2.1 and 2.2 now give K A K+ A.

(K

4) LetA /K, using (K

3) we have K A= K. Hence C n(K {A}) =C n((KA) {A}) or K+ A= (K A) + A.

(K5) if Let A. Weakening and 2.1 give A and A (K A) +A. As B L, {A, A} B , K A= (K A) + A= K.only if Let (K A) + A= K.Thus (K A) + A Aand (K A) + A (A A).Using D.T. K A (A A) andK A (A (A A)).But second statement is same as K A ((A A) A) because (A B) (B A). Using modus ponens we getK A A. (K4) now implies A.

(K

6) Let K A= (K A) + A. D.T. gives (K A) (A ). But A B B A, using (K6) gives (K B) = (K A) (A ). Weakening gives(KB)+B (A ) which in turn yields (KB)+B as A (KB)+B.Thus K A K B.Similarly K B K A giving K A= K B.

(K7) Let K A B = (K (A B)) + (A B).(A B) (K (A B)) (D.T.)Now A (A B) (A B)By (K7) and (K6),K(AB)K(AB) K((AB)(AB)) =K A.If we show that (A B) K (A B) then proof will be complete as aboverelation will give (A B) K A. But as ((A B) ) (A (B ))applying modus ponens twice will yield ((K A) + A) + B = K A + B.So (A B) K (A B) needs to be proved. Using (K2) and (K5) we haveK (A B) KK (A B) + (A B). Thus ((A B) ) K (A B) + (A B) which on application of D.T. gives ((A B) ) K (A B) as((A B) ) ((A B) ((A B) )).

(K8) Let B / KA and KA+ B = ((K A) + A) + B . Converse of D.T.and D.T. along with A (A B) A give (A B) / K A and(A B) K A= K (A B) A respectively.Now using (K8) withA =A B andB =Awe get (A B) K (A

B) A K (A B). Finally using modus ponens K (A B) + (A B) =K (A B) as (A B) (A B).

Similar to R(), contraction can be defined using a revision function. Contractionrepresents knowledge left after removal of some particular information let us say A froma belief set K. Equivalently, this can be seen as the information remaining insideK afterconsistent addition ofAto K. The consistent addition ofAis essentially revision andensures removal ofA from the knowledge base. This way of obtaining a contraction is calledHarper-Identity and is written as

K A= K K A (2.3)

We call the function above C() i.e. contraction C produced from a revision * .Fol-lowing theorem ensures thatC() is indeed a contraction function.


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2.5. INTERRELATIONS 13

Theorem 2.5.2 If a revision function * satisfies (K1) t o ( K6) then C(*) satisfies(K1) to ( K6). Furthermore, if * also satisfies (K7) and (K8) then C(*) satisfies(K7) and (K8) respectively.

Proof

(K1) K A= K K Ais a belief set as intersection of two belief sets is also a beliefset. IfK K A then K, K Aas K K A K, K Aand both Kand K A are beliefs sets. Hence we have K K A.

(K2) K A= K K A K.

(K3) Let A /K or A /K. Using (K4), K A= K+ (A). But KK+ (A)giving K A= K K+ (A) =K.

(K

4) Let A and assume that A K A = K K A. ThusA K A. Using(K2), A K A and we have K A = K. (K5) now gives A or A

which is a contradiction and hence A /K A.

(K5) Let K, we need to show (KK A) +A. (A ) impliesA K. Using Converse of D.T. its sufficient to prove A K A. Butthat is fine as A (A ) and A K A because of (K2).

(K6) If A B then A B. From (K6) we have K A= K B which givesK A= K K A= K K B = K B.

(K7) K A K B which means K, K A, K B. Taking A = (A B)

and B

= (A) and using (K

7) we get K((A B) A) = K A K(A B) +A. Similarly we can get K B K(A B) +B. UsingD.T., A , B K(A B). Now C() already satisfies (K5) so K (K K (A B)) + A B. And we have A B K (A B).But {A B , A , B } and hence K (A B). Thus K K (A B) =K (A B).

(K8) Let A /K (A B). Which meansA /K or A /K (A B). IfA /K thenK (A B) K= K A. So consider the case whenA KbutA /K (A B).Taking A = A B, B = A and using (K8) we have K (A B) +A K (A B) A= K A. But K (A B) K (A B) + Aand we haveK (A B) =K K (A B) K K A= K A.


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Chapter 3

Constructing Contraction and

Revision

In previous chapter contraction and revision function were investigated indirectly by meansof a number of postulates imposed on the functions. It was also shown in that if it is possibleto give explicit construction of a contraction function, then this also yields a revision functionby means of Levi-Identity (or vice-versa through the Harper-identity). Thus epistemicchange on belief sets is completely determined by contraction function (or revision function).The goal of this chapter is to present three constructions for contraction function leadingto single proposal.

3.1 Epistemic Entrenchment

If one can rank beliefs such that a preference among the beliefs can be induced then acontraction function can be constructed. One such ranking (an ordering) is called EpistemicEntrenchment ranking. This ranking is particular to the belief setK under considerationand is a binary relation satisfying following postulates-

(EE1) is transitive.

(EE2) IfA B thenB A

(EE3) EitherA B A or A B B

(EE4) K=K thenB A for all B if and only ifA /K

(EE5) IfA B for all B thenA

Now given an epistemic entrenchment ranking , contraction function can be defined as

K A=

{B K: A < A B}, if AK, otherwise

(3.1)

The symbol < is strict preference i.e. A < B (A B and A B) A B as is atotal order. A similar method can be used in the reverse way. Given a contraction functionan epistemic entrenchment ranking can be obtained as

A B if A /K (A B)or A B (3.2)

Following theorem proves that above ordering is indeed a epistemic entrenchment ranking.

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16 CHAPTER 3. CONSTRUCTING CONTRACTION AND REVISION

If A then {B K : A < A B} = {B K : A / K A (A B) = K A and(A B) K A}. We will show that ifB Kthen (A /K A and (A B) K A)(A /KA and B KA). B K(KA)+A (using (K5)) implies AB KA.

Thus ifA B K Athen (A B) (A B) =B K A. IfB K Athen obviously(A B) K A.

If A then A KA. That is (KA) +A = KA. Using (K5) we haveK(K A) + A= (K A). Along with (K2) we get (K A) =K.

For the second part we can define the function as in equation 3.1. We only need to showthat this function is a contraction.

(K1) Suppose K A for some L. By compactness of there exist 1, 2, . . ,nsuch that (1 2 .. n) . To show K A we have to show that Kand eitherA < A or A. SupposeA, thenA i > Afor alli. Hence we haveA (1 2 .. n)> Aas by (EE2)A (1 2 .. n) = (A 1) .. (A n)

which is equal to some A k (because of (EE3)) and for which A k > A. Now(EE2) also gives A (1 2 .. n) A which with (EE1) givesA < A .So for n 1 above proof works. If n = 0 then hence A and by (EE2),CA for all C L. A and (EE5) give C > A for some C. (EE1) now givesA > A.

(K2) Follows from the definition.

(K3) As A /K, K=K. Hence (EE4) gives for all B L, B A. Now if K then(EE4) gives a Csuch that C or > CA. Using (EE1) and (EE2) we haveA > Aor A > A.

(K4) A A hence A= A which means that A A > A is not possible.

(K5) Let K, using inverse of DT we have to show that A K A. By theequation 3.1 we need to prove that eitherA < A (A) or A. But Kso doesA . Let us assume thatA and show A < A (A ). Since A (A ),by (EE2) we have CA (A ) for all C L. A not being a tautology satisfiesA < C for some C by (EE5). Thus by (EE1) we get A < A (A ).

(K6) If A B then by (EE2) we haveA = B andA C=B C. ThusA < A CB < B C. The equality ofK AandK B now follows from the definition.

(K7) We have to show that K A K B K (A B). Let K AandK B

i.e. A < A and B < B . Thus by (EE2), (EE1) we get A B < A andA B < B . Now (EE3) gives that (A ) (B ) = (A ) or (B ). Using(EE1), A B A, hence A > A and we have K A.

3.2 Partial Meet Contraction

Partial meet contraction is very intuitive way of defining contraction functions. At the firststep a contractionK Awith a reasonableA must not containA. So we can start thinking


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3.2. PARTIAL MEET CONTRACTION 17

of beliefs which dont entail A or maximal subset m of K which fail to imply A. As aformalization think of these subsets as m satisfying following three properties-

(i) m K

(ii) A /C n(m)

(iii) For any m such that m m K, A C n(m)

The definition gives very much what we desire, like m is closed and addition of anythingin K\m to m gives K(proofs later). Surprisingly these subsets also satisfy K(1 6),[12, pp 6263] calls them maxichoice contractions. Here we can implement our basic rulethat logical inferences alone are insufficient for structuring the belief change i.e. there hasto be a preference some where. We will see how further restrictions lead us to contractionfunctions.

Let us denote all maximal subsets of K which fail to imply A in K by KA and

M(K) =AKKA. Note that if A / K then we take KA = {K}. For contractionK A, instead of somem inKAwe will choose some preferred elements ofKAand taketheir intersection. Let Sdenotes the selection function for choosing the preferred elementsin KA. Following definition may lead us to our contraction function.

K A=

S(KA) ifA

K otherwise (3.3)

where, S(KA) KA

If function above is a contraction then it is called Partial Meet Contraction. Differentdefinitions of selection function Scan lead us to different functions. But we will see that

the most appropriate Sis when it chooses the top elements in KA based on a reflexiveand transitive preference relation on all maximal sets M(K) ofK.

S(KA) ={m KA: m m, m KA} (3.4)

Following results will help us in proving the theorem 5.3.1 which tells that partial meetcontraction function defined above is indeed a contraction.

Lemma 3.2.1 Let m is inKAfor some A in K, then m is closed andC n(m {}) =Kwhere K\ m.

Proof Let m and assume / m. Now m implies K as m K and K isclosed. Hencem m {} K, which implies m {} A or m A. But thenm A (as m ) gives contradiction.For this we already have C n(m {}) Kand only KC n(m {}) has to be proved.Let K\m, we show that A m and hence any set containing A and m(inparticularC n(m {})) will imply . By contradiction suppose A /m which meansm m {A } K as (A ) and K. Thus (A ) A m. But((A ) A) A gives that A m, a contradiction.

Second part of above result is very crucial. It says that all the beliefs in K outside mbehave similarly w.r.t. m. Suppose m is inKAfor some A, andB K\ mthen additionof some in K \ mto m will give all ofKand in particularB . Thus one can check thatmwill also satisfy all the three conditions for the membership ofKB. Here is the formalizedresult-


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Lemma 3.2.2 LetA K andm KA thenB K\ m, m KB.

Using this result we compare K(A B), KA and KB. One can easily see that if

m K(A B) then it is in KA or KB because the second condition A B / mimplies A / m or B / m. What if m is in KA alone? The first and second conditionfor membership ofKA B are satisfied. Now our previous result implies third condition.Suppose m m K, then in addition to A Cn(m), C n(m) =Kis also there implyingthat B C n(m) (as B K). Hence A B is inC n(m). We present this result and thenmove on to our main theorem.

Lemma 3.2.3 Let A and B are in K thenK(A B)KA KB.

Theorem 3.2.1 Let be a reflexive and transitive relation on M(K). Then functiondefined through 3.3 withSas in 3.4, satisfies (K1)-(K8).

Proof

(K1) This holds because each maximal subset m in S(KA) is closed and intersectionof closed sets is closed.



(K4) Suppose A K A. This can only happen only under the alternate definition of3.3 whenK= K Aand A.

(K5) Let K. Assume A so that first definition of 3.3 is used, we show that(A ) K A. Using second part of the lemma 3.2.1 with = A we getm KA,Cn(m {A}) = K. ThusA m, K implying that A K A=

S(KA). Hence (K A) + A.

(K6) Use,A B implies KA= KB.

(K7) IfAor B then obviously this holds, ifAthenKA= Kand KB = K(AB)(A B B). So let us assume A and B . We have to show (

S(KA))

(

S(KB)) S(K(A B)) . But S(KA KB) = S(KA) S(KB)

S(KA) S(KB). Hence (S(KA KB)) (S(KA) S(KB)). Now byproperty of sets (S(KA)S(KB)) = (S(KA))(S(KB)) and by lemma5.3.3 we get (

S(KA)) (

S(KB)) =

(S(KA) S(KB))

(S(KA

KB)) =

S(KA B).

(K8) Let A / K(A B). Thus there exists an m K(A B) such that m m, m K(A B) and A / m. Note that m KA as A / m. Now weshow

S(K(A B))

S(KA) by showingS(KA) S(K(A B)). Ifm

S(KA) thenm m m, m KABimplying thatm m, m K(AB)due to transitivity of. Hencem S(K(AB)) givingS(KA) S(K(AB)).

In the next section we present another approach for constructing contraction function.


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3.3. SYSTEM OF SPHERES 19

3.3 System of Spheres

This method was employed by grove [9]. All the notations are borrowed from [8, section

4.5]. In this method we consider maximal consistent sets which are same as models as ourbasic units. A set M L is called maximal consistent if -

(i) Mis consistent i.e. B L s.t. B and B M.

(ii) If / M then M {} is inconsistent.

Let for a set of beliefs K, [K] denotes all maximal consistent extensions of K. Where aset of beliefs Mbeing maximal means . We use M to denote all maximal consistent sets.Thus-

[K] ={M M: KM} (3.5)

For defining a contraction K AofKw.r.t. A, a system of spheres S is considered. Each

element ofSis a subset ofM i.e. it contains some maximal consistent sets. In addition Ssatisfies following properties-

(i) Sis totally ordered by. That is for any S, S S either S S or S S.

(ii) [K] is the minimum ofS i.e. ifS S then [K] S.

(iii) M S.

(iv) If A is a belief and there is any sphere in S intersecting [A] then there is a smallestsphere inS intersecting [A]. We will use SA for this smallest sphere.

Given the definition of S we can define KA as those formulae of K that are near to

consequences ofA. Formally-

K A=

([K] C(A)), ifA

K otherwise where C(B) = [B] SB (3.6)

Simplifying the definition we get-

K A= (

[K]) (

C(A)) =K (

C(A)) (3.7)

as K =

[K]. We can see using Harper-Identity that

C(A) is nothing but revision ofKw.r.t. A. Before moving on to our main result we prove some small results.

Lemma 3.3.1 IfA B then [A] [B] andSB SA.

Proof Let A B then ifM[A] thenB Mimplying thatM[B]. Also SA [A] isnon empty implies SA [B] is not empty which further implies SB SA by virtue of (iv)condition onS.

Lemma 3.3.2 [A B] = [A] [B]

Proof A, B (A B) implies [A] [B] [A B] using the previous result. For theother way we will use following property of maximal consistent sets -

A B M then eitherA M orB M

If M [A B ] then either A M or B M, implying M [A] or M [B] i.e.M[A] [B].


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Theorem 3.3.1 Let K-A be as in 3.7, then it satisfiesK(1 8).

Proof

(K1) Use closeness of maximal consistent sets.

(K 2) Follows from the definition.

(K3) SupposeA /K, thusK {A}is consistent (if not then K {A}= K implyingA Kor A K). Thus there is a maximal consistent extension ofK{A} whichis in [K] and [A] hence S[A] = [K], implyingK

(C(A)). Thus K A= K.

(K4) LetA. Thus [A] is not empty. Hence A /C(A), implyingA /K A.

(K5) Let K, we prove that A KA. Now K implies A K.

Also if M [A] then A M as A (A ) and A M. ThusA C(A) implying A K A.

(K6) Use,A B implies [A][B].

(K7) We have to show K(

C(A)) (

C(B)) = K(

(C(A) C(B))) K(

(C((A B)))). We will show that C((A B)) C(A) C(B). So,

C((AB)) =S(AB)[(AB)] =SAB[AB] =SAB([A][B]) =(SAB [A]) (SAB [B]). But then A, B (A B) and previousresults imply (SAB [A]) (SAB [B]) (SA [A]) (SB [B]) =C(A) C(B).

(K

8) LetA /K(AB) =K(C(AB)), we have to show K(C(AB))K (

C(A)). Assume A K, then there is an M C((A B)) = S(AB)

[(A B)] such that A /M. But thenM [(A B)] = [A] [B] and A / MimpliesM[A]. ThusS(AB) [A] is non-empty, implying SA S(AB). AlsoA (A B) impliesSAB SA. Thus we get SA= S(AB)which along with[A] [(AB)] gives C(A) =SA[A] =S(AB)[A] S(AB)[(AB)] =C((A B)). Hence we get

C((A B))

C(A).

3.4 Implementation of Belief Revision

This section describes how a computer implementation of a contraction or revision func-tion can be achieved. With every implementation there is always an overhead of provingtheorems efficiently. Even if one ignores that, from computational point of view, it seemsimpossible to deal directly with belief sets. For implementation, methods of partial meetcontraction and spheres are completely out of question. These methods use closed setsas their basic units which are simply too big. Epistemic entrenchment (EE) seems to beplausible choice because of explicit definition of a contraction. But then difficulty arise asordering requires ranks on typically infinite number of sentences. Also EE ordering is lostin process of change, and as a consequence the iteration of change functions is not naturallysupported. In this chapter we describe a computational model which uses finite partialentrenchment(FPE) ranking [1, chapter 15], [13]. This ranking is essentially a finite repre-sentation of EE ranking. After describing FPE we give a PROLOG based implementationof contraction functions.


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3.4. IMPLEMENTATION OF BELIEF REVISION 21

3.4.1 Finite Partial Entrenchment Ranking

FPE grades the content of a finite knowledge base according to its epistemic importance.

Formally this ranking maps a finite set of sentences to rational numbers. The higher thevalue assigned to a sentence the more firmly held it is, or the more entrenched it is.

Definition 3.4.1 A FPE ranking is a function B from a finite subset of sentences intothe interval [0, 1] such that the following conditions are satisfied for all domain(B) =dom(B):(i) { dom(B) :B()< B()} if is not a tautology(ii) If , thenB() = 0(iii) B() = 1 if and only if

Here 1st and 2nd conditions only put restrictions on tautologies and contradictory propo-sitions. The first condition says that a sentence assigned higher value than an arbitrary

sentence, do not entail . The sentences mapped to numbers greater than zero representthe explicit beliefs, and their logical closure represents its implicit beliefs-

Definition 3.4.2 Let B denotes family of all FPE rankings. Then for some B in B,exp(B) ={ dom(B) :B()> 0} andcontent(B) =C n(exp(B)).

Let be a nontautological sentence and B be a finite partial entrenchment ranking thendegree of acceptance of is-

degree(B, ) =

largest j such that{ exp(B) :B() j} if content(B)0 otherwise

Theorem below, tells how a FPE ranking can generate an EE ordering using degree ofacceptance. The proof is straight forward.

Theorem 3.4.1 Let B B and , be propositions. Define B by B iff , ordegree(B, ) degree(B, ). ThenB is an EE ordering related to content(B).

Thus we have solved the problem of defining ordering on an infinite set by defining it onits finite representation. But how one can design the contraction or revision using all this isyet be seen. Actually instead of only defining contractions we will setup a general functionwhich revises the degree of given proposition . So by changing degree to 0 we can employcontraction and in general we can move sentences up or down as the preference changes.For this adjustments are defined, which take a FPE B a proposition , a rational in [0, 1]and give a new FPE in which degree() is i.

Definition 3.4.3 The adjustment of a finite partial entrenchment rankingB is a function such that

B (, i) =

B(, i) if i degree(B, )(B(, 0))+(, i) otherwise

where

B(, i)() =

i if degree(B, ) =degree(B, )and B()> iB() otherwise

for all dom(B)

B+(, i)() = B() if B()> ii if orB() i < degree(B, )degree(B, ) otherwise


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for all dom(B) {}

With this definition we try to implement adjustment function in PROLOG.

3.4.2 A PROLOG Implementation

For calculating degree we use the function below

FUNCTION degree(B, a)

degree(B,a) max(B)+1if dom(B) then bottom min(B)else bottom min(B)-1do degree(B, a) degree(B,a)-1until degree(B,a) = bottom or {b dom(B):B(b) degree(B,a)} a

Then the adjustment function is defined as in the definition. We achieve negation ()of a proposition in PROLOG by failure. Function not(X) fails ifXis success

not(not(X)):- X.not(X):- X,!,fail;true

For OR and AND ; and, are used respectively. Implication is achieved by the definitionsymbol : , suppose we want to include a b as our belief then b : a is loaded in thedatabase. So wheneverb is searched for success and it reaches this proposition, the successofa is checked. Ifa is a success then b also becomes success. For adjustment function weconsulted function definition given on page 204 of [1]. Here is the PROLOG code for mainfunctions

adjust(B,A,I):-

length(B,N1),Len1 is N1*1.0,

enterAll(B),removeAll(B),

findplace(B,Len1,A,I,Ranki,B1),

length(B1,N),Len is N*1.0,

rank(B1,A,Ranka),

newB(B1,Ranka*2,Degreea),

(Degreea>=I,!,

movedown(B1,A,Ranka,Ranki,NewB)

;

moveup(B1,A,Ranka,Ranki,NewB)

),

write(NewB).

movedown(B,A,Ranka,Ranki,NewB):-

removeA(B,Ranka*2-1,A,NewB1),addA(NewB1,Ranki*2-1,A,NewB2),

forD(B,NewB2,0.0,A,Ranka,Ranki,NewB).

moveup(B,A,Ranka,Ranki,NewB):-

rank(B,not(A),RankNa),

lengthExp(B,Lenexp),

length(B,Len),

(RankNa=


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4.2. A SEQUENT CALCULUS 25

(1) p q p(2) {p q, q r} (p r)(3) p (p q) q

(4) p p(5) p p(6) p (p q)(7) p p(8) p palong with the deduction theorem

If {} then ( )or

Similar to classical logic, three-valued logic is compact. The definition of consistency issame i.e. a consistent set mustnt entail a proposition and its negation (). The maximalconsistent sets are consistent and inclusion of any formula outside it makes it inconsistent.Similar to classical logic, every consistent set has a maximal consistent extension([5, page467]). Finally the usual completeness and soundness results hold here.

During the study of three-valued logic we thought of derivability in detail. We also lookedfor other possible axiomatizations. In the end we were able to derive a sequent calculusfor this logic. Following sections will presents the sequent calculus. For proving theoremsin this logic in general we generalized automated deduction methods for the classical caseand implemented them. Many theorems in later sections are proved using these automatedmethods. In the following sections we will also present the automated proof methods forthree-valued logic.

4.2 A Sequent Calculus

In [10, Section 3.3] gentzen sequents were used for presenting a sequent calculus. For thislogic we consider sequents of a different type. Sequents will be of the form

;

with , , finite. Premises(assumptions) are of two types i.e. formulae in are differentfrom those in . One can think of as of lesser credibility than . The set of conclusions however has only one status. Below we present the rule for our sequent calculus.

Axiom Rule:

, ; , ,

Logical Rules:

, , ;

, ; (L)

; , , ,

; , (R)

; , , ; , , ; , ,

; , (M)

;

, ;

; , (L) ;

,

, ; , (R)


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26 CHAPTER 4. THREE-VALUED LOGIC AND BELIEF REVISION

; , , , ; , , ; ,

; , (M)

; , ; ,

, ; (L)

, ; , , , ; , ,

; , (R)

; , , ; , , , ; , , ; ,

; , (M)

; ,

, ; (L)

; ,

; , (M)

, ;

; , (R)

4.2.1 Truth of a Sequent

Having defined the sequents we need to give semantics associated with them. Let ; with , & finite, be a sequent andv be a valuation. Then ; is said to be trueunder v and written v |= ; if and only if there is a wff , or such thatv() = 0, 1/2 or 1 respectively. A valid sequent is true under all valuations and is denotedas |= ; .

4.2.2 Provability of a Sequent

A sequent ; is provable if there exists a derivation where ; is root and

instances of Axiom Rule are the leaves. Root is obtained by gradual application of sequentcalculus rules on leaves. A provable sequent is written as ; .

4.2.3 Equivalence

Soundness:If; then |= ; .Proof Since the sequent is provable, the derivation using the rules is finite and all thebranches start from axiom rule. Now any valuation v can assign only three values{0,1/2,1}to the repeating formula in the instance of axiom rule and hence according to our def-inition the sequent is true under all valuations. Now all other sequent calculus rules areconstructed in a manner that, premise sequent is true under a valuation if and only if theconclusion is true under it. Thus using the derivation we can conclude that our provablesequent is also true under all valuations.

Completeness:If |= ; then; .ProofLetSbe the set of sequents s.t. = and fbe the strategy used for deriva-tion i.e. ftakes a sequent inSto a non-proposition in . Now given the sequent; build the derivation tree Twhose root is this sequent and nodes are formedusing the rules in sequent calculus. Formally if ; /S then it is a leaf otherwisechildren of ; are the premises of the rule corresponding to = f(; ).Since , and are finite T is finite. So we only need to prove that leaves of T arenothing but instances of axiom rule. On the contrary let us assume that there is a leaf1;

1

1 such that 1

1

1 = . As rules are defined for all possible occur-rences of a given type of formula this leaf can only have propositional letters. Now all


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For

+( ) ( ) ( )+ + + +

For

+( ) ( ) ( ) + + + + +

4.3.2 Tertiary Decision Diagrams

Tertiary Decision diagram of a wffis a tree in which each vertex is a propositional variableand has three down going edges each for three possible valuations. We redefine the methods

A

T

T

F

B

U T

A=F

A=U

A=T

B=T

B=UB=F

Figure 4.1: TDD of (p q), U is for truth value 1/2

in [10, Chapter 2, page 63] to get TDD. We have a recursive function Convert(p) whichtakes a wff p and returns its TDD. Convert(p) uses four other routines

(1) TDDneg(TDDp)- If TDDp is TDD of p then this function returns TDD ofp

(2) TDDor(TDDp,TDDq) - This returns TDD ofp q

(3) TDDand(TDDp,TDDq) - This return TDD ofp q

(4) TDDimp(TDDp,TDDq) - This returns TDD ofp q

(5) TDDmake(p, TDD1, TDD2, TDD3) - This function returns a TDD whose top node isp and other three TDDs are its children. This function additionally checks that allthe three TDDs are not equal, if they come out equal then it returns one of the TDDsdiscarding p.

Each of the first four routines use TDDmake to join two trees. For example TDDordoes the following:

TDDor(T,) =T, TDDor(F, ) TDD, TDDor(U,U)=U


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4.5 Interrelations

Changes in epistemic states modelled as expansion, revision and contraction are interrelated.

Revision of a belief set K with respect to a sentence A can be seen as removal of anyincompatible information A and then addition ofA along with all logical consequences.In 3valued logic the Levi-identity takes the following form-

K A= (K A) + A (4.3)

Let us call the revision function aboveR(). Following theorem tells that functionR()defined through Levi - identity is indeed a revision function.

Theorem 4.5.1 If a contraction function satisfies (K1) to (K6) thenR()satisfies(K1) to (K6). Furthermore if also satisfies (K7) and (K8) then R() satisfies

(K7) and (K8) respectively.

Proof

(K1) From definition.

(K2) A A hence (K A) + A A.

(K3) From (K2), K A Kand hence (K A) + A K+ A.

(K4) LetA /K. Using (K3), K A= Kand hence K A + A= K+ A.

(K5) if Let A. ThusA, A (K A) + A. But thenK A= (K A) + A= Kas B L, {A, A} B.only if Let (K A) + A = K. Thus in particular (K A) + A A. UsingD.T. we get K A (A A) which with the theorem (A A) A givesK A A.

(K6) Use A B if and only ifA B and (K6).

(K7) Let K A B =K (A B) + (A B). Thus (A B) K (A B) = K (A B) as De Morgan holds for , i.e. (A B) (A B).Also (A B) (A B) A. So we will show that (A B)

K (A B) which by (K

7) will give (A B) K (A B) K(A B) K ((A B) (A B)) =K (A). From where we canget (K A + A B) = (K A + A) + B.(AB) K (A B) can be shown by using (K5), (A B) K (A B) KK (A B) + (A B). Applying (D.T.) ((A B) ((A B) )) K (A B). But ((A B) ((A B))) ((A B) ). Hence we get (A B) K (A B).

(K8) Assume B /K A + A and let (K A + A) + B. From the assumptionwe get (A B) /K A. Also (A B) K A= K (A B) Abecause A ((A B)A). Using (K8) withA = (A B) andB =Awe get K A K (A B). Hence we get K (A B) + (A B) andour proof is complete as (A B) (A B).


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4.6. EPISTEMIC ENTRENCHMENT 31

Similar to Levi identity, Harper Identity takes the following form-

K A= K K (A) (4.4)

Let us call revision function defined above C().

Theorem 4.5.2 If a revision function * satisfies (K1) t o ( K6) then C(*) satisfies(K1) to ( K6). Furthermore, if * also satisfies (K7) and (K8) then C(*) satisfies(K7) and (K8) respectively.

Proof

(K1) Intersection of two belief sets is also a belief set.

(K2) Follows from definition.

(K3) A /K implies A /K as A A. Using (K4) we get K A= K+ A

which means K K A= K.

(K4) LetA KA= KKA. Thus bothAand A KAgivingKA= K.But then (K5) gives A.

(K5) Let K, we need to show (KK A) +A. (A ) impliesA Kwhich implies (A ) K. So our proof will be complete after showing(A ) K A. But that is fine as A (A ) andA K Abecauseof (K2).

(K6) Use A B iffA B and (K6).

(K7) Following results will be assumed during the proof:(i) (A B) (A B)(ii) ((A B) A) A.Now K A K B means K, K A, K B. TakingA = (A B)and B = (A) and using (K7) we get K ((A B) A) = K A K (A B) + A. Similarly we can get K B K (A B) + B. UsingD.T., A , B K (A B). Now C() already satisfies (K5) so K(K K (A B)) + A B. And we have (A B) K (A B).But {(A B ), (A ), (B )} because = 0 or 1/2 demandsthat one of the premises is not equal to 1 and hence K (A B). Thus K K (A B) =K (A B).

(K8) LetA /K (A B). Which meansA /K or A /K (A B). IfA /K thenK (A B) K=K A. So consider the case when A KbutA /K (A B).Taking A = A B, B = A and using (K8) we have K (A B) + A K (A B) A= K A. ButK (A B) K (A B) + Aand we haveK (A B) =K K (A B) K K A= K A.

4.6 Epistemic Entrenchment

In 3-valued logic the epistemic entrenchment doesnt change significantly. Epistemic en-trenchment is mostly independent of nature of the inference relation. It only sees it as apreference relation over the language. We will see that the earlier definition of contractionin terms of Epistemic entrenchment works here.


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Theorem 4.6.1 Let be a contraction function and letKis a belief set. Then the ordering defined in 3.2 is an epistemic entrenchment ranking.

ProofFollows line by line similar to the classical case.

Theorem 4.6.2 Let K be a belief set. For every contraction function ofKthere is an epis-temic entrenchment(related to K) such that 3.1 is true. Conversely for every epistemicentrenchment related to K, there is a contraction function satisfying 3.1.

Proof For the first part theorem 4.6.1 gives the required epistemic entrenchment ranking.We only need to show that 3.1 holds for this ranking.If A then {B K : A < A B} = {B K : A / KA (A B) = KAand (A B) KA}. We will show that if B K then the constraint (A / KAand (A B) K A) on the above set is same as (A / K A and B K A) which

is same as KA.If B KA then obviously (A B) KA. For the other way (B B) and B K implies B K and K A = K A. Using (K5) asK(KA)+Aimplies (A B) KA. Now (AB) (AB)implies (A B) K A. But then {(A B), (A B)} B as forB = 0, 1/2 B = 0 implying that both the premises cant be 1 and hence B K A.If A then A KA. That is (KA) +A = KA. Using (K5) we haveK(K A) + A= (K A). Along with (K2) we get (K A) =K.

For the second part we can define the function as in equation 3.1. We only need to showthat this function is a contraction. All the postulates go through easily except the ( K5).Below is the proof-

(K

5) Let K, we will show that (A ) K A. By the equation 3.1 we need toprove that either A < A (A ) or A. Note that K so is (A ). Let usassume that A and show A < A (A ). Since A (A ), by (EE2) wehave C A (A ) for all C L. A not being a tautology satisfies A < C forsome C by (EE5). Thus by (EE1) we get A < A (A ).


As in the classical case this method also leads us to contraction functions. Retractionof a belief can be modelled through maximal subsets of K which fail to imply K. Frompractical view point that is very much expected because addition of beliefs with possibleconnotation wont make any differences as far as belief change is concerned. With theassumed background we move on to proving theorem similar to 5.3.1. But before that wemake a small comment on why the lemma 3.2.1 should hold here. In the proof of lemma3.2.1 we make use of modus ponens just after deduction theorem, here similar result canbe obtained with on which modus ponens works. Also in the second part we use modusponens on ((A ) A) A, which works here because following weaker result holds((A ) A) A.

Theorem 4.7.1 Let be a reflexive and transitive relation on M(K). Then functiondefined through equation 3.3 withSas in equation 3.4, satisfies (K1)-(K8).

Proof


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4.8. SYSTEM OF SPHERES 33

(K1) We show that each m KA is closed i.e. it is a belief set. And we already knowthat intersection of closed sets is closed. Supposem now suppose / m thusm m {} K. Hence m {} A or m ( A), but then m A as m

giving contradiction.



(K4) Suppose A K A. This can only happen when K=K A and A.

(K5) Let K. We show that (A ) K A, suppose not then there is a m S(KA) such that (A ) / m. Thusm m {(A )} K. Which implies((A ) A) m, but then ((A ) A) A further implies A m leadingto contradiction. So (A ) K Aand hence K A + A.

(K6) Use, A B implies KA= KB.

(K7) This proof goes similar to the classical case as in proof of theorem because (A B) B holds here.

(K8) Here too we dont use anything new, hence classical proof goes through.

4.8 System of Spheres

System of spheres derive contraction by maximal consistent sets. Since three-valued logicdistinguishes betweenAand Awe have to be careful of definition of consistent sets. We

see that p p but converse is not true. If we define consistency ofK L as

B L such thatB andB K

then there is no difference. One can prove that this definition is equivalent to classicaldefinition of consistency. If we look at maximal consistent set M then also they satisfymost properties of classical maximal consistent sets. Like ifM then M. Hence wedont need to alter the definition of maximal consistent sets. We need to change definitionof contraction. Since{p, p}is inconsistent, its sufficient to take beliefs in K Awhich arenear to consequences ofA. The following modification as expected serves the purpose-

K A= ([K]) (C(A)) =K (C(A)) (4.5)Before proving that above function is a contraction, we prove that maximal consistent setssatisfy the property-

A B M then eitherA M orB M

Proof Let A B M but A, B / M. Hence M{A} and M{B} are inconsistent. ThusA B M. But (A B) (AB) implies (A B) M implying M isinconsistent, a contradiction. Besides this results concerningSA and [A] also hold. One caneasily show that ifA B thenSB SA and [A] [B]. Here comes our final theorem-

Theorem 4.8.1 LetK A be as in 4.5, then it satisfiesK(1 8).


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Proof

(K1) Since maximal consistent sets are closed here, there intersection is also closed.


(K3) SupposeA /K, thusK{A} is consistent (if not thenK{A}= K, implyingA KorA K(as A A)). Thus there is a maximal consistent extensionofK {A} which is in [K] and [A] hence S[A] = [K], implyingK

(C(A)).

Thus K A= K.

(K4) LetA. Thus [A] is not empty. Hence A /C(A), implyingA /K A.

(K5) Let K, we prove that A KA. Now K implies A K.Also if M [A] then A M as A (A ) and A M. Thus

A C(A) implyingA K A.

(K6) Use [A] = [B] as A is same as (A B).

(K7) The classical proof goes through in our 3-valued context as [A B] = [A] [B] and (A B) (A B).

(K8) The classical proof goes through here too.

4.9 Interpretation and Applicability

In earlier sections we noticed how consistency and closure of belief sets have helped us inmodifying the theory of belief change so that it gets a sleek structure in three-valued logic.In this section we discuss the interpretations of the obtained results. First of all we realizethat A is the central agent around which all the changes happened. In classical logicA and A were same meaning belief in A does not happen. But here since they are notsame we have to find new interpretations. We know that A is stronger than A. Sointerpretation ofA remains the same. But what aboutA? We guess the interpretationofA as possibly A doesnt hold. If we assume this interpretation then the immediatequestion is why {A, A} is inconsistent. The answer is, if A is in my belief set then ibelieve thatA must happen. Now in such a situation showing a positive confidence in Aby means ofA is clearly unacceptable. Its like ignoring ones beliefs. The existence ofpossiblity notion in the form ofA motivated us to look for alternative axiomatizationof 3-valued logic. In [4] and [2] we find close relationships between S5 which has modalitiesand three-valued logic. Following theorem from [4]([2]) tells us that interpretation ofAiscorrect. And in general we can give possibility connotation to any belief.

Theorem 4.9.1 A Wajsberg algebra< X, , , 1> is equivalent to a 3-valued Lukasiewiczalgebra< A, , , , , M, 0, 1> via the transformation.a b (Ma b) (M b a)a b (a b) ba B (a b)M a a a0 1


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4.10. GENERALIZATION TON-VALUED LOGICS 35

This theorem is interesting because algebra is in terms of a new connective M. In this alge-bra ourAtakes the formMAwhich means possibly A does not happen, if interpretationofMis taken as it is possible that. Our next theorem which is borrowed from [2] ensures

that M behaves exactly like possibility connective.

Theorem 4.9.2 3-valued Lukasiewicz logic is embeddable into S5.

Embeddable means for every wff in 3-valued logic there is a translation of inS5 such that is a theorem in 3-valued logic if and only if

is a theorem in S5. Theembedding maps the connective Mof Lukasiewicz algebra to possibility connective M ofS5. Hence our guess is perfect and our beliefs can be made vague by saying M Ainstead ofA.

After knowing the interpretation there are numerous uses of this kind of belief change.In fact, since we dont loose anything as far as belief change in classical background is con-cerned, one should think of it as fundamental. In day-today life we actually have possibility

notions in our beliefs.

4.10 Generalization to n-valued Logics

The belief change in 3-valued logics can be extended ton-valued logics. n-valued logics takethe n truth values Ln= {0, 1/(n 1),..., (n 2)/(n 1), 1}. Interpretation of connectivesis same as that of 3-valued logic. In most of the 3-valued logic proofs, we noticed theadjustments in classical proofs by replacing by and by. We also observed thatmodus ponens with produces same results as with . So we expect our 3-valued resultsto generalize for n-valued logics where we take A A (A (A ...(A

n2 times A))..)

withA B A (A (A ...(A n1 times

B))..). Following result will help us in embedding

classical logic into any n-valued logic. Since embedding is in terms of and , it is veryuseful in generalization.

Lemma 4.10.1 Suppose is a formula inL and is formed by replacing byandby. Also supposenth lift of a classical valuationv2 is a valuationvn

based onLn suchthatv2(p) vn

(p) v2(p) + (n 2)/(n 1), for each propositional variable - p. Then forany,

v2() vn() v2() + (n 2)/(n 1) (4.6)

ProofWe proceed by induction on length ||of. When||= 1 then is a propositionalvariable, for which inequality holds by assumption. Suppose inequality holds for any suchthat || m. We prove that inequality holds for||= m + 1:

(i) . Ifv2() = 0 then we only need to show vn() (n 2)/(n 1). v2() =

1 v2() = 1 and || = m gives 1 vn() v2() + (n 2)/(n 1) that isvn

() = 1. Hence vn() =vn

() = 0 (n 2)/(n 1).

Ifv2() = 1 then we show that vn() = 1. v2() = 1 1 = 0, thus 0 vn

()(n 2)/(n 1). Hence vn() =vn() = 1.

(ii) . If v2( ) = 0 then v2() = 1 and v2() = 0. Using induction wehave vn

() = 1 and vn() (n 2)/(n 1). Since vn(b c) =vn(c), whenever

vn(b) = 1, we get vn() =vn( ) =vn() (n 2)/(n 1).


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If v2( ) = 1 then if v2() = 0 then 0 vn() (n 2)/(n 1). Butvn(b c) = 1 ifvn(b)< 1, thus vn

() = 1. Ifv2() = 1 then v2() = 1 and hencevn

() = 1 and vn() = 1. Since vn(b) = 1 implies vn(b c) = vn(c), we have

vn() =vn() = 1.

(iii) or . Follows easily from the definitionsv( ) =max(v(), v()) andv( ) =min(v(), v()).

Theorem 4.10.1 is a theorem inn-valued logic if and only if is a theorem in classicallogic.

ProofWe use completeness of any n-valued logic. Some formula is a theorem if it is 1 onall valuations. Only if part follows immediately because any theorem in n-valued logic isa classical theorem: any classical valuation(a valuation on L2) is also a valuation on Ln,

and hence if something is true on any valuation on Ln

then it is true on any valuation onL2. Thus is a theorem in classical logic, where {, }are equivalent to {, }. Hence

and thus is a theorem.For the if part we show that is 1 on all valuations vn based on Ln. Supposevn is

some valuation based on Ln. Then there exists a classical valuation v2 whose nth lift is

vn, that is v2(p) vn(p) v2(p) + (n 2)/(n 1) for all propositional variables p. Nowusing previous lemma we have v2()vn() (n 2/(n 1)). Sincev2() = 1 we getvn(

) = 1 and hence is true on all valuations vn based on Ln, that is to say is a

theorem in n-valued logic. Now all the results obtained in 3-valued logic can be generalized to n valued logic.

Thus if we do belief changes withAinstead ofAthen we get all the belief change results

in anyn-valued logic. Making comparisons with existing approaches like [7] where belnapslogic is used might be interesting.


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Chapter 5

Rough Belief Revision

In this chapter we will see how rough logic can be used to do belief change. In 3-valued

logic we were able to provide room for vagueness by realizing the notion of possibility ofa belief. There a beliefA can have a weaker assertion M(A). Here in rough logic we dontintroduce new assertions of a belief, instead we make them weaker by taking incompleteevidences as beliefs. So a belief may fail as its truth was dependent upon some incompleteinformation. Most definitions and initial results are borrowed from [3].

5.1 Rough Truth

The notion of rough truth was introduced in [15] as a part of the first formal proposal onreasoning with rough sets. It was proposed to reflect inductive truth, i.e. truth relative toour present state of knowledge, and one that, with gain of knowledge, leads to total, deduc-tive truth. This sense of gradualness finds an expression in, possibly, the only qualitativeversion of approximate orsoft truth, as opposed to other quantitative definitions foundin, e.g., probabilistic, multi-valued or fuzzy logics.

It has generally been accepted that the propositional aspects of rough set theory areadequately expressed by the modal system S5. AnS5(Kripke) model (X ,R,) is essentiallyan approximation space (X, R) (X is some set with equivalence relation R on it) whereX=, with the function interpreting every well-formed formula(wff) ofS5 as a rough setin (X, R). IfL, Mdenote the necessity and possibility connectives respectively, a modal wffL(M ), representing definitely (possibly) , is interpreted by as the lower (upper)

approximation() {x X : all x

related to x are in (), i.e. if xRx

then x

()}(() (()c)c) of the set (). In general for a subset A ofXwe can define lower(A)

and upper(A) of as:

A=

{[x] : x X, [x] A}A=

{[x] : x X, [x] A=}

where [x] is equivalence class of x

Now with all the above background, a wff may be termed roughly true in M (X ,R,), if() = X. The notion of truth can be extended torough validity which canbe further used to define rough consequenceand rough (in)consistencey. The language ofrough logic LR is that of a normal modal propositional logic. In the following is any setof wffs, , any wffs ofLR.

37


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38 CHAPTER 5. ROUGH BELIEF REVISION

5.1.1 The semantics

Definition 5.1.1 AnS5-modelM (X,R,) is arough model of, if and only if every

member of is roughly true inM, i.e. () =X.

Definition 5.1.2 is a rough semantic consequence of (denoted|) if and only ifevery rough model of is a rough model of. If is empty, is said to beroughly valid,written |.

We present now the logic ofLRwhich will serve as the base logic for rough belief change.As already said the language is of a normal modal logic and let S5 denotes the derivabilityrelation inS5. We consider two rules of inference:

R1. R2. M M

whereS5M M M M

The consequence relation defining the system LR is given as follows.

Definition 5.1.3 is arough consequence of (denoted R) if and only if there is asequence1,...,n() such that eachi(i= 1,...,n) is either (i) a theorem ofS5, or (ii)a member of, or (iii) derived from some of1,...,i1 by R1 or R2.If is empty, is said to be a rough theorem, writtenR.

Some derived rules of inference:

DR1. DR2. (M )L

S5 (M )L

DR3. M DR4. M M M

DR5. M DR6.

M

DR7. M DR8. S5

M M M M

Some immediate results are -

Theorem 5.1.1 (Deduction) For any, , , if {} R thenR .

Theorem 5.1.2 (Soundness & completeness) R if and only if|.

Theorem 5.1.3 LR is paraconsistent.

We state the following without proof. One may remark though, that the proof ofTheorem 5.1.4 uses only S5 properties, and that of Theorem 5.1.6 uses the rule of inferenceR2 (in fact, DR3).


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5.2. ROUGH BELIEF CHANGE 39

Theorem 5.1.4 is roughly consistent if and only if it has a rough model.

Theorem 5.1.5 R if and only ifM ={M : } S5M

Theorem 5.1.6 If is not roughly consistent thenR for every wff.

Theorem 5.1.7 If {M } Rfor every wff, thenR.

Observation 5.1.1

The classical rules of Modus Ponens and Necessitation fail to be sound with respect to therough truth semantics. The rule{, } R is not sound either, butR2 is soundand suffices for our purpose.

Interestingly, the soundness result establishes that the converse of the deduction theorem

is not true e.g. R p Lp, but {p} R Lp, p being any propositional variable.However, the converse does go through, i.e. R implies {} R , ifis a modal wff, i.e. is of the formL or M , for some wff this is because ofDR2.

R (M)L. So there is no difference between the modal and non-modal wffs interms of the object-level implication. However, as just noted, {}RL in general indicating that the meta-level implicationR does make this distinction.

5.2 Rough belief change

In classical belief revision, the base language is assumed ([12]) to be closed under theBoolean operators of negation, conjunction, disjunction, and implication. The underly-ing consequence relation is supraclassical (includes classical consequence) and satisfies cut,deduction theorem, monotonicity and compactness.

In contrast, we consider the modal language ofLR as the base. We follow the classicalline for the rest of the definitions. A belief set is a set Kof wffs such that Cn(K) = K.For a pair (K, A), there is a unique belief set K A (K A) representing rough revision(contraction) ofKwith respect to A. The new belief set can be defined through a new setof eight basic postulates (that follow). The expansionK+ AofKby the wffA is the beliefsetC n(K {A}). It is expected that rough contraction/revision by two roughly equal [14]beliefs would lead to identical belief sets. To express this, we make use of the rough equality

connective [11] inS5: A B (LA LB) (M A M B).Postulates for rough revision

(K1) For any wff A and belief set K, K Ais a belief set.(K2) A K A.(K3) K A K+ A.(K4) If AK, then K+ A K A.(K5) If K Ais not roughly consistent, then RA.(K5) IfK A is roughly inconsistent, then RA.(K6) If RA B , thenK A= K B.(K7) K (M A M B) K A + B.(K8) If BK Athen K A + B K (M A M B).


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Postulates for rough contraction(K1) For any wff A and belief set K, K Ais a belief set.(K2) K A K.

(K3) If A K, thenK A= K.(K4) If A K A, thenRA.(K4) IfLA K A, then RA.(K5) K(K A) + A, ifA is of the form LB or M B for some wffB .(K6) If RA B , thenK A= K B.(K7) K A K B K (M A M B).(K8) If A K (M A M B) thenK (M A M B) K A.

The major consideration here is to preserve rough consistency during belief change. Theidea, expectedly, is that ifK+ A is roughly consistent, it could itself serve as K A. Letus notice the difference with the classical scenario: suppose K Cn({p}), p being anypropositional variable. ThenK+ p is roughly consistent, and so it is K p itself. But,classically, Kp K+ p. Since we also have the notion of rough inconsistency, there isthe option of avoiding such inconsistency during belief change. It is thus that there are twoversions of postulates involving consistency preservation.

K1 and K1 express the constraint of deductive closure. K2, 3 and K2, 3 are self-explanatory. AK implies consistency and hence rough consistency ofK+ A, so that, inview of the previous remarks, K4 is justifiable in the rough context. In K5, we stipulatethatK Ais generally roughly consistent, except in the case whenA is roughly valid, i.e.in no situation definitely A holds (though possibly A may hold). K4 again stipulatesthat, in general, A KA, except when A is possible in all situations. K4 couldappear more relevant: definitelyA may follow from our beliefs despite contraction by Aonly ifA is, in every situation, possible. The controversial recovery postulate K5 in [6] isadmitted here, only in the case of contraction with a definable/describable[14],[15] belief,i.e. A such that S5 A LA. The last two axioms express the relationship of changefunctions with respect to beliefs and their conjunctions. The failure of soundness of theclassical conjunction rule {A, B} R A B here, necessitates a modification in the AGMK7, K8, K7, K8.

Observation 5.2.1 K5 impliesK5 andK4 impliesK4.

The following interrelationships between rough contraction and revision are then ob-served, if the LeviandHarper identities[12] are used.

Theorem 5.2.1 Let the Levi identity give, i.e. K A (K A) + A,where the contraction function satisfies only K1 8 (K5 is not assumed). ThensatisfiesK1 8, thusK4 is also satisfied.

Proof

(K1 4) Follow easily.

(K5) SupposeK Ais not roughly consistent. By Theorem 5.1.6,K ARB , for any wffB inLR. In particular, K A(Cn((K A) {A})RM B, andK ARM B,for any wff B. By using deduction theorem and DR4, K A R A. Hence byK4,RA.


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5.2. ROUGH BELIEF CHANGE 41

(K6) This can be proved by observing that R A B if and only ifR A B, andby using K6.

(K

7) First note that K A = K M A, by Observation 5.1.1(c) and K

6 ( is aspecial case of). ThusK A= K ((M A M B) (M A M B)). *Secondly, because ofDR5, 6 and R2,K+ M A M B = C n(K {A, B}), for any K. **Let K M A M B, i.e. K (M A M B) {M A M B} R. By deductiontheorem,K (M A M B)RM A M B . SoM A M B K (M A M B) K K (M A M B) {M A M B}, the last by K5, as M A M Bis equivalent to a modal wff. It is then easy to see that K(M A M B) RM A M B . So M A M B K (M A M B) K (M A M B)K((M A M B) (M A M B)), by K7. Therefore, using *, K A R

(M A M B) . As (M A M B) is also equivalent to a modal wff, by converse ofdeduction theorem (cf. Observation 5.1.1(b)), K A {M A M B} R. By (**),K A {A, B}(=K A + B)R.

(K8) Suppose BK A (K A) + A, i.e. K A {A} RB. Thus K A {A} RM A M B. Let K A + B = K A + {A, B}. Using * and K

8, wehave K A= K ((M A M B) (M A M B)) =K (M A M B) (M AM B)) K ((M A M B)). Thus, by **, C n(K ((M A M B)) {A, B}) =K ((M A M B)) + M A M B = K (M A M B).

Theorem 5.2.2 Let be given by the Harper identity, i.e. K A K K A.(a) If the revision function satisfiesK14,K5 (K5will also be satisfied) andK68,

then satisfiesK1 8 (so K4 is satisfied).(b) If the revision function satisfiesK18, then satisfiesK13,K4 andK58.

Proof

(K1 3, 6) Follow easily.

((a), K4) As A K A (by K2) as well as A K A (assumption), K A isinconsistent, and hence roughly so. Thus by K5, RA. It follows that RA.

((b), K4) LA, A K A, by assumption and K2. So, using S5 properties, M KA S5 LA as well as M K A S5 LA, implying that K A is not roughly

consistent. By K5, RA. Thus RA.

(K5) LetA be LB, for some wffB . If K, it can be shown thatA K K A.By DR2, (K A) + A.

(K7) Using K6 we haveK A= K M A= K ((M A M B) M A). *By * and K7,KA= K((M AM B)M A) K(M AM B)+M A. Nowlet K A K B. Using deduction theorem,K (M A M B)RM A .Similarly, K (M A M B) R M B . Observe that S5 MM M .Thus by DR7, 8, K (M A M B) R (M A M B) M . As K

2 holds and(M A M B) is equivalent to a modal wff, K (M A M B) R M , by DR2.Finally, by DR5, we have K (M A M B).


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(K8) Suppose A K(M A M B) KK (M A M B). So A K, or A K (M A M B). In the former case, as K2, 3 already hold, we trivially obtainK (M A M B) K=K A. Consider the latter case.

Using DR5, (M A) K (M A M B). By K8, K (M A M B) + M AK((M A M B) M A) = K M A = K A. So K (M A M B) K (M A M B) + M A K A, andK8 holds.


In previous section we directly used Levi and Harper identity to investigate rough beliefchange. Investigation led us to incorporate some changes in AGM postulates. Since ourresults dont lead to a clear picture of the belief change, we start from a basic approach ofconstructing contraction function. We can then identify new properties of the constructed

function and hence modify the AGM postulates in a systematic manner.Similar to classical construction of contraction from maximal subsets ofKwhich fail togive A, we can construct partial meet contractions in rough background. If the definitionof maximal subsets ofKis taken as in section 3.2 then we can show that following holds

Lemma 5.3.1 Let m is inKAfor some A in K, then m is closed andC n(m {}) =Kwhere K\ m.

Proof Let m R and assume / m. Now m R implies m R M which im-plies M K as m K and K is closed. Hence m m {M } K, which impliesm {M } R A or mR M A. But then mR A, as mR M and M is modal,

which gives contradiction. Hence m.For this we already have C n(m {}) Kand only KC n(m {}) has to be proved.Let K\ m, we show that M A M m and hence any set containing M A and m(in particularC n(m {})) will implyM (). By contradiction suppose M A M /mwhich means m m {M A M } K as S5 M (M A M ) and (M ) K.Thus (M A M ) M A m. But S5 ((M A M ) M A) M A gives thatM A(A) m, a contradiction.

Using second part of the above result, we again get that all the beliefs in K outside mbehave identically with respect to m. Supposem is in KA for some A, and B K\ mthen addition of some in K\ mto m will give all ofKand in particular B . Thus one can

check thatm will satisfy all the three conditions for membership ofKB. Thus we have-

Lemma 5.3.2 LetA K andm KA thenB K\ m, m KB.

Using this result we compareK(M AM B),KAandKB. Note that {M AM B} RR {A, B}. One can easily see that ifm K(M A M B) then it is in KA or KBbecause the second condition M A M B / m implies A / m or B / m. What ifm is inKA? The first and second condition for membership ofKM A M B are satisfied andour previous lemma implies the third condition - Suppose m m K, then in additionto A C n(m), Cn(m) = K is also there implying that B C n(m) (as B K). HenceM A M B is in C n(m). We present this result and then move on to new postulates.

Lemma 5.3.3 Let A and B are in K thenK(M A M B) =KA KB.


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5.3. PARTIAL MEET CONTRACTION 43

In the given background of rough logic we make some changes in the earlier postulates.The procedure is to use the same definition of contraction using the selection functionSasin equation 3.4 and then identifying the new properties. We consider the classical form of

the recovery postulate, and also find some changes in K6, 7, 8

(K1) K Ais a belief set(K2) K A K(K3) If A /K thenK A= K(K4) If RA thenA /K A(K5) K(K A) + A(K6) If A R RB thenK A= K B(K7) K A K B K (M A M B)(K8) If A /K (M A M B) thenK (M A M B) K A

Theorem 5.3.1 Let be a reflexive and transitive relation on M(K). Then functiondefined through equation 3.3 with consequence relationR, satisfies (K

1)-(K8).

Proof

(K1) This holds because each maximal subset m in S(KA) is closed and intersectionof closed sets is closed.


(K3) IfA /K then KA= {K} and hence K A= K.

(K4) IfRA thenKA is non-empty. Hence A /K A.

(K5) Let K. Assume RA so that first definition of 3.3 is used, we show that(M A ) KA. Using second part of the lemma 5.3.1 with = A we getm KA,Cn(m {A(orMA)}) = K. Thus M A m, K implyingM A K A=

S(KA). Hence (K A) + A.

(K6) Use, A RB and B RA implies KA= KB.

(K7) If R A or R B then obviously this holds, if R A then K A = K andKB = K(M A M B)(M A M B R R B). So let us assume RA and

RB. We have to show (S(KA)) (S(KB)) S(K(M A M B)) . ButS(KA KB) =S(KA) S(KB) S(KA) S(KB). Hence

(S(KA

KB))

(S(KA)S(KB)). Now by property of sets

(S(KA)S(KB)) =(

S(KA))(

S(KB)) and by lemma 5.3.3 we get (

S(KA))(

S(KB)) =(S(KA) S(KB))

(S(KA KB)) =

S(KM A M B).

(K8) Let A /K (M A M B). Thus there exists an m K(M A M B) such thatm m, m K(M AM B) andA /m. Note thatm KAasA /m. Now weshow

S(K(M AM B))

S(KA) by showingS(KA) S(K(M AM B)).

If m S(KA) then m m m, m KM A M B implying that m m, m K(M A M B) due to transitivity of. Hence m S(K(M A M B))

giving S(KA) S(K(M A M B)).


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5.4 Rough Consistency

In rough logic we look at following types of consistencies. A subset ofLR is:

Roughly consistent ifM ={M A: A } is S5 consistent that is there is no B LRsuch that MS5B and MS5B.Absolutely consistent if there is a B LR such that RB.Negation consistentif there is no B inLR such that RB and RB.Modal negation consistentif there is no B LR such that R M B and RM B.

Note that in S5, consistency is same as classical consistency where absolute and nega-tion consistency coincide. Also in rough logic negation consistency implies modal negationconsistency but not otherwise.

Lemma 5.4.1 InLR, rough consistency, absolute consistency and modal negation consis-tency are equivalent.

ProofSuppose is not roughly consistent then M is S5 inconsistent, that is to say forall B LR M S5 B. Hence M S5 M B and we have R B for all B in LR that isto say is not absolutely consistent. If is not absolutely consistent then obviously it isnot modal negation consistent. Finally suppose is not modal negation consistent. ThusRM B and RM B for someB LR. HenceMS5M M B and MS5 MM B,which is same as M S5 M B, M B. ThusM is not S5 consistent or is not roughlyconsistent.

Lemma 5.4.2 Let LR.(i) {A} RB , B LR if and only ifRM A.

(ii) {M A} RB , B LR if and only ifRA.

Proof (i) if part follows directly from equivalence of absolute consistency and modal nega-tion consistency. For only if part {A} R B if and only if M {M A} S5 M B.Thus in particular M {M A} S5 M A. Using D.T., M S5 M A M A, butM A M AS5M Aand hence MS5M A(MM A). Hence RM A.(ii) Replacing A byM A in first part we get {M A} R B, B LR if and only ifRMM A. Since MM A LM A M A and M AR RA, the result follows.

5.5 Maximal Consistent Sets

In rough logic maximal consistent sets are those roughly consistent sets which become notroughly consistent on adding anything outside them. Formally, Mr LR is maximal con-sistent if:(i) Mr is roughly consistent and(ii) IfA /Mr then Mr {A} is not roughly consistent.

Following properties of rough maximal consistent sets are notable:

(a) Mr is closed : Suppose MrRA andA /Mr. ThusMr {A}is not roughly consistent.Hence M Mr {M A} is S5 inconsistent, that is M Mr S5M A. But then MRAgives M M S5 M A which with M Mr S5 M A implies that Mr is not roughlyconsistent - a contradiction.


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5.6. REVISION FROM SYSTEM OF SPHERES 45

(b) For every B LR exactly one of M B and M B is in Mr : Obviously both M BandM B can not be inMr as it is (also) modally negation consistent. Now suppose,M B /Mr. ThusM Mr{M M B(or MB)} is notS5 consistent and we haveM Mr S5

M B(or MM B). Which implies MrRM B or M B Mr as Mr is closed.

(c) For every B LR exactly one ofLB and LB is in Mr : TakingB as B in (b), weget exactly one ofLB(MB) or LB(MB) is in Mr.

(d) Every roughly consistent set has a maximal consistent extension : Let {0, 1, 2, 3,...}be an enumeration ofLR. Construct 0, 1, ..., i,... as:0:= i+1 := i {i}if i {i}is roughly consistent, i otherwise. Take Mr =

i0i.

Note that each i is roughly consistent. Since Mr, we only need to provethat Mr is maximal consistent. First Mr is roughly consistent: Suppose not thenM Mr S5B , B, thus by compactness there is a finite subset ofMr which gives both

B andB. Now ifk is the maximum index of formulae contained in the finite subsetthenMk+1S5 B, B meaning k+1is not roughly consistent - a contradiction. Forproving second condition, letA /Mr. SoA = ifor somei and i+1does not containi. Thus i+1 = i and i {i} is not roughly consistent. ThusMr {i} is alsonot roughly consistent as i Mr.

(e) IfK is closed and roughly consistent then K =

[K] : Since Kis roughly consistent[K] is non-empty, thus obviously K

[K]. For the other side, we show that for all

A /K,A /

[K]. We first showK+ M Ais roughly consistent : IfK+ M Ais notroughly consistent then M K {MM A(or M A)} is S5 inconsistent implying thatM K S5 M A which means K R A or A K - a contradiction. Now maximal

consistent extension ofK+ M Ais in [K] and does not contain M A(or A) implyingthat A /

[K].

(f) [M A M B] = [A] [B] : Straight from the fact that{M A M B} R R {A, B}.

5.6 Revision from System of Spheres

Using system of spheres we can obtain revision function. We define the revision as:

K A=

C(A), if [A]=

Kwhere, C(A) =SA [A] (5.1)

Note that [A] is empty means A has no maximal consistent extension. That is possible onlywhen{A}is not roughly consistent which is same as RM A. Based on new properties ofmaximal consistent sets we make changes in the classical postulates for revision. Followingpostulates fit best in rough context:

(K1) K Ais a belief set(K2) A K A(K3) K A K+ A(K4) If M A /K then K+ A K A(K5) K A= K if and only ifRM A(K6) If A R RB thenK A= K B(K7) K (M A M B) (K A) + B(K8) If M B /K Athen (K A) + B K (M A M B)


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We prove that revision function defined in equation 5.1 satisfies all the above postulates.

Theorem 5.6.1 LetK be a belief set and its revision with respect to a beliefA is defined

as in equation 5.1 then it satisfies all the above postulates.

Proof

(K1) C(A) contains maximal consistent sets which are closed. So there intersection hasto be closed, and hence K A is a belief set.

(K2) If [A] is non-empty then each maximal consistent set in C(A) is also in [A] and hencecontainsA, otherwise when [A] is empty K A= K which contains A.

(K3) IfK+ A= K then obviously K A K+ A, otherwise when K+ A=K thenit is absolutely consistent and hence roughly consistent. Thus K+A has a maximal

consistent extension Mr. Since Mr contains both K and A, [K] [A] = and wehave SA= [K]. Thus C(A) = [K] [A]. But t

belief revision in non-classical logics

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