lost in translation: language independence in propositional logic – application to belief change

Artificial Intelligence 206 (2014) 1–24

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Artificial Intelligence

www.elsevier.com/locate/artint

Lost in translation: Language independence in propositionallogic – application to belief change ✩

Pierre Marquis a,∗, Nicolas Schwind b

a CRIL CNRS UMR8188, Université d’Artois, rue Jean Souvraz, F-62307 Lens, Franceb National Institute of Informatics, 2-1-2 Hitotsubashi, Chiyoda-ku, 101-8430 Tokyo, Japan

a r t i c l e i n f o a b s t r a c t

Article history:Received 22 October 2012Received in revised form 25 September2013Accepted 27 September 2013Available online 3 October 2013

Keywords:Language independenceBelief change

While propositional logic is widely used as a representation framework for many AIapplications, the concept of language independence in the propositional setting has notreceived much attention so far. In this paper, we define language independence for apropositional operator as robustness w.r.t. symbol translation. We motivate the need tofocus on symbol translations of restricted types, introduce and study several families oftranslations of interest, and provide a number of characterization results. We also identifythe computational complexity of recognizing symbol translations from those families. Thenwe investigate the robustness of belief merging, belief revision and belief update operatorsw.r.t. translations of different types. It turns out that some rational merging/revision/updateoperators are not guaranteed to offer the most basic (yet non-trivial) form of languageindependence.

© 2013 Published by Elsevier B.V.

1. Introduction

In propositional logic, propositional symbols are the formal counterparts of propositions which are not analyzed (i.e.,decomposed) within the language of the logic but can be arbitrarily sophisticated nevertheless. For instance, an (informal)proposition like “John’s house is located north of the station” can be represented by a propositional symbol like jn. Providedthat “north” means either “north-west” or “north-east”, the same informal proposition can also be represented by slightlymore complex formulae, like jnw ⊕ jne where ⊕ denotes the exclusive-or connective, and jnw means “John’s house islocated north-west of the station” and jne means “John’s house is located north-east of the station”.

The problem of which informal propositions of interest should be associated with propositional symbols is not ruled bylogic since this is mainly a domain-dependent modeling issue. Thus, if “John’s house is located north-west of the station” isalso a proposition of interest, then representing “John’s house is located north of the station” as jnw ⊕ jne is an alternativeto the representation of it by jn; on the other hand, one may consider that jn is a representation of “John’s house islocated north of the station”, and then represent “John’s house is located north-west of the station” as jn ∧ jw where jwis associated with “John’s house is located west of the station”.

Representing information as propositional formulae raises some expressiveness limitations. An informal proposition like“John’s house is located north of the station” also refers to a number of terms like “John”, “house”, “station”, “being north of”used to build it, but which cannot be represented within propositional logic. Furthermore, when this informal propositionis stated, a number of implicit propositions which are logical consequences of it can be considered: “in the city whereJohn lives there is a station”, “John lives in a house”, etc. Of course, whatever the informal proposition is represented as

✩ This is an extended and revised version of the paper entitled “Lost in Translation: Language Independence in Propositional Logic – Application to BeliefRevision and Belief Merging” published in the proceedings of IJCAI’11, pages 1002–1007.

* Corresponding author.E-mail addresses: [email protected] (P. Marquis), [email protected] (N. Schwind).

0004-3702/$ – see front matter © 2013 Published by Elsevier B.V.http://dx.doi.org/10.1016/j.artint.2013.09.005

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2 P. Marquis, N. Schwind / Artificial Intelligence 206 (2014) 1–24

jn or as jnw ⊕ jne, it is impossible to derive from the representation any of the implicit propositions above, because suchpropositions cannot be expressed as Boolean combinations of jn, jnw or jne. These limitations are not problematic if noneof those propositions is of specific interest (and, again, this is not a logical concern but a domain-dependent one). Whatis important from a logical point of view is to make a representation choice so that every proposition of interest can berepresented as a formula based on the chosen propositional symbols (otherwise the corresponding propositional languageis not expressive enough w.r.t. the propositions of interest). Several representation choices are usually possible withoutquestioning such an expressiveness issue.

In artificial intelligence, propositional formulae are used to build propositional structures which represent propositionalinformation. In applications where propositional logic is enough from a representation perspective, information processingcan typically be modeled as a combination of queries (e.g., deducing some facts from the available beliefs or inferring someplausible conclusions from them) and transformations (e.g., forgetting some facts, revising or updating some beliefs, etc.).Propositional queries and propositional transformations can be formalized as propositional operators, which are mappingsfrom propositional structures to propositional structures.

In this paper, the language independence issue for propositional operators is addressed. Intuitively, a propositional oper-ator Ω is language-independent if and only if whenever the representation language is modified in such a way that symbolsin the original language correspond to formulae in the target language, if the input of Ω is modified so as to reflect thelanguage change, then the output of Ω should be modified accordingly. Thus, language change can be modeled by symboltranslations (i.e., substitutions), and language independence for a propositional operator is the faculty for it to be robustw.r.t. symbol translations.

For instance, assume that mn represents the fact that “Mary’s house is located north of the station”. If it is known that“John’s house is located north of the station” ( jn) and that “John’s house is located north of the station” if and only if “Mary’shouse is located north of the station” ( jn ⇔ mn) since Mary and John live together, then one can deduce that “Mary’s houseis located north of the station”: mn is a logical consequence of jn ∧ ( jn ⇔ mn). If we consider another representation choice,where “John’s house is located north of the station” is represented by jnw ⊕ jne, then the same conclusion follows: from( jnw ⊕ jne)∧ (( jnw ⊕ jne) ⇔ mn), the formula mn can still be deduced. Accordingly, the deduction operator is robust whenthe symbol translation σ such that σ( jn) = jnw ⊕ jne is applied to the formula jn ∧ ( jn ⇔ mn).

Clearly, such a notion of language independence should not be confused with the notion of syntax independence, whichreflects the ability of an operator to be robust w.r.t. substitution of formulae (or sets of formulae) by equivalent formulae(or sets), a topic which got much attention in artificial intelligence. Quite surprisingly, the language independence issue inpropositional logic has not received so far a systematic investigation, despite the fundamental nature of the issue and thesignificance of propositional languages in artificial intelligence.

This paper is intended to fill the gap. The main contributions are as follows. First of all, language independence isdefined as robustness w.r.t. symbol translations; we provide a number of characterization results for symbol translations,including a model-based characterization of translations in terms of injective, surjective relations over worlds. Then wemotivate the need to focus on symbol translations of restricted types, i.e., satisfying some additional properties, whichleads to refined notions of language independence. A model-based characterization for each of the refined notions is given,and we make precise how those notions are logically connected. We also identify the complexity of recognizing symboltranslations satisfying properties of interest. Finally, the robustness of propositional belief change operators (revision, updateand merging) w.r.t. translations of different types is investigated within our language independence setting. It turns out thatonly limited forms of language independence are satisfied in the general case by belief change operators. Especially, weshow that rational revision (resp. update, merging) operators, i.e., those satisfying the AGM (resp. KM, IC) postulates, are notguaranteed to satisfy the most basic (yet non-trivial) form of language independence. We have also derived characterizationresults based on language independence for the distance-based IC merging operator �dD ,Σ (where the drastic distance dD

and sum as an aggregation function are used), as well as for the corresponding AGM revision and KM update operators; theobtained results show those rational change operators as the most robust operators in terms of language independence.

The rest of the paper is organized as follows. After some formal preliminaries in Section 2, the notion of symbol trans-lation and some characterization results are pointed out in Section 3. In Section 4 the notion of language independence isdefined as robustness w.r.t. symbol translations, and several refinements of it are presented. Some complexity results areprovided in Section 5. In Section 6, belief change operators are put in the setting and the level of language independenceachieved by those operators is identified. Some related work is discussed in Section 7, before the conclusion. For ease ofreading, proofs are reported in a final Appendix A.

2. Formal preliminaries

We consider a finite set PS of propositional symbols. For any subset X of PS, PROPX denotes the propositional languageover X defined in the usual way from X and a finite set of connectives, including the Boolean constants � (true) and ⊥(false) and the standard connectives ¬, ∨, ∧, ⇒, ⇔, ⊕. For every formula α, α+ denotes α and α− denotes ¬α. For everyformula α, Var(α) is the set of propositional symbols occurring in α.

Let X ⊆ PS. An interpretation ω (or a world) over X is a mapping from X to {0,1}, which can also be viewed an elementof 2X , i.e., as a subset of X such that for each x ∈ X, x belongs to it if and only if ω(x) = 1. A canonical term associated withω is a term over X (i.e., a conjunction of literals over X) such that for each x ∈ X, x occurs as a literal of the term if ω(x) = 1,

P. Marquis, N. Schwind / Artificial Intelligence 206 (2014) 1–24 3

else ¬x occurs in it. To avoid heavy expressions, given a set X of propositional symbols, the notions of interpretation over X ,element of 2X and canonical term will be confounded throughout the paper. For instance, if X = {p,q}, then the world ωover X such that ω(p) = 1 and ω(q) = 0 is also viewed as the set {p} and as the term p ∧ ¬q. When a total, strict order <

over X is provided, any interpretation ω is also represented as a binary sequence. Thus, provided that p < q, the world ωabove is also represented as the sequence 10.

The notion of satisfaction of a formula is the usual one. |� denotes logical entailment and ≡ denotes logical equivalence.A formula is said to be complete w.r.t. X iff it admits exactly one model over X , i.e., it is equivalent to a canonical termover X . Mod(α) denotes the set of models over PS of a formula α.

A formula α is said to be generated from a set C of connectives and a set Σ of formulae iff α ∈ Σ or there exists aconnective c ∈ C of arity n > 0 such that α ≡ c(α1, . . . ,αn) where each αi (i ∈ 1, . . . ,n) is generated from C and Σ . {¬,∧}is said to be functionally complete w.r.t. X given a set Σ of formulae iff for every propositional formula α generated from{¬,∧} and X , there exists a propositional formula β generated from {¬,∧} and Σ such that β ≡ α.

The following definition recalls some properties of binary relations which will be considered in the following:

Definition 1 (Properties of binary relations). Let E, F be two sets and let R be a binary relation over E × F . R is said to be:

• injective if and only if(∀e, e′ ∈ E , ∀ f ∈ F , if R(e, f ) and R(e′, f ) then e = e′);

• functional if and only if(∀e ∈ E , ∀ f , f ′ ∈ F , if R(e, f ) and R(e, f ′) then f = f ′);

• surjective if and only if (∀ f ∈ F , ∃e ∈ E such that R(e, f ));• left-total if and only if (∀e ∈ E , ∃ f ∈ F such that R(e, f )).

R is said to be a one-to-one correspondence if and only if R is an injective, functional, surjective, left-total relation.

Lastly, |E| denotes the cardinality of any set E .In the following, we are interested in the robustness to symbol translations offered by propositional operators over

propositional structures:

Definition 2 (Propositional structure). A propositional structure Σ over a set X ⊆ PS of propositional symbols is a vector ofsets of propositional formulae from PROPX (only finite vectors and finite sets are considered). If n is the dimension of thevector, Σ is an n-propositional structure.

Sets of formulae are interpreted conjunctively, so that two (finite) sets of formulae α and β are equivalent, denotedα ≡ β , when the conjunction of formulae of α is equivalent to the conjunction of formulae of β . While a belief base isstandardly defined as a (possibly infinite) set of formulae, it is considered as a finite set of formulae in the following.

Two n-propositional structures Σ = 〈α1, . . . ,αn〉 and Σ ′ = 〈α′1, . . . ,α

′n〉 are equivalent, denoted Σ ≡ Σ ′ iff ∀i ∈ 1, . . . ,n,

we have αi ≡ α′i .

Definition 3 (Propositional operator). A propositional operator Ω is a mapping associating a propositional structure with apropositional structure.

Many propositional queries and transformations can be viewed as propositional operators; here are a few examples ofoperators considered in the paper:

Example 1.

• ded (deduction): 〈{α1, . . . ,αn}, {β}〉 �→ 〈{�}〉 if β is a logical consequence of the belief base {α1, . . . ,αn}, and 〈{⊥}〉otherwise;

• ◦ (belief revision): 〈{α1, . . . ,αn}, {β}〉 �→ 〈{β1, . . . , βm}〉 where {β1, . . . , βm} is the belief base resulting from revising thebelief base {α1, . . . ,αn} by β using the belief revision operator ◦;

• � (belief update): 〈{α1, . . . ,αn}, {β}〉 �→ 〈{β1, . . . , βm}〉 where {β1, . . . , βm} is the belief base resulting from updating thebelief base {α1, . . . ,αn} by β using the belief update operator �;

• � (belief merging): 〈{α11 , . . . ,α1

n(1)}, . . . , {αp1 , . . . ,α

pn(p)}, {μ}〉 �→ 〈{β1, . . . , βm}〉 where {β1, . . . , βm} is the belief base re-

sulting from merging the vector of belief bases 〈{α11 , . . . ,α1

n(1)}, . . . , {αp1 , . . . ,α

pn(p)}〉 using the belief merging operator

� when μ is the integrity constraint.


3. On translations

3.1. Definition

The key notion for defining notions of language independence is that of symbol translation; in the rest of the paper,when not stated explicitly, X and Y denote subsets of PS.

Definition 4 (Translation). A (symbol) translation σ is a mapping from X to PROPY , which is extended to a morphism (alsodenoted σ ) from propositional structures over X to propositional structures over Y , defined inductively by (for any positiveinteger n):

• σ(�) = � and σ(⊥) = ⊥;• for every connective c of arity n, σ(c(α1, . . . ,αn)) = c(σ (α1), . . . , σ (αn)),• σ({α1, . . . ,αn}) = {σ(α1), . . . , σ (αn)};• σ(〈α1, . . . ,αn〉) = 〈σ(α1), . . . , σ (αn)〉.

PROPXY is the set of all symbol translations from X to PROPY . Two symbol translations σ1, σ2 of PROPX

Y are said to beequivalent if and only if ∀x ∈ X , σ1(x) ≡ σ2(x). When X = {x1, . . . , xn} we also sometimes use the notation σ : x1 �→σ(x1); . . . ; xn �→ σ(xn) to define a symbol translation σ ∈ PROPX

Y .

There is no need to assume any connection between X and Y , i.e., we may have X ∩ Y = ∅, but, conversely, this is notforbidden (e.g., we may have X = Y ). For instance, if σ is such that X = {p,q}, Y = {q, r}, σ(p) = q and σ(q) = ¬q ∨ r, wehave σ(p ∧ ¬q) = q ∧ ¬(¬q ∨ r) ≡ q ∧ ¬r. Finally, for every Z ⊆ PS such that X ⊆ Z , every symbol translation σ from PROPX

Yis naturally extended as a symbol translation still denoted σ from PROPZ

Y ∪Z such that ∀z ∈ Z \ X , σ(z) = z.A fundamental result in propositional logic when dealing with such symbol translations is the well-known substitution

theorem, which results directly from the truth functionality of connectives and can be found in many textbooks (e.g., [1]):

Proposition 1 (Substitution theorem). Let α, β be two formulae from PROPX . If α ≡ β then for every symbol translation σ of PROPXY ,

we have σ(α) ≡ σ(β).

The following corollary presents alternative ways to state this theorem:

Corollary 1.

1. Let {α1, . . . ,αn} be a finite set of formulae from PROPX , and let β be a formula from PROPX . We have {α1, . . . ,αn} |� β if andonly if for every symbol translation σ of PROPX

Y , we have

σ({α1, . . . ,αn}

) |� σ(β).

2. |� α if and only if for every symbol translation σ of PROPXY , we have |� σ(α). Similarly, |� ¬α if and only if for every symbol

translation σ of PROPXY , we have |� ¬σ(α).

Let us illustrate point 1 of Corollary 1 by considering the example provided in the introduction. We have jn ∧ ( jn ⇔mn) |� mn. Let σ : jn �→ ( jnw ⊕ jne). Then σ( jn ∧ ( jn ⇔ mn)) = ( jnw ⊕ jne) ∧ (( jnw ⊕ jne) ⇔ mn) implies σ(mn) = mn.

Obviously enough, in the general case, there is no direct logical connections between a formula α and the associatedtranslated formula σ(α). They can be logically independent, like α = p and σ(α) = q when σ : p �→ q is considered. It canalso be the case that one of them implies the other one (consider α = p, σ1 : p �→ p ∧ q and σ2 : p �→ p ∨ q) or that theformulae are jointly inconsistent (consider α = p and σ : p �→ ¬p). Actually, the only direct logical connections between αand σ(α) are obtained in the specific case α is valid or α is contradictory (and are given by point 2 of Corollary 1). Seealso the forthcoming Proposition 4.

Nevertheless, some indirect logical connections between α and σ(α) exist, as stated by the following proposition:

Proposition 2. Let α be a formula from PROPX . Let σ be a symbol translation from X to PROPY . We have∧

x∈X (x ⇔ σ(x)) |� α ⇔σ(α).

This proposition shows that α and σ(α) are equivalent modulo the theory where each symbol in X is equivalent to itsimage by σ .


Fig. 1. An injective, surjective relation R over 2{x,y} × 2{p,q,r} , a formula α = ¬x ∨ ¬y from PROP{x,y} and its translation σR (α) ≡ ¬p ∨ (q ⊕ r).

3.2. A model-based characterization of symbol translations

We investigate now how a symbol translation looks like from a model-based point of view. We provide a characterizationof every symbol translation σ from X to PROPY (modulo equivalence) in terms of a binary relation over 2X × 2Y .

Definition 5 (Semantical mapping). The semantical mapping Ψ of symbol translations is the mapping associating with everysymbol translation σ of PROPX

Y the binary relation Rσ ⊆ 2X × 2Y such that for every ω ∈ 2X , ω′ ∈ 2Y , (ω,ω′) ∈ Rσ if andonly if ω′ |� σ(ω). Rσ (ω) denotes the set {ω′ ∈ 2Y | Rσ (ω,ω′)}.

Given a symbol translation σ the relation Rσ provides us with a canonical representation of σ from a model-based pointof view. Indeed, one can easily check that for every symbol translation σ from PROPX

Y , for every interpretation ω over X ,we have

σ(ω) ≡∨{

ω′ ∣∣ ω′ ∈ Rσ (ω)}.

The following model-based characterization of symbol translations shows that Ψ is a one-to-one correspondence; ac-cordingly, its inverse mapping Ψ −1 associates any injective, surjective relation R on 2X × 2Y with a symbol translationσR = Ψ −1(R):

Proposition 3. The semantical mapping Ψ of symbol translations is a one-to-one correspondence from PROPXY / ≡, the set of all symbol

translations (modulo equivalence), to the set of all injective, surjective relations over 2X × 2Y . Its inverse mapping Ψ −1 associates forevery subsets X, Y of PS, an injective, surjective relation R over 2X × 2Y with a representative symbol translation, denoted σR , suchthat for every propositional symbol x ∈ X,

σR(x) =∨{

ω′ ∈ 2Y∣∣ ω′ ∈ R(ω),ω |� x

}.

When R is not functional, the associated symbol translation σR “refines” some worlds over X by making each one ofthem correspond to several worlds over Y . By contrast, when R is not left-total, σR “discards” some worlds over X , namelyno world over Y corresponds to them.

Example 2. Let X = {x, y}, Y = {p,q, r} and let R be the injective, surjective relation R over 2X × 2Y defined asR({x, y}) = {{p,q, r}, {p}}, R({x}) = {{p,q}}, R({y}) = {{p, r}, {q, r}, {q}, {r}, {}} and R({}) = {}. Then by Proposition 3, R pro-vides a canonical representation of the symbol translation Ψ −1(R) = σR : x �→ p ∧ (q ∨ ¬r); y �→ ¬p ∨ ¬q ∨ r. One can easilycheck that for every formula α from PROPX , σR(α) ≡ ∨{R(ω) | ω |� α}. Fig. 1 provides a graphical representation of therelation R , a formula α = ¬x ∨ ¬y from PROPX and its translation σR(α) ≡ ¬p ∨ (q ⊕ r) from PROPY .

4. On language independence

4.1. Language independence based on translations

We are now ready to define language independence as robustness w.r.t. symbol translations satisfying some properties:


Fig. 2. A P -language independent propositional operator.

Definition 6 (Language independence). Let P be a property on symbol translations from X to PROPY . A propositional operatorΩ associating an m-propositional structure with an n-propositional structure is P -language independent if and only if forevery n-propositional structure Σ over X , for every translation σ PROPX

Y satisfying P , we have

Ω(σ(Σ)

) ≡ σ(Ω(Σ)

).

When P always holds (i.e., P is a valid condition), every symbol translation σ can be considered, and Ω is said to be�-language independent.

Fig. 2 provides a graphical representation of the behavior of a language-independent propositional operator.In general, P characterizes what should be preserved (or dually what can be lost) when a symbol translation is applied.

Indeed, replacing in a uniform way propositional symbols by formulae may lead to question a number of properties whichhold for propositional symbols but can be lost when formulae are considered. Especially, unlike general formulae, propo-sitional symbols are neither inconsistent nor valid, conjunction of possibly negated but distinct propositional symbols isalways consistent, and they are complete formulae, i.e., propositional symbols are enough to generate a full propositionallanguage over them whenever a functionally complete set of connectives (as {¬,∧}) is considered.

Formally, we focus on the following properties:

Definition 7 (Properties of symbol translations). A symbol translation σ from X = {x1, . . . , xn} to PROPY satisfies:

• SIN (symbol insensitivity) if and only if σ is a bijection from X to Y ;• TPR (triviality prevention) if and only if ∀x ∈ X , σ(x) is consistent but not valid;• AIP (atom independence preservation) if and only if

∀s1, . . . , sn ∈ {+,−},∧

xi∈X σ(xi)si is consistent;

• BAP (belief amount preservation) if and only if for every complete formula α from PROPX , σ(α) is a complete formulafrom PROPY .

• FCP (functional completeness preservation) if and only if {¬,∧} is functionally complete w.r.t. Y given {σ(x) | x ∈ X};• REV (reversibility) if and only if there exists a symbol translation θ from Y to PROPX such that ∀x ∈ X , θ(σ (x)) ≡ x.

�-language independence is hard to be satisfied since it leads to consider as admissible the symbol translations whichdo not preserve anything from the input propositional structure, except trivialities (tautologies and contradictions) whichare always kept. Indeed, we have the following result:

Proposition 4. Let α be a formula of PROPX and β a formula of PROPY . If α is neither valid nor inconsistent, then there exists a symboltranslation σ ∈ PROPX

Y such that σ(α) ≡ β .

This motivates the need to restrict the set of admissible translations.SIN-language independence asks the choice of the symbols names to be non-significant. This independence property is

highly expected. However, as we will see in the following, it is not always guaranteed for revision/merging/update operators.The rationale for TPR is to set aside translations which leads to replace a symbol, which is in essence neither valid nor

inconsistent, by a formula which would be valid or inconsistent.The principle of AIP is as follows: at start, all the symbols of PS are logically independent, which means that ∀s1, . . . , sn ∈

{+,−},∧

xi∈PS xsii is consistent; for instance, if PS = {x, y}, then ¬x ∧ ¬y is consistent. If another representation choice is

made, such an independence should be preserved. For instance, x �→ p ∨ q; y �→ ¬p ∨ q does not satisfy AIP since ¬(p ∨ q)∧¬(¬p ∨ q) is inconsistent; contrastingly, x �→ p; y �→ q ∨ r satisfies AIP.

BAP-language independence requires admissible translations to leave unchanged the number of models of formulae, i.e.,every formula admits the same number of models in both the initial and the target languages. For example, x �→ p ⇔ q; y �→p satisfies BAP, while x �→ p; y �→ q ∨ r does not.

FCP asks for the preservation of functional completeness. At start, taking any fully expressive set of connectives (like{¬,∧}) proves enough to represent every Boolean function over any subset X of PS. When another representation choice ismade, it is expected that any functionally complete set of connectives proves enough to represent every Boolean function


over Y given that the set of “atoms” is now {σ(x) | x ∈ X}. As an example, consider again x �→ p ∨q; y �→ ¬p ∨q; this symboltranslation does not satisfy FCP since p cannot be defined using {¬,∧} when the set of “atoms” is {p ∨ q,¬p ∨ q}; on theother hand, x �→ p ⇔ q; y �→ p satisfies FCP: both p and q can be defined using {¬,∧} when the set of atoms is {p ⇔ q, p};this is obvious for p, and for q, we have q ≡ (p ⇔ q) ⇔ p.1 Clearly enough, {¬,∧} is functionally complete w.r.t. Y given{σ(x) | x ∈ X} if and only if for every formula β of PROPY , there exists a formula α of PROP{σ(x)|x∈X} such that α ≡ β . Thisis equivalent to state that for every y ∈ Y , there exists a formula αy of PROP{σ(x)|x∈X} such that αy ≡ y (i.e., Y is definablefrom {σ(x) | x ∈ X}), which is in its turn equivalent to state that for every y ∈ Y , there exists a formula αy of PROPX suchthat σ(αy) ≡ y.

The idea underlying REV is the one of reversibility. The intuition is that nothing important is lost in a symbol translationwhen the translation can be reversed, which requires the target language of the translation to be at least as expressiveas the initial language. For instance, x �→ p ∨ q; y �→ ¬p ∨ q does not satisfy REV since the system of equations {p ∨ q ≡x,¬p ∨q ≡ y} with variables p and q has no solution over PROP{x,y} . Contrastingly, σ : x �→ p ⇔ q; y �→ p satisfies REV sinceθ : p �→ y,q �→ y ⇔ x is such that θ(σ (x)) ≡ x and θ(σ (y)) ≡ y.

4.2. Connections between notions of language independence

Clearly enough, none of the properties above is trivial in the sense that it would be satisfied by every (or by no) symboltranslation. Thus, the identity translation σ : x �→ x satisfies all the properties above, while σ : x �→ ⊥, y �→ p ∨ q satisfiesnone of them.

Discovering connections between notions of language independence boils down to study the dependencies between theproperties of translations.

First, while based on quite different intuitions, it turns out that REV and AIP are the same property:

Proposition 5. REV and AIP are equivalent.

Now, most of the remaining dependencies w.r.t. the other properties can be easily derived from the following propositionwhich provides, in a similar fashion as to Proposition 3, for every property P on symbol translations listed above, a model-based characterization of symbol translations satisfying P , or more precisely, for any subsets X, Y of PS, a characterizationof the quotient set PROPX

Y / ≡ satisfying P in terms of some specific injective, surjective relations over 2X × 2Y .

Proposition 6. Let σ be a symbol translation from PROPXY and Rσ be its semantical mapping:

1. σ satisfies AIP if and only if Rσ is left-total;2. σ satisfies BAP if and only if Rσ is a one-to-one correspondence;3. σ satisfies FCP if and only if Rσ is functional;4. σ satisfies SIN if and only if Rσ is a one-to-one correspondence which is decomposable, i.e., there exists a bijection ηRσ from X

to Y defined by ∀x ∈ X, ∀y ∈ Y , ηRσ (x) = y if and only if {Rσ (ω) | ω ∈ 2X ,ω |� x} = {ω′ | ω′ ∈ 2Y ,ω′ |� y};5. σ satisfies TPR if and only if ∀x ∈ X, ∃ω,ω′ ∈ 2X , ω |� x, ω′ |� ¬x, Rσ (ω) �= ∅ and Rσ (ω′) �= ∅.

From Proposition 6, one can easily derive some of the following dependencies between properties of symbol translations:

Proposition 7. SIN implies BAP. BAP implies FCP. BAP also implies REV, which implies TPR. Each implication is strict (i.e., the conversedoes not hold). Moreover, if a symbol translation satisfies REV and FCP, then it satisfies BAP. Lastly, FCP does not imply and is notimplied by any of REV or TPR.

These logical implications between properties of symbol translations are exhaustively represented through the implica-tion graph given in Fig. 3, where for the sake of clarity, arrows coming from the reflexivity and the transitivity of implicationare omitted.

Other notable consequences of Proposition 6 are listed in the following corollary:

Corollary 2. Let σ be a symbol translation of PROPXY .

1. σ satisfies AIP (or, equivalently, REV) only if |X |� |Y |;2. σ satisfies BAP only if |X | = |Y |;

1 Note that this property of functional completeness preservation is slightly more demanding than the one reported in [2], where {¬,∧} is asked tofunctionally complete w.r.t.

⋃x∈X Var(σ (x)) (and not w.r.t. to its superset Y ) given {σ(x) | x ∈ X}; thus, if X = {x}, then σ : x �→ p satisfies FCP in the sense

of [2] whatever Y , while it does not satisfy FCP in the sense of this paper when Y �= {p} (for instance, when Y = {p,q}). This additional requirement isharmless since one may always choose Y to be equal to its restriction

⋃x∈X Var(σ (x)) without loss of generality (i.e., without any change to {σ(Σ) | Σ is

a propositional structure over X}).


Fig. 3. Implication graph of properties of symbol translations.

3. σ satisfies FCP only if |X |� |Y |;4. if |X | = |Y | then AIP, BAP and FCP are equivalent.

The first point comes from the fact that when R is an injective, surjective relation over 2X × 2Y which is left-total, thenR−1 is a surjective mapping from 2Y to 2X , hence |X |� |Y |. The second point is obvious since a one-to-one correspondencebetween two finite sets 2X and 2Y may exist only if the two sets have the same number of elements, i.e., precisely when|X | = |Y |. The third point comes from the fact that when R is an injective, surjective relation over 2X × 2Y which isfunctional, then R−1 is an injective mapping from 2Y to 2X , hence |X | � |Y |. The last point comes from the fact that everyinjective, surjective relation over 2X × 2Y where X and Y are finite sets such that |X | = |Y |, is left-total if and only if it isfunctional.

To conclude this section, let us now turn to the concept of syntax independence. Syntax independence at the beliefbase level has received much attention so far (see e.g., [3–6]). It can be stated as follows when propositional operators areconsidered:

Definition 8 (Syntax independence). A propositional operator Ω which associates a m-propositional structure with a givenn-propositional structure is said to be syntax-independent if and only if whenever 〈α1, . . . ,αn〉 ≡ 〈α′

1, . . . ,α′n〉, we also have

Ω(〈α1, . . . ,αn〉) ≡ Ω(〈α′1, . . . ,α

′n〉).

It turns out that the concept of syntax independence and the concept of P -language independence (whatever P amongthe choices we considered) are logically independent, in the sense that one can find propositional operators satisfying bothproperties, while other operators satisfy none of them and still other operators satisfy only one of them:

Proposition 8. The concept of syntax independence and the concept of P -language independence for propositional operators arelogically independent.

5. The complexity of recognizing translations

In this section, we consider the problem of determining whether a given symbol translation satisfies one of the propertiesconsidered in the previous section. Such an issue has to be addressed each time a representation change is expected, givena propositional operator which is known as P -language independent. Indeed, a symbol translation is acceptable in this caseonly if it satisfies P .

First, we show that FCP amounts to a definability issue (see [7] for details):

Proposition 9. σ : {x1, . . . , xn} → PROPY satisfies FCP if and only if∧n

i=1(zi ⇔ σ(xi)) defines Y in terms of {z1, . . . , zn}, where{z1, . . . , zn} ∩ Y = ∅.

We have also derived the results reported in Proposition 10 (symbol translations σ ∈ PROPXY are supposed to be repre-

sented extensionally as 〈X, Y , {(x, σ (x)) | x ∈ X}〉). This proposition shows that the complexity of deciding whether a givensymbol translation satisfies one of the properties of interest varies with the property under consideration between (deter-ministic) polynomial time and the second level of the polynomial hierarchy.

Proposition 10.

1. Determining whether a given symbol translation σ ∈ PROPXY satisfies SIN is in P.

2. Determining whether a given symbol translation σ ∈ PROPXY satisfies TPR is NP-complete.

3. Determining whether a given symbol translation σ ∈ PROPXY satisfies FCP is coNP-complete in the general case. The problem is in

P if |X | < |Y |.4. Determining whether a given symbol translation σ ∈ PROPX

Y satisfies BAP is coNP-complete in the general case. The problem is inP if |X | �= |Y |.


5. Determining whether a given symbol translation σ ∈ PROPXY satisfies REV (or, equivalently, AIP) is Π

p2 -complete. The problem is

coNP-complete if |X | = |Y |, and is in P if |X | > |Y |.

Interestingly, the problem of determining whether a given symbol translation σ ∈ PROPXY satisfies REV is a specific case of

the following problem of propositional matching (given 2n formulae α1, β1, . . . ,αn, βn , determining whether there exists asubstitution θ from X = ⋃n

i=1 Var(αi) to PROPY with Y = ⋃ni=1 Var(βi) such that for i ∈ 1, . . . ,n, θ(αi) ≡ βi ). Hence, point 5

of Proposition 10 shows that the problem of propositional matching is at least as hard as the problem of Boolean unificationwith constants [8].

6. Language independence for belief change

6.1. Preliminaries on belief revision, belief merging and belief update

Belief merging operators aim at defining a belief base (the merged base) which represents the beliefs of a group of agentsgiven their individual belief bases, and some integrity constraints. Formally, a (belief) base K denotes the set of beliefs ofan agent, it is a finite set of propositional formulae, interpreted conjunctively. A profile E = 〈K1, . . . , Kn〉 is a finite andnon-empty vector of consistent bases representing the beliefs from the group of n agents involved in the merging process.When E = 〈K1, . . . , Kn〉, the dimension of E is n. The assumption that all bases forming a profile are consistent is standard[9], since an inconsistent belief base is often considered to provide no information for the merging process. In general,agents are expected to play equivalent roles in the merging process, so that a profile E = 〈K1, . . . , Kn〉 is also viewed as amulti-set {K1, . . . , Kn}. The integrity constraints μ are represented by a formula. � denotes the union operator for multi-sets,\ denotes the difference operator for multi-sets, and ≡ the equivalence of profiles (two multi-sets of bases are equivalentwhen there is a bijection between them so that each base from a profile is equivalent to its image in the other profile).∧

E denotes the conjunction of the belief bases of E , i.e.,∧

E = ∧Ki∈E Ki . Then, the notation ω |� E stands for ω |� ∧

E .Lastly, a preorder � over a set S is a reflexive, transitive relation over S . For each preorder �, � denotes the correspondingindifference relation and < the corresponding strict ordering.

A merging operator � is a mapping associating a formula μ (representing the integrity constraints) and a profile E witha new base �μ(E). Let us recall the standard logical properties which are expected for belief merging operators [9]:

Definition 9 (IC merging operator). A merging operator � is an IC merging operator if and only if for every formula μ, for everyprofile E , E1, E2 and for every consistent belief base K1, K2, it satisfies the following postulates (so-called IC postulates):

(IC0) �μ(E) |� μ;(IC1) If μ is consistent, then �μ(E) is consistent;(IC2) If

∧E ∧ μ is consistent, then �μ(E) ≡ ∧

E ∧ μ;(IC3) If E1 ≡ E2 and μ1 ≡ μ2, then �μ1 (E1) ≡ �μ2 (E2);(IC4) If K1 |� μ, K2 |� μ and �μ(〈K1, K2〉) ∧ K1 is consistent, then �μ(〈K1, K2〉) ∧ K2 is consistent;(IC5) �μ(E1) ∧ �μ(E2) |� �μ(E1 � E2);(IC6) If �μ(E1) ∧ �μ(E2) is consistent, then �μ(E1 � E2) |� �μ(E1) ∧ �μ(E2);(IC7) �μ1 (E) ∧ μ2 |� �μ1∧μ2 (E);(IC8) If �μ1 (E) ∧ μ2 is consistent, then �μ1∧μ2 (E) |� �μ1 (E) ∧ μ2.

Each IC merging operator corresponds to a syncretic assignment [9]:

Definition 10 (Syncretic assignment). A syncretic assignment is a mapping which associates with every profile E a total pre-order �E

2 over worlds and such that for every profile E , E1, E2 and for every consistent belief base K1, K2, it satisfies thefollowing conditions:

(1) If ω |� E and ω′ |� E , then ω �E ω′;(2) If ω |� E and ω′ �|� E , then ω <E ω′;(3) If E1 ≡ E2, then �E1=�E2 ;(4) If ω |� K1, then ∃ω′ |� K2 ω′ �〈K1,K2〉 ω;(5) If ω �E1 ω′ and ω �E2 ω′ , then ω �E1�E2 ω′;(6) If ω <E1 ω′ and ω �E2 ω′ , then ω <E1�E2 ω′ .

Theorem 1. (See [9].) A merging operator � is an IC merging operator if and only if there exists a syncretic assignment associatingevery profile E with a total preorder �E such that for every formula μ,

Mod(�μ(E)

) = min(Mod(μ),�E

).

2 When E = 〈K 〉 consists of a single base K , we sometimes write �K instead of �〈K 〉 in order to alleviate the notations.


Distance-based merging operators are characterized by a distance (actually, a pseudo-distance is enough, i.e., triangu-lar inequality is not mandatory) between worlds and an aggregation function f (a mapping associating with a tuple ofnon-negative real numbers a non-negative real number) [10]:

Definition 11 (Distance-based merging operator). Let d be a distance between worlds and f be an aggregation function. Themerging operator �d, f is defined for every profile E and every formula μ by Mod(�

d, fμ (E)) = min(Mod(μ),�E ) where the

total preorder �E over worlds induced by E is defined by:

• ω �E ω′ if and only if d(ω, E)� d(ω′, E);• d(ω, E) = f K∈E (d(ω, K ));• d(ω, K ) = minω′ |�K d(ω,ω′).3

Usual distances are dD , the drastic distance (dD(ω,ω′) = 0 if ω = ω′ and 1 otherwise), which corresponds to the infinity-norm distance, also known as Chebyshev distance, and dH the Hamming distance (dH (ω,ω′) = n if ω and ω′ differ on nvariables), which corresponds to the 1-norm distance, also referred to as the Manhattan distance. Note that some distance-based operators are not IC merging ones (conditions of non-decreasingness, minimality, identity, symmetry, composition anddecomposition must be satisfied by f in the general case, see [10] for details), but taking advantage of “usual” aggregationfunctions as Σ , Gmax (leximax), etc. leads to IC merging operators.

Belief revision operators can be viewed as belief merging operators restricted to singleton profiles, so that a revisionoperator ◦ is a mapping associating a base K and a formula μ with a new base K ◦ μ. Note that in contrast with ICmerging operators which consider that all the bases forming a profile are consistent, revision operators do not require basesK to be consistent; however, this slight difference between belief merging operators and belief revision operators does notaffect the results presented in the rest of this section.

AGM postulates [11,12] for belief revision are direct counterparts of IC postulates for singleton profiles. Stricto sensu,the following postulates are not the AGM ones but those pointed out by Katsuno and Mendelzon in [12]; since they areequivalent to the AGM ones when a finite propositional logic is considered and belief sets are represented by the deductiveclosures of formulae, this misuse of language is harmless; it also prevents from any confusion with the KM postulatesdedicated to belief update:

Definition 12 (AGM revision operator). A revision operator ◦ is an AGM revision operator if and only if for every formula μ,μ1, μ2, for every base K , K1, K2, it satisfies the following postulates:

(R1) K ◦ μ |� μ;(R2) If K ∧ μ is consistent, then K ◦ μ ≡ K ∧ μ;(R3) If μ is consistent, then K ◦ μ is consistent;(R4) If K1 ≡ K2 and μ1 ≡ μ2, then K1 ◦ μ1 ≡ K2 ◦ μ2;(R5) (K ◦ μ1) ∧ μ2 |� K ◦ (μ1 ∧ μ2);(R6) If (K ◦ μ1) ∧ μ2 is consistent, then K ◦ (μ1 ∧ μ2) |� (K ◦ μ1) ∧ μ2.

For instance, the AGM postulate (R2) (resp. (R3)) for belief revision corresponds to the IC postulate (IC2) (resp. (IC1)).Then, as to the case of IC merging operators, AGM revision operators can be characterized in terms of total preorders overworlds. Indeed, each AGM revision operator corresponds to a faithful assignment [12], that is, a syncretic assignment (cf.Definition 10) restricted to singleton profiles.4

Definition 13 (Faithful assignment). A faithful assignment is a mapping which associates with every base K a preorder �K

over worlds and such that for every base K , K1, K2, it satisfies the following conditions:

(1) If ω |� K and ω′ |� K , then ω �K ω′;(2) If ω |� K and ω′ �|� K , then ω <K ω′;(3) If K1 ≡ K2, then �K1=�K2 .

Theorem 2. (See [12].) A revision operator ◦ is an AGM revision operator if and only if there exists a faithful assignment associatingevery base K with a total preorder �K over worlds such that for every formula μ,

Mod(K ◦ μ) = min(Mod(μ),�K

).

3 When K is inconsistent, since min is associative, d(ω, K ) is defined as the maximum distance between any pair of worlds, which exists since this setis finite – this value is the neutral element for min as expected.

4 The conditions that a syncretic assignment has to satisfy are more demanding than those that faithful assignments must satisfy (see [9] for details).Especially, it is not the case that every faithful assignment can be extended to a syncretic assignment.


Noteworthy, AGM revision operators include Dalal’s revision operator ◦Dal [11] and the drastic revision operator ◦D .Dalal’s revision operator ◦Dal (respectively, the drastic revision operator ◦D ) corresponds for any aggregation function f tothe distance-based merging operator �dH , f (resp. �dD , f ) when restricted to singleton profiles.5 The corresponding faithfulassignments associate respectively with every base K the total preorders �H

K and �DK over worlds such that ω �H

K ω′ if andonly if dH (ω, K ) � dH (ω′, K ) and ω �D

K ω′ if and only if dD(ω, K ) � dD(ω′, K ). In a simpler way, ◦D can be defined forevery base K and every formula μ as:

K ◦D μ ≡{

K ∧ μ if K ∧ μ is consistent,

μ otherwise.

A last family of change operators that we consider in this paper is that of belief update operators [13]. Like a revisionoperator, an update operator � is a mapping associating a base K and a formula μ with a new base K � μ. However,update operators perform a model-wise change, whereas belief revision operators make a change at the whole base level(see [13] for a discussion). The expected behavior for update operators is captured through the following set of KM postu-lates [13]:

Definition 14 (KM update operator). An update operator � is a KM update operator if and only if for every formula μ, μ1,μ2, for every base K , K1, K2, it satisfies the following postulates:

(U1) K � μ |� μ;(U2) If K |� μ, then K � μ ≡ K ;(U3) If K is consistent and μ is consistent, then K � μ is consistent;(U4) If K1 ≡ K2 and μ1 ≡ μ2, then K1 � μ1 ≡ K2 � μ2;(U5) (K � μ1) ∧ μ2 |� K � (μ1 ∧ μ2);(U6) If (K � μ1) |� μ2 and (K � μ2) |� μ1, then K � μ1 ≡ K � μ2;(U7) If K is a complete base, then (K � μ1) ∧ (K � μ2) |� K � (μ1 ∨ μ2);(U8) (K1 ∨ K2) � μ ≡ (K1 � μ) ∨ (K2 � μ).

We have the following characterization theorem for KM update operators in terms of partial preorders over worlds:

Theorem 3. (See [12].) An update operator � is a KM update operator if and only if there exists a faithful assignment associating everyinterpretation ω with a preorder �ω such that for every base K and formula μ,

Mod(K � μ) =⋃ω|�K

min(Mod(μ),�ω

).

KM update operators include Forbus’ operator �F [14], which is the distance-based update operator based on the Ham-ming distance, and the drastic update operator �D , which is the distance-based update operator based on the drasticdistance. Accordingly, the corresponding faithful assignments associate respectively with every world ω the total pre-orders �H

ω and �Dω over worlds such that ω′ �H

ω ω′′ if and only if dH (ω′,ω) � dH (ω′′,ω) and ω′ �Dω ω′′ if and only if

dD(ω′,ω)� dD(ω′′,ω). In a simpler way, the drastic update operator �D can be defined for every base K and every formulaμ as:

K �D μ ≡{

K if K |� μ,

μ otherwise.

6.2. Language independence for belief change

As a case study, we have investigated how some belief change operators from the literature are language-independent.As Proposition 4 suggests, �-language independence is hardly achieved by operators which have to deal with jointly

inconsistent (but typically individually consistent) propositional information, while avoiding trivialization. Surprisingly,KM postulates for update operators are compatible with �-language independence, as stated by the following proposi-tion:

Proposition 11. The KM drastic update operator �D is �-language independent.

However, no similar result holds for AGM revision operators and IC merging operators. Actually, focusing on translationssatisfying TPR or FCP is not enough as well, since AGM revision operators and IC merging operators are nor TPR-languageindependent, nor FCP-language independent; more precisely:

5 Here, when K is not consistent, we set dD (ω, K ) = dH (ω, K ) = 0.


Proposition 12. Let P be a property on symbol translations which does not imply AIP, i.e., there exists a symbol translation satisfyingP but not satisfying AIP. Then no belief revision operator ◦ satisfying (R2) and (R3) and no belief merging operator � satisfying (IC1)

and (IC2) is P-language independent.

Hence, for AGM revision operators and IC merging operators one needs to consider forms of language independencethat are at least as restrictive as AIP-language independence. We first investigate whether AIP-language independence iscompatible with AGM and IC postulates. The following proposition gives a positive answer to this question; actually, itturns out that there is exactly one AGM revision operator (resp. exactly one IC merging operator, exactly one KM updateoperator) that is AIP-language independent; furthermore, this AGM revision operator (resp. IC merging operator, KM up-date operator) is the unique AGM revision operator (resp. IC merging operator, KM update operator) that is BAP-languageindependent:

Proposition 13.

1. The distance-based merging operator �dD ,Σ , based on the drastic distance dD and on the aggregation function Σ , is the only ICmerging operator that is AIP-language independent; it is also the only IC merging operator that is BAP-language independent;

2. The drastic revision operator ◦D is the only AGM revision operator that is AIP-language independent; it is also the only AGMrevision operator that is BAP-language independent;

3. The drastic update operator �D is the only KM update operator that is AIP-language independent; it is also the only KM updateoperator that is BAP-language independent.

Note that point 1 of Proposition 13 actually holds for every merging operator �dD , f , based on the drastic distance dD

and on an aggregation function f satisfying non-decreasingness, minimality, identity, symmetry and strict monotony, sinceall these operators are equal to �dD ,Σ [15].

Proposition 13 can be considered as a representation theorem for the three operators �dD ,Σ , ◦D , and �D . Note that in[16] Konieczny et al. provide a first full axiomatic characterization of the distance-based merging operator �dD ,Σ in termsof independence property to rationalization by expansion. Indeed, �dD ,Σ is the only IC merging operator � satisfying thefollowing property: for every profile 〈K1, . . . , Km〉, for every formula μ,

�μ

(〈K1, . . . , Km〉) ≡ �μ

(〈K1 ∧ μ, . . . , Km ∧ μ〉).Proposition 13 provides a further characterization of the distance-based operator �dD ,Σ in terms of language independence.It shows that, under IC postulates, the notion of independence to rationalization by expansion and the one of AIP-languageindependence (or, BAP-language independence) coincide.

Let us now go back to the language independence issue. Since there is only one belief revision operator, one beliefmerging operator and one belief update operator that is BAP-language independent, in order to investigate how other beliefchange operators behave, one needs to consider a weaker form of language independence. As SIN-language independenceasks the choice of the symbols names to be non-significant, this property is highly expected for belief change operators.However, quite surprisingly, SIN-language independence is not always guaranteed by rational operators:

Proposition 14. There exist belief revision (resp. belief merging, belief update) operators which satisfy all AGM postulates (resp. all ICpostulates, all KM postulates) but are not SIN-language independent.

This calls for three further axioms expressing such a form of language independence ((SIN-R) for belief revision, (SIN-M)for belief merging, and (SIN-U) for belief update), stating respectively that for every bijective symbol translation σ from PSto PS:

(SIN-R) σ(K ◦ μ) ≡ σ(K ) ◦ σ(μ);(SIN-M) σ(�μ(E)) ≡ �σ(μ)(σ (E));(SIN-U) σ(K � μ) ≡ σ(K ) � σ(μ).

Those axioms complement the ones related to syntax independence ((R4) for belief revision, (U4) for belief updateand (IC3) for belief merging) and seem highly desirable. Actually, the distance-based belief change operators, basedon “standard distances” (like the drastic distance and the Hamming one), satisfy those axioms. More generally, whenPS = {x1, . . . , xn} contains n atoms, let a distance d over interpretations over X be decomposable when there existsa mapping fd : Rn → R symmetric in each argument and a mapping gd : {0,1} × {0,1} → R such that d(ω1,ω2) =fd(gd(ω1(x1)),ω2(x1)), . . . , gd(ω1(xn),ω2(xn)). We have:

Proposition 15. Every belief change operator based on a decomposable distance is SIN-language independent.


Table 1The subclasses of rational belief change operators satisfying different forms of language independence.

IC merging operator AGM revision operator KM update operator

SIN-language independence {�d, f | d is decomposable} ⊆ O ⊂ IC {◦d | d is decomposable} ⊂ O ⊂ AGM {�d | d is decomposable} ⊂ O ⊂ KMBAP-language independence O = {�dD ,Σ } O = {◦D } O = {�D }AIP-language independence O = {�dD ,Σ } O = {◦D } O = {�D }FCP-language independence O = ∅ O = ∅ O = {�D }TPR-language independence O = ∅ O = ∅ O = {�D }�-language independence O = ∅ O = ∅ O = {�D }

Clearly enough, both dD and dH are decomposable; for the drastic distance dD , we can take fdD = max and gdD (x, y) = 0if x = y, = 1 otherwise; for the Hamming distance dH , we can take fdH = Σ and gdH = gdD . Thus, as a direct consequenceof Proposition 15, Dalal’s revision operator ◦Dal and Forbus’ update operator �F are SIN-language independent.

The following proposition shows that the converse of Proposition 15 does not hold for AGM revision operators and KMupdate operators; that is, there exist rational belief revision/update operators that are not based on a decomposable distanceand are SIN-language independent. We actually prove the stronger result:

Proposition 16. There exist belief revision (resp. belief update) operators which satisfy all AGM postulates (resp. all KM postulates),are not distance-based operators but satisfy (SIN-R) (resp. (SIN-U)).

The question remains open for belief merging operators which satisfy all IC postulates, i.e., it is not known whether thereexist IC merging operators which are not based on a decomposable distance and satisfy (SIN-M). In fact, as far as we knowthe existence of IC merging operators which are not distance-based is still an open issue; however, given any property P onrevision/update operators, in general it is easy to exhibit an AGM revision operator (resp. a KM revision operator) which isnot distance-based and satisfy P ; this is due to the weakness of conditions (1–3) of the faithful assignment which do notrequire any dependence between the preorders �K and �K ′ , for any base K , K ′ with K �≡ K ′ .

Table 1 summarizes the results presented in the section with a systematic picture of the dependencies between rationalbelief change operators and the different forms of language independence of interest in this paper; for each family ofbelief change operators and each property P on symbol translations, it provides some conditions on the sets O of rationalchange operators satisfying P -language independence. In the table, IC (resp. AGM, KM) denotes the set of all IC merging(resp. AGM revision, KM update) operators. The obtained characterization results are complete ones, except for SIN-languageindependence.

7. Other related work

Independence is well-known as a key notion in many domains within artificial intelligence, especially graphical modelsfor representing beliefs (e.g., Bayes nets) or preferences (e.g., CP-nets or GAI-nets). In propositional logic, since Belnap’s workabout relevance logic in the sixties, a number of concepts of independence (also referred to as irrelevance, separability, etc.)has been pointed out in the literature (see [17,18] for a survey). Qualitative independence, i.e., how the fact of learninga new piece of evidence individually affects previous beliefs, has been studied as well for a while [19]. None of thesenotions of independence in propositional logic are directly related to a form of language independence (in all the workscited above, no symbol translation is considered), and none of those papers is concerned with the language independenceissue of propositional belief change operators.

Makinson [20] studied a notion of relevance (canonical relevance) which is expected to be respected in belief change(roughly, contracting a belief base/set K with some piece of information α should preserve every consequence β of K whenα is irrelevant to β). While canonical relevance is syntax-independent in the usual sense, Makinson shows that it is notlanguage-independent; revisiting Makinson’s counter-example in the light of our work, we can state more precisely thatcanonical relevance is SIN-language independent, but it is not REV-language independent. Though language independenceis considered in this paper in the context of belief change, the focus is on canonical relevance; unlike our paper, no inves-tigation of the concept of language independence is reported and no study of robustness w.r.t. symbol translations of manybelief change operators is provided.

More closely related to our work is [21]. In this paper, Jaeger points out a very general setting for representation in-dependence and investigates the representation independence of a number of non-monotonic inference rules. The notionof representation independence of inference relations pointed out by Jaeger is anchored on the concept of representationalvariants. When instantiated to classical propositional logic and assuming that so-called “abstract interpretations” are lim-ited to symbol translations (which is a natural choice in this case, see [21]), representational variants can be defined asfollows: two propositional formulae φ ∈ PROPY1 and ψ ∈ PROPY2 are representational variants w.r.t. σ : X → PROPY1 andτ : X → PROPY2 if and only if for any formula α ∈ PROPX , we have φ |� σ(α) if and only if ψ |� τ (α). Then an infer-ence relation � is representation independent whenever two representational variants φ ∈ PROPY1 and ψ ∈ PROPY2 w.r.t.σ : X → PROPY1 and τ : X → PROPY2 have the same consequences modulo the corresponding “abstract interpretations” σand τ , i.e., for any formula α ∈ PROPX , we have φ � σ(α) if and only if ψ � τ (α).


Interestingly, Jaeger points out the need to restrict the set of admissible “abstract interpretations” to make the conceptof representation independence meaningful (this is reminiscent to the restriction on symbol translations we considered).

Despite this similarity, Jaeger’s work and the present one differ in a number of aspects. Thus, Jaeger’s setting is notrestricted to classical propositional logic but is suited to more general logical systems. In addition, even when instantiatedto classical propositional logic and lifted to propositional operators, Jaeger’s representation independence does not coincidewith our concept of language independence. Indeed, let us state that two n-propositional structures Σ = 〈α1, . . . ,αn〉 overY1 and Σ ′ = 〈α′

1, . . . ,α′n〉 over Y2 are representational variants w.r.t. σ : X → PROPY1 and τ : X → PROPY2 if and only for

each i ∈ {1, . . . ,n}, αi and α′i are representational variants w.r.t. σ and τ ; and let us state that a propositional operator

Ω is representation independent if and only if whenever Σ and Σ ′ are representation independent w.r.t. σ and τ , Ω(Σ)

and Ω(Σ ′) are also representation independent w.r.t. σ and τ . It turns out that this concept of representational inde-pendence does not coincide with language independence. Thus, consider the propositional operator Ω¬ which associatesevery 1-propositional structure 〈α〉 with the 1-propositional structure 〈¬α〉; obviously enough, Ω¬ is �-language inde-pendent; however, it is not representation independent: take Σ = 〈�〉 and Σ ′ = 〈x〉, and consider the symbol translationsσ : ∅ → PROPY1 and τ : ∅ → PROPY2 . There are only two distinct formulae α ∈ PROP∅ up to logical equivalence, namely �and ⊥. Since � |� �, x |� �, � �|� ⊥, and x �|� ⊥, � and x are representational variants in Jaeger’s sense. Accordingly, Σ andΣ ′ are representational variants w.r.t. σ and τ . However, Ω¬(Σ) and Ω¬(Σ ′) are not representational variants w.r.t. σ andτ since ¬� |� ⊥ while ¬x �|� ⊥.

Finally, while Jaeger investigates the representational independence of many non-monotonic inference relations (includ-ing forms of closed world reasoning, rational closure, and propositional probabilistic inference), he does not focus on thelanguage independence issue for belief change operators. This makes another significant difference between his work andthe present work.

8. Conclusion

This paper is centered on the concept of language independence in finite propositional logic. The main contributions areas follows:

• We have defined language independence as robustness w.r.t. symbol translations.• We have provided a model-based characterization of symbol translations in terms of injective, surjective relations over

worlds.• We have motivated the need to focus on symbol translations of different types, i.e., satisfying some additional properties,

which leads to several notions of language independence. We have provided a model-based characterization for eachone of these properties, and we have shown how they are logically connected.

• We have identified the complexity of recognizing symbol translations satisfying properties of interest.• We have investigated the robustness of belief change operators w.r.t. translations of different types. Only limited forms

of language independence are satisfied in the general case. We have shown that the distance-based operator �dD ,Σ ,where the drastic distance dD and sum as an aggregation function are used, as well as its corresponding drastic revisionand update operators are the most robust rational operators in terms of language independence, since they are the onlyones that are AIP-language independent (resp. BAP-language independent). Moreover, even SIN-language independenceis not guaranteed for rational belief change operators.

This work paves the way for three main directions for further research. A first one concerns belief change; deriving“constructive” representation theorems characterizing the class of operators satisfying SIN-language independence would beuseful. A second direction consists in investigating the “degree” of language independence of other propositional operatorsfrom the literature (for instance, abduction, circumscription and other forms of closed-world reasoning). A third one consistsin determining how our results can be lifted to other propositional settings.

Acknowledgements

We would like to thank Andreas Herzig for pointing out Ref. [21], and the anonymous reviewers for their helpful com-ments.

Appendix A. Proofs

Proposition 2. Let α be a formula from PROPX . Let σ be a symbol translation from X to PROPY . We have∧

x∈X (x ⇔ σ(x)) |� α ⇔σ(α).

Proof.∧

x∈X (x ⇔ σ(x)) is equivalent to∧

x∈X (x ∧ σ(x)) ∨ (¬x ∧ ¬σ(x)). Distributing∧

over ∨, given the substitutiontheorem, we get that

∧x∈X (x ⇔ σ(x)) is equivalent to

∨ω∈2PS (ω ∧ σ(ω)). Now, given an interpretation ω, there are two

cases:


• ω |� α. Then from point 1 of Corollary 1, we also have σ(ω) |� σ(α). Hence, ω ∧ σ(ω) |� α ∧ σ(α) |� α ⇔ σ(α).• ω |� ¬α. Then from point 1 of Corollary 1, we also have σ(ω) |� ¬σ(α). Hence, ω ∧ σ(ω) |� ¬α ∧ ¬σ(α) |� α ⇔ σ(α).

Whatever the case, ω ∧ σ(ω) |� α ⇔ σ(α). Hence, we have∨

ω∈2PS (ω ∧ σ(ω)) |� α ⇔ σ(α). This concludes the proof. �Proposition 3. The semantical mapping Ψ of symbol translations is a one-to-one correspondence from PROPX

Y / ≡, the set of all symboltranslations (modulo equivalence), to the set of all injective, surjective relations over 2X × 2Y . Its inverse mapping Ψ −1 associates forevery subsets X, Y of PS, an injective, surjective relation R over 2X × 2Y with a representative symbol translation, denoted σR , suchthat for every propositional symbol x ∈ X,

σR(x) =∨{

ω′ ∈ 2Y∣∣ ω′ ∈ R(ω),ω |� x

}.

Proof. The proof takes advantage of the following lemma:

Lemma 1. Let σ : X → PROPY , α ∈ PROPX and ω′ |� σ(α). There is one and only one interpretation ω ∈ 2X such that ω′ |� σ(ω);moreover, ω |� α.

Proof. Let σ : X → PROPY , α ∈ PROPX and ω′ |� σ(α). From the substitution theorem σ(α) ≡ σ(∨{ω | ω |� α}), that is,

σ(α) ≡ ∨{σ(ω) | ω |� α}. Hence, ω′ |� ∨{σ(ω) | ω |� α}. Since all the distinct models ω, ω′′ of α are jointly inconsistent,from Corollary 1, it is also the case that for all such pairs ω, ω′′ , their images by σ are jointly inconsistent (i.e., σ(ω) ∧σ(ω′′) |� ⊥). Therefore, from ω′ |� ∨{σ(ω) | ω |� α}, we get that there exists a model ω of α such that ω′ |� σ(ω). Supposethat there exists another interpretation ω′′ |� α such that ω′ |� σ(ω′′). Again, since ω ∧ ω′′ |� ⊥, from Corollary 1, we getthat σ(ω) ∧ σ(ω′′) |� ⊥. This conflicts with the fact that ω′ |� σ(ω) and ω′ |� σ(ω′′). Thus, ω is the unique model of αsuch that ω′ |� σ(ω). Finally, suppose that a model ω′′ of ¬α satisfies ω′ |� σ(ω′′). Then we would get from Corollary 1that σ(ω′′) |� σ(¬α). Hence by transitivity of entailment, we would have ω′ |� σ(¬α). But we know that ω′ |� σ(ω), andσ(¬α) ∧ σ(α) ≡ σ(¬α ∧ α) ≡ σ(⊥) ≡ ⊥ is inconsistent, and ω′ should be a model of it, contradiction. �

We now prove Proposition 3:From Lemma 1 there is a unique interpretation ω ∈ 2X such that ω′ |� σ(ω). Therefore Ψ (σ ) = Rσ is an injective,

surjective relation. We now show that Ψ is injective and surjective:(Injectivity of Ψ ) Let σ ,σ ′ two symbol translations from X to PROPY , which are not equivalent. Then there exists x ∈ X

such that σ(x) �≡ σ ′(x). Hence, we have σ(x) �|� σ ′(x) or σ ′(x) �|� σ(x). Assume that σ(x) �|� σ ′(x) (the remaining case issymmetric). Then there exists ω′ ∈ 2Y such that ω′ |� σ(x) and ω′ �|� σ ′(x) (or equivalently, ω′ |� σ ′(¬x)). From Lemma 1∃ω1 ∈ 2X such that ω1 |� x and ω′ |� σ(ω1), hence ∃ω1 |� x such that ω′ ∈ Rσ (ω1). From Lemma 1 again, ∃ω2 ∈ 2X suchthat ω2 |� ¬x and ω′ |� σ ′(ω2), hence ∃ω2 |� ¬x such that ω′ ∈ Rσ ′ (ω2) (where, of course, Rσ ′ stands for Ψ (σ ′)). Sinceω1 |� x and ω2 |� ¬x, ω1 and ω2 are distinct interpretations. Since ω1 is the unique interpretation such that ω′ |� σ(ω1),we have ω′ �|� σ(ω2), i.e., ω′ /∈ Rσ (ω2). The fact that ω′ ∈ Rσ ′ (ω2) enables to conclude that Rσ �= Rσ ′ .

(Surjectivity of Ψ ) Let R ′ be an injective, surjective relation on 2X × 2Y and let σR ′ be the symbol translation from Xto PROPY such that for every propositional symbol x ∈ 2X , σR ′(x) = ∨{ω′ ∈ 2Y | ω′ ∈ R ′(ω),ω |� x}. We show now thatRσR′ (= Ψ (σR ′)) = R ′ . Let ω ∈ 2X . Let X+

ω = {x ∈ X | ω |� x} and X−ω = {x ∈ X | ω |� ¬x}. We have σR ′ (ω) ≡ ∧

x∈X+ω{σR ′ (x)} ∧∧

x∈X−ω{σR ′(¬x)} ≡ ∧

x∈X+ω{σR ′(x)} ∧ ∧

x∈X−ω{¬σR ′(x)} ≡ ∧

x∈X+ω{∨{ω′ ∈ 2Y | ω′ ∈ R ′(ω′′),ω′′ |� x}} ∧ ∧

x∈X−ω{¬∨{ω′ ∈ 2Y |

ω′ ∈ R ′(ω′′),ω′′ |� x}}. Now, let ω∗ ∈ 2Y . We have ω∗ |� σR ′(ω) if and only if (ω∗ |� ∧x∈X+

ω{∨{ω′ ∈ 2Y | ω′ ∈ R ′(ω′′),ω′′ |�

x}} and ω∗ |� ∧x∈X−

ω{¬∨{ω′ ∈ 2Y | ω′ ∈ R ′(ω′′),ω′′ |� x}}) if and only if (∀x ∈ X+

ω , ω∗ ∈ ⋃{R ′(ω′′) | ω′′ |� x} and ∀x ∈ X−ω ,

ω∗ /∈ ⋃{R ′(ω′′) | ω′′ |� x}) if and only if (ω∗ ∈ ⋃{R ′(ω′′) | ω′′ |� ∧x∈X+

ωx} and ω∗ /∈ ⋃{R ′(ω′′) | ω′′ |� ∨

x∈X−ω

x}). But sinceR ′ is an injective, surjective relation we get that ω∗ /∈ ⋃{R ′(ω′′) | ω′′ |� ∨

x∈X−ω

x} if and only if ω∗ ∈ ⋃{R ′(ω′′) | ω′′ |�∧x∈X−

ω¬x} (the if part comes from the injectivity of R ′ , the only if part comes from the surjectivity of R ′). Hence, we

have ω∗ |� σR ′ (ω) if and only if (ω∗ ∈ ⋃{R ′(ω′′) | ω′′ |� ∧x∈X+

ωx} and ω∗ ∈ ⋃{R ′(ω′′) | ω′′ |� ∧

x∈X−ω

¬x}) if and only ifω∗ ∈ ⋃{R ′(ω′′) | ω′′ |� ∧

x∈X+ω

x ∧ ∧x∈X−

ω¬x} if and only if ω∗ ∈ ⋃{R ′(ω′′) | ω′′ = ω} if and only if ω∗ ∈ R ′(ω). We got that

ω∗ ∈ R ′(ω) if and only if ω∗ |� σR ′ (ω). Furthermore, by definition of RσR′ , we have ω∗ |� σR ′(ω) if and only if ω∗ ∈ RσR′ (ω).Hence, ω∗ ∈ R ′(ω) if and only if ω∗ ∈ RσR′ (ω). Therefore, RσR′ = R ′ , which concludes the proof. �Proposition 4. Let α be a formula of PROPX and β a formula of PROPY . If α is neither valid nor inconsistent, then there exists a symboltranslation σ ∈ PROPX

Y such that σ(α) ≡ β .

Proof. We take advantage of the semantical mapping of symbol translations (cf. Definition 5). Let α be a formula fromPROPX that is consistent and not valid, and let β be a formula from PROPY . Let ω,ω′ be two (distinct) interpretationsover X such that ω |� α and ω′ �|� α. We construct the binary relation R on 2X × 2Y as follows: R(ω) = Mod(β), R(ω′) =Mod(¬β) and for every interpretation ω′′ over X distinct from ω and ω′ , R(ω′′) = ∅. R is obviously an injective, surjective


relation, so that from Proposition 3 there is a symbol translation Ψ −1(R) = σR that satisfies σR(ω) ≡ β , i.e., that satisfiesσR(α) ≡ β . �Proposition 5. REV and AIP are equivalent.

Proof. First of all, we show that we can assume without loss of generality that the input symbol translation σ : X → PROPYis such that X ∩ Y = ∅. Indeed, let us prove that a symbol translation σ satisfies REV iff the symbol translation σ ′ : X →PROPY ′ given by ∀x ∈ X , σ ′(x) = σ(x)′ satisfies REV as well (note that Y ′ is such that X ∩Y ′ = ∅). By construction, σ ′ is equalto the restriction over X of the symbol translation τ ◦ σ where τ : Y → PROP′

Y is the bijective symbol translation such that∀y ∈ Y , τ (y) = y′ . If σ ′ is such that there exists a symbol translation θ ′ from Y ′ to PROPX such that ∀x ∈ X , θ ′(σ ′(x)) ≡ x,then we have ∀x ∈ X , θ ′(τ (σ (x))) ≡ x; accordingly, the symbol translation θ = θ ′ ◦ τ is such that ∀x ∈ X , θ(σ (x)) ≡ x. Theother way around, if σ is such that there exists a symbol translation θ from Y to PROPX such that ∀x ∈ X , θ(σ (x)) ≡ x,then we also have ∀x ∈ X , θ(τ−1(τ (σ (x)))) ≡ x; this shows that the symbol translation θ ′ = θ ◦ τ−1 is such that ∀x ∈ X ,θ ′(σ ′(x)) ≡ x.

Suppose now that σ : X → PROPY is such that X ∩ Y = ∅. By definition, σ satisfies REV iff there exists a symbol trans-lation θ from Y to PROPX such that ∀xi ∈ X = {x1, . . . , xn}, θ(σ (xi)) ≡ xi iff there exists a symbol translation θ from Y toPROPX such that ∀xi ∈ X , θ(σ (xi)) ⇔ xi is valid. Since ∀xi ∈ X , Var(σ (xi)) ⊆ Y and X ∩ Y = ∅, this is also equivalent tostating that there exists a symbol translation θ from Y to PROPX such that ∀xi ∈ X , θ(σ (xi) ⇔ xi) is valid. This holds iffthere exists a symbol translation θ from Y to PROPX such that θ(

∧xi∈X (σ (xi) ⇔ xi)) is valid.

Now, for each y j ∈ Y , the restriction y j �→ θ(y j) of θ can be viewed as a Skolem function, so that there existsa symbol translation θ from Y to PROPX such that θ(

∧xi∈X (σ (xi) ⇔ xi)) is valid iff the quantified Boolean formula

∀X .(∃Y .(∧

xi∈X (σ (xi) ⇔ xi))) is valid (see e.g. [22] for details). This holds iff the quantified Boolean formula (with freevariables X)

∃Y .

( ∧xi∈X

(σ(xi) ⇔ xi

))

is valid. Now,∧

xi∈X (σ (xi) ⇔ xi) is equivalent to∨

s1,...,sn∈{+,−}(∧

xi∈X (σ (xi)si ∧ xsi

i )). Hence, ∃Y .(∧

xi∈X (σ (xi) ⇔ xi)) isequivalent to

∃Y .

( ∨s1,...,sn∈{+,−}

( ∧xi∈X

(σ(xi)

si ∧ xsii

))),

which in turn is equivalent to∨

s1,...,sn∈{+,−} ∃Y .(∧


i )), since existential quantifications “distributes over”

disjunctions (see e.g. [23] for details). Finally, ∃Y .∧


i ) is equivalent to

∧xi∈X

(xsi

i ∧ ∃Y .

( ∧xi∈X

σ(xi)si

)).

Since Var(∧

xi∈X σ(xi)si ) ⊆ Y , the formula ∃Y .(

∧xi∈X σ(xi)

si ) is equivalent to � if∧

xi∈X σ(xi)si is consistent and to ⊥

otherwise (see e.g. [23]). Thus,∨

s1,...,sn∈{+,−} ∃Y .(∧


i )) is equivalent to∨

s1,...,sn∈{+,−}∧

xi∈X xsii iff for each

s1, . . . , sn ∈ {+,−},∧

xi∈X σ(xi)si is consistent. The fact that

∨s1,...,sn∈{+,−}

∧xi∈X xsi

i is valid completes the proof. �Proposition 6. Let σ be a symbol translation from PROPX

Y and Rσ be its semantical mapping:

1. σ satisfies AIP if and only if Rσ is left-total;2. σ satisfies BAP if and only if Rσ is a one-to-one correspondence;3. σ satisfies FCP if and only if Rσ is functional;4. σ satisfies SIN if and only if Rσ is a one-to-one correspondence which is decomposable, i.e., there exists a bijection ηRσ from X

to Y defined by ∀x ∈ X, ∀y ∈ Y , ηRσ (x) = y if and only if {Rσ (ω) | ω ∈ 2X ,ω |� x} = {ω′ | ω′ ∈ 2Y ,ω′ |� y};5. σ satisfies TPR if and only if ∀x ∈ X, ∃ω,ω′ ∈ 2X , ω |� x, ω′ |� ¬x, Rσ (ω) �= ∅ and Rσ (ω′) �= ∅.

Proof.

1. AIP. Let σ be a symbol translation from X = {x1, . . . , xn} to PROPY satisfying AIP. Then ∀s1, . . . , sn ∈ {+,−},∧

xi∈X σ(xi)si

is consistent, thus σ(∧

xi∈X xsii ) is consistent, so ∀ω ∈ 2X , Rσ (ω) �= ∅. Therefore, Rσ is left-total. Conversely, let X, Y ⊆ PS

and R be an injective, surjective, left-total relation on 2X × 2Y . For every ω ∈ 2X , R(ω) �= ∅ since R is left-total. Thusfor every ω ∈ 2X , σR(ω) ≡ ∨{ω′ | ω′ ∈ R(ω)} is consistent. Furthermore ∀s1, . . . , sn ∈ {+,−},

∧xi∈X σR(xi)

si , which is

equivalent to σR(∧

xi∈X xsii ), is equivalent to σR(ω) for the interpretation ω ∈ 2X which is equivalent to

∧xi∈X xsi

i . Thus∀s1, . . . , sn ∈ {+,−},

∧x ∈X σR(xi)

si is consistent. Therefore, σR satisfies AIP.
i


2. BAP. Let σ be a symbol translation from X to PROPY satisfying BAP. For every complete formula α from PROPX , σ(α)

is a complete formula from PROPY , thus for every interpretation ω ∈ 2X , σ(ω) admits exactly one model, i.e., Rσ (ω) isa singleton. This means that Rσ is left-total and functional. Conversely, let X, Y ⊆ PS and R be a one-to-one correspon-dence on 2X × 2Y . For every complete formula α from PROPX , let ω be its unique model. Since R is one-to-one, R(ω) isa singleton. Thus for every complete formula α, σR(α) is a complete formula from PROPY . Therefore, σR satisfies BAP.

3. FCP. Let σ be a symbol translation from X = {x1, . . . , xn} to PROPY satisfying FCP. Let ω′1,ω

′2 ∈ 2Y such that ω′

1 �=ω′

2. Since σ satisfies FCP, there is a formula α1 from PROPX such that σ(α1) ≡ ω′1 and there is a formula α2 from

PROPX such that σ(α2) ≡ ω′2. Then since ω′

1 �= ω′2, we have σ(α1) ∧ σ(α2) |� ⊥. Now, by Lemma 1 (see the proof of

Proposition 3), since ω′1 |� σ(α1) and ω′

2 |� σ(α2), there is a unique interpretation ω1 ∈ 2X such that ω′1 |� σ(ω1), i.e.,

such that ω′1 ∈ Rσ (ω1), and there is a unique interpretation ω2 ∈ 2X such that ω′

2 |� σ(ω2), i.e., such that ω′2 ∈ Rσ (ω2);

moreover, ω1 |� α1 and ω2 |� α2. Towards a contradiction, assume that ω1 = ω2. Then ω1 |� α1 ∧ α2. Thus σ(ω1) |�σ(α1) ∧ σ(α2) |� ⊥. This contradicts ω′

1 |� σ(ω1). Hence, ω1 �= ω2. This shows that Rσ is functional. Conversely, letX, Y ⊆ PS and R be an injective, surjective, functional relation on 2X × 2Y . Since R is surjective and functional, for everyinterpretation ω′ ∈ 2Y , there exists a unique interpretation R−1(ω′) ∈ 2X such that (R−1(ω′),ω′) ∈ R . Accordingly, ω′ isequivalent to σR(R−1(ω′)). Furthermore, since R−1(ω′) ∈ 2X , we have that R−1(ω′) is equivalent to

∧ni=1 xsi

i for somes1, . . . , sn ∈ {+,−}. As a consequence, σR(R−1(ω′)) is equivalent to

∧ni=1 σR(xi)

si for some s1, . . . , sn ∈ {+,−}, whichshows that σR(R−1(ω′)) (hence, ω′) is definable from {σR(x) | x ∈ X}. Finally, let β be any formula from PROPY . Sinceevery model ω′ of β is definable from {σR(x) | x ∈ X}, this is also the case of their disjunction. Thus, β is definable from{σR(x) | x ∈ X}, which shows that σR satisfies FCP.

4. SIN. Let σ be a symbol translation from X = {x1, . . . , xn} to PROPY satisfying SIN. For every interpretation ω ∈ 2X , ω isequivalent to

∧xi∈X xsi

i , for some s1, . . . , sn ∈ {+,−}. Hence σ(ω) ≡ σ(∧

xi∈X xsii ) ≡ ∧

xi∈X σ(xi)si ≡ ∧

yi∈Y ysii , that is, an

element of 2Y . Hence, for every ω ∈ 2X , Rσ (ω) is a singleton. This means that Rσ is left-total and functional. Therefore,Rσ is a one-to-one correspondence. Now from Corollary 1, ∀x ∈ X , ∀ω ∈ 2X , we have ω |� x iff σ(ω) |� σ(x). Thus thereexists a bijection ηRσ = σ from X to Y such that ∀x ∈ X , ∀y ∈ Y , σ(x) = y if and only if {Rσ (ω) | ω ∈ 2X ,ω |� x} ={ω′ | ω′ ∈ 2Y ,ω′ |� y}. Conversely, let X, Y ⊆ PS and R be a one-to-one correspondence on 2X × 2Y such that there isa bijection ηR from X to Y defined by ∀x ∈ X , ∀y ∈ Y , ηR(x) = y if and only if {R(ω) | ω ∈ 2X ,ω |� x} = {ω′ | ω′ ∈2Y ,ω′ |� y}. By definition of σR , we have ∀x ∈ X , σR(x) = ∨{ω′ ∈ 2Y | ω′ ∈ R(ω) and ω |� x}. Since each R(ω) is asingleton, this is equal to

∨{R(ω) and ω |� x}. By definition of ηR , {R(ω) and ω |� x} = {ω′ ∈ 2Y | ω′ |� η(x)}. HenceσR(x) = ∨{ω′ ∈ 2Y | ω′ |� η(x)}. But

∨{ω′ ∈ 2Y | ω′ |� η(x)} is equivalent to η(x). Thus σR is equivalent to ηR . Since ηR

is a bijection from X to Y , σR is also a bijection from X to Y , thus σR satisfies SIN.5. TPR. Let σ be a symbol translation from X to PROPY satisfying TPR. Then for every x ∈ X , σ(x) is consistent and not

valid. Let x ∈ X . There exist two interpretations ω1,ω2 from 2Y such that ω1 |� σ(x) and ω2 |� ¬σ(x). By Lemma 1(see the proof of Proposition 3), there exist two interpretations ω,ω′ from 2X , ω |� x, ω′ |� ¬x, such that ω1 |� σ(ω)

and ω2 |� σ(ω′). Therefore, for every x ∈ X , there exist two interpretations ω,ω′ ∈ 2X , ω |� x, ω′ |� ¬x, such thatRσ (ω) �= ∅ and Rσ (ω′) �= ∅. Conversely, let X, Y ⊆ PS and R be an injective, surjective relation on 2X × 2Y such that∀x ∈ X , ∃ω,ω′ ∈ 2X , ω |� x, ω′ |� ¬x, R(ω) �= ∅ and R(ω′) �= ∅. Let x ∈ X . Then there exist two interpretations ω,ω′from 2X such that ω |� x, ω′ |� ¬x and such that σR(ω) and σR(ω′) are both consistent. Yet from point 1 of Corollary 1,σR(ω) |� σR(x) and σR(ω′) |� σR(¬x), this shows that σR(x) and σR(¬x) are both consistent, or equivalently, that σR(x)is consistent and not valid. Therefore, σR satisfies TPR. �

Proposition 7. SIN implies BAP. BAP implies FCP. BAP also implies REV, which implies TPR. Each implication is strict (i.e., the conversedoes not hold). Moreover, if a symbol translation satisfies REV and FCP, then it satisfies BAP. Lastly, FCP does not imply and is notimplied by any of REV or TPR.

Proof.

• SIN implies BAP. Direct from Proposition 6. The implication is strict: let X = {x}, Y = {p} and σ ∈ PROPXY such that

σ(x) = ¬p. Clearly, σ satisfies BAP but does not satisfy SIN.• BAP implies FCP. Direct from Proposition 6. The implication is strict: let X = {x, y}, Y = {p} and σ ∈ PROPX

Y such thatσ(x) = � and σ(y) = p. Clearly, σ satisfies FCP but does not satisfy BAP.

• BAP implies REV . BAP implies AIP from Proposition 6. And from Proposition 5, AIP and REV are equivalent properties.• REV implies TPR. Let σ be a symbol translation from PROPX

Y satisfying AIP. From Proposition 5, σ satisfies REV. Then Rσ

is a left-total relation over 2X × 2Y . Let x ∈ X . Then ∀ω ∈ 2X such that ω |� x, Rσ (ω) �= ∅. Moreover, ∀ω′ ∈ 2X such thatω′ |� ¬x, Rσ (ω′) �= ∅. Since ω �= ω′ and Rσ is injective, we have Rσ (ω)∩ Rσ (ω′) = ∅, thus Rσ (ω) �= 2Y . Therefore, fromProposition 6 σ satisfies TPR. The implication is strict: let X = {x, y}, Y = {p}, and σ ∈ PROPX

Y such that σ(x) = p, andσ(y) = ¬p. σ satisfies TPR but does not satisfy AIP.

• If a symbol translation satisfies REV and FCP, then it satisfies BAP. Direct from Proposition 6.• FCP does not imply REV and is not implied by REV . If FCP would imply REV , then it would imply TPR as well, since REV

implies TPR, but FCP does not imply TPR. On the other hand, let X = {x}, Y = {p,q} and σ ∈ PROPXY such that σ(x) = p.

σ satisfies REV but it does not satisfy FCP.


• FCP does not imply TPR and is not implied by TPR. Indeed, let X = {x}, Y = ∅ and σ ∈ PROPXY such that σ(x) = ⊥. σ

trivially satisfies FCP but does not satisfy TPR. Now, let X = {x}, Y = {p,q} and σ ∈ PROPXY such that σ(x) = p ∨ q. σ

satisfies TPR but does not satisfy FCP. �Proposition 8. The concept of syntax independence and the concept of P -language independence for propositional operators arelogically independent.

Proof. Given Proposition 7, it is enough to point out (1) a propositional operator which is both syntax-independent and�-language independent, (2) a propositional operator which is syntax-independent but not SIN-language independent, (3) apropositional operator which is �-language independent but not syntax-independent, and finally (4) a propositional opera-tor which neither is syntax-independent nor is SIN-language independent.

1. ded (the deduction operator) is both syntax-independent and �-language independent.2. Just consider the AGM revision operator ◦ given in the proof of Proposition 14.3. Consider for instance the propositional operator Ω that associates with every 1-propositional structure 〈{α}〉 the

1-propositional structure 〈{�}〉 if α is a conjunctive formula of the form β ∧ γ where β , γ are propositional formulae,and 〈{⊥}〉 otherwise. Ω obviously is �-language independent but it is not syntax-independent.

4. Consider for instance the propositional operator Ω that associates with every 1-propositional structure 〈{α}〉 the1-propositional structure 〈{�}〉 if α = �, and 〈{p}〉 otherwise, where p ∈ PS. Then Ω neither is syntax-independentnor is SIN-language independent. �

Proposition 9. σ : {x1, . . . , xn} → PROPY satisfies FCP if and only if∧n

i=1(zi ⇔ σ(xi)) defines Y in terms of {z1, . . . , zn}, where{z1, . . . , zn} ∩ Y = ∅.

Proof. {¬,∧} is functionally complete w.r.t. Y given {σ(x) | x ∈ X} iff for every y ∈ Y , there exists a formula αy generatedfrom {¬,∧} and {σ(x) | x ∈ X} as the set of “atoms” such that αy ≡ y. Under the assumption that

∧ni=1(zi ⇔ σ(xi)) holds,

this is equivalent to state that for every y ∈ Y there exists a formula αy generated from {¬,∧} and {z1, . . . , zn} as the setof atoms such that αy ≡ y. Each αy can be viewed as a definition of y in terms of {z1, . . . , zn} in

∧ni=1(zi ⇔ σ(xi)). �

Proposition 10.

1. Determining whether a given symbol translation σ ∈ PROPXY satisfies SIN is in P.

2. Determining whether a given symbol translation σ ∈ PROPXY satisfies TPR is NP-complete.

3. Determining whether a given symbol translation σ ∈ PROPXY satisfies FCP is coNP-complete in the general case. The problem is in

P if |X | < |Y |.4. Determining whether a given symbol translation σ ∈ PROPX

Y satisfies BAP is coNP-complete in the general case. The problem is inP if |X | �= |Y |.

5. Determining whether a given symbol translation σ ∈ PROPXY satisfies REV (or, equivalently, AIP) is Π

p2 -complete. The problem is

coNP-complete if |X | = |Y |, and is in P if |X | > |Y |.

Proof. Let X = {x1, . . . , xn}.

1. SIN. Obvious.2. TPR. For each x ∈ X , σ(x) is consistent and not valid iff one can find a model of it and a counter-model of it. In order to

determine whether σ satisfies TPR it is enough to guess for each x ∈ X an interpretation from 2Y which satisfies σ(x)and an interpretation from 2Y which satisfies ¬σ(x); this can be easily done in non-deterministic polynomial time.We prove NP-hardness by exhibiting a polynomial reduction from the satisfiability problem. Let α be any propositionalformula. Let us associate with it in polynomial time the symbol translation σ : {x} → Var(α) ∪ {y} with y /∈ Var(α) suchthat σ(x) = α ∧ y. α is consistent if and only if σ satisfies TPR.

3. FCP. Proposition 9 gives a polynomial reduction of this problem to the definability one. Since the latter is in coNP[7] and coNP is closed under polynomial reduction, the membership of the former to coNP follows. We provecoNP-hardness by exhibiting a polynomial reduction from the unsatisfiability problem. Consider a propositional for-mula α from PROPX ; let us associate with it in polynomial time the symbol translation σ from PROPX

Y whereX = Var(α) ∪ {p,q} (with Var(α) ∩ {p,q} = ∅), Y = X , such that ∀x ∈ Var(α), σ(x) = x, and σ(p) = (p ∧ α) ∨ q, andσ(q) = (q ∧ α) ∨ p. If α is inconsistent, then every y ∈ Y is definable from {σ(x) | x ∈ X}. Indeed, in such a case, forevery y ∈ Y , if y ∈ Var(α), then the formula y from PROP{σ(x) | x ∈ X} is such that y ≡ y, and in the remaining casewhen y = p (resp. y = q) we can find σ(q) (resp. σ(p)) in PROP{σ(x) | x ∈ X} such that p ≡ σ(q) (resp. q ≡ σ(p)).Accordingly, if α is inconsistent then σ satisfies FCP. Now, if α is consistent, then σ does not satisfy FCP. Indeed, takingadvantage of Proposition 9, it is enough to prove that Σ = ∧n

i=1(zi ⇔ xi) ∧(zp ⇔ ((p ∧ α) ∨ q)) ∧ (zq ⇔ ((q ∧ α) ∨ p))

does not define p in terms of {z1, . . . , zn, zp, zq}. To do so, it is enough to prove that ∃{x1, . . . , xn, p,q}.(Σ ∧ p) does not


imply ¬∃{x1, . . . , xn, p,q}.(Σ ∧¬p) (see Corollary 9 in [7]). This is the case because when α is consistent, the implicantzp ∧ zq of ∃{x1, . . . , xn, p,q}.(Σ ∧ p) does not imply the implicate ¬zp ∨ ¬zq ∨ ¬αz of ¬∃{x1, . . . , xn, p,q}.(Σ ∧ ¬p),where αz is the formula obtained from α by replacing each xi by zi (i ∈ 1, . . . ,n). Thus, α is inconsistent if and only ifσ satisfies FCP, which concludes the proof.The fact that the problem is in P if |X | < |Y | directly comes from Corollary 2, since σ satisfies FCP only if |X |� |Y |.

4. BAP. The membership to coNP directly follows from Corollary 2 and from point 3 of this proposition, since σ satisfiesBAP if it satisfies FCP and |X | = |Y |. We prove coNP-hardness by exhibiting a polynomial reduction from the unsatisfia-bility problem. Consider a propositional formula α from PROPX ; let us associate with it in polynomial time the symboltranslation σ from PROPX

Y where X = Var(α) ∪ {new} (with new /∈ Var(α)), Y = X , such that ∀x ∈ Var(α), σ(x) = x, andσ(new) = α ∨ new. If α is inconsistent, then ∀ω ∈ 2X , we have σ(ω) ≡ ω. If α is consistent, then there exists ω ∈ 2Var(α)

such that ω |� α. ω ∧ ¬new corresponds to an interpretation from 2X such that σ(ω ∧ ¬new) ≡ ω ∧ ¬α ∧ ¬new ≡ ⊥.Thus, α is inconsistent if and only if σ satisfies BAP.The fact that the problem is in P if |X | �= |Y | directly comes from Corollary 2, since σ satisfies BAP only if |X | = |Y |.

5. REV . We take advantage of the equivalence given by Proposition 5. For membership to Πp2 , one considers the comple-

mentary problem (determine whether σ does not satisfy AIP) and shows that it belongs to Σp2 . Consider the following

algorithm: guess s1, . . . , sn ∈ {+,−} then check using one call to an NP-oracle that∧

xi∈X σ(xi)si is inconsistent. Clearly

enough, this non-deterministic algorithm with oracle NP solves the complementary problem, showing that it belongs toΣ

p2 . We prove Π

p2 -hardness by considering a reduction from the validity problem for quantified Boolean formulas (QBFs)

of the form ∀A.(∃B.α) where A = {a1, . . . ,an} and B = {b1, . . . ,bm} are two disjoint sets of propositional atoms andVar(α) = A ∪ B . Consider such a quantified Boolean formula (QBF). We associate with this QBF in polynomial time thesymbol translation σ ∈ PROPX

Y with X = {x0, x1, . . . , xn}, Y = A ∪ B ∪{new} and new /∈ A ∪ B , defined by σ(x0) = α ∧ newand for each i ∈ 1, . . . ,n, σ(xi) = ai . σ satisfies REV (or equivalently AIP) iff ∀s0, . . . , sn ∈ {+,−},

∧xi∈X σ(xi)

si is con-sistent iff ∀s0 ∈ {+,−} ∀A (α ∧ new)s0 is consistent iff ∀A.(¬α ∨ ¬new) is consistent and ∀A (α ∧ new) is consistent.Clearly enough, the QBF ∀A.(¬α ∨ ¬new) is consistent (every interpretation ω such that ω(new) = 0 is a model of it).Finally, the QBF ∀A.(α ∧ new) is consistent iff the QBF ∀A.α is consistent iff the QBF ∀A.(∃B.α) is valid. The fact thatthe latter problem is Π

p2 -complete completes the proof.

The fact that the problem is coNP-complete if |X | = |Y | comes from Corollary 2. Indeed, if |X | = |Y |, REV and BAP areequivalent properties. The problem is in P if |X | > |Y |, since from Corollary 2 σ satisfies REV only if |X |� |Y |. �

Proposition 11. The KM drastic update operator �D is �-language independent.

Proof. Let σ be any symbol translation from PROPXY , K be a base from PROPX and μ be a formula from PROPX . By definition

of �D , we have K �D μ ≡ K if K |� μ, ≡ μ otherwise; hence, we have σ(K ) �D σ(μ) ≡ σ(K ) if σ(K ) |� σ(μ), ≡ σ(μ)

otherwise. Yet from point 1 of Corollary 1, we have K |� μ if and only if σ(K ) |� σ(μ). Thus, σ(K ) �D σ(μ) ≡ σ(K ) ifK |� μ, ≡ σ(μ) otherwise. Therefore, σ(K ) �D σ(μ) ≡ σ(K �D μ). �Proposition 12. Let P be a property on symbol translations which does not imply AIP, i.e., there exists a symbol translation satisfyingP but not satisfying AIP. Then no belief revision operator ◦ satisfying (R2) and (R3) and no belief merging operator � satisfying (IC1)

and (IC2) is P-language independent.

Proof. Let PS = {x1, . . . , xn}, and let σ be a symbol translation from PROPXY that does not satisfy AIP. This means that

∃s1, . . . , sn ∈ {+,−},∧

xi∈PS σ(xi)si is inconsistent, or equivalently, σ(

∧xi∈PS xsi

i ) is inconsistent. Thus, there is a consistentformula ασ such that σ(ασ ) is inconsistent. Now, consider a belief revision operator ◦ satisfying the rationality postulates(R2) and (R3) and a belief merging operator � satisfying the rationality postulates (IC1) and (IC2). For every consistentformula α, (R2) implies that α ◦ � ≡ α and (IC2) implies that �(〈α〉)� ≡ α. Thus, both α ◦ � and �(〈α〉)� are consistentwhen α is consistent. If ◦ (resp. �) would satisfy P-language independence with P a property on symbol translations whichdoes not imply AIP, then for every σ satisfying P but not satisfying AIP, (R3) (resp. (IC1)) would imply that σ(ασ ) ◦σ(�) ≡σ(ασ ) (resp. �(σ(〈ασ 〉))σ (�) ≡ σ(ασ )) is consistent, which is not the case. �Proposition 13.

1. The distance-based merging operator �dD ,Σ , based on the drastic distance dD and on the aggregation function Σ , is the only ICmerging operator that is AIP-language independent; it is also the only IC merging operator that is BAP-language independent;

2. The drastic revision operator ◦D is the only AGM revision operator that is AIP-language independent; it is also the only AGMrevision operator that is BAP-language independent;

3. The drastic update operator �D is the only KM update operator that is AIP-language independent; it is also the only KM updateoperator that is BAP-language independent.

Proof. In this proof, we denote by φM a formula whose models are those of the set M . To alleviate heavy notations, forevery base K , K ′ , K \ K ′ will stand for φMod(K )\Mod(K ′) .


1. We first show that the distance-based merging operator �dD ,Σ is AIP-language independent.Let X ⊆ PS, E be a profile of bases containing formulae from PROPX , μ be a formula from PROPX and σ be a symboltranslation from PROPX

Y satisfying AIP.

By construction, the models of �dD ,Σμ (E) are precisely the models of μ satisfying a maximal number of bases in the

profile E .Let ω′ be an interpretation over Y such that ω′ |� σ(μ); since μ is equivalent to the disjunction of all its canonicalterms associated with its models, there must exist an interpretation ω over X such that ω′ |� σ(ω).Consider now any ω be an interpretation over X , and ω′ an interpretation over Y such that ω′ |� σ(ω). Then (A) forevery base K in the profile E , ω′ |� σ(K ) if and only if ω |� K . Indeed, if ω |� K , then from point 1 of Corollary 1, wehave σ(ω) |� σ(K ), and by transitivity of |�, we also have ω′ |� σ(K ); and if ω |� ¬K , then from point 1 of Corollary 1,we have σ(ω) |� ¬σ(K ), and by transitivity of |�, we also have ω′ |� ¬σ(K ). Similarly, we can show that ω′ |� σ(μ) ifand only if ω |� μ.Accordingly, if ω is a model of μ satisfying a maximal number (say k) of bases of E , then every model ω′ of σ(ω) isa model of σ(μ) satisfying k bases of σ(E); furthermore, there is no model ω′′ of σ(μ) satisfying at least k + 1 basesof σ(E): if this were the case, then there would exist a model ω′′′ of μ such that ω′′ |� σ(ω′′′), and because of (A),ω′′′ would satisfy at least k + 1 bases of E , contradicting the fact that no model of μ satisfies more than k bases of E .Hence, if ω |� �

dD ,Σμ (E), then every model ω′ of σ(ω) satisfies ω′ |� �

dD ,Σσ(μ) (σ (E)).

The other way around, let ω′ be a model of σ(μ) satisfying a maximal number (say k) of bases of σ(E). Let ω be anymodel of μ such that ω′ |� σ(ω). From (A), we know that ω satisfies k bases of E . Furthermore, there is no modelω′′′ of μ satisfying at least k + 1 bases of E: indeed, consider any model ω′′ of σ(ω′′′) (since σ satisfies AIP, σ(ω′′′)is necessarily consistent, hence such a model ω′′ exists). Since ω′′′ |� μ, we have σ(ω′′′) |� σ(μ) and by transitivity of|�, ω′′ is a model of σ(μ). Because of (A), ω′′ would satisfy at least k + 1 bases of σ(E), contradicting the fact thatno model of σ(μ) satisfies more than k bases of σ(E). Hence, if ω′ |� �

dD ,Σσ(μ) (σ (E)), then for every ω |� μ such that

ω′ |� σ(ω), we have ω |� �dD ,Σμ (E).

Altogether, we get that σ(�dD ,Σμ (E)) ≡ �

dD ,Σσ(μ) (σ (E)), showing that the distance-based merging operator �dD ,Σ is

AIP-language independent.Now, let � be an IC merging operator that is BAP-language independent. Let us first show that for every consistentbase K , for every interpretation ω,ω′ , ω <K ω′ implies ω |� K . Let K be a consistent base, ω,ω′ two interpretationssuch that ω <K ω′ . Since �K is a preorder, ω <K ω′ implies that ω and ω′ are distinct interpretations. Towards acontradiction, suppose ω �|� K . Let R be a relation on 2P S × 2P S defined as R(ω) = {ω′}, R(ω′) = {ω} and for everyother interpretation ω′′ , R(ω′′) = {ω′′}. R obviously is a one-to-one correspondence. Thus from point 2 of Proposition 6,R is the image by the semantical mapping Ψ of a symbol translation σR that satisfies BAP. Suppose that ω′ |� K . Underthe assumption that ω �|� K , using condition (2) of the syncretic assignment we get that ω′ <K ω, which contradictsω <K ω′ . Hence, under the assumption ω �|� K , we must have ω′ �|� K . Given the definition of R , we have σR(K ) ≡ K andσR(φ{ω,ω′}) ≡ φ{ω,ω′} . Now, since � is BAP-language independent, we have σR(�φ{ω,ω′ }(〈K 〉)) ≡ �σR (φ{ω,ω′ })(σR(〈K 〉)), orequivalently, σR(�φ{ω,ω′} (〈K 〉)) ≡ �φ{ω,ω′ }(〈K 〉). Since �φ{ω,ω′}(〈K 〉) ≡ ω (from the fact that ω <K ω′ and from Theorem 1)we get σR(�φ{ω,ω′ }(K )) ≡ σR(ω) ≡ ω, this contradicts σR(ω) ≡ ω′ , thus this contradicts the initial assumption ω �|�K . Therefore, for every consistent base K , for every interpretation ω,ω′ , ω <K ω′ implies ω |� K . Taking this lastresult together with conditions (1) and (2) of the syncretic assignment, we get for every consistent base K , for everyinterpretation ω,ω′ ,

ω <〈K 〉 ω′ if and only if ω |� K and ω′ �|� K . (1)

The equivalence reported at Eq. (1) shows that every IC merging operator that is BAP-language independent correspondsto a syncretic assignment that ranks all the worlds over at most two plausibility levels for any singleton profile.We intend now to prove that for every consistent base K , K ′ , for every interpretation ω, ω′ , if ω <K ω′ and ω′ <K ′ ω,then ω �〈K ,K ′〉 ω′ . Let ω,ω′ be two interpretations, K , K ′ be two consistent bases. Assume that ω <K ω′ and ω′ <K ′ ω.Towards a contradiction, suppose that ω′ <〈K ,K ′〉 ω (the remaining case can be treated in a similar way due to the factthat (ω, K ), and (ω′, K ′) play symmetric roles in the statement to be proven). From Eq. (1) above, ω ∈ Mod(K )\Mod(K ′)and ω′ ∈ Mod(K ′) \ Mod(K ). Thus we have K �|� K ′ and K ′ �|� K . There are two remaining cases:• Case 1: assume K ∧ K ′ |� ⊥. Let Mod(K ) = {ω1, . . . ,ωp}. Let R be a relation on 2PS × 2PS defined as R(ωi) = {ωi+1}

for every i ∈ {1, . . . , p − 1}, R(ωp) = {ω1} and for every interpretation ω′′ �|� K , R(ω′′) = {ω′′}. Note that ω is one ofthe ωi (for i ∈ {1, . . . , p}) and that R(ω′) = {ω′}. R is obviously a one-to-one correspondence. Thus from point 2 ofProposition 6, R is the image by the semantical mapping Ψ of a symbol translation σR that satisfies BAP.

Our assumption is that ω <K ω′ , ω′ <K ′ ω and ω′ <〈K ,K ′〉 ω. Since � is BAP-language independent, we have for every

j � 1, σj

R(ω) <σ

jR (K )

σj

R(ω′), σj

R(ω′) <σ

jR (K ′) σ

jR(ω) and σ

jR(ω′) <〈σ j

R (K ),σj

R (K ′)〉 σj

R(ω), where σj

R = σR ◦ · · · ◦ σR︸︷︷︸j

. Yet by

construction of R , on the one hand we have for every j � 1, σj

R(K ) ≡ K , σj

R(K ′) ≡ K ′ and σj

R(ω′) ≡ ω′; on the other

hand we have for every ωi |� K , there exists j � 1 such that σj(ω) ≡ ωi . This implies that for every ωi |� K , ωi <〈K 〉 ω′ ,
R


ω′ <〈K ′〉 ωi and ω′ <〈K ,K ′〉 ωi . Hence, there is an interpretation ω′ |� K ′ such that for every interpretation ωi |� K ,ω′ <〈K ,K ′〉 ωi . However, since ω′ |� K ′ , condition (4) of the syncretic assignment requires that there exists ωi |� K suchthat ωi �〈K ,K ′〉 ω′ , contradiction.• Case 2: assume K ∧ K ′ �|� ⊥. Notice that since K �|� K ′ and K ′ �|� K , each of K \ K ′ and K ′ \ K is a consistent base.

From Eq. (1), we know that ω <〈K\K ′〉 ω′ and ω′ <〈K ′\K 〉 ω. Thus, from Case 1 above we deduce that ω �〈K ,K ′\K 〉 ω′(i), ω �〈K ′,K\K ′〉 ω′ (ii) and ω �〈K\K ′,K ′\K 〉 ω′ (iii). On the one hand, from (i), (ii) and from condition (5) of thesyncretic assignment we get that ω �〈K ,K ′,K\K ′,K ′\K 〉 ω′ . On the other hand, from (iii), from the initial assumptionthat ω′ <〈K ,K ′〉 ω and from condition (6) of the syncretic assignment we get that ω′ <〈K ,K ′,K\K ′,K ′\K 〉 ω, contradiction.Therefore, we have for every consistent base K , K ′ and every interpretation ω,ω′ ,

if ω <〈K 〉 ω′ and ω′ <〈K ′〉 ω, then ω �〈K ,K ′〉 ω′. (2)

Let us now show that for every interpretation ω,ω′ , every profile E and every consistent base K , if ω <E ω′ , thenω �E�〈K 〉 ω′ . We prove it by induction on the dimension n of E:

• Base case (n = 1): let ω <〈K 〉 ω′ . Then we have ω �〈K 〉 ω′ . Let K ′ be a consistent base, if ω �〈K ′〉 ω′ , then by condition(5) of the syncretic assignment we get ω �〈K ,K ′〉 ω′ . In the remaining case, i.e., if ω′ <〈K ′〉 ω, then using Eq. (2) weget ω �〈K ,K ′〉 ω′ , which implies that ω �〈K 〉 ω′ .

• Inductive step (n > 1): The implication to be proven, i.e., if ω <E ω′ , then ω �E�〈K 〉 ω′ , is equivalent to its contra-positive, which is equivalent to: if ω <E�〈K 〉 ω′ then ω �E ω′ . Thus, the induction hypothesis can be stated as: forevery base K ′ of E , ω <E ω′ then ω �E\〈K ′〉 ω′ . Consider ω and ω′ such that ω <E�〈K 〉 ω′ . Since �E�〈K 〉 is total, thisis equivalent to ω′ �E�〈K 〉 ω. Using condition (5) of the syncretic assignment, this implies that (ω′ �E ω or ω′ �K ω),or equivalently, (ω <E ω′ or ω <〈K 〉 ω′). If ω <E ω′ then ω �E ω′ holds, which is the expected conclusion. Henceit remains to consider the case when ω ≮E ω′ (which is equivalent to ω′ �E ω) and ω <〈K 〉 ω′ . If ω′ �E ω thenthere are two possible cases: ω′ �E ω and ω′ <E ω. In the first case ω′ �E ω, we have ω �E ω′ holds, which isthe expected conclusion. In the second case ω′ <E ω, by induction hypothesis, we have that for every base K ′ ofE , ω′ �E\〈K ′〉 ω. Furthermore, when ω′ <E ω, condition (5) on the syncretic assignment requires that there existsa base K ′′ of E such that ω′ <〈K ′′〉 ω. Exploiting the induction hypothesis with K ′ = K ′′ , we get that ω′ �E\〈K ′′〉 ω.Using Eq. (2), from ω <〈K 〉 ω′ and ω′ <〈K ′′〉 ω, we derive that ω �〈K ,K ′′〉 ω′ , which implies that ω′ �〈K ,K ′′〉 ω. Usingcondition (5) on the syncretic assignment, from ω′ �E\〈K ′′〉 ω and ω′ �〈K ,K ′′〉 ω, we derive that ω′ �E�〈K 〉 ω, whichcontradicts ω <E�〈K 〉 ω′ . Hence, when ω <E�〈K 〉 ω′ holds, it cannot be the case that ω′ <E ω. Stated otherwise, wemust also have ω �E ω′ in the second case. This is the expected conclusion: for every interpretation ω,ω′ , for everyprofile E and every consistent base K ,

if ω <E ω′, then ω �E�〈K 〉 ω′. (3)

Now, for every profile E and every interpretation ω let us denote |ω(E)| = |{Ki base of E | ω |� Ki}|, i.e., the numberof belief bases in E which are satisfied by ω. We intend to prove that for every profile E and every interpretationω,ω′ , ω <E ω′ if and only if |ω(E)| > |ω′(E)|. We prove it by induction on the dimension n of E:

• Base case (n = 1): using Eq. (1), ω <〈K 〉 ω′ if and only if ω |� K and ω′ �|� K , this directly leads us to |ω(E)| > |ω′(E)|.• Base case (n = 2): let us show that ω <〈K ,K ′〉 ω′ if and only if (ω �〈K 〉 ω′ and ω <〈K ′〉 ω′) or (ω <〈K 〉 ω′ and ω �〈K ′〉

ω′).(Only If ) Direct from condition (6) of the syncretic assignment.(If ) Let ω <〈K ,K ′〉 ω′ . Towards a contradiction, assume ω′ <〈K 〉 ω. Assuming ω′ �〈K ′〉 ω (respectively, ω <〈K ′〉 ω′) con-tradicts ω <〈K ,K ′〉 ω′ by condition (5) of the syncretic assignment (respectively, by Eq. (2)). Hence, ω �〈K 〉 ω′ . Since Kand K ′ play symmetric roles in the equivalence to be proven, we can prove in a similar way that ω �〈K ′〉 ω′ . Now,assuming ω �〈K 〉 ω′ and ω �〈K ′〉 ω′ contradicts ω <〈K ,K ′〉 ω′ by condition (5) of the syncretic assignment. Hence,(ω �〈K 〉 ω′ and ω <〈K ′〉 ω′) or (ω <〈K 〉 ω′ and ω �〈K ′〉 ω′).Since ω <〈K ,K ′〉 ω′ if and only if (ω �〈K 〉 ω′ and ω <〈K ′〉 ω′) or (ω <〈K 〉 ω′ and ω �〈K ′〉 ω′), then by Eq. (1) andcondition (2) on the syncretic assignment we deduce that ω <〈K ,K ′〉 ω′ if and only if |ω(〈K , K ′〉)| > |ω′(〈K , K ′〉)|.

• Inductive step (n � 2): assume that for every profile E such that |E| = k with k � n, we have ω <E ω′ if and only if|ω(E)| > |ω′(E)|. Let E be a profile of dimension n + 1. We need to show that ω <E ω′ if and only if |ω(E)| > |ω′(E)|.(Only If ) Assume that ω <E ω′ . Let K be a base of E:– Assume ω <〈K 〉 ω′ . Since ω <E ω′ , then using the contrapositive of the implication reported at Eq. (3), we have

ω �E\〈K 〉 ω′ . By induction hypothesis, we have |ω(〈K 〉)| > |ω′(〈K 〉)| and |ω(E \〈K 〉)| � |ω′(E \〈K 〉)|. Hence, |ω(E)| >|ω′(E)|.

– Assume ω �〈K 〉 ω′ . Since ω <E ω′ , then by condition (5) of the syncretic assignment we have ω <E\〈K 〉 ω′ . Byinduction hypothesis, we have |ω(〈K 〉)| = |ω′(〈K 〉)| and |ω(E \ 〈K 〉)| > |ω′(E \ 〈K 〉)|. Hence, |ω(E)| > |ω′(E)|.

– Assume ω′ <〈K 〉 ω. Since ω <E ω′ , then by condition (5) of the syncretic assignment there exists a belief baseK ′ ∈ E such that ω <〈K ′〉 ω′ . By Eq. (2), we get ω �〈K ,K ′〉 ω′ . And by condition (5) of the syncretic assignment, wenecessarily have ω <E\〈K ,K ′〉 ω′ . By induction hypothesis, we have |ω(〈K , K ′〉)| = |ω′(〈K , K ′〉)| and |ω(E \〈K , K ′〉)| >|ω′(E \ 〈K , K ′〉)|. Hence, |ω(E)| > |ω′(E)|.


(If ) Assume that |ω(E)| > |ω′(E)|. This means that there exists a belief base K ∈ E such that ω |� K and ω′ �|� K . Fromcondition (2) on the syncretic assignment, we have ω <〈K 〉 ω′ . We get |ω(E \ 〈K 〉)| � |ω′(E \ 〈K 〉)|, and by inductionhypothesis, ω �E\〈K 〉 ω′ . Hence, by condition (6) of the syncretic assignment we get ω <E ω′ .Therefore, we have for every profile E and every interpretation ω,ω′ ,

ω <E ω′ if and only if∣∣ω(E)

∣∣ >∣∣ω′(E)

∣∣. (4)

Lastly, it is easy to verify that the syncretic assignment satisfying the equivalence at Eq. (4) above is associated withthe distance-based merging operator �dD ,Σ . Hence, we have just proved that �dD ,Σ is the only IC merging operatorthat is BAP-language independent.

2. The drastic AGM revision operator ◦D is obviously AIP-language independent and BAP-language independent, since itcorresponds to the distance-based merging operator �dD ,Σ restricted to singleton profiles (i.e., K ◦D μ ≡ �

dD ,Σμ (〈K 〉)),

and �dD ,Σ is AIP-language independent and BAP-language independent (cf. point 1 of this proof). Now, Eq. (1) (cf.point 1 of this proof) also holds for any AGM revision operator that is BAP-language independent. Indeed, to proveEq. (1), we only used conditions (1) and (2) of the syncretic assignment of the IC merging operator, plus the assumptionthat this operator is BAP-language independent. The point is that conditions (1) and (2) of the faithful assignment forAGM revision operators are equivalent to conditions (1) and (2) of the syncretic assignment for IC merging operators.Thus, Eq. (1) also holds for any AGM revision operator ◦ that is BAP-language independent. Now, from Eq. (1), condition(1) of the faithful assignment and Theorem 2 [12] (which requires that the faithful assignment associated with an AGMrevision operator ◦ maps every base K to a total preorder �K ), the faithful assignment associated with ◦ must besuch that for every base K , and interpretations ω, ω′ , ω �K ω′ iff (ω |� K and ω′ |� K ) or (ω �|� K and ω′ �|� K ), andω <K ω′ iff ω |� K and ω′ �|� K . Thus, whatever the BAP-language independent AGM revision operator ◦, there is aunique total preorder �K associated with K by the faithful assignment corresponding to ◦. This shows that every AGMrevision operator which is BAP-language independent coincides with ◦D . Hence this is also the case for AIP-languageindependence: every AGM revision operator which is AIP-language independent coincides with ◦D .

3. From Proposition 11, the drastic update operator �D is obviously AIP-language independent and BAP-language indepen-dent. Moreover, it is the only KM update operator that is BAP-language independent. Indeed, Eq. (1) (cf. point 1 of thisproof) also holds for any KM update operator that is BAP-language independent. Indeed, to prove Eq. (1), we only usedconditions (1) and (2) of the syncretic assignment of the IC merging operator, plus the assumption that this operator isBAP-language independent. Especially, we did not take advantage of the fact that �K is total. Furthermore, conditions (1)and (2) of the faithful assignment for AGM revision operators (or for KM update operators) are equivalent to conditions(1) and (2) of the syncretic assignment for IC merging operators. Thus, Eq. (1) also holds for any KM update operatorthat is BAP-language independent. Once this is stated, let us consider any KM update operator � that is BAP-languageindependent. From Theorem 3 [12], for any base K and formula μ, we have Mod(K � μ) = ⋃

ω|�K min(Mod(μ),�ω)

= ⋃ω|�K Mod(ω � μ). For every pair of distinct interpretations ω and ω′ , by condition (2) of the faithful assignment,

we have ω <ω ω′ . Hence if ω |� μ, then ω � μ ≡ ω ≡ ω �D μ. In the remaining case (i.e., if ω �|� μ), then from Eq. (1),for every pair ω′ and ω′′ of models of μ, we have ω′ ≮ω ω′′ and ω′′ ≮ω ω′ , which means that ω and ω′ are eitherindifferent w.r.t. �ω (i.e., ω′ �ω ω′′) or incomparable w.r.t. �ω (i.e., ω′ �ω ω′′ and ω′′ �ω ω′). Whatever the case, wehave min({ω′,ω′′},�ω) = {ω′,ω′′}. Thus, when ω �|� μ, we have min(Mod(μ),�ω) = Mod(μ). This shows that whenω �|� μ, we have ω � μ ≡ μ ≡ ω �D μ, and this completes the proof. �

Proposition 14. There exist belief revision (resp. belief merging, belief update) operators which satisfy all AGM postulates (resp. all ICpostulates, all KM postulates) but are not SIN-language independent.

Proof. Consider a belief revision operator ◦ and a belief update operator �, both associated with a faithful assignment suchthat:

10 <10 01 <10 11 <10 00,

01 <01 11 <01 10 <01 00

when PS = {a,b}. Now, let σ : a �→ b; b �→ a be a bijective substitution over PS. Let α = a ∧ ¬b and β = b. We have α ◦ β ≡α � β ≡ ¬a ∧ b. We also have σ(α) ≡ ¬a ∧ b and σ(β) ≡ a. We finally have σ(α) ◦ σ(β) ≡ σ(α) � σ(β) ≡ a ∧ b. This is notequivalent to σ(α ◦ β) or σ(α � β), which are both equivalent to a ∧ ¬b.

The proof is similar for a belief merging operator, considering a profile with a unique base α and β as integrity con-straints. Especially, what remains to be done is to show that there exists a syncretic assignment which is “compatible” withthe faithful assignment above. A syncretic assignment is a mapping which associates with every profile a complete preorderover the interpretations (while a faithful assignment associates with every base a complete preorder over the interpreta-tions). The conditions that a syncretic assignment have to satisfied are more demanding than those that faithful assignmentsmust satisfy (see [9] for details). Especially, it is not the case that every faithful assignment can be extended to a syncreticassignment. Let us show that this is nevertheless the case for the assignment considered above. Recall that every mergingoperator based on a pseudo-distance and the aggregation function Σ is an IC merging operator [9]. Consider for instance any


mapping d from 2PS × 2PS to the natural numbers satisfying d(ω,ω) = 0 for every interpretation ω, d(ω1,ω2) = d(ω2,ω1)

for every pair of interpretations ω1, ω2, and:

• d(01,11) = 2,• d(01,10) = 3,• d(00,01) = d(10,11) = 4,• d(00,10) = 5,• d(00,11) = 6.

By construction, d is a pseudo-distance (in fact, it is a distance, i.e., the triangular inequality is also satisfied by d).Consider any admissible aggregation function (e.g., Σ ). Now it is easy to check that we have:

• d(10,10) = 0 < d(10,01) = 3 < d(10,11) = 4 < d(10,00) = 5,• d(01,01) = 0 < d(01,11) = 2 < d(01,10) = 3 < d(01,00) = 4.

Accordingly, we have �d,Σb ({a∧¬b}) ≡ ¬a∧b, and �

d,Σa ({b ∧¬a}) ≡ a∧b. Clearly, for σ : a �→ b; b �→ a a bijective symbol

translation from PS to PS, α = a ∧ ¬b and β = b, we do not have σ(�d,Σβ (〈α〉)) equivalent to �

d,Σσ(β)

(〈σ(α)〉). �Proposition 15. Every belief change operator based on a decomposable distance is SIN-language independent.

Proof. Let σ be a symbol translation satisfying SIN, i.e., a bijection from X to Y . We can associate with σ a symbol transla-tion σ ′ from PS to PS, with PS = {x1, . . . , xn} defined by: for every x ∈ X , σ ′(x) = σ(x), and for every x ∈ PS \ X , σ ′(x) = τ (x),where τ is any bijection from PS \ X to PS \ Y . It is easy to see that for every formula α from PROPX , σ(α) = σ ′(α). More-over, σ ′ is a bijection from PS to PS, so it satisfies SIN. Therefore, a propositional operator ω is SIN-language independentiff for every n-propositional structure Σ over X , for every bijection σ ′ from PS to PS, we have Ω(σ ′(Σ)) ≡ σ ′(Ω(Σ)).

From point 4 of Proposition 6, Rσ ′ is a bijection from 2PS to 2PS , so for proving the result, it is enough to show that forevery pair of interpretations ω1, ω2 over PS, we have d(ω1,ω2) = d(σ ′(ω1),σ

′(ω2)). Since d is decomposable, there existsa mapping fd : Rn →R which is symmetric in each argument and a mapping gd : {0,1} × {0,1} → R such that d(ω1,ω2) =fd(gd(ω1(x1),ω2(x1)), . . . , gd(ω1(xn),ω2(xn))). Thus, we have to show that fd(gd(ω1(x1),ω2(x1)), . . . , gd(ω1(xn),ω2(xn))) =d(σ ′(ω1),σ

′(ω2)) = fd(gd(σ′(ω1)(x1),σ

′(ω2)(x1)), . . . , gd(σ′(ω1)(xn),σ ′(ω2)(xn))). Slightly abusing the notations, let us

also note σ ′ the permutation over {1, . . . ,n} such that for any i ∈ 1, . . . ,n, σ(i) = j if and only if σ ′(xi) = x j . Now,for every xi ∈ PS, if σ ′(xi) = x j , we have ω1(xi) = σ ′(ω1)(x j) = σ ′(ω1)(xσ ′(i)) and ω2(xi) = σ ′(ω2)(x j) = σ ′(ω2)(xσ ′(i)).As a consequence, d(ω1,ω2) = fd(gd(ω1(x1),ω2(x1)), . . . , gd(ω1(xn),ω2(xn))) = fd(gd(σ

′(ω1)(xσ(1)), σ′(ω2)(xσ(1))), . . . ,

gd(σ′(ω1)(xσ(n)), σ

′(ω2)(xσ ′(n)))). For every pair of interpretations ω1, ω2 over PS, we have d(ω1,ω2) = d(σ ′(ω1),σ′(ω2)).

The fact that σ ′ is a permutation and that fd is symmetric in each argument completes the proof. �Proposition 16. There exist belief revision (resp. belief update) operators which satisfy all AGM postulates (resp. all KM postulates),are not distance-based operators but satisfy (SIN-R) (resp. (SIN-U)).

Proof. Let PS = {a,b} and consider a belief revision operator ◦ and a belief update operator �, both associated with a faithfulassignment such that:

10 <10 11 <10 00 <10 01,

01 <01 00 <01 11 <01 10

and such that for every base K such that K �≡ a ∧ ¬b and K �≡ ¬a ∧ b, �K is a two-level total preorder satisfying conditions(1–3) of the faithful assignment, that is, for every interpretation ω,ω′ , ω <K ω′ if and only if ω |� K and ω′ �|� K . Wehave to show that for every base K and every interpretation ω,ω′ , ω �K ω′ implies σ(ω) �σ(K ) σ (ω′) for every symboltranslation σ which satisfy SIN. Yet since |PS| = 2, there is only one bijection from X to Y (with X, Y ⊆ PS) which is notthe identity function, i.e., σ : a �→ b; b �→ a. Then, let ω,ω′ be two interpretations and K be a base such that ω �K ω′ . IfK ≡ a ∧ ¬b or K ≡ ¬a ∧ b, then one can see from the definition of �10 and �01 that σ(ω) �σ(K ) σ (ω′) by enumerating allcases. Otherwise, K �≡ a ∧ ¬b and K �≡ ¬a ∧ b. Then there are two cases:

• ω <K ω′: in this case ω |� K and ω′ �|� K . From point 1 of Corollary 1 we get that σ(ω) |� σ(K ) and σ(ω′) �|� σ(K ).Therefore, σ(ω) <σ(K ) σ (ω′).

• ω �K ω′: then two cases must be considered:(1) ω |� K and ω′ |� K , and(2) ω �|� K and ω′ �|� K .


In case (1), from point 1 of Corollary 1 we get that σ(ω),σ (ω′) |� σ(K ), so σ(ω) �σ(K ) σ (ω′). In case (2), since σ(K ) is atwo-level preorder satisfying conditions (1–3) of the faithful assignment, we have σ(ω) �σ(K ) σ (ω′).

It remains to show that ◦ and � are not distance-based ones. Assume towards a contradiction that ◦, � are based ona pseudo-distance d. We know that �00 and �11 are two-level total preorders satisfying conditions (1–3) of the faithfulassignment, so we have d(01,00) = d(10,00) and d(01,11) = d(10,11). Yet from the definition of �10 and �01 we get thatd(10,11) < d(10,00) and d(01,11) > d(01,00). Contradiction. �References

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lost in translation: language independence in propositional logic – application to belief change

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