Paraconsistent belief revision does not coincide with expansion!

In [24], Tanaka proposed Grove’s sphere systems in order to deal with an AGM-like paraconsistent belief revision theory for some paraconsistent logics, includ-ing da Costa’s paraconsistent logicsCn, forn≥1 (see [6]). For the latter, he shows a kind of trivializing property, which can be reformulated as follows. Ob-serve that each calculusCn is an special case ofLFIs in which the consistency operator ◦n is not primitive, but it is defined by means of a suitable combi-nation of conjunctions and negations (see [2, Subsection 3.7]). For instance,

◦1α=def ¬(α∧ ¬α) inC1. Thus:

Lemma([24, Lemma 4])

Let K be a belief set in C_n, and let αbe a sentence. If ◦nα /∈K thenK∗α=

K+α.

This means that, in the really interesting situations, that is, when the sen-tenceαto be added is not consistent (or it is not ‘well-behaved’, according to da Costa’s terminology) in the given belief setK, then the proposed revision method coincides with expansion. Similar results of collapse of revision with expansion holds for the other paraconsistent logics analyzed in [24]. This is in line with Priest’s considerations mentioned at the beginning of this section and motivates [11] to claim that the definition of a paraconsistent belief revision system which does not collapse with expansion constitutes a challenge to be tackled:

“Defining a distinct [to expansion] revision operation using para-consistent logic, thus, remains an open question.”

(P. Girard and K. Tanaka, [11, page 3])

The limiting result pointed out by [24] in the Lemma above does not hold, in general, in our setting: letLbembCor any extension of it as in Definition 3.6.

Letαbe a sentence and letK be a belief set inL such that◦α /∈K. Suppose aditionally that◦¬α /∈K(inCbrthis is a consequence of the fact that◦α /∈K),

¬α6∈Cn(∅) and¬α∈K. For instance, takeαas a propositional variablepand K=Cn({¬p}). Then ¬α∈(K+α)\(K∗α) in AGMp or AGM◦, where the internal revision∗can be taken as extensional or not. This constitutes a family of counterexamples to Tanaka’s Lemma. A particular instance is the following:

Example 7.3 Recall Example 7.2. It is easy to see that◦p /∈KifLismbC.¹⁰ Indeed, it is enough to consider the valuation v of Example 7.2. From this,

¬p∈(K+p)\(K∗p)in AGMp and AGM◦. This means thatK+p6=K∗pin AGMp and AGM◦, despite ◦p /∈K. This holds for the internal revision∗being extensional or not.

Thus, the very general approach to paraconsistent belief revision provided by AGMp and AGM◦allows us to overcome limiting results such as the ones found in [24]. This is justified by the fact that his construction presented for each logic (based on a special case of Grove’s spheres systems, in which each theory is finitely axiomatizable) satisfies the AGM-like postulates, but the converse result is not proved. That is, a representation theorem for each operation is missing. Being so, the constructions are too specific, whence results as the collapse of revision with expansion mentioned above could be expected. In contrast, our proposal is based on AGM-like postulates, on the one hand, and a family of constructions based on selection functions, on the other, obtaining a representation theorem for each operation. With this general framework at hand, it is possible to analyze the relationship between paraconsistency and belief change from a broad perspective, avoiding such trivializing results.

10Note that inmbCciw(and its extensions) the sentences∼pand◦p∧¬pare interderivable.

Thus, given that∼p∈Kit follows that◦p∈K.

8 Final Remarks

This papers studies, from a very general perspective, AGM-like systems of belief change based on paraconsistent logics. Two basic proposals were developed with full technical details: AGMp and AGM◦. The first one is oriented to supraclassi-cal paraconsistent logics, that its, expansions of the classisupraclassi-cal propositional logic CPLby adding (at least) a paraconsistent negation¬satisfying (at least) the law of excluded middle. It is shown that revision can be defined from contrac-tion, as usual, by means of the Levi identity (namely,K∗α=def(K÷ ¬α) +α), and basically these operations are the same as in AGM, by changing the classical negation∼by the paraconsistent negation¬; however, some minor adjustements are required. The real novelty of AGMp is that, capitalizing on the features of the paraconsistent negations, it is possible to define, for belief sets, revisions from contractions by means of thereverse Levi identityintroduced by Hansson only for belief bases, namely: K~α=def (K+α)÷ ¬α.

The second paradigm proposed here, AGM◦, is specifically designed for the Logics of Formal Inconsistency (LFIs), in which aconsistencyoperator◦allows us to recover all the classical inferences (including theexplosion law) within the logic. The idea behind AGM◦ is that, if ◦α ∈ K then α cannot be retracted fromK(similar to what happens whenαis a theorem of the underlying logic).

Thus, only ‘unsecure’ or ‘unreliable’ information is subject to change. Besides contraction, both internal and external revisions are defined in AGM◦by means of (direct and reverse) Levi identity. Of course both the postulates and the concrete constructions by means of selection functions must be adapted to this setting.

Finally, Hansson’s consolidation and semi-revision for belief bases are ex-tended to belief sets, capitalizing once again on the non-explosiveness of the paraconsistent negation ¬. Additionally, the ‘classical’ Levi identity (that is, w.r.t. the classical negation) was also considered for AGMp and AGM◦in order to define ‘classical’ internal revision (even though based on paraconsistent log-ics). A few toy examples show that paraconsistent revision do not necessarily coincides with plain expansion, as it is claimed by some authors.

Both systems AGMp and AGM◦ presented in this paper apprehend the dynamics of contradictory theories, particularly represented by the operators of external revisionandsemi-revision. Furthermore, AGM◦provides to the Logics of Formal Inconsistency an intuitive interpretation for the formal consistency connective, as an epistemic attitude.

Acknowledgements: The authors would like to thank to Sven Ove Hansson and Eduardo Ferm´e for the interesting and fruitful discussions generated during their visit to Campinas on May 2016, as well as for their useful comments on this work. The authors are also grateful to the annonymous referees for their corrections and suggestions, which help to improve the quality of the final version of this paper. The main ideas of this paper were presented, in a previous form, in [28]. A preliminary version of the main results presented in this paper

is contained in the pre-print [27], as well as in the PhD thesis [25], supported by a scholarship from The National Council for Scientific and Technological Development (CNPq), Brazil. This research was financed by The State of S˜ao Paulo Research Foundation (FAPESP), Brazil, Thematic Project number 2010-/51038-0 (“LogCons”). Testa was additionaly supported by FAPESP, Brazil, postdoctoral fellowship number 2014/22119-2. Coniglio was also supported by CNPq, Brazil, individual research grant number 308524/2014-4.

Appendix: Proofs of the main results

Firstly, some general properties about tarskian logics and remainder sets will be stated in order to prove the representation theorems. Recall the definition of tarskian and standard propositional logics:

Definition(Tarskian and standard logics)

A logicLdefined over a languageLand with a consequence relation`istarskian if it satisfies the following properties, for everyX∪Y ∪ {α} ⊆L:

(i) ifα∈X thenX `α;

(ii) ifX `αandX ⊆Y thenY `α;

(iii) ifX `αandY `β for every β∈X thenY `α.

A tarskian logicL isfinitaryif it satisfies the following:

(iv) ifX `αthen there exists a finite subsetX0 ofX such that X0`α.

A tarskian logicL defined over a propositional language L generated by a sig-nature from a set of propositional variables is calledstructuralif it satisfies the following property:

(v) if X `αthenσ[X]`σ(α), for every substitutionσ of formulas for vari-ables.

A propositional logic isstandardif it is tarskian, finitary and structural (see [30]).

All the logics considered in this paper are standard, and so ‘logic’ will stand for ‘standard propositional logic’. The consequence operatorCn:℘(L)→℘(L) associated to a logicL is defined as follows: Cn(X) ={α∈L : X`α}.

Lemma A(Distributivity)

Let L be a logic with a classical disjunction ∨ and a classical conjunction ∧.

That is, for every set of formulasX∪ {α, β}: Cn(X∪ {α})∩Cn(X∪ {β}) = Cn(X∪ {α∨β}), andCn(X∪ {α, β}) =Cn(X∪ {α∧β}). Then, the following distributivitylaw holds: if∅ 6=Xi=Cn(Xi)(fori= 1,2) then

Cn(X1∪ {α})∩Cn(X2∪ {α}) =Cn((X1∩X2)∪ {α})

for every formulaα.

Proof. By monotonicity of Cn, Cn((X1∩X2)∪ {α}) ⊆ Cn(Xi ∪ {α}) (for i= 1,2). ThusCn((X1∩X2)∪ {α})⊆Cn(X1∪ {α})∩Cn(X2∪ {α}). Now, let β∈Cn(X1∪{α})∩Cn(X2∪{α}). Thenβ∈Cn(Xi∪{α}) and so, by finitariness and monotonicity ofCn, there exists a finite set of formulasFi⊆Xi such that β ∈ Cn(Fi∪ {α}), for i = 1,2. By monotonicity of Cn, it can be assumed that each Fi has at least one element. Since L has a classical conjunction ∧ then there exists a formulaαi (the conjunction of the elements ofFi) such that β∈Cn({αi} ∪ {α}), fori= 1,2. SinceLhas a classical disjunction∨it follows that β ∈Cn({α1∨α₂} ∪ {α}). But α₁∨α₂ ∈Cn(X₁)∩Cn(X₂) =X₁∩X₂,

henceβ ∈Cn((X₁∩X₂)∪ {α}). 2

Lemma B Let L be a logic with a classical disjunction∨ (see Lemma A) and with a negation¬such that`α∨¬αinL, for every formulaα. LetX∪{α} ⊆L. Then,

X, α` ¬α implies X` ¬α.

Proof. Suppose thatX, α` ¬α. SinceL is a tarskian logic thenX,¬α` ¬α.

By the basic property of disjunction∨(see Lemma A) it follows thatX, α∨¬α`

¬α. But `α∨ ¬αby hypothesis and thenX ` ¬α, sinceL is tarskian. 2 Lemma C IfX ∈K⊥α, thenX ∈T h(L).

Proof. LetX ∈K⊥α. If β ∈Cn(X)\X then α∈Cn(X∪ {β}). SinceL is tarskian, this implies thatα∈Cn(X), a contradiction. ThenX =Cn(X) and

soX ∈T h(L). 2

Next result, a fundamental one, is an adaptation to the present framework of a well-known result due to Lindembaum- Los.

Lemma D(Upper-bound)

LetK be a belief set inL andα∈K. IfX ⊆K is such thatα6∈Cn(X), then there is a setX⁰∈K⊥αsuch that X ⊆X⁰.

Proof. First, assuming that the languageLis denumerable, let us arrange the sentences ofK into a sequence β₁, β₂, . . . (if L is not denumerable, the proof above must be extended in order to use transfinite induction). LetX =X₀and for eachn≥0 we defineX_n+1 as follows:

Xn+1=

Xn ifα∈Cn(Xn∪ {βn+1})

Xn∪ {βn+1} otherwise .

By construction, for every n, α6∈Cn(Xn). Let X⁰ =S

nXn. It is easy to verify thatX⊆X⁰⊆K. By compactness, ifα∈Cn(X⁰) thenα∈Cn(X⁰⁰) for some finiteX⁰⁰⊆K. It follows thatα∈Cn(Xn) for some n, a contradiction.

Thenα6∈Cn(X⁰). Moreover, ifβ ∈Kandβ6∈X⁰then, in particular,β6∈X_n+1 where n+ 1 is such that β =β_n+1. This means that α∈ Cn(X_n∪ {β}), by construction, and soα∈Cn(X⁰∪ {β}), by monotonicity. Thus,X⁰∈K⊥α. 2

Corollary E

LetK be a belief set inL. Thenα∈K\Cn(∅)if and only ifK⊥α6=∅.

Proof. ‘Only if’ part: Let α ∈ K\Cn(∅) and take X = ∅ in Lemma D. It guarantees thatK⊥α6=∅.

‘If’ part: By Definition 2.3 of remainder, ifα /∈Korα∈Cn(∅) thenK⊥α=∅.

2

Theorem 4.6

An operation∗:T h(L)×L→T h(L)is an AGMp internal revision overLiff it is an internal partial meet revision operator overL, that is: there is an AGMp selection functionγinLsuch thatK∗α= Tγ(K,¬α)

+α= (K÷_γ¬α) +α, for everyK andα.

Proof.

(construction⇒ postulates)

Letγbe an AGMp selection function, and defineK∗α= Tγ(K,¬α) +α for every (K, α)∈T h(L)×L. We have to prove that∗ satisfies the postulates for internal AGMp partial meet revision of Definition 4.5.

The satisfaction of the∗closure,∗successand∗inclusionpostulates follows from the construction (and the definition ofγ).

∗vacuity: Suppose that¬α6∈K. Henceγ(K,¬α) ={K}(by Corollary E and Definition 4.3). From this,K∗α= T

γ(K,¬α)

+α=K+α.

∗non-contradiction: Suppose that ¬α /∈ Cn(∅). If K⊥¬α 6= ∅ then ∅ 6=

γ(K,¬α)⊆K⊥¬αand so ¬α /∈K⁰ =Tγ(K,¬α). By Lemma B,¬α /∈ K⁰+α=K∗α. On the other hand, ifK⊥¬α=∅thenγ(K,¬α) ={K}

and so K∗α =K+α. By Corollary E, ¬α /∈K or ¬α∈ Cn(∅). But

¬α /∈Cn(∅), hence¬α /∈K. By Lemma B,¬α /∈K+α=K∗α.

∗relevance: Letβ∈K\K∗α. Thenβ /∈ Tγ(K,¬α)

+αwhence there exists X ∈K⊥¬αsuch thatβ6∈X. By definition of∗,K∩(K∗α)⊆K∩(X+α).

Observe thatX ⊆K∩(X+α). Suppose that there existsψ∈(K∩(X+ α))\X. Then X, ψ ` ¬α(since X ∈K⊥¬α). But X, α` ψ, therefore X, α` ¬α. ThenX ` ¬αby Lemma B, a contradiction. This means that X =K∩(X+α) and soK∩(K∗α)⊆X =Cn(X)⊆K. From the fact thatX ∈K⊥¬α, it follows that¬α /∈Cn(X) and¬α∈Cn(X) +β, since β∈K\X.

(postulates⇒ construction)

Let∗be an operator satisfying the postulates of Definition 4.5 and consider a functionγ:T h(L)×L−→℘(T h(L))\ {∅} defined as follows:

(i) Suppose that (K, β)∈T h(L)×Lis such thatβ=¬αfor some α. Then

γ(K, β) =

( {X ∈K⊥¬α : K∩(K∗α)⊆X} ifK⊥¬α6=∅, {K} otherwise .

(ii) Otherwise (that is, ifβ6=¬α), let

γ(K, β) =

( K⊥β ifK⊥β6=∅, {K} otherwise .

It will be proven that (1)γis an AGMp selection function, and (2)K∗α= Tγ(K,¬α)

+αfor every (K, α).

1. If K⊥¬α =∅ then γ(K,¬α) = {K}. IfK⊥¬α6= ∅, it must be proven that γ(K,¬α) 6=∅. By Corollary E, ¬α∈ K\Cn(∅). Then, by ∗non-contradiction, ¬α 6∈ K∗α. From this, ¬α 6∈ K ∩(K ∗α) ⊆ K. By Lemma D, there exists X⁰ ∈ K⊥¬αsuch thatK∩(K∗α)⊆X⁰, hence X⁰∈γ(K,¬α) and thenγ(K,¬α)6=∅.

2. Firstly it will be proven thatK∗α⊆ T

γ(K,¬α)

+α. By construction, K∩(K∗α)⊆Tγ(K,¬α). Hence, K∩(K∗α)

+α⊆ Tγ(K,¬α) +α.

Since the logicLsatisfies the hypothesis of Lemma A, andKandK∗αare non-empty closed theories ofL, it follows that K∩(K∗α)

+α= (K+ α)∩((K∗α) +α). From this, (K+α)∩((K∗α) +α)⊆ Tγ(K,¬α)

+α.

By ∗success and ∗inclusion, (K∗α) +α= K∗α⊆K+α. From this, K∗α⊆ Tγ(K,¬α)

+α. In order to prove the other inclusion, suppose by absurd thatβ∈ T

γ(K,¬α)

\(K∗α). SinceT

γ(K,¬α)⊆K then β ∈ K\(K ∗α). By ∗relevance, there exists X such that K ∩(K ∗ α) ⊆ Cn(X) ⊆ K, ¬α /∈ Cn(X), and ¬α ∈ Cn(X) +β. From this,

¬α∈K\Cn(∅). By Corollary E,K⊥¬α6=∅ and so γ(K,¬α) ={X ∈ K⊥¬α : K∩(K∗α)⊆X}. By Lemma D and Lemma C, there exists X⁰ ∈ K⊥¬α such that K∩(K∗α) ⊆ Cn(X) ⊆ X⁰ = Cn(X⁰). This means that X⁰ ∈γ(K,¬α) and soTγ(K,¬α)⊆X⁰. Thus, β∈X⁰. But

¬α∈Cn(X)+β, then¬α∈Cn(X⁰), a contradiction (sinceX⁰∈K⊥¬α).

ThereforeTγ(K,¬α)⊆K∗α. From this, Tγ(K,¬α)

+α⊆(K∗α) + α=K∗α, by∗success.

2

Theorem 4.8

An operation ~ : T h(L)×L → T h(L) is an AGMp external revision over L iff it is an external partial meet revision operator over L, that is: there is an AGMp selection functionγ in L such thatK~α=Tγ(K+α,¬α), for every K andα.

Proof. (construction⇒ postulates)

~closure: By the definition of~.

~success: Suppose that (K +α)⊥¬α 6= ∅ and let X ∈ (K+α)⊥¬α such that α 6∈ X. Consider X⁰ = X ∪ {α}. Since X ⊂ X⁰ ⊆ K+α then

¬α∈Cn(X⁰), by item (iii) of Definition 2.3, that is, X, α ` ¬α. Hence

X ` ¬αby Lemma B. But this contradicts the fact that ¬α6∈ Cn(X), by item (ii) of Definition 2.3. Henceα∈X for everyX ∈(K+α)⊥¬α.

Thus, if (K+α)⊥¬α6= ∅ then α∈ T

γ(K+α,¬α) = K~α. Now, if (K+α)⊥¬α=∅ then α∈ T

γ(K+α,¬α) =K~α, since in this case γ(K+α,¬α) ={K+α}, by Definition 5.10 (and obviouslyα∈K+α).

~inclusion: Clearly K~α = Tγ(K+α,¬α) ⊆ K+α, by definitions 2.3 and 4.3.

~non-contradiction: Supose that¬α∈K~α= (K+α)÷_γ¬α. By Theorem 4.4 and÷success (see Definition 4.2), it follows that¬α∈Cn(∅).

~relevance: Letβ ∈K\((K+α)÷γ¬α). Hence (K+α)⊥¬α6=∅(otherwise K~α= (K+α)÷γ ¬α=K+α and thenK\((K+α)÷γ ¬α) = ∅, a contradiction). Then there exists X ∈ γ(K+α,¬α) ⊆ (K+α)⊥¬α such that β 6∈ X. By Lemma C, X = Cn(X). By definition of ~, K~α⊆X ⊆K+α. LetX⁰ =Cn(X∪ {β}). HenceX ⊂X⁰ ⊆K+α (since β ∈ K). By Definition 2.3, ¬α /∈ Cn(X) and X⁰ ` ¬α, that is, X, β` ¬α.

~pre-expansion: (K+α)~α= ((K+α) +α)÷γ¬α= (K+α)÷γ¬α=K~α.

(postulates⇒ construction)

Let~be an operator satisfying the postulates of Definition 4.7 and consider a functionγ:T h(L)×L−→℘(T h(L))\ {∅}defined as follows:

(i) Suppose that (K, β)∈ T h(L)×L is such that β =¬α for some α, where α∈K. Then

γ(K, β) =

( {X ∈K⊥¬α : K~α⊆X} ifK⊥¬α6=∅, {K} otherwise . (ii) Otherwise (that is, ifβ6=¬α, orβ =¬αbutα /∈K), let

γ(K, β) =

( K⊥β ifK⊥β6=∅, {K} otherwise .

We will prove that (1) it is an AGMp selection function (recall Definition 4.3), and (2)K~α=T

γ(K+α,¬α).

1. It is obvious that item (ii) of the definition of γ characterizes a selection function in the sense of Definition 4.3. For item (i), suppose thatα∈K (and so K = K+α). Clearly γ(K,¬α) ⊆ K⊥¬α when K⊥¬α 6= ∅, and γ(K,¬α) = {K} otherwise. It remains to prove that γ(K,¬α) 6=∅ if K⊥¬α 6=∅. Then, suppose that K⊥¬α 6=∅. Hence ¬α 6∈ Cn(∅) by Corollary E. By~non-contradiction it is the case that¬α6∈K~α. By

~closureand~inclusion,¬α6∈K~α=Cn(K~α)⊆K+α=K. Hence, by Lemma D, there existsX ∈K⊥¬αsuch thatK~α⊆X (observe that

K⊥¬α6=∅ implies, by Corollary E, that ¬α∈K and so Lemma D can be applied). ThenX ∈γ(K,¬α) and so γ(K,¬α)6=∅ifK⊥¬α6=∅.

2. Now let us prove thatK~α= (K+α)÷ ¬α=T

γ(K+α,¬α).

(a) Suppose that (K+α)⊥¬α6=∅. Then¬α∈K+α, by Corollary E.

ClearlyK~α⊆Tγ(K+α,¬α) by definition ofγ. Letβ6∈K~α.

We have to prove that there exists X ∈ γ(K +α,¬α) such that β 6∈X. If β 6∈K+αthen β 6∈X for anyX ∈γ(K+α,¬α) (since every X ∈γ(K+α,¬α) is contained in K+α). Suppose now that β ∈ K +α. By ~pre-expansion β 6∈ (K+α)~α and then, by

~relevance, there exists Z such that K~α= (K+α)~α⊆Z ⊆ (K+α) +α = K+α, ¬α 6∈ Cn(Z) and ¬α ∈ Cn(Z) +β. By Lemma D there existsX∈(K+α)⊥¬αsuch thatK~α⊆Z ⊆X.

Hence X ∈γ(K+α,¬α). Since ¬α∈ Cn(Z) +β, thenX, β ` ¬α and henceX 0β (otherwiseX ` ¬α). Thenβ 6∈X as required. It proves thatK~α=Tγ(K+α,¬α) if (K+α)⊥¬α6=∅.

(b) Finally suppose that (K+α)⊥¬α = ∅. Then T

γ(K+α,¬α) = K+α, by definition of γ, whence K~α⊆ K+αby ~inclusion.

On the other hand, if there exists β ∈(K+α)\(K~α) then, by

~pre-expansionand~relevance, there is a setX ⊆K+αsuch that

¬α 6∈ Cn(X) (hence ¬α 6∈ Cn(∅)) but ¬α ∈ Cn(X) +β (hence

¬α∈ K+α). By Corollary E, (K+α)⊥¬α 6=∅, a contradiction.

ThenK~α=K+α=Tγ(K+α,¬α).

2

Theorem 5.11

An operation ÷:T h(L)×L −→T h(L) satisfies the postulates of AGM◦ con-traction iff there exists an AGM◦ selection function ς in L such that K÷α= Tς(K, α), for everyK andα.

Proof. (construction⇒ postulates)

÷closure: Since everyX ∈K⊥αis a closed theory (by Lemma C) andKitself is a closed theory, thenK÷_ςα=Tς(K, α) is a closed theory, since the intersection of closed theories is also closed.

÷success: Suppose thatα /∈Cn(∅) and ◦α /∈K. If K⊥α=∅ thenα /∈K, by Corollary E. Then α /∈Tς(K, α) since, in this case,ς(K, α) ={K}. On the other hand, if K⊥α6= ∅ then ∅ 6=ς(K, α) ⊆K⊥α. But α /∈ X for everyX ∈K⊥αand so α /∈Tς(K, α).

÷inclusion: Follows directly from the construction.

÷failure: Follows directly from the construction.

÷relevance: Ifβ ∈K\(K÷ςα) then there existsX ∈ς(K, α) such thatβ /∈X.

By definition of K⊥α and by Lemma C,K÷ςα ⊆ X =Cn(X) ⊆K, α /∈Cn(X) andα∈Cn(X∪ {β}) =Cn(X) +β.

(postulates⇒ construction)

Let ÷ be an operator satisfying the postulates for AGM◦ contraction of Definition 5.9 and letς be the following function:

ς(K, α) =

( {X ∈K⊥α : K÷α⊆X} if◦α /∈KandK⊥α6=∅,

{K} otherwise .

We have to prove that 1) ς is an AGM◦ selection function, and 2) K÷α = Tς(K, α).

1. The fact thatς(K, α)⊆T h(L) follows directly by construction. If◦α /∈K and K⊥α 6=∅ then α ∈ K and α /∈ Cn(∅), by Corollary E. Then, the

÷successand ÷inclusionpostulates guarantee that α /∈K÷α⊆K. By Lemma D, there existsX such thatK÷α⊆X∈K⊥α, whenceς(K, α)6=

∅. On the other hand, if◦α∈K orK⊥α=∅ thenς(K, α) ={K}.

2. Note that K÷α ⊆T

ς(K, α) = K÷ςα by construction. Suppose that β∈T

ς(K, α)\(K÷α). Then,β ∈K\(K÷α), since T

ς(K, α)⊆Kby definition. By÷relevance, there existsX such thatK÷α⊆Cn(X)⊆K and α /∈ Cn(X), but α ∈ Cn(X) +β. Thus, α ∈ K\(K ÷α). By Lemma D, there existsX⁰∈K⊥αsuch thatX ⊆X⁰, whenceK⊥α6=∅.

By ÷failure,◦α /∈K. Thus, by definition ofς(K, α),X⁰∈ς(K, α). From this, β ∈X⁰. But α∈Cn(X) +β, hence α∈Cn(X⁰) +β. This means thatα∈Cn(X⁰), a contradiction. ThereforeK÷α=Tς(K, α).

2

Theorem 5.13

An operation∗:T h(L)×L−→T h(L)overLsatisfies the postulates for internal partial meet AGM◦ revision (see Definition 5.12) if and only if there exists an AGM◦ selection functionς inLsuch thatK∗α= T

ς(K,¬α)

+α, for every K andα.

Proof.

(construction⇒ postulates)

Let ς be an AGM◦ selection function (recall Definition 5.10), and define K∗α = T

ς(K,¬α)

+α for every (K, α) ∈ T h(L)×L. We have to prove that∗ satisfies the postulates for internal AGM◦ partial meet revision of Defi-nition 5.12.

The satisfaction of the∗closure,∗successand∗inclusionpostulates follows from the construction (and the definition ofς).

∗vacuity: Suppose that ¬α 6∈ K. This implies that ς(K,¬α) = {K}, by Corollary E and Definition 5.10. Therefore K∗α= Tς(K,¬α)

+α= K+α.

∗non-contradiction: Suppose that ¬α /∈Cn(∅) and ◦¬α /∈ K. If K⊥¬α6=∅ then∅ 6=ς(K,¬α)⊆K⊥¬αand so¬α /∈K⁰=T

ς(K,¬α). By Lemma B,

¬α /∈K⁰+α=K∗α. On the other hand, ifK⊥¬α=∅thenς(K,¬α) = {K}and soK∗α=K+α. By Corollary E,¬α /∈K or¬α∈Cn(∅). But

¬α /∈Cn(∅), hence¬α /∈K. By Lemma B,¬α /∈K+α=K∗α.

∗failure: If◦¬α∈ K then K⁰ =T

ς(K,¬α) is K, by definition ofς. Hence K∗α=K⁰+α=K+α.

∗relevance: It is proved as in the proof of Theorem 4.6 for the corresponding postulate.

(postulates⇒ construction)

Let∗be an operator satisfying the postulates of AGM◦internal revision and consider a functionς :T h(L)×L−→℘(T h(L))\ {∅}defined as follows:

(i) Suppose that (K, β)∈T h(L)×Lis such thatβ=¬αfor some α. Then

ς(K, β) =

( {X∈K⊥¬α : K∩(K∗α)⊆X} if◦¬α6∈K andK⊥¬α6=∅,

{K} otherwise .

(ii) Otherwise (that is, ifβ6=¬α), let

ς(K, β) =

( K⊥β ifK⊥β6=∅, {K} otherwise .

It will be proven that (1) ς is an AGM◦ selection function (recall Defini-tion 5.10), and (2)K∗α= T

ς(K,¬α)

+αfor every (K, α).

1. If K⊥¬α = ∅ then ς(K,¬α) = {K}. If K⊥¬α 6=∅, it must be proven that ς(K,¬α) 6= ∅. By Corollary E, ¬α ∈ K\ Cn(∅). If ◦¬α ∈ K then ς(K,¬α) = {K} 6= ∅. If ◦¬α /∈ K then, by ∗non-contradiction,

¬α 6∈K∗α. From this, ¬α 6∈ K∩(K∗α) ⊆ K. By Lemma D, there existsX⁰ ∈K⊥¬αsuch thatK∩(K∗α)⊆X⁰, henceX⁰∈ς(K,¬α) and thenς(K,¬α)6=∅.

2. Firstly it will be proven that K∗α ⊆ Tς(K,¬α)

+α. By the very construction, K∩(K∗α) ⊆ T

ς(K,¬α). Hence, K∩(K∗α) +α ⊆ Tς(K,¬α)

+α. Since the logicLsatisfies the hypothesis of Lemma A, andKandK∗αare non-empty closed theories ofL, it follows that K∩ (K∗α)

+α= (K+α)∩((K∗α) +α). From this, (K+α)∩((K∗α) +α)⊆ Tς(K,¬α)

+α. By∗successand∗inclusion, (K∗α)+α=K∗α⊆K+α.

From this,K∗α⊆ T

ς(K,¬α)

+α. In order to prove the other inclusion, there are two cases to analyze:

(a) If◦¬α∈Kthen, by∗failure,K∗α=K+α. Using Corollary E and definition ofς it follows thatς(K,¬α) ={K}, hence T

ς(K,¬α) + α=K+α=K∗α.

(b) If ◦¬α /∈ K, suppose by absurd that β ∈ T

ς(K,¬α)

\(K∗α).

Then β ∈ K\(K∗α), since T

ς(K,¬α) ⊆ K . Using ∗relevance, there existsX such thatK∩(K∗α)⊆Cn(X)⊆K,¬α /∈Cn(X), and ¬α∈ Cn(X) +β. Thus, ¬α∈ K\Cn(∅). Using Corollary E it follows that K⊥¬α 6= ∅, whence ς(K,¬α) = {X ∈ K⊥¬α : K∩(K∗α)⊆X}. By Lemma D and Lemma C, there existsX⁰ ∈ K⊥¬αsuch that K∩(K∗α)⊆Cn(X)⊆X⁰ =Cn(X⁰). But then X⁰ ∈ ς(K,¬α) and so Tς(K,¬α)⊆X⁰. As a consequence of this, β ∈X⁰. But¬α∈Cn(X) +β, then¬α∈Cn(X⁰), a contradiction (sinceX⁰ ∈K⊥¬α). This means thatTς(K,¬α)⊆K∗α. Therefore,

Tς(K,¬α)

+α⊆(K∗α) +α=K∗α, by∗success.

2

Theorem 5.15

An operation ~ : T h(L)×L −→ T h(L) over L satisfies the postulates for external partial meet AGM◦revision (see Definition 5.14) iff there is an AGM◦

selection functionς inLsuch thatK~α=T

ς(K+α,¬α), for everyK andα.

Proof. (construction⇒ postulates)

~closure: It follows as in the proof of Theorem 5.11.

~success: Suppose that◦¬α∈K or (K+α)⊥¬α=∅. Then,ς(K+α,¬α) = {K+α}by definition and soα∈K+α=Tς(K+α,¬α). Now, suppose that◦¬α /∈Kand (K+α)⊥¬α6=∅. LetX∈(K+α)⊥¬α, and suppose by absurd that α /∈ X = Cn(X). Then X ⊂ X ∪ {α} ⊆ K+α and so X, α ` ¬α (by item (iii) of Definition 2.3). But then X ` ¬α, by Lemma B, a contradiction. Thereforeα∈X for everyX ∈(K+α)⊥¬α and soα∈T

ς(K+α,¬α).

~inclusion: It follows by construction.

~non-contradiction: Suppose that ¬α /∈ Cn(∅) and ◦¬α /∈ K+α. If (K+ α)⊥¬α=∅thenς(K+α,¬α) ={K+α}and¬α /∈K+α, by Corollary E.

This implies thatTς(K+α,¬α) =K+α, whence¬α /∈Tς(K+α,¬α).

On the other hand, if (K+α)⊥¬α6=∅ then ∅ 6=ς(K+α,¬α)⊆(K+ α)⊥¬α. This implies that¬α /∈Tς(K+α,¬α).

~failure: Suppose that ◦¬α ∈ K+α. By Definition 5.10, ς(K +α,¬α) = {K+α}. ThenT

ς(K+α,¬α) =K+α.

~relevance: Letβ∈K\Tς(K+α,¬α). Therefore (K+α)⊥¬α6=∅(otherwise, Tς(K+α,¬α) =K+αand so K\Tς(K+α,¬α) =K\(K+α) =∅, a contradiction). Hence, there exists X ∈ς(K+α,¬α) ⊆(K+α)⊥¬α such thatβ /∈X. By construction and Lemma C,Tς(K+α,¬α)⊆X = Cn(X)⊆K+α. LetX⁰=X∪ {β}. ThereforeX ⊂X⁰ ⊆K+αby the fact thatβ ∈K. Then¬α∈Cn(X⁰) and hence ¬α∈Cn(X) +β, while

¬α /∈Cn(X).

~pre-expansion: T

ς((K+α) +α,¬α) =T

ς(K+α,¬α).

(postulates⇒ constructions)

Let ~ be an operator satisfying the postulates of AGM◦ external revision and consider a functionς :T h(L)×L−→℘(T h(L))\ {∅}defined as follows:

(i) Suppose that (K, β)∈ T h(L)×L is such that β =¬α for some α, where α∈K. Then

ς(K, β) =

( {X∈K⊥¬α : K~α⊆X} if◦¬α /∈K andK⊥¬α6=∅,

{K} otherwise .

(ii) Otherwise (that is, ifβ6=¬α, orβ =¬αbutα /∈K), let

ς(K, β) =

( K⊥β ifK⊥β6=∅, {K} otherwise .

We have to prove that (1)ς is an AGM◦selection function (see Definition 5.10), and (2)K~α=Tς(K+α,¬α).

1. By considering the case (ii) of the construction of ς, it is clear that the conditions of Definition 5.10 are fullfilled. Now, suppose that α ∈ K (hence K = K+α) and let us analyze the definition of ς(K,¬α). If K⊥¬α =∅ then ς(K,¬α) ={K} as required. It remains to prove that

∅ 6=ς(K,¬α) ifK⊥¬α6=∅, and that ς(K,¬α)⊆K⊥¬αif, additionally,

◦¬α /∈K. The latter holds by the very definition ofς. Suppose then that K⊥¬α 6= ∅; by Corollary E, ¬α 6∈ Cn(∅), and ¬α ∈ K. If ◦¬α ∈ K then ς(K,¬α) ={K} 6=∅. Finally, if ◦¬α 6∈K then ς(K,¬α) ={X ∈ K⊥¬α : K~α⊆X}. By~non-contradiction,¬α6∈K~α. By~closure and~inclusion,¬α6∈K~α=Cn(K~α)⊆K+α=K. By Lemma D (recalling that ¬α∈K), there existsX ∈K⊥¬αsuch that K~α⊆X.

This means thatX ∈ς(K,¬α), whenceς(K,¬α)6=∅.

2. Suppose firstly that◦¬α /∈K+αand (K+α)⊥¬α6=∅. Then¬α∈K+α and ¬α /∈ Cn(∅). By construction, K~α⊆ Tς(K+α,¬α). In order to prove the converse inclusion, let β /∈ K ~α. It is enough to prove that there exists X ∈ ς(K+α,¬α) such that β 6∈ X. If β 6∈ K+α then β 6∈ X for every X ∈ ς(K+α,¬α) (since X ⊆ K+α for every X ∈ς(K+α,¬α)). Suppose now that β ∈K+α. By ~pre-expansion, β /∈(K+α)~αand then, by~relevance, there existsZsuch thatK~α= (K+α)~α ⊆ Cn(Z) ⊆ (K +α) +α = K +α, ¬α /∈ Cn(Z) and

¬α∈Cn(Z) +β. By Lemma D (taking into account that ¬α∈K+α), there exists X ∈ (K+α)⊥¬α such thatK~α⊆Cn(Z)⊆ X. Hence, X ∈ ς(K +α,¬α). Since ¬α ∈ Cn(Z) +β, then ¬α ∈ X +β and therefore β /∈ X = Cn(X) (otherwise, ¬α ∈ Cn(X)). It follows that β /∈Tς(K+α,¬α). From this it is concluded thatK~α=Tς(K+α,¬α).

Finally, suppose that◦¬α∈K+αor (K+α)⊥¬α=∅. By construction, it follows thatT

ς(K+α,¬α) =K+α. HenceK~α⊆T

ς(K+α,¬α), by~inclusion. There are two case to analyze:

(a) If◦¬α∈K+αthen, by~failure,K~α=K+α=T

ς(K+α,¬α).

(b) If (K+α)⊥¬α=∅then, by Corollary E,¬α /∈K+αor¬α∈Cn(∅).

Suppose (by absurd) that there exists β ∈(K+α)\(K~α). By

~pre-expansion,β ∈(K+α)\((K+α)~α). By~relevance, there exists X ⊆ K+α such that ¬α /∈ Cn(X) and ¬α ∈ Cn(X) +β.

But then ¬α∈K+αand¬α /∈Cn(∅), a contradiction. From this K~α=K+α=Tς(K+α,¬α).

2 Theorem 4.14(Representation for extensional AGMp contraction) Proof. The proof is similar to the one for standard AGM, with minor differences with respect to÷extensionality, on the one hand, and the first property of the selection function, on the other. Let us concentrate on these differences:

(construction⇒ postulates)

Letγbe a general AGMp selection function. In order to prove thatK÷γα= Tγ(K, α) satisfies ÷extensionality, suppose that α≡L β and ¬α≡L ¬β. By Definition 4.13,γ(K, α) =γ(K, β). Being so,K÷γα=K÷γβ.

(postulates⇒ construction)

Suppose now that÷is an extensional AGMp contraction overL, and define a functionγas follows:

γ(K, α) =

( {X ∈K⊥α : K÷α⊆X} ifK⊥α6=∅, {K} otherwise .

We have to prove that 1)γ is a general AGMp selection function, and 2)K÷ α=Tγ(K, α). For 1), it is enough to prove that item 1 of Definition 4.13 is satisfied (since the other properties are proved as in classical AGM, using AGM-compliance). Thus, suppose thatα≡Lβ and¬α≡L¬β. ThenK⊥α=K⊥β, andK÷α=K÷β, by÷extensionality. This means thatγ(K, α) =γ(K, β).

The proof of 2) is as in classical AGM, using AGM-compliance (observe that

÷extensionalityis not used here). 2

Theorem 4.15 (Representation for extensional AGMp internal revi-sion)

Proof. It is similar to the one given for Theorem 4.6, with the following changes:

(construction⇒ postulates)

Given a general AGMp selection function γ, it is necessary to prove that K∗α= Tγ(K,¬α)

+αsatisfies, additionally, ∗extensionality. Thus, sup-pose that α ≡L β and ¬α ≡L ¬β. Then ¬¬α ≡L α ≡L β ≡L ¬¬β and so γ(K,¬α) =γ(K,¬β), by Definition 4.13. From this, K∗α= Tγ(K,¬α)

+

α= T

γ(K,¬β)

+α= T

γ(K,¬β)

+β =K∗β.

(postulates⇒ construction)

Let∗be an operator satisfying the postulates of an external AGMp internal revision and consider a functionγ :T h(L)×L−→℘(T h(L))\ {∅}defined as follows:

γ(K, α) =

( {X ∈K⊥α : K∩(K∗ ¬α)⊆X} ifK⊥α6=∅, {K} otherwise .

With a proof similar to that for Theorem 4.6, it can be seen that 1)γis an AGMp selection function, and 2)K∗α= Tγ(K,¬α)

+αfor every (K, α).

To see 1), the only detail to be taken into account is that α ≡L ¬¬α (and so¬α≡L ¬¬¬α), whenceK∗α=K∗ ¬¬αby∗extensionality. Finally, using again∗extensionality, it is immediate to prove thatγis, in fact, a general AGMp selection function.

2 Theorem 4.16 (Representation for extensional AGMp external revi-sion)

Proof. It is similar to the one given for Theorem 4.8, with minor changes.

These changes are analogous to the ones given in the proof of Theorem 4.15.

The details are left to the reader. 2

Theorem 5.20(Representation for extensional AGM◦ contraction) Proof. It is similar to the one given for Theorem 5.11, with minor changes.

These changes are analogous to the ones given in the respective proof of the-orems 4.15 and 4.16. For instance, in order to see that, given an extensional AGM◦contraction÷, the induced mappingς is a general AGM◦selection func-tion, it is enough to observe the following: ifα≡Lβ,¬α≡L¬β and◦α≡L◦β then K ÷α = K ÷β, K⊥α = K⊥β, and ◦α ∈ K iff ◦β ∈ K. From this ς(K, α) =ς(K, β). The details are left to the reader. 2 Theorem 5.21 (Representation for extensional AGM◦ internal revi-sion)

Proof. It is similar to the one given for Theorem 5.13, with minor changes.

These changes are analogous to the ones given in the proof of Theorem 4.15.

For instance, in order to see that, given an extensional AGM◦ internal revision

∗, the induced mappingς is a general AGM◦selection function, observe that ς must be defined as follows:

ς(K, α) =

( {X ∈K⊥α : K∩(K∗ ¬α)⊆X} ifK⊥α6=∅, {K} otherwise .

Now, it is enough to observe the following: if α ≡L β, ¬α ≡L ¬β and

◦α ≡L ◦β then ¬¬α ≡L ¬¬β and so K∗ ¬α =K∗ ¬β. On the other hand,

K⊥α = K⊥β, and ◦α ∈ K iff ◦β ∈ K. From this ς(K, α) = ς(K, β). The

details are left to the reader. 2

Theorem 5.22 (Representation for extensional AGM◦ external revi-sion)

Proof. It is similar to the one given for Theorem 5.15, with minor changes.

These changes are analogous to the ones given in the proof of theorem 5.21. 2 The upper-bound lemma (Lemma D) as well as its corollary (Corollary E) can be generalized for sets, taking into account Definition 6.4.

Lemma FIf X ∈K⊥A then X ∈T h(L). Proof. It is analogous to the proof

of Lemma C. 2

Lemma G (Upper-bound for sets)

Let K be a belief set in L, and ∅ 6=A⊂L such thatK∩A6=∅. Let X ⊆K such thatCn(X)∩A=∅. Then, there existsX⁰∈K⊥PAsuch that X⊆X⁰. Proof. It is analogous to the proof of Lemma D, but now Xn+1 is defined as follows:

X_n+1=

Xn ifCn(Xn∪ {βn+1})∩A6=∅ Xn∪ {βn+1} otherwise

2 Corollary HLet K be a belief set inL, and ∅ 6=A⊂L. Then: K⊥PA6=∅ if and only if K∩A 6= ∅ and Cn(∅)∩A = ∅. Proof. If K∩A 6= ∅ and Cn(∅)∩A = ∅ take X = ∅ in Lemma G. The converse follows by the very

definition ofK⊥PA. 2

Recall that a setX in a normal paraconsistent logic is contradictory ifα∧

¬α∈Cn(X) for some formulaα. Then:

Corollary ILetK be a belief set in a normal paraconsistent logicL, and let Ω_K be the set of contradictory sentences of K (recall Definition 6.5). Then:

K⊥PΩK6=∅ if and only ifK is contradictory. Proof. Immediate from

Corol-lary H and the definitions. 2

Theorem 6.7

An operation! :T h(L)−→T h(L)over a normal paraconsistent logic Lsatisfies the postulates of Definition 6.3 iff there exists a consolidation functionγ^f inL (in the sense of Definition 6.6) such thatK! =T

γ^f(K) for every belief setK inL.

Proof.

(construction⇒ postulates)

closure: By Lemma F, everyX ∈K⊥PΩK is a closed theory, andKitself is a closed theory. From this,T

γ^f(K) is a closed theory, since the intersection of closed theories is also closed.

inclusion: It follows by construction.

non-contradiction: Suppose thatK 6=L. IfK⊥PΩ_K 6=∅ then∅ 6=γ^f(K)⊆ K⊥PΩ_K. Let X ∈ K⊥PΩ_K. Then, Tγ^f(K)

∩Ω_K ⊆ X∩Ω_K = ∅.

On the other hand, ifK⊥_PΩ_K =∅ thenγ^f(K) ={K}. By Corollary I, K∩Ω_K =∅. Thus Tγ^f(K)

∩Ω_K =K∩Ω_K=∅.

failure: It follows from the definition ofγ^f.

relevance: Letβ∈K\Tγ^f(K). Then,Tγ^f(K)6=Kand so, by construction, K⊥PΩK 6= ∅. Thus, there exists X ∈ γ^f(K) ⊆ K⊥PΩK such that β /∈X. By construction and by Lemma F, T

γ^f(K)⊆X=Cn(X)⊆K.

LetX⁰ =X∪ {β}. ThenX ⊂Cn(X⁰)⊆K by the fact that β ∈K. By Definition 6.4, ΩK∩Cn(X⁰)6=∅, that is, ΩK∩(Cn(X) +β)6=∅.

(postulates⇒ construction) Consider the following function:

γ^f(K) =

( {X ∈K⊥PΩK : K!⊆X} ifK6=LandK⊥PΩK 6=∅,

{K} otherwise .

We must prove that (1) γ^f is a consolidation function in the sense of Defini-tion 6.6, and (2)K! =Tγ^f(K).

1. It follows by construction thatγ^f(K)⊆K⊥PΩKifK6=LandK⊥PΩK 6=

∅, andγ^f(K){K}otherwise. It remains to prove thatγ^f(K)6=∅whenever K 6= L and K⊥PΩK 6=∅. Observe that, if K 6= L and K⊥PΩK 6= ∅, then ΩK∩Cn(K!) =∅, by non-contradiction and closure. Byinclusion, K! ⊆ K. Then, by Lemma G, there exists X ∈ K⊥PΩK such that K!⊆X. It follows that X∈γ^f(K) and thenγ^f(K)6=∅.

2. It follows by construction and by inclusion that K! ⊆γ^f(K). We must show that γ^f(K) ⊆ K!. To this end, it is sufficient to show that, if β /∈K! thenβ /∈Tγ^f(K). Thus, let β /∈K!. Ifβ /∈K thenβ /∈γ^f(K) trivially. Now, suppose that β ∈ K. By relevance, there exists X such that K!⊆Cn(X)⊆K,Cn(X)∩Ω_K =∅, butCn(X∪ {β})∩Ω_K 6=∅.

By Lemma G and Lemma F, there existsX⁰∈K⊥_PΩ_K such that K!⊆ Cn(X)⊆X⁰ =Cn(X⁰). Hence,X⁰∈γ^f(K). Since Ω_K∩Cn(X∪{β})6=∅ it follows that β /∈ Cn(X⁰) = X⁰ (otherwise ΩK ∩Cn(X⁰) 6=∅). From this,β /∈Tγ^f(K).

2

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