Which Semantics for Neighbourhood Semantics?

Which Semantics for Neighbourhood Semantics?

Carlos Areces

INRIA Nancy, Grand Est, France

Diego Figueira

INRIA, LSV, ENS Cachan, France

Abstract

In this article we discuss two alternative proposals for neighbourhood semantics (which we callstrict andlooseneighbourhood semantics, N= andN⊆ respectively) that have been previously introduced in the literature. Our main tools are suitable notions of bisimulation. While an elegant notion of bisim-ulation exists forN⊆, the required bisimulation for

N=is rather involved. We propose a simple exten-sion ofN= with a universal modality that we call

N=(E), which comes together with a natural notion of bisimulation. We also investigate the complexity of the satisfiability problem forN⊆andN=(E).

1

Epistemic Logic and Neighbourhood

Semantics

Epistemic logic, the logic that studies notions like agents’s beliefsandknowledge, is an important and long-standing area of research in artificial intelligence [Faginet al., 1995].

In epistemic logic, the formula[α]ϕis used to represent that agentαbelieves or knows thatϕis the case. When the agentαis understood by context, or when we are not inter-ested on modelling the behaviour of different agents at the same time, we will usually write[ ]ϕinstead of[α]ϕ. In the rest of this article we will discuss the case for a single agent. By adding the[ ]operator to classical propositional logic, we can already express a number of interesting properties. For example, the formula[ ](ϕ∧ψ) →([ ]ϕ∧[ ]ψ)intuitively says that if an agent believes or knows the conjunction of two factsϕandψ, then it also knows bothϕandψ.

Epistemic logic is usually considered a member of the large family of modal logics [Blackburn et al., 2001], and as we will discuss in this article, it shares with them many of their properties (e.g., characterizations in terms of bisimulations, good computational behavior, etc.). But, as it is well known (see [Vardi, 1986]), semantics specified over standard Kripke models in terms of possible worlds and accessibility rela-tions [Blackburnet al., 2001] have some undesired epistemic properties. The reason is that even the weakest logic definable in terms of Kripke models (i.e., the modal logicKdefined as

∗

Work partially done while Master’s student at Departamento de Computaci´on, FCEyN, UBA, Argentina.

the set of those modal formulas valid on the class of all Kripke structures) may be already too strong for modeling knowl-edge or belief. For example, Kripke semantics makes valid all instances of the formula scheme([ ]ϕ∧[ ](ϕ→ψ))→[ ]ψ and the inference rule|=ϕthen|= [ ]ϕ,while both are unin-tuitive under an epistemic interpretation of[ ]. If we read[ ]as knowledge, they would require an agent to know all tautolo-gies, and to be able to effectively draw all consequences of its knowledge. This is what is calledthe logical omniscience problem, and it was already discussed in [Hintikka, 1975].

[Vardi, 1986] proposes to adopt a different semantics, originally introduced in [Montague, 1968; 1970] and [Scott, 1968] in a different setup. This alternative semantics uses the notion ofneighbourhoodsto define the meaning of the epis-temic operator. The intuitive idea is that an agent’s knowledge is characterized not by a set of possible worlds but rather by an explicit set of propositions known by the agent. More pre-cisely, we can defineepistemic structuresas follows.

Definition 1(Epistemic Structure). Anepistemic structureis

a tupleM=hW, N,k · kM_iwhere • W 6=∅is a set (ofpossible worlds).

• N is a function mapping elements fromW to sets of subsets ofW (i.e.,N(w)⊆ ℘(W), for eachwinW). We will usually writeNwinstead ofN(w)and callNw theneighbourhoodsofw.

• k · kM_{is an assignment function from the set of}

propo-sitional symbol to subsets ofW (i.e.,kpkM _{⊆ W, for}

each propositional symbolp∈PROP). ₃ Throughout the paper, letMbe the modelhW, N,k · kM_i

andM0_{be the modelhW}0_{, N}0_{,k · k}M0

i.

Notice that, instead of the standard accessibility relation between worlds in a Kripke model, an epistemic structure specifies, for each worldw ∈ W, a set of setsNw. As we can represent a propositionP by a set of possible worlds, the intuition is that ifP∈NwthenP is known inw.

Given a modelMwe can extendk · kM_{to all formulas in}

the language. The boolean cases are standard:

k¬ϕkM_=W\kϕkM_{, kϕ∧ψk}M_=kϕkM_{∩ kψk}M_.

The[ ]operator, on the other hand, has been defined in two different ways in the literature, giving origin to two different

operators that we will note[⊆]and[=]:

k[=]ϕkM _{= {w| ∃X ∈N}

wsuch thatX =kϕkM},

k[⊆]ϕkM _{= {w| ∃X ∈N}

wsuch thatX ⊆ kϕkM}. The[=]operator is the most widely used, and it is the one originally introduced in [Vardi, 1986]. We will call this se-mantics thestrict neighbourhood semanticsN=, because for

[=]ϕto be true atw,kϕkM_{should be one of the}

neighbour-hoods ofw, i.e.,kϕkM_∈N

w. The[⊆]operator on the other hand, is slightly weaker. We only requirekϕkM_tocoverat

least one of the neighbourhoods ofwand not to exactly co-incide with it. We will call this weaker semantics theloose neighbourhood semanticsN⊆, and it has been mentioned by van Benthem in a number of articles (e.g., [Aiello and van Benthem, 2002]).

It is easy to see that the two semantics are indeed different:

Example 2. Consider the epistemic structure

1 p 2 p

M:

M=h{1,2},{N1={{2}}, N2={}},{kpkM={1,2}}i. Clearly1∈ k[⊆]pkM_{but16∈ k[=]pk}M_{, askpk}M_6={2}.

Actually, the following property can easily be proved.

Proposition 3. In any epistemic structure M, and for any

formulaϕ,k[=]ϕkM_{⊆ k[⊆]ϕk}M_.

We will investigate the difference between these two oper-ators by means of bisimulations. In the next section we will show that while an elegant notion of bisimulation exists for

N⊆, the required bisimulation forN=is rather involved. We will propose in Section 3 a simple extension ofN= that we callN=(E), which comes together with a natural notion of bisimulation. Finally, we close the paper in Section 4 investi-gating the complexity of the satisfiability problem ofN⊆and N=(E), and show that they are both NP-complete.

2

Bisimulations for

N

and

N

It is easy to define an adequate notion of bisimulation forN⊆.

Definition 4(N⊆-bisimulation). LetMandM0_{be two}

epis-temic structures. AnN⊆-bisimulationbetweenMandM0is

a non-empty relationZ ⊆W ×W0such that ifxZx0then:

Prop: ∀p∈PROP,x∈ kpkM

iffx0∈ kpkM0

.

Zig: ∀T ∈Nx∃T0∈Nx00s.t.:∀w0∈T0∃w∈Ts.t.wZw0.

Zag: ∀T0∈Nx00∃T ∈Nxs.t.:∀w∈T ∃w0∈T0s.t.wZw0. 3

We will use the notationhM, wiwhenwis an element of the epistemic structureM. Then we can writehM, wi ↔⊆ hM0_{, w}0_{iif there is anN}

⊆-bisimulation betweenMandM0

linking w and w0_{. N⊆-bisimulations satisfy the expected}

properties: they preserve formulas inN⊆, and exactly

char-acterize formula equivalence on finite models.

Definition 5 (Pointwise equivalence). Given hM, wi and

hM0_{, w}0_{iwe say that they arepointwise equivalentfor a}

se-mantics S (notation: hM, wi ≡S hM0_{, w}0_{i) iff ∀ϕ ∈ S :}

w∈ kϕkM_iffw0 _{∈ kϕk}M0

. 3

Proposition 6. Given epistemic structuresMandM0_then:

hM, wi ↔⊆ hM0, w0iimplieshM, wi ≡N⊆ hM

0_{, w}0_{i. If} MandM0_{are finite then the converse also holds.}

The case forN=is more complex. Given that the semantic condition forN= is very similar toN⊆ we could try to use

the following definition:

Definition 7(SymmetricN=-bisimulation). LetMandM0

be two epistemic structures. AsymmetricN=-bisimulation betweenMandM0 _{is a non-empty relationZ ⊆W ×W}0

such that ifxZx0then:

Prop: ∀p∈PROP,x∈ kpkM

iffx0∈ kpkM0

.

Zig: ∀T ∈Nx∃T0∈Nx00 that verifies:

Zig 1: ∀w0∈T0∃w∈Ts.t.wZw0, and

Zig 2: ∀w0∈W0\T0 ∃w∈W\T s.t.wZw0.

Zag: ∀T0∈N_x00∃T ∈Nxthat verifies:

Zag 1: ∀w∈T ∃w0∈T0s.t.wZw0, and

Zag 2: ∀w∈W\T ∃w0∈W0\T0s.t.wZw0. 3 We will writehM, wi ↔s

=hM0, wiif there is a symmetric

N=-bisimulation betweenMandM0_linkingwandw0_.

Note that Definition 7 is an extension of the definition of

N⊆-bisimulation with the symmetric conditions Zig 2and Zag 2. Now, symmetricN=-bisimulations do preserve satis-fiability ofN=formulas (so do isomorphism, for that matter), but we will argue that it is too strong, and that they do not match the limited expressive power ofN=.

Proposition 8. There exist finite epistemic structures which

are pointwise equivalent and not symmetricN=-bisimilar. Proof. Let us consider the following two models:

w0 w1 p

M: M0: w00 w02 p q w10 p That is,M=h{w0, w1}, N,k · kM_{i, wherekpk}M _={w1},

Nw0 = {{w1}}. AndM

0 _{= h{w}0

0, w01, w02}, N0,k · kM 0

i, wherekpkM0 _{= {w}0

1, w20}, kqkM 0

= {w02}, and Nw00

0 =

{{w₁0},{w₁0, w₂0}}. w0 andw00satisfy the same formulas in

N=but, even thoughMandM0are finite, it is not possible to define a symmetricN=-bisimulation that links them .

Actually MandM0 _{in Proposition 8 are even}

differenti-ated, a notion we will introduce now, and that will be useful to define the adecuate notion of bisimulation forN=.

Definition 9(Differentiation). The classDofdifferentiated

epistemic structures is defined as

D={M | ∀w∈W ∀T ∈Nw∃ϕs.t.kϕkM=T}. I.e., D is the class of epistemic structures where every neighbourhood is the extension of a formula. We say that Tisdifferentiable inMif for someϕ,kϕkM_{=T, and that}

ϕis acharacteristic formulaofT. ThedifferentiationofM

is the structureMd_{=hW, N}d_{,k · k}M_i

Nwd ={T |T ∈NwandT differentiable inM}. I.e.,Ndis obtained by eliminating all sets inNwhich are not

differentiable. 3

It should be clear that the operation of differentiation does not affect satifiability of N= formulas as stated next (see [Areces and Figueira, 2008] for details).

Proposition 10. ∀M,∀ϕ∈ N=:kϕkM=kϕkM

d

Given Proposition 10, we can define the notion of N =-bisimulation only for differentiated epistemic structures.

Definition 11 (N=-bisimulation). Let M andM0 _{be two}

differentiated epistemic structures. AnN=-bisimulation be-tweenMandM0_{is a non-empty relationZ⊆W×W}0_such

that ifwZw0then:

Prop: ∀p∈PROP,w∈ kpkM

iffw0∈ kpkM0 Zig: ∀T∈Nw∃T0∈Nw00that verifies:

Zig 1: ∀w0∈T0∃w∈T s.t.wZw0, and

Zig 2: ∀X0∈/N_w00differentiable inM0s.t.T0₍X0⊆W0, then there arew0∈X0\T0, w∈W\T s.t.wZw0

Zag: ∀T0enN_w00∃T ∈Nwthat verifies:

Zag 1: ∀w∈T∃w0∈T0s.t.wZw0and

Zag 2: ∀X /∈ Nw differentiable inMs.t.T ( X ⊆ W, then there arew∈X\T, w0∈W0\T0s.t.wZw0 3

We write hM, wi ↔= hM0, w0i if there is an N =-bisimulation betweenMandM0_linkingwandw0_{. We first}

check thatN=-bisimulations preserve satisfiability.

Proposition 12. Given epistemic structuresMandM0_then:

hM, wi ↔=hM0, w0iimplieshM, wi ≡N= hM

0_{, w}0_i.

Proof. The Boolean cases are easy. We only discuss the case

[=]ψ. LetZbe anN=-bisimulation linkingwandw0_. [⇒]Ifw∈ k[=]ψkM_{, thenkψk}M_∈N

w. By conditionZig 1, there must be aT0 ∈N_w00 such that∀x0∈T0∃x∈ kψkM such thatxZx0. Letx0∈T0 ∈N_w00, asx∈ kψkMandxZx0, by inductive hypothesis:x0∈ kψkM0

. HenceT0 ⊆ kψkM0 . To show thatkψkM0 _{∈ N}0

w0, notice first that every world inkψkM0

\T0 is not bisimilar to any other inW \ kψkM

because if any of these worlds s0 was bisimilar to another s ∈ W \ kψkM_{, it would mean thats 6∈ kψk}M _buts0 _∈ kψkM0_{, which by inductive hypothesis is absurd. Now, if}

T0 = kψkM0

we are done, ifT0 ( kψkM 0

, we know by bisimulation conditionZig 2(takingX0askψkM0

), that nec-essarilyT0_∪(kψkM0 _\T0_{) ∈ N}0

w0, forT0 ∈ N_w00 and all elements inkψkM0

\T0 are not bisimilar to any other from W \ kψkM_{. AsT}0 _{⊂ kψk}M0_{, thenT}0_∪(kψkM0 _\T0_{) =} kψkM0

, this set is therefore differentiable (by the formulaψ), sokψkM0

∈N_w00, i.e.,w0 ∈ k[=]ψkM 0

.

[⇐]The converse is analogous usingZag 1andZag 2. OverfinitemodelsN=equivalence is precisely character-ized by the notion ofN=-bisimilarity as stated next.

Proposition 13. Let M and M0 _{be finite, differentiable}

epistemic structures then: hM, wi ≡N= hM

0_{, w}0_iimplies hM, wi ↔=hM0, w0i.

Proof. The required bisimulationZ is defined as wZw0 _iff hM, wi ≡N= hM

0_{, w}0_{i. By definition, (w, w}0_{) ∈ Z and}

henceZis non-empty.Propalso holds by definition. Zig. Assume thatZigdoes not hold, i.e.,wZw0,T ∈Nw yet ∀T0 ∈ N_w00 either Zig 1 or Zig 2 fail. Let N_w00 =

{T₁0, . . . , T_t0} ∪ {S₁0, . . . , S_l0}where all the elements from the first set failZig 1and those from the second failZig 2. Let T ={h1, . . . , hn}andW \T ={¯h1, . . . ,¯hm}.

Setψ0 := ϕT whereϕT is a characteristic formula ofT. Asw∈ k[=]ψ0kM_,w0 _{∈ k[=]ψ0k}M0

andkψ0kM0

∈ N_w00. Let us repeat the following reasoning argument taking as an invariant that at stepr, kψrkM

0

∈ N_w00. We show that in every step,kψrkM

0

(kψr+1kM 0

. In a finite number of steps q ≤ #W0 _{we will obtainψ}

q s.t. w ∈ k[=]ψqkM butw0 6∈

k[=]ψqkM 0

, which contradicts the hypothesis. At stepr: Ifkψr−1kM

0

=T_k0. Then there ish0_k ∈T_k0 s.t. for every hi ∈T there isψik wherehi ∈ kψkikMbuth0k 6∈ kψ

k ikM

0 . Letψr := ψr−1∧(W

n j=1ψ

k

j). It can be easily showed that w∈ k[=]ψrkM. Ifw0 6∈ k[=]ψrkM

0

, absurd. This is some-thing that will happen if we haver = #W0, ask[=]ψkM0 will necessarily be the empty set. If not, iterate taking into account thatkψrkM

0

( kψr−1kM 0

(because h0_k no longer satisfiesψr), and besideskψr−1kM=kψrkM=T.

If kψr−1kM 0

= S_k0. Then, there exists X¯0 = S_k0 ∪ {x¯01, . . . ,x¯0z} ∈/ Nw00 such that S_k0 ∩ {x¯0₁, . . . ,x¯0_z} =∅, dif-ferentiable by its characteristic formula (let it be η), where for every x¯0

i ∈ X¯0 \ Sk0 andh¯j ∈ W \ T there is a ψ¯ki,j where¯hj∈ kψ¯i,jk kMbutx¯0i6∈ kψ¯i,jk kM

0

. Letψr :=ψr−1∨

(η∧Wz i=1

Vm j=1¬ψ¯

k

i,j). As there is no elementx∈W \T whereM, x|=Vm

j=0¬ψ¯ k

i,jfor anyi, then there is no element x ∈ W \T whereM, x |= Wz

i=1 Vm

j=1¬ψ¯ k

i,j. This means thatkψr−1kM = kψrkM = T. But everyx0 ∈ kψr−1kM

0

verifiesM0_{, x}0 _|=ψ

r, andkη∧W z i=1

Vm

j=1¬ψ¯i,jk kM 0

=X0. ThenkψrkM

0

=X0. And as mentioned before,kψrkM 0

/

∈

N_w00. At this point we can stop iterating since we know that w∈ k[=]ψrkMbutw0 6∈ k[=]ψrkM

0

. Which is absurd. TheZagcondition is proved similarly.

The finiteness condition of Prop. 13 cannot be dropped.

Lemma 14. Proposition 13 can fail in infinite models.

Proof. Let us consider the following two models:

w0

h p1p2p3. . . w1 p1

w2 p1p2 w3 p1p2p3

.. .

M:

w₀0

w0₁ p1 w0₂ p1p2 w0₃ p1p2p3

.. .

M0:

I.e., M = hW, N,k · kM_{i, where W = {w}

i | i ≥

0} ∪ {h}, Nw0 = {{wi | i > 0} ∪ {h}}, N

0

a = ∅ for a 6= w₀0, and ∀i > 0 : kpikM = {w1, . . . , wi, h}. And

M0 _{= hW}0_{, N}0_{,k · k}M0_{i, where W}0 _{= {w}0

i | i ≥ 0}, N_w00

0 ={{w

0

i | i > 0}},Na0 =∅fora6=w00, and∀i > 0 :

kpikM 0

= {w0

1, . . . , wi0}. hM, w0i 6↔= hM0, w00ias his not bisimilar toanyelement inNw0

0 (henceZagis not

satis-fied). However,w0andw00satisfy exactly the same formulas ofN=: for anyϕ, letpk be the proposition with biggestkin ϕ, thenhcannot be distinguished fromwkbyϕ.

3

Extensions to

N

In Section 2 we said that the notion of symmetric N =-bisimulations that we introduced in Def. 7, though natural, was too strong forN=. On the other hand, the notion ofN =-bisimulations that we introduced in Def. 11 did seemed to match the expressive power ofN=but it was involved.

We can solve this dilemma by strengthening the expres-sive power ofN=to exactly match the behavior of symmet-ric N=-bisimulations. Paying some consideration to Defi-nition 7, we may notice that the problem boils down to the logic’s (in)ability to distinguish the subset relation between neighbourhoods. We can check then that he following binary operator_Owould do the job:

kϕ_OψkM _{= {w| ∃X}

1, X2∈Nws.t.X16=X2, X1=kϕkM_{, X2=kϕ∨ψk}M_}.

If we call N=(O)the logic that we obtain by extending

N=with this operator, we have a perfect match for symmetric

N=-bisimulations.

Proposition 15. Given two epistemic structures M and

M0 _{then hM, wi ↔}s

= hM0, w0iimplies hM, wi ≡N=(O)

hM0_{, w}0_{i. Moreover, ifMandM}0 _{are finite, then the}

con-verse also holds.

Proof. The proof is similar, but slighly more complex (and lengthier), than the previous results and we refer to the tech-nical report [Areces and Figueira, 2008] for the details.

The trouble now is thatOseems rather artificial, and it is difficult to find a suitable epistemic interpretation for it. We have not sorted out the trouble withN=.

Luckily, we are now very close to a solution: we propose to extendN=with the existential modalityEinstead ofO. The existential modalityEis defined as

kEϕkM₌

W ifkϕkM_6=∅ ∅ otherwise

Let us callN=(E)the languageN= extended with theE operator. The first thing we note is that_Ocan be expressed in

N=(E):kϕOψkM=k[=]ϕ∧[=](ϕ∨ψ)∧E(ψ∧ ¬ϕ)kM. Moreover, we can naturally adjust symmetric N =-bisimulation toN=(E).

Definition 16 (symmetric total N=-bisimulation). Let M

and M0 _{be two epistemic structures. A total symmetric} N=-bisimulationbetween these models is asymmetricN= -bisimulationZ ⊆W ×W0 such that the domain ofZ coin-cides withW and the range ofZcoincides withW0. 3 That symmetric totalN=-bisimulation preserve validity of formulas inN=(E)is easily checked.

Proposition 17. LetM,M0 _{be two epistemic structures,Z}

a total symmetricN=-bisimulation between them,w ∈ W, w0∈W0andwZw0. Then,hM, wi ≡N=(E)hM

0_{, w}0_i.

Proposition 18. LetMandM0 _{be two finite differentiable}

epistemic structures, such that hM, wi ≡N=(E) hM

0_{, w}0_i.

Then there existsZ, a total symmetricN=-bisimulation be-tweenMandM0_{such thatwZw}0_.

Proof. DefinexZx0 iffhM, xi ≡N=(E) hM

0_{, x}0_{i. We must}

prove that this is indeed a bisimulation. By definition,

(w, w0_{) ∈ Z henceZ is non-empty and Propholds. We}

must now check the other conditions of definition 16. Zig. We will proceed with the same strategy as in the proof of Proposition 13. Assume thatZigdoes not hold, and let N_w00 ={T₁0, . . . , T_t0} ∪ {S₁0, . . . , S_l0}where allT_i0 failZig 1 and allS0_ifailZig 2. LetT = {h1, . . . , hn}andW \T =

{¯h1, . . . ,h¯m}.

Set ψ0 := ϕT the characteristic formula of T. As w ∈

k[=]ψ0kM_{, thenw}0 _{∈ k[=]ψ0k}M0

and kψ0kM0

∈ N_w00. Again we will take as an invariant that at stepr,kψrkM

0

∈

N_w00. We show that in every step, kψrkM 0

( kψr+1kM 0

. In a finite number of stepsq ≤ #W0_{we will obtain ψ}

q s.t. w∈ k[=]ψqkMbutw0 6∈ k[=]ψqkM

0

, which contradicts the hypothesis. At stepr:

Ifkψr−1kM 0

=T_k0. Then there existsh0_k ∈T_k0 such that for everyhi ∈Tthere is aψki wherehi∈ kψikkMbuth0k 6∈

kψk_ikM0

. Letψr = ψr−1∧(W n j=1ψ

k

j). We state thatw ∈

k[=]ψrkM. Ifw06∈ k[=]ψrkM 0

, absurd. If not, iterate noting thatkψrkM

0

(kψr−1kM 0

(becauseh0_kdoes not satisfiesψr any longer) and therefore #kψrkM

0

< #kψr−1kM 0

, and besideskψr−1kM=kψrkM=T.

Ifkψr−1kM 0

=S0_k. Then there exists¯h0 ∈W0\S_k0 such that for every¯hi∈W \T there is aψ¯ikwhere¯hi 6∈ kψ¯kikM but¯h0 ∈ kψ¯k_ikM0_{. Letϕ}k ₌Vm

i=0ψ¯ k

i,ψr =ψr−1∨ϕk. If

kψrkM 0

/

∈N_w00, thenw∈ k[=]ψrkMbutw0 6∈ k[=]ψrkM 0

, a contradiction. Otherwise, letη = E(ϕk _{∧ ¬ψ}

r−1), then w0∈ kηkM0

butw6∈ kηkM_{. Again a contradiction.}

Zagis proved similarly.

Finally, we should show that Z is total. Suppose that it is not. Without loss of generality, suppose that there exists v ∈ W that is not bisimilar to any other element of W0. LetW0 _={w0

0, . . . , w0p}. By definition of bisimulation, this means thatv is not equivalent to anyworld in W0. Then, for everyw0_i there isψi s.t.v ∈ kψikMbutw0i 6∈ kψikM

0 . Letψ = Vp

i=0ψi. Thenw ∈ kEψkM butw0 6∈ kEψkM 0

, contradicting thatwandw0are pointwise equivalent.

It is easy to see thatN⊆is not more expressive thanN=(E), asN⊆ cannot express properties of ‘unconnected’ states in

the model. We can use bisimulations to prove that alsoN=(E) is not more expressive thanN⊆(we refer again to [Areces and

Figueira, 2008] for details).

Proposition 19. N=(E)andN⊆are incomparable in terms

of expressive power.

4

Complexity Analysis

In this section we will discuss the complexity of N⊆ and N=(E). It was already shown in [Vardi, 1989] that the com-plexity of satisfiabily forN= isNP-complete. We will first show that we can use that result to proveNP-completeness of

N⊆. We will then prove that even when we extendN=with theEoperator, the complexity remains inNP.

Definition 20(Schema Mand Supplementation). The for-mula schemaMis[ ](ϕ∧ψ) → [ ]ϕ∧[ ]ψ. It corresponds to the following conditions over a class of epistemic structures. If a model satisfiesMfor arbitraryϕandψthen ifX ⊆ Y andX ∈NwthenY ∈Nw.

LetMbe an epistemic structure. Thesupplementationof

M, denotedM+_{, is the structurehW, N}+_{,k · k}M_{iin which}

Na+ is the superset closure ofNa, for everyainW. That is, for everya ∈ W andX ⊆ W,X ∈ N+

a if and only if

Y ⊆Xfor someY ∈Na. 3

In what follows, we will need to compare satisfiability of formulas inN⊆andN=. On the semantic side, we will write |=⊆ for the satisfaction relation ofN⊆, and|==for satisfac-tion ofN=. On the syntactic side we will always write[ ]for the modal operator, and interpreted according to the indicated semantics. The following result is fairly straightforward and can be proved by induction on the complexity of the formula.

Proposition 21. For every formulaϕand epistemic structure

M,kϕkM

⊆ =kϕkM

+

= .

Corollary 22. LetS be the class of supplemented models.

For every formulaϕ∈ N⊆and epistemic structureM ∈ S: M |=⊆ ϕif and only ifM |==ϕ.

Proof. EveryM ∈ SverifiesM+_{=Mby idempotence of} supplementation. It only remains to apply Prop. 21.

Proposition 23. The satisfaction problem ofN⊆is inNP.

Proof. In [Vardi, 1989] we can find theNPalgorithm for the satisfaction problem ofN=restricted to the classS(the class of supplemented models, denoted asε_{3}in the literature).

To check whether ϕis satisfiable in N⊆, we can feed ϕ

as input of the NP Turing machine that solves the satisfac-tion problem forN=over the class of supplemented models. If it answersyes, then there must be a supplemented model

M+_{such thatM}+_{, w|=}

=ϕ. We can then instantiate Propo-sition 21 with M+ _{and, as M}++ _{= M}+_{, conclude that}

M+_{, w|=}

⊆ϕ. Then,ϕis satisfiable inN⊆. If it answersno,

let us supposead absurdumthat∃Msuch thatw ∈ kϕkM ⊆ .

By Proposition 21 this implies thatw∈ kϕkM+

= . But as the algorithm has answeredno, there cannot be a supplemented modelM+ _{that satisfiesϕunderN=. Absurd. Then, ϕis} unsatisfiable underN⊆.

Next we show anNPalgorithm for satisfiability of formu-las inN=(E). But first some necessary definitions.

Definition 24(Valuation and Universal Valuation). A

valua-tionfor ϕ ∈ N=(E)is any functionν : sub(ϕ)∪ {⊥} →

{0,1} such that, and: i) ν(ψ) = 1 iff ν(∼ψ) = 0; ii) if ν(Eψ) = 0 then ν(ψ) = 0; iii) ν(⊥) = 0; and iv) ν(ψ1∨ψ2) = 1iffν(ψ1) = 1orν(ψ2) = 1. Here,sub(ϕ)

stands for the set of subformulas ofϕclosed under∼, where we define∼(ψ) =ξifψ=¬ξ, and∼(ψ) =¬ψotherwise.

Auniversal valuationforϕis any functionµthat fixes the values of all universal subformulas. Lettinguniv(ϕ) ={Eψ|

Eψ ∈ sub(ϕ)}, thenµ : univ(ϕ) → {0,1}. We say that a valuationν forϕiscompatiblewith a universal valuationµ iff they agree in the value of all elements ofuniv(ϕ). 3

We first describe the procedure intuitively and then give a precise algorithm. Given a formulaϕ, the algorithm describes thevaluationspresent in the model that satisfies it. The proce-dure will first fix auniversal valuationµfor the truth value for eachEψsubformula. All other valuations must be compati-ble withµ. We first guess a valuationνsuch thatν(ϕ) = 1. If ν([=]ψ1) = 1 andν([=]ψ2) = 0 for some pair of sub-formulas, this means thatkψ1kis a neighbourhood butkψ1k

is not. Then there must be an element where the truth value ofψ1and ofψ2differs. We require then a valuationν0 such that ν(ψ1) = 1andν(ψ2) = 0(or viceversa). Recursively, the algorithm searches for new valuations satisfying similar demands. We can restrict it to always look for valuations not yet tried, to ensure termination.

In the code below, calls towitness(ψ1, ψ2, R)find a val-uation that distinguishes ψ1 from ψ2. The valuations (that can be seen as elements from the model) are always main-tained in theRvariable. The first call towitness(line 2 of sat) just asks for for the existence of a valuation that makes ϕtrue. We will also need witnesses for all formulasψsuch thatµ(Eψ) = 1as the second part ofsatdescribes.

Algorithm:sat(ϕ)

1: guess a universal valuationµforϕ

2: R←witnessϕ µ(ϕ,⊥,∅)

3: for allEψ∈sub(ϕ)s.t.µ(Eψ) = 1do 4: ifthere is noν∈Rwhereν(ψ) = 1then 5: R←witnessϕµ(ψ,⊥, R)

6: end if 7: end for 8: returnR

Algorithm:witnessϕ

µ(ψ1, ψ2, R)

1: guess a valuationνforϕcompatible withµs.t.:ν(ψ1) = 1and

ν(ψ2) = 0, orν(ψ1) = 0andν(ψ2) = 1

2: ifthere is not such a valuationthen 3: printUNSATand exit

4: end if 5: R←R∪ {ν}

6: for all [=]η1,[=]η2 ∈ sub(ϕ) s.t. ν([=]η1) = 1 and

ν([=]η2) = 0do

7: ifthere is noν0 ∈Rsuch thatν0(η1) = 1andν0(η2) = 0,

orν0(η1) = 0andν0(η2) = 1then

8: R←witnessϕµ(η1, η2, R)

9: end if 10: end for 11: returnR

Complexity. The algorithm makes as many recursive calls

as elements are inRwhen it ends. Each element inν ∈Ris the witness forψ1, ψ2 such thatν(ψ1) = 0andν(ψ2) = 1

or viceversa. We should also note that there are not more than two witnesses for a certain pair ψ1, ψ2, thanks to the condition of line 4 insatand 7 in witness. We will then have at mostO(|ϕ|2_{)elements inR, so the number of calls} to thewitness algorithm is polynomial. Checking that an assignment is a valuation, and that it is compatible withµis polynomial. Then the body ofwitnesstakes onlyNPtime.

Proof. We build a model based on the elements fromR re-turned by the algorithm. LetMbe the model whereW =R, andkpkM _{={ν ∈ R | p ∈ sub(ϕ)∧ν(p) = 1}. We}

de-fine Nν as the smallest set such that, ifν([=]ψ) = 1, then

{ν0 | ν0(ψ) = 1} ∈ Nν. Our claim is thatτ ∈ kϕkM, for someτ ∈ W such thatτ(ϕ) = 1. Note that there should exist suchτ because of the first call towitnesswithϕas a parameter (line 2 ofsat). Furthermore, we will prove that for everyψ∈sub(ϕ)andν ∈W:ν ∈ kψkM_{iffν(ψ) = 1. We}

proceed by induction on the length of the formula. It is easy to check the base case of propositional variables.

The modal case. Suppose now that ν ∈ k[=]ψkM_{, with}

[=]ψ∈ sub(ϕ). ThenkψkM ∈Nν. By IH,kψkM ={ν0 | ν0(ψ) = 1}, so{ν0|ν0(ψ) = 1} ∈Nν.

By construction of Nν, any element of it is a set of the form{ν0 |ν0(η) = 1}, caused by the existence of a formula

[=]η ∈ sub(ϕ)such that ν([=]η) = 1. We then have that

kψkM _{= {ν}0 _{| ν}0_{(ψ) = 1} = {ν}0 _{| ν}0_{(η) = 1} ∈ N}

ν. Suppose ad absurdum thatν([=]ψ) = 0. Asν([=]η) = 1and ν([=]ψ) = 0, when the algorithm was on the step whereν was added it must have either 1. added a new elementν0such thatν0(η) = 1andν0(ψ) = 0(or viceversa), or 2. checked the existence of such an element inR. Either way there is an elementν0 ∈ Rsuch that it is in one but not in the other of the two sets: {ν0 | ν0(η) = 1},{ν0 | ν0(ψ) = 1}. But then they cannot be equal. Absurd. Hence,ν([=]ψ) = 1.

The converse is simpler. Let us supposeν([=]ψ) = 1, then{ν0 | ν0(ψ) = 1} ∈Nν by construction. Applying IH,

kψkM_∈N

νand thusν ∈ k[=]ψkM.

The universal modality.Ifν ∈ kEψkMwithEψ∈sub(ϕ),

then there is an elementν0 such thatν0 ∈ kψkM_{. Applying}

IH this happens iffν0(ψ) = 1. By definition, ν andν0 are compatible with theµreturned by the algorithm.

Suppose ad absurdum that ν(Eψ) = 0. Asν andµare compatible, andµis fixed for all the calls to thewitness al-gorithm, then we have that0 =ν(Eψ) =µ(Eψ) =ν0(Eψ). By definition ofvaluationwe have that ifν0_{(Eψ) = 0then}

ν0(ψ) = 0, which is absurd. Henceν(Eψ) = 1.

For the converse implication suppose ν(Eψ) = 1. Then µ(Eψ) = 1and the procedure will have added an elementν0 such thatν0(ψ) = 1(lines 3, 4 ofsat). By IH,ν0 ∈ kψkM_,

and henceν∈ kEψkM_=W.

Proposition 26. Ifϕis satisfiable thensat(ϕ)succeeds.

Proof. LetMbe such thatw ∈ kϕkM_{. ForEψ ∈ sub(ϕ),}

defineµ(Eψ) = 1iffw∈ kEψkM, and for eachv∈W and ψ∈sub(ϕ), letνvbe such thatνv(ψ) = 1iffv ∈ kψkM. It is easy to see thatνvis a valuation, and that it is compatible withµ. LetS ={νv | v ∈W}. The idea is that each time the algorithm needs to guess a valuation, it picks one fromS. Suppose however that at a certain point,witness(ψ1, ψ2) cannot find a suitable valuation among S. If the witness function was called from a valuationνv ∈ S, then –by def-inition of the algorithm– the call must have been made be-cause νv([=]ψ1) = 1 and νv([=]ψ2) = 0 (or viceversa). And then –by construction of νv– it must be the case that v ∈ k[=]ψ1kM andv 6∈ k[=]ψ2kM (or viceversa). This means thatkψ1kM _∈N

v andkψ2kM 6∈Nv (or viceversa),

in other words kψ1kM _{6= kψ2k}M_{. Let v}0 _{∈ (kψ1k}M _∪ kψ2kM_)\(kψ1kM_{∩ kψ2k}M_{). Thenν}

v0 ∈ S should have worked as a guessing for the algorithm. Absurd.

Corollary 27. The satisfiability problem forN=(E)isNP

-complete.

5

Conclusions

Neighbourhood semantics is widely used in epistemic logic, but sometimes it is not clearexactly which semanticsis in-volved. As we discussed in this paper, two alternative pro-posals can be found in the literature.

We showed thatN=, the semantics originally introduced in [Vardi, 1986] does not seem to have a nice characteriza-tion in terms of bisimulacharacteriza-tion, and we proposedN=(E)as a natural extension. TheEoperator can be used to model infor-mation which is globally true in an epistemic structure, hence the extension seems to be well motivated for the application. Moreover, a simple notion of bisimulation exists forN=(E) and its satisfiability problem remainsNP-complete. We also showed that satisfiability forN⊆isNP-complete mostly as a

corollary of results from [Vardi, 1986].

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