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Sugeno integral in a finite Boolean algebra
Fuzzy Sets and Systems 159 (2008) 1709 – 1718 www.elsevier.com/locate/fss
Sugeno integral in a finite Boolean algebra Agnès Rico Equipe MA2 D du LIRIS, Université Claude Bernard Lyon I, 43 bld du 11 novembre 1918, 69622 Villeurbanne, France Received 11 March 2007; received in revised form 31 October 2007; accepted 1 November 2007 Available online 17 November 2007
Abstract The aim of this paper is to provide a representation theorem of the Sugeno integral when the evaluation scale is a finite Boolean algebra. This result is a generalisation of a result proved when the evaluation scale is totally ordered. A major difficulty is the renunciation of the comonotonic functions which have to be replaced by the co-included functions. So we need to care about the properties satisfied by the Sugeno integral when the evaluation scale is a finite totally ordered set. To begin we show that, when the scale is totally ordered, the co-included functions are solely needed to characterise the Sugeno integral and that they are less constraining than the comonotonic functions classically used. Next we focus on finite Boolean algebra, and in this new context we define the Sugeno integral and we present the properties still satisfied. To end this article, we present a representation theorem of the Sugeno integral on a finite Boolean algebra. © 2007 Elsevier B.V. All rights reserved. Keywords: Sugeno integral; Boolean algebra; Capacity; Multicriteria decision
1. Introduction In decision making an essential purpose is to rank objects evaluated according to some criteria. One major difficulty is due to criteria which can have different natures. So, for each criterion we can have different scales. In order to aggregate partial scores, we make the scale commensurable. Approximately, if we denote C = {c1 , . . . , cN } the set of criteria and {E1 , . . . , EN } the different scales where Ei is the scale according to the criterion ci , each element of Ei can be compared to each element of Ej for all i, j ∈ {1, . . . , N}. In the literature, to characterise the fuzzy integrals, the Choquet integral or the Sugeno integral, all the criteria are assumed to share the same scale E. Moreover E is supposed to be a totally ordered set [11,4]. To be more precise, one major aim is to rank, in an evaluation scale E, mappings X from some non-empty set C to an evaluation scale E. In multicriteria theory, X is usually called an object, C is the set of criteria and X(c) denotes the partial score of X with respect to criterion c ∈ C. We denote V the set of all functions X : C → E. In practice, sometimes, we need to evaluate objects with a scale that is not totally ordered and many authors are interested about the use of the fuzzy integrals on a lattice [7,1]. To illustrate our motivation, take as an example, the industrial context in which objects can be evaluated by measures taken by detectors. When the given measures and the differences between them are smaller than the detector’s sensibility we cannot compare the result [3]. In such a context, we have a partial order and we need to rank in an evaluation scale that is a lattice. Next, to evaluate the involved objects, all the measures have to be aggregated to give a score to each object. Technically speaking, unlike the Choquet E-mail address: [email protected]. 0165-0114/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2007.11.008
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integral which needs arithmetic operators, the definition of the Sugeno integral only uses the minimum operator and the maximum operator. Consequently, the Sugeno integral seems to be the most suitable when the evaluation scale E is a lattice. A question which arises is: what well-known properties in the totally ordered context can be adapted in the lattice context? After making liberal allowance for the properties and the representation theorem about the Sugeno integral proved when the evaluation scale is totally ordered [8,6], to begin the generalisation we propose to work on a Boolean algebra. 1 This new situation implies renouncing of the notion of comonotonic functions. This property is replaced by the co-included functions defined in this article. As the aim of this article is to generalise a representation theorem about the Sugeno integral we need to care about this new concept. Rather, we precise that, when the scale is totally ordered, comonotonic functions and co-included functions are not the same and co-included functions are solely needed to characterise the Sugeno integral. To bring out the generalisation we keep notations used when the scales are totally ordered: we consider a finite Boolean algebra E, a finite set of criteria C and we denote V the set of all functions from C to the evaluation scale E and a functional from V to E. The aim of this paper is to present some properties to be verified by in order for to be a Sugeno integral. It generalises some well-known results when the evaluation scale is a totally ordered set [4,11]. The paper is organised as follows: Section 2 deals with the classical Sugeno integral. After the notations, we recall some results proved when E is totally ordered. Next, we give a representation theorem of the Sugeno integral with co-included functions when E is totally ordered. Section 3 presents the finite Boolean algebras and their properties used in this paper. In Section 4, after notations and preliminary result, we present the properties that are still satisfied when the Sugeno integral is defined on a Boolean algebra. To end this section we prove our main result: given a finite Boolean algebra E, a finite set of criteria C, V the set of all functions from C to the evaluation scale E and a functional from V to E, we present some properties to be verified by in order for to be a Sugeno integral. Section 5 is devoted to some concluding remarks. 2. The Sugeno integral on a finite totally ordered set 2.1. Notations To begin with, we introduce the notations used in this section: • C = {c1 , . . . , cN } is a finite set. • P(C) is the set of subsets of C. • (L, ∨, ∧) is a finite totally ordered set with a maximal element 1 and a minimal element 0. The minimum operator is denoted by ∧ and the maximum operator by ∨. • V = {X: C → L} is the set of all functions from C to L. • A total order, , is classically defined on L by ∀, ∈ L, if and only if one of the equivalent properties holds: ∧ = or ∨ = . • For every subset A of P(C), we denote 1A the characteristic function of A: 1A : C →L 1 if c ∈ A, c → 0 else. 2.2. The Sugeno integral characterised with comonotonic functions The aim of this section is to recall some well-known definitions and properties about the Sugeno integral when the evaluation scale L is a totally ordered set. Definition 1. • A capacity v is a set function v: P(C) → L which satisfies v(∅) = 0, v(C) = 1 and ∀A, B ⊆ C, A ⊆ B ⇒ v(A)v(B). • A possibility is a capacity such that ∀A, B ⊆ C, v(A ∪ B) = v(A) ∨ v(B). • A necessity is a capacity such that ∀A, B ⊆ C, v(A ∩ B) = v(A) ∧ v(B).
1 Which is a special lattice.
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Definition 2. Let X: C → L be a function and v be a capacity on C. The Sugeno integral of X with respect to v is: Sv (X) = ∧ v(X ), ∈L
where v(X ) = v({c ∈ C|X(c) }). Definition 3. Two functions X, Y ∈ V are comonotonic if and only if ∀ci , cj ∈ C,
X(ci ) < X(cj ) ⇒ Y (ci ) Y (cj ).
When the evaluation scale L is a totally ordered set, the Sugeno integral is defined and characterised, see for example [11,5,10]. The following interesting result can be found in [4]. Theorem 4. Let V be the set of functions from C to L and be a functional from V to L. satisfies the following properties: (1) ∀ ∈ L, ( ∧ 1C ) = , (2) ∀ ∈ L, ∀X ∈ V , ( ∧ X) = ∧ (X), (3) X, Y comonotonic implies (X ∨ Y ) = (X) ∨ (Y ), if and only if is a Sugeno integral. 2.3. The Sugeno integral characterised with the co-included functions One major difficulty to characterise the Sugeno integral on a Boolean algebra is the comonotonic property. To put it more thorough, comonotonic functions have to be replaced by co-included functions. In order to be precise, in this section, we present a representation theorem of the Sugeno integral with the co-included functions when the scale is totally ordered. First we introduce the co-included functions and we compare them with the comonotonic functions. Definition 5. Two functions X, Y ∈ V are co-included if and only if ∀ ∈ L we have {X } ⊆ {Y } or {Y } ⊆ {X }. Note that if two functions X and Y are such that X Y , or Y X, then obviously they are co-included. Proposition 6. If X, Y ∈ V are two comonotonic functions, then X and Y are co-included functions. But if X, Y ∈ V are two co-included functions then X and Y are not necessary comonotonic functions. Proof. • First let us prove that if X, Y ∈ V are two comonotonic functions, then X and Y are co-included functions. Let X, Y ∈ V be two comonotonic functions and suppose that ∃l ∈ L such that {X l}{Y l} and {Y l}{X l}. In such a case we have ◦ ∃c1 ∈ C such that Y (c1 ) l and X(c1 ) < l, ◦ ∃c2 ∈ C such that X(c2 ) l and Y (c2 ) < l. 2 So X and Y are not comonotonic functions because Y (c1 ) > Y (c2 ) and X(c1 ) < X(c2 ). • The following example shows that the converse is not true. Let L = {0, e, 1} be a finite totally ordered set with 0 < e < 1 and C = {c1 , c2 } be a set of criteria. Let us define two functions X and Y : C → L as follows: X Y
c1 0 1
c2 e e
2 These inequalities are satisfied because in this section L is a finite totally ordered set.
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◦ We have X(c1 ) < X(c2 ) and Y (c1 ) > Y (c2 ) so X and Y are not comonotonic functions. ◦ But according to the following inclusions, X and Y are co-included functions {X 0} = {Y 0} = C, {X e} = {c2 } ⊆ {Y e} = C, {X 1} = ∅ ⊆ {Y 1} = C.
To end this section and in order to bring out our generalisation proved at the end of the article, we are going to prove that co-included functions are necessary and sufficient to characterise the Sugeno integral when L is totally ordered. Theorem 7. Let L be a finite totally ordered set, V be the set of all functions from a finite set C to L and be a functional from V to L. satisfies the following properties: • (1) ∀ ∈ L, ( ∧ 1C ) = , • (2) ∀ ∈ L, X ∈ V , ( ∧ X) = ∧ (X), • (3 ) X, Y ∈ V = co-included ⇒ (X ∨ Y ) = (X) ∨ (Y ) if and only if is a Sugeno integral. Proof. • First we prove that Sugeno integral satisfies the three properties (1), (2) and (3 ). First and second properties are induced from Theorem 4 so we care about the third one. We consider X, Y two coincluded functions. The Sugeno integral is increasing so Sv (X) ∨ Sv (Y ) Sv (X ∨ Y ) and we just have to show that Sv (X ∨ Y )Sv (X) ∨ Sv (Y ). Let us choose in L. As X and Y are co-included functions either {X } ⊆ {Y } or {Y } ⊆ {X } holds. ◦ If {X } ⊆ {Y }, then {X ∨ Y } ⊆ {Y } because {X ∨ Y } ⊆ {X } ∪ {Y }. Consequently, ∧ v(X ∨ Y ) ∧ v(Y )Sv (Y )Sv (Y ) ∨ Sv (X). ◦ If {Y } ⊆ {X }, then similarly we obtain ∧ v(X ∨ Y ) Sv (Y ) ∨ Sv (X). So for all ∈ L, ∧ v(X ∨ Y )Sv (Y ) ∨ Sv (X) which entails Sv (X ∨ Y )Sv (Y ) ∨ Sv (X). • Next, we prove that if satisfies the three properties then is a Sugeno integral. Let X be a function from C to L and {xi }i∈{1,...,N} be its values. Without loss of generality we can suppose x1 · · · xN . Hence we have X=
N
xi ∧ 1Ai
where Ai = {X xi }.
(1)
i=1
We denote Xi = xi ∧ 1Ai . For all i ∈ {1, . . . , N − 1} functions Xi and N j =i+1 Xj are comonotonic so they are co-included. Finite re-iteration on N i and property (3 ) imply (X) = i=1 (xi ∧ 1Ai ). Then property (2) allows us to write (X) = N i=1 xi ∧ (1Ai ). To conclude we have to prove that ∀A ⊆ C, v(A) = (1A ) defines a capacity on P(C): ◦ v(∅) = (0 ∧ 1C ) = 0, ◦ v(C) = (1 ∧ 1C ) = 1, ◦ if A ⊆ B, then v(A) = (1A ) (1B ) = v(B) because 1A and 1B are co-included functions. The aim of this article is to present the same representation theorem when the evaluation scale is a finite Boolean algebra: this will be achieved in Theorem 26. The proof of Theorem 26 will be similar to the proof of Theorem 7 with necessary adaptations. First let us recall some lattice properties given in [2,9].
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3. Lattice definitions and properties A lattice is a partially ordered set (or poset) in which all nonempty finite subsets have both a supremum ∨ and an infimum ∧. This article cares about bounded lattices with a greatest element denoted by 1 and a smallest element denoted by 0. Definition 8. A lattice (E, ∨, ∧) is • distributive if ∀a, b, c ∈ E,
a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)
and
a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c),
• complemented if ∀a ∈ E,
∃a ∈ E such that a ∧ a = 0
and a ∨ a = 1.
A distributive and complemented lattice is called a Boolean lattice. In such a lattice the complement of each element is unique, and we can define an order by a b ⇔ a ∧ b = a, ∀a, b ∈ E. We write (E, , ∧, ∨, 0, 1, ) and we speak about Boolean algebra. The well-known example is the lattice P() of all subsets of a set , with union ∪, intersection ∩ and complementarity. Proposition 9. In a distributive lattice ∀a, b, ca ∧ b = a ∧ c
and a ∨ b = a ∨ c imply b = c.
Proof. See [2] or [9]. Definition 10. An element p of a lattice is a ∨-irreducible (or irreducible) element if p is not the least element 0 and p = x ∨ y implies p = x or p = y. Note that in a totally ordered set each element other than 0 is ∨-irreducible. Definition 11. Let N be a subset of E and a be an element of E. A decomposition of a denoted a = p∈N p, is irredundant if for every p0 ∈ N we have p∈N\p0 p < a. If p∈N\p0 p = a then p0 is superfluous. Proposition 12. In a finite distributive lattice, every element has an irredundant decomposition with ∨-irreducible elements and this decomposition is unique. Proof. See [9]. Lemma 13. In a distributive lattice, if p is ∨ -irreducible and p a ∨ b, Proof. See [9].
then p a or p b.
Definition 14. Let a be an element in E, a is an atom if and only if a = 0 and e a ⇒ e = 0 or e = a. In other words, an atom is a minimal element of E\{0}. Note that if a, b are atoms such that a = b, then a ∧ b = 0. To end this subsection, we recall this essential proposition. Proposition 15. All finite Boolean algebras are isomorphic to P(X), the set of subsets of X where X is a finite set. Proof. See [9].
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4. The Sugeno integral on a finite Boolean algebra This section states and proves our main result: a Sugeno integral representation theorem with the co-included functions when the evaluation scale E is a Boolean algebra. To do this first we present definitions and a first result: atoms are sufficient to define the Sugeno integral on a Boolean algebra. Next we present some properties that are still satisfied by the Sugeno integral in this new context. To end this section, after technical results our main theorem is proved. 4.1. Notation, definitions and first results In this section we suppose (E, ∨, ∧) to be a finite Boolean algebra. So from Proposition 15, E is isomorphic to one P(), the Boolean algebra of all subsets of a finite set . We denote E∨ the set of ∨-irreducible elements in E which are named {a1 , . . . , ap } = E∨ . In such a case E∨ is also the set of atoms. E is bounded: 1 is the greatest element and 0 is the smallest one. C = {c1 , . . . , cN } is always a finite set, V = {X: C → E} is the set of functions from C to E and for every subset A of P(C), we denote 1A the characteristic function of A defined as follows: 1A : C →E 1 if c ∈ A, c → 0 else. In such a case, capacities, possibilities, necessities and the Sugeno integral can be defined as follows: Definition 16. • A capacity v is a set function v: P(C) → E which satisfies ◦ v(∅) = 0, ◦ v(C) = 1, ◦ ∀A, B ⊆ C, A ⊆ B ⇒ v(A) v(B). • A possibility is a capacity such that ∀A, B ⊆ C, v(A ∪ B) = v(A) ∨ v(B). • A necessity is a capacity such that ∀A, B ⊆ C, v(A ∩ B) = v(A) ∧ v(B). Definition 17. Let X: C → E be a function, the Sugeno integral of X with respect to a capacity v is Sv (X) = ∧ v(X ). ∈E
Proposition 18. Let E be a Boolean algebra, E∨ be its atoms set (or irreducible elements of E), C be a finite set, X be a function from C to E and v be a capacity on P(C). The Sugeno integral of X with respect to v is Sv (X) = ak ∧ v(X ak ). ak ∈E∨
∨ Proof. For the proof we denote Sv (X) = ak ∈E∨ ak ∧ v(X ak ). ∨ ∨ We have Sv (X) = Sv (X) ∨ ( ∈E / ∨ ∧ v(X )). So we have only to prove that ∈E / ∨ ∧ v(X )) Sv (X). Let ∈ E\E∨ be not ∨-irreducible in E, has got an irredundant decompositionwith ∨-irreducible elements: = ni=1 ai for some ai ∈ E∨ . We have ∧ v(X ) = (a1 ∨ · · · ∨ an ) ∧ v(X ) = k∈{1,...,n} ak ∧ v(X ). ∀k ∈ {1, . . . , n}, the inclusion {X } ⊆ {X ak } implies v({X }) v({X ak }). So ∀k ∈ {1, . . . , n}, ak ∧ v({X }) is smaller than ak and v({X ak }), and this implies ak ∧ v({X }) ak ∧ v({X ak }). Concluding, we have: ak ∧ v(X ak ) Sv∨ (X). ∀k ∈ {1, . . . , n}, ak ∧ v({X }) k∈{1,...,n}
Sv∨ (X) is greater than all ak ∧v({X }) so Sv∨ (X) is greater than the lower upper bound and then ∧v(X ) Sv∨ (X) for all ∈ E\E∨ . ∨ ∨ In conclusion, ∈E / ∨ ∧ v(X ))Sv (X) and Sv (X) = Sv (X) .
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To end this subsection we define, in the finite Boolean algebra context, the concept of co-included functions. Definition 19. Let X and Y be two functions in V = {X: C → E}. X and Y are co-included functions if and only if ∀ ∈ E,
{X } ⊆ {Y } or
{Y } ⊆ {X }.
4.2. Properties of the Sugeno integral In this section we present some Sugeno integral properties which remain satisfied when E is a Boolean algebra. Proposition 20. ∀A ⊆ C, Sv (1A ) = v(A).
Proof. According to definition 17, Sv (1A ) = ∈E ∧ v(1A ). If = 0 then {1A } = C, and if = 0, then {1A } = A. So we have Sv (1A ) = ∈E\{0} ∧ v(A) = v(A) ∧ ∈E\{0} = v(A) ∧ 1 = v(A). Proposition 21. On a finite Boolean algebra, the Sugeno integral with respect to a capacity is increasing. Proof. Let X, Y be two functions in V such that X Y . X Y ⇒ ∀ ∈ E, {X } ⊆ {Y } ⇒ ∀ ∈ E, v({X })v({Y }) ⇒ ∀ ∈ E, ∧ v({X }) ∧ v({Y }) ⇒ ∀ ∈ E, ∧ v({X }) Sv (Y ) ⇒ Sv (X)Sv (Y ). Property 22. ∀ ∈ E, Sv ( ∧ 1C ) = . Proof. ∀ ∈ E,
Sv ( ∧ 1C ) =
ak ∧ v( ∧ 1C ak )
ak ∈E∨
=
ak
ak ∈E∨ |ak
= .
Property 23. (1) ∀X ∈ V , ∀ ∈ E, Sv ( ∧ X) = ∧ Sv (X). (2) ∀X ∈ V , ∀ ∈ E, Sv ( ∨ X) = ∨ Sv (X). Proof. • Let us prove the first property. The Sugeno integral is an increasing function so ∀X ∈ V , ∀ ∈ E, Sv ( ∧ X) ∧ Sv (X). We just need to prove that ∀X ∈ V , ∀ ∈ E, ∧ Sv (X)Sv ( ∧ X). According to Proposition 18, we have ⎛ ⎞ ak ∧ v(X ak )⎠ ∧ Sv (X) = ∧ ⎝ =
ak ∈E∨
( ∧ ak ∧ v(X ak )).
ak ∈E∨
If ak then ∧ ak = 0 so ∧ Sv (X) = ak ∈E∨ |ak
(ak ∧ v(X ak )).
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The inequality ak implies {X ak } = { ∧ X ak }which entails ∧ Sv (X) = ak ∈E∨ |ak (ak ∧ v( ∧ X ak )) Sv ( ∧ x). • Let us prove the second property. The Sugeno integral is an increasing function so ∀X ∈ V , ∀ ∈ E, ∨ Sv (X) Sv ( ∨ X). We just need to prove that ∀X ∈ V , ∀ ∈ E, ∨ Sv (X) Sv ( ∨ X). According to Proposition 18, we have ak ∧ v( ∨ X ak )). Sv ( ∨ X) = ak ∈E∨
◦ If ak , then v( ∨ X ak ) = v(C) = 1. ◦ If ak , then { ∨ X ak } = {X ak }: {X ak } ⊆ { ∨ X ak } is always satisfied and, if c satisfies ∨ X(c)ak then, according to Lemma 13, X(c)ak because ak and ak is an atom. So we have ak ∧ v( ∨ X ak )) ak ∧ v( ∨ X ak )) Sv ( ∨ X) = ak ∈E∨ |ak
= =
ak ∈E∨ |ak
ak ∧ v( ∨ X ak ))
ak ∈E∨ |ak
ak ∧ v(X ak ))
ak ∈E∨ |ak
∨ Sv (X). So we have Sv ( ∨ X) ∨ Sv (X).
Property 24. Let v be a capacity and X, Y ∈ V be co-included functions, then (1) Sv (X ∨ Y ) = Sv (X) ∨ Sv (Y ). (2) Sv (X ∧ Y ) = Sv (X) ∧ Sv (Y ). Proof. Let X, Y ∈ V be two co-included functions. (1) According to Proposition 21 the Sugeno integral is an increasing function, so Sv (X) ∨ Sv (Y ) Sv (X ∨ Y ). Let us prove the converse inequality. Let be a ∨-irreducible element of E. • If {X } ⊆ {Y }, then as is a ∨-irreducible element, {X ∨Y } ⊆ {Y }∪{X } = {Y }. Consequently, ∧v(X ∨Y ) ∧v(Y ) Sv (Y ) Sv (X)∨ Sv (Y ). • If {Y } ⊆ {X }, similarly we obtain ∧ v(X ∨ Y ) Sv (X) ∨ Sv (Y ). So for all ∈ E∨ , ∧ v(X ∨ Y )Sv (X) ∨ Sv (Y ) which entails Sv (X ∨ Y ) Sv (X) ∨ Sv (Y ). (2) ⎡ ⎤ ⎡ ⎤ ak ∧ v(X ak ))⎦ ∧ ⎣ ak ∧ v(Y ak ))⎦ Sv (X) ∧ Sv (Y ) = ⎣ ak ∈E∨
=
ak ∈E∨
ak ∧ (v(X ak ) ∧ v(Y ak ))
(2)
ak ∈E∨
because in a Boolean algebra, if ak , aj ∈ E∨ , ak = aj then ak ∧ aj = 0. As X, Y ∈ V are co-included functions, for each ak ∈ E∨ we have either {X ak } ⊆ {Y ak } or the converse. • If ak is such that {X ak } ⊆ {Y ak } then if c ∈ C satisfies X(c) ak then Y (c)ak and X(c) ∧ Y (c)ak so {X ak } ⊆ {X ∧ Y ak }. We always have {X ∧ Y ak } ⊆ {X ak } so {X ak } = {X ∧ Y ak }. So in this case v(X ak ) = v(X ∧ Y ak ) and v(X ak ) ∧ v(Y ak ) = v(X ∧ Y ak ).
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• If ak is such that {Y ak } ⊆ {X ak } then similarly we prove that {Y ak } = {X ∧ Y ak }. And in this case we have v(Y ak ) = v(X ∧ Y ak ) and v(X ak ) ∧ v(Y ak ) = v(X ∧ Y ak ). So as a conclusion, Sv (X) ∧ Sv (Y ) = ak ∈E∨ ak ∧ v(X ∧ Y ak ) = Sv (X ∧ Y ). Property 25. (1) If v is a possibility, then ∀X, Y ∈ V , Sv(X ∨ Y ) = Sv (X) ∨ Sv (Y ). (2) If v is a necessity, then ∀X, Y ∈ V , Sv(X ∧ Y ) = Sv (X) ∧ Sv (Y ). Proof. (1) Let v be a possibility and X, Y ∈ V be two functions. Sv (X ∨ Y ) =
ak ∧ v(X ∨ Y ak ).
ak ∈E∨
ak is an atom so according to Lemma 13 {X ∨ Y ak } = {X ak } ∪ {Y ak } and as v is a possibility, we have v(X ∨ Y ak ) = v(X ak ) ∨ v(Y ak ). Hence, Sv (X ∨ Y ) = ak ∧ [v(X ak ) ∨ v(Y ak )]. ak ∈E∨
E is a finite distributive lattice, so we have Sv (X ∨ Y ) = Sv (X) ∨ Sv (Y ). (2) Let v be a necessity and X, Y ∈ V be two functions. ak ∧ v(X ∧ Y ak ). Sv (X ∧ Y ) = ak ∈E∨
{X ∧ Y ak } = {X ak } ∩ {Y ak } and as v is a necessity, we have v(X ∧ Y ak ) = v(X ak ) ∧ v(Y ak ). Hence, ak ∧ [v(X ak ) ∧ v(Y ak )], Sv (X ∧ Y ) = ak ∈E∨
which entails Sv (X ∧ Y ) = Sv (X) ∧ Sv (Y ) because we can use the equality given by Eq. (2) written in the second part in the proof of the Property 24. 4.3. A characterisation of the Sugeno integral on a finite Boolean algebra We cannot use the same proof as the one made when the scale is a totally ordered set because the functions in this new context do not have the same decomposition. But the idea of the proof is the same; to be more precise a function belonging to V as a decomposition with co-included functions and we use the properties belonging to the representation theorem. Theorem 26. Let C be a finite set, E be a finite Boolean algebra and E∨ = {a1 , . . . , ap } be its atoms set. We denote V the set of all functions from C to E and a functional from V to E. satisfies the following properties: (1) ∀ ∈ E, ( ∧ 1C ) = , (2) ∀ ∈ E, ∀X ∈ V , ( ∧ X) = ∧ (X), (3) X, Y co-included ⇒ (X ∨ Y ) = (X) ∨ (Y ) if and only if is a Sugeno integral. Proof. In the previous section we have already proved that the Sugeno integral satisfies the properties (1)–(3). Let us prove that if a functional satisfies these three properties then it is a Sugeno integral. p Let X: C → E be a function, it can easily be proved that X = i=1 ai ∧ 1Ai where Ai = {c ∈ C/X(c)ai }. p Hence: (X) = ( i=1 ai ∧ 1Ai ). p ∀i0 ∈ {1, . . . , p}, functions ai0 ∧ 1Ai0 and i=i0 +1 ai ∧ 1Ai are co-included.
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p For any ∈ E, we have to compare {ai0 ∧ 1Ai0 } and { i=i0 +1 ai ∧ 1Ai }. Remember that, when E is a finite Boolean algebra, the atoms are the ∨-irreducible elements. p • If = 0, then {ai0 ∧ 1Ai0 } = { i=i0 +1 ai ∧ 1Ai } = C. p • If = ai0 , then {ai0 ∧ 1Ai0 } = Ai0 and { i=i0 +1 ai ∧ 1Ai } = ∅. • If = 0, ai0 , then {ai0 ∧ 1Ai0 } = ∅. p So ∀ ∈ E we have {ai0 ∧ 1Ai0 } ⊆ { i=i0 +1 ai ∧ 1Ai } or the opposite. By iteration of property (3), we have p p (X) = i=1 (ai ∧ 1Ai ). Then the second property permits us to write (X) = i=1 ai ∧ (1Ai ). Similarly as in the proof made when E is a totally ordered set, v(.) defined on P(C) by ∀A ⊆ C, v(A) = (1A ) is a capacity. So is a Sugeno integral with respect to v. 5. Concluding remarks When the evaluation scale is a totally ordered set, we can permute operators ∨ and ∧ in the Sugeno integral expression. Then we obtain a second representation theorem for the Sugeno integral swapping ∧ and ∨. This result seems more difficult to obtain as we expected and it will be interested to care about it. To conclude it is a challenging question to know what remains true in our work when E is a distributive lattice and not necessary a finite Boolean algebra. Acknowledgements We thank the anonymous referees for very valuable comments. References [1] A. Ban, I. Fechete, Componentwise decomposition of some lattice-valued fuzzy integrals, Inform. Sci. 177 (2007) 1430–1440. [2] G. Birkhoff, Lattice Theory, third ed., American Mathematical Society, Providence, RI, 1967. [3] S. Bonnevay, M. Lamure, I. Lavallé, N. Nicoloyannis, Aide à la prise de décision et systèmes multi-agents, in: Intelligence artificielle et complexité, Paris, September 1998. [4] L. de Campos, M. Lamata, S. Moral, A unified approach to define fuzzy integrals, Fuzzy Sets and Systems 39 (1991) 75–90. [5] D. Dubois, J.-L. Marichal, H. Prade, M. Roubens, R. Sabbadin, The use of the discrete Sugeno integral in decision making: a survey, Internat. J. Uncertainty Fuzziness Knowledge-Based Systems 5 (2001) 539–561. [6] D. Dubois, H. Prade, R. Sabbadin, Qualitative decision theory with Sugeno integrals, in: M. Grabisch, T. Murofushi, M. Sugeno (Eds.), Fuzzy Measures and Integrals—Theory and Applications, Physica Verlag, Heidelberg, 2000, pp. 314–332. [7] X. Liu, G. Zhang, Lattice-valued fuzzy measure and lattice valued fuzzy integral, Fuzzy Sets and Systems 62 (1994) 319–332. [8] J.L. Marichal, On Sugeno integral as an aggregation function, Fuzzy Sets and Systems 114 (2000) 347–365. [9] D. Ponasse, J-C. Carrega, Algèbre et topologie booléennes, Masson, Paris, 1979. [10] A. Rico, Modélisation des préférences pour l’aide à la décision par l’intégrale de Sugeno, Ph.D. Thesis, Université Paris 1 Panthéon Sorbonne, 2002. [11] M. Sugeno, Theory of fuzzy integrals and its applications, Ph.D. Thesis, Tokyo Institute of Technology, 1974.
Sugeno integral in a finite Boolean algebra Agnès Rico Equipe MA2 D du LIRIS, Université Claude Bernard Lyon I, 43 bld du 11 novembre 1918, 69622 Villeurbanne, France Received 11 March 2007; received in revised form 31 October 2007; accepted 1 November 2007 Available online 17 November 2007
Abstract The aim of this paper is to provide a representation theorem of the Sugeno integral when the evaluation scale is a finite Boolean algebra. This result is a generalisation of a result proved when the evaluation scale is totally ordered. A major difficulty is the renunciation of the comonotonic functions which have to be replaced by the co-included functions. So we need to care about the properties satisfied by the Sugeno integral when the evaluation scale is a finite totally ordered set. To begin we show that, when the scale is totally ordered, the co-included functions are solely needed to characterise the Sugeno integral and that they are less constraining than the comonotonic functions classically used. Next we focus on finite Boolean algebra, and in this new context we define the Sugeno integral and we present the properties still satisfied. To end this article, we present a representation theorem of the Sugeno integral on a finite Boolean algebra. © 2007 Elsevier B.V. All rights reserved. Keywords: Sugeno integral; Boolean algebra; Capacity; Multicriteria decision
1. Introduction In decision making an essential purpose is to rank objects evaluated according to some criteria. One major difficulty is due to criteria which can have different natures. So, for each criterion we can have different scales. In order to aggregate partial scores, we make the scale commensurable. Approximately, if we denote C = {c1 , . . . , cN } the set of criteria and {E1 , . . . , EN } the different scales where Ei is the scale according to the criterion ci , each element of Ei can be compared to each element of Ej for all i, j ∈ {1, . . . , N}. In the literature, to characterise the fuzzy integrals, the Choquet integral or the Sugeno integral, all the criteria are assumed to share the same scale E. Moreover E is supposed to be a totally ordered set [11,4]. To be more precise, one major aim is to rank, in an evaluation scale E, mappings X from some non-empty set C to an evaluation scale E. In multicriteria theory, X is usually called an object, C is the set of criteria and X(c) denotes the partial score of X with respect to criterion c ∈ C. We denote V the set of all functions X : C → E. In practice, sometimes, we need to evaluate objects with a scale that is not totally ordered and many authors are interested about the use of the fuzzy integrals on a lattice [7,1]. To illustrate our motivation, take as an example, the industrial context in which objects can be evaluated by measures taken by detectors. When the given measures and the differences between them are smaller than the detector’s sensibility we cannot compare the result [3]. In such a context, we have a partial order and we need to rank in an evaluation scale that is a lattice. Next, to evaluate the involved objects, all the measures have to be aggregated to give a score to each object. Technically speaking, unlike the Choquet E-mail address: [email protected]. 0165-0114/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2007.11.008
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integral which needs arithmetic operators, the definition of the Sugeno integral only uses the minimum operator and the maximum operator. Consequently, the Sugeno integral seems to be the most suitable when the evaluation scale E is a lattice. A question which arises is: what well-known properties in the totally ordered context can be adapted in the lattice context? After making liberal allowance for the properties and the representation theorem about the Sugeno integral proved when the evaluation scale is totally ordered [8,6], to begin the generalisation we propose to work on a Boolean algebra. 1 This new situation implies renouncing of the notion of comonotonic functions. This property is replaced by the co-included functions defined in this article. As the aim of this article is to generalise a representation theorem about the Sugeno integral we need to care about this new concept. Rather, we precise that, when the scale is totally ordered, comonotonic functions and co-included functions are not the same and co-included functions are solely needed to characterise the Sugeno integral. To bring out the generalisation we keep notations used when the scales are totally ordered: we consider a finite Boolean algebra E, a finite set of criteria C and we denote V the set of all functions from C to the evaluation scale E and a functional from V to E. The aim of this paper is to present some properties to be verified by in order for to be a Sugeno integral. It generalises some well-known results when the evaluation scale is a totally ordered set [4,11]. The paper is organised as follows: Section 2 deals with the classical Sugeno integral. After the notations, we recall some results proved when E is totally ordered. Next, we give a representation theorem of the Sugeno integral with co-included functions when E is totally ordered. Section 3 presents the finite Boolean algebras and their properties used in this paper. In Section 4, after notations and preliminary result, we present the properties that are still satisfied when the Sugeno integral is defined on a Boolean algebra. To end this section we prove our main result: given a finite Boolean algebra E, a finite set of criteria C, V the set of all functions from C to the evaluation scale E and a functional from V to E, we present some properties to be verified by in order for to be a Sugeno integral. Section 5 is devoted to some concluding remarks. 2. The Sugeno integral on a finite totally ordered set 2.1. Notations To begin with, we introduce the notations used in this section: • C = {c1 , . . . , cN } is a finite set. • P(C) is the set of subsets of C. • (L, ∨, ∧) is a finite totally ordered set with a maximal element 1 and a minimal element 0. The minimum operator is denoted by ∧ and the maximum operator by ∨. • V = {X: C → L} is the set of all functions from C to L. • A total order, , is classically defined on L by ∀, ∈ L, if and only if one of the equivalent properties holds: ∧ = or ∨ = . • For every subset A of P(C), we denote 1A the characteristic function of A: 1A : C →L 1 if c ∈ A, c → 0 else. 2.2. The Sugeno integral characterised with comonotonic functions The aim of this section is to recall some well-known definitions and properties about the Sugeno integral when the evaluation scale L is a totally ordered set. Definition 1. • A capacity v is a set function v: P(C) → L which satisfies v(∅) = 0, v(C) = 1 and ∀A, B ⊆ C, A ⊆ B ⇒ v(A)v(B). • A possibility is a capacity such that ∀A, B ⊆ C, v(A ∪ B) = v(A) ∨ v(B). • A necessity is a capacity such that ∀A, B ⊆ C, v(A ∩ B) = v(A) ∧ v(B).
1 Which is a special lattice.
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Definition 2. Let X: C → L be a function and v be a capacity on C. The Sugeno integral of X with respect to v is: Sv (X) = ∧ v(X ), ∈L
where v(X ) = v({c ∈ C|X(c) }). Definition 3. Two functions X, Y ∈ V are comonotonic if and only if ∀ci , cj ∈ C,
X(ci ) < X(cj ) ⇒ Y (ci ) Y (cj ).
When the evaluation scale L is a totally ordered set, the Sugeno integral is defined and characterised, see for example [11,5,10]. The following interesting result can be found in [4]. Theorem 4. Let V be the set of functions from C to L and be a functional from V to L. satisfies the following properties: (1) ∀ ∈ L, ( ∧ 1C ) = , (2) ∀ ∈ L, ∀X ∈ V , ( ∧ X) = ∧ (X), (3) X, Y comonotonic implies (X ∨ Y ) = (X) ∨ (Y ), if and only if is a Sugeno integral. 2.3. The Sugeno integral characterised with the co-included functions One major difficulty to characterise the Sugeno integral on a Boolean algebra is the comonotonic property. To put it more thorough, comonotonic functions have to be replaced by co-included functions. In order to be precise, in this section, we present a representation theorem of the Sugeno integral with the co-included functions when the scale is totally ordered. First we introduce the co-included functions and we compare them with the comonotonic functions. Definition 5. Two functions X, Y ∈ V are co-included if and only if ∀ ∈ L we have {X } ⊆ {Y } or {Y } ⊆ {X }. Note that if two functions X and Y are such that X Y , or Y X, then obviously they are co-included. Proposition 6. If X, Y ∈ V are two comonotonic functions, then X and Y are co-included functions. But if X, Y ∈ V are two co-included functions then X and Y are not necessary comonotonic functions. Proof. • First let us prove that if X, Y ∈ V are two comonotonic functions, then X and Y are co-included functions. Let X, Y ∈ V be two comonotonic functions and suppose that ∃l ∈ L such that {X l}{Y l} and {Y l}{X l}. In such a case we have ◦ ∃c1 ∈ C such that Y (c1 ) l and X(c1 ) < l, ◦ ∃c2 ∈ C such that X(c2 ) l and Y (c2 ) < l. 2 So X and Y are not comonotonic functions because Y (c1 ) > Y (c2 ) and X(c1 ) < X(c2 ). • The following example shows that the converse is not true. Let L = {0, e, 1} be a finite totally ordered set with 0 < e < 1 and C = {c1 , c2 } be a set of criteria. Let us define two functions X and Y : C → L as follows: X Y
c1 0 1
c2 e e
2 These inequalities are satisfied because in this section L is a finite totally ordered set.
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◦ We have X(c1 ) < X(c2 ) and Y (c1 ) > Y (c2 ) so X and Y are not comonotonic functions. ◦ But according to the following inclusions, X and Y are co-included functions {X 0} = {Y 0} = C, {X e} = {c2 } ⊆ {Y e} = C, {X 1} = ∅ ⊆ {Y 1} = C.
To end this section and in order to bring out our generalisation proved at the end of the article, we are going to prove that co-included functions are necessary and sufficient to characterise the Sugeno integral when L is totally ordered. Theorem 7. Let L be a finite totally ordered set, V be the set of all functions from a finite set C to L and be a functional from V to L. satisfies the following properties: • (1) ∀ ∈ L, ( ∧ 1C ) = , • (2) ∀ ∈ L, X ∈ V , ( ∧ X) = ∧ (X), • (3 ) X, Y ∈ V = co-included ⇒ (X ∨ Y ) = (X) ∨ (Y ) if and only if is a Sugeno integral. Proof. • First we prove that Sugeno integral satisfies the three properties (1), (2) and (3 ). First and second properties are induced from Theorem 4 so we care about the third one. We consider X, Y two coincluded functions. The Sugeno integral is increasing so Sv (X) ∨ Sv (Y ) Sv (X ∨ Y ) and we just have to show that Sv (X ∨ Y )Sv (X) ∨ Sv (Y ). Let us choose in L. As X and Y are co-included functions either {X } ⊆ {Y } or {Y } ⊆ {X } holds. ◦ If {X } ⊆ {Y }, then {X ∨ Y } ⊆ {Y } because {X ∨ Y } ⊆ {X } ∪ {Y }. Consequently, ∧ v(X ∨ Y ) ∧ v(Y )Sv (Y )Sv (Y ) ∨ Sv (X). ◦ If {Y } ⊆ {X }, then similarly we obtain ∧ v(X ∨ Y ) Sv (Y ) ∨ Sv (X). So for all ∈ L, ∧ v(X ∨ Y )Sv (Y ) ∨ Sv (X) which entails Sv (X ∨ Y )Sv (Y ) ∨ Sv (X). • Next, we prove that if satisfies the three properties then is a Sugeno integral. Let X be a function from C to L and {xi }i∈{1,...,N} be its values. Without loss of generality we can suppose x1 · · · xN . Hence we have X=
N
xi ∧ 1Ai
where Ai = {X xi }.
(1)
i=1
We denote Xi = xi ∧ 1Ai . For all i ∈ {1, . . . , N − 1} functions Xi and N j =i+1 Xj are comonotonic so they are co-included. Finite re-iteration on N i and property (3 ) imply (X) = i=1 (xi ∧ 1Ai ). Then property (2) allows us to write (X) = N i=1 xi ∧ (1Ai ). To conclude we have to prove that ∀A ⊆ C, v(A) = (1A ) defines a capacity on P(C): ◦ v(∅) = (0 ∧ 1C ) = 0, ◦ v(C) = (1 ∧ 1C ) = 1, ◦ if A ⊆ B, then v(A) = (1A ) (1B ) = v(B) because 1A and 1B are co-included functions. The aim of this article is to present the same representation theorem when the evaluation scale is a finite Boolean algebra: this will be achieved in Theorem 26. The proof of Theorem 26 will be similar to the proof of Theorem 7 with necessary adaptations. First let us recall some lattice properties given in [2,9].
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3. Lattice definitions and properties A lattice is a partially ordered set (or poset) in which all nonempty finite subsets have both a supremum ∨ and an infimum ∧. This article cares about bounded lattices with a greatest element denoted by 1 and a smallest element denoted by 0. Definition 8. A lattice (E, ∨, ∧) is • distributive if ∀a, b, c ∈ E,
a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)
and
a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c),
• complemented if ∀a ∈ E,
∃a ∈ E such that a ∧ a = 0
and a ∨ a = 1.
A distributive and complemented lattice is called a Boolean lattice. In such a lattice the complement of each element is unique, and we can define an order by a b ⇔ a ∧ b = a, ∀a, b ∈ E. We write (E, , ∧, ∨, 0, 1, ) and we speak about Boolean algebra. The well-known example is the lattice P() of all subsets of a set , with union ∪, intersection ∩ and complementarity. Proposition 9. In a distributive lattice ∀a, b, ca ∧ b = a ∧ c
and a ∨ b = a ∨ c imply b = c.
Proof. See [2] or [9]. Definition 10. An element p of a lattice is a ∨-irreducible (or irreducible) element if p is not the least element 0 and p = x ∨ y implies p = x or p = y. Note that in a totally ordered set each element other than 0 is ∨-irreducible. Definition 11. Let N be a subset of E and a be an element of E. A decomposition of a denoted a = p∈N p, is irredundant if for every p0 ∈ N we have p∈N\p0 p < a. If p∈N\p0 p = a then p0 is superfluous. Proposition 12. In a finite distributive lattice, every element has an irredundant decomposition with ∨-irreducible elements and this decomposition is unique. Proof. See [9]. Lemma 13. In a distributive lattice, if p is ∨ -irreducible and p a ∨ b, Proof. See [9].
then p a or p b.
Definition 14. Let a be an element in E, a is an atom if and only if a = 0 and e a ⇒ e = 0 or e = a. In other words, an atom is a minimal element of E\{0}. Note that if a, b are atoms such that a = b, then a ∧ b = 0. To end this subsection, we recall this essential proposition. Proposition 15. All finite Boolean algebras are isomorphic to P(X), the set of subsets of X where X is a finite set. Proof. See [9].
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4. The Sugeno integral on a finite Boolean algebra This section states and proves our main result: a Sugeno integral representation theorem with the co-included functions when the evaluation scale E is a Boolean algebra. To do this first we present definitions and a first result: atoms are sufficient to define the Sugeno integral on a Boolean algebra. Next we present some properties that are still satisfied by the Sugeno integral in this new context. To end this section, after technical results our main theorem is proved. 4.1. Notation, definitions and first results In this section we suppose (E, ∨, ∧) to be a finite Boolean algebra. So from Proposition 15, E is isomorphic to one P(), the Boolean algebra of all subsets of a finite set . We denote E∨ the set of ∨-irreducible elements in E which are named {a1 , . . . , ap } = E∨ . In such a case E∨ is also the set of atoms. E is bounded: 1 is the greatest element and 0 is the smallest one. C = {c1 , . . . , cN } is always a finite set, V = {X: C → E} is the set of functions from C to E and for every subset A of P(C), we denote 1A the characteristic function of A defined as follows: 1A : C →E 1 if c ∈ A, c → 0 else. In such a case, capacities, possibilities, necessities and the Sugeno integral can be defined as follows: Definition 16. • A capacity v is a set function v: P(C) → E which satisfies ◦ v(∅) = 0, ◦ v(C) = 1, ◦ ∀A, B ⊆ C, A ⊆ B ⇒ v(A) v(B). • A possibility is a capacity such that ∀A, B ⊆ C, v(A ∪ B) = v(A) ∨ v(B). • A necessity is a capacity such that ∀A, B ⊆ C, v(A ∩ B) = v(A) ∧ v(B). Definition 17. Let X: C → E be a function, the Sugeno integral of X with respect to a capacity v is Sv (X) = ∧ v(X ). ∈E
Proposition 18. Let E be a Boolean algebra, E∨ be its atoms set (or irreducible elements of E), C be a finite set, X be a function from C to E and v be a capacity on P(C). The Sugeno integral of X with respect to v is Sv (X) = ak ∧ v(X ak ). ak ∈E∨
∨ Proof. For the proof we denote Sv (X) = ak ∈E∨ ak ∧ v(X ak ). ∨ ∨ We have Sv (X) = Sv (X) ∨ ( ∈E / ∨ ∧ v(X )). So we have only to prove that ∈E / ∨ ∧ v(X )) Sv (X). Let ∈ E\E∨ be not ∨-irreducible in E, has got an irredundant decompositionwith ∨-irreducible elements: = ni=1 ai for some ai ∈ E∨ . We have ∧ v(X ) = (a1 ∨ · · · ∨ an ) ∧ v(X ) = k∈{1,...,n} ak ∧ v(X ). ∀k ∈ {1, . . . , n}, the inclusion {X } ⊆ {X ak } implies v({X }) v({X ak }). So ∀k ∈ {1, . . . , n}, ak ∧ v({X }) is smaller than ak and v({X ak }), and this implies ak ∧ v({X }) ak ∧ v({X ak }). Concluding, we have: ak ∧ v(X ak ) Sv∨ (X). ∀k ∈ {1, . . . , n}, ak ∧ v({X }) k∈{1,...,n}
Sv∨ (X) is greater than all ak ∧v({X }) so Sv∨ (X) is greater than the lower upper bound and then ∧v(X ) Sv∨ (X) for all ∈ E\E∨ . ∨ ∨ In conclusion, ∈E / ∨ ∧ v(X ))Sv (X) and Sv (X) = Sv (X) .
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To end this subsection we define, in the finite Boolean algebra context, the concept of co-included functions. Definition 19. Let X and Y be two functions in V = {X: C → E}. X and Y are co-included functions if and only if ∀ ∈ E,
{X } ⊆ {Y } or
{Y } ⊆ {X }.
4.2. Properties of the Sugeno integral In this section we present some Sugeno integral properties which remain satisfied when E is a Boolean algebra. Proposition 20. ∀A ⊆ C, Sv (1A ) = v(A).
Proof. According to definition 17, Sv (1A ) = ∈E ∧ v(1A ). If = 0 then {1A } = C, and if = 0, then {1A } = A. So we have Sv (1A ) = ∈E\{0} ∧ v(A) = v(A) ∧ ∈E\{0} = v(A) ∧ 1 = v(A). Proposition 21. On a finite Boolean algebra, the Sugeno integral with respect to a capacity is increasing. Proof. Let X, Y be two functions in V such that X Y . X Y ⇒ ∀ ∈ E, {X } ⊆ {Y } ⇒ ∀ ∈ E, v({X })v({Y }) ⇒ ∀ ∈ E, ∧ v({X }) ∧ v({Y }) ⇒ ∀ ∈ E, ∧ v({X }) Sv (Y ) ⇒ Sv (X)Sv (Y ). Property 22. ∀ ∈ E, Sv ( ∧ 1C ) = . Proof. ∀ ∈ E,
Sv ( ∧ 1C ) =
ak ∧ v( ∧ 1C ak )
ak ∈E∨
=
ak
ak ∈E∨ |ak
= .
Property 23. (1) ∀X ∈ V , ∀ ∈ E, Sv ( ∧ X) = ∧ Sv (X). (2) ∀X ∈ V , ∀ ∈ E, Sv ( ∨ X) = ∨ Sv (X). Proof. • Let us prove the first property. The Sugeno integral is an increasing function so ∀X ∈ V , ∀ ∈ E, Sv ( ∧ X) ∧ Sv (X). We just need to prove that ∀X ∈ V , ∀ ∈ E, ∧ Sv (X)Sv ( ∧ X). According to Proposition 18, we have ⎛ ⎞ ak ∧ v(X ak )⎠ ∧ Sv (X) = ∧ ⎝ =
ak ∈E∨
( ∧ ak ∧ v(X ak )).
ak ∈E∨
If ak then ∧ ak = 0 so ∧ Sv (X) = ak ∈E∨ |ak
(ak ∧ v(X ak )).
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The inequality ak implies {X ak } = { ∧ X ak }which entails ∧ Sv (X) = ak ∈E∨ |ak (ak ∧ v( ∧ X ak )) Sv ( ∧ x). • Let us prove the second property. The Sugeno integral is an increasing function so ∀X ∈ V , ∀ ∈ E, ∨ Sv (X) Sv ( ∨ X). We just need to prove that ∀X ∈ V , ∀ ∈ E, ∨ Sv (X) Sv ( ∨ X). According to Proposition 18, we have ak ∧ v( ∨ X ak )). Sv ( ∨ X) = ak ∈E∨
◦ If ak , then v( ∨ X ak ) = v(C) = 1. ◦ If ak , then { ∨ X ak } = {X ak }: {X ak } ⊆ { ∨ X ak } is always satisfied and, if c satisfies ∨ X(c)ak then, according to Lemma 13, X(c)ak because ak and ak is an atom. So we have ak ∧ v( ∨ X ak )) ak ∧ v( ∨ X ak )) Sv ( ∨ X) = ak ∈E∨ |ak
= =
ak ∈E∨ |ak
ak ∧ v( ∨ X ak ))
ak ∈E∨ |ak
ak ∧ v(X ak ))
ak ∈E∨ |ak
∨ Sv (X). So we have Sv ( ∨ X) ∨ Sv (X).
Property 24. Let v be a capacity and X, Y ∈ V be co-included functions, then (1) Sv (X ∨ Y ) = Sv (X) ∨ Sv (Y ). (2) Sv (X ∧ Y ) = Sv (X) ∧ Sv (Y ). Proof. Let X, Y ∈ V be two co-included functions. (1) According to Proposition 21 the Sugeno integral is an increasing function, so Sv (X) ∨ Sv (Y ) Sv (X ∨ Y ). Let us prove the converse inequality. Let be a ∨-irreducible element of E. • If {X } ⊆ {Y }, then as is a ∨-irreducible element, {X ∨Y } ⊆ {Y }∪{X } = {Y }. Consequently, ∧v(X ∨Y ) ∧v(Y ) Sv (Y ) Sv (X)∨ Sv (Y ). • If {Y } ⊆ {X }, similarly we obtain ∧ v(X ∨ Y ) Sv (X) ∨ Sv (Y ). So for all ∈ E∨ , ∧ v(X ∨ Y )Sv (X) ∨ Sv (Y ) which entails Sv (X ∨ Y ) Sv (X) ∨ Sv (Y ). (2) ⎡ ⎤ ⎡ ⎤ ak ∧ v(X ak ))⎦ ∧ ⎣ ak ∧ v(Y ak ))⎦ Sv (X) ∧ Sv (Y ) = ⎣ ak ∈E∨
=
ak ∈E∨
ak ∧ (v(X ak ) ∧ v(Y ak ))
(2)
ak ∈E∨
because in a Boolean algebra, if ak , aj ∈ E∨ , ak = aj then ak ∧ aj = 0. As X, Y ∈ V are co-included functions, for each ak ∈ E∨ we have either {X ak } ⊆ {Y ak } or the converse. • If ak is such that {X ak } ⊆ {Y ak } then if c ∈ C satisfies X(c) ak then Y (c)ak and X(c) ∧ Y (c)ak so {X ak } ⊆ {X ∧ Y ak }. We always have {X ∧ Y ak } ⊆ {X ak } so {X ak } = {X ∧ Y ak }. So in this case v(X ak ) = v(X ∧ Y ak ) and v(X ak ) ∧ v(Y ak ) = v(X ∧ Y ak ).
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• If ak is such that {Y ak } ⊆ {X ak } then similarly we prove that {Y ak } = {X ∧ Y ak }. And in this case we have v(Y ak ) = v(X ∧ Y ak ) and v(X ak ) ∧ v(Y ak ) = v(X ∧ Y ak ). So as a conclusion, Sv (X) ∧ Sv (Y ) = ak ∈E∨ ak ∧ v(X ∧ Y ak ) = Sv (X ∧ Y ). Property 25. (1) If v is a possibility, then ∀X, Y ∈ V , Sv(X ∨ Y ) = Sv (X) ∨ Sv (Y ). (2) If v is a necessity, then ∀X, Y ∈ V , Sv(X ∧ Y ) = Sv (X) ∧ Sv (Y ). Proof. (1) Let v be a possibility and X, Y ∈ V be two functions. Sv (X ∨ Y ) =
ak ∧ v(X ∨ Y ak ).
ak ∈E∨
ak is an atom so according to Lemma 13 {X ∨ Y ak } = {X ak } ∪ {Y ak } and as v is a possibility, we have v(X ∨ Y ak ) = v(X ak ) ∨ v(Y ak ). Hence, Sv (X ∨ Y ) = ak ∧ [v(X ak ) ∨ v(Y ak )]. ak ∈E∨
E is a finite distributive lattice, so we have Sv (X ∨ Y ) = Sv (X) ∨ Sv (Y ). (2) Let v be a necessity and X, Y ∈ V be two functions. ak ∧ v(X ∧ Y ak ). Sv (X ∧ Y ) = ak ∈E∨
{X ∧ Y ak } = {X ak } ∩ {Y ak } and as v is a necessity, we have v(X ∧ Y ak ) = v(X ak ) ∧ v(Y ak ). Hence, ak ∧ [v(X ak ) ∧ v(Y ak )], Sv (X ∧ Y ) = ak ∈E∨
which entails Sv (X ∧ Y ) = Sv (X) ∧ Sv (Y ) because we can use the equality given by Eq. (2) written in the second part in the proof of the Property 24. 4.3. A characterisation of the Sugeno integral on a finite Boolean algebra We cannot use the same proof as the one made when the scale is a totally ordered set because the functions in this new context do not have the same decomposition. But the idea of the proof is the same; to be more precise a function belonging to V as a decomposition with co-included functions and we use the properties belonging to the representation theorem. Theorem 26. Let C be a finite set, E be a finite Boolean algebra and E∨ = {a1 , . . . , ap } be its atoms set. We denote V the set of all functions from C to E and a functional from V to E. satisfies the following properties: (1) ∀ ∈ E, ( ∧ 1C ) = , (2) ∀ ∈ E, ∀X ∈ V , ( ∧ X) = ∧ (X), (3) X, Y co-included ⇒ (X ∨ Y ) = (X) ∨ (Y ) if and only if is a Sugeno integral. Proof. In the previous section we have already proved that the Sugeno integral satisfies the properties (1)–(3). Let us prove that if a functional satisfies these three properties then it is a Sugeno integral. p Let X: C → E be a function, it can easily be proved that X = i=1 ai ∧ 1Ai where Ai = {c ∈ C/X(c)ai }. p Hence: (X) = ( i=1 ai ∧ 1Ai ). p ∀i0 ∈ {1, . . . , p}, functions ai0 ∧ 1Ai0 and i=i0 +1 ai ∧ 1Ai are co-included.
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p For any ∈ E, we have to compare {ai0 ∧ 1Ai0 } and { i=i0 +1 ai ∧ 1Ai }. Remember that, when E is a finite Boolean algebra, the atoms are the ∨-irreducible elements. p • If = 0, then {ai0 ∧ 1Ai0 } = { i=i0 +1 ai ∧ 1Ai } = C. p • If = ai0 , then {ai0 ∧ 1Ai0 } = Ai0 and { i=i0 +1 ai ∧ 1Ai } = ∅. • If = 0, ai0 , then {ai0 ∧ 1Ai0 } = ∅. p So ∀ ∈ E we have {ai0 ∧ 1Ai0 } ⊆ { i=i0 +1 ai ∧ 1Ai } or the opposite. By iteration of property (3), we have p p (X) = i=1 (ai ∧ 1Ai ). Then the second property permits us to write (X) = i=1 ai ∧ (1Ai ). Similarly as in the proof made when E is a totally ordered set, v(.) defined on P(C) by ∀A ⊆ C, v(A) = (1A ) is a capacity. So is a Sugeno integral with respect to v. 5. Concluding remarks When the evaluation scale is a totally ordered set, we can permute operators ∨ and ∧ in the Sugeno integral expression. Then we obtain a second representation theorem for the Sugeno integral swapping ∧ and ∨. This result seems more difficult to obtain as we expected and it will be interested to care about it. To conclude it is a challenging question to know what remains true in our work when E is a distributive lattice and not necessary a finite Boolean algebra. Acknowledgements We thank the anonymous referees for very valuable comments. References [1] A. Ban, I. Fechete, Componentwise decomposition of some lattice-valued fuzzy integrals, Inform. Sci. 177 (2007) 1430–1440. [2] G. Birkhoff, Lattice Theory, third ed., American Mathematical Society, Providence, RI, 1967. [3] S. Bonnevay, M. Lamure, I. Lavallé, N. Nicoloyannis, Aide à la prise de décision et systèmes multi-agents, in: Intelligence artificielle et complexité, Paris, September 1998. [4] L. de Campos, M. Lamata, S. Moral, A unified approach to define fuzzy integrals, Fuzzy Sets and Systems 39 (1991) 75–90. [5] D. Dubois, J.-L. Marichal, H. Prade, M. Roubens, R. Sabbadin, The use of the discrete Sugeno integral in decision making: a survey, Internat. J. Uncertainty Fuzziness Knowledge-Based Systems 5 (2001) 539–561. [6] D. Dubois, H. Prade, R. Sabbadin, Qualitative decision theory with Sugeno integrals, in: M. Grabisch, T. Murofushi, M. Sugeno (Eds.), Fuzzy Measures and Integrals—Theory and Applications, Physica Verlag, Heidelberg, 2000, pp. 314–332. [7] X. Liu, G. Zhang, Lattice-valued fuzzy measure and lattice valued fuzzy integral, Fuzzy Sets and Systems 62 (1994) 319–332. [8] J.L. Marichal, On Sugeno integral as an aggregation function, Fuzzy Sets and Systems 114 (2000) 347–365. [9] D. Ponasse, J-C. Carrega, Algèbre et topologie booléennes, Masson, Paris, 1979. [10] A. Rico, Modélisation des préférences pour l’aide à la décision par l’intégrale de Sugeno, Ph.D. Thesis, Université Paris 1 Panthéon Sorbonne, 2002. [11] M. Sugeno, Theory of fuzzy integrals and its applications, Ph.D. Thesis, Tokyo Institute of Technology, 1974.