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Ordering based on uninorms
Information Sciences 330 (2016) 315–327
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Information Sciences journal homepage: www.elsevier.com/locate/ins
Ordering based on uninorms Ümit Ertu˘grul a, M. Nesibe Kesicio˘glu b,∗, Funda Karaçal a a b
Department of Mathematics, Karadeniz Technical University, Trabzon 61080, Turkey Department of Mathematics, Recep Tayyip Erdo˘gan University, Rize 53100, Turkey
a r t i c l e
i n f o
Article history: Received 30 April 2015 Revised 27 September 2015 Accepted 6 October 2015 Available online 19 October 2015
a b s t r a c t In this paper, an order induced by uninorms on bounded lattices is given and discussed. By defining such an order, the T-partial order is extended to a more general form. The relationship between the order induced by the underlying t-norm and t-conorm and the order induced by a uninorm is studied. Finally, the order is extended to the order induced by n-uninorms on chains.
Keywords: Uninorm Bounded lattice Partial order
© 2015 Elsevier Inc. All rights reserved.
1. Introduction Uninorms on the unit interval [0, 1] by Yager and Rybalov [20], are one of the important aggregation operators having interesting structures that generalize the notions of t-norms and t-conorms [8,22], with more applications like fuzzy logic, expert systems, neural networks, fuzzy system modeling [7,11,21]. The generalization problem of any logical operators on the unit interval [0, 1] to a complete lattice has been an attractive problem for many researchers [5,9,14,19,24,25]. Recently, the order generating problem from logical operators and the properties of new algebraic structure has gained interest [2,15–18]. In [16], a partial order defined by means of t-norms bounded lattices has been introduced
x T y :⇔ T (, y) = x, for some ∈ L, where L is a bounded lattice, T is a t-norm on L and x, y ∈ L. This partial order T is called a T-partial order on L. An order by means of t-conorms on bounded lattices can be defined in similar way. As a further work, one of the most interesting problems is to define an order generated by uninorms on bounded lattices. In this sense, the pre-order notion from uninorms was firstly given by Hlin˘ená et al. [13], but the order notion generated by uninorms is not known yet in the literature. In this paper, we define an order induced by uninorms on bounded lattices. Because the structure of uninorms is a special combination of t-norms and t-conorms, with this order, we extend the T-partial (S-partial) order T (S ) to a more general form. The paper is organized as follows: We shortly recall some basic notions and results in Section 2. In Section 3, we give an order U induced by a uninorm on bounded lattice L. We determine the relationship between the order induced by a uninorm and the order on the lattice. Also, we deeply investigate the order obtained from the underlying t-norms and t-conorms of uninorms on bounded lattices. We show that when the underlying t-norm and t-conorm of a uninorm on a bounded lattice L are divisible, the
∗
Corresponding author. Tel.: +90 464 2236126; fax: +90 464 2234019. E-mail addresses: [email protected] (Ü. Ertu˘grul), [email protected] (M.N. Kesicio˘glu), [email protected] (F. Karaçal).
http://dx.doi.org/10.1016/j.ins.2015.10.019 0020-0255/© 2015 Elsevier Inc. All rights reserved.
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order U coincides with the order on L. So, we conclude that the orders generated by representable and idempotent uninorms coincide with the order on the bounded lattice L. We determine the relationship between the algebraic structures obtained from the orders induced by a t-norm and its N-dual t-conorm. To give an example, we show that a bounded lattice L need not be a lattice w.r.t. the order induced by U . Also, we give a necessary and sufficient condition making a bounded lattice L also a lattice w.r.t. the order U . Under the given conditions, we determine a subset of L which is a lattice w.r.t. the order U . In Section 4, we define the order obtained from 2-uninorms on chains. In this respect, we give a more general form than the order U induced by U on a chain. Further, this generalization is extended to the n-uninorms on chains. 2. Notations, definitions and a review of previous results Definition 1 [14]. Let (L, ≤, 0, 1) be a bounded lattice. An operation U: L2 → L is called a uninorm on L, if it is commutative, associative, increasing with respect to the both variables and has a neutral element e ∈ L. In this study, the notation U (e) will be used for the set of all uninorms on L with neutral element e ∈ L. Definition 2 [11]. Let U be a uninorm on the unit square with neutral element e ∈ ]0, 1[. U is called almost continuous if it is continuous on [0, 1]2 {(0, 1), (1, 0)}. Theorem 1 [10]. A uninorm U with neutral element e ∈ ]0, 1[ is representable if and only if it is continuous on [0, 1]2 {(0, 1), (1, 0)}. Definition 3 [16,23]. An operation T (S) on a bounded lattice L is called a triangular norm (triangular conorm) if it is commutative, associative, increasing with respect to the both variables and has a neutral element 1 (0). Definition 4 [16,17]. A t-norm T (or a t-conorm S) on a bounded lattice L is divisible if the following condition holds: For all x, y ∈ L with x ≤ y there is z ∈ L such that x = T (y, z) (or y = S(x, z)). Definition 5 [12]. Let U be a uninorm on a bounded lattice L with a neutral element e ∈ L. An element x ∈ L is called an idempotent element of U if U (x, x) = x. In the paper, HU (HT and HS , respectively) denotes the set of all idempotent elements of a uninorm U (a t-norm T and a t-conorm S, respectively). Definition 6 [16,17,23]. (i) A t-norm T ( or t-conorm S) on a bounded lattice L is called ∨-distributive if for every a, b1 , b2 ∈ L
T (a, b1 ∨ b2 ) = T (a, b1 ) ∨ T (a, b2 )(or S(a, b1 ∨ b2 ) = S(a, b1 ) ∨ S(a, b2 )) holds. (ii) A t-norm T (or t-conorm S) on a complete lattice L is called infinitely ∨-distributive if for every a ∈ L and subset {bτ |τ ∈ Q} of L
T a,
τ ∈Q
bτ
=
T (a, bτ ) or S a,
τ ∈Q
τ ∈Q
bτ
=
S(a, bτ )
τ ∈Q
holds. An (infinitely) ∧-distributive t-norm (t-conorm) is defined in a similar way. Definition 7 ([3,18,23]). Let (L, ≤, 0, 1) be a bounded lattice. A decreasing function N: L → L is called a negation if N(0) = 1 and N(1) = 0. A negation N on L is called strong if it is an involution, i.e., N(N(x)) = x, for all x ∈ L. Definition 8 ([3]). Let T be a t-norm on a bounded lattice L and N a strong negation on L. The t-conorm S defined by
S(x, y) := N(T (N(x), N(y))), x, y ∈ L is called a N-dual t-conorm to T on L. Definition 9 ([16]). Let L be a bounded lattice, T a t-norm on L. The order defined by
x T y :⇔ T (, y) = x for some ∈ L is called a T -partial order (triangular order) for t-norm T. Similarly, the notion S− partial order can be defined as follows: Definition 10. Let L be a bounded lattice, S be a t-conorm on L. The order defined by is called a S-partial order for t-conorm S:
x S y :⇔ S(, x) = y for some ∈ L is called a S-partial order for t-conorm S. Note that many properties satisfied for T -partial order are also satisfied for S-partial order.
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3. The order U Definition 11. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e). Define the following relation, for x, y ∈ L, as
x U y :⇔
⎧ ⎨if x, y ∈ [0, e] ⎩
if x, y ∈ [e, 1]
if
(x, y) ∈ L
∗
and there exist
k ∈ [0, e] such that U (k, y) = x or,
and there exist ∈ [e, 1] such that U (x, ) = y
or,
(1)
and x ≤ y,
where Ie = {x ∈ L | x e} and = [0, e] × [e, 1] ∪ [0, e] × Ie ∪ [e, 1] × [0, e] ∪ [e, 1] × Ie ∪ Ie × [0, e] ∪ Ie × [e, 1] ∪ Ie × Ie . Here, note that the notation x||y denotes that x and y are incomparable. L∗
Proposition 2. The relation U defined in (1) is a partial order on L. Proof. (i) If x ∈ [0, e] or x ∈ [e, 1], then since U (x, e) = x x U x. Let x ∈ [0, e] and x ∈ [e, 1]. Since x ≤ x, U , we have that x U x. So, the relation U satisfies the reflexivity. (ii) Let x U y and y U x for elements x, y ∈ L. Let x and y be in [0, e]. Since x U y and y U x, there exist elements 1 , 2 in [0, e] such that
U (1 , y) = x and U (2 , x) = y. By using the monotonicity of U, we have that
x = U (1 , y) ≤ U (e, y) = y and
y = U (2 , x) ≤ U (e, x) = x. Thus, x = y. Let x and y be in [e, 1]. In this case, the proof is similar to the case of x, y ∈ [0, e]. Now, let x, y ∈ L∗ . Since x U y and y U x, it is obtained that x ≤ y and y ≤ x, whence x = y. So, the antisymmetry property holds. (iii) Let x U y and y U z for elements x, y, z ∈ L. Then, the following cases do not occur:
x ∈ [0, e], y ∈ [e, 1] and z||e or; x ∈ [0, e], y||e and z ∈ [0, e] or; x ∈ [e, 1], y ∈ [e, 1] and z||e or; x ∈ [e, 1], y||e and for all z ∈ L or; x||e, y ∈ [0, e] and for all z ∈ L or; x||e, y ∈ [e, 1] and z ∈ [0, e] or; x||e, y ∈ [e, 1] and z||e or; x||e, y||e and z ∈ [0, e]. The remain possible cases are as follows: 3.1. x ∈ [0, e] 3.1.1. y ∈ [0, e] 3.1.1.1. z ∈ [0, e] Since x U y and y U z, there exist 1 , 2 of [0, e] such that U (y, 1 ) = x and U (2 , z) = y. Then,
x = U (1 , y) = U (1 , U (2 , z)) = U (U (1 , 2 ), z) Let := U(1 , 2 ). It follows ∗ ≤ e from 1 , 2 ≤ e. So, since x = U (∗ , z), ∗ ≤ e, it is obtained that x U z. 3.1.1.2. z ∈ [0, e] Since y U z, it is clear that y ≤ z. Also, since x U y, there exists an element ≤ e such that U (, y) = x. It follows x U z from x = U (, y) ≤ U (e, y) = y ≤ z, it is obtained that x U z. 3.1.2. y ∈ [e, 1] 3.1.2.1. z ∈ [0, e] Since y U z, it is clear that e ≤ y ≤ z ≤ e. Then, it must be that y = z = e. U (z, x) = U (e, x) = x, x U z. 3.1.2.2. z ∈ [e, 1] Since x ≤ e ≤ z, it is clear that x U z. ∗
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Fig. 1. (L, ≤).
3.1.3. y||e 3.1.3.1. z ∈ [0, e] Since x U y and y U z, it is clear that x ≤ y and y ≤ z. So, it follows x U z from x ≤ z. 3.2. x ∈ [e, 1] 3.2.1. y ∈ [0, e] In this case, since x U y, it must be that e ≤ x ≤ y ≤ e. Then, x = y = e. Also, since y U z, we have that x = y U z. 3.2.2. y ∈ [e, 1] 3.2.2.1. z ∈ [0, e] Since y U z, it must be that e ≤ y ≤ z ≤ e, whence y = z = e. By x U y, there exists an element k ∈ [e, 1] such that U (x, k) = y. It follows immediately to x U z from x = U (x, e) ≤ U (x, k) = y = z. 3.2.2.2. z ∈ [e, 1] Since x U y and y U z, there exist 1 , 2 of [e, 1] such that U (1 , x) = y and U (2 , y) = z. Then, we have that
z = U (2 , y) = U (2 , U (1 , x)) = U (U (2 , 1 ), x), where U(2 , 1 ) ∈ [e, 1], whence x U z. 3.3. x||e. 3.3.1. y ∈ [e, 1] 3.3.1.1. z ∈ [e, 1] Since y U z, there exists an element ∈ [e, 1] such that U (y, ) = z. x U y gives that x ≤ y. Thus, we have that
x ≤ y = U (y, e) ≤ U (y, ) = z, whence x U z. 3.3.2. y||e 3.3.2.1. z ∈ [0, e] Since x U y and y U z, it is clear that x ≤ y and y ≤ z. Thus, we obtain that x U z. So, the transitivity holds. Proposition 3. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e). If x U y for any x, y ∈ L, then x ≤ y. Proof. Let x U y for any x, y ∈ L. If x, y ∈ [0, e], then there exists an element ≤ e such that
U (, y) = x. Since x = U (, y) ≤ U (e, y) = y, we have that x ≤ y. Let x, y ∈ [e, 1]. Then, there exists an element k ≥ e such that
U (k, x) = y. Since x = U (e, x) ≤ U (k, x) = y, we have that x ≤ y. Let (x, y) ∈ L∗ . Since x U y, it is clear that x ≤ y. Remark 1. The converse of Proposition 3 may not be satisfied. For example: Consider the lattice (L = {0, a, b, d, e, 1}, ≤, 0, 1) whose lattice diagram is displayed in Fig. 1.
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319
Table 1 The uninorm U on L. U
0
a
b
e
d
1
0 a b e d 1
0 0 0 0 d 1
0 0 0 a d 1
0 0 0 b d 1
0 a b e d 1
d d d d 1 1
1 1 1 1 1 1
Fig. 2. (L, U ).
Define the function U: L2 → L as in Table 1. It is clear that the function U is a uninorm on L with the neutral element e. Although a ≤ b, aU b since there doesn’t exist an element k ∈ [0, e] such that a = U (k, b). The order U on L has its diagram as follows (see Fig. 2). Proposition 4. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e). Then, (L, U ) is a bounded partially ordered set. Proof. It is clear that (L, U ) is a partially ordered set by Proposition 2. Let x ∈ [0, e]. Since U (0, x) ≤ U (0, e) = 0, we have that U (0, x) = 0. Let := 0 ≤ e. Since U (, x) = 0, ≤ e, it is obtained that 0U x. Let x ∈ [0, e]. Then, it follows 0U x from 0 ≤ x. So, for any x ∈ L, 0U x. That 1 is the greatest element with respect to U is shown in a similar way. Remark 2. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e). (i) Note that the order U coincides with the order T (S ), when e = 1 (e = 0). (ii) Even if (L, ≤, 0, 1) is a chain, the partially ordered set (L, U ) may not be a chain. For example consider L = [0, 1] and take the smallest uninorm Ue : [0, 1]2 → [0, 1] with a neutral element e ∈ ]0, 1[ defined as follows [12]:
Ue (x, y) =
⎧ ⎨0
max (x, y)
(x, y) ∈ [0, e]2 , (x, y) ∈ [e, 1]2 ,
min (x, y)
otherwise.
⎩
Then, 4e and 2e are not comparable w.r.t. U . If 2e U 4e , by Proposition 3, it would be 2e ≤ 4e , a contradiction. Suppose that 4e U 2e . Then, there exists an element ≤ e such that U (, 2e ) = 4e . But, by the definition of the uninorm Ue , 0 = Ue (, 2e ) = 4e , which is a contradiction. So, 4e and 2e are not comparable w.r.t. U . (iii) In the paper, for any subset X of L, in the whole study, X U (X U ) denotes the set of the upper (lower) bounds of X w.r.t. U . Also, for any a, b ∈ L, a∧U b (a∨U b) denotes the greatest (least) element of the lower (upper) bounds w.r.t. U , if there exists. Similar notations will be used for the orders T and S . Remark 3. L need not be a lattice w.r.t. U . For example:
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let L = [0, 1] and take the uninorm U := Umin (T nM ,S
Umin (T nM ,SM , 1 ) (x, y) = 2
1 M, 2 )
⎧ ⎪ ⎪ 0 ⎪ ⎪ ⎨
: [0, 1]2 → [0, 1] defined as follows [4]:
1 2 1 (x, y) ∈ 0, and x + y ≤ , 2 2 1 2 (x, y) ∈ , 1 ,
max (x, y) ⎪ ⎪ ⎪ ⎪ ⎩ min (x, y)
2
otherwise.
Then, ([0, 1], U ) is a meet semi-lattice, but not join semi-lattice. Firstly, let us show that ([0, 1], U ) is a meet semi-lattice. Let x, y ∈ [0, 1] be arbitrary. If x and y are comparable w.r.t. U , it is clear that the infimum x∧U y of x and y w.r.t. U exists. Suppose that x and y are not comparable w.r.t. U . Then, it is obvious that x, y = 0, 1 and x = y. In this case, let us show that x ∧U y = 0. Assume that x ∧U y = k = 0. • Let x, y ∈ [0, 12 ] and x + y ≤ y. Thus, it must be that k ≤
1 2 . Then, it follows k U x and k U y from x ∧U y = k. By Proposition 3, we have 1 1 1 2 since x ∈ [0, 2 ]. By k U x and k U y, there exist 1 , 2 of [0, 2 ] such that
that k ≤ x and k ≤
U (1 , x) = k = 0 and U (2 , y) = k = 0. Then, 1 + x > have that
1 2
and 2 + y >
1 2,
otherwise we obtain that U (1 , x) = k = 0 and U (2 , y) = k = 0, a contradiction. Then, we
U (1 , x) = min (1 , x) = k and U (2 , y) = min (2 , y) = k. Since U (1 , x) = min (1 , x) = k, either k = x or 1 = k. If k = x, it would be x U y by U (2 , y) = k = x, a contradiction. Then, it must be 1 = k. Similarly, it must be that 2 = k. Since k + x = 1 + x > 12 , k + y = 2 + y > 12 and x + y ≤ 12 , we obtain that k > 12 − x ≥ y. Since k = U (2 , y) = U (k, y) = min (k, y) = y, we have a contradiction again. So, x ∧U y = 0 when x, y ∈ [0, 12 ] and x + y ≤ 12 . • Let x, y ∈ [0, 12 ] and x + y > 12 . For any x, y ∈ [0, 1], suppose that x ≤ y. Then, since U (x, y) = min (x, y) = x, we have that x U y. Similarly, if y ≤ x, we get that y U x. Therefore, x and y are comparable w.r.t. U , which contradicts to our assumption. • Let x, y ∈ [ 21 , 1] and x ≤ y. If we take as := y ≥ 12 , since U (, x) = max (, x) = max (y, x) = y, we have that x U y, which is a contradiction because of x and y are not comparable. Similarly, we have that y U x when y ≤ x, a contradiction. • Let be neither x, y ∈ [0, 12 ] nor x, y ∈ [ 12 , 1]. If x ≤ y, then, we have that x U y. Similarly, when y ≤ x, we get that y U x. This contradicts to the fact that x and y are not comparable w.r.t. U . So, we obtain that if x and y are not comparable w.r.t. U , then it must be that x ∧U y = 0. That is, ([0, 1], U ) is a meet semi-lattice. Now, let us show that ([0, 1], U ) is not a join semi-lattice. It can be easily shown that 14 and 18 are not comparable w.r.t. U . By the definition of U , for any k > 1 2
Let t ≤
1 2,
since
1 4
U k and
and t ∈ { 14 , 18 } . Then, 41 U t and
1 8
U
Then, it must be that 1 + t >
1 2
U (1 , t ) = min (1 , t ) =
min (1 , t ) = 1 =
If we take 1 =
U
1
and 2 + t >
4
,t
1 2
such that
1 2.
By the definition of the uninorm, we have that
1 1 and U (2 , t ) = min (2 , t ) = . 4 8 1 8
U
1 4 , contradiction. If we suppose min (2 , t )
=t =
1 8 , we would have
1 1 and min (2 , t ) = 2 = . 4 8 1 2 , we have that t 1 1 8 , we obtain that 4
and 2 + t > 1 4
U
U t. Thus, there exist two elements 1 , 2 ≤
If min (1 , t ) = t = 14 , it would be 18 = U (2 , 14 ), whence a similar contradiction. So, it must be
1 2
U k, it is clear that k ∈ { 14 , 18 } .
1 1 and U (2 , t ) = . 4 8
U (1 , t ) =
Since 1 + t >
1 8
and 2 =
1 1 ,t = and U 4 8
3 8 . So, if t U t and 18
>
∈ { 14 , 18 } , then it must be that t > U
3 8 . Conversely, suppose that t
U t since
1 = . 8
Thus, t ∈ { 14 , 18 } . Then, we have that
1 1 ,
4 8
U
U
=
3 ,1 . 8
Since the least element of the set ] 38 , 1] w.r.t. U does not exist, we have that ([0, 1], U ) is not a join semi-lattice.
>
3 8.
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Suppose U is a uninorm with a neutral element e ∈ ]0, 1[. Define two functions TU and SU on the unit square as follows:
TU (x, y) :=
U (ex, ey) , x, y ∈ [0, 1] e
(2)
SU (x, y) :=
U (e + (1 − e)x, e + (1 − e)y) − e , x, y ∈ [0, 1] 1−e
(3)
and
Proposition 5 ([11]). For any uninorm U with neutral element e ∈ ]0, 1[, TU is a t-norm and SU is a t-conorm. TU given in (2) is called the underlying t-norm of U and SU given in (3) is called the underlying t-conorm of U. Proposition 6 ([14]). Let (L, ≤, 0, 1) be a bounded lattice, and U a uninorm with a neutral element e ∈ L. Then (i) T ∗ = U |[0,e]2 : [0, e]2 → [0, e] is a t-norm on [0, e].
(ii) S∗ = U |[e,1]2 : [e, 1]2 → [e, 1] is a t-conorm on [e, 1]. Proposition 7. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e) such that e ∈ L{0, 1}. Then, T∗ and S∗ are divisible if and only if U =≤. Proof. By Proposition 3, if a U b for any a, b ∈ L, then a ≤ b. Conversely, let a ≤ b. Suppose that a, b ∈ [0, e]. Since T∗ is divisible, it is clear that T ∗ =≤. That is, a T ∗ b. Then, there exists an element ∈ [0, e] such that T ∗ (, b) = a. Thus, we have that
U (, b) = U
|[0,e] (, b) = T ∗ (, b) = a, ∈ [0, e].
This shows that a U b. Let a, b ∈ [e, 1]. Since S∗ is divisible, it is obvious that S∗ =≤. Then, there exists an element ∗ ∈ [e, 1] such that S∗ (∗ , a) = b. Thus, we have that
U (∗ , a) = U
|[e,1] (∗ , a) = S∗ (∗ , a) = b, ∗ ∈ [e, 1].
This means that a U b. Now, assume that a, b ∈ [0, e] and a, b ∈ [e, 1]. Then, it is clear that a U b. Thus, it is obtained that the order U coincides the order ≤, when T∗ and S∗ are divisible. Conversely, suppose that U =≤. Let a ≤ b for a, b ∈ [0, e]. Then, a U b. Since a, b ∈ [0, e], it is clear that a T ∗ b, whence there exists an element ∈ [0, e] such that T ∗ (, a) = b. This shows that T∗ is divisible. Similar argument is true for S∗ . Corollary 8. Let U: [0, 1]2 → [0, 1] be a uninorm with a neutral element e ∈ ]0, 1[. Then, TU and SU are continuous if and only if U =≤. Moreover, if U is an idempotent uninorm on a bounded lattice L, then it must be that T ∗ = T∧ and S∗ = S∨ . So, the order U coincides with the order ≤ . Corollary 9. Let U: [0, 1]2 → [0, 1] be a uninorm with a neutral element e ∈ ]0, 1[. If U is representable, then U =≤. Proof. If U is representable, by Theorem 1, U is continuous on [0, 1]2 {(0, 1), (1, 0)}, that is, U is almost continuous. By Proposition 5 of [11], TU and SU are continuous. Then, by Proposition 7, we obtain that U =≤. Remark 4. The converse of Corollary 9 need not be true. That is, even if U =≤, U may not be representable. Indeed, if we take an arbitrary idempotent uninorm U, it is clear that U =≤ by Corollary 8, but it is not representable. Proposition 10. Let T: L2 → L be a t-norm on a bounded lattice L and S the N-dual t-conorm of T. Then, (L, T ) is a meet (join) semi-lattice if and only if (L, S ) is a join (meet) semi-lattice. Proof. Let (L, T ) be a meet semi-lattice. Since S is the N-dual t-conorm of T, for any x, y ∈ L
S(x, y) = N(T (N(x), N(y))), where N: L → L is a strong negation. Since (L, T ) is a meet semi-lattice, there exists the infimum of N(x) and N(y) with respect to the order T . Let
N(x) ∧T N(y) = k. Then, kT N(x) and kT N(y). There exist elements k1 and k2 of L such that
T (N(x), k1 ) = k and T (N(y), k2 ) = k. Since N is strong, we can reorganize above equalities as follows:
N(T (N(x), N(N(k1 )))) = N(k) and N(T (N(y), N(N(k2 )))) = N(k). So, we have that
S(x, N(k1 )) = N(k) and S(y, N(k2 )) = N(k).
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If we take ∗1 := N(k1 ) and ∗2 := N(k2 ), we get that
x S N(k) and y S N(k), which shows that N(k) ∈ {x, y}S .
Let m ∈ {x, y}S be arbitrary. Then, xS m and yS m. Thus, there exist elements m1 , m2 ∈ L such that
S(x, m1 ) = m and S(y, m2 ) = m. Since N(T (N(x), N(m1 ))) = S(x, m1 ) = m and N(T (N(y), N(m2 ))) = S(y, m2 ) = m, by the strongness of N, we have that
T (N(x), N(m1 )) = N(m) and T (N(y), N(m2 )) = N(m). Thus, N(m) T N(x) and N(m) T N(y), whence N(m) ∈ {N(x), N(y)} . Since N(x) ∧T N(y) = k, N(m) T k. Then, there exists T
an element of L such that T (, k) = N(m). If we use the strongness of N again, we have that m = N(T (N(N()), N(N(k)))) = S(N(), N(k)), whence N(k)S m. Thus, x ∨S y = N(k). Conversely, it can be easily shown that if (L, S ) is a join semi-lattice, then (L, T ) is a meet semi-lattice in a similar way.
Corollary 11. Let T: L2 → L be a t-norm on a bounded lattice L and S the N-dual t-conorm of T. Then, (L, T ) is a bounded lattice if and only if (L, S ) is a bounded lattice. Proposition 12. Let L be a lattice and U ∈ U (e) such that e ∈ L{0, 1} is comparable with all elements of L. Then, ([0, e], T ∗ ) and ([e, 1], S∗ ) are lattices if and only if (L, U ) is a lattice. Proof. Suppose that ([0, e], T ∗ ) and ([e, 1], S∗ ) are lattices. Let x, y ∈ [0, e] be arbitrary. Since ([0, e], T ∗ ) is a lattice, x ∨T ∗ y and x ∧T ∗ y exist. Let
x ∨T ∗ y = k ∈ [0, e] and x ∧T ∗ y = m ∈ [0, e]. Now, let us show that x ∨U y = k and x ∧U y = m. Since x ∨T ∗ y = k, x T ∗ k and y T ∗ k. Then, there exist 1 , 2 of [0, e] such that
T ∗ (1 , k) = x and T ∗ (2 , k) = y. Thus, U (1 , k) = U |[0,e] (1 , k) = T ∗ (1 , k) = x and U (2 , k) = U |[0,e] (2 , k) = T ∗ (2 , k) = y, whence x U k and y U k, that is,k ∈
{x, y}U .
Let t ∈ {x, y}U be arbitrary. Then, x U t and y U t. Since e is comparable with the elements of L, either t ≤ e or e ≤ t. If t ≤ e, then there exist 1 , 2 of [0, e] such that
x = U (1 , t ) = U
|[0,e] (1 , t ) = T ∗ (1 , t ) and y = U (2 , t ) = U |[0,e] (2 , t ) = T ∗ (2 , t ).
Then, xT∗ t and yT∗ t, that is, we have that t ∈ {x, y}T ∗ . Since x ∨T ∗ y = k, k T ∗ t. Thus, there exists an element s ∈ [0, e] such that
U (t, s) = U
|[0,e] (t, s) = T ∗ (t, s) = k.
Therefore, k U t. Let e ≤ t. Then, k ≤ t. Since k ∈ [0, e] and t ∈ [e, 1], U , we get that k U t. So, x ∨U y = k. Similarly, it can be shown that x ∧U y = m. Now, let x, y ∈ [e, 1]. Since ([e, 1], S∗ ) is a lattice, there exist
x ∨S∗ y = k∗ ∈ [e, 1] and x ∧S∗ y = m∗ ∈ [e, 1]. By similar arguments to the first part of the proof, it can be shown that x ∨U y = k∗ and x ∧U y = m∗ . Suppose that x, y ∈ [0, e] and x, y ∈ [e, 1]. Since e is comparable with the elements of L, it must be either x ≤ e and e ≤ y or y ≤ e and e ≤ x. Suppose that x ≤ e and e ≤ y. Then, it is clear that x ≤ y, whence we obtain x U y. In this case,
x ∨U y = y and x ∧U y = x. Let e ≤ x and y ≤ e. Similarly, we get that
x ∨U y = x and x ∧U y = y. That is, if x, y ∈ [0, e] and x, y ∈ [e, 1], then
x ∨U y = x ∨ y and x ∧U y = x ∧ y exist. So, (L, U ) is a lattice. Conversely, if (L, U ) is a lattice, it is clear that ([0, e], T ∗ ) and ([e, 1], S∗ ) are lattices. Corollary 13. Let L be a chain, U ∈ U (e) such that e ∈ L{0, 1} and S∗ be an N-dual t-conorm of T∗ . ([0, e], T ∗ ) is a lattice if and only if (L, U ) is a lattice.
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Fig. 3. (L, ≤). Table 2 The uninorm U on L. U
0
a
b
t
e
1
0 a b t e 1
0 0 0 t 0 1
0 0 0 t a 1
0 0 0 t b 1
t t t 1 t 1
0 a b t e 1
1 1 1 1 1 1
Table 3 The t-norm T∗ . T∗
0
a
b
e
0 a b e
0 0 0 0
0 0 0 a
0 0 0 b
0 a b e
Corollary 14. Let U be a uninorm on [0, 1] with neutral element e ∈ ]0, 1[. If the underlying t-norm and t-conorm are TD and SD respectively, then ([0, 1], U ) is a lattice. Proof. Let the underlying t-norm of U be TD . Proposition 3.4. in [16], ([0, e], TD ) is a lattice. By Corollary 11 gives that ([e, 1], SD ) is a lattice. So, by Proposition 12, ([0, 1], U ) is a lattice. Remark 5. As a generalization of Proposition 7 and Corollary 14, if we take the underlying t-norm of U as Tλ , where Tλ ’s are the families of Schweizer–Sklar, Hamacher, Frank, Yager, Aczél–Alsina, Dombi, Sugeno–Weber and Mayor–Torens t-norms and the underlying t-conorm of U as the dual t-conorms of Tλ ’s, then we can easily see that ([0, 1], U ) is a lattice. Remark 6. If we drop the condition given in Proposition 12, that is, if e is not comparable with all elements of L, then the claim need not be satisfied as in the following example. Example 1. Consider the lattice (L, ≤, 0, 1) given its lattice diagram as in Fig. 3. Take the uninorm U on L as in Table 2. Then, T∗ and its lattice diagram are as in Table 3 and Fig. 4. Also, t-conorm S∗ and its lattice diagram are in Table 4 and Fig. 5. Finally, the order U is depicted in Fig. 6. As it is easily seen in the figures, although ([0, e], T ∗ ) and ([e, 1], S∗ ) are lattices, (L, U ) is not. Proposition 15. Let L be a chain and U ∈ U (e) such that e ∈ L{0, 1} and HU be the set of all idempotent elements of U. Then, (HU , U ) is a chain.
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Fig. 4. ([0, e], T ∗ ). Table 4 The t-conorm S∗ . S∗
e
1
e 1
e 1
1 1
Fig. 5. ([e, 1], S∗ ).
Fig. 6. (L, U ).
Proof. Let a, b ∈ HU . Since L is a chain, for any a, b ∈ L either a ≤ b or b ≤ a. Suppose that a ≤ b. If a, b ∈ [0, e], then
a = U (a, a) ≤ U (a, b) ≤ U (a, e) = a, whence we have that U (a, b) = a, that is a U b. If a, b ∈ [e, 1], then
b = U (e, b) ≤ U (a, b) ≤ U (b, b) = b, whence we obtain that U (a, b) = b, which shows that a U b. Let (a, b) ∈ [0, e]2 and (a, b) ∈ [e, 1]2 . Since a ≤ b, it must be that a ≤ e ≤ b. So a U b. Proposition 16. Let L be a bounded lattice, U ∈ U (e) such that e ∈ L{0, 1} and let HT ∗ and HS∗ the sets of all idempotent elements of T∗ and S∗ , respectively. If T∗ is ∨-distributive and S∗ is ∧-distributive, then (HT ∗ ∪ HS∗ , U ) is a bounded lattice. Proof. Let a, b ∈ HT ∗ ∪ HS∗ be arbitrary. Suppose that a, b ∈ HT ∗ . Then,
U (a, a) = T ∗ (a, a) = a and U (b, b) = T ∗ (b, b) = b. Since U (a, b) = U (a, b), a, b ∈ [0, e], it is clear that U(a, b)U a and U(a, b)U b. Thus, U (a, b) ∈ {a, b} . Let ∈ {a, b} . Then, U a and U b. Thus, there exist elements 1 , 2 of [0, e] such that
U (1 , a) = = U (2 , b).
U
U
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325
Table 5 The uninorm U on L. U
0
a
b
c
d
e
1
0 a b c d e 1
0 0 0 0 0 0 0
0 a a a d d 1
0 a b b d e 1
0 a b c d e 1
0 d d d 1 1 1
0 d e e 1 1 1
0 1 1 1 1 1 1
Also,
U (, a) = U (U (1 , a), a) = U (1 , U (a, a)) = U (1 , a) = and U (, b) = U (U (2 , b), b) = U (2 , U (b, b)) = U (2 , b) = . Since U (, U (a, b)) = U (U (, a), b) = U (, b) = , it is obtained that U U(a, b). Thus, a ∧U b = U (a, b). Since a, b ∈ [0, e], it is clear that a∨b ≤ e. Also, since a = T ∗ (a, a) ≤ T ∗ (a ∨ b, a ∨ b) and similarly b = T ∗ (b, b) ≤ T ∗ (a ∨ b, a ∨ b), we have that a ∨ b ≤ T ∗ (a ∨ b, a ∨ b) ≤ T ∗ (a ∨ b, e) = a ∨ b. Thus, T ∗ (a ∨ b, a ∨ b) = a ∨ b, whence a ∨ b ∈ HT ∗ . Moreover, a ∨ b ∈ {a, b}U . Indeed, since
a = T ∗ (a, a) ≤ T ∗ (a ∨ b, a) ≤ T ∗ (e, a) = a and b = T ∗ (b, b) ≤ T ∗ (a ∨ b, b) ≤ T ∗ (e, b) = b, we have that
a = T ∗ (a ∨ b, a) = U (a ∨ b, a) and b = T ∗ (a ∨ b, b) = U (a ∨ b, b). Then, it is obtained that a U a∨b and b U a∨b. Thus, a ∨ b ∈ {a, b}U . Let k ∈ {a, b}U . Then, a U k and b U k. If k ∈ [0, e], then there exist elements 1 , 2 of [0, e] such that
T ∗ (1 , k) = U (1 , k) = a and T ∗ (2 , k) = U (2 , k) = b. Since T∗ is ∨-distributive, we have that
a ∨ b = T ∗ (k, 1 ) ∨ T ∗ (k, 2 ) = T ∗ (k, 1 ∨ 2 ) = U (k, 1 ∨ 2 ), whence a∨b U k. Let k ∈ [0, e]. Since a U k and b U k, we have that a ≤ k and b ≤ k. Thus, a∨b ≤ k. Therefore, we obtain that a∨b U k. So, a ∨U b = a ∨ b. Let a, b ∈ HS∗ . It can be easily shown that a ∨U b = U (a, b) and a ∧U b = a ∧ b. Let a ∈ HT ∗ and b ∈ HS∗ . Then, it is clear that a ≤ e ≤ b. By the definition of U , we have that a U b. So, a ∧U b = a and a ∨U b = b. Similarly, if a ∈ HS∗ and b ∈ HT ∗ , then a ∧U b = b and a ∨U b = a. So, (HT ∗ ∪ HS∗ , U ) is a lattice. Moreover, since 0 ∈ HT ∗ and 1 ∈ HS∗ , (HT ∗ ∪ HS∗ , U ) is a bounded lattice. Corollary 17. Let L be a complete lattice and U ∈ U (e) such that e ∈ L{0, 1}. If T∗ is an infinitely ∨-distributive t-norm and S∗ is an inifinitely ∧-distributive t-conorm , then (HT ∗ ∪ HS∗ , U ) is a complete lattice. Remark 7. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e) such that e ∈ L{0, 1}. Since U =≤ when T∗ and S∗ are divisible , it is clear that the monotonicity w.r.t. U holds. But, in general, U need not satisfy the monotonicity w.r.t. the order U . For example: consider the lattice L = {0, a, b, c, d, e, 1} with 0 < a < b < c < d < e < 1 and define the function U: L2 → L as in Table 5. It can be easily shown that U is a uninorm on L with the neutral element c. It is clear that a U b since U (b, a) = a for a ≤ c. If U preserves the monotonicity with respect to U , then
d = U (a, e) U U (b, e) = e. But, there does not exist any element t ≥ c such that
U (d, t ) = e, whence dU e. So, it is not satisfied the monotonicity w.r.t. U .
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4. Ordering of n-uninorms A 2-uninorm (introduced by Akella [1]) is an operation which is increasing, associative and commutative on the unit interval with an absorbing element separating two subintervals having their own neutral elements. The importance of 2- uninorms is due to being the generalization of both nullnorms and uninorms. In this section, we define the order obtained from 2-uninorms on chains. By this way, we give more general form of the order U given in (1). This generalization gives a more general extension of our order by using n-uninorm introduced by Akella in [1]. Definition 12 ([6]). Let (L, ≤, 0, 1) be a chain. An operator F: L2 → L is called 2-uninorm if it is commutative, associative, increasing with respect to both variables and fulfilling
∀x ≤ k F (e, x) = x and ∀x ≥ k F ( f, x) = x, where e, k, f ∈ L with 0 ≤ e ≤ k ≤ f ≤ 1. By Uk(e, f) we denote the class of all 2-uninorms on chains. Definition 13. Let U2 ∈ Uk(e, f) . Define the following relation: For every x, y ∈ L,
⎧ ∃ ≤ e ⎪ ⎪ ⎪ ⎪ ∃ ⎨ m ∈ [e, k] x U 2 y :⇔ ∃n ∈ [k, f ] ⎪ ⎪ ⎪ ∃ p ∈ [ f, 1] ⎪ ⎩ x ≤ y,
such that U 2 (, y) = x,
when
x, y ∈ [0, e] or,
such that U 2 (x, m) = y,
when
such that U (y, n) = x,
when x, y ∈ [k, f ]
or,
such that U 2 (x, p) = y,
when
x, y ∈ [ f, 1]
or,
2
x, y ∈ [e, k] or, (4)
otherwise.
Proposition 18. The relation U 2 defined in (4) is a partial order on L. Proof. The proof is trivial by Proposition 2.
Remark 8. Let U2 ∈ Uk(e, f) on a chain L. Note that, • If 0 = k ≤ f = 1, then 2-uninorm U2 is a triangular norm T with neutral element f = 1. In this case, the orders U 2 and T coincide. • If 0 = e ≤ k = 1, then 2-uninorm U2 is a triangular conorm S with neutral element e = 0. In this case, the orders U 2 and S coincide. • If 0 = k ≤ f ≤ 1 or 0 ≤ e ≤ k = 1, then 2-uninorm U2 is a uninorm U with neutral element f or e. In this case, the orders U 2 and U coincide. • If e = 0 and f = 1, then 2-uninorm U2 is a nullnorm V with zero element k. In this case, the orders U 2 and V coincide. Definition 14 ([1]). Let (L, ≤, 0, 1) be a chain. Then, {e1 , e2 , . . . , en }z1 ,z2 ,...,zn−1 is called an n-neutral element of V if V (ei , x) = x for all x ∈ [zi−1 , zi ] for 0 = z0 < z1 < · · · < zn = 1 and ei ∈ [zi−1 , zi ], i = 1, 2, . . . , n. Definition 15 ([1]). Let L be a chain and a binary operator Un on L, is an n-uninorm if it is associative, monotone, non-decreasing in each variable and commutative and has an n-neutral element {e1 , e2 , . . . , en }z1 ,z2 ,...,zn−1 . Similarly, the order given in (4) can be generalized for n-uninorms as follows. Proposition 19. Let L be a chain and Un be an n-uninorm on L with an n-neutral element {e1 , e2 , . . . , en }z1 ,z2 ,...,zn−1 , i = 1, 2, . . . , n. Then, the relation given in (5)
⎧ ⎨∃ ∈ [zi−1 , ei ] x U n y :⇔ ∃m ∈ [ei , zi ] ⎩ x ≤ y,
such that U n (, y) = x, such that U (m, x) = y, n
when when
x, y ∈ [zi−1 , ei ] x, y ∈ [ei , zi ]
or, or,
(5)
otherwise,
is a partial order on L. 5. Conclusions We have introduced and discussed a partial order on a bounded lattice L from a uninorm on L. So, we have extended the T-partial (S-partial) order to a more general form. Also, we have determined the relationship between the algebraic structures obtained from the orders induced by a t-norm and its N-dual t-conorm. According to the underlying t-norm and t-conorm of a uninorm, we have characterized the order induced by the uninorm. We have studied the relationship between the algebraic structures obtained from the orders induced by the underlying t-norm and t-conorm and induced by a uninorm. Moreover, we have introduced a partial order by means of 2-uninorms and n-uninorms on chains. So, we have given the partial orders induced by operators which are more general than uninorms and nullnorms on chains.
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Acknowledgment The authors are very grateful to all referees for their helpful comments and valuable suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25]
P. Akella, Structure of n-uninorms, Fuzzy Sets Syst. 158 (2007) 1631–1651. E. Asıcı, ¸ F. Karaçal, On the T-partial order and properties, Inf. Sci. 267 (2014) 323–333. ´ M. Baczynski, B. Jayaram, Fuzzy implications, in: Studies in Fuzziness and Soft Computing, 231, Springer, Berlin, Heidelberg, 2008. G. Beliakov, A. Pradera, T. Calvo, Aggregation functions: a guide for practitioners, in: Studies in Fuzziness and Soft Computing, 221, Springer, Berlin, Heidelberg, 2007. G. Birkhoff, Lattice Theory, 3rd ed., Providence, 1967. P. Dryga´s, E. Rak. Distributivity equation in the class of 2-uninorms, Fuzzy Sets Syst. http://dx.doi.org/10.1016/j.fss.2015.02.014. D. Dubois, H. Prade, Fundamentals of Fuzzy Sets, Kluwer Academic Publisher, Boston, 2000. D. Dubois, .H. Prade, A review of fuzzy set aggregation connectives, Inf. Sci. 36 (1985) 85–121. U. Ertu˘grul, F. Karaçal, R. Mesiar, Modified ordinal sums of triangular norms and triangular conorms on bounded lattices, Int. J. Intell. Syst. 30 (2015) 807–817. J. Fodor, B. De Baets, A single-point characterization of representable uninorms, Fuzzy Sets Syst. 202 (2012) 89–99. J. Fodor, R. Yager, A. Rybalov, Structure of uninorms, Int. J. Uncertain. Fuzziness Knowl.-Based Syst. 5 (1997) 411–427. M. Grabisch, J.-L. Marichal, R. Mesiar, E. Pap, Aggregation Functions, Cambridge University Press, 2009. D. Hlin˘ená, M. Kalina, P. Král, Pre-orders and orders generated by conjunctive uninorms, Inf. Process. Manage. Uncertain. Knowl.-Based Syst. Commun. Comput. Inf. Sci. 444 (2014) 307–316. F. Karaçal, R. Mesiar, Uninorms on bounded lattices, Fuzzy Sets Syst. 261 (2015) 33–43. F. Karaçal, E. Asıcı, ¸ Some notes on T-partial order, J. Inequal. Appl. 2013 (2013) 219. F. Karaçal, M.N. Kesicio˘glu, A T-partial order obtained from t-norms, Kybernetika 47 (2011) 300–314. M.N. Kesicio˘glu, F. Karaçal, R. Mesiar, Order-equivalent triangular norms, Fuzzy Sets Syst. 268 (2015) 59–71. M.N. Kesicio˘glu, R. Mesiar, Ordering based on implications, Inf. Sci. 276 (2014) 377–386. M.N. Kesicio˘glu, On the property of T-distributivity, Fixed Point Theory Appl. 2013 (2013) 32. R.R. Yager, A. Rybalov, Uninorm aggregation operators, Fuzzy Sets Syst. 80 (1996) 111–120. R.R. Yager, Uninorms in fuzzy system modelling, Fuzzy Sets Syst. 122 (2001) 167–175. R.R. Yager, Aggregation operators and fuzzy systems modelling, Fuzzy Sets Syst. 67 (1994) 129–145. Z. Ma, W.M. Wu, Logical operators on complete lattices, Inf. Sci. 55 (1991) 77–97. Z.D. Wang, J.X. Fang, Residual operators of left and right uninorms on a complete lattice, Fuzzy Sets Syst. 160 (2009a) 22–31. Z.D. Wang, J.X. Fang, Residual coimplicators of left and right uninorms on a complete lattice, Fuzzy Sets Syst. 160 (2009b) 2086–2096.
Contents lists available at ScienceDirect
Information Sciences journal homepage: www.elsevier.com/locate/ins
Ordering based on uninorms Ümit Ertu˘grul a, M. Nesibe Kesicio˘glu b,∗, Funda Karaçal a a b
Department of Mathematics, Karadeniz Technical University, Trabzon 61080, Turkey Department of Mathematics, Recep Tayyip Erdo˘gan University, Rize 53100, Turkey
a r t i c l e
i n f o
Article history: Received 30 April 2015 Revised 27 September 2015 Accepted 6 October 2015 Available online 19 October 2015
a b s t r a c t In this paper, an order induced by uninorms on bounded lattices is given and discussed. By defining such an order, the T-partial order is extended to a more general form. The relationship between the order induced by the underlying t-norm and t-conorm and the order induced by a uninorm is studied. Finally, the order is extended to the order induced by n-uninorms on chains.
Keywords: Uninorm Bounded lattice Partial order
© 2015 Elsevier Inc. All rights reserved.
1. Introduction Uninorms on the unit interval [0, 1] by Yager and Rybalov [20], are one of the important aggregation operators having interesting structures that generalize the notions of t-norms and t-conorms [8,22], with more applications like fuzzy logic, expert systems, neural networks, fuzzy system modeling [7,11,21]. The generalization problem of any logical operators on the unit interval [0, 1] to a complete lattice has been an attractive problem for many researchers [5,9,14,19,24,25]. Recently, the order generating problem from logical operators and the properties of new algebraic structure has gained interest [2,15–18]. In [16], a partial order defined by means of t-norms bounded lattices has been introduced
x T y :⇔ T (, y) = x, for some ∈ L, where L is a bounded lattice, T is a t-norm on L and x, y ∈ L. This partial order T is called a T-partial order on L. An order by means of t-conorms on bounded lattices can be defined in similar way. As a further work, one of the most interesting problems is to define an order generated by uninorms on bounded lattices. In this sense, the pre-order notion from uninorms was firstly given by Hlin˘ená et al. [13], but the order notion generated by uninorms is not known yet in the literature. In this paper, we define an order induced by uninorms on bounded lattices. Because the structure of uninorms is a special combination of t-norms and t-conorms, with this order, we extend the T-partial (S-partial) order T (S ) to a more general form. The paper is organized as follows: We shortly recall some basic notions and results in Section 2. In Section 3, we give an order U induced by a uninorm on bounded lattice L. We determine the relationship between the order induced by a uninorm and the order on the lattice. Also, we deeply investigate the order obtained from the underlying t-norms and t-conorms of uninorms on bounded lattices. We show that when the underlying t-norm and t-conorm of a uninorm on a bounded lattice L are divisible, the
∗
Corresponding author. Tel.: +90 464 2236126; fax: +90 464 2234019. E-mail addresses: [email protected] (Ü. Ertu˘grul), [email protected] (M.N. Kesicio˘glu), [email protected] (F. Karaçal).
http://dx.doi.org/10.1016/j.ins.2015.10.019 0020-0255/© 2015 Elsevier Inc. All rights reserved.
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order U coincides with the order on L. So, we conclude that the orders generated by representable and idempotent uninorms coincide with the order on the bounded lattice L. We determine the relationship between the algebraic structures obtained from the orders induced by a t-norm and its N-dual t-conorm. To give an example, we show that a bounded lattice L need not be a lattice w.r.t. the order induced by U . Also, we give a necessary and sufficient condition making a bounded lattice L also a lattice w.r.t. the order U . Under the given conditions, we determine a subset of L which is a lattice w.r.t. the order U . In Section 4, we define the order obtained from 2-uninorms on chains. In this respect, we give a more general form than the order U induced by U on a chain. Further, this generalization is extended to the n-uninorms on chains. 2. Notations, definitions and a review of previous results Definition 1 [14]. Let (L, ≤, 0, 1) be a bounded lattice. An operation U: L2 → L is called a uninorm on L, if it is commutative, associative, increasing with respect to the both variables and has a neutral element e ∈ L. In this study, the notation U (e) will be used for the set of all uninorms on L with neutral element e ∈ L. Definition 2 [11]. Let U be a uninorm on the unit square with neutral element e ∈ ]0, 1[. U is called almost continuous if it is continuous on [0, 1]2 {(0, 1), (1, 0)}. Theorem 1 [10]. A uninorm U with neutral element e ∈ ]0, 1[ is representable if and only if it is continuous on [0, 1]2 {(0, 1), (1, 0)}. Definition 3 [16,23]. An operation T (S) on a bounded lattice L is called a triangular norm (triangular conorm) if it is commutative, associative, increasing with respect to the both variables and has a neutral element 1 (0). Definition 4 [16,17]. A t-norm T (or a t-conorm S) on a bounded lattice L is divisible if the following condition holds: For all x, y ∈ L with x ≤ y there is z ∈ L such that x = T (y, z) (or y = S(x, z)). Definition 5 [12]. Let U be a uninorm on a bounded lattice L with a neutral element e ∈ L. An element x ∈ L is called an idempotent element of U if U (x, x) = x. In the paper, HU (HT and HS , respectively) denotes the set of all idempotent elements of a uninorm U (a t-norm T and a t-conorm S, respectively). Definition 6 [16,17,23]. (i) A t-norm T ( or t-conorm S) on a bounded lattice L is called ∨-distributive if for every a, b1 , b2 ∈ L
T (a, b1 ∨ b2 ) = T (a, b1 ) ∨ T (a, b2 )(or S(a, b1 ∨ b2 ) = S(a, b1 ) ∨ S(a, b2 )) holds. (ii) A t-norm T (or t-conorm S) on a complete lattice L is called infinitely ∨-distributive if for every a ∈ L and subset {bτ |τ ∈ Q} of L
T a,
τ ∈Q
bτ
=
T (a, bτ ) or S a,
τ ∈Q
τ ∈Q
bτ
=
S(a, bτ )
τ ∈Q
holds. An (infinitely) ∧-distributive t-norm (t-conorm) is defined in a similar way. Definition 7 ([3,18,23]). Let (L, ≤, 0, 1) be a bounded lattice. A decreasing function N: L → L is called a negation if N(0) = 1 and N(1) = 0. A negation N on L is called strong if it is an involution, i.e., N(N(x)) = x, for all x ∈ L. Definition 8 ([3]). Let T be a t-norm on a bounded lattice L and N a strong negation on L. The t-conorm S defined by
S(x, y) := N(T (N(x), N(y))), x, y ∈ L is called a N-dual t-conorm to T on L. Definition 9 ([16]). Let L be a bounded lattice, T a t-norm on L. The order defined by
x T y :⇔ T (, y) = x for some ∈ L is called a T -partial order (triangular order) for t-norm T. Similarly, the notion S− partial order can be defined as follows: Definition 10. Let L be a bounded lattice, S be a t-conorm on L. The order defined by is called a S-partial order for t-conorm S:
x S y :⇔ S(, x) = y for some ∈ L is called a S-partial order for t-conorm S. Note that many properties satisfied for T -partial order are also satisfied for S-partial order.
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3. The order U Definition 11. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e). Define the following relation, for x, y ∈ L, as
x U y :⇔
⎧ ⎨if x, y ∈ [0, e] ⎩
if x, y ∈ [e, 1]
if
(x, y) ∈ L
∗
and there exist
k ∈ [0, e] such that U (k, y) = x or,
and there exist ∈ [e, 1] such that U (x, ) = y
or,
(1)
and x ≤ y,
where Ie = {x ∈ L | x e} and = [0, e] × [e, 1] ∪ [0, e] × Ie ∪ [e, 1] × [0, e] ∪ [e, 1] × Ie ∪ Ie × [0, e] ∪ Ie × [e, 1] ∪ Ie × Ie . Here, note that the notation x||y denotes that x and y are incomparable. L∗
Proposition 2. The relation U defined in (1) is a partial order on L. Proof. (i) If x ∈ [0, e] or x ∈ [e, 1], then since U (x, e) = x x U x. Let x ∈ [0, e] and x ∈ [e, 1]. Since x ≤ x, U , we have that x U x. So, the relation U satisfies the reflexivity. (ii) Let x U y and y U x for elements x, y ∈ L. Let x and y be in [0, e]. Since x U y and y U x, there exist elements 1 , 2 in [0, e] such that
U (1 , y) = x and U (2 , x) = y. By using the monotonicity of U, we have that
x = U (1 , y) ≤ U (e, y) = y and
y = U (2 , x) ≤ U (e, x) = x. Thus, x = y. Let x and y be in [e, 1]. In this case, the proof is similar to the case of x, y ∈ [0, e]. Now, let x, y ∈ L∗ . Since x U y and y U x, it is obtained that x ≤ y and y ≤ x, whence x = y. So, the antisymmetry property holds. (iii) Let x U y and y U z for elements x, y, z ∈ L. Then, the following cases do not occur:
x ∈ [0, e], y ∈ [e, 1] and z||e or; x ∈ [0, e], y||e and z ∈ [0, e] or; x ∈ [e, 1], y ∈ [e, 1] and z||e or; x ∈ [e, 1], y||e and for all z ∈ L or; x||e, y ∈ [0, e] and for all z ∈ L or; x||e, y ∈ [e, 1] and z ∈ [0, e] or; x||e, y ∈ [e, 1] and z||e or; x||e, y||e and z ∈ [0, e]. The remain possible cases are as follows: 3.1. x ∈ [0, e] 3.1.1. y ∈ [0, e] 3.1.1.1. z ∈ [0, e] Since x U y and y U z, there exist 1 , 2 of [0, e] such that U (y, 1 ) = x and U (2 , z) = y. Then,
x = U (1 , y) = U (1 , U (2 , z)) = U (U (1 , 2 ), z) Let := U(1 , 2 ). It follows ∗ ≤ e from 1 , 2 ≤ e. So, since x = U (∗ , z), ∗ ≤ e, it is obtained that x U z. 3.1.1.2. z ∈ [0, e] Since y U z, it is clear that y ≤ z. Also, since x U y, there exists an element ≤ e such that U (, y) = x. It follows x U z from x = U (, y) ≤ U (e, y) = y ≤ z, it is obtained that x U z. 3.1.2. y ∈ [e, 1] 3.1.2.1. z ∈ [0, e] Since y U z, it is clear that e ≤ y ≤ z ≤ e. Then, it must be that y = z = e. U (z, x) = U (e, x) = x, x U z. 3.1.2.2. z ∈ [e, 1] Since x ≤ e ≤ z, it is clear that x U z. ∗
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Fig. 1. (L, ≤).
3.1.3. y||e 3.1.3.1. z ∈ [0, e] Since x U y and y U z, it is clear that x ≤ y and y ≤ z. So, it follows x U z from x ≤ z. 3.2. x ∈ [e, 1] 3.2.1. y ∈ [0, e] In this case, since x U y, it must be that e ≤ x ≤ y ≤ e. Then, x = y = e. Also, since y U z, we have that x = y U z. 3.2.2. y ∈ [e, 1] 3.2.2.1. z ∈ [0, e] Since y U z, it must be that e ≤ y ≤ z ≤ e, whence y = z = e. By x U y, there exists an element k ∈ [e, 1] such that U (x, k) = y. It follows immediately to x U z from x = U (x, e) ≤ U (x, k) = y = z. 3.2.2.2. z ∈ [e, 1] Since x U y and y U z, there exist 1 , 2 of [e, 1] such that U (1 , x) = y and U (2 , y) = z. Then, we have that
z = U (2 , y) = U (2 , U (1 , x)) = U (U (2 , 1 ), x), where U(2 , 1 ) ∈ [e, 1], whence x U z. 3.3. x||e. 3.3.1. y ∈ [e, 1] 3.3.1.1. z ∈ [e, 1] Since y U z, there exists an element ∈ [e, 1] such that U (y, ) = z. x U y gives that x ≤ y. Thus, we have that
x ≤ y = U (y, e) ≤ U (y, ) = z, whence x U z. 3.3.2. y||e 3.3.2.1. z ∈ [0, e] Since x U y and y U z, it is clear that x ≤ y and y ≤ z. Thus, we obtain that x U z. So, the transitivity holds. Proposition 3. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e). If x U y for any x, y ∈ L, then x ≤ y. Proof. Let x U y for any x, y ∈ L. If x, y ∈ [0, e], then there exists an element ≤ e such that
U (, y) = x. Since x = U (, y) ≤ U (e, y) = y, we have that x ≤ y. Let x, y ∈ [e, 1]. Then, there exists an element k ≥ e such that
U (k, x) = y. Since x = U (e, x) ≤ U (k, x) = y, we have that x ≤ y. Let (x, y) ∈ L∗ . Since x U y, it is clear that x ≤ y. Remark 1. The converse of Proposition 3 may not be satisfied. For example: Consider the lattice (L = {0, a, b, d, e, 1}, ≤, 0, 1) whose lattice diagram is displayed in Fig. 1.
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319
Table 1 The uninorm U on L. U
0
a
b
e
d
1
0 a b e d 1
0 0 0 0 d 1
0 0 0 a d 1
0 0 0 b d 1
0 a b e d 1
d d d d 1 1
1 1 1 1 1 1
Fig. 2. (L, U ).
Define the function U: L2 → L as in Table 1. It is clear that the function U is a uninorm on L with the neutral element e. Although a ≤ b, aU b since there doesn’t exist an element k ∈ [0, e] such that a = U (k, b). The order U on L has its diagram as follows (see Fig. 2). Proposition 4. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e). Then, (L, U ) is a bounded partially ordered set. Proof. It is clear that (L, U ) is a partially ordered set by Proposition 2. Let x ∈ [0, e]. Since U (0, x) ≤ U (0, e) = 0, we have that U (0, x) = 0. Let := 0 ≤ e. Since U (, x) = 0, ≤ e, it is obtained that 0U x. Let x ∈ [0, e]. Then, it follows 0U x from 0 ≤ x. So, for any x ∈ L, 0U x. That 1 is the greatest element with respect to U is shown in a similar way. Remark 2. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e). (i) Note that the order U coincides with the order T (S ), when e = 1 (e = 0). (ii) Even if (L, ≤, 0, 1) is a chain, the partially ordered set (L, U ) may not be a chain. For example consider L = [0, 1] and take the smallest uninorm Ue : [0, 1]2 → [0, 1] with a neutral element e ∈ ]0, 1[ defined as follows [12]:
Ue (x, y) =
⎧ ⎨0
max (x, y)
(x, y) ∈ [0, e]2 , (x, y) ∈ [e, 1]2 ,
min (x, y)
otherwise.
⎩
Then, 4e and 2e are not comparable w.r.t. U . If 2e U 4e , by Proposition 3, it would be 2e ≤ 4e , a contradiction. Suppose that 4e U 2e . Then, there exists an element ≤ e such that U (, 2e ) = 4e . But, by the definition of the uninorm Ue , 0 = Ue (, 2e ) = 4e , which is a contradiction. So, 4e and 2e are not comparable w.r.t. U . (iii) In the paper, for any subset X of L, in the whole study, X U (X U ) denotes the set of the upper (lower) bounds of X w.r.t. U . Also, for any a, b ∈ L, a∧U b (a∨U b) denotes the greatest (least) element of the lower (upper) bounds w.r.t. U , if there exists. Similar notations will be used for the orders T and S . Remark 3. L need not be a lattice w.r.t. U . For example:
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let L = [0, 1] and take the uninorm U := Umin (T nM ,S
Umin (T nM ,SM , 1 ) (x, y) = 2
1 M, 2 )
⎧ ⎪ ⎪ 0 ⎪ ⎪ ⎨
: [0, 1]2 → [0, 1] defined as follows [4]:
1 2 1 (x, y) ∈ 0, and x + y ≤ , 2 2 1 2 (x, y) ∈ , 1 ,
max (x, y) ⎪ ⎪ ⎪ ⎪ ⎩ min (x, y)
2
otherwise.
Then, ([0, 1], U ) is a meet semi-lattice, but not join semi-lattice. Firstly, let us show that ([0, 1], U ) is a meet semi-lattice. Let x, y ∈ [0, 1] be arbitrary. If x and y are comparable w.r.t. U , it is clear that the infimum x∧U y of x and y w.r.t. U exists. Suppose that x and y are not comparable w.r.t. U . Then, it is obvious that x, y = 0, 1 and x = y. In this case, let us show that x ∧U y = 0. Assume that x ∧U y = k = 0. • Let x, y ∈ [0, 12 ] and x + y ≤ y. Thus, it must be that k ≤
1 2 . Then, it follows k U x and k U y from x ∧U y = k. By Proposition 3, we have 1 1 1 2 since x ∈ [0, 2 ]. By k U x and k U y, there exist 1 , 2 of [0, 2 ] such that
that k ≤ x and k ≤
U (1 , x) = k = 0 and U (2 , y) = k = 0. Then, 1 + x > have that
1 2
and 2 + y >
1 2,
otherwise we obtain that U (1 , x) = k = 0 and U (2 , y) = k = 0, a contradiction. Then, we
U (1 , x) = min (1 , x) = k and U (2 , y) = min (2 , y) = k. Since U (1 , x) = min (1 , x) = k, either k = x or 1 = k. If k = x, it would be x U y by U (2 , y) = k = x, a contradiction. Then, it must be 1 = k. Similarly, it must be that 2 = k. Since k + x = 1 + x > 12 , k + y = 2 + y > 12 and x + y ≤ 12 , we obtain that k > 12 − x ≥ y. Since k = U (2 , y) = U (k, y) = min (k, y) = y, we have a contradiction again. So, x ∧U y = 0 when x, y ∈ [0, 12 ] and x + y ≤ 12 . • Let x, y ∈ [0, 12 ] and x + y > 12 . For any x, y ∈ [0, 1], suppose that x ≤ y. Then, since U (x, y) = min (x, y) = x, we have that x U y. Similarly, if y ≤ x, we get that y U x. Therefore, x and y are comparable w.r.t. U , which contradicts to our assumption. • Let x, y ∈ [ 21 , 1] and x ≤ y. If we take as := y ≥ 12 , since U (, x) = max (, x) = max (y, x) = y, we have that x U y, which is a contradiction because of x and y are not comparable. Similarly, we have that y U x when y ≤ x, a contradiction. • Let be neither x, y ∈ [0, 12 ] nor x, y ∈ [ 12 , 1]. If x ≤ y, then, we have that x U y. Similarly, when y ≤ x, we get that y U x. This contradicts to the fact that x and y are not comparable w.r.t. U . So, we obtain that if x and y are not comparable w.r.t. U , then it must be that x ∧U y = 0. That is, ([0, 1], U ) is a meet semi-lattice. Now, let us show that ([0, 1], U ) is not a join semi-lattice. It can be easily shown that 14 and 18 are not comparable w.r.t. U . By the definition of U , for any k > 1 2
Let t ≤
1 2,
since
1 4
U k and
and t ∈ { 14 , 18 } . Then, 41 U t and
1 8
U
Then, it must be that 1 + t >
1 2
U (1 , t ) = min (1 , t ) =
min (1 , t ) = 1 =
If we take 1 =
U
1
and 2 + t >
4
,t
1 2
such that
1 2.
By the definition of the uninorm, we have that
1 1 and U (2 , t ) = min (2 , t ) = . 4 8 1 8
U
1 4 , contradiction. If we suppose min (2 , t )
=t =
1 8 , we would have
1 1 and min (2 , t ) = 2 = . 4 8 1 2 , we have that t 1 1 8 , we obtain that 4
and 2 + t > 1 4
U
U t. Thus, there exist two elements 1 , 2 ≤
If min (1 , t ) = t = 14 , it would be 18 = U (2 , 14 ), whence a similar contradiction. So, it must be
1 2
U k, it is clear that k ∈ { 14 , 18 } .
1 1 and U (2 , t ) = . 4 8
U (1 , t ) =
Since 1 + t >
1 8
and 2 =
1 1 ,t = and U 4 8
3 8 . So, if t U t and 18
>
∈ { 14 , 18 } , then it must be that t > U
3 8 . Conversely, suppose that t
U t since
1 = . 8
Thus, t ∈ { 14 , 18 } . Then, we have that
1 1 ,
4 8
U
U
=
3 ,1 . 8
Since the least element of the set ] 38 , 1] w.r.t. U does not exist, we have that ([0, 1], U ) is not a join semi-lattice.
>
3 8.
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321
Suppose U is a uninorm with a neutral element e ∈ ]0, 1[. Define two functions TU and SU on the unit square as follows:
TU (x, y) :=
U (ex, ey) , x, y ∈ [0, 1] e
(2)
SU (x, y) :=
U (e + (1 − e)x, e + (1 − e)y) − e , x, y ∈ [0, 1] 1−e
(3)
and
Proposition 5 ([11]). For any uninorm U with neutral element e ∈ ]0, 1[, TU is a t-norm and SU is a t-conorm. TU given in (2) is called the underlying t-norm of U and SU given in (3) is called the underlying t-conorm of U. Proposition 6 ([14]). Let (L, ≤, 0, 1) be a bounded lattice, and U a uninorm with a neutral element e ∈ L. Then (i) T ∗ = U |[0,e]2 : [0, e]2 → [0, e] is a t-norm on [0, e].
(ii) S∗ = U |[e,1]2 : [e, 1]2 → [e, 1] is a t-conorm on [e, 1]. Proposition 7. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e) such that e ∈ L{0, 1}. Then, T∗ and S∗ are divisible if and only if U =≤. Proof. By Proposition 3, if a U b for any a, b ∈ L, then a ≤ b. Conversely, let a ≤ b. Suppose that a, b ∈ [0, e]. Since T∗ is divisible, it is clear that T ∗ =≤. That is, a T ∗ b. Then, there exists an element ∈ [0, e] such that T ∗ (, b) = a. Thus, we have that
U (, b) = U
|[0,e] (, b) = T ∗ (, b) = a, ∈ [0, e].
This shows that a U b. Let a, b ∈ [e, 1]. Since S∗ is divisible, it is obvious that S∗ =≤. Then, there exists an element ∗ ∈ [e, 1] such that S∗ (∗ , a) = b. Thus, we have that
U (∗ , a) = U
|[e,1] (∗ , a) = S∗ (∗ , a) = b, ∗ ∈ [e, 1].
This means that a U b. Now, assume that a, b ∈ [0, e] and a, b ∈ [e, 1]. Then, it is clear that a U b. Thus, it is obtained that the order U coincides the order ≤, when T∗ and S∗ are divisible. Conversely, suppose that U =≤. Let a ≤ b for a, b ∈ [0, e]. Then, a U b. Since a, b ∈ [0, e], it is clear that a T ∗ b, whence there exists an element ∈ [0, e] such that T ∗ (, a) = b. This shows that T∗ is divisible. Similar argument is true for S∗ . Corollary 8. Let U: [0, 1]2 → [0, 1] be a uninorm with a neutral element e ∈ ]0, 1[. Then, TU and SU are continuous if and only if U =≤. Moreover, if U is an idempotent uninorm on a bounded lattice L, then it must be that T ∗ = T∧ and S∗ = S∨ . So, the order U coincides with the order ≤ . Corollary 9. Let U: [0, 1]2 → [0, 1] be a uninorm with a neutral element e ∈ ]0, 1[. If U is representable, then U =≤. Proof. If U is representable, by Theorem 1, U is continuous on [0, 1]2 {(0, 1), (1, 0)}, that is, U is almost continuous. By Proposition 5 of [11], TU and SU are continuous. Then, by Proposition 7, we obtain that U =≤. Remark 4. The converse of Corollary 9 need not be true. That is, even if U =≤, U may not be representable. Indeed, if we take an arbitrary idempotent uninorm U, it is clear that U =≤ by Corollary 8, but it is not representable. Proposition 10. Let T: L2 → L be a t-norm on a bounded lattice L and S the N-dual t-conorm of T. Then, (L, T ) is a meet (join) semi-lattice if and only if (L, S ) is a join (meet) semi-lattice. Proof. Let (L, T ) be a meet semi-lattice. Since S is the N-dual t-conorm of T, for any x, y ∈ L
S(x, y) = N(T (N(x), N(y))), where N: L → L is a strong negation. Since (L, T ) is a meet semi-lattice, there exists the infimum of N(x) and N(y) with respect to the order T . Let
N(x) ∧T N(y) = k. Then, kT N(x) and kT N(y). There exist elements k1 and k2 of L such that
T (N(x), k1 ) = k and T (N(y), k2 ) = k. Since N is strong, we can reorganize above equalities as follows:
N(T (N(x), N(N(k1 )))) = N(k) and N(T (N(y), N(N(k2 )))) = N(k). So, we have that
S(x, N(k1 )) = N(k) and S(y, N(k2 )) = N(k).
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If we take ∗1 := N(k1 ) and ∗2 := N(k2 ), we get that
x S N(k) and y S N(k), which shows that N(k) ∈ {x, y}S .
Let m ∈ {x, y}S be arbitrary. Then, xS m and yS m. Thus, there exist elements m1 , m2 ∈ L such that
S(x, m1 ) = m and S(y, m2 ) = m. Since N(T (N(x), N(m1 ))) = S(x, m1 ) = m and N(T (N(y), N(m2 ))) = S(y, m2 ) = m, by the strongness of N, we have that
T (N(x), N(m1 )) = N(m) and T (N(y), N(m2 )) = N(m). Thus, N(m) T N(x) and N(m) T N(y), whence N(m) ∈ {N(x), N(y)} . Since N(x) ∧T N(y) = k, N(m) T k. Then, there exists T
an element of L such that T (, k) = N(m). If we use the strongness of N again, we have that m = N(T (N(N()), N(N(k)))) = S(N(), N(k)), whence N(k)S m. Thus, x ∨S y = N(k). Conversely, it can be easily shown that if (L, S ) is a join semi-lattice, then (L, T ) is a meet semi-lattice in a similar way.
Corollary 11. Let T: L2 → L be a t-norm on a bounded lattice L and S the N-dual t-conorm of T. Then, (L, T ) is a bounded lattice if and only if (L, S ) is a bounded lattice. Proposition 12. Let L be a lattice and U ∈ U (e) such that e ∈ L{0, 1} is comparable with all elements of L. Then, ([0, e], T ∗ ) and ([e, 1], S∗ ) are lattices if and only if (L, U ) is a lattice. Proof. Suppose that ([0, e], T ∗ ) and ([e, 1], S∗ ) are lattices. Let x, y ∈ [0, e] be arbitrary. Since ([0, e], T ∗ ) is a lattice, x ∨T ∗ y and x ∧T ∗ y exist. Let
x ∨T ∗ y = k ∈ [0, e] and x ∧T ∗ y = m ∈ [0, e]. Now, let us show that x ∨U y = k and x ∧U y = m. Since x ∨T ∗ y = k, x T ∗ k and y T ∗ k. Then, there exist 1 , 2 of [0, e] such that
T ∗ (1 , k) = x and T ∗ (2 , k) = y. Thus, U (1 , k) = U |[0,e] (1 , k) = T ∗ (1 , k) = x and U (2 , k) = U |[0,e] (2 , k) = T ∗ (2 , k) = y, whence x U k and y U k, that is,k ∈
{x, y}U .
Let t ∈ {x, y}U be arbitrary. Then, x U t and y U t. Since e is comparable with the elements of L, either t ≤ e or e ≤ t. If t ≤ e, then there exist 1 , 2 of [0, e] such that
x = U (1 , t ) = U
|[0,e] (1 , t ) = T ∗ (1 , t ) and y = U (2 , t ) = U |[0,e] (2 , t ) = T ∗ (2 , t ).
Then, xT∗ t and yT∗ t, that is, we have that t ∈ {x, y}T ∗ . Since x ∨T ∗ y = k, k T ∗ t. Thus, there exists an element s ∈ [0, e] such that
U (t, s) = U
|[0,e] (t, s) = T ∗ (t, s) = k.
Therefore, k U t. Let e ≤ t. Then, k ≤ t. Since k ∈ [0, e] and t ∈ [e, 1], U , we get that k U t. So, x ∨U y = k. Similarly, it can be shown that x ∧U y = m. Now, let x, y ∈ [e, 1]. Since ([e, 1], S∗ ) is a lattice, there exist
x ∨S∗ y = k∗ ∈ [e, 1] and x ∧S∗ y = m∗ ∈ [e, 1]. By similar arguments to the first part of the proof, it can be shown that x ∨U y = k∗ and x ∧U y = m∗ . Suppose that x, y ∈ [0, e] and x, y ∈ [e, 1]. Since e is comparable with the elements of L, it must be either x ≤ e and e ≤ y or y ≤ e and e ≤ x. Suppose that x ≤ e and e ≤ y. Then, it is clear that x ≤ y, whence we obtain x U y. In this case,
x ∨U y = y and x ∧U y = x. Let e ≤ x and y ≤ e. Similarly, we get that
x ∨U y = x and x ∧U y = y. That is, if x, y ∈ [0, e] and x, y ∈ [e, 1], then
x ∨U y = x ∨ y and x ∧U y = x ∧ y exist. So, (L, U ) is a lattice. Conversely, if (L, U ) is a lattice, it is clear that ([0, e], T ∗ ) and ([e, 1], S∗ ) are lattices. Corollary 13. Let L be a chain, U ∈ U (e) such that e ∈ L{0, 1} and S∗ be an N-dual t-conorm of T∗ . ([0, e], T ∗ ) is a lattice if and only if (L, U ) is a lattice.
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323
Fig. 3. (L, ≤). Table 2 The uninorm U on L. U
0
a
b
t
e
1
0 a b t e 1
0 0 0 t 0 1
0 0 0 t a 1
0 0 0 t b 1
t t t 1 t 1
0 a b t e 1
1 1 1 1 1 1
Table 3 The t-norm T∗ . T∗
0
a
b
e
0 a b e
0 0 0 0
0 0 0 a
0 0 0 b
0 a b e
Corollary 14. Let U be a uninorm on [0, 1] with neutral element e ∈ ]0, 1[. If the underlying t-norm and t-conorm are TD and SD respectively, then ([0, 1], U ) is a lattice. Proof. Let the underlying t-norm of U be TD . Proposition 3.4. in [16], ([0, e], TD ) is a lattice. By Corollary 11 gives that ([e, 1], SD ) is a lattice. So, by Proposition 12, ([0, 1], U ) is a lattice. Remark 5. As a generalization of Proposition 7 and Corollary 14, if we take the underlying t-norm of U as Tλ , where Tλ ’s are the families of Schweizer–Sklar, Hamacher, Frank, Yager, Aczél–Alsina, Dombi, Sugeno–Weber and Mayor–Torens t-norms and the underlying t-conorm of U as the dual t-conorms of Tλ ’s, then we can easily see that ([0, 1], U ) is a lattice. Remark 6. If we drop the condition given in Proposition 12, that is, if e is not comparable with all elements of L, then the claim need not be satisfied as in the following example. Example 1. Consider the lattice (L, ≤, 0, 1) given its lattice diagram as in Fig. 3. Take the uninorm U on L as in Table 2. Then, T∗ and its lattice diagram are as in Table 3 and Fig. 4. Also, t-conorm S∗ and its lattice diagram are in Table 4 and Fig. 5. Finally, the order U is depicted in Fig. 6. As it is easily seen in the figures, although ([0, e], T ∗ ) and ([e, 1], S∗ ) are lattices, (L, U ) is not. Proposition 15. Let L be a chain and U ∈ U (e) such that e ∈ L{0, 1} and HU be the set of all idempotent elements of U. Then, (HU , U ) is a chain.
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Fig. 4. ([0, e], T ∗ ). Table 4 The t-conorm S∗ . S∗
e
1
e 1
e 1
1 1
Fig. 5. ([e, 1], S∗ ).
Fig. 6. (L, U ).
Proof. Let a, b ∈ HU . Since L is a chain, for any a, b ∈ L either a ≤ b or b ≤ a. Suppose that a ≤ b. If a, b ∈ [0, e], then
a = U (a, a) ≤ U (a, b) ≤ U (a, e) = a, whence we have that U (a, b) = a, that is a U b. If a, b ∈ [e, 1], then
b = U (e, b) ≤ U (a, b) ≤ U (b, b) = b, whence we obtain that U (a, b) = b, which shows that a U b. Let (a, b) ∈ [0, e]2 and (a, b) ∈ [e, 1]2 . Since a ≤ b, it must be that a ≤ e ≤ b. So a U b. Proposition 16. Let L be a bounded lattice, U ∈ U (e) such that e ∈ L{0, 1} and let HT ∗ and HS∗ the sets of all idempotent elements of T∗ and S∗ , respectively. If T∗ is ∨-distributive and S∗ is ∧-distributive, then (HT ∗ ∪ HS∗ , U ) is a bounded lattice. Proof. Let a, b ∈ HT ∗ ∪ HS∗ be arbitrary. Suppose that a, b ∈ HT ∗ . Then,
U (a, a) = T ∗ (a, a) = a and U (b, b) = T ∗ (b, b) = b. Since U (a, b) = U (a, b), a, b ∈ [0, e], it is clear that U(a, b)U a and U(a, b)U b. Thus, U (a, b) ∈ {a, b} . Let ∈ {a, b} . Then, U a and U b. Thus, there exist elements 1 , 2 of [0, e] such that
U (1 , a) = = U (2 , b).
U
U
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325
Table 5 The uninorm U on L. U
0
a
b
c
d
e
1
0 a b c d e 1
0 0 0 0 0 0 0
0 a a a d d 1
0 a b b d e 1
0 a b c d e 1
0 d d d 1 1 1
0 d e e 1 1 1
0 1 1 1 1 1 1
Also,
U (, a) = U (U (1 , a), a) = U (1 , U (a, a)) = U (1 , a) = and U (, b) = U (U (2 , b), b) = U (2 , U (b, b)) = U (2 , b) = . Since U (, U (a, b)) = U (U (, a), b) = U (, b) = , it is obtained that U U(a, b). Thus, a ∧U b = U (a, b). Since a, b ∈ [0, e], it is clear that a∨b ≤ e. Also, since a = T ∗ (a, a) ≤ T ∗ (a ∨ b, a ∨ b) and similarly b = T ∗ (b, b) ≤ T ∗ (a ∨ b, a ∨ b), we have that a ∨ b ≤ T ∗ (a ∨ b, a ∨ b) ≤ T ∗ (a ∨ b, e) = a ∨ b. Thus, T ∗ (a ∨ b, a ∨ b) = a ∨ b, whence a ∨ b ∈ HT ∗ . Moreover, a ∨ b ∈ {a, b}U . Indeed, since
a = T ∗ (a, a) ≤ T ∗ (a ∨ b, a) ≤ T ∗ (e, a) = a and b = T ∗ (b, b) ≤ T ∗ (a ∨ b, b) ≤ T ∗ (e, b) = b, we have that
a = T ∗ (a ∨ b, a) = U (a ∨ b, a) and b = T ∗ (a ∨ b, b) = U (a ∨ b, b). Then, it is obtained that a U a∨b and b U a∨b. Thus, a ∨ b ∈ {a, b}U . Let k ∈ {a, b}U . Then, a U k and b U k. If k ∈ [0, e], then there exist elements 1 , 2 of [0, e] such that
T ∗ (1 , k) = U (1 , k) = a and T ∗ (2 , k) = U (2 , k) = b. Since T∗ is ∨-distributive, we have that
a ∨ b = T ∗ (k, 1 ) ∨ T ∗ (k, 2 ) = T ∗ (k, 1 ∨ 2 ) = U (k, 1 ∨ 2 ), whence a∨b U k. Let k ∈ [0, e]. Since a U k and b U k, we have that a ≤ k and b ≤ k. Thus, a∨b ≤ k. Therefore, we obtain that a∨b U k. So, a ∨U b = a ∨ b. Let a, b ∈ HS∗ . It can be easily shown that a ∨U b = U (a, b) and a ∧U b = a ∧ b. Let a ∈ HT ∗ and b ∈ HS∗ . Then, it is clear that a ≤ e ≤ b. By the definition of U , we have that a U b. So, a ∧U b = a and a ∨U b = b. Similarly, if a ∈ HS∗ and b ∈ HT ∗ , then a ∧U b = b and a ∨U b = a. So, (HT ∗ ∪ HS∗ , U ) is a lattice. Moreover, since 0 ∈ HT ∗ and 1 ∈ HS∗ , (HT ∗ ∪ HS∗ , U ) is a bounded lattice. Corollary 17. Let L be a complete lattice and U ∈ U (e) such that e ∈ L{0, 1}. If T∗ is an infinitely ∨-distributive t-norm and S∗ is an inifinitely ∧-distributive t-conorm , then (HT ∗ ∪ HS∗ , U ) is a complete lattice. Remark 7. Let (L, ≤, 0, 1) be a bounded lattice and U ∈ U (e) such that e ∈ L{0, 1}. Since U =≤ when T∗ and S∗ are divisible , it is clear that the monotonicity w.r.t. U holds. But, in general, U need not satisfy the monotonicity w.r.t. the order U . For example: consider the lattice L = {0, a, b, c, d, e, 1} with 0 < a < b < c < d < e < 1 and define the function U: L2 → L as in Table 5. It can be easily shown that U is a uninorm on L with the neutral element c. It is clear that a U b since U (b, a) = a for a ≤ c. If U preserves the monotonicity with respect to U , then
d = U (a, e) U U (b, e) = e. But, there does not exist any element t ≥ c such that
U (d, t ) = e, whence dU e. So, it is not satisfied the monotonicity w.r.t. U .
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4. Ordering of n-uninorms A 2-uninorm (introduced by Akella [1]) is an operation which is increasing, associative and commutative on the unit interval with an absorbing element separating two subintervals having their own neutral elements. The importance of 2- uninorms is due to being the generalization of both nullnorms and uninorms. In this section, we define the order obtained from 2-uninorms on chains. By this way, we give more general form of the order U given in (1). This generalization gives a more general extension of our order by using n-uninorm introduced by Akella in [1]. Definition 12 ([6]). Let (L, ≤, 0, 1) be a chain. An operator F: L2 → L is called 2-uninorm if it is commutative, associative, increasing with respect to both variables and fulfilling
∀x ≤ k F (e, x) = x and ∀x ≥ k F ( f, x) = x, where e, k, f ∈ L with 0 ≤ e ≤ k ≤ f ≤ 1. By Uk(e, f) we denote the class of all 2-uninorms on chains. Definition 13. Let U2 ∈ Uk(e, f) . Define the following relation: For every x, y ∈ L,
⎧ ∃ ≤ e ⎪ ⎪ ⎪ ⎪ ∃ ⎨ m ∈ [e, k] x U 2 y :⇔ ∃n ∈ [k, f ] ⎪ ⎪ ⎪ ∃ p ∈ [ f, 1] ⎪ ⎩ x ≤ y,
such that U 2 (, y) = x,
when
x, y ∈ [0, e] or,
such that U 2 (x, m) = y,
when
such that U (y, n) = x,
when x, y ∈ [k, f ]
or,
such that U 2 (x, p) = y,
when
x, y ∈ [ f, 1]
or,
2
x, y ∈ [e, k] or, (4)
otherwise.
Proposition 18. The relation U 2 defined in (4) is a partial order on L. Proof. The proof is trivial by Proposition 2.
Remark 8. Let U2 ∈ Uk(e, f) on a chain L. Note that, • If 0 = k ≤ f = 1, then 2-uninorm U2 is a triangular norm T with neutral element f = 1. In this case, the orders U 2 and T coincide. • If 0 = e ≤ k = 1, then 2-uninorm U2 is a triangular conorm S with neutral element e = 0. In this case, the orders U 2 and S coincide. • If 0 = k ≤ f ≤ 1 or 0 ≤ e ≤ k = 1, then 2-uninorm U2 is a uninorm U with neutral element f or e. In this case, the orders U 2 and U coincide. • If e = 0 and f = 1, then 2-uninorm U2 is a nullnorm V with zero element k. In this case, the orders U 2 and V coincide. Definition 14 ([1]). Let (L, ≤, 0, 1) be a chain. Then, {e1 , e2 , . . . , en }z1 ,z2 ,...,zn−1 is called an n-neutral element of V if V (ei , x) = x for all x ∈ [zi−1 , zi ] for 0 = z0 < z1 < · · · < zn = 1 and ei ∈ [zi−1 , zi ], i = 1, 2, . . . , n. Definition 15 ([1]). Let L be a chain and a binary operator Un on L, is an n-uninorm if it is associative, monotone, non-decreasing in each variable and commutative and has an n-neutral element {e1 , e2 , . . . , en }z1 ,z2 ,...,zn−1 . Similarly, the order given in (4) can be generalized for n-uninorms as follows. Proposition 19. Let L be a chain and Un be an n-uninorm on L with an n-neutral element {e1 , e2 , . . . , en }z1 ,z2 ,...,zn−1 , i = 1, 2, . . . , n. Then, the relation given in (5)
⎧ ⎨∃ ∈ [zi−1 , ei ] x U n y :⇔ ∃m ∈ [ei , zi ] ⎩ x ≤ y,
such that U n (, y) = x, such that U (m, x) = y, n
when when
x, y ∈ [zi−1 , ei ] x, y ∈ [ei , zi ]
or, or,
(5)
otherwise,
is a partial order on L. 5. Conclusions We have introduced and discussed a partial order on a bounded lattice L from a uninorm on L. So, we have extended the T-partial (S-partial) order to a more general form. Also, we have determined the relationship between the algebraic structures obtained from the orders induced by a t-norm and its N-dual t-conorm. According to the underlying t-norm and t-conorm of a uninorm, we have characterized the order induced by the uninorm. We have studied the relationship between the algebraic structures obtained from the orders induced by the underlying t-norm and t-conorm and induced by a uninorm. Moreover, we have introduced a partial order by means of 2-uninorms and n-uninorms on chains. So, we have given the partial orders induced by operators which are more general than uninorms and nullnorms on chains.
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Acknowledgment The authors are very grateful to all referees for their helpful comments and valuable suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25]
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