Note on “On the extension of nullnorms and uninorms to fuzzy truth values” [Fuzzy Sets Syst. 352 (2018) 92-118]

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.1 (1-19)

Available online at www.sciencedirect.com

ScienceDirect Fuzzy Sets and Systems ••• (••••) •••–••• www.elsevier.com/locate/fss

Discussion

Note on “On the extension of nullnorms and uninorms to fuzzy truth values” [Fuzzy Sets Syst. 352 (2018) 92-118] Wei Zhang a,b , Bao Qing Hu a,b,∗ a School of Mathematics and Statistics, Wuhan University, Wuhan 430072, PR China b Computational Science, Hubei Key Laboratory, Wuhan University, Wuhan 430072, PR China

Received 3 June 2019; received in revised form 21 August 2019; accepted 7 September 2019

Abstract Xie recently extended nullnorms and uninorms to fuzzy truth values and investigated whether the extended nullnorms and extended uninorms form respectively type-2 nullnorms and type-2 uninorms on the algebra of fuzzy truth values. In this note, we show by counterexamples that there exist some flaws in a previous paper by Xie [Fuzzy Sets Syst. 352 (2018) 92-118], such as Corollaries 2.1, 3.1, 4.1 and 4.2, Lemmas 3.3, 3.4, 3.5, 3.6, 3.7 and 3.8, Propositions 3.4, 3.5, 3.10, 3.11, 4.1 and 4.4, Theorems 3.1, 3.3, 4.1 and 4.3, and then we provide the correct versions. In addition, Proposition 3.7 in that paper is correct, but its proof contains a flaw. We point out the flaw and give a more general result than Proposition 3.7 in that paper. © 2019 Elsevier B.V. All rights reserved. Keywords: Nullnorms; Uninorms; Fuzzy truth values; Extended nullnorms; Extended uninorms; Type-2 fuzzy sets

1. Introduction Type-2 fuzzy sets were introduced by Zadeh in 1975 [31], as an extension of type-1 fuzzy sets, and have been widely investigated in mathematical theory (see, e.g. [9–11,17,18,25,32]) and engineering application. For example, type-2 fuzzy sets have applications in approximation [28], control [1,5,24], decision making [4], clustering [16], databases [19] and so on. The membership grades of type-2 fuzzy sets are functions from [0, 1] to [0, 1], which are referred to as fuzzy truth values. Recently, operations on type-2 fuzzy sets become a hot research topic, such as type-2 t-norms (see [7,13,14,20,26]), aggregation operations (see [21,22,27]), negations [12,23] and type-2 implications [8,28]. In particular, nullnorms [2] and uninorms [30] are aggregation operations [3] with neutral elements and absorbing elements on [0, 1], respectively. They are generalizations of t-norms and t-conorms as well. Naturally, in [29] Xie extended nullnorms and uninorms to fuzzy truth values and investigated whether the extended nullnorms and extended uninorms form respectively type-2 DOI of original article: https://doi.org/10.1016/j.fss.2018.03.002. * Corresponding author at: School of Mathematics and Statistics, Wuhan University, Wuhan 430072, PR China.

E-mail addresses: [email protected] (W. Zhang), [email protected] (B.Q. Hu). https://doi.org/10.1016/j.fss.2019.09.009 0165-0114/© 2019 Elsevier B.V. All rights reserved.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.2 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

2

nullnorms and type-2 uninorms on the algebra of fuzzy truth values. The work in [29] is interesting, but some of the results are incorrect. This note is organized as follows. In Section 2, we recall some essential concepts and properties. In Section 3, we point out the incorrect results in [29] by counterexamples and provide the correct versions in turn. In addition, Proposition 3.7 in [29] is correct, but its proof contains a flaw. We point out the flaw and give a more general result than Proposition 3.7 in [29]. In the final section, our research is concluded. 2. Preliminary In this section, we recall some essential requirements which are used in the sequel. The notations and terminologies used in this paper are as consistent as possible with those in [29]. 2.1. Operations on [0, 1] In this subsection, we recall some basic knowledge about t-norms, t-conorms, nullnorms and uninorms. Definition 2.1 ([15,29]). 1) A binary operation T : [0, 1]2 → [0, 1] is called a triangular norm (t-norm for short) if it is commutative, associative, increasing in each variable and has 1 as neutral element, i.e., T (x, 1) = x for any x ∈ [0, 1]. 2) A binary operation S : [0, 1]2 → [0, 1] is called a triangular conorm (t-conorm for short) if it is commutative, associative, increasing in each variable and has 0 as neutral element, i.e., S(x, 0) = x for any x ∈ [0, 1]. 3) A t-norm T (or t-conorm S) is said to be continuous if it is continuous with respect to each variable. 4) A t-norm T (or t-conorm S) is said to be positive if T (x, y) = 0 ⇔ x = 0 or y = 0 (or S(x, y) = 1 ⇔ x = 1 or y = 1). The following are some basic t-norms and t-conorms: TM (x, y) = min(x, y),

SM (x, y) = max(x, y),

TP (x, y) = xy,

SP (x, y) = x + y − xy,

TL (x, y) = (x + y − 1) ∨ 0,  0, if (x, y) ∈ [0, 1)2 , TD (x, y) = x ∧ y, otherwise,

SL (x, y) = (x + y) ∧ 1,  1, if (x, y) ∈ (0, 1]2 , SD (x, y) = x ∨ y, otherwise.

Among the above t-norms and t-conorms, TM , TP , SM and SP are continuous and positive. Definition 2.2 (See Definition 2.2 in [29]). A nullnorm is a binary operator F : [0, 1]2 → [0, 1], which is commutative, associative, increasing in each variable and there exists some element k ∈ [0, 1] (called absorbing element of F ) such that F (k, x) = k for all x ∈ [0, 1], and F (0, x) = x for all x  k and F (1, x) = x for all x  k. Clearly, F is a t-norm if k = 0, and a t-conorm if k = 1. If k ∈ (0, 1), then F is called proper nullnorm. Theorem 2.1 (See Theorem 2.1 in [29]). If a nullnorm F has an absorbing element k different from 0, 1, then ⎧ x y ⎪ if (x, y) ∈ [0, k]2 , ⎨kSF ( k , k ), y−k x−k F (x, y) = k + (1 − k)TF ( 1−k , 1−k ), if (x, y) ∈ [k, 1]2 , ⎪ ⎩ k, otherwise, where TF and SF are a t-norm and a t-conorm, respectively. For any nullnorm F , we denote it by F = (k, SF , TF ), where TF and SF are called the underlying t-norm and underlying t-conorm of F , respectively. In particular, F = (k, ∨, ∧) is the unique idempotent nullnorm with absorbing element k.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.3 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

3

Definition 2.3 (See Definition 2.3 in [29]). A binary function U : [0, 1]2 → [0, 1] is called a uninorm if it is commutative, associative, increasing in each variable and there exists some element e ∈ [0, 1] (called neutral element of U ) such that U (x, e) = x for all x ∈ [0, 1]. Obviously, uninorm U is a t-norm if e = 1, and a t-conorm if e = 0. U is called proper uninorm if its neutral element e ∈ (0, 1). Theorem 2.2 ([6,15]). Let U be a uninorm with neutral element e ∈ [0, 1]. 1) If U (0, 1) = 0 and e ∈ (0, 1), then U (·, 1) is continuous on [0, e) if and only if there exists a t-norm TU and a t-conorm SU such that ⎧ x y ⎪ if (x, y) ∈ [0, e]2 , ⎨eTU ( e , e ), y−e x−e U (x, y) = e + (1 − e)SU ( 1−e , 1−e ), if (x, y) ∈ [e, 1]2 , ⎪ ⎩ min(x, y), otherwise. 2) If U (0, 1) = 1 and e ∈ (0, 1), then U (·, 0) is continuous on (e, 1] if and only if there exists a t-norm TU and a t-conorm SU such that ⎧ x y ⎪ if (x, y) ∈ [0, e]2 , ⎨eTU ( e , e ), y−e x−e U (x, y) = e + (1 − e)SU ( 1−e , 1−e ), if (x, y) ∈ [e, 1]2 , ⎪ ⎩ max(x, y), otherwise. In Theorem 2.2, the set of uninorms in case 1) is denoted by Umin and the set of uninorms in case 2) by Umax . A uninorm U is denoted by (e, TU , SU ) if it is in Umin or Umax , where TU and SU are called the underlying t-norm and underlying t-conorm of U , respectively. In this work, we only consider uninorms that are in Umin or Umax . More types of uninorms can be found in [6,15]. 2.2. Fuzzy truth values In this subsection, we introduce fuzzy truth values. Fuzzy truth values are functions from [0, 1] to [0, 1]. The set of all fuzzy truth values is denoted by F , that is, F = {f | f : [0, 1] → [0, 1]}. Definition 2.4 ([14,29]). Let ∗ be a binary operation on [0, 1]. We assume that operation ∗ : F 2 → F as  (f ∗ g)(z) = (f (x) ∧ g(y)),



∅ = 0 and define the extended (1)

z=x∗y

where x, y, z ∈ [0, 1] and f, g ∈ F . In particular, we denote ∗ = if ∗ = ∧ and denote ∗ =  if ∗ = ∨. Namely,  (f g)(z) = (f (x) ∧ g(y)),

(2)

z=x∧y

(f  g)(z) =



(f (x) ∧ g(y)).

(3)

z=x∨y

In order to facilitate the operations  and , we introduce two auxiliary operations. For f ∈ F , let   f L (x) = f (y), f R (x) = f (y). yx

(4)

yx

Obviously, f = f L if and only if f is an increasing function and f = f R if and only if f is a decreasing function. Denote the height of f by

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.4 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

4

f sup =



f (x).

x∈[0,1]

It is clear that f L (1) = f sup and f R (0) = f sup for any f ∈ F . Denote f LR = (f L )R and f RL = (f R )L . One can check that f LR = f RL and they are constant functions with value f sup . For each f, g ∈ F , the pointwise join f ∨ g and the pointwise meet f ∧ g are respectively given by (f ∨ g)(x) = f (x) ∨ g(x) and (f ∧ g)(x) = f (x) ∧ g(x) for all x ∈ [0, 1]. Moreover, the pointwise order relation  in F is given by f  g if f (x)  g(x) for all x ∈ [0, 1]. Clearly, f  f L and f  f R for all f ∈ F . In [25], Walker and Walker proved that for each f, g ∈ F , f g = (f ∧ g R ) ∨ (f R ∧ g) = (f ∨ g) ∧ (f R ∧ g R ),

(5)

f  g = (f ∧ g ) ∨ (f ∧ g) = (f ∨ g) ∧ (f ∧ g ).

(6)

L

L

L

L

Here are some basic properties. Corollary 2.1 ([9,25]). Let f, g, h ∈ F . The following properties hold: 1) f f = f, f  f = f (idempotent laws). 2) f g = g f, f  g = g  f (commutative laws). 3) f (g h) = (f g) h, f  (g  h) = (f  g)  h (associative laws). 4) f (f  g) = f  (f g). 5) (f g)R = f R g R = f R ∧ g R , (f g)L = f L g L = (f L ∧ g LR ) ∨ (f LR ∧ g L ). 6) (f  g)L = f L  g L = f L ∧ g L , (f  g)R = f R  g R = (f R ∧ g RL ) ∨ (f RL ∧ g R ). Definition 2.5 (See Definition 2.5 in [29]). A fuzzy truth value f ∈ F is said to be 1) normal if f (x) = 1. The set of all normal fuzzy truth values is denoted by FN . x∈[0,1]

2) convex on [a, b] if for all a  x  z  y  b, f (z)  f (x) ∧ f (y), where [a, b] ⊆ [0, 1]. Analogously one can define convexity in an open subinterval of [0, 1]. If [a, b] = [0, 1], then it also called convex for short. The set of all convex fuzzy truth values is denoted by FC . Let us denote FN C as the set of all convex and normal fuzzy truth values. In [29], the author frequently used the undefined terms that h is convex on some half-open subinterval of [0, 1]. For example, in Corollary 2.1 of [29], the terms “h is convex on (0, 1]” and “h is convex on [0, 1)” were used. We imitate the above definition and extend it to any subinterval of [0, 1]. Definition 2.6. A fuzzy truth value f ∈ F is said to be convex on /a, b/ if for all x, y, z ∈ /a, b/ with x  z  y, f (z)  f (x) ∧ f (y), where /a, b/ is a subinterval of [0, 1]. It is well-known that f = f L ∧ f R if f is convex [25]. Next, we introduce the definition of non-convex elements (resp. convex elements) of f . Definition 2.7 ([32,33]). Let f ∈ F and x ∈ [0, 1]. Then x is called a non-convex element of f if f L (x) ∧ f R (x) > f (x), x is called a convex element of f if f L (x) ∧ f R (x) = f (x). The set of all non-convex elements of f is denoted by C c (f ) and the set of all convex elements of f is denoted by C(f ). Obviously, f is convex if C(f ) = [0, 1] (or C c (f ) = ∅). The following remark illustrates the relationship between Definitions 2.6 and 2.7. Remark 2.1. For each f ∈ F , if /a, b/ ⊆ C(f ), then f is convex on /a, b/. But not vice verse. Next, we show this claim. Let x, y, z ∈ /a, b/ such that x  z  y. Since /a, b/ ⊆ C(f ), we have f (x) ∧ f (y) ∧ f (z) = f L (x) ∧ f R (x) ∧ f L (y) ∧ f R (y) ∧ f L (z) ∧ f R (z) = f (x) ∧ f (x) ∧ f (y) ∧ f (y) L

R

= f (x) ∧ f (y),

L

R

by Definition 2.7

by monotonicity of f L and f R

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.5 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

5

which implies f (z)  f (x) ∧ f (y). Therefore, f is convex on /a, b/. Conversely, let  1, if x = 0, 1, f (x) = 0, otherwise. Denote the above function by f01 . By Definition 2.6, we know that f01 is convex on any subinterval of [0, 1] except itself. By Definition 2.7, we have C(f01 ) = {0, 1}. Therefore, for the interval (0, 1], f01 is convex on it, but it is not a subset of C(f01 ). Note that, in the proof of Corollary 2.1 in [29] Xie claimed that if h is convex on (0, 1], then h is convex on [0, 1]. However, f01 is a counterexample. In fact, 0, 1 ∈ C(f ) for all f ∈ F . Therefore, f is convex (on [0, 1]) if (0, 1) ⊆ C(f ) ⊆ [0, 1]. 2.3. Type-2 operations In this subsection, we introduce type-2 nullnorms and type-2 uninorms that were defined by Xie in [29]. In [25], two partial orders and  on F are respectively defined by f g if f g = f and f  g if f  g = g. In general, partial orders and  are not the same. However, they coincide in A ⊆ F if (A, , ) is a lattice. For example, because FN C is a distributive lattice (see [25]), the partial orders and  coincide in FN C . Next, we introduce two notations throughout this paper. For any constant k ∈ [0, 1], we define fuzzy truth value k as  1, if x = k, k(x) = 0, otherwise, and denote k as the constant function with value k. Definition 2.8 (See Definition 2.6 in [29]). Let A = (A, 0, 1, , ), where A ⊆ F . A function  : A2 → A is called a type-2 nullnorm over A, if it is commutative, associative, increasing in each variable with at least one of the partial orders and  and there exists k ∈ A, called the absorbing element of , such that f  k = k for all f ∈ A, f  0 = f for all f k (or f  k) and f  1 = f for all k f (or k  f ). Definition 2.9 (See Definition 2.7 in [29]). Let A = (A, 0, 1, , ), where A ⊆ F . A function • : A2 → A is called a type-2 uninorm over A, if it is commutative, associative, increasing in each variable with at least one of the partial orders and  and there exists e ∈ A, called the neutral element of •, such that f • e = f for all f ∈ A. 3. Counterexamples and corrections In what follows, All t-norms T and t-conorms S are required to be continuous, including underlying t-norms and underlying t-conorms of nullnorms (resp. uninorms). In this section, we show by counterexamples that Corollaries 2.1, 3.1, 4.1 and 4.2, Lemmas 3.3, 3.4, 3.5, 3.6, 3.7 and 3.8, Propositions 3.4, 3.5, 3.10, 3.11, 4.1 and 4.4, Theorems 3.1, 3.3, 4.1 and 4.3 in [29] contain some flaws. And then we provide the correct versions. In addition, Proposition 3.7 in [29] is correct, but its proof contains a flaw. We point out the flaw and give a more general result than Proposition 3.7 in [29]. 3.1. The case of t-norms and t-conorms In this subsection, we show by counterexample that Corollary 2.1 in [29] contains some flaws and then we provide the correct version. Corollary 2.1 in [29]: Let T be a positive t-norm and S be a positive t-conorm. A ⊆ FN and f, g, h ∈ A. (i). ((f g) T h)(z) = ((f T h) (g T h))(z) or

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.6 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

6

((f  g) T h)(z) = ((f T h)  (g T h))(z) for any f, g ∈ A and z > 0 if and only if h is convex on (0, 1]. (ii). ((f g) S h)(z) = ((f S h) (g S h))(z) or ((f  g) S h)(z) = ((f S h)  (g S h))(z) for any f, g ∈ A and z < 1 if and only if h is convex on [0, 1). The following is a counterexample for Corollary 2.1 in [29]. Example 3.1. Let T = ∧. That is T = . It follows from the associativity of that ((f g) T h)(z) = ((f T h) (g T h))(z), for any f, g ∈ F and any z > 0. That is to say, even if h is not convex on (0, 1], this equality also holds. Therefore, the necessity of Corollary 2.1 (i) in [29] does not necessarily hold. Similarly, also the necessity of Corollary 2.1 (ii) in [29] does not necessarily hold. Note that, Theorem 2.4 in [29] played a key role in [29]. In fact, it has proved by Walker and Walker in Theorem 63 in [25]. However, Xie did not use it correctly. This theorem is as follows: Theorem 3.1 ([25]). Let T be a t-norm and S be a t-conorm. The following distributive laws hold for all f, g ∈ F if and only if h is convex: (f g) T h = (f T h) (g T h), (f g) S h = (f S h) (g S h),

(f  g) T h = (f T h)  (g T h), (f  g) S h = (f S h)  (g S h).

In fact, the following is a more exact expression of the above theorem. Theorem 3.2. Let T be a t-norm and S be a t-conorm. Then the following hold: 1) (f  g) T h = (f T h)  (g T h) for all f, g ∈ F if and only if h is convex. 2) (f g) S h = (f S h) (g S h) for all f, g ∈ F if and only if h is convex. 3) If h is convex, then for all f, g ∈ F , (f g) T h = (f T h) (g T h) and (f  g) S h = (f S h)  (g S h). Proof. It follows from the proof of Theorem 63 in [25]. 2 According to Example 3.1, we know that even if (f g) T h = (f T h) (g T h) and (f  g) S h = (f S h)  (g S h) for all f, g ∈ F , h is not necessarily convex. The following corollary is a more exact expression of Corollary 2.1 in [29]. Corollary 3.1 (The correction of Corollary 2.1 in [29]). Let T be a t-norm and S be a t-conorm, and let h ∈ F . 1) If h is convex on (0, 1], then ((f g) T h)(z) = ((f T h) (g T h))(z) for any f, g ∈ F and any z > 0. 2) If h is convex on [0, 1), then ((f  g) S h)(z) = ((f S h)  (g S h))(z) for any f, g ∈ F and any z < 1. Further, if T and S are positive, then 3) ((f  g) T h)(z) = ((f T h)  (g T h))(z) for any f, g ∈ F and any z > 0 if and only if h is convex. 4) ((f g) S h)(z) = ((f S h) (g S h))(z) for any f, g ∈ F and any z < 1 if and only if h is convex. Proof. We only prove statements 1) and 3). The proofs of statements 2) and 4) are similar to the proofs of statements 1) and 3), respectively.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.7 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

1) Let f, g ∈ F and z > 0. On the one hand,  ((f T h) (g T h))(z) =

7

(f (p) ∧ g(q) ∧ h(s) ∧ h(t))

T (p,s)∧T (q,t)=z





(f (p) ∧ g(q) ∧ h(s) ∧ h(t))

T (p,s)∧T (q,t)=z s=t



=

(f (p) ∧ g(q) ∧ h(t))

T (p∧q,t)=z

= ((f g) T h)(z). On the other hand, for any p, s, q, t with T (p, s) ∧ T (q, t) = z > 0, we have p, s, q, t ∈ (0, 1]. By the monotonicity of T , we get T (p ∧ q, s ∧ t)  T (p, s) ∧ T (q, t)  T (p ∧ q, s ∨ t). Further, because of the continuity of T , there exists a y ∈ [s ∧ t, s ∨ t] ⊆ (0, 1] such that T (p ∧ q, y) = T (p, s) ∧ T (q, t). Since h is convex on (0, 1], it follows from Definition 2.6 that h(y)  h(s ∧ t) ∧ h(s ∨ t) = h(s) ∧ h(t) and thus  ((f T h) (g T h))(z) = (f (p) ∧ g(q) ∧ h(s) ∧ h(t)) T (p,s)∧T (q,t)=z





(f (p) ∧ g(q) ∧ h(y))

T (p∧q,y)=z

= ((f g) T h)(z). Therefore, ((f g) T h)(z) = ((f T h) (g T h))(z). 3) Necessity: Since T is a positive t-norm, by calculations, one obtains that for each f, g, h ∈ F ,  ((f  g) T h)(0) = (f (p) ∧ g(q) ∧ h(y)) T (p∨q,y)=0

=[



(f (p) ∧ g(q) ∧ h(y))] ∨ [

p∨q=0, y∈[0,1]



(f (p) ∧ g(q) ∧ h(y))]

y=0, p∨q∈[0,1]

= [f (0) ∧ g(0) ∧ hsup ] ∨ [f sup ∧ g sup ∧ h(0)] and ((f T h)  (g T h))(0)  = (f (p) ∧ g(q) ∧ h(s) ∧ h(t)) T (p,s)∨T (q,t)=0

=[



(f (p) ∧ g(q) ∧ h(s) ∧ h(t))] ∨ [

p=q=0, s,t∈[0,1]

∨[



(f (p) ∧ g(q) ∧ h(s) ∧ h(t))]

p=t=0, s,q∈[0,1]

(f (p) ∧ g(q) ∧ h(s) ∧ h(t))] ∨ [

s=q=0, p,t∈[0,1]

 

(f (p) ∧ g(q) ∧ h(s) ∧ h(t))]

s=t=0, p,q∈[0,1]

= [f (0) ∧ g(0) ∧ hsup ] ∨ [f (0) ∧ h(0) ∧ g sup ] ∨ [g(0) ∧ h(0) ∧ f sup ] ∨ [f sup ∧ g sup ∧ h(0)] = [f (0) ∧ g(0) ∧ hsup ] ∨ [f sup ∧ g sup ∧ h(0)]. Further, for each f, g ∈ F , because ((f  g) T h)(z) = ((f T h)  (g T h))(z) for any z > 0, we get (f  g) T h = (f T h)  (g T h). Therefore, it follows from Theorem 3.2 1) that h is convex. Sufficiency: Since h is convex, by Theorem 3.2 1), we have that (f  g) T h = (f T h)  (g T h) for any f, g ∈ F . Therefore, ((f  g) T h)(z) = ((f T h)  (g T h))(z) for any f, g ∈ F and any z > 0. 2 Here are explanations about the changes from Corollary 2.1 in [29] to Corollary 3.1. Remark 3.1. 1) For the statement 3) (resp. the statement 4)) of Corollary 3.1, if the condition that h is convex is changed to that h is convex on (0, 1] (resp. [0, 1)), then its sufficiency does not necessarily hold.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.8 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

8

For example, let T = ∧ and h = f01 . Lemma 10 in [25] proved that for all f, g, h ∈ F , the following equalities hold: (f  g) h = (f R ∧ g RL ∧ h) ∨ (f RL ∧ g R ∧ h) ∨ (f ∧ g L ∧ hR ) ∨ (f L ∧ g ∧ hR ) (f h)  (g h) = (f ∧ g R

RL

∨ (f ∧ g

∧ h) ∨ (f

RL

RL

L

R

L

∧ h ∧ h ) ∨ (f RL ∧ g ∧ hR ∧ hL ). R

(7)

∧ g ∧ h) ∨ (f ∧ g ∧ h ) ∨ (f ∧ g ∧ h ) R

L

R

(8)

In particular, taking f (x) = x, g(x) = 1 − x for all x ∈ [0, 1], it is clear that ((f  g) T h)(0.4) = 0.4 < 0.6 = ((f T h)  (g T h))(0.4). A similar example can be given for the statement 4) of Corollary 3.1. 2) Let A ⊆ FN . Then the claims “((f  g) T h)(z) = ((f T h)  (g T h))(z) for any f, g ∈ A and any z > 0 if and only if h is convex” and “((f g) S h)(z) = ((f S h) (g S h))(z) for any f, g ∈ A and any z > 0 if and only if h is convex” are incorrect. For example, let A = {0, f01 , 1}. one can check that the subalgebra (A, , ) is a lattice. Let T = ∧ and h = f01 . According to Corollary 7.8.9 in [10], we know that (A, , ) is distributive. Consequently, ((f  g) T h)(z) = ((f T h) (g T h))(z) for any f, g ∈ A and any z > 0. Therefore, the necessity of the first claim does not necessarily hold. A similar example can be given for the second claim. 3.2. The case of proper nullnorms In this subsection, we show by counterexamples that Corollary 3.1, Lemmas 3.3, 3.4, 3.5, 3.6, 3.7 and 3.8, Propositions 3.4, 3.5, 3.10 and 3.11, Theorems 3.1 and 3.3 in [29] contain some flaws. And then we provide the correct versions. In addition, Proposition 3.7 in [29] is correct, but its proof contains a flaw. We point out the flaw and give a more general result than Proposition 3.7 in [29]. Proposition 3.1. Let F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, and let f, g ∈ F . Then for each z ∈ [0, k), ((f  g) F h)(z) = ((f ◦ φ −1  g ◦ φ −1 ) SF h ◦ φ −1 )(φ(z)), ((f F h)  (g F h))(z) = ((f ◦ φ −1 SF h ◦ φ −1 )  (g ◦ φ −1 SF h ◦ φ −1 ))(φ(z)), where φ(x) = xk , x ∈ [0, k]. And for each z ∈ (k, 1], ((f g) F h)(z) = ((f ◦ ϕ −1 g ◦ ϕ −1 ) TF h ◦ ϕ −1 )(ϕ(z)), ((f F h) (g F h))(z) = ((f ◦ ϕ −1 TF h ◦ ϕ −1 ) (g ◦ ϕ −1 TF h ◦ ϕ −1 ))(ϕ(z)), where ϕ(x) =

x−k 1−k , x

∈ [k, 1].

Proof. For the case z ∈ (k, 1], it follows from the proof of Lemma 3.3 in [29]. Similarly, we can prove the case z ∈ [0, k). 2 Lemma 3.3 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where TF is positive. ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and z ∈ (k, 1] if and only if h is convex on (k, 1]. The following is a counterexample for Lemma 3.3 in [29]. Example 3.2. Let A = FN , F = (0.5, ∨, ∧) and F be its extension. For any f, g ∈ A and any z ∈ (k, 1], by Proposition 3.1 we have . ((f g) F h)(z) = [(f ◦ ϕ −1 g ◦ ϕ −1 ) h ◦ ϕ −1 ](ϕ(z)) = D1 and . ((f F h) (g F h))(z) = [(f ◦ ϕ −1 h ◦ ϕ −1 ) (g ◦ ϕ −1 h ◦ ϕ −1 )](ϕ(z)) = E1 , where ϕ(x) = x−k 1−k , x ∈ [k, 1]. Because of the associativity of , we know that even if h is not convex on (k, 1], always D1 = E1 . Therefore, the necessity of Lemma 3.3 in [29] does not necessarily hold.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.9 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

9

Before correcting these flaws in Lemma 3.3 in [29], we need to introduce the following properties. Proposition 3.2. Let h ∈ F , k ∈ (0, 1), φ(x) = xk , x ∈ [0, k] and ϕ(x) = 1) If h is convex on [0, k), then h ◦ φ −1 is convex on [0, 1). 2) If [0, k) ⊆ C(h), then h ◦ φ −1 is convex. 3) If h is convex on (k, 1], then h ◦ ϕ −1 is convex on (0, 1]. 4) If (k, 1] ⊆ C(h), then h ◦ ϕ −1 is convex.

x−k 1−k ,

x ∈ [k, 1].

Proof. We only prove statements 1) and 2). The proofs of statements 3) and 4) are similar to the proofs of statements 1) and 2), respectively. 1) Let x, y, z ∈ [0, 1) such that x  z  y. Clearly, 0  φ −1 (x)  φ −1 (z)  φ −1 (y) < k. Since h is convex on [0, k), by Definition 2.6, we have h(φ −1 (z))  h(φ −1 (x)) ∧ h(φ −1 (y)), i.e., h ◦ φ −1 (z)  h ◦ φ −1 (x) ∧ h ◦ φ −1 (y). Therefore, h ◦ φ −1 is convex on [0, 1). 2) Let z ∈ [0, 1). By simple calculations,    (h ◦ φ −1 )L (z) = (h ◦ φ −1 )(y) = h(ky) = h(t) = yz

yz

tkz



h(t) = hL (φ −1 (z)).

tφ −1 (z)

Similarly, we get (h ◦ φ −1 )R (z) = hR (φ −1 (z)). Since φ −1 (z) ∈ [0, k) ⊆ C(h), it follows from Definition 2.7 that (h ◦ φ −1 )L (z) ∧ (h ◦ φ −1 )R (z) = hL (φ −1 (z)) ∧ hR (φ −1 (z)) = h(φ −1 (z)) = h ◦ φ −1 (z), i.e., z ∈ C(h ◦ φ −1 ). Therefore, [0, 1) ⊆ C(h ◦ φ −1 ), that is, h ◦ φ −1 is convex.

2

Lemma 3.1 (The correction of Lemma 3.3 in [29]). Let h ∈ F , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension. If h is convex on (k, 1], then ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ F and any z ∈ (k, 1]. Proof. Let f, g ∈ F and z ∈ (k, 1]. According to Proposition 3.1, we have . ((f g) F h)(z) = [(f ◦ ϕ −1 g ◦ ϕ −1 ) TF h ◦ ϕ −1 ](ϕ(z)) = D and . ((f F h) (g F h))(z) = [(f ◦ ϕ −1 TF h ◦ ϕ −1 ) (g ◦ ϕ −1 TF h ◦ ϕ −1 )](ϕ(z)) = E, −1 is convex on where ϕ(x) = x−k 1−k , x ∈ [k, 1]. Since h is convex on (k, 1], by Proposition 3.2 3), we have that h ◦ ϕ (0, 1]. Therefore, by Corollary 3.1 1) we get D = E, i.e., ((f F h) (g F h))(z) = ((f g) F h)(z). 2

Lemma 3.4 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension. If ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and z ∈ [0, k), then h is convex on [0, k). The following is a counterexample for Lemma 3.4 in [29]. Example 3.3. Let  1, if x = 0, 0.3, h(x) = 0, otherwise, A = {h}, F = (0.5, ∨, ∧) and F be its extension. Obviously, h is not convex on [0, 0.5). However, by the idempotency of , we have that ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and z ∈ [0, 0.5). Therefore, Lemma 3.4 in [29] is false. Before correcting these flaws in Lemma 3.4 in [29], we need to introduce the following properties.

JID:FSS AID:7737 /DIS

10

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.10 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

Proposition 3.3. Let h ∈ F , A ⊆ F , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension. 1) For each z ∈ [0, k), if 0, 1 ∈ A and ((f g) F h)(z) = ((f F h) (g F h))(z) for any f, g ∈ A, then z ∈ C(h). 2) For each z ∈ (k, 1], if 1, 1 ∈ A and ((f  g) F h)(z) = ((f F h)  (g F h))(z) for any f, g ∈ A, then z ∈ C(h). Proof. We only provide the proof of the statement 1), the proof of the statement 2) being analogous. 1) Let z ∈ [0, k). Then clearly, ((0 1) F h)(z) = ((0 F h) (1 F h))(z). In the proof of Lemma 3.4 in [29], Xie proved that (I ) = ((0 1) F h)(z) = h(z) and (I I ) = ((0 F h) (1 F h))(z) = hL (z) ∧ hR (z). From (I ) = (I I ), we get h(z) = hL (z) ∧ hR (z), i.e., z ∈ C(h).

2

Lemma 3.2 (The correction of Lemma 3.4 in [29]). Let h ∈ F , A ⊆ F such that 0, 1 ∈ A, F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension. If ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and any z ∈ [0, k), then [0, k) ⊆ C(h). Proof. Let z ∈ [0, k). It follows from Proposition 3.3 1) that z ∈ C(h). Thus, [0, k) ⊆ C(h).

2

Proposition 3.4 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where TF is positive. ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and z ∈ [0, k) ∪ (k, 1], then h is convex on [0, k) and (k, 1]. For Proposition 3.4 in [29], according to Examples 3.2 and 3.3, we know that it is incorrect. And in fact, after minor modifications to its conditions, we can only get that [0, k) ⊆ C(h) as the result of Lemma 3.2. Lemma 3.5 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where SF = SM . ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and z ∈ [0, k) if and only if h is convex on [0, k). For Lemma 3.5 in [29], Example 3.3 shows that its necessity does not necessarily hold. And the following example shows that its sufficiency does not necessarily hold. Example 3.4. Let f (x) = 1 − x, x ∈ [0, 1], ⎧  ⎪ ⎨2x, if x ∈ [0, 0.5], 1, g(x) = 0, if x ∈ (0.5, 1), and h(x) = ⎪ 0, ⎩ 1, if x = 1,

if x = 0, 0.5, otherwise,

and let A = {f, g, h}, F = (0.5, ∨, ∧) and F be its extension. Clearly, h is convex on [0, k). By calculations, ((f g) F h)(0.4) = (f F h)(0.4) = (f  h)(0.4) = 0.6 and



((f F h) (g F h))(0.4) =

(f (p) ∧ g(q) ∧ h(s) ∧ h(t))

F (p,s)∧F (q,t)=0.4



= [(

F (p,s)=0.4

∨ [(



F (q,t)=0.4

Also,



(f (p) ∧ h(s))) ∧ (

(g(q) ∧ h(t)))]

F (q,t)0.4

(g(q) ∧ h(t))) ∧ (



(f (p) ∧ h(s)))].

F (p,s)0.4

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.11 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••



(f (p) ∧ h(s)) = (f  h)(0.4) = 0.6,

F (p,s)=0.4

and

 F (q,t)=0.4

(g(q) ∧ h(t)) = (g  h)(0.4) = 0.8,



(g(q) ∧ h(t)) 

F (q,t)0.4



11



(g(q) ∧ h(t)) = 1,

F (q,t)0.4 q=t=0.5

(f (p) ∧ h(s)) 

F (p,s)0.4



(f (p) ∧ h(s)) = 1.

F (p,s)0.4 p=0, s=0.5

Thus ((f F h) (g F h))(0.4) = 0.8 > 0.6 = ((f g) F h)(0.4). Therefore, the sufficiency of Lemma 3.5 in [29] does not necessarily hold. Lemma 3.3 (The correction of Lemma 3.5 in [29]). Let h ∈ F , A ⊆ F such that 0, 1 ∈ A, F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where SF = SM . ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and any z ∈ [0, k) if and only if [0, k) ⊆ C(h). Proof. Necessity: It follows from Lemma 3.2. Sufficiency: Let f, g ∈ A and z ∈ [0, k). In the proof of Lemma 3.5 in [29], Xie proved that . ((f g) F h)(z) = [(f ◦ φ −1 g ◦ φ −1 ) SM h ◦ φ −1 ](φ(z)) = M and . ((f F h) (g F h))(z) = [(f ◦ φ −1 SM h ◦ φ −1 ) (g ◦ φ −1 SM h ◦ φ −1 )](φ(z)) = N, where φ(x) = xk , x ∈ [0, k]. Since [0, k) ⊆ C(h), by Proposition 3.2 2), we know that h ◦ φ −1 is convex. Therefore, by Corollary 3.1 4) we have M = N , i.e., ((f F h) (g F h))(z) = ((f g) F h)(z). 2 Proposition 3.5 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where SF = SM and TF is positive. ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and z ∈ [0, k) ∪ (k, 1] if and only if h is convex on [0, k) and (k, 1]. For Proposition 3.5 in [29], Examples 3.2 and 3.3 show that its necessity does not necessarily hold. And Example 3.4 shows that its sufficiency does not necessarily hold. Proposition 3.4 (The correction of Proposition 3.5 in [29]). Let h ∈ F , A ⊆ F , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where SF = SM . If [0, k) ⊆ C(h) and h is convex on (k, 1], then ((f F h)

(g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and any z ∈ [0, k) ∪ (k, 1]. Proof. It follows from Lemma 3.1 and the sufficiency of Lemma 3.3.

2

Theorem 3.1 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where SF = SM and TF is positive. ((f F h) (g F h))(z) = ((f g) F h)(z) for any f, g ∈ A and z ∈ [0, 1] if and only if h is convex on [0, k) and (k, 1]. For Theorem 3.1 in [29], Examples 3.2 and 3.3 show that its necessity does not necessarily hold. And Example 3.4 shows that its sufficiency does not necessarily hold. Theorem 3.3 (The correction of Theorem 3.1 in [29]). Let h ∈ F , A ⊆ F , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where SF = SM and TF is positive. If [0, k) ⊆ C(h) and h is convex on (k, 1], then (f F h) (g F h) = (f g) F h for any f, g ∈ A. Proof. It follows from Proposition 3.3 in [29] and Proposition 3.4.

2

Proposition 3.7 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where SF = SM and TF is positive. If f, g are normal and convex on [0, k) and (k, 1], then f F g is normal and convex on [0, k) and (k, 1]. And

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.12 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

12

⎧ −1 −1 ⎪ ⎨((f ◦ ϕ ) TF (g ◦ ϕ ))(ϕ(z)), (f F g)(z) = (f  g)(z) = (f (z) ∨ g(z)) ∧ f L (z) ∧ g L (z), ⎪ ⎩ L (f (k) ∨ g L (k)) ∧ (f R (k) ∨ g R (k)),

z > k, z < k, z = k,

where ϕ(x) = x−k 1−k , x ∈ [k, 1]. Proposition 3.7 in [29] is correct, but its proof contains some flaws. For example, in the proof of Proposition 3.7 in [29] Xie claimed that f ◦ ϕ −1 and g ◦ ϕ −1 are convex on (0, 1] and hence convex on [0, 1]. Obviously, by Remark 2.1, we know that the claim is incorrect. In fact, we can get a more general result than Proposition 3.7 in [29]. For this purpose we need to introduce the following propositions. Proposition 3.5. Let F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, and let f, g ∈ F . If f, g are convex on [0, k) and (k, 1], then f F g is convex on [0, k) and (k, 1]. Proof. Let x, y, z ∈ (k, 1] such that x  z  y. By calculations, (f F g)(x) ∧ (f F g)(y) = [



(f (x1 ) ∧ g(x2 ))] ∧ [

F (x1 ,x2 )=x



=



(f (y1 ) ∧ g(y2 ))]

F (y1 ,y2 )=y

(f (x1 ) ∧ f (y1 ) ∧ g(x2 ) ∧ g(y2 )).

F (x1 ,x2 )=x F (y1 ,y2 )=y

Since F (x1 , x2 ) = x > k and F (y1 , y2 ) = y > k, we have x1 , x2 , y1 , y2 ∈ (k, 1]. Also, because F is continuous and F (x1 , x2 ) = x  z  y = F (y1 , y2 ), there exist z1 ∈ [x1 ∧ y1 , x1 ∨ y1 ] and z2 ∈ [x2 ∧ y2 , x2 ∨ y2 ] such that z = F (z1 , z2 ). Further, because f, g are convex on (k, 1], we obtain that f (z1 )  f (x1 ∧ y1 ) ∧ f (x1 ∨ y1 ) = f (x1 ) ∧ f (y1 ) and g(z2 )  g(x2 ∧ y2 ) ∧ g(x2 ∨ y2 ) = g(x2 ) ∧ g(y2 ). Therefore,   (f (x1 ) ∧ f (y1 ) ∧ g(x2 ) ∧ g(y2 ))  (f (z1 ) ∧ g(z2 )) = (f F g)(z), F (x1 ,x2 )=x F (y1 ,y2 )=y

F (z1 ,z2 )=z

that is, (f F g)(x) ∧ (f F g)(y)  (f F g)(z). Thus f F g is convex on (k, 1]. Similarly, we can prove that f F g is convex on [0, k). 2 Proposition 3.6. Let F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, and let f, g ∈ F . Then ⎧ z < k, ((f ◦ φ −1 ) SF (g ◦ φ −1 ))(φ(z)), ⎪ ⎪ ⎪ ⎨ z > k, ((f ◦ ϕ −1 ) TF (g ◦ ϕ −1 ))(ϕ(z)), (f F g)(z) = ⎪ (f L (k) ∧ g R (k)) ∨ (f R (k) ∧ g L (k)) ∨ ((f ◦ φ −1 SF g ◦ φ −1 )(1)) ⎪ ⎪ ⎩ z = k, ∨((f ◦ ϕ −1 TF g ◦ ϕ −1 )(0)), where φ(x) = xk , x ∈ [0, k] and ϕ(x) =

x−k 1−k , x

∈ [k, 1]. Further, if SF and TF are positive, then

(f F g)(k) = (f L (k) ∧ g R (k)) ∨ (f R (k) ∧ g L (k)) = (f L (k) ∨ g L (k)) ∧ (f R (k) ∨ g R (k)). Proof. Let z ∈ [0, k). Then F (x, y) = z ⇔ SF (φ(x), φ(y)) = φ(z). In fact, if F (x, y) = z < k, we have x, y < k. By Theorem 2.1, we get z = F (x, y) = kSF ( xk , yk ). Thus SF ( xk , yk ) = kz i.e., SF (φ(x), φ(y)) = φ(z). On the other hand, if SF (φ(x), φ(y)) = φ(z) < 1, we have φ(x), φ(y) < 1, i.e., x, y < k. Thus F (x, y) = kSF ( xk , yk ) = kSF (φ(x), φ(y)) = kφ(z) = z.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.13 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

Therefore,



((f ◦ φ −1 ) SF (g ◦ φ −1 ))(φ(z)) =

13

(f ◦ φ −1 (u) ∧ g ◦ φ −1 (v))

SF (u,v)=φ(z)



=

(f (x) ∧ g(y)) by denoting x = φ −1 (u) and y = φ −1 (v)

SF (φ(x),φ(y))=φ(z)

=



(f (x) ∧ g(y))

F (x,y)=z

= (f F g)(z), i.e., (f F g)(z) = ((f ◦ φ −1 ) SF (g ◦ φ −1 ))(φ(z)). Similarly, for z ∈ (k, 1], we can prove that (f F g)(z) = ((f ◦ ϕ −1 ) TF (g ◦ ϕ −1 ))(ϕ(z)). In the proof of Proposition 3.8 in [29], Xie proved that (f F g)(k) = (f L (k) ∧ g R (k)) ∨ (f R (k) ∧ g L (k)) ∨ ((f ◦ φ −1 SF g ◦ φ −1 )(1)) ∨ ((f ◦ ϕ −1 TF g ◦ ϕ −1 )(0)). Further, if SF and TF are positive, in the proof of Proposition 3.7 in [29] Xie proved that (f F g)(k) = (f L (k) ∧ R g (k)) ∨ (f R (k) ∧ g L (k)) = (f L (k) ∨ g L (k)) ∧ (f R (k) ∨ g R (k)). 2 Proposition 3.7 (The generalization of Proposition 3.7 in [29]). Let F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, and let f, g ∈ F . If f, g are normal and convex on [0, k) and (k, 1], then f F g is normal and convex on [0, k) and (k, 1]. And ⎧ z < k, ((f ◦ φ −1 ) SF (g ◦ φ −1 ))(φ(z)), ⎪ ⎪ ⎪ ⎨((f ◦ ϕ −1 ) (g ◦ ϕ −1 ))(ϕ(z)), z > k, TF (f F g)(z) = L R R L −1 −1 ⎪ (f (k) ∧ g (k)) ∨ (f (k) ∧ g (k)) ∨ ((f ◦ φ SF g ◦ φ )(1)) ⎪ ⎪ ⎩ z = k, ∨((f ◦ ϕ −1 TF g ◦ ϕ −1 )(0)), where φ(x) = xk , x ∈ [0, k] and ϕ(x) =

x−k 1−k , x

∈ [k, 1].

Proof. In order to get the result, by using Propositions 3.5 and 3.6, we only need to prove that if f, g are normal, then f F g is normal. In fact, this claim has been proved by Torres-Blanc et al. in Proposition 6 in [22]. 2 Similarly, we can get the following facts. Lemma 3.6 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where SF is positive. ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ A and z ∈ [0, k) if and only if h is convex on [0, k). Lemma 3.4 (The correction of Lemma 3.6 in [29]). Let h ∈ F , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension. If h is convex on [0, k), then ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ F and any z ∈ [0, k). Lemma 3.7 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension. If ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ A and z ∈ (k, 1], then h is convex on (k, 1]. Lemma 3.5 (The correction of Lemma 3.7 in [29]). Let h ∈ F , A ⊆ F such that 1, 1 ∈ A, F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension. If ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ A and any z ∈ (k, 1], then (k, 1] ⊆ C(h). Proposition 3.10 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where SF is positive. ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ A and z ∈ [0, k) ∪ (k, 1], then h is convex on [0, k) and (k, 1]. Proposition 3.10 in [29] is false. And in fact, after minor modifications to its conditions, we can only get that (k, 1] ⊆ C(h) as the result of Lemma 3.5.

JID:FSS AID:7737 /DIS

14

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.14 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

Lemma 3.8 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension. If TF = TM , then ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ A and z ∈ (k, 1] if and only if h is convex on (k, 1]. Lemma 3.6 (The correction of Lemma 3.8 in [29]). Let h ∈ F , A ⊆ F such that 1, 1 ∈ A, F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where TF = TM . ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ A and any z ∈ (k, 1] if and only if (k, 1] ⊆ C(h). Proposition 3.11 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where TF = TM and SF is positive. ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ A and z ∈ [0, k) ∪ (k, 1] if and only if h is convex on [0, k) and (k, 1]. Proposition 3.8 (The correction of Proposition 3.11 in [29]). Let h ∈ F , A ⊆ F , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where TF = TM . If h is convex on [0, k) and (k, 1] ⊆ C(h), then ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ A and any z ∈ [0, k) ∪ (k, 1]. Theorem 3.3 in [29]: Let A ⊆ FN , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where TF = TM and SF is positive. ((f F h)  (g F h))(z) = ((f  g) F h)(z) for any f, g ∈ A and z ∈ [0, 1] if and only if h is convex on [0, k) and (k, 1]. Theorem 3.4 (The correction of Theorem 3.3 in [29]). Let h ∈ F , A ⊆ F , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension, where TF = TM and SF is positive. If h is convex on [0, k) and (k, 1] ⊆ C(h), then (f F h)  (g F h) = (f  g) F h for any f, g ∈ A. Corollary 3.1 in [29]: Let A = (A, 0, 1, , ) with A ⊆ F , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension over A. (i). If A ⊆ FCN , then F is a type-2 nullnorm over A with absorbing element k. (ii). If SF = SM and TF is positive, then F is a type-2 nullnorm with absorbing element k if and only if A is the set of normal fuzzy truth values which are convex on [0, k) and (k, 1]. (iii). If TF = TM and SF is positive, then F is a type-2 nullnorm with absorbing element k if and only if A is the set of normal fuzzy truth values which are convex on [0, k) and (k, 1]. The following is a counterexample for statements (ii) and (iii) of Corollary 3.1 in [29]. Example 3.5. Let A be the set of normal fuzzy truth values which are convex on [0, k) and (k, 1], F = (0.5, ∨, ∧) and F be its extension, and let f, g, h be as in Example 3.4. One can check that f, g, h ∈ A and f g. However, (f F h)(0.4) = 0.6 < 0.8 = ((f F h) (g F h))(0.4), i.e., f F h  g F h. Similarly, there exist f1 , g1 , h1 ∈ A and f1  g1 such that f1 F h1  g1 F h1 . Therefore, F is not a type-2 nullnorm over A. Corollary 3.2 (The correction Corollary 3.1 in [29]). Let A = (A, 0, 1, , ) with A ⊆ F , F = (k, SF , TF ) be a type-1 nullnorm with k ∈ (0, 1) and F be its extension over A. 1) If A ⊆ FCN , then F is a type-2 nullnorm over A with absorbing element k. 2) If SF = SM , TF is positive and A = {h ∈ FN : [0, k) ⊆ C(h) and h is convex on (k, 1]}, then F is a type-2 nullnorm with absorbing element k. 3) If TF = TM , SF is positive and A = {h ∈ FN : (k, 1] ⊆ C(h) and h is convex on [0, k)}, then F is a type-2 nullnorm with absorbing element k. Proof. 1) It follows from Corollary 3.1 (i) in [29]. 2) In order to prove F is a type-2 nullnorm (with absorbing element k) over A, by using Propositions 3.1 and 3.2 in [29] and Theorem 3.3, we only need to prove the closure property (i.e., f F g ∈ A for all f, g ∈ A). Let f, g ∈ A. According to Proposition 6 in [22], we have f F g ∈ FN . Also, by Proposition 3.5, we have that f F g is convex on (k, 1]. Next, we prove that [0, k) ⊆ C(f F g) as follows.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.15 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

15

Let z ∈ [0, k). Since SF = SM and [0, k) ⊆ C(f ), C(g), by calculations, we have (f F g)L (z) ∧ (f F g)R (z) = (f L F g L )(z) ∧ (f R F g R )(z)

by Proposition 4 in [22]

= (f  g )(z) ∧ (f  g )(z)

by Proposition 3.6

= f (z) ∧ g (z) ∧ [(f (z) ∧ g

(z)) ∨ (f RL (z) ∧ g R (z))]

L

L

L

R

L

R

R

RL

= f (z) ∧ g (z) ∧ (f (z) ∨ g (z)) L

L

R

R

byf

RL

(z) = g

RL

by Corollary 2.1 6)

(z) = 1

= (f (z) ∧ f (z) ∧ g (z)) ∨ (f (z) ∧ g (z) ∧ g (z)) L

R

L

L

= (f (z) ∧ g L (z)) ∨ (f L (z) ∧ g(z)) = (f  g)(z)

L

R

by z ∈ C(f ), C(g)

by Eq. (6)

= (f F g)(z), i.e., z ∈ C(f F g). Therefore, f F g ∈ A. 3) Similarly, we can prove the closure of F in A. Further, by using Propositions 3.1 and 3.2 in [29] and Theorem 3.4, one can get that F is a type-2 nullnorm with absorbing element k. 2 Example 3.1 in [29] pointed out that if F = (k, SF , TF ) is a type-1 nullnorm with SF = SM and TF = TM and A is the set of normal fuzzy truth values which are convex on [0, k) and (k, 1], then F is a type-2 nullnorm over A. However, by Example 3.5 we know that Example 3.1 in [29] is incorrect. 3.3. The case of proper uninorms In this subsection, we show by counterexamples that Corollaries 4.1 and 4.2, Propositions 4.1 and 4.4, Theorems 4.1 and 4.3 in [29] contain some minor flaws and then we provide the correct versions. Proposition 4.1 in [29]: Let A ⊆ F , U = (e, TU , SU ) ∈ Umax be a type-1 uninorm with neutral element e ∈ (0, 1) and U be its extension. ((f U h)  (g U h))(z) = ((f  g) U h)(z) for any f, g ∈ A and z ∈ [0, e] if and only if h is convex on [0, e]. The following is a counterexample for Proposition 4.1 in [29]. Example 3.6. Let  1, if x = 0, 0.3, h(x) = 0, otherwise, A = {h}, U = (0.5, ∧, ∨) ∈ Umax and U be its extension. Obviously, h is not convex on [0, e]. However, by the idempotency of , we have that ((f U h)  (g U h))(z) = ((f  g) U h)(z) for any f, g ∈ A and z ∈ [0, e]. Therefore, the necessity of Proposition 4.1 in [29] is false. Before correcting these flaws in Proposition 4.1 in [29], we need to introduce the following properties. Proposition 3.9. Let h ∈ F , A ⊆ F , U = (e, TU , SU ) be a type-1 uninorm with neutral element e ∈ (0, 1) and U be its extension, and let   1, if x ∈ [0, e], 1, if x ∈ [e, 1], fe (x) = fe  = 0, otherwise, 0, otherwise. 1) If U ∈ Umax , e, fe ∈ A and ((f U h)  (g U h))(z) = ((f  g) U h)(z) for any f, g ∈ A and any z ∈ [0, e], then h is convex on [0, e]. 2) If U ∈ Umin , e, fe ∈ A and ((f U h) (g U h))(z) = ((f g) U h)(z) for any f, g ∈ A and any z ∈ [e, 1], then h is convex on [e, 1]. Proof. We only provide the proof of the statement 1), the proof of the statement 2) being analogous.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.16 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

16

1) Let z ∈ [0, e]. Clearly, ((e U h)  (fe U h))(z) = ((e  fe ) U h)(z). By calculations,  (I ) = ((e  fe ) U h)(z) = (e U h)(z) = h(y) = h(z) U (e,y)=z

and (I I ) = ((e U h)  (fe U h))(z)  (h(s) ∧ h(t)) = U (e,s)∨U (q,t)=z

=[



s=z, U (q,t)z

= [h(z) ∧ Also,

 U (q,t)z

 U (q,t)=z

(h(s) ∧ h(t))]

sz, U (q,t)=z





h(t)] ∨ [hL (z) ∧

U (q,t)z

h(t) 



(h(s) ∧ h(t))] ∨ [



h(t)].

U (q,t)=z

h(t) = h(z). It is easy to show that

U (q,t)z q=e, t=z

h(t) =



h(t).

zte



In fact, because of the monotonicity of U , we have z = U (q, t)  U (e, t) = t  e. So

h(t) 

U (q,t)=z



h(t). On

zte

the other hand, for each t with z  t  e (i.e., TU (0, et ) = 0  ez  et = TU (1, et )  1), because of thecontinuity of TU , there exists a q ∈ [0, e] such that ez = TU ( qe , et ), which implies U (q, t) = eTU ( qe , et ) = z. So h(t)  U (q,t)=z  h(t). zte

Therefore,



(I I ) = h(z) ∨ [hL (z) ∧

h(t)] = hL (z) ∧

zte



From (I ) = (I I ), we have h(z) = hL (z) ∧



h(t).

zte

h(t). Based on this result, we prove that h is convex on [0, e].

zte

Let x, y, z ∈ [0, e] such that x  z  y. Then    h(x) ∧ h(y) ∧ h(z) = hL (x) ∧ h(t1 ) ∧ hL (y) ∧ h(t2 ) ∧ hL (z) ∧ h(t) xt1 e

= h (x) ∧ L



xt1 e

yt2 e

h(t1 ) ∧ h (y) ∧ L



zte

h(t2 )

yt2 e

= h(x) ∧ h(y), i.e., h(z)  h(x) ∧ h(y). Therefore, by Definition 2.6, we know that h is convex on [0, e].

2

Proposition 3.10 (The correction of Proposition 4.1 in [29]). Let h ∈ F , A ⊆ F such that e, fe ∈ A, U = (e, TU , SU ) ∈ Umax be a type-1 uninorm with neutral element e ∈ (0, 1) and U be its extension. ((f U h)  (g U h))(z) = ((f  g) U h)(z) for any f, g ∈ A and any z ∈ [0, e] if and only if h is convex on [0, e]. Proof. Necessity: It follows from Proposition 3.9. Sufficiency: It follows from the sufficiency of Proposition 4.1 in [29]. 2 Theorem 4.1 in [29]: Let A ⊆ F , U = (e, TU , SU ) ∈ Umax be a type-1 uninorm with neutral element e ∈ (0, 1) and SU = SM , and U be its extension. Then ((f U h)  (g U h))(z) = ((f  g) U h)(z) for any f, g ∈ A and z ∈ [0, 1] if and only if h is convex on [0, e]. For Theorem 4.1 in [29], Example 3.6 is a counterexample.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.17 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

17

Theorem 3.5 (The correction of Theorem 4.1 in [29]). Let h ∈ F , A ⊆ F such that e, fe ∈ A, U = (e, TU , SU ) ∈ Umax be a type-1 uninorm with neutral element e ∈ (0, 1) and SU = SM , and U be its extension. Then (f U h)  (g U h) = (f  g) U h for any f, g ∈ A if and only if h is convex on [0, e]. Proof. Necessity: It follows from Proposition 3.9. Sufficiency: It follows from the sufficiency of Theorem 4.1 in [29]. 2 Corollary 4.1 in [29]: Let A = (A, 0, 1, , ) with A ⊆ F , U = (e, TU , SU ) ∈ Umax be a type-1 uninorm and U be its extension. (i). If SU = SM , then U is a type-2 uninorm on A with neutral element e if and only if A is the set of fuzzy truth values convex on [0, e]. (ii). If A ⊆ FC and any fuzzy truth value in A equals 1 at e, then U is a type-2 uninorm on A with neutral element e. The following is a counterexample for the statement (i) of Corollary 4.1 in [29]. Example 3.7. Let  1, if x = 0, e, f0e (x) = 0, otherwise, A = {0, e, 1, f0e }, U = (0.5, ∧, ∨) ∈ Umax and U be its extension. Obviously, f0e is not convex on [0, e]. And it is easy to check that U is a type-2 uninorm on A with neutral element e. Therefore, for the statement (i) of Corollary 4.1 in [29], its necessity is false. Corollary 3.3 (The correction of Corollary 4.1 in [29]). Let A = (A, 0, 1, , ) with A ⊆ F , U = (e, TU , SU ) ∈ Umax be a type-1 uninorm and U be its extension over A. 1) If SU = SM and A is the set of fuzzy truth values which are convex on [0, e], then U is a type-2 uninorm on A with neutral element e. 2) If A ⊆ FC and any fuzzy truth value in A equals 1 at e, then U is a type-2 uninorm on A with neutral element e. Proof. It follows from Corollary 4.1 in [29].

2

The following can be similarly obtained for U = (e, TU , SU ) ∈ Umin . Proposition 4.4 in [29]: Let U = (e, TU , SU ) ∈ Umin be a type-1 uninorm with neutral element e ∈ (0, 1) and U be its extension over A. Then ((f U h) (g U h))(z) = ((f g) U h)(z) for any f, g ∈ A and z ∈ [e, 1] if and only if h is convex on [e, 1]. Proposition 3.11 (The correction of Proposition 4.4 in [29]). Let h ∈ F , A ⊆ F such that e, fe ∈ A, U = (e, TU , SU ) ∈ Umin be a type-1 uninorm with neutral element e ∈ (0, 1) and U be its extension. ((f U h) (g U h))(z) = ((f g) U h)(z) for any f, g ∈ A and any z ∈ [e, 1] if and only if h is convex on [e, 1]. Theorem 4.3 in [29]: Let U = (e, TU , SU ) ∈ Umin be a type-1 uninorm with TU = TM and U be its extension. Then (f U h) (g U h) = (f g) U h for any f, g ∈ A if and only if h is convex on [e, 1]. Theorem 3.6 (The correction of Theorem 4.3 in [29]). Let h ∈ F , A ⊆ F such that e, fe ∈ A, U = (e, TU , SU ) ∈ Umin be a type-1 uninorm with neutral element e ∈ (0, 1) and TU = TM , and U be its extension. Then (f U h) (g U h) = (f g) U h for any f, g ∈ A if and only if h is convex on [e, 1]. Corollary 4.2 in [29]: Let A = (A, 0, 1, , ), U = (e, TU , SU ) ∈ Umin be a type-1 uninorm and U be its extension. (i). If TU = TM , then U is a type-2 uninorm on A with neutral element e if and only if A is the set of fuzzy truth values convex on [e, 1].

JID:FSS AID:7737 /DIS

18

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.18 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

Table 1 The list of incorrect results in [29] and the theorems (propositions, lemmas, ...) of their correct forms. Incorrect results in [29]

Counterexamples

Correct versions

Corollary 2.1 Lemma 3.3 Lemma 3.4 Proposition 3.4 Lemma 3.5 Proposition 3.5 Theorem 3.1 Proposition 3.7 Lemma 3.6 Lemma 3.7 Proposition 3.10 Lemma 3.8 Proposition 3.11 Theorem 3.3 Corollary 3.1 Example 3.1 Proposition 4.1 Theorem 4.1 Corollary 4.1 Proposition 4.4 Theorem 4.3 Corollary 4.2

Example 3.1 Example 3.2 Example 3.3 Examples 3.2 and 3.3 Examples 3.3 and 3.4 Examples 3.2, 3.3 and 3.4 Examples 3.2, 3.3 and 3.4 Remark 2.1

Corollary 3.1 Lemma 3.1 Lemma 3.2 Lemma 3.2 Lemma 3.3 Proposition 3.4 Theorem 3.3 Proposition 3.7 Lemma 3.4 Lemma 3.5 Lemma 3.5 Lemma 3.6 Proposition 3.8 Theorem 3.4 Corollary 3.2

Example 3.5 Example 3.5 Example 3.6 Example 3.6 Example 3.7

Proposition 3.10 Theorem 3.5 Corollary 3.3 Proposition 3.11 Theorem 3.6 Corollary 3.4

(ii). If A ⊆ FC and any fuzzy truth value in A equals 1 at e, then U is a type-2 uninorm on A with neutral element e. Corollary 3.4 (The correction of Corollary 4.2 in [29]). Let A = (A, 0, 1, , ) with A ⊆ F , U = (e, TU , SU ) ∈ Umin be a type-1 uninorm and U be its extension over A. 1) If TU = TM and A is the set of fuzzy truth values which are convex on [e, 1], then U is a type-2 uninorm on A with neutral element e. 2) If A ⊆ FC and any fuzzy truth value in A equals 1 at e, then U is a type-2 uninorm on A with neutral element e. 4. Conclusions For Corollaries 2.1, 3.1, 4.1 and 4.2, Lemmas 3.3, 3.4, 3.5, 3.6, 3.7 and 3.8, Propositions 3.4, 3.5, 3.10, 3.11, 4.1 and 4.4, Theorems 3.1, 3.3, 4.1 and 4.3 in [29], this note illustrates by counterexamples that these results are incorrect and provides the correct versions in turn. In addition, Proposition 3.7 in [29] is correct, but its proof contains a flaw. We point out the flaw and give a more general result than Proposition 3.7 in [29]. To intuitively show our modifications to incorrect results in [29], please refer to Table 1. Acknowledgement The work described in this paper was supported by grants from the National Natural Science Foundation of China (Grant nos. 11971365 and 11571010). References [1] A.T. Azar, Overview of type-2 fuzzy logic systems, Int. J. Fuzzy Syst. Appl. 2 (4) (2012) 1–28. [2] T. Calvo, B. De Baets, J. Fodor, The functional equations of Frank and Alsina for uninorms and nullnorms, Fuzzy Sets Syst. 120 (3) (2001) 385–394. [3] T. Calvo, A. Kolesárová, M. Komorníková, R. Mesiar, Aggregation operators: properties, classes and construction methods, Stud. Fuzziness Soft Comput. 97 (2002) 3–104.

JID:FSS AID:7737 /DIS

[m3SC+; v1.304; Prn:19/09/2019; 14:31] P.19 (1-19)

W. Zhang, B.Q. Hu / Fuzzy Sets and Systems ••• (••••) •••–•••

19

[4] S.M. Chen, L.W. Lee, Fuzzy decision-making based on likelihood-based comparison relations, IEEE Trans. Fuzzy Syst. 18 (3) (2010) 613–628. [5] S. Coupland, R. John, Geometric type-1 and type-2 fuzzy logic systems, IEEE Trans. Fuzzy Syst. 15 (1) (2007) 3–15. [6] J.C. Fodor, R.R. Yager, A. Rybalov, Structure of uninorms, Int. J. Uncertain. Fuzziness Knowl.-Based Syst. 5 (1997) 411–427. [7] Z. Gera, J. Dombi, Exact calculations of extended logical operations on fuzzy truth values, Fuzzy Sets Syst. 159 (2008) 1309–1326. [8] Z. Gera, J. Dombi, Type-2 implications on non-interactive fuzzy truth values, Fuzzy Sets Syst. 159 (2008) 3014–3032. [9] J. Harding, C. Walker, E. Walker, Lattices of convex normal functions, Fuzzy Sets Syst. 159 (2008) 1061–1071. [10] J. Harding, C. Walker, E. Walker, The Truth Value Algebra of Type-2 Fuzzy Sets: Order Convolutions of Functions on the Unit Interval, CRC Press, 2016. [11] J. Harding, C. Walker, E. Walker, Overview of lattices of convex normal functions, Int. J. Intell. Syst. 31 (2016) 257–275. [12] P. Hernández, S. Cubillo, C. Torres-Blanc, Negations on type-2 fuzzy sets, Fuzzy Sets Syst. 252 (2014) 111–124. [13] P. Hernández, S. Cubillo, C. Torres-Blanc, On t-norms for type-2 fuzzy sets, IEEE Trans. Fuzzy Syst. 23 (4) (2015) 1155–1163. [14] B.Q. Hu, C.K. Kwong, On type-2 fuzzy sets and their t-norm operations, Inf. Sci. 255 (2014) 58–81. [15] E.P. Klement, R. Mesiar, E. Pap, Triangular Norms, Kluwer Academic Publishers, 2000. [16] O. Linda, M. Manic, General type-2 fuzzy C-means algorithm for uncertain fuzzy clustering, IEEE Trans. Fuzzy Syst. 20 (5) (2012) 883–897. [17] J.M. Mendel, R.I. John, Type-2 fuzzy sets made simple, IEEE Trans. Fuzzy Syst. 10 (2) (2002) 117–127. [18] M. Mizumoto, K. Tanaka, Some properties of fuzzy sets of type-2, Inf. Control 31 (1976) 312–340. [19] A. Niewiadomski, On finity, countability, cardinalities, and cylindric extensions of type-2 fuzzy sets in linguistic summarization of databases, IEEE Trans. Fuzzy Syst. 18 (3) (2010) 532–545. [20] J.T. Starczewski, Extended triangular norms, Inf. Sci. 179 (2009) 742–757. [21] Z. Takáˇc, Aggregation of fuzzy truth values, Inf. Sci. 271 (2014) 1–13. [22] C. Torres-Blanc, S. Cubillo, P. Hernández-Varela, Aggregation operators on type-2 fuzzy sets, Fuzzy Sets Syst. 324 (2017) 74–90. [23] C. Torres-Blanc, S. Cubillo, P. Hernández-Varela, New negations on the membership functions of type-2 fuzzy sets, IEEE Trans. Fuzzy Syst. (2018), https://doi.org/10.1109/TFUZZ.2018.2879033. [24] C. Wagner, H. Hagras, Toward general type-2 fuzzy logic systems based on zSlices, IEEE Trans. Fuzzy Syst. 18 (4) (2010) 637–660. [25] C. Walker, E. Walker, The algebra of fuzzy truth values, Fuzzy Sets Syst. 149 (2005) 309–347. [26] C. Walker, E. Walker, T-norms for type-2 fuzzy sets, in: 2006 IEEE International Conference on Fuzzy Systems, Canada, 2006, pp. 1235–1239. [27] C.Y. Wang, Notes on aggregation of fuzzy truth values, Inf. Sci. 296 (2015) 119–127. [28] C.Y. Wang, Type-2 fuzzy implications and fuzzy-valued approximation reasoning, Int. J. Approx. Reason. 102 (2018) 108–122. [29] A.F. Xie, On the extension of nullnorms and uninorms to fuzzy truth values, Fuzzy Sets Syst. 352 (2018) 92–118. [30] R.R. Yager, A. Rybalov, Uninorm aggregation operators, Fuzzy Sets Syst. 80 (1) (1996) 111–120. [31] L. Zadeh, The concept of a linguistic variable and its application to approximate reasoning, Inf. Sci. 8 (1975) 199–249. [32] W. Zhang, B.Q. Hu, A note on the lattice structure for subalgebras of the algebra of truth values of type-2 fuzzy sets, Fuzzy Sets Syst. (2019), https://doi.org/10.1016/j.fss.2019.05.011. [33] W. Zhang, X.P. Wang, Note on the absorption laws in the algebra of truth values of type-2 fuzzy sets, Fuzzy Sets Syst. 332 (2018) 111–115.