A note on the lattice structure for subalgebras of the algebra of truth values of type-2 fuzzy sets

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A note on the lattice structure for subalgebras of the algebra of truth values of type-2 fuzzy sets 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 4 January 2019; received in revised form 28 April 2019; accepted 21 May 2019

Abstract The algebra of truth values of type-2 fuzzy sets is not a lattice, but some of its subalgebras may be lattices. In this paper, we give a necessary and sufficient condition under which subalgebras of the truth value algebra of type-2 fuzzy sets form lattices. Further, we prove that if a certain subalgebra is a lattice, then it is isomorphic to an appropriate subalgebra whose all elements are convex functions with the same height. Based on these results, if a certain subalgebra is a lattice, the equivalent characterizations of the partial order in it are given. © 2019 Elsevier B.V. All rights reserved. Keywords: Type-2 fuzzy sets; Non-convex functions; Subalgebras; Sublattices; Complete lattices; Partial orders

1. Introduction and preliminary Type-2 fuzzy sets were introduced by Zadeh in 1975, as an extension of type-1 fuzzy sets, and have been heavily investigated both as a mathematical object and as a tool in applications (see, e.g. [8,10–12]). The algebra of truth values of type-2 fuzzy sets is a set of all mappings from the unit interval into itself, with operations given by the various convolutions of the pointwise operations [10]. From [9] we know the truth value algebra of type-2 fuzzy sets does not form a lattice since the absorption laws fail. To study the truth value algebra and its subalgebras, the absorption laws of the algebra play a key role. In particular, the subalgebra of all convex normal functions satisfies the absorption laws, thus it is a lattice, and in fact, is a complete lattice. So far, this subalgebra has been studied extensively (see, e.g. [1–3,7]). However, we focus on the more general situation. Recently, Zhang and Wang [13] presented a necessary and sufficient condition under which the absorption laws hold for a subalgebra of the algebra of truth values of type-2 fuzzy sets. On this basis, this paper investigates the lattice structure of such subalgebra. For convenience, subalgebras of the truth value algebra are called sublattices if the subalgebras form lattices. In Section 2, this paper gives the equivalent characterization of a sublattice and the equivalent characterizations of the par* 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.2019.05.011 0165-0114/© 2019 Elsevier B.V. All rights reserved.

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tial order in it, respectively, and proves that there exists a monomorphism from a sublattice to an appropriate complete sublattice, where this complete sublattice is equal to or isomorphic to the subalgebra of all convex normal functions. As applications, we discuss whether a sublattice containing non-convex functions is complete. Moreover, the example of complete sublattices and non-complete sublattices containing non-convex functions is given, respectively. In the final section, our researches are concluded. Next, we recall some essential requirements which are used in the sequel. The truth values of type-2 fuzzy sets are also known as fuzzy truth values. Let us denote the set of all fuzzy truth values by [0, 1][0,1] . That is, [0, 1][0,1] = {f | f : [0, 1] → [0, 1]}. Definition 1.1 ([2,10]). On [0, 1][0,1] , define operations , ,  , 0, 1 as follows: (1) (f  g)(x) = sup{f (y) ∧ g(z) : y ∨ z = x}, (2) (f  g)(x) = sup{f (y) ∧ g(z) : y ∧ z = x}, (3) f  (x) =sup{f (y) : 1 − y = x} = f (1 − x), 1, if x = 0, 1, if x = 1, (4) 0(x) = and 1(x) = 0, if x = 0, 0, if x = 1, where ∨ and ∧ are maximum and minimum operations, respectively, in the unit interval [0, 1]. The algebra ([0, 1][0,1] , , , , 0, 1) is the basic algebra of truth values for type-2 fuzzy sets, and has been extensively studied. See [2–5,9,10], for example. In what follows, we denote M = ([0, 1][0,1] , , ). Definition 1.2 ([2,10]). For f ∈ M, let f L and f R be the elements of M defined by f L (x) = sup{f (y) : y  x}, f R (x) = sup{f (y) : y  x}. Denote the height of f by f sup = sup{f (x) : x ∈ [0, 1]}. Obviously, f L (1) = f sup and f R (0) = f sup for any f ∈ M. Denote f LR = (f L )R and f RL = (f R )L . Then by Definition 1.2, one can check that f LR is a constant function with value f sup . Although we are interested in the algebra M, the set [0, 1][0,1] also has two pointwise operators ∨ and ∧ on it. For any f, g ∈ [0, 1][0,1] , the pointwise join f ∨ g and the pointwise meet f ∧ g are respectively defined 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  on [0, 1][0,1] is given by f  g if f (x)  g(x) for all x ∈ [0, 1]. It is clear that ([0, 1][0,1] , ∨, ∧) is a distributive lattice [10]. Proposition 1.1 ([6,10]). The following hold for all f, g ∈ M. (1) f  f L ; f  f R ; f  f LR . (2) (f L )L = f L ; (f R )R = f R ; f LR = f RL . (3) (f L ∧ f R )L = f L ; (f L ∧ f R )R = f R . (4) If f  g, then f L  g L and f R  g R . Theorem 1.1 ([2,10]). The following hold for all f, g ∈ M: f  g = (f ∧ g L ) ∨ (f L ∧ g) = (f ∨ g) ∧ (f L ∧ g L ), f  g = (f ∧ g R ) ∨ (f R ∧ g) = (f ∨ g) ∧ (f R ∧ g R ). Corollary 1.1 ([2,10]). Let f, g, h ∈ M. Some basic properties of M are as follows. (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)L = f L  g L = f L ∧ g L , (f  g)R = f R  g R = (f R ∧ g LR ) ∨ (f LR ∧ g R ). (6) (f  g)R = f R  g R = f R ∧ g R , (f  g)L = f L  g L = (f L ∧ g RL ) ∨ (f RL ∧ g L ).

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Fig. 1. Example of a non-convex function.

A subalgebra of an algebra is a nonempty subset of the algebra that is closed under the operations of the containing algebra. So when we say that A is a subalgebra of M, we mean that A is a subset of M that is closed under the operators  and . It is well-known that an algebra with two binary operations  and  is a lattice if it satisfies the properties (1)-(3) in Corollary 1.1 and the absorption laws (i.e., f  (f  g) = f  (f  g) = f ). Therefore, by Corollary 1.1 (1)-(4), we know that a subalgebra A of M is a lattice if f  (f  g) = f and g  (g  f ) = g for any f, g ∈ A. In this case, we say that A is a sublattice [7], meaning that it is a subalgebra of M and it is a lattice under the operators  and . 2. Sublattices In this section, we propose an equivalent characterization of sublattices and point out that any sublattice is isomorphic to a corresponding sublattice whose all elements are convex functions with the same height. Further, we give the equivalent characterizations of the partial order in each sublattice. As applications, the example of complete sublattices and non-complete sublattices containing non-convex functions is given, respectively. 2.1. Non-convex functions Definition 2.1 ([10]). A function f ∈ M is convex if for all x, y, z ∈ [0, 1] for which x  y  z, we have f (y)  f (x) ∧ f (z). Equivalently, f is convex if f = f L ∧ f R . Denote C as the subalgebra containing all convex functions of M. Let f ∈ M. Then by Definition 2.1, we have f = f L ∧ f R if f is a non-convex function. Clearly, f  f L ∧ f R . Therefore, if f is a non-convex function, then f L ∧ f R  f , i.e., there exists x ∈ [0, 1] such that f L (x) ∧ f R (x) > f (x). Definition 2.2 ([13]). Let f ∈ M 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). Let us denote the set of all non-convex elements of f by C ∗ (f ), and the set of all convex elements of f by C(f ), respectively. For instance, let f ∈ M as Fig. 1. Then C ∗ (f ) = (a, b) ∪ {c} and C(f ) = [0, a] ∪ [b, c) ∪ (c, 1]. Proposition 2.1 ([13]). The following hold for all f, h, l ∈ M. (1) C ∗ (f ) ∩ C(f ) = ∅, C ∗ (f ) ∪ C(f ) = [0, 1]. (2) If sup{f (x) : x ∈ [0, 1]} = a, and f (x0 ) = a for x0 ∈ [0, 1], then x0 ∈ C(f ). (3) If C(f ) ⊆ C(h) and C(f ) ⊆ C(l), then C(f ) ⊆ C(h  l) and C(f ) ⊆ C(h  l). Remark 2.1. By Definition 1.2, we have that f (0) = f L (0) = f L (0) ∧ f R (0) and f (1) = f R (1) = f L (1) ∧ f R (1), which implies 0, 1 ∈ C(f ). That is, C ∗ (f ) ⊆ (0, 1) for any f ∈ M.

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Corollary 2.1. If f, g ∈ M, then C(f ) ∩ C(g) ⊆ C(f  g) and C(f ) ∩ C(g) ⊆ C(f  g). Proof. It is straightforward from Proposition 2.1(3).

2

Remark 2.2. In general, C(f  g) = C(f  g), C(f  g)  C(f ) ∪ C(g) and C(f  g)  C(f ) ∪ C(g). For the first statement, we take that   1, if x = 0, 0.5, 1, if x = 0.5, 1, f (x) = and g(x) = 0, otherwise, 0, otherwise. Obviously, C(f ) = {0} ∪ [0.5, 1] and C(g) = [0, 0.5] ∪ {1}, as well as f  g = g and f  g = f . Thus C(f  g) = C(f  g). For the last two statements, we take that ⎧ ⎧ 5x, if x ∈ [0, 0.2], 4x, if x ∈ [0, 0.2], ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨2 − 5x, ⎨ if x ∈ (0.2, 0.4], 1.6 − 4x, if x ∈ (0.2, 0.4], f (x) = and g(x) = ⎪ ⎪ 5x − 2, if x ∈ (0.4, 0.6], 4x − 1.6, if x ∈ (0.4, 0.6], ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ 2.5 − 2.5x, if x ∈ (0.6, 1], 2 − 2x, if x ∈ (0.6, 1]. Clearly, C(f ) = [0, 0.2] ∪ [0.6, 1] = C(g). One can check that 0.24 ∈ C(f  g) and 0.56 ∈ C(f  g). Thus C(f  g)  C(f ) ∪ C(g) and C(f  g)  C(f ) ∪ C(g). 2.2. The lattice structure Theorem 2.1 ([13]). Let f, g ∈ M. Then f  (f  g) = f if and only if f  g LR and g(x)  f (x) for all x ∈ C ∗ (f ). It is clear that f  g LR is equivalent to f sup  g sup for any f, g ∈ M. As an immediate consequence of Theorem 2.1, we get a necessary and sufficient condition under which the subalgebra A of M forms a lattice. Corollary 2.2. Let A be a subalgebra of M. Then A is a sublattice if and only if for any f, g ∈ A, f sup = g sup as well as f (x)  g(x) for all x ∈ C ∗ (g) and g(x)  f (x) for all x ∈ C ∗ (f ). Proof. ⇒ Suppose that A is a sublattice. Then for any f, g ∈ A, we get that f  (f  g) = f and g  (g  f ) = g. This together with Theorem 2.1 implies that f sup = g sup , and f (x)  g(x) for all x ∈ C ∗ (g), and g(x)  f (x) for all x ∈ C ∗ (f ). ⇐ Obviously. 2 From Corollary 2.2, we know that if A is a sublattice, then all functions in A have the same height. In this case, the same height is referred to as the height of sublattice A. When the height of sublattice A equal to 0, A = {0} where 0(x) = 0 for all x ∈ [0, 1]. This case is trivial. Thus, in what follows, we only consider sublattices whose height is greater than 0. For any m ∈ (0, 1], denote Nm = {f ∈ M : f sup = m} and Lm = Nm ∩ C = {f ∈ M : f sup = m, f = f L ∧ f R }. Note that a function f ∈ M is normal if f sup = 1. From Theorem 12 in [4], we know that Nm is a subalgebra of M, further, these subalgebras are isomorphic to each other. Also, from Corollary 18 in [4], we know that subalgebra Lm is a complete lattice. Obviously, these sublattices Lm are isomorphic to each other. In particular, L1 containing all normal and convex functions of M is an interesting subalgebra that has been extensively studied. (See [2,6,7] for more information about the sublattice L1 .) Theorem 2.2. Let A be a sublattice with height m. Then  : A → Lm defined by (f ) = f L ∧ f R for all f ∈ A is a monomorphism from A to Lm . Proof. For any h ∈ A, using Proposition 1.1(3), we get ((h))LR = (hL ∧ hR )LR = hLR , i.e., ((h))sup = m. Also, ((h))L ∧ ((h))R = hL ∧ hR = (h). Thus (h) ∈ Lm . As a consequence,  is a mapping from A to Lm .

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Suppose that (h) = (l) for any h, l ∈ A. Since A is a sublattice, by Corollary 2.2 and Definition 2.2 and Proposition 2.1(1), we get l(x)  h(x) < (h)(x) = (l)(x) = l(x) if x ∈ C ∗ (h)\C ∗ (l). This is a contradiction. Thus C ∗ (h)\C ∗ (l) = ∅, i.e., C ∗ (h) ⊆ C ∗ (l). In a similar way, it can prove that C ∗ (l)\C ∗ (h) = ∅, i.e., C ∗ (l) ⊆ C ∗ (h). Thus C ∗ (h) = C ∗ (l). Subsequently, if x ∈ C ∗ (h), then l(x)  h(x)  l(x) by Corollary 2.2, i.e., h(x) = l(x). If x ∈ C(h), then h(x) = (h)(x) = (l)(x) = l(x). Consequently, h = l. Therefore, the mapping  is injective. To prove the mapping  is a monomorphism from A to Lm , it remains to show that it is a homomorphism. Suppose h, l ∈ A. Then using Corollary 1.1(5), (h  l) = (h  l)L ∧ (h  l)R = (hL ∧ l L ) ∧ [(hR ∧ l LR ) ∨ (hLR ∧ l R )] = (hL ∧ hR ∧ l L ) ∨ (hL ∧ l L ∧ l R ). On the other hand, using Theorem 1.1 and Proposition 1.1(3), (h)  (l) = (hL ∧ hR )  (l L ∧ l R ) = [(hL ∧ hR ) ∨ (l L ∧ l R )] ∧ (hL ∧ hR )L ∧ (l L ∧ l R )L = [(hL ∧ hR ) ∨ (l L ∧ l R )] ∧ hL ∧ l L = (hL ∧ hR ∧ l L ) ∨ (hL ∧ l L ∧ l R ). So  preserves . In a similar way, by Theorem 1.1 and Corollary 1.1(6), it can show that  preserves . Therefore,  is a monomorphism from A to Lm . 2 As an immediate consequence of Theorem 2.2, we know that if A is a sublattice, then A is isomorphic to (A). As applications, we discuss whether a sublattice containing non-convex functions is complete. Let L(g) = {f ∈ M : f sup = g sup , f (x) = g(x) f or all x ∈ C ∗ (g), C ∗ (f ) ⊆ C ∗ (g)}.

(1)

For each function g = 0, Zhang and Wang proved in Theorem 2.2 of [13] that the set L(g) is a sublattice. Obviously, the sublattice L(g) contains non-convex functions if C ∗ (g) = ∅. The following is a concrete example. Example 2.1. Let  0, if x ∈ [0.4, 0.8], g(x) = 1, otherwise. Then (L(g), , ) is a non-complete sublattice containing non-convex functions. Obviously, we only need to prove non-completeness of L(g). Suppose that (L(g), , ) is complete. Taking a family of functions ⎧ ⎪ if x ∈ [0.4, 0.8], ⎨0, fi (x) = 1, i ∈ [0, 0.4), if x = i, ⎪ ⎩ 0.9, otherwise, one can check that fi ∈ L(g). By the definition of , we have  1, if x = i, (fi )(x) = i ∈ [0, 0.4). 0.9, otherwise, Since (L(g)) ⊆ L1 , by simple calculation (the details about the join of any family of functions in L1 can be seen Theorem 16 in [2]), we get  1, if x = 0.4, ( (fi ))(x) = 0.9, otherwise. i∈[0,0.4)



Denote fsup =



i∈[0,0.4)

fi . According to Theorem 2.2,  is an isomorphism from L(g) to (L(g)), which implies

fsup (x)  ((fsup ))(x) = (



i∈[0,0.4)

(fi ))(x) = 0.9 when x = 0.4. Also, because fsup ∈ L(g), we have (fsup )sup =

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g sup = 1, which implies fsup (0.4) = 1. Moreover, because C ∗ (g) = [0.4, 0.8], by Eq. (1), we have fsup (0.4) = g(0.4) = 0. This is a contradiction. Therefore, the sublattice (L(g), , ) is not complete. Note that a sublattice containing non-convex functions may be complete. In order to illustrate this claim, we need to introduce the partial orders in M. Definition 2.3 ([7,10]). Let f, g ∈ M. Then, f  g if f  g = g and f  g if f  g = f . Both f  g and f  g are partial orders on M. In [6], Harding, Walker and Walker gave in Proposition 3.2.5 that a subalgebra A of M satisfies the absorption laws (i.e., A is a sublattice) if and only if  = . We write briefly  instead of  and  if  = . Corollary 2.3. Let A be a sublattice. Then for any f, g ∈ A, f  g if and only if (f )  (g). Proof. It is straightforward from Definition 2.3 and Theorem 2.2.

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Proposition 2.2. Let A be a sublattice. Then for any f, g ∈ A, f  g if and only if g L  f L and f R  g R . Proof. ⇒ Suppose f  g, so that f  g = g and f  g = f . By Corollary 1.1(5), g L = (f  g)L = f L ∧ g L , i.e., g L  f L . By Corollary 1.1(6), f R = (f  g)R = f R ∧ g R , i.e., f R  g R . ⇐ Suppose that g L  f L and f R  g R . To prove f  g, it suffices to show f  g = g. It follows from Theorem 1.1 that f  g = (f ∨ g) ∧ f L ∧ g L = (f ∨ g) ∧ g L = (f ∧ g L ) ∨ g since g L  f L . If x ∈ C ∗ (g), then by Corollary 2.2, f (x) ∧ g L (x)  f (x)  g(x), which implies (f  g)(x) = g(x). If x ∈ C(g), then f (x) ∧ g L (x)  f R (x) ∧ g L (x)  g R (x) ∧ g L (x) = g(x) since f R  g R , which implies that (f  g)(x) = g(x). Thus f  g = g. 2 In particular, there is a class of complete sublattices containing non-convex functions. Proposition 2.3. Let g ∈ M such that C ∗ (g) = (0, 1). Then the sublattice (L(g), , ) is complete and for any family fi and the of functions fi ∈ L(g), i ∈  where  is an index set (finite, or infinite), the least upper bound fsup = greatest lower bound finf = ⎧ ⎪ fi (0), ⎪ ⎪ ⎨i∈ fsup (x) = g(x), ⎪  ⎪ ⎪ fi (1), ⎩



i∈

 fi are respectively given by

i∈

i∈

⎧ ⎪ fi (0), ⎪ ⎪ ⎨i∈ if x ∈ (0, 1), and finf (x) = g(x), ⎪  ⎪ ⎪ if x = 1, fi (1), ⎩ if x = 0,

if x = 0, if x ∈ (0, 1), if x = 1.

i∈

 Proof. For any family of functions fi ∈ L(g), i ∈ , we have (fi )sup = g sup . Put fsup as above. If g(x) = g sup , x∈(0,1)   then (fsup )sup = fsup (0) ∨ fsup (1) ∨ g(x) = g sup . If g(x) < g sup , then fi (0) = g sup or fi (1) = g sup for x∈(0,1) x∈(0,1)   all i ∈ . This implies fi (0) = g sup or fi (1) = g sup , hence (fsup )sup = g sup . Also, by Remark 2.1 we know i∈

i∈

C ∗ (fsup ) ⊆ (0, 1). As a consequence, fsup∈ L(g).  fi (0)  fi (0) and fi (1)  fi (1) = fsup (1), as well as (fsup )L (1) = Obviously, for any i ∈ , fsup (0) = i∈

i∈

g sup = (fi )L (1) and (fi )R (0) = g sup = (fsup )R (0). Then for any i ∈ , since fsup (x) = fi (x) for all x ∈ (0, 1), we get (fsup )L  (fi )L and (fi )R  (fsup )R . By Proposition 2.2, we have fi  fsup for all i ∈ , which implies fi  fsup . On the other hand, let f ∈ L(g) such that fi  f for all i ∈ . It follows from Proposition 2.2 that



i∈

for any i ∈ , f (0)  fi (0) and fi (1)  f (1), which implies f (0)  fsup (0) and fsup (1)  f (1). Hence fsup  f . fi = fsup . Also, for finf , in a similar way, it can prove that finf = fi ∈ L(g). Therefore, the sublattice Therefore,



i∈

(L(g), , ) is complete.



2

i∈

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The following is a concrete example. Example 2.2. Let ⎧ ⎪ if x = 0, 1, ⎨0, g(x) = 1 − 2x, if x ∈ (0, 0.5], ⎪ ⎩ 2x − 1, if x ∈ (0.5, 1). Then one check that g sup = 1 and C ∗ (g) = (0, 1). Put ⎧ ⎪ if x = 0, ⎨a, fab (x) = g(x), if x ∈ (0, 1), ⎪ ⎩ b, if x = 1, where a, b ∈ [0, 1]. Therefore, L(g) = {fab : a, b ∈ [0, 1]} is a complete sublattice containing non-convex functions. 3. Conclusions In this paper, we give the equivalent characterization of a sublattice of M and the equivalent characterizations of the partial order in it, respectively, and prove that any sublattice A is isomorphic to (A). In the future work, we will concentrate on the equivalent characterizations of complete sublattices. Acknowledgements The work described in this paper was supported by grants from the National Natural Science Foundation of China (Grant nos. 11571010 and 61179038). References [1] J. Harding, C. Walker, E. Walker, On complete sublattices of the algebra of truth values of type-2 fuzzy sets, in: 2007 IEEE International Fuzzy Systems Conference, London, July 23–26, 2007, pp. 1–5. [2] J. Harding, C. Walker, E. Walker, Lattices of convex normal functions, Fuzzy Sets Syst. 159 (2008) 1061–1071. [3] J. Harding, C. Walker, E. Walker, Convex normal functions revisited, Fuzzy Sets Syst. 161 (2010) 1343–1349. [4] J. Harding, C. Walker, E. Walker, Partial orders on the truth value algebra of finite type-2 fuzzy sets, in: Joint IFSA World Congress and NAFIPS Annual Meeting, Canada, June 24–28, 2013, pp. 163–168. [5] J. Harding, C. Walker, E. Walker, Partial orders on fuzzy truth value algebras, Int. J. Uncertain. Fuzziness Knowl.-Based Syst. 23 (2015) 193–219. [6] 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. [7] J. Harding, C. Walker, E. Walker, Overview of lattices of convex normal functions, Int. J. Intell. Syst. 31 (2016) 257–275. [8] J. Mendel, R. John, Type-2 fuzzy sets made simple, IEEE Trans. Fuzzy Syst. 10 (2002) 117–127. [9] M. Mizumoto, K. Tanaka, Some properties of fuzzy sets of type-2, Inf. Control 31 (1976) 312–340. [10] C. Walker, E. Walker, The algebra of fuzzy truth values, Fuzzy Sets Syst. 149 (2005) 309–347. [11] C. Walker, E. Walker, Type-2 operations on finite chains, Fuzzy Sets Syst. 236 (2014) 33–49. [12] L. Zadeh, The concept of a linguistic variable and its application to approximate reasoning, Inf. Sci. 8 (1975) 199–249. [13] 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.