Note on the absorption laws in the algebra of truth values of type-2 fuzzy sets

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Fuzzy Sets and Systems ••• (••••) •••–•••

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www.elsevier.com/locate/fss

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Short communication

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Note on the absorption laws in the algebra of truth values of type-2 fuzzy sets ✩

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Wei Zhang, Xue-ping Wang ∗

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College of Mathematics and Software Science, Sichuan Normal University, Chengdu, Sichuan 610066, People’s Republic of China

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Received 6 December 2016; received in revised form 10 March 2017; accepted 4 May 2017

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Abstract This note gives a necessary and sufficient condition under which the absorption laws in the algebra of truth values of type-2 fuzzy sets are true. © 2017 Elsevier B.V. All rights reserved.

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Keywords: Type-2 fuzzy set; Non-convex element; Convex element; Absorption law

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1. Introduction 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 for use in applications (see, e.g. [6,8–10]). The truth value algebra of type-2 fuzzy sets is a set of all mappings of the unit interval into itself, with operations given by various convolutions of the pointwise operations [8]. From [7] we know the truth value algebra doesn’t 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. Nowadays, to the best of our knowledge, it seems that the study of conditions under which the absorption laws of the algebra are true has not yet been attached by any author except every element of the truth value algebra is convex (see, e.g. [4,7,8]). This paper will give a necessary and sufficient condition under which the absorption laws of the truth value algebra of type-2 fuzzy sets are hold. As applications, we construct a subalgebra of the truth value algebra of type-2 fuzzy sets which is a lattice. Definition 1.1 ([1,8]). 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},

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✩

Supported by National Natural Science Foundation of China (No. 61573240). * Corresponding author. Fax: +86 28 84761393. E-mail addresses: [email protected] (W. Zhang), [email protected] (X.-p. Wang).

http://dx.doi.org/10.1016/j.fss.2017.05.006 0165-0114/© 2017 Elsevier B.V. All rights reserved.

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W. Zhang, X.-p. Wang / Fuzzy Sets and Systems ••• (••••) •••–•••

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Fig. 1. Examples of operations  and .

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Fig. 2. Examples of f L and f R .

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(3) ¬f (x) = sup{f (y) : 1 − y = x} = f (1 − x), (4)   1, if x = 1, 1, if x = 0, 1(x) = and 0(x) = 0, if x = 1, 0, if x = 0,

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where ∨ and ∧ are maximum and minimum operations, respectively, in the unit interval [0, 1].

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The algebra ([0, 1][0,1] , , , ¬, 0, 1) is the basic algebra of truth values for type-2 fuzzy sets, and has been studied extensively. See [1,2,7,8], for example. In what follows, we denote M = ([0, 1][0,1] , , ). Fig. 1 shows examples illustrating f  g, f  g for two functions f and g in M, respectively. Definition 1.2 ([1,8]). For f ∈ M, let f L and f R be the elements of M defined by

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f (x) = sup{f (y) : y  x}, and

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f R (x) = sup{f (y) : y  x}.

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

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

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Corollary 1.1 ([1,8]). Let f, g, h ∈ M. Some basic properties of M are as follows.

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Theorem 1.1 ([1,8]). The following hold for all f, g ∈ M:

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Let f LR = (f L )R for any f ∈ M. Then by Definition 1.2, one can check that f LR = f RL . We show examples of f L and f R from a given function f ∈ M in Fig. 2.

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(1) (2) (3) (4)

f f f f

 f = f , f  f = f (idempotent laws).  g = g  f , f  g = g  f (commutative laws).  (g  h) = (f  g)  h, f  (g  h) = (f  g)  h (associative laws).  (f  g) = f  (f  g).

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Fig. 3. Example of a non-convex function f .

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(5) (f (f (f (f

 g)L = f L  g L = f L ∧ g L ,  g)R = f R  g R = f R ∧ g R ,  g)R = f R  g R = (f R ∧ g LR ) ∨ (f LR ∧ g R ),  g)L = f L  g L = (f L ∧ g RL ) ∨ (f RL ∧ g L ).

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Definition 1.3 ([8]). 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).

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Let f ∈ M. By Proposition 1.1, 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).

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(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).

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Proposition 2.1. The following hold for all f, h, l ∈ M.

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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. Let f ∈ M be as Fig. 3. Then C ∗ (f ) = (a, b) ∪ (c, d) and C(f ) = [0, a] ∪ [b, c] ∪ [d, 1].

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Definition 2.1. 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).

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2. Absorption laws

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Proposition 1.1 ([8]). A function f ∈ M is convex if and only if f = f L ∧ f R .

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Proof. We just prove (3) since the proofs of both (1) and (2) are trivial. Let x ∈ C(f ). Then by Definition 2.1 we have that hL (x) ∧ hR (x) = h(x), l L (x) ∧ l R (x) = l(x) since C(f ) ⊆ C(h) and C(f ) ⊆ C(l). Thus by Corollary 1.1(5) (h  l)L (x) ∧ (h  l)R (x) = (hL (x) ∧ l L (x)) ∧ [(hLR (x) ∧ l R (x)) ∨ (hR (x) ∧ l LR (x))] = [hL (x) ∧ l L (x) ∧ l R (x)] ∨ [hL (x) ∧ hR (x) ∧ l L (x)] = [h (x) ∧ l(x)] ∨ [h(x) ∧ l (x)] = (h  l)(x). L

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Therefore, C(f ) ⊆ C(h  l). Similarly, we have C(f ) ⊆ C(h  l).

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Lemma 2.1. Let f, g ∈ M. If f  (f  g) = f , then f  g LR .

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Proof. From Theorem 1.1 and Corollary 1.1, we have that

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

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

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= (f L ∧ f R ∧ g) ∨ (f ∧ g R ) ∨ (f ∧ g RL ) ∨ (f ∧ g L )

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= (f L ∧ f R ∧ g) ∨ (f ∧ g RL ).

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If f  g LR , then there exists an element x ∈ [0, 1] such that f (x) > g LR (x)  g(x). Since (f L ∧ f R )(x)  f (x), we further have (f  (f  g))(x) = (f L ∧ f R ∧ g)(x) ∨ (f ∧ g LR )(x)

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= g(x) ∨ g

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LR

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LR

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Theorem 2.1. 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 ).

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Proof. ⇒: Let f  (f  g) = f . Then by Lemma 2.1, we know f  g LR , which deduces that f  (f  g) = (f L ∧ f R ∧ g) ∨ f . So that (f L ∧ f R ∧ g)  f . The last inequality implies either f L (x) ∧ f R (x)  f (x) or g(x)  f (x) for all x ∈ [0, 1]. By Definition 2.1, we know f L (x) ∧ f R (x) > f (x) for all x ∈ C ∗ (f ). Therefore, g(x)  f (x) for all x ∈ C ∗ (f ). ⇐: Suppose that g(x)  f (x) for all x ∈ C ∗ (f ). Then f L (x) ∧ f R (x) ∧ g(x)  g(x)  f (x) for all x ∈ C ∗ (f ). If x∈ / C ∗ (f ), then f L (x) ∧ f R (x) = f (x), and f L (x) ∧ f R (x) ∧ g(x) = f (x) ∧ g(x)  f (x). Thus f L (x) ∧ f R (x) ∧ g(x)  f (x) for all x ∈ [0, 1], i.e., f L ∧ f R ∧ g  f . Therefore, f  (f  g) = (f L ∧ f R ∧ g) ∨ (f ∧ g RL ) = f since f  g LR . 2

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Proof. If

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for all f ∈ L(m, g). Thus by Proposition 2.1(1) and Definition 2.1, we have

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C ∗ (f ) = ∅ if and only if C(f ) = [0, 1] if and only if f (x) = f L (x) ∧ f R (x) for all x ∈ [0, 1].

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where m ∈ M is a constant function with value m. If L(m, g) = ∅, then L(m, g) is a subalgebra of M, further, L(m, g) is a lattice under operators  and .

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L(m, g) = {f ∈ M : f LR = m, f (x) = g(x) for all x ∈ C ∗ (g), C ∗ (f ) ⊆ C ∗ (g)}

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Theorem 2.2. Let m ∈ (0, 1] and g ∈ M be such that sup{g(x) : x ∈ [0, 1]}  m. Let

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For f ∈ M, the height of f is sup{f (x) : x ∈ [0, 1]} (see [3]). Next, by Theorem 2.1, we can construct a subalgebra of M.

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(x) < f (x),

which contradicts f  (f  g) = f . Thus f  g LR .

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Thus

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L(m, g) = {f ∈ M : f LR = m, f = f L ∧ f R },

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that is, L(m, g) is a subset of M consisting of all the convex functions of height m. By Corollary 18 of [3], L(m, g) is a complete lattice under operators  and . Next, suppose that C ∗ (g) = ∅. Let h, l ∈ L(m, g) since L(m, g) = ∅. Then by Corollary 1.1(5) we have (h  l)LR = hLR  l LR = m  m = m. C ∗ (g),

Since h, l ∈ L(m, g), we have that for any x ∈ h(x) = g(x) = l(x) which implies g(x)  l L (x). Thus by Theorem 1.1, (h  l)(x) = (h(x) ∨ l(x)) ∧ hL (x) ∧ l L (x) = g(x), i.e., ∗

(h  l)(x) = g(x) for all x ∈ C (g).

hL (x)

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and g(x) 

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(2)

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Because C ∗ (h) ⊆ C ∗ (g), C ∗ (l) ⊆ C ∗ (g), we have C(g) ⊆ C(h), C(g) ⊆ C(l) by Proposition 2.1(1). Thus, by Proposition 2.1(3), we have C(g) ⊆ C(h  l). Then C ∗ (h  l) ⊆ C ∗ (g).

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Formulae (1), (2) and (3) mean that h  l ∈ L(m, g). Similarly, we can prove that h  l ∈ L(m, g). Therefore, L(m, g) is a subalgebra of M. From (1), (2) and (3) in Corollary 1.1, to prove that the subalgebra L(m, g) is a lattice under operations  and , it suffices to show that h  (h  l) = h and h  (h  l) = h are true for all h, l ∈ L(m, g). Indeed, hLR = m = l LR . Thus by Definition 1.2, h  hLR = l LR , i.e., h  l LR . Now, let x ∈ C ∗ (h). Then from C ∗ (h) ⊆ C ∗ (g), we have l(x) = g(x) = h(x), i.e., l(x) = h(x) for all x ∈ C ∗ (h). Therefore, by Theorem 2.1, h  (h  l) = h. Moreover, we have h  (h  l) = h by Corollary 1.1(4). 2

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All our results are still valid while M = (I J , , ) in which I is a complete chain and J is a bounded chain with involution.

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References

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The authors thank the referees for their valuable comments and suggestions.

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Acknowledgements

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3. Conclusion

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Note that from Corollary 7.8.9 of [5], we further know that L(m, g) in Theorem 2.2 is distributive. Also, from the proof of Theorem 2.2, it is easy to see that {f ∈ M : C ∗ (f ) ⊆ C ∗ (g)} for any g ∈ M is a subalgebra of M. However, L = {f ∈ M : f LR = m, C ∗ (f ) ⊆ C ∗ (g)} is not a lattice under operations  and  generally. For instance, let m = 1, ⎧ ⎧ if x = 0, ⎪ ⎪1, ⎪ ⎪ 1, if x = 0, ⎨ ⎨0.5, if x = 0.5, and g(x) = 1, if x = 1, Then C ∗ (g) = (0, 1). Let h = g, and l(x) = Then one can check ⎪ ⎪1, if x = 1, ⎩ ⎪ ⎪ 0, otherwise. ⎩ 0, otherwise. that h, l ∈ L, and h  l = l = h  l implies h  (h  l) = l = h. Therefore, (L, , ) is not a lattice.

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