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Characterizations of some fuzzy prefilters (filters) in EQ-algebras.

1. Introduction

Every many-valued logic is uniquely determined by the algebraic properties of the structure of its truth values. At present, it is generally accepted that, in fuzzy logic, the algebraic structure should be a residuated lattice, possibly fulfilling some additional properties. BL-algebras, MTL-algebras, MV-algebras, and so forth are the best known classes of residuated lattices [1-3]. Note that the typical operations on these algebras are multiplication [C] and implication [right arrow] which are closely tied by adjointness property.

Fuzzy type theory [4, 5] whose basic connective is a fuzzy equality was developed as a counterpart of the classical higher-order logic (type theory in which identity is a basic connective; see [6]). Since the algebra of truth values is no longer a residuated lattice, a specific algebra called an EQ-algebra [7] for fuzzy type theory was proposed by Novak and De Baets. LQ-algebras are interesting and important algebras from many points of view. First, the above residuated lattices based logical algebras are all particular cases of LQ-algebras. Second, the adjointness property which strictly couples [encircled dot] and [right arrow] on residuated lattices based logical algebras is relaxed. Indeed, in LQ-algebras, [right arrow] is defined directly from fuzzy equality by the formula x [right arrow] y = (x A y) ~ x. But the fuzzy equality ~ cannot be reconstructed from the implication [right arrow] in LQ-algebras in general. Third, LQ-algebras open a possibility to develop a fuzzy logic with a noncommutative conjunction but a single implication only [7]. From these points of view, it is meaningful to study LQ-algebras.

The filter theory plays an important role in studying logical algebras. From logic point of view, various filters have natural interpretation as various sets of provable formulas. Up to now, some types of filters on special residuated lattices based logical algebras have been widely studied and some important results are obtained [8-12]. It is proved that Boolean filters and implicative filters coincide together in BL-algebras [11, 13], and that implicative filters are positive implicative filters in BL-algebras, but they all are equivalent in MV-algebras [8,14]. Moreover, the properties of filters have a strong influence on the structure properties of algebras.

The sets of provable formulas in corresponding inference systems from the point of view of uncertain information can be described by fuzzy filters of those algebraic semantics. At present, many authors studied some kinds of fuzzy filters of various logics algebras [8, 9, 13-19]. Liu and Li [13-15] introduced and studied the fuzzy filters, fuzzy implicative filters, fuzzy Boolean filters, and fuzzy positive implicative filters in BL-algebras and [R.sub.0]-algebras, respectively. Ma et al. established generalized fuzzy filter theory of BL-algebras in [20-23]. Recently, Zhang et al. have extended the notions of special fuzzy filters to general residuated lattices [24, 25].

Since residuated lattices (BE-algebras, MV-algebras, MTL-algebras, and [R.sub.0]-algebras) are particular types of EQ-algebras, it is natural and meaningful to extend some notions of residuated lattices to EQ-algebras and establish more general theories in EQ-algebras for studying the common properties of the abovementioned algebras. For EQ-algebras, the notions of prefilters (which coincide with filters in residuated lattices) and prime prefilters were proposed and some of their properties were obtained [26]. In [27], implicative and positive implicative prefilters of EQ-algebras have been studied and some interesting results have been obtained.

The aim of this paper is to characterize some types of fuzzy prefilters (filters) and to discuss the relationships among these notions in EQ-algebras. Moreover, using our obtained results about various fuzzy filters in EQ-algebras, we further develop the classical filter theory in EQ-algebras and the fuzzy filter theory in residuated lattices. In this paper, several characterizations of fuzzy positive implicative prefilters (filters), fuzzy implicative prefilters (filters), and fuzzy fantastic prefilters (filters) in EQ-algebras are derived. Moreover, the relations among these fuzzy prefilters (filters) are considered. As applications of our obtained results, we give some new results about classical filters in EQ-algebras and some related results about fuzzy filters in residuated lattices. Based on the results obtained in this paper, it will lay a foundation for providing an algebraic tool in considering many problems in fuzzy logic.

2. Preliminaries

In the section, we present some definitions and results about EQ-algebras that will be used in the sequel.

Definition 1 (see [7, 26]). An algebra (L, [conjunction], [encircled dot], ~, 1) of type (2,2,2,0) is called an EQ-algebra if it satisfies the following axioms:

(E1) (L, [conjunction], 1) is a [conjunction]-semilattice with top element 1. We set x [less than or equal to] y if and only if x [conjunction] y = x, as usual;

(E2) (L, [encircled dot], 1) is a commutative monoid and [encircled dot] is isotone with respect to [less than or equal to];

(E3) x ~ x = 1 (reflexivity axiom);

(E4) ((x [conjunction] y) ~ z) [encircled dot] (s ~ x) [less than or equal to] z ~ (s [conjunction] y) (substitution axiom);

(E5) (x ~ y) [encircled dot] (s ~ t) [less than or equal to] (x ~ s) ~ (y ~ t) (congruence axiom);

(E6) (x [conjunction] y [conjunction] z) ~ x [less than or equal to] (x [conjunction] j) ~ x (monotonicity axiom);

(E7) x [encircled dot] y [less than or equal to] x ~ y (boundedness axiom), for all x, y, z, s, t [member of] L.

Let L be an EQ-algebra. For all x [member of] L, we put [??] = x ~ 1. If L contains a bottom element 0, then we may define the unary operation [logical not] on L by [logical not]x = x ~ 0.

Definition 2 (see [7]). An EQ-algebra L is called

(1) a good EQ-algebra if [??] = x for all x [member of] L,

(2) a separated EQ-algebra if x ~ y = 1 implies that x = y for all x, y [member of] L,

(3) a residuated EQ-algebra if (x [encircled dot] y) [conjunction] z = x [encircled dot] y if and only if x [conjunction] ((y [conjunction] z) ~ y) = x for all x, y, z [member of] L,

(4) an involutive EQ-algebra if it contains a bottom element 0 and for all x [member of] L it holds that [conjunction][conjunction] = x.

Let L be an EQ-algebra. For all x, y [member of] L, we put x [right arrow] y = (x [conjunction] y) ~ x.

The derived operation [right arrow] is called an implication.

Remark 3. Let (L, [conjunction], [disjunction], [encircled dot], [??], 0, 1) be a residuated lattice. For any x, y [member of] L, we define x ~ y = (x [??] y) [conjunction] (y [??] x). Then, (L, [conjunction], [encircled dot], ~, 1) is a residuated EQ-algebra [7]. It is easily proved that x [right arrow] y = (x [conjunction] y) ~ x = x [??] y. In general, a residuated EQ-algebra may not be a residuated lattice [26]. This shows that residuated lattices are proper classes of EQ-algebras [27].

The following properties are well known to hold in EQ-algebras, we summarize them as follows.

Lemma 4 (see [7, 26]). Let (L, [conjunction], [encircled dot], ~, 1) be an EQ-algebra. Then, the following properties hold:

(1) x ~ y = y ~ x, x ~ y [less than or equal to] x [right arrow] y,

(2) x [less than or equal to] 1 ~ x = 1 [right arrow] x [less than or equal to] y [right arrow] x,

(3) x [encircled dot] y [less than or equal to] x [conjunction] y [less than or equal to] x, y,

(4) (x ~ y) [encircled dot] (y ~ z) [less than or equal to] x ~ z,

(5) (x [right arrow] y) [encircled dot] (y [right arrow] z) [less than or equal to] x [right arrow] z,

(6) x ~ y [less than or equal to] (x ~ z) ~ (y ~ z),

(7) x [right arrow] y [less than or equal to] (y [right arrow] z) [right arrow] (x [right arrow] z),

(8) x [right arrow] y [less than or equal to] (z [right arrow] x) [right arrow] (z [right arrow] y),

(9) x ~ y [less than or equal to] (x [conjunction] z) ~ (y [conjunction] z),

(10) x [right arrow] y [less than or equal to] (x [conjunction] z) [right arrow] (y [conjunction] z),

(11) x [right arrow] y = x [right arrow] (x [conjunction] y),

(12) if x [less than or equal to] y, then x [right arrow] y = 1, x ~ y = y [right arrow] x,

(13) if x [less than or equal to] y, then y [right arrow] z [less than or equal to] x [right arrow] z, z [right arrow] x [less than or equal to] z [right arrow] y,

(14) if L contains a bottom element 0, then [logical not]0 = 1, [logical not]x = x [right arrow] 0,

(15) if L is a residuated EQ-algebra, then x [encircled dot] y [right arrow] z [less than or equal to] x [right arrow] (y [right arrow] z), for any x, y, z [member of] L.

Lemma 5 (see [7, 26]). Let L be a good EQ-algebra. Then, the following properties hold, for any x, y, z [member of] L:

(1) x [less than or equal to] (x [right arrow] y) [right arrow] y,

(2) x [right arrow] (y [right arrow] z) = y [right arrow] (x [right arrow] z),

(3) x [encircled dot] (x [right arrow] y) [less than or equal to] y,

(4) x [encircled dot] (x [right arrow] y) [less than or equal to] x [conjunction] y.

Lemma 6 (see [26]). Let L be an EQ-algebra. Then, the following are equivalent:

(1) an EQ-algebra L is residuated;

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(2) the EQ-algebra L is good and x [right arrow] y [less than or equal to] (x [encircled dot] z) [right arrow] (y [encircled dot] z) for any x, y, z [member of] L;

(3) the EQ-algebra L is good and x [less than or equal to] y [right arrow] (x [encircled dot] y) for any x, y [member of] L.

In what follows, we recall the notions of some prefilters (filters) in EQ-algebras.

Definition 7 (see [26]). A nonempty subset F of an EQ-algebra L is called a prefilter of L if it satisfies, for any x, y [member of] L,

(F1) 1 [member of] F,

(F2) x [member of] F; x [right arrow] y [member of] F imply that y [member of] F, a prefilter F is called a filter if it satisfies, for any x, y, z [member of] L,

(F3) x [right arrow] y [member of] F imply (x [encircled dot] z) [right arrow] (y [encircled dot] z) [member of] F.

A prefilter F of an EQ-algebra L is called proper if and only if F [not equal to] L.

Definition 8 (see [27]). A prefilter F of an EQ-algebra L is called a positive implicative prefilter of L if it satisfies, for any x, y, z [member of] L,

(F4) if x [right arrow] (y [right arrow] z) [member of] F and x [right arrow] y [member of] F, then x [right arrow] z [member of] F.

If F is a filter and it satisfies (F4), then F is called a positive implicative filter of L.

Definition 9 (see [27]). A nonempty subset F of an EQ-algebra L is called an implicative prefilter of L if it satisfies (F1) and

(F5) z [right arrow] ((x [right arrow] y) [right arrow] x) [member of] F and z [member of] F imply that x e F for any x, y, z [member of] L.

An implicative prefilter F is called an implicative filter if it satisfies (F3).

Now, we review the concept of fuzzy sets; see [28].

A fuzzy set in L is a mapping [mu]: L [right arrow] [0,1].

Let [mu] be a fuzzy set in L. For all t [member of] [0,1], the set [[mu].sub.t] = [x [member of] L | [mu](x) [greater than or equal to] t] is called a level subset of [mu].

Let F be a nonempty subset, we denote the characteristic function of F by [[chi].sub.F].

For convenience, for any a, b [member of] [0,1], we denote max[a, b] and min[fl, b] by a [disjunction] b and a [conjunction] b, respectively.

For any fuzzy sets [mu], v in L, we define [mu] [subset or equal to] v if and only if [mu](x) [less than or equal to] v(x) for all x [member of] L.

3. Fuzzy Prefilters (Filters) in EQ-Algebras

In this section, we introduce the notion of fuzzy prefilters (filters) in EQ-algebras and give some of their properties that will be used in the sequel.

In what follows, let L denote an EQ-algebra unless otherwise is specified.

Definition 10. Let [mu] be a fuzzy set in L. [mu] is called a fuzzy prefilter of L if it satisfies

(FF1) [mu](1) [greater than or equal to] [mu](x) for all x [member of] L,

(FF2) [mu](y) [greater than or equal to] [mu](x [right arrow] y) for a x, y [member of] L.

A fuzzy prefilter [mu] is called a fuzzy filter if it satisfies

(FF3) [mu]((x [encircled dot] z) [right arrow] (y [encircled dot] z)) [greater than or equal to] [mu](x [right arrow] y) for ally, z [member of] L.

The following examples show that fuzzy prefilters (filters) in EQ-algebras exist.

Example 11. Let L = [0, a, b, 1] be a chain with Cayley tables as Table 1

Then, (L, [conjunction], [encircled dot], ~, 1) is an EQ-algebra in [27]. Define a fuzzy set [mu] in L as follows: [mu](1) = 0.8, [mu](b) = 0.6, and [mu](0) = [mu](a) = 0.4. One can check that [mu] is a fuzzy prefilter of L.

Example 12. Let L = [0, a, b, 1] be a chain with Cayley tables as Table 2.

Then, (L, [conjunction], [encircled dot], ~, 1) is an EQ-algebra. Define a fuzzy set [mu] in L as follows: [mu](1) = [mu](b) = 0.8 and [mu](0) = [mu](a) = 0.4. It is easy to check that [mu] is a fuzzy filter of L.

Proposition 13. Let p be a fuzzy prefilter of L. For all x, y [member of] L, if x [less than or equal to] y, then [mu](x) [less than or equal to] [mu](y); that is, [mu] is order-preserving.

Proof. It is easy and omitted.

Remark 14. From Lemma 6, in a residuated EQ-algebra, we know that x [right arrow] y [less than or equal to] (x [encircled dot] z) [right arrow] (y [encircled dot] z) for any x, y, z [member of] L. Combining Proposition 13, we can obtain that fuzzy prefilters and fuzzy filters coincide in residuated EQ-algebras.

By using the level prefilters (filters) of an EQ-algebra, we can characterize fuzzy prefilters (filters).

Theorem 15. Let [mu] be a fuzzy set in L. [mu] is a fuzzy prefilter (filter) of L if and only if, for all t [member of] [0,1], [[mu].sub.t] is either empty or a prefilter (filter) of L.

Proof. The proof is straightforward.

Corollary 16. Let F be a nonempty subset of L. F is a prefilter (filter) if and only if [[chi].sub.F] is a fuzzy prefilter (filter) of L.

Next, we discuss some properties of fuzzy filters in an EQ-algebra, which will be used in the sequel.

Proposition 17. Let L be an EQ-algebra and let p be a fuzzy filter of L; for all x, y, z [member of] L, the following hold:

(1) [mu](x [encircled dot] y) = [mu](x) [conjunction] [mu](y),

(2) [mu](x [right arrow] z) [greater than or equal to] [mu](x [right arrow] y) [conjunction] [mu](y [right arrow] z).

Proof. (1) Since x [encircled dot] y [less than or equal to] x [conjunction] y [less than or equal to] x, y, then [mu](x [mid dot] y) [less than or equal to] [mu](x) [conjunction] [mu](y) by Proposition 13. From y [less than or equal to] 1 [right arrow] y, it follows that [mu](y) [less than or equal to] [mu](1 [right arrow] y). Since [mu] is a fuzzy filter of L, then [mu]((x [encircled dot] 1) [right arrow] (x [encircled dot] y)) [greater than or equal to] [mu](1 [right arrow] y) by (FF3); that is, [mu](x [right arrow] (x [encircled dot] y)) [greater than or equal to] [mu](1 [right arrow] y). By (FF2), we get [mu](x [encircled dot] y) [greater than o] [mu](x) [conjunction] [mu](x [right arrow] (x [encircled dot] y)). Hence, we have [mu](x [encircled dot] y) [greater than or equal to] [mu](x) [conjunction] [mu](x [right arrow] (x [encircled dot] y)) [greater than or equal to] [mu](x) [conjunction] [mu](1 [right arrow] y) [greater than or equal to] [mu](x) [conjunction] [mu](y). Therefore, [mu](x [encircled dot] y) = [mu](x) [conjunction] [mu](y).

(2) Since (x [right arrow] y) [encircled dot] (y [right arrow] z) [less than or equal to] x [right arrow] z, we have [mu](x [right arrow] z) [greater than or equal to] [mu]((x [right arrow] y) [encircled dot] (y [right arrow] z)) by Proposition 13. From (1), it follows that [mu]((x [right arrow] y) [encircled dot] (y [right arrow] z)) = [mu](x [right arrow] y) [conjunction] [mu](y-z). Therefore, [mu](x [right arrow] z) [greater than or equal to] [mu](x [right arrow] y) [conjunction] [mu](y [right arrow] z).

4. Fuzzy Positive Implicative Prefilters (Filters)

In the section, we introduce the notion of fuzzy positive implicative prefilters (filters) in EQ-algebras and give some of their characterizations. Moreover, applying these characterizations, we obtain some new results about positive implicative filters in EQ-algebras.

Definition 18. Let [mu] be a fuzzy prefilter in L. [mu] is called a fuzzy positive implicative prefilter of L if it satisfies

(FF4) [mu](x [right arrow] z) [greater than or equal to] [mu](x [right arrow] (y [right arrow] z)) [conjunction] [mu](x [right arrow] y) for all x, y, z [member of] L.

A fuzzy filter [mu] of L is called a fuzzy positive implicative filter if it satisfies (FF4).

For better understanding of the above definition, we illustrate it by the following example.

Example 19. Let L be the EQ-algebra and let [mu] be the fuzzy set of L defined in Example 12. One can see that [mu] is a fuzzy positive implicative filter of L.

The following result gives a characterization of fuzzy positive implicative prefilters by fuzzy prefilters.

Theorem 20. Let p be a fuzzy prefilter in L. [mu] is a fuzzy positive implicative prefilter of L if and only if [[mu].sup.a] : L [right arrow] [0,1] is a fuzzy prefilter in L, where [[mu].sup.a](x) = [mu](a [right arrow] x) for all a, x [member of] L.

Proof. Suppose that [mu] is a fuzzy positive implicative prefilter of L. Since a [right arrow] 1 = 1, then [mu](a [right arrow] 1) = [mu](1). It follows that [[mu].sup.a](1) = [mu](a [right arrow] 1) = [mu](1). From a [right arrow] x [less than or equal to] 1, we have p(a [right arrow] x) [less than or equal to] [mu](1); that is, [[mu].sup.a](x) [less than or equal to] [mu](1). Then, [[mu].sup.a](x) [less than or equal to] [[mu].sup.a](1). On the other hand, since [mu] is a fuzzy positive implicative prefilter of L, then [mu](a [right arrow] y) [greater than or equal to] [mu](a [right arrow] (x [right arrow] y)) [conjunction] [mu](a [right arrow] x); that is, [[mu].sup.a](y) [greater than or equal to] [[mu].sup.a](x [right arrow] y) [conjunction] [[mu].sup.a](x). Therefore, pa is a fuzzy prefilter in L.

Conversely, for any a [member of] L, since [[mu].sup.a] is a fuzzy prefilter in L, then [[mu].sup.x] is a fuzzy prefilter in L. It follows that [[mu].sup.x](z) > [[mu].sup.x](y [right arrow] z) A [[mu].sup.x](y). By the definition of [[mu].sup.x], we have [rho](x [right arrow] z) [greater than or equal to] [rho](x [right arrow] (y [right arrow] z)) [conjunction] [mu](x [right arrow] y). Therefore, p is a fuzzy positive implicative prefilter of L.

As a consequence of Theorem 20, we have the following properties of [[mu].sup.a].

Proposition 21. Let [mu] be a fuzzy positive implicative prefilter of L. Then, for any a [member of] L, [[mu].sup.a] is the fuzzy prefilter containing [mu].

Proof. Assume that [mu] is a fuzzy positive implicative prefilter of L. By Theorem 20, we obtain that pa is a fuzzy prefilter. For any x [member of] L, x [less than or equal to] a [right arrow] x. It follows from Proposition 13 that [mu](x) [less than or equal to] [mu](a [right arrow] x); that is, [mu](x) [less than or equal to] [[mu].sup.a](x). And so [mu] [subset or equal to] [[mu].sup.a]. Therefore, pa is the fuzzy prefilter containing [mu].

Proposition 22. Let [mu], v be two fuzzy prefilters of L. Then, for any a, b [member of] L, the following statements hold:

(1) [[mu].sup.a] = [mu] if and only if [mu](a) = [mu](1),

(2) a [less than or equal to] b implies that [[mu].sup.b] [subset or equal to] [[mu].sup.a],

(3) [mu] [subset or equal to] v implies that [[mu].sup.a] [subset or equal to] [v.sup.a],

(4) [([mu] [intersection] v).sup.a] = [[mu].sup.a] [intersection] [v.sup.a], [([mu] [union] v).sup.a] = [[mu].sup.a] [union] [v.sup.a].

Proof. (1) Suppose that [[mu].sup.a] = [mu]. Then, [mu](a) = [[mu].sup.a](a) = [mu](a [right arrow] a) = [mu](1); hence, [mu](a) = [mu](1).

Conversely, assume that p(a) = p(1). For any x [member of] I, since x [less than or equal to] a [right arrow] x and [mu] is a fuzzy prefilter, we have [mu](x) [less than or equal to] [mu](a [right arrow] x) = [[mu].sup.a](x). Hence, [mu] [subset or equal to] [[mu].sup.a]. On the other hand, since [mu] is a fuzzy prefilter, then [mu](x) [greater than or equal to] [mu](a [right arrow] x) [conjunction] [mu](a) = [mu](a [right arrow] x) [conjunction] [mu](1) = [mu](a [right arrow] x) for all x [member of] L. It follows that [mu](x) [greater than or equal to] [[mu].sup.a](x); that is, [mu] [contains or equal to] [[mu].sup.a]. Consequently, we obtain [[mu].sup.a] = [mu].

(2) Suppose that a [less than or equal to] b. For any x [member of] L, then b [right arrow] x [less than or equal to] a [right arrow] x. Since [mu] is a fuzzy prefilter, we have [mu](b [right arrow] x) [less than or equal to] [mu](a [right arrow] x). It follows that [[mu].sup.b](x) [less than or equal to] [[mu].sup.a](x). So [[mu].sup.b] [subset or equal to] [[mu].sup.a].

(3) and (4) It is easy to prove them, and we hence omit the details.

In the following, we give some equivalent conditions of fuzzy positive implicative prefilters (filters) for further discussions.

Theorem 23. Let p be a fuzzy prefilter (filter) in L. The following are equivalent:

(1) [mu] is a fuzzy positive implicative prefilter (filter) of L,

(2) [mu](x [right arrow] y) [greater than or equal to] [mu](x [right arrow] (x [right arrow] y)) for all x, y [member of] L,

(3) [mu](x [right arrow] y) = [mu](x [right arrow] (x [right arrow] y)) for all x, y [member of] L.

Proof. (1) [??] (2) Suppose that [mu] is a fuzzy positive implicative prefilter of L. Then, [mu](x [right arrow] y) [greater than or equal to] [mu](x [right arrow] (x [right arrow] y)) [conjunction] [mu](x [right arrow] x) follows from Definition 18. Consequently, we have [mu](x [right arrow] y) [greater than or equal to] [mu](x [right arrow] (x [right arrow] y)).

(2) [??] (3) From x [right arrow] y [less than or equal to] x [right arrow] (x [right arrow] y), it follows that [mu](x [right arrow] y) [less than or equal to] [mu](x [right arrow] (x [right arrow] y)), which together with (2) leads to [mu](x [right arrow] y) = p(x [right arrow] (x [right arrow] y)).

(3) [??] (1) By Lemma 4 (7), we have x [right arrow] (y [right arrow] z) [less than or equal to] ((y [right arrow] z) [right arrow] (x [right arrow] z)) [right arrow] (x [right arrow] (x [right arrow] z)) and x [right arrow] y [less than or equal to] (y [right arrow] z) [right arrow] (x [right arrow] z). Since [mu] is a fuzzy prefilter in L, then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Therefore, p is a fuzzy positive implicative prefilter of L.

Theorem 24. Let p be a fuzzy set in L. [mu] is a fuzzy positive implicative prefilter (filter) of L if and only if, for all t [member of] [0,1], [[mu].sub.t] is either empty or a positive implicative prefilter (filter) of L.

Proof. The proof is easy, and we hence omit the details.

Corollary 25. Let F be a nonempty subset of L. F is a positive implicative prefilter (filter) if and only if yp is a fuzzy positive implicative prefilter (filter).

Now, we continue to characterize fuzzy positive implicative filters.

Theorem 26. Let L be an EQ-algebra and let [mu] be a fuzzy filter. Then, [mu] is a fuzzy positive implicative filter of L if and only if [mu](x [conjunction](x [right arrow] y) [right arrow] y) = [mu](1) for all x, y [member of] L.

Proof. Since [mu] is a fuzzy positive implicative filter of L, then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. For any [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Conversely, by Proposition 17, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. It follows from Lemma 4 that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Therefore, p is a fuzzy positive implicative filter of L.

In order to get some related results about fuzzy filters in residuated lattices (see Corollary 66), we characterize fuzzy positive implicative filters in a residuated EQ-algebra.

Theorem 27. Let L be a residuated EQ-algebra and p be a fuzzy filter. The following are equivalent:

(1) [mu] is a fuzzy positive implicative filter of L,

(2) [mu](x [right arrow] (x [encircled dot] x))=[mu](1) for all x [member of] L,

(3) [mu](x [conjunction] y [right arrow] (x [encircled dot] y)) = [mu](1) for all x, y [member of] L,

(4) [mu](x [conjunction](x [right arrow] y) [right arrow] (x [encircled dot] y)) = [mu](1) for all x, y [member of] L.

Proof. (1) [??] (2) Since L is a residuated EQ-algebra, we have 1 = (x [encircled dot] x) [right arrow] (x [encircled dot] x) [less than or equal to] x [right arrow] (x [right arrow] (x [encircled dot] x)) by Lemma 4 (15). By Proposition 13, we get [mu](1) [right arrow] [mu](x [right arrow] (x [right arrow] (x [encircled dot] x))), which implies that [mu](x [right arrow] (x [right arrow] (x [encircled dot] x))) = [mu](1). From Definition 18, it follows that [mu](x [right arrow] (x [encircled dot] x)) [greater than or equal to] [mu](x [right arrow] (x [right arrow] (x [encircled dot] x))) [conjunction] [mu](x [right arrow] x) = [mu](x [right arrow] (x [right arrow] (x [encircled dot] x))) = [mu](1). Consequently, we have [mu](x [right arrow] (x [mu] x)) = [mu](1).

Conversely, by Proposition 17, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. By Lemma 5 (3), we have y [encircled dot] (y [right arrow] z) [less than or equal to] z. Then, [mu]((y [encircled dot] (y [right arrow] z)) [right arrow] z) = [mu](1). It follows that [mu](x [right arrow] z) [greater than or equal to] [mu]((x [encircled dot] x) [right arrow] (y [encircled dot] (y [right arrow] z))). Hence, we obtain that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] by Proposition 17. Since p is a fuzzy filter, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] by (FF3). Combining them, we get [mu](x [right arrow] z) [greater than or equal to] [mu](x [right arrow] (y [right arrow] z)) [conjunction] [mu](x [right arrow] y). Therefore, p is a fuzzy positive implicative filter of L.

(1) [??] (3) Suppose that [mu] is a fuzzy positive implicative filter of L. [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

(3) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

(4) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Therefore, [mu] is a fuzzy positive implicative filter by (2).

Theorem 28. Let L be a residuated EQ-algebra and [mu] be a fuzzy filter. The following are equivalent:

(1) [mu] is a fuzzy positive implicative filter of L,

(2) [mu]((x [encircled dot] y) [right arrow] z) = [right arrow] ((x [conjunction] y) [right arrow] z) for all x, y, z [member of] L,

(3) [mu](x [conjunction](x [right arrow] y) [right arrow] (x [conjunction] y)) = [mu](1) for all x, y [member of] L,

(4) [mu](x [conjunction](x [right arrow] y) [right arrow] (y [conjunction] (y [right arrow] x))) = [mu](1) for all x, y [member of] L.

Proof. (1) [??] (2) Suppose that [mu] is a fuzzy positive implicative filter. We have [mu](y [conjunction] (y [right arrow] z) [right arrow] z) = [mu](1) by Theorem 26. Since L is a residuated EQ-algebra, then (x & y)[right arrow]z [less than or equal to] x[right arrow](y[right arrow]z) for all x, y, z [member of] L by Lemma 4 (15). So, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. By Proposition 17, we get [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. On the other hand, since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Conversely, since L is a residuated EQ-algebra, then it is a good EQ-algebra. It follows from Lemma 5 (3) that we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. By Theorem 26, we have [mu] is a fuzzy positive implicative filter of L.

(1) [??] (3) Suppose that [mu] is a fuzzy positive implicative filter. Then, we have [mu](x [conjunction](x[right arrow]y)[right arrow](x [encircled dot] y)) = [mu](1) for all x, y [member of] L by Theorem 27 (4). From y [less than or equal to] x [right arrow] y, it follows that x [encircled dot] y [less than or equal to] x [encircled dot] (x[right arrow]y). Then, we have x [conjunction](x[right arrow]y) [right arrow] (x [encircled dot] y) [less than or equal to] x [conjunction] (x[right arrow]y)[right arrow]x [encircled dot] (x[right arrow]y). Hence, we get [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Moreover, by Lemma 5, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

(3) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

(4) [??] (1) By (4), we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Then, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since L is a residuated EQ-algebra, we get y < x[right arrow](x [encircled dot] y) by Lemma 6 (3). It follows that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Therefore, [mu] is a fuzzy positive implicative filter of L by Theorem 27 (3).

As an application of Theorems 27 and 28, we can obtain some new characterizations of classical positive implicative filters in residuated EQ-algebras.

Corollary 29. Let L be a residuated EQ-algebra and let F be a filter of L. Thefollowing are equivalent:

(1) F is a positive implicative filter of L,

(2) x[right arrow](x [encircled dot] x) [member of] F for all x [member of] L,

(3) x [conjunction] y[right arrow](x [encircled dot] y) [member of] F for all x, y [member of] L,

(4) x [conjunction] (x[right arrow]y)[right arrow](x [encircled dot] y) [member of] F for all x, y [member of] L,

(5) x [conjunction](x[right arrow]y)[right arrow]y [member of] F for all x, y [member of] L,

(6) x [conjunction] (x[right arrow]y)[right arrow](x [conjunction] y) [member of] F for all x, y [member of] L,

(7) x [conjunction](x[right arrow]y)[right arrow](y [conjunction] (y[right arrow]x)) e F for all x, y [member of] L,

(8) x [conjunction](x[right arrow]y)[right arrow]y [member of] F for all x, y [member of] L.

The extension property for fuzzy positive implicative prefilters is obtained from the following proposition.

Proposition 30. Let L be a good EQ-algebra and let [mu] and v be two fuzzy prefilters which satisfy [mu] [subset or equal to] v, [mu](1) = v(1). If [mu] is a fuzzy positive implicative prefilter, then so is v.

Proof. We set t = x[right arrow](x[right arrow]y); then, x[right arrow](x[right arrow](t [right arrow] y)) = t[right arrow](x[right arrow](x[right arrow]y)) = 1.Since [mu] is a fuzzy positive implicative prefilter, we have [mu](x[right arrow](t[right arrow]y)) = [mu](x [right arrow](x [right arrow](t[right arrow]y))) = [mu](1) by Theorem 23. Then, [mu](x[right arrow](t [right arrow] y)) = [mu](1) = v(1). From [mu] [subset or equal to] v, it follows that v(x[right arrow](t [right arrow] y)) > [mu](x[right arrow](t[right arrow]y)) = v(1). So, v(x[right arrow](t[right arrow]y)) = v(1). Since v is a fuzzy prefilter, then v(x[right arrow]y) [greater than or equal to] v(t[right arrow](x [right arrow] y)) [conjunction]v(t). It follows that v(x[right arrow]y) [greater than or equal to] v(1) [conjunction] v(t) = v(t) = v(x[right arrow](x[right arrow]y)). By Theorem 23, we get that v is a fuzzy positive implicative prefilter.

Proposition 31. Let L be an EQ-algebra and let [mu] and v be two fuzzy filters which satisfy [mu] [subset or equal to] v, [mu](1) = v(1). If [mu] is a fuzzy positive implicative filter, then so is v.

Proof. It follows from Theorem 26.

5. Fuzzy Implicative Prefilters (Filters)

In this section, we introduce the notion of fuzzy implicative prefilters (filters) in EQ-algebras and discuss some of their properties.

Definition 32. Let [mu] be a fuzzy set in L. [mu] is called a fuzzy implicative prefilter of L if it satisfies

(FF1) [mu](1) [greater than or equal to] [mu](x) for all x [member of] L,

(FF5) [mu](x) [greater than or equal to] [mu](z[right arrow]((x[right arrow]y)[right arrow]x)) [conjunction] [mu](z) for all x, y, z [member of] L.

A fuzzy implicative prefilter [mu] of L is called a fuzzy implicative filter if it satisfies (FF3).

The following example shows that fuzzy implica prefilters in EQ-algebras exist.

Example 33. Let L = {0, a, b1} be a chain with Cayley tables as Table 3.

Then, (L, [disjunction], [encircled dot], 1) is an EQ-algebin [2Define a fuzzy set p in L as follows: p= p(a) p(b) p(c)= m and p(0) = n, where 0 [less than or equal to] n < m [less than or equal to] 1. Oncan check that [mu] is a fuzzy implicative prefilter of L.

Lemma 34. Let p be a fuzzy implicative prefilter of L. For all x, y [member of] L, if x [less than or equal to] y, then [mu](x) [less than or equal to] [m](y); that is, p is order preserving.

Proof. Let p be a fuzzy implicative prefilter of L. By Definition 32, we have [mu](y) [greater than or equal to] [mu](x[right arrow]((y[right arrow]y) [right arrow] y)) [disjunction] [mu](x). From (y[right arrow]y)[right arrow]y = 1[right arrow]y [greater than or equal to] y, it follows that x[right arrow]((y[right arrow]y)[right arrow]y) [greater than or equal to] x[right arrow]y. If x [less than or equal to] y, then x[right arrow]y = 1, which implies that x[right arrow]((y[right arrow]y)[right arrow]y) = 1. Therefore, we get [mu](y) [greater than or equal to] [mu](1) [disjunction] [mu](x) = [mu](x); that is, [mu] is order-preserving.

The following result describes the relationship between fuzzy implicative prefilters (filters) and fuzzy prefilters (filters).

Theorem 35. Each fuzzy implicative prefilter of L is a fuzzy prefilter.

Proof. Suppose that [mu] is a fuzzy implicative prefilter of L. Then, [mu](1) [greater than or equal to] [mu](x) for all x [member of] L. From y [less than or equal to] 1 [right arrow]y, we get that x[right arrow]y [less than or equal to] x[right arrow](1[right arrow]y). By Lemma 34, we have [mu](x[right arrow]y) [less than or equal to] [mu](x[right arrow](1[right arrow]y)). It follows from Definition 32 that [mu](y) [greater than or equal to] [mu](x[right arrow]((y[right arrow]1)[right arrow]y)) [disjunction] [mu](x) = [mu](x[right arrow](1[right arrow]y)) [disjunction] [mu](x) [greater than or equal to] [mu](x[right arrow]y) [disjunction][mu](x), which implies that (FF2) holds. Therefore, p is a fuzzy prefilter of L.

In the following example, we show that the converse of above theorem is not correct.

Example 36. Let L be the EQ-algebra defined in Example 12. Define a fuzzy set p in L as follows: [mu](1) = 0.9 and [mu](0) = [mu](a) = [mu](b) = 0.7. One can check that [mu] is a fuzzy prefilter of L. However, since [mu](a) < [mu](1) [disjunction] [mu](1[right arrow]((a[right arrow]0)[right arrow]a)), then p is not a fuzzy implicative prefilter of L.

Our next aim is to show some characterizations of fuzzy prefilters (filters).

Theorem 37. Let [mu] be a fuzzy prefilter (filter) in L. The following are equivalent:

(1) [mu] is a fuzzy implicative prefilter (filter) of L,

(2) [mu](x) [greater than or equal to] [mu]((x[right arrow]y)[right arrow]x) for all x, y [member of] L,

(3) [mu](x) = [mu]((x[right arrow]y)[right arrow]x) for all x, y [member of] L.

Proof. (1) [??] (2) Suppose that p is a fuzzy implicative prefilter of L. Then, [mu](1) [greater than or equal to] [mu](x) for all x [member of] L. From Definition 32, we have [mu](x) [greater than or equal to] [mu](1[right arrow]((x[right arrow]y) -x)) [disjunction] [mu](1) = [mu](1[right arrow]((x[right arrow]y)[right arrow]x)). Since (x [right arrow] y)[right arrow]x [less than or equal to] 1[right arrow]((x[right arrow]y)[right arrow]x)), then [mu](1[right arrow]((x [right arrow] y) [right arrow] x)) [greater than or equal to] [mu]((x[right arrow]y)[right arrow]x). Consequently, we have [mu](x) [greater than or equal to] [mu](1[right arrow]((x[right arrow]y)[right arrow]x)) [greater than or equal to] [mu]((x[right arrow]y)[right arrow]x).

(2) [??] (3) From x [less than or equal to] (x[right arrow]y)[right arrow]x, it follows that [mu](x) [disjunction] [mu]((x[right arrow]y)[right arrow]x). Combining (2), we get [mu](x) = [mu]((x[right arrow]y)[right arrow]x).

(3) [??] (1) Let p be a fuzzy prefilter in L. Then, [mu](1) [greater than or equal to] [mu](x) for all r [member of] L. By Definition 10, we have [mu]((x[right arrow]y)[right arrow]x) [greater than or equal to] [mu](z) [disjunction] [mu](z[right arrow]((x[right arrow]y)[right arrow]x)). Combining (3), we obtain [mu](x) [greater than or equal to] [mu](z) [disjunction] [mu](z[right arrow]((x[right arrow]y)[right arrow]x)). By Definition 32, we have [mu] is a fuzzy implicative prefilter of L.

As a consequence of Theorem 37, we have the following corollary.

Corollary 38. Let L be an EQ-algebra with a bottom element 0 and let [mu] be a fuzzy prefilter (filter) in L. The following are equivalent:

(1) [mu] is a fuzzy implicative prefilter (filter) of L,

(2) [mu](x) [greater than or equal to] [mu]([logical not]x[right arrow]x) for all x [member of] L,

(3) [mu](x) = [mu]([logical not]x[right arrow]x) for all x [member of] L.

Proof. (1) [??] (2) Assume that [mu] is a fuzzy implicative prefilter of L. Then, [mu](x) [greater than or equal to] [mu]((x[right arrow]0)[right arrow]x) = [mu]([right arrow] x[right arrow]x) by Theorem 37.

(2) [??] (3) Since x [less than or equal to] [right arrow] x[right arrow]x, then [mu](x) [less than or equal to] [mu]([right arrow]x[right arrow]x) as [mu] is a fuzzy prefilter in L. Combining [mu](x) [greater than or equal to] [mu]([right arrow] x[right arrow]x), we get [mu](x) = [mu]([right arrow] x[right arrow]x).

(3) [??] (1) From x[right arrow]0 [less than or equal to] x[right arrow]y, it follows that (x [right arrow] y)[right arrow]x [less than or equal to] [right arrow] x[right arrow]x. Then, p((x[right arrow]y)[right arrow]x) [less than or equal to] [mu]([right arrow] x [right arrow] x) as [mu] is a fuzzy prefilter. Combining (3), we get [mu](x) [greater than or equal to] [mu]((x[right arrow]y)[right arrow]x). Therefore, we have that p is a fuzzy implicative prefilter of L by Theorem 37.

Theorem 39. Let p be a fuzzy set in L. [mu] is a fuzzy implicative prefilter (filter) of L if and only if, for all t [member of] [0,1], [[mu].sub.t] is either empty or an implicative prefilter (filter) of L.

Proof. The proof is easy, and we hence omit the details.

Corollary 40. Let F be a nonempty subset of L. F is an implicative prefilter (filter) if and only if [[chi].sub.F] is a fuzzy implicative prefilter (filter).

Definition 41. Let p be a fuzzy prefilter in L. We call that p has weak exchange principle if it satisfies for all x, y, z [member of] L [mu](x [right arrow] (y [right arrow] z)) = [mu](y [right arrow] (x [right arrow] z)).

Example 42. From Lemma 5, we know that any fuzzy prefilter p of a good EQ-algebra L has weak exchange principle.

Example 43. Let L be the EQ-algebra and [mu] be the fuzzy set of L defined in Example 12. One can check that L is not a good EQ-algebra and [mu] is a fuzzy prefilter of L, which has weak exchange principle.

We have the following characterizations of fuzzy implicative prefilters in an EQ-algebra with a bottom element 0.

Theorem 44. Let L be an EQ-algebra with a bottom element 0 and let [mu] be a fuzzy prefilter with weak exchange principle. The following are equivalent:

(1) [mu] is a fuzzy implicative prefilter of L,

(2) [mu](x [right arrow] z) [greater than or equal to] [mu](x [right arrow] ([logical not]z [right arrow] y)) [conjunction] [mu](y [right arrow] z) for all x, y, z [member of] L,

(3) [mu](x [right arrow] z) [greater than or equal to] [mu](x [right arrow] ([logical not] z [right arrow] z)) for all x, z [member of] L,

(4) [mu](x [right arrow] z) = [mu](x [right arrow] ([logical not] z [right arrow] z)) for all x, z [member of] L.

Proof. (1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. From Definition 10, it follows that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Notice that [logical not](x [right arrow] z) [less than or equal to] [logical not]z, and we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since [mu] is a fuzzy implicative prefilter of L, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] by Corollary 38. Consequently, we obtain [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Conversely, suppose that p satisfies [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since p is a fuzzy prefilter, then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. It follows that [mu](x) [greater than or equal to] [mu]([logical not]x [right arrow] x). By Corollary 38, we get that p is a fuzzy implicative prefilter of L.

(1) [??] (3) Suppose that p is a fuzzy implicative prefilter of L. Then, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1) by (2). Therefore, we obtain [mu](x [right arrow] z) [greater than or equal to] [mu](x [right arrow] ([logical not]z [right arrow] z).

(3) [??] (4) From z [less than or equal to] [logical not]z [right arrow] z, it follows that x [right arrow] z [less than or equal to] x [right arrow] ([logical not]z [right arrow] z). Then, [mu](x [right arrow] z) [less than or equal to] [mu](x [right arrow] ([logical not]z [right arrow] z)) as [mu] is a fuzzy prefilter. Combining (3), we get [mu](x [right arrow] z) = [mu](x [right arrow] ([logical not]z [right arrow] z)).

(4) [??] (1) Let p be a fuzzy prefilter which satisfies [mu](x [right arrow] z) = [mu](x [right arrow] ([logical not]z [right arrow] z)) for all x, z [member of] L. Then, [mu](1 [right arrw] x) = [mu](1 [right arrow] ([logical not]x [right arrow] x)). Since [mu] is a fuzzy prefilter, then [mu](x) [greater than or equal to] [mu](1 [right arrow] x) [conjunction] [mu](1) = [mu](1 [right arrow] x). From [conjunction]x [right arrow] x [less than or equal to] 1 [right arrow] ([logical not]x [right arrow] x), it follows that [mu](1 [right arrow] ([logical not]x [right arrow] x)) [greater than or equal to] [mu]([logical not]x [right arrow] x). Consequently, we obtain [mu](x) [greater than or equal to] [mu]([logical not]x [right arrow] x). Using Corollary 38, we get that p is a fuzzy implicative prefilter of L.

6. The Relations among Special Fuzzy Prefilters (Filters)

In this section, we introduce fuzzy fantastic prefilters (filters) in EQ-algebras and discuss the relations among various fuzzy prefilters (filters). Particularly, we find some conditions under which a fuzzy implicative prefilter (filter) is equivalent to a fuzzy positive implicative prefilter (filter). Moreover, as applications, we obtain some relations among corresponding classical filters in EQ-algebras and some related results about fuzzy filters in residuated lattices.

Definition 45. Let F be a prefilter in L. F is called a fantastic prefilter of L if it satisfies the following:

(F6) y [right arrow] x [member of] F implies that ((x [right arrow] y) [right arrow] y) [right arrow] x [member of] F for all x, y [member of] L.

If F is a filter and it satisfies (F6), then it is called a fantastic filter of L.

It is time to give some examples of fantastic prefilters.

Example 46. Let L = {0, a, b, 1} be a chain with Cayley tables as Table 4.

Then, (L, [conjunction], [encircled dot], ~, 1) is an EQ-algebra. One can check that F = {1} is a fantastic prefilter.

Definition 47. Let [mu] be a fuzzy prefilter in L. p is called a fuzzy fantastic prefilter of L if it satisfies

(FF6) [mu](((x[right arrow]y)[right arrow]y)[right arrow]x) [greater than or equal to] [mu](y[right arrow]x) for all x, y [member of] L.

A fuzzy filter [mu] of L is called a fuzzy fantastic filter if it satisfies (FF6).

Example 48. Let (L, [conjunction], [encircled dot], ~, 1) be an EQ-algebra in Example 46. Define a fuzzy set [mu] in L as follows: [mu](1) = 0.7 and [mu](0) = [mu](a) = [mu](b) = 0.4. One can check that [mu] is a fuzzy fantastic prefilter of L.

Next, we derive a characterization of fuzzy fantastic prefilters (filters).

Theorem 49. Let [mu] be a fuzzy prefilter (filter) in L. Then, [mu] is a fuzzy fantastic prefilter (filter) of L if and only if [mu](((x [right arrow] y) [right arrow] y)[right arrow]x) [greater than or equal to] [mu](z[right arrow](y[right arrow]x)) [conjunction] [mu](z) for all x, y, z [member of] L.

Proof. Suppose that [mu] is a fuzzy fantastic prefilter of L. From Definition 45, it follows that [mu](((x[right arrow]y)[right arrow]y)[right arrow]x) [greater than or equal to] [mu](y[right arrow]x) for all x, y [member of] L. Since [mu] is a fuzzy prefilter in L, then [mu](y[right arrow]x) [greater than or equal to] [mu](z[right arrow](y[right arrow]x)) [conjunction] [mu](z). Therefore, we obtain [mu](((x[right arrow]y)[right arrow]y)[right arrow]x) [greater than or equal to] [mu](z[right arrow](y[right arrow]x)) [conjunction] [mu](z).

Conversely, suppose that p satisfies [mu](((x[right arrow]y) [right arrow] y) [right arrow] x) [less than or equal to] [mu](z[right arrow](y[right arrow]x)) [conjunction] [mu](z) for all x, y, z [member of] L. Taking z = 1, we obtain [mu](((x[right arrow]y)[right arrow]y)[right arrow]x)) [greater than or equal to] [mu](1[right arrow](y[right arrow]x)) [conjunction] [mu](1) = [mu](1[right arrow](y[right arrow]x)). From y[right arrow]x [less than or equal to] 1[right arrow](y[right arrow]x), it follows that [mu](y[right arrow]x) [less than or equal to] [mu](1[right arrow](y[right arrow]x)). Consequently, we obtain [mu](((x[right arrow]y) [right arrow] y) [right arrow] x) [greater than arrow] [mu](y[right arrow]x). Therefore, [mu] is a fuzzy fantastic prefilter of L.

Theorem 50. Let p be a fuzzy set in L. [mu] is a fuzzy fantastic prefilter (filter) of L if and only if, for all t [member of] [0,1], [[mu].sub.t] is either empty or a fantastic prefilter (filter) of L.

Proof. The proof is easy, and we hence omit the details.

Corollary 51. Let F be a nonempty subset of L. F is a fantastic prefilter (filter) if and only if yp is a fuzzy fantastic prefilter (filter).

In what follows, we pay attention to the relations among various special fuzzy prefilters (filters).

First, the relationship between fuzzy implicative prefilters and fuzzy fantastic prefilters can be described by the following theorem.

Theorem 52. Each fuzzy implicative prefilter (filter) with weak exchange principle is a fuzzy fantastic prefilter (filter) in an EQ-algebra.

Proof. Suppose that [mu] is a fuzzy implicative prefilter with weak exchange principle. From x [less than or equal to] ((x[right arrow]y)[right arrow]y)[right arrow]x, it follows that (((x[right arrow]y)[right arrow]y)[right arrow]x)[right arrow]y [less than or equal to] x[right arrow]y. This implies that ((((x[right arrow]y)[right arrow]y)[right arrow]x)[right arrow]y) [right arrow] (((x [right arrow] y)[right arrow]y)[right arrow]x) [greater than or equal to] (x[right arrow]y)[right arrow](((x [right arrow] y)[right arrow]y)[right arrow]x). Since [mu] is a fuzzy implicative prefilter, then [mu](((((x[right arrow]y)[right arrow]y)[right arrow]x)[right arrow]y)[right arrow](((x[right arrow]y) [right arrow] y)[right arrow]x)) [greater than or equal to] [mu]((x[right arrow]y)[right arrow](((x[right arrow]y)[right arrow]y)[right arrow]x)) by Lemma 34. We can obtain that [mu]((x[right arrow]y)[right arrow](((x [right arrow] y) [right arrow] y)[right arrow]x)) = [mu](((x[right arrow]y)[right arrow]y)[right arrow]((x [right arrow] y)[right arrow]x)) as [mu] has weak exchange principle. From y[right arrow]x [less than or equal to] ((x[right arrow]y)[right arrow]y)[right arrow]((x[right arrow]y)[right arrow]x), it follows that [mu](y[right arrow]x) [less than or equal to] [mu](((x[right arrow]y)[right arrow]y)[right arrow]((x[right arrow]y)[right arrow]x)). Together with them, we obtain [mu](((((x[right arrow]y)[right arrow]y) [right arrow] x)[right arrow]y)[right arrow](((x[right arrow]y)[right arrow]y)[right arrow]x)) [greater than or equal to] [mu](y[right arrow]x). On the other hand, since [mu] is a fuzzy implicative prefilter, then [mu](((x[right arrow]y)[right arrow]y)[right arrow]x) [greater than or equal to] [mu](1[right arrow]((((x[right arrow]y) [right arow] y)[right arrow]x)[right arrow]y)[right arrow](((x[right arrow]y)[right arrow]y)[right arrow]x)) [conjunction] [mu](1) = [mu](1[right arrow]((((x[right arrow]y)[right arrow]y)[right arrow]x)[right arrow]y)[right arrow](((x [right arrow] y)[right arrow]y)[right arrow]x)) [greater than or equal to] [mu](((((x[right arrow]y)[right arrow]y)[right arrow]x) [right arrow] y) [right arrow] (((x[right arrow]y)[right arrow]y)[right arrow]x)). Consequently, we obtain [mu](((x[right arrow]y)[right arrow]y)[right arrow]x) [greater than or equal to] [mu](y[right arrow]x). Therefore, we get that p is a fuzzy fantastic prefilter by Definition 45.

From the following example, we can see that the converse of Theorem 52 may not be true.

Example 53. Let L be the EQ-algebra and let [mu] be the fuzzy set of L defined in Example 48. We know that [mu] is a fuzzy fantastic prefilter of L. But it is not a fuzzy implicative prefilter of L, since [mu](b) < [mu]((b[right arrow]0)[right arrow]b).

Next, the following theorems show the relationship between fuzzy implicative prefilters (filters) and fuzzy positive implicative prefilters (filters).

Theorem 54. Each fuzzy implicative prefilter with weak exchange principle is a fuzzy positive implicative prefilter in an EQ-algebra.

Proof. Suppose that [mu] is a fuzzy implicative prefilter of L. Then, [mu] is a fuzzy prefilter of L by Theorem 35. It follows that [mu]((x[right arrow](x[right arrow]z)) [greater than or equal to] [mu](x[right arrow]y) [conjunction] [mu]((x[right arrow]y)[right arrow](x [right arrow] (x [right arrow] z))). Since y[right arrow](x[right arrow]z) [less than or equal to] (x[right arrow]y)[right arrow](x [right arrow] (x [right arrow] z)), then [mu](y[right arrow](x[right arrow]z)) [less than or equal to] [mu]((x[right arrow]y) [right arrow] (x [right arrow] (x [right arrow] z))). In a similar way, we get that [mu](x [right arrow] (x [right arrow] z)) [less than or equal to] [mu](((x[right arrow]x)[right arrow]z)[right arrow](x[right arrow]z)). Hence, we obtain [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] as [mu] has weak exchange principle. From [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since [mu] is a fuzzy implicative prefilter of L, then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Consequently, we obtain [mu](x[right arrow]z) [greater than or equal to] [mu](x[right arrow]y) [conjunction] [mu](x[right arrow](y[right arrow]z)). Therefore, [mu] is a fuzzy positive implicative prefilter.

Theorem 55. Each fuzzy implicative filter is a fuzzy positive implicative filter in an EQ-algebra.

Proof. From x [conjunction](x[right arrow]y) [less than or equal to] x and x[conjunction](x[right arrow]y) [less than or equal to] x[right arrow]y, it follows that x [conjunction](x[right arrow]y) [less than or equal to] x[right arrow]y [less than or equal to] (x [conjunction](x[right arrow]y))[right arrow]y. Then, we have ((x [conjunction] (x[right arrow]y))[right arrow]y)[right arrow]y [less than or equal to] (x [conjunction] (x [right arrow] y)) [right arow] y, which implies that ((x [conjunction] (x[right arrow]y))[right arrow]y) [right arrow] y) [right arrow] ((x [conjunction](r [right arrow] y)) [right arrow] y) = 1. Hence, we obtain [mu](((x [conjunction] (x [right arrow] y)) [right arrow] y) [right arrow] y) [right arrow] ((x [conjunction] (x [right arrow] y)) [right arrow] y)) = [mu](1). Since [mu] is a fuzzy implicative filter of L, then [mu](((x [conjunction] (x [right arrow] y)) [right arrow] y) [right arrow] y) [right arrow] ((x [conjunction](x [right arrow] y)) [right arrow] y)) = [mu]((x [conjunction] (x [right arrow] y)) [right arrow] y) by Theorem 37. It follows that [mu]((x [conjunction] (x [right arrow] y)) [right arrow] y) = [mu](1). Therefore, we get that [mu] is a fuzzy positive implicative filter by Theorem 26.

The following example shows that the converse of the above theorem may not be true.

Example 56. Let L be the EQ-algebra and let [mu] be the fuzzy set of L defined in Example 19. We know that [mu] is a fuzzy positive implicative filter of L. But it is not a fuzzy implicative filter of L, since [mu](a) < [mu]((a [right arrow] 0) [right arrow] a).

The following result displays the relations among fuzzy positive implicative prefilters (filters), fuzzy implicative prefilters (filters), and fuzzy fantastic prefilters (filters). Also, it provides a condition under which the converse of Theorems 54 and 55 can be true.

Theorem 57. Let L be a good EQ-algebra. Then, [mu] is a fuzzy implicative prefilter (filter) if and only if [mu] is both a fuzzy positive implicative prefilter (filter) and a fuzzy fantastic prefilter (filter) of L.

Proof. Suppose that [mu] is a fuzzy implicative prefilter. Since L is a good EQ-algebra, then [mu] has weak exchange principle. Hence, we have that [mu] is a fuzzy fantastic prefilter of L by Theorem 54. Moreover, using Theorem 55, we get that [mu] is a fuzzy positive implicative prefilter.

Conversely, suppose that [mu] is both a fuzzy positive implicative prefilter and a fuzzy fantastic prefilter of L. Since p is a fuzzy positive implicative prefilter, then [mu]((x [right arrow] y) [right arrow] y) [greater than or equal to] [mu]((x [right arrow] y) [right arrow] ((x [right arrow] y) [right arrow] y)) by Theorem 23. Notice that L is a good EQ-algebra; we obtain x [less than or equal to] (x [right arrow] y) [right arrow] y by Lemma 5 (2). Then, (x [right arrow] y) [right arrow] x [less than or equal to] (x [right arrow] y) [right arrow] ((x [right arrow] y) [right arrow] y). It follows that [mu]((x [right arrow] y) [right arrow] x) [less than or equal to] [mu]((x [right arrow] y) [right arrow] ((x [right arrow] y) [right arrow] y)). Consequently, we obtain [mu]((x [right arrow] y) [right arrow] y) [greater than or equal to] [mu]((x [right arrow] y) [right arrow] x). On the other hand, since [mu] is a fuzzy fantastic prefilter of L, we get [mu](((x [right arrow] y) [right arrow] y) [right arrow] x)) [greater than or equal to] [mu](y [right arrow] x) by Definition 47. From (x [right arrow] y) [right arrow] x [less than or equal to] y [right arrow] x, it follows that [mu](y [right arrow] x) [greater than or equal to] [mu]((x [right arrow] y) [right arrow] x), which implies that [mu](((x [right arrow] y) [right arrow] y) [right arrow] x) [greater than or equal to] [mu]((x [right arrow] y) [right arrow] x). Moreover, since [mu] is a fuzzy prefilter of L, we have [mu](x) [greater than or equal to] [mu](((x [right arrow] y) [right arrow] y) [right arrow] x) [conjunction] [mu]((x [right arrow] y) [right arrow] y) and [mu]((x [right arrow] y) [right arrow] x) [greater than or equal to] [mu](z [right arrow] ((x [right arrow] y) [right arrow] x)) [conjunction] [mu](z). Consequently, we obtain [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Therefore, [mu] is a fuzzy implicative prefilter of L by Definition 32.

Combining Corollaries 25, 40 and 51 and Theorem 57 and taking the special case of fuzzy prefilters (filters), we can obtain the relations among positive implicative prefilters (filters), implicative prefilters (filters), and fantastic prefilters (filters) as corollaries.

Corollary 58. Let L be a good EQ-algebra and let F be a nonempty subset of L. Then, F is an implicative prefilter (filter) if and only if F is both a positive implicative prefilter (filter) and a fantastic prefilter (filter) of L.

Since residuated EQ-algebras are good EQ-algebras, in view of Corollary 58, we have the following.

Corollary 59. Let L be a residuated EQ-algebra and let F be a fantastic filter of L. Then, F is an implicative filter if and only if F is a positive implicative filter of L.

Corollary 60. Let L be a residuated EQ-algebra and let F be a positive implicative filter of L. Then, F is an implicative filter if and only if F is a fantastic filter of L.

The above results sufficiently show that fuzzy filters are a useful tool to obtain results on classical filters. Moreover, the above results further develop the classical filter theory in EQ-algebras.

In what follows, we continue to find the conditions under which a fuzzy positive implicative prefilter (filter) is a fuzzy implicative prefilter (filter).

Lemma 61. Let F be a positive implicative prefilter of an EQalgebra L. Then, for all a [member of] L, [F.sub.a] = {x [member of] L|a [right arrow] x [member of] F} is the least prefilter containing F and a.

Proof. Suppose that F is a positive implicative prefilter of L. Since a [right arrow] 1 = 1 [member of] F, then 1 [member of] [F.sub.a]. For all x, y [member of] L, if x, x [right arrow] y [member of] [F.sub.a], then a [right arrow] x [member of] F and a [right arrow] (x [right arrow] y) [member of] F. Since F is a positive implicative prefilter, we have a [right arrow] y [member of] F; that is, y [member of] [F.sub.a]. Therefore, [F.sub.a] is a prefilter in L.

For any x [member of] L, if x [member of] F, since x [less than or equal to] a [right arrow] x, and F is a prefilter, we have a [right arrow] x [member of] F; that is, x [member of] [F.sub.a]. Hence, F [subset or equal to] [F.sub.a]. Moreover, since a [right arrow] a = 1 [member of] F, then a [member of] [F.sub.a]. Therefore, F [union] {a} [subset or equal to] [F.sub.a]. Now, let G be a prefilter of L such that F [union] {a} [subset or equal to] G; then, for any x [member of] [F.sub.a], we have a [right arrow] x [member of] F [subset or equal to] G. Since a [member of] G and G is a prefilter in L, we have x [member of] G; that is, [F.sub.a] [subset or equal to] G. Therefore, [F.sub.a] is the least prefilter containing F and a.

A prefilter F of an EQ-algebra L is called maximal if and only if it is proper and no proper prefilter of L strictly contains F; that is, for each prefilter, G [not equal to] F; if F [subset or equal to] G then G = L.

Theorem 62. Let [mu] be a fuzzy prefilter with weak exchange principle in an EQ-algebra L. The following are equivalent:

(1) [mu] is a fuzzy implicative prefilter and [[mu].sub.[mu]1] is a maximal prefilter in L,

(2) [mu] is a fuzzy positive implicative prefilter and [[mu].sub.[mu](1)] is a maximal prefilter in L,

(3) [mu] satisfies that [mu](x) [not equal to] [mu](1) and [mu](y)[not equal to][mu](1) can imply that [mu](x [right arrow] y) = [mu](1) and [mu](y [right arrow] x) = [mu](1) for all x, y [member of] L.

Proof. (1) [??] (2) It follows from Theorem 54.

(2) [??] (3) Suppose that [mu](x)[not equal to][mu](1) and [mu](y) [not equal to] [mu](1); thus, x [member of] [[mu].sub.[mu](1)] and y [not member of] [[mu].sub.[mu](1)]. Since [mu] is a fuzzy positive implicative prefilter of L, we can easily prove that [[mu].sub.[mu](1)] is a positive implicative prefilter in L. It follows from Lemma 61 that [F.sup.y] = {t [member of] L | y[right arrow]t [member of] [[mu].sub.[mu](1)]} is the least prefilter containing [[mu].sub.[mu](1)] and y. Notice that is a maximal prefilter in L; we get [F.sub.y] = L. Hence, x [member of] [F.sub.y]; that is, y[right arrow]x [member of] [[mu].sub.[mu](1)], which implies that [mu](y[right arrow]x) = [mu](1). In a similar way, we can get [mu](x[right arrow]y) = [mu](1).

(3) [??] (1) Assume on the contrary that p is not a fuzzy implicative prefilter of L. Then, by Theorem 37, there exist x, y [member of] L such that [mu](x) < [mu]((x[right arrow]y)[right arrow]x). Hence, [mu](x) [not equal to] [mu](1). Now, we consider two cases: either [mu](y) = [mu](1) or [mu](y) [not equal to] [mu](1).

If [mu](y) = [mu](1), since [mu](y) [less than or equal to] [mu](x[right arrow]y), then [mu](x [right arrow] y) = [mu](1). Since [mu] is a fuzzy prefilter in L, then [mu](x) [greater than or equal to] [mu]((x[right arrow]y)[right arrow]x) [conjunction] [mu](x[right arrow]y) = [mu]((x[right arrow]y)[right arrow]x).

If [mu](y) = [mu](1), combining [mu](x) = [mu](1), we have [mu](x [right arrow] y) = [mu](1) by assumption. Similarly, we have [mu](x) [greater than or equal to] [mu]((x [right arrow] y)[right arrow]x).

In any case, we have a contradiction. Therefore, [mu] is a fuzzy implicative prefilter of L. It follows from Theorem 54 that [mu] is a fuzzy positive implicative prefilter of L. Hence, [[mu].sub.[mu](1)] is a positive implicative prefilter in L. By Lemma 61, we have that [F.sub.a] = {x [member of] L | a[right arrow]x [member of] [[mu].sub.[mu](1)]} is the least prefilter containing [[mu].sub.[mu](1)] and a.

In order to prove that [[mu].sub.[mu](1)] is a maximal prefilter in L, it is sufficient to show that, for all a [member of] L-[[mu].sub.[mu](1)], [F.sub.a] = {x [member of] L | a [right arrow] x [member of] [[mu].sub.[mu](1)]} = L For all t [member of] L, if t [member of] [[mu].sub.[mu](1)], then t [member of] [F.sub.a] = {x [member of] L | a[right arrow]x [member of] [[mu].sub.[mu](1)]}. If t [not member of] [[mu].sub.[mu](1)], then [mu](t) [not equal to] [mu](1). Since a [not member of] [[mu].sub.[mu](1)], then [mu](a) [not equal to] [mu](1). It follows from (3) that [mu](a[right arrow]t) = [mu](1); that is, a[right arrow]t [member of] [[mu].sub.[mu](1)], which implies that t [member of] [F.sub.a] = {x [member of] L | a[right arrow]x [member of] [[mu].sub.[mu](1)]}. In any case, we have [F.sub.a] = L. Therefore, [[mu].sub.[mu](1)] is a maximal prefilter in L. This completes the proof.

Corollary 63. Let L be a good EQ-algebra. Suppose that [mu] is a fuzzy prefilter and [[mu].sub.[mu](1)] is a maximal prefilter in L. Then, [mu] is a fuzzy implicative prefilter of L if and only if it is a fuzzy positive implicative prefilter of L.

Next, we further find the conditions under which a fuzzy positive implicative filter is equivalent to a fuzzy implicative filter.

Theorem 64. Let L be a good EQ-algebra with a bottom element 0 and let [mu] be a fuzzy positive implicative filter. Then, [mu] is a fuzzy implicative filter if and only if it satisfies [mu]([logical not][logical not]x) = [mu](x) for all x [member of] L.

Proof. Suppose that p is a fuzzy implicative filter. From [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. By Corollary 38, we have [mu](x) = [mu]([logical not]x[right arrow]x). Consequently, we obtain [mu](x) [greater than or equal to] [mu]([logical not][logical not]x). Since L is a good EQ-algebra with a bottom element 0, then x [less than or equal to] (x[right arrow]0)[right arrow]0 by Lemma 5; that is, x [less than or equal to] [logical not][logical not]x. It follows that [mu](x) [less than or equal to] [mu]([logical not][logical not] x). Therefore, we get [mu]([logical not][logical not]x) = [mu](x).

Conversely, suppose that [mu] is a fuzzy positive implicative filter and satisfies [mu]([logical not][logical not]x) = [mu](x) for all x [member of] L. Since [logical not]x [right arrow] x [less than or equal to] (x[right arrow]0)[right arrow]([logical not] x[right arrow]0) = [logical not] x[right arrow]([logical not]x[right arrow]0), then [mu]([logical not]x[right arrow]([logical not]x[right arrow]0)) [greater than or equal to] [mu]([logical to]x[right arrow]x). It follows from Theorem 23 that [mu]([logical not]x[right arrow]0) = [mu]([logical not]x[right arrow]([logical not]x[right arrow]0)). So we have [mu]([logical not] [logical not]x) [greater than or equal to] [mu]([logical not]x[right arrow]x). Combining [mu] ([logical not] [logical not]x) = [mu](x), we obtain [mu](x)[greater than or equal to][mu]([logical not]x[right arrow]x). Therefore, [mu] is a fuzzy implicative filter of L by Corollary 38.

Notice that an involutive EQ-algebra L satisfies [logical not] [logical not]x = x for all x [member of] L. We have the following.

Corollary 65. Let L be a good involutive EQ-algebra. Then, [mu] is a fuzzy implicative filter of L if and only if it is a fuzzy positive implicative filter of L.

Since involutive residuated lattices are involutive residuated EQ-algebras, which are good involutive EQ-algebras, we can obtain some related results about fuzzy filters in residuated lattices.

Corollary 66. Let L be an involutive residuated lattice and let [mu] be a fuzzy filter of L. Then, the following are equivalent:

(1) [mu] is a fuzzy implicative filter of L,

(2) [mu] is a fuzzy positive implicative filter of L,

(3) [[mu].sup.a] is a fuzzy filter for all a e L,

(4) [mu](x[right arrow]y) [greater than or equal to] [mu](x[right arrow](x[right arrow]y)) for all x, y [member of] L,

(5) [mu](x[right arrow]y) = [mu](x[right arrow](x[right arrow]y)) for all x, y [member of] L,

(6) [mu](x [conjunction] (x[right arrow]y)[right arrow]y) = [mu](1) for all x, y [member of] L,

(7) [mu](x[right arrow](x [encircled dot] x))=[mu](1) for all x [member of] L,

(8) [mu](x [conjunction] y[right arrow](x [encircled dot] y)) = [mu](1) for all x, y [member of] L,

(9) [mu](x [conjunction](x[right arrow]y)[right arrow](x [conjunction] y)) = [mu](1) for all x, y [member of] L,

(10) [mu]((x [encircled dot] y)[right arrow]z) = [mu]((x [conjunction] y)[right arrow]z) for all x, y, z [member of] L,

(11) [mu](x [conjunction](x[right arrow]y)[right arrow](x [conjunction] y)) = [mu](1) for all x, y [member of] L,

(12) [mu](x [conjunction] (x[right arrow]y)[right arrow](y [conjunction] (y[right arrow]x))) = [mu](1) for all x, y [member of] L,

(13) [mu](x) [greater than or equal to] [mu]((x[right arrow]y)[right arrow]x) for all x, y [member of] L,

(14) [mu](x) = [mu]((x[right arrow]y)[right arrow]x) for all x, y [member of] L,

(15) [mu](x) [greater than or equal to] [mu]([logical not]x[right arrow]x) for all x [member of] L,

(16) [mu](x) = [mu]([logical not]x[right arrow]x) for all x [member of] L,

(17) [mu](x[right arrow]z) [greater than or equal to] [mu](x[right arrow]([logical not]z[right arrow]y)) [conjunction] [mu](y[right arrow]z) for all x, y, z [member of] L,

(18) [mu](x[right arrow]z) [greater than or equal to] [mu](x[right arrow]([logical not]z[right arrow]z)) for all x, z [member of] L,

(19) [mu](x[right arrow]z) = [mu](x[right arrow]([logical not] z[right arrow]z)) for all x, z [member of] L.

7. Conclusion

In this paper, we present several characterizations of some fuzzy prefilters (filters) in EQ-algebras. Using characterizations of these fuzzy prefilters (filters), we mainly consider the relations among special fuzzy prefilters (filters). We find some conditions under which a fuzzy positive implicative prefilter (filter) is equivalent to a fuzzy implicative prefilter (filter). As applications of our obtained results, we give some new characterizations about classical filters in EQ-algebras and some related results about fuzzy filters in residuated lattices. In our future work, we will introduce the notion of states on EQ-algebras and discuss the relations between fantastic filters and states on EQ-algebras.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

http://dx.doi.org/10.1155/2014/829527

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Xiao Long Xin, Peng Fei He, and Yong Wei Yang

Department of Mathematics, Northwest University, Xi'an 710127, China

Correspondence should be addressed to Xiao Long Xin; xlxin@nwu.edu.cn

Received 19 March 2014; Accepted 22 April 2014; Published 5 May 2014

Academic Editor: Hee S. Kim

TABLE 1

[encircled dot]     0       a       b       1

0                   0       0       0       0
a                   0       0       0       a
b                   0       0       0       b
1                   0       a       b       1

~                   0       a       b       1

0                   1       a       a       a
a                   a       1       b       b
b                   a       b       1       1
1                   a       b       1       1

[right              0       a       b       1
arrow]

0                   1       1       1       1
a                   a       1       1       1
b                   a       b       1       1
1                   a       b       1       1

TABLE 2

[encircled dot]     0       a       b       1

0                   0       0       0       0
a                   0       a       a       a
b                   0       a       b       b
1                   0       a       b       1

~                   0       a       b       1

0                   1       a       0       0
a                   0       1       a       a
b                   0       a       1       1
1                   0       a       1       1

[right              0       a       b       1
arrow]

0                   1       1       1       1
a                   0       1       1       1
b                   0       a       1       1
1                   0       a       1       1

TABLE 3

[encircled dot]     0       a       b       c       1

0                   0       0       0       0       0
a                   0       0       0       0       a
b                   0       0       0       0       b
c                   0       0       0       0       c
1                   0       a       b       c       1

~                   0       a       b       c       1

0                   1       a       0       0       0
a                   0       1       b       b       b
b                   0       b       1       c       c
c                   0       b       c       1       1
1                   0       b       c       1       1

[right              0       a       b       c       1
arrow]

0                   1       1       1       1       1
a                   0       1       1       1       1
b                   0       b       1       1       1
c                   0       b       c       1       1
1                   0       b       c       1       1

TABLE 4

[encircled dot]    0       a       b       1

0                  0       0       0       0
a                  0       0       0       a
b                  0       0       a       b
1                  0       a       b       1

~                  0       a       b       1

0                  1       b       a       0
a                  b       a       b       a
b                  a       b       1       b
1                  0       a       b       1

[right             0       a       b       1
arrow]

0                  1       1       1       1
a                  b       1       1       1
b                  a       b       1       1
1                  0       a       b       1
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Author:Xin, Xiao Long; He, Peng Fei; Yang, Yong Wei
Publication:The Scientific World Journal
Article Type:Report
Date:Jan 1, 2014
Words:13387
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