Fuzzy relation equations in pseudo BL-algebras

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  • ABSTRACT

    Bandler and Kohout investigated the solvability of fuzzy relation equations with inf-implication compositions in complete lattices. Perfilieva and Noskova investigated the solvability of fuzzy relation equations with inf-implication compositions in BL-algebras. In this paper, we investigate various solutions of fuzzy relation equations with inf-implication compositions in pseudo BL-algebras.


  • KEYWORD

    Pseudo BL-algebras , inf-implication compositions , fuzzy relation equations

  • 1. Introduction

    Sanchez [1] introduced the theory of fuzzy relation equations with various types of compositions: max-min, min-max, and min-α. Fuzzy relation equations with new types of compositions (continuous t-norm and residuated lattice) have been developed [2-5]. In particular, Bandler and Kohout [6] investigated the solvability of fuzzy relation equations with inf-implication compositions in complete lattices. Perfilieva and Noskova investigated the solvability of fuzzy relation equations with inf-implication compositions in BL-algebras. In contrast, noncommutative structures play an important role in metric spaces and algebraic structures (groups, rings, quantales, and pseudo BL-algebras) [7-15]. Georgescu and Iorgulescu [12] introduced pseudo MV-algebras as the generalization of MV-algebras. Georgescu and Leustean [11] introduced generalized residuated lattice as a noncommutative structure.

    In this paper, we investigate various solutions of fuzzy relation equations with inf-implication compositions AiR = Bi and AiR = Bi in pseudo BL-algebras.

    2. Preliminaries

    Definition 2.1. [11] A structure (L, ?, ?, ⊙, →, ⇒, ?, ⊥) is called apseudo BL-algebra if it satisfies the following conditions:

    (A1) (L, ?, ?, →, ?, ⊥) is bounded where ? is the universal upper bound and ⊥ denotes the universal lower bound;

    (A2) (L, ⊙, ?) is a monoid;

    (A3) it satisfies a residuation, i.e.,

    a ⊙ b ≤ c iff a ≤ b → c iff b ≤ a ⇒ c.

    (A4) ab = (ab) ⊙ a = a ⊙ (ab).

    (A5) (ab) ? (ba) = ? and (ab) ? (ba) = ?.

    We denote a0 = a →⊥ and a = a ⇒⊥.

    A pseudo BL-chain is a linear pseudo BL-algebra, i.e., a pseudo BL-algebra such that its lattice order is total.

    In this paper, we assume that (L, ?, ?, ⊙, →, ⇒, ?, ⊥) is a pseudo BL-algebra.

    Lemma 2.2. [11] For each x, y, z, xi, yiL, we have the following properties:

    (1) If yz, (xy) ≤ (xz), xyxz, and zxyx for →∈ {→, ⇒}.

    (2) xyxyxy.

    (3) (xy) → z = x → (yz) and (xy) ⇒ z = y ⇒ (xz).

    (4) x → (yz)= y ⇒ (xz) and x ⇒ (yz)= y → (xz).

    (5) x ⊙ (xy) ≤ y and (xy) ⊙ xy.

    (6) x ⊙ (yz)=(xy) ∨ (xz) and (xy) ⊙ z = (xz) ∨ (yz).

    (7) xy = ? iff xy iff xy = ?

    3. Fuzzy Relation Equations in Pseudo BL-Algebras

    Theorem 3.1. Let a =(a1,a2, ..., an) ∈ Ln and bL. We define two equations with respect to an unknown x = (x1, ..., (xn) ∈ Ln as

    image
    image

    Then, (1) (I) is solvable iff it has the least solution y = (y1, ..., yn) ∈ Ln such that yj = baj, j =1, ..., n.

    (2) (II) is solvable iff it has the least solution x =((x1, ..., (xn) ∈ Ln such that xj = ajb, j =1, ..., n.

    (3) If (I) is solvable, then

    image

    (4) If (II) is solvable, then

    image

    Proof. (1) (⇒) Let x =((x1, ..., (xn) be a solution of (I). Since

    image

    Moreover,

    image

    Thus, y = (ba1, ..., ban) is the least solution.

    (?) It is trivial.

    (3) Let x =(x1, ..., xn) denote a solution of (I). Then, b =

    image

    (2) and (4) are similarly proved as (1) and (3), respectively.

    Theorem 3.2. Let L denote a pseudo BL-chain in equations (I) and (II) of Theorem 3.1.

    (1) If b ? ? and

    image

    with B = {ajk |1 ≤ km, b = (ajk)*}, then

    image

    is a maximal solution of (II). Moreover, if x is a solution of (II), there exists k∈{jk |1 ≤ km} such that

    xjk = 0, j = k, xjajb, jk

    where there exists xjkX such that x ≤ xjk.

    (2) If b?? and

    image

    with B = {ajk |1 ≤ km, b = (ajk)0}, then

    image

    is a maximal solution of (I). Moreover, if x is a solution of (I), there exists k ∈ {jk |1 ≤ km} such that

    xjk = 0, j = k, xj ≥ b ⊙aj, j ≠ k

    where there exists xjkX such that x ≤ xjk.

    Proof. (1) (⇒)

    image

    is a solution of (II) because

    image

    Let x ≥ xjk be a solution of (II). Then,

    image

    with xjkajkb and

    image

    Since b ? 1,

    image

    Since L is linear, ajk >xjk . Since

    image

    we have

    xjk = ajk ∧ xjk = ajk ⊙ (ajk ⇒ xjk) = ajk ⊙ b = ajk ⊙ (ajk ⇒⊥)= ⊥.

    Thus, x = xjk.

    image

    is a maximal solution of (II).

    Let x =(x1, ..., xn) be a solution of (II). Since

    image

    by the linearity of L, there exists a family K = {jk | ajkB, ajk ⇒⊥ = b, 1 ≤ km} such that

    image

    , because by linearity of L, ajkB, (aj) >b implies that

    image

    For kK, since ak ⇒⊥ = akxk = b ≠ ? and L is linear, ak >xk and akb = ak ⊙ (akxk)= ak ⊙ (ak ⇒ ⊥) = ⊥ akxk = xk. Then,

    image

    (?) It is trivial.

    (2) It is similarly proved as (1).

    Example 3.3. Let K = {(x, y) ∈ R2 | x > 0} denote a set, and we define an operation ? : K × KK as follows:

    (x1,y1) ? (x2,y2)=(x1x2,x1y2 + y1).

    Then, (K, ?) is a group with e = (1, 0),

    image

    We have a positive cone P = {(a, b) ∈ R2 | a =1,b ≥ 0, or a> 1} because PP?1 = {(1, 0)}, PPP, (a, b)?1P ⊙ (a, b)= P, and PP?1 = K. For (x1,y1), (x2,y2) ∈ K, we define

    (x1,y1) ≤ (x2,y2) ⇔ (x1,y1)?1 ⊙ (x2,y2) ∈ P, (x2,y2) ⊙ (x1,y1)?1Px1x2 or x1 = x2,y1y2.

    Then, (K, ≤?) is a lattice-group with totally order ≤. (ref. [1])

    The structure

    image

    is a Pseudo BL-chain where

    image

    is the least element and ? = (1, 0) is the greatest element from the following statements:

    image

    Furthermore, we have (x, y)=(x, y)?? =(x, y)?? from:

    image

    (1) An equation is defined as

    image

    Since

    image

    by Theorem 3.1(3), it is not solvable.

    (2) An equation is defined as

    image

    Since

    image

    X = {x = ((x1,y1), (x2,y2), ⊥) or x = ((x1,y1), ⊥, (x3,y3)) | (x1,y1), (x2,y2), (x3,y3) ≥ ⊥}

    is a solution set of (I).

    M = {(?, ?, ⊥), (?, ⊥, ?)} is a maximal solution family of (I).

    (3) An equation is defined as

    image

    Since

    image

    by Theorem 3.1(3), it is not solvable.

    (4) An equation is defined as

    image

    Since

    image

    X = {x = ((x1,y1), (x2,y2), ⊥) | (x1,y1), (x2,y2) ≥ ⊥} is a solution family of (II). (?, ?, ⊥) is a maximal solution of (II).

    Definition 3.4. Let L denote a pseudo BL-chain. L satisfies the right conditional cancellation law if

    ? ? a ⊙ x ≤ a ⊙ y ⇒ x ≤ y.

    L satisfies the left conditional cancellation law if

    ? ? x ⊙ a ≤ y ⊙ a ⇒ x ≤ y.

    Theorem 3.5. Let L denote a pseudo BL-chain in two equations (I) and (II) of Theorem 3.1.

    Then, (1) If L satisfies the right conditional cancellation law b ? ? and

    image

    with B = {ajk |1 ≤ km, b ? (ajk)*}, then

    image

    is a maximal solution family of (II). Moreover, if x is a solution of (II), there exists a family K = {jk | ajkB, ajkxjk = b, 1 ≤ km} such that

    xk = ak ⊙ b, k ∈ K, xj ≥ aj ⊙ b, j ? K

    where there exists xjkX such that x ≤ xjk.

    (2) If L satisfies the left conditional cancellation law b ? ? and

    image

    with B = {ajk | 1 ≤ km, b= (ajk)0}, then

    image

    is a maximal solution of (I). Moreover, if x is a solution of (I), there exists k ∈ {jk | 1 ≤ km} such that

    xk = b⊙ak, j = k, xj ≥ b⊙aj, j ≠ k

    where there exists xjkX such that x ≤ xjk.

    Proof. (1)

    image

    is a solution of (II) because

    image

    Let x ≥ xjk denote a solution of (II). Then,

    image

    with xjkajkb and

    image

    Since b ? 1,

    image

    Since L is linear, ajkxjk. Thus,

    xjk = ajk ∧ xjk = ajk ⊙ (ajk ⇒ xjk ) = ajk ⊙ b.

    Therefore, x = xjk.

    image

    is a maximal solution of (II).

    Let x =(x1, ..., xn) denote a solution of (II). Since

    image

    by the linearity of L, there exists a family K = {jk | ajkB, ajkxjk = b, 1 ≤ km} such that

    image

    because ajkB, (aj)0b implies that

    image

    For kK, since akxk = b ≠ ? and L is linear, ak >xk and akb = ak ⊙ (akxk)= akxk = xk. For jK, since ajxjb, xjajb. Hence,

    xk = ak ⊙ b, k ∈ K, xj ≥ aj ⊙ b, j ? K

    (?) It is trivial.

    (2) It is similarly proved as (1).

    Example 3.6. The structure

    image

    is defined as that in Example 3.3. Then, L satisfies the right conditional cancellation law because

    ⊥ ? (a, b) ⊙ (x1,y1) ≤ (a, b) ⊙ (x2,y2)

    (⇔)⊥ ? (ax1, ay1 + b) ≤ (ax2, ay2 + b)

    (⇒)ax1 = ax2, ay1 + b ≤ ay1 + b, or ax1 ? ax2

    (⇒)x1 = x2,y1 ≤ y1, or x1 ? x2

    (⇒)(x1,y1)

    ≤ (x2,y2).

    Similarly, L satisfies the left conditional cancellation law.

    (1) An equation is defined as

    image

    Since

    image

    is a maximal solution of (II) because

    image

    X = {x = ((x1, y1), (x2, y2), ⊥ | (x1, y1), (x2, y2) ≥ ⊥} is a solution set of (II).

    (2) An equation is defined as

    image

    Since

    image

    and

    image

    and

    image

    are maximal solutions of (II) because

    image

    is a solution set of (II).

    (3) An equation is defined as

    image

    Since

    image

    X = {x = ((x1, y1), (x2, y2), ⊥) or x = (x1, y1), ⊥, (x3, y3)) | (x1, y1), (x2, y2), (x3, y3) ≥ ⊥}

    is a solution set of (I).

    Theorem 3.7. Let ai =(ai1,ai2, ..., ain) ∈ Ln and biL. We define two equations with respect to an unknown x = (x1, ..., xn) ∈ Ln as

    image
    image

    Then, (1) (III) is solvable iff it has the least solution x = (x1, ..., xn) ∈ Ln such that

    image

    (2) (IV) is solvable iff it has the least solution x = (x1, ..., xn) ∈ Ln such that

    image

    (3) If (III) is solvable, then

    image

    (4) If (IV) is solvable, then

    image

    (5) If (III) (resp. (IV)) is solvable and x1, ..., xm is a solution of each ith equation, i =1, 2, ..., m, then

    image

    is a solution of (III) (resp. (IV)). Moreover, if each solution xi of the ith equation is maximal, any maximal solution x of (III) (resp. (IV)) is

    image

    Proof. (1) (⇒) Let y =(y1, ..., yn) denote a solution of (III). Since

    image

    Moreover,

    image

    Then,

    image

    Substitute

    image

    Thus, (x1, ..., xn) is the least solution.

    (?) It is trivial.

    (3)

    image

    (2) and (4) are similarly proved as (1) and (3), respectively.

    5) Let xi =(xi1, ..., xin) denote a solution of the ith equation in (III) and

    image

    Then,

    image

    Moreover,

    image

    Hence,

    image

    Therefore,

    image

    is a solution of (III).

    Moreover, if xi is a maximal solution of the ith equation in (III), then

    image

    is a solution of (III). Let y = (y1, ..., yn) denote a solution of (III). Then, y ≤ xi for each i = 1, ...m. Then,

    image

    Hence, x is a maximal solution of (III).

    Example 3.8. The structure

    image

    is defined as that in Example 3.3.

    (1) An equation is defined as

    image
    image

    is a solution set.

    image

    is a maximal solution set.

    (2) An equation is defined as

    image

    is a solution set.

    image

    is a maximal solution set.

    image

    is a solution set of (1) and (2).

    image

    is a maximal solution set of (1) and (2).

    (3) An equation is defined as

    image
    image

    is a solution set.

    image

    is a maximal solution set.

    (4) An equation is defined as

    image
    image

    is a solution set.

    image

    is a maximal solution set.

    image

    is a solution set of (3) and (4).

    image

    is a maximal solution set of (3) and (4).

    4. Conclusion

    Bandler and Kohout [6] investigated the solvability of fuzzy relation equations with inf-implication compositions in complete lattices. Perfilieva and Noskova investigated the solvability of fuzzy relation equations with inf-implication compositions in BL-algebras. In this paper, we investigated various solutions of fuzzy relation equations with inf-implication compositions in pseudo BL-algebras.

    In the future, we will investigate various solutions of fuzzy relation equations with sup-compositions in pseudo BL-algebras and other algebraic structures.

      >  Conflict of Interest

    No potential conflict of interest relevant to this article was reported.

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