Fuzzy relation equations in pseudo BLalgebras
 Author: Kim Yong Chan
 Organization: Kim Yong Chan
 Publish: International Journal of Fuzzy Logic and Intelligent Systems Volume 13, Issue3, p208~214, 25 Sep 2013

ABSTRACT
Bandler and Kohout investigated the solvability of fuzzy relation equations with infimplication compositions in complete lattices. Perfilieva and Noskova investigated the solvability of fuzzy relation equations with infimplication compositions in BLalgebras. In this paper, we investigate various solutions of fuzzy relation equations with infimplication compositions in pseudo BLalgebras.

KEYWORD
Pseudo BLalgebras , infimplication compositions , fuzzy relation equations

1. Introduction
Sanchez [1] introduced the theory of fuzzy relation equations with various types of compositions: maxmin, minmax, and min
α . Fuzzy relation equations with new types of compositions (continuous tnorm and residuated lattice) have been developed [25]. In particular, Bandler and Kohout [6] investigated the solvability of fuzzy relation equations with infimplication compositions in complete lattices. Perfilieva and Noskova investigated the solvability of fuzzy relation equations with infimplication compositions in BLalgebras. In contrast, noncommutative structures play an important role in metric spaces and algebraic structures (groups, rings, quantales, and pseudo BLalgebras) [715]. Georgescu and Iorgulescu [12] introduced pseudo MValgebras as the generalization of MValgebras. Georgescu and Leustean [11] introduced generalized residuated lattice as a noncommutative structure.In this paper, we investigate various solutions of fuzzy relation equations with infimplication compositions
A_{i} ⇒R =B_{i} andA_{i} →R =B_{i} in pseudo BLalgebras.2. Preliminaries
Definition 2.1. [11] A structure (L , ？, ？, ⊙, →, ⇒, ？, ⊥) is calledapseudo BLalgebra 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)
a ？b = (a →b ) ⊙a =a ⊙ (a ⇒b ).(A5) (
a →b ) ？ (b →a ) = ？ and (a ⇒b ) ？ (b ⇒a ) = ？.We denote
a ^{0} =a →⊥ anda ^{？} =a ⇒⊥.A pseudo BLchain is a linear pseudo BLalgebra, i.e., a pseudo BLalgebra such that its lattice order is total.
In this paper, we assume that (
L , ？, ？, ⊙, →, ⇒, ？, ⊥) is a pseudo BLalgebra.Lemma 2.2. [11] For eachx ,y ,z ,x_{i} ,y_{i} ∈L , we have the following properties:(1) If
y ≤z , (x ⊙y ) ≤ (x ⊙z ),x →y ≤x →z , andz →x ≤y →x for →∈ {→, ⇒}.(2)
x ⊙y ≤x ∧y ≤x ∨y .(3) (
x ⊙y ) →z =x → (y →z ) and (x ⊙y ) ⇒z =y ⇒ (x ⇒z ).(4)
x → (y ⇒z )=y ⇒ (x →z ) andx ⇒ (y →z )=y → (x ⇒z ).(5)
x ⊙ (x ⇒y ) ≤y and (x →y ) ⊙x ≤y .(6)
x ⊙ (y ∨z )=(x ⊙y ) ∨ (x ⊙z ) and (x ∨y ) ⊙z = (x ⊙z ) ∨ (y ⊙z ).(7)
x →y = ？ iffx ≤y iffx ⇒y = ？3. Fuzzy Relation Equations in Pseudo BLAlgebras
Theorem 3.1. Leta =(a _{1},a _{2}, ...,a_{n} ) ∈L^{n} andb ∈L . We define two equations with respect to an unknown x = (x _{1}, ..., (x_{n} ) ∈L^{n} asThen, (1) (I) is solvable iff it has the least solution y = (
y _{1}, ...,y_{n} ) ∈L^{n} such thaty_{j} =b ⊙a_{j} ,j =1, ...,n .(2) (II) is solvable iff it has the least solution x =((
x _{1}, ..., (x_{n} ) ∈L^{n} such thatx_{j} =a_{j} ⊙b ,j =1, ...,n .(3) If (I) is solvable, then
(4) If (II) is solvable, then
Proof . (1) (⇒) Let x =((x _{1}, ..., (x_{n} ) be a solution of (I). SinceMoreover,
Thus, y = (
b ⊙a _{1}, ...,b ⊙a_{n} ) is the least solution.(？) It is trivial.
(3) Let x =(
x _{1}, ...,x_{n} ) denote a solution of (I). Then,b =(2) and (4) are similarly proved as (1) and (3), respectively.
Theorem 3.2. LetL denote a pseudo BLchain in equations (I) and (II) of Theorem 3.1.(1) If
b ？ ？ andwith
B = {a_{jk} 1 ≤k ≤m ,b = (a_{jk} )*}, thenis a maximal solution of (II). Moreover, if
x is a solution of (II), there existsk ∈{j_{k} 1 ≤k ≤m } such thatx_{jk} = 0,j =k ,x_{j} ≥a_{j} ⊙b ,j ≠k where there exists x_{jk} ∈
X such that x ≤ x_{jk}.(2) If
b ？？ andwith
B = {a_{jk} 1 ≤k ≤m ,b = (a_{jk} )^{0}}, thenis a maximal solution of (I). Moreover, if
x is a solution of (I), there existsk ∈ {j_{k} 1 ≤k ≤m } such thatxjk = 0, j = k, xj ≥ b ⊙aj, j ≠ k
where there exists x_{jk} ∈
X such that x ≤ x_{jk}.Proof . (1) (⇒)is a solution of (II) because
Let x ≥ x_{jk} be a solution of (II). Then,
with
x_{jk} ≥a_{jk} ⊙b andSince
b ？ 1,Since
L is linear,a_{jk} >x_{jk} . Sincewe have
xjk = ajk ∧ xjk = ajk ⊙ (ajk ⇒ xjk) = ajk ⊙ b = ajk ⊙ (ajk ⇒⊥)= ⊥.
Thus, x = x_{jk}.
is a maximal solution of (II).
Let x =(
x _{1}, ...,x_{n} ) be a solution of (II). Sinceby the linearity of
L , there exists a familyK = {j_{k} a_{jk} ∈B ,a_{jk} ⇒⊥ =b , 1 ≤k ≤m } such that, because by linearity of
L ,a_{jk} ？B , (a_{j} )^{？} >b implies thatFor
k ∈K , sincea_{k} ⇒⊥ =a_{k} ⇒x_{k} =b ≠ ？ andL is linear,a_{k} >x_{k} anda_{k} ⊙b =a_{k} ⊙ (a_{k} ⇒x_{k} )=a_{k} ⊙ (a_{k} ⇒ ⊥) = ⊥a_{k} ∧x_{k} =x_{k} . Then,(？) It is trivial.
(2) It is similarly proved as (1).
Example 3.3. LetK = {(x ,y ) ∈R ^{2} x > 0} denote a set, and we define an operation ？ :K ×K →K as follows:(x1,y1) ？ (x2,y2)=(x1x2,x1y2 + y1).
Then, (
K , ？) is a group withe = (1, 0),We have a positive cone
P = {(a ,b ) ∈R ^{2} a =1,b ≥ 0, ora > 1} becauseP ∩P ^{？1} = {(1, 0)},P ⊙P ⊂P , (a ,b )^{？1} ⊙P ⊙ (a ,b )=P , andP ∪P ^{？1} =K . For (x _{1},y _{1}), (x _{2},y _{2}) ∈K , we define(
x _{1},y _{1}) ≤ (x _{2},y _{2}) ⇔ (x _{1},y _{1})^{？1} ⊙ (x _{2},y _{2}) ∈P , (x _{2},y _{2}) ⊙ (x _{1},y _{1})^{？1} ∈P ⇔x _{1} ？x _{2} orx _{1} =x _{2},y _{1} ≤y _{2}.Then, (
K , ≤？) is a latticegroup with totally order ≤. (ref. [1])The structure
is a Pseudo BLchain where
is the least element and ？ = (1, 0) is the greatest element from the following statements:
Furthermore, we have (
x ,y )=(x ,y )^{？？} =(x ,y )^{？？} from:(1) An equation is defined as
Since
by Theorem 3.1(3), it is not solvable.
(2) An equation is defined as
Since
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
Since
by Theorem 3.1(3), it is not solvable.
(4) An equation is defined as
Since
X = {x = ((x _{1},y _{1}), (x _{2},y _{2}), ⊥)  (x _{1},y _{1}), (x _{2},y _{2}) ≥ ⊥} is a solution family of (II). (？, ？, ⊥) is a maximal solution of (II).Definition 3.4. LetL denote a pseudo BLchain.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. LetL denote a pseudo BLchain in two equations (I) and (II) of Theorem 3.1.Then, (1) If
L satisfies the right conditional cancellation lawb ？ ？ andwith
B = {a_{jk} 1 ≤k ≤m ,b ？ (a_{jk} )^{*}}, thenis a maximal solution family of (II). Moreover, if
x is a solution of (II), there exists a familyK = {j_{k} a_{jk} ∈B ,a_{jk} ⇒x_{jk} =b , 1 ≤k ≤m } such thatxk = ak ⊙ b, k ∈ K, xj ≥ aj ⊙ b, j ？ K
where there exists x_{jk} ∈
X such that x ≤ x_{jk}.(2) If
L satisfies the left conditional cancellation lawb ？ ？ andwith
B = {a_{jk}  1 ≤k ≤m ,b = (a_{jk} )^{0}}, thenis a maximal solution of (I). Moreover, if x is a solution of (I), there exists
k ∈ {j_{k}  1 ≤k ≤m } such thatxk = b⊙ak, j = k, xj ≥ b⊙aj, j ≠ k
where there exists x_{jk} ∈
X such that x ≤ x_{jk}.Proof . (1)is a solution of (II) because
Let x ≥ x_{jk} denote a solution of (II). Then,
with
x_{jk} ≥a_{jk} ⊙b andSince
b ？ 1,Since
L is linear,a_{jk} ？x_{jk} . Thus,xjk = ajk ∧ xjk = ajk ⊙ (ajk ⇒ xjk ) = ajk ⊙ b.
Therefore, x = x_{jk}.
is a maximal solution of (II).
Let x =(
x _{1}, ...,x_{n} ) denote a solution of (II). Sinceby the linearity of
L , there exists a familyK = {j_{k} a_{jk} ∈B ,a_{jk} ⇒x_{jk} =b , 1 ≤k ≤m } such thatbecause
a_{jk} ？B , (a_{j} )^{0} ≥b implies thatFor
k ∈K , sincea_{k} ⇒x_{k} =b ≠ ？ andL is linear,a_{k} >x_{k} anda_{k} ⊙b =a_{k} ⊙ (a_{k} ⇒x_{k} )=a_{k} ∧x_{k} =x_{k} . Forj ？K , sincea_{j} ⇒x_{j} ≥b ,x_{j} ≥a_{j} ⊙b . 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 structureis 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
Since
is a maximal solution of (II) because
X = {x = ((x _{1},y _{1}), (x _{2},y _{2}), ⊥  (x _{1},y _{1}), (x _{2},y _{2}) ≥ ⊥} is a solution set of (II).(2) An equation is defined as
Since
and
and
are maximal solutions of (II) because
is a solution set of (II).
(3) An equation is defined as
Since
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 =(a _{i}a _{i1},a _{i2}, ...,a_{in} ) ∈L^{n} andb_{i} ∈L . We define two equations with respect to an unknown x = (x _{1}, ...,x_{n} ) ∈L^{n} asThen, (1) (III) is solvable iff it has the least solution x = (
x _{1}, ...,x_{n} ) ∈L^{n} such that(2) (IV) is solvable iff it has the least solution x = (
x _{1}, ...,x_{n} ) ∈L^{n} such that(3) If (III) is solvable, then
(4) If (IV) is solvable, then
(5) If (III) (resp. (IV)) is solvable and x_{1}, ..., x
_{m} is a solution of each ith equation,i =1, 2, ...,m , thenis a solution of (III) (resp. (IV)). Moreover, if each solution
x_{i} of the ith equation is maximal, any maximal solution x of (III) (resp. (IV)) isProof . (1) (⇒) Let y =(y _{1}, ...,y_{n} ) denote a solution of (III). SinceMoreover,
Then,
Substitute
Thus, (
x _{1}, ...,x_{n} ) is the least solution.(？) It is trivial.
(3)
(2) and (4) are similarly proved as (1) and (3), respectively.
5) Let
x_{i} =(x _{i1}, ...,x_{in} ) denote a solution of the ith equation in (III) andThen,
Moreover,
Hence,
Therefore,
is a solution of (III).
Moreover, if
x_{i} is a maximal solution of the ith equation in (III), thenis a solution of (III). Let y = (
y _{1}, ...,y_{n} ) denote a solution of (III). Then, y ≤ x_{i} for eachi = 1, ...m . Then,Hence, x is a maximal solution of (III).
Example 3.8. The structureis defined as that in Example 3.3.
(1) An equation is defined as
is a solution set.
is a maximal solution set.
(2) An equation is defined as
is a solution set.
is a maximal solution set.
is a solution set of (1) and (2).
is a maximal solution set of (1) and (2).
(3) An equation is defined as
is a solution set.
is a maximal solution set.
(4) An equation is defined as
is a solution set.
is a maximal solution set.
is a solution set of (3) and (4).
is a maximal solution set of (3) and (4).
4. Conclusion
Bandler and Kohout [6] investigated the solvability of fuzzy relation equations with infimplication compositions in complete lattices. Perfilieva and Noskova investigated the solvability of fuzzy relation equations with infimplication compositions in BLalgebras. In this paper, we investigated various solutions of fuzzy relation equations with infimplication compositions in pseudo BLalgebras.
In the future, we will investigate various solutions of fuzzy relation equations with supcompositions in pseudo BLalgebras and other algebraic structures.
> Conflict of Interest
No potential conflict of interest relevant to this article was reported.