# On Fuzzy Positive Implicative Filters inBE-Algebras

Authors: Sun Shin Ahn; Jeong Soon Han
Journal: The Scientific World Journal
 (2014)

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## Abstract

We study several degrees in defining a fuzzy positive implicative filter, which is a generalization of a fuzzy filter inBE-algebras.


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## Body

## 1. Introduction

In [1], H. S. Kim and Y. H. Kim introduced the notion of a BE-algebra. Ahn and So [2, 3] introduced the notion of ideals in BE-algebras. Ahn et al. [4] fuzzified the concept of BE-algebras and investigated some of their properties. Jun and Ahn [5] provided several degrees in defining a fuzzy implicative filter.In this paper, we study several degrees in defining a fuzzy positive implicative filter, which is a generalization of a fuzzy filter inBE-algebras.


## 2. Preliminaries

We recall some definitions and results discussed in [1–3].An algebra(X;*,1) of type (2, 0) is called a BE-algebra if(BE1)
x*x=1 for all x∈X,(BE2)
x*1=1 for all x∈X,(BE3)
1*x=x for all x∈X,(BE4)
x*(y*z)=y*(x*z), for all x,y,z∈X (exchange).We introduce a relation “≤” on a BE-algebra X by x≤y, if and only if x*y=1. A nonempty subset S of a BE-algebra X is said to be asubalgebra of X, if it is closed under the operation “*.” By noticing that x*x=1, for all x∈X, it is clear that 1∈S. A BE-algebra (X;*,1) is said to be self-distributive, if x*(y*z)=(x*y)*(x*z), for all x,y,z∈X.Definition 1 (see [1]).
Let(X;*,1) be a BE-algebra and let F be a nonempty subset of X. Then, F is called afilter of X if (F1)
1∈F,(F2)
x*y∈F and x∈F imply y∈F,for all x,y,z∈X.
A nonempty subsetF of a BE-algebra X is called an implicative filter of X if it satisfies (F1) and (F3)
x*(y*z)∈F and x*y∈F imply x*z∈F, for all x,y,z∈X.Example 2 (see [1]).
LetX:={1,a,b,c,d,0} be a BE-algebra with the following table:
(1)*1abcd011abcd0a11accdb111cccc1ab1abd11a11a0111111
Then F1:={1,a,b} is a filter of X, but F2:={1,a} is not a filter of X, since a*b∈F2 and a∈F2, but b∉F2.Proposition 3.
Let(X;*,1) be a BE-algebra and let F be a filter of X. If x≤y and x∈F, for any y∈X, then y∈F.Proposition 4.
Let(X;*,1) be a self-distributive BE-algebra. Then, the following hold, for any x,y,z∈X: (i)
ifx≤y,
thenz*x≤z*yandy*z≤x*z;(ii)
y*z≤(z*x)*(y*x);(iii)
y*z≤(x*y)*(x*z).ABE-algebra (X;*,1) is said to be transitive if it satisfies Proposition 4 (iii).Definition 5 (see [5]).
A fuzzy subsetμ of a BE-algebra X is called a fuzzy filter of X, if it satisfies, for all x,y∈X, (d1)
μ(1)≥μ(x),(d2)
μ(x)≥min⁡{μ(y*x),μ(y)}. A fuzzy subset μ of a BE-algebra X is called a fuzzy implicative filter of X if it satisfies (d1) and (d3)
μ(x*z)≥min⁡{μ(x*(y*z)),μ(x*y)}, for all x,y,z∈X.Definition 6 (see [5]).
LetF be a nonempty subset of a BE-algebra X which is not necessarily a filter of X. One says that a subset G of X is an enlarged filter of X related to F, if it satisfies the following:(1)
F is a subset of G,(2)
1∈G,(3)
(∀x,y∈X)(∀x∈F)(x*y∈F⇒y∈G).


## 3. Fuzzy Positive Implicative Filters ofBE-Algebras with Degrees in (0,1]

X denotes a BE-algebra unless specified otherwise.Definition 7.
A nonempty subsetF of X is called a positive implicative filter of a BE-algebra X if it satisfies (F1) and (F4)
x*((y*z)*y)∈F and x∈F imply y∈F, for all x,y,z∈X.
Note that every positive implicative filter of aBE-algebra X is a filter of X.Example 8.
(1) LetX:={1,a,b,c,d} be a self-distributive BE-algebra ([1]) with the following table:
(2)*1abcd11abcda11bcdb1a1ccc11b1bd11111
Then {1,b} is an implicative filter of X but not a positive implicative filter of X, since b*((a*d)*a)=1,b∈{1,b} and a∉{1,b}.
(2) Consider aBE-algebra X:={1,a,b,c,d,0} as in Example 2. Then {1} is a filter of X but not an implicative filter of X, since d*(a*0)=1∈{1},d*a=1∈{1}, and d*0=a∉{1}. Also, it is not a positive implicative filter of X, since 1*((a*b)*a)=1∈{1},1∈{1} and a∉{1}.
(3) LetX:={1,a,b,c} be a self-distributive BE-algebra ([5]) with the following table:
(3)*1abc11abca11bcb1a1cc11b1
Then {1,b} is an implicative filter of X and {1,a,b} is a positive implicative filter of X.Definition 9.
A fuzzy subsetμ of a BE-algebra X is called a fuzzy positive implicative filter of X, if it satisfies (d1) and (d4)
μ(y)≥min⁡{μ(x*((y*z)*y)),μ(x)}, for all x,y∈X.Definition 10.
LetF be a nonempty subset of a BE-algebra X which is not necessarily a positive implicative filter of X. One says that a subset G of X is an enlarged positive implicative filter of X related to F, if it satisfies the following:(1)
F is a subset of G,(2)
1∈G,(3)
(∀x,y,z∈X)(∀x∈F)(x*((y*z)*y)∈F⇒y∈G).
Obviously, every positive implicative filter is an enlarged positive implicative filter ofX related to itself. Note that there exists an enlarged positive implicative filter of X related to any nonempty subset F of X.Example 11.
Consider aBE-algebra X={1,a,b,c,d} which is given in Example 8 (1). Note that F:={1,b} is not a positive implicative filter. Then, G:={1,a,b,c} is an enlarged positive implicative filter of X related to F and G is not a positive implicative filter of X, since b*((d*d)*d)=c∈G,b∈G and d∉G.Proposition 12.
LetF be a nonempty subset of a BE-algebra X. Every enlarged positive implicative filter of X related to F is an enlarged filter of X related to F.Proof.
LetG be an enlarged positive implicative filter of X related to F. By putting z:=1 in Definition 10 (3), we have
(4)(∀x,y∈X)(x*((y*1)*y)=x*y∈F,s((y*1)*y)x∈F⟹y∈G).
Hence, G is an enlarged filter of X related to F.The converse of Proposition12 is not true in general as seen in the following example.Example 13.
LetX:={1,a,b,c} be a transitive BE-algebra ([2]) with the following table:
(5)*1abc11abca11aab111ac11a1
Let F:={1} and G:={1,c}. Then G is an enlarged filter of F but it is not an enlarged positive implicative filter of F, since 1*((a*c)*a)=1∈F,1∈F and a∉G.
In what follows letλ and κ be members of (0,1], and let n and k denote a natural number and a real number, respectively, such that k<n, unless otherwise specified.Definition 14 (see [5]).
A fuzzy subsetμ of a BE-algebra X is called a fuzzy filter of X with degree (λ,κ), if it satisfies the following:(e1)
(∀x∈X)(μ(1)≥λμ(x)),(e2)
(∀x,y∈X)(μ(x)≥κmin⁡{μ(y*x),μ(y)}).
A fuzzy subsetμ of a BE-algebra X is called a fuzzy implicative filter of X with degree (λ,κ), if it satisfies (e1) and(e3)
(∀x,y,z∈X)(μ(x*z)≥κmin⁡{μ(x*(y*z)),μ(x*y)}).Definition 15.
A fuzzy subsetμ of a BE-algebra X is called a fuzzy positive implicative filter of X with degree (λ,κ), if it satisfies (e1) and(e4)
(∀x,y,z∈X)(μ(y)≥κmin⁡{μ(x*((y*z)*y)),μ(x)}).Proposition 16 (see [5]).
Every fuzzy filter of aBE-algebra X with degree (λ,κ) satisfies the following assertions:(i)
(∀x,y∈X)(μ(x*y)≥λκμ(y));(ii)
(∀x,y∈X)(y≤x⇒μ(x)≥λκμ(y));(iii)
(∀x,y,z∈X)(x≤y*z⇒μ(z)≥min⁡{κμ(y),λκ2μ(x)}).Note that ifλ≠κ, then a fuzzy positive implicative filter with degree (λ,κ) may not be a fuzzy positive implicative filter with degree (κ,λ) and vice versa. Obviously, every fuzzy positive implicative filter is a fuzzy positive implicative filter with degree (λ,κ), but the converse may not be true.Example 17.
Consider a self-distributiveBE-algebra X={1,a,b,c} which is given in Example 8 (3). Define a fuzzy subset μ:X→[0,1] by
(6)μ=(1abc0.70.40.80.4).
Then, μ is a fuzzy implicative filter of X with degree (2/5,3/5) and a fuzzy filter of X with degree (2/5,3/5), but it is neither a fuzzy filter of X nor a fuzzy positive implicative filter of X with degree (2/5,3/5) since
(7)μ(1)=0.7≱0.8=μ(b),μ(a)=0.4≱0.42=35×0.7μ(a)=35×μ(1)μ(a)=35×min⁡{μ(1*((a*c)*a))=μ(1),μ(1)}.Example 18.
LetX:={1,a,b,c} be a BE-algebra ([2]) with the following table:
(8)*1abc11abca11bcb1111c1ac1
Define a fuzzy subset μ:X→[0,1] by
(9)μ=(1abc0.70.80.40.4).
Then, μ is a fuzzy positive implicative filter of X with degree (3/5,2/5). But it is neither a fuzzy filter of X nor a fuzzy positive implicative filter of X with degree (2/5,3/5) since
(10)μ(1)=0.7≱0.8=μ(a),μ(c)=0.4≱0.42=35×0.7μ(c)=35×μ(1)μ(c)=35×min⁡{μ(a*((c*b)*c))=μ(1),μ(a)}.
Also, it is not a fuzzy implicative filter ofX with degree (2/5,3/5) since
(11)μ(c*b)=μ(c)=0.4≱0.42=35×0.7=35×μ(1)=35×min⁡{μ(c*(c*b))sssssssssss(c*(c*b))=μ(1),μ(c*c)=μ(1)}.Proposition 19.
Ifμ is a fuzzy positive implicative filter of a BE-algebra X with degree (λ,κ), then μ is a fuzzy filter of X with degree (λ,κ).Proof.
By puttingz:=y in (e4), we have
(12)μ(y)≥κmin⁡{μ(x*((y*y)*y)),μ(x)}=κmin⁡{μ(x*y),μ(x)}
for any x,y∈X. Thus, μ is a fuzzy filter of X with degree (λ,κ).The converse of Proposition19 is not true in general (see Example 17).Note that a fuzzy filter with degree(λ,κ) is a fuzzy filter if and only if (λ,κ)=(1,1).Proposition 20.
Letμ be a fuzzy positive implicative filter of a BE-algebra X with degree (λ,κ). Then, the following holds:
(13)(∀x,y∈X)(μ(x)≥κλμ((x*y)*x)).Proof.
Assume thatμ is a fuzzy positive implicative filter of a BE-algebra X with degree (λ,κ) and let x,y∈X. Using (e4) and (e1), we have
(14)μ(x)≥κmin⁡{μ(1*((x*y)*x)),μ(1)}=κmin⁡{μ((x*y)*x),λμ((x*y)*x)}=κλμ((x*y)*x).
This completes the proof.Proposition 21.
Letμ be a fuzzy filter of a BE-algebra X with degree (λ,κ) satisfying
(15)(∀x,y∈X)(μ(x)≥μ((x*y)*x)).
Then, μ is a positive implicative filter of X with degree (λ,κ).Proof.
Letx,y,z∈X. Using (e2), we have
(16)μ(y)≥μ((y*z)*y)≥κmin⁡{μ(x*((y*z)*y)),μ(x)}.Thus,μ is a positive implicative filter of a BE-algebra X with degree (λ,κ).Corollary 22.
Letμ be a fuzzy filter of X. Then, μ is a fuzzy positive implicative filter of X, if and only if
(17)(∀x,y∈X)(μ(x)≥μ((x*y)*x)).Proof.
It follows from Propositions20 and 21.Proposition 23.
Every fuzzy positive implicative filter of aBE-algebra X with degree (λ,κ) satisfies the following assertions:(i)
(∀x,y∈X)(μ(x*y)≥λκμ(y));(ii)
(∀x,y∈X)(x≤y⇒μ(y)≥λκμ(x)).Proof.
It follows from Propositions16 and 19.Corollary 24.
Letμ be a fuzzy positive implicative filter of a BE-algebra X with degree (λ,κ). If λ=κ, then(i)
(∀x,y∈X)(μ(x*y)≥λ2μ(y)),(ii)
(∀x,y∈X)(x≤y⇒μ(y)≥λ2μ(x)).Proposition 25.
LetX be a self-distributive BE-algebra X. Let μ be a fuzzy positive implicative filter of X with degree (λ,κ). Then,
(18)(∀x,y∈X)(μ((y*x)*x)≥λ2κ3μ(y)).Proof.
Letx,y∈X. Let μ be a fuzzy positive implicative filter of X with degree (λ,κ). By Proposition 19, μ is a fuzzy filter of X with degree (λ,κ). Since x≤(y*x)*x, using Proposition 4 (i), we have ((y*x)*x)*y≤x*y. Hence,
(19)(x*y)*y≤(y*x)*((x*y)*x)=(x*y)*((y*x)*x)≤(((y*x)*x)*y)*((y*x)*x).
Using Propositions 16 and 23, we have
(20)μ((y*x)*x)≥min⁡{κμ(((y*x)*x)*y),λκ2μ((x*y)*y)}≥min⁡{κ2λμ(y),(λκ2)κλμ(y)}=κ2λmin⁡{μ(y),λκμ(y)}=κ2λ(λκ)μ(y)=λ2κ3μ(y).
This completes the proof.Definition 26.
([6]) Let X be a BE-algebra. X is said to be commutative if the following identity holds:(C)
(x*y)*y=(y*x)*x; that is, x∨y=y∨x, where x∨y=(y*x)*x, for all x,y∈X.Theorem 27.
LetX be a commutative self-distributive BE-algebra. Every fuzzy positive implicative filter of X with degree (λ,κ) is a fuzzy implicative filter of X with degree (λ,κ3λ2).Proof.
Letμ be a fuzzy positive implicative filter of X with degree (λ,κ). By Proposition 19, μ is a fuzzy filter of X with degree (λ,κ). Using (BE4) and Proposition 4 (iii), we obtain (x*(y*z))*((x*y)*(x*(x*z)))=1, for any x,y,z∈X. Hence, by Proposition 16 (iii), we have μ(x*(x*z))≥min⁡{κμ(x*(y*z)),λκ2μ(x*y)}. On the other hand, using (BE4) and (C), we obtain
(21)((x*z)*z)*(x*z)=x*(((x*z)*z)*z)=x*((z*(x*z))*(x*z))=x*(1*(x*z))=x*(x*z).
Using Proposition 20, we have
(22)μ(x*z)≥κλμ(((x*z)*z)*(x*z))=κλμ(x*(x*z))≥κλmin⁡{κμ(x*(y*z)),λκ2μ(x*y)}=κ2λmin⁡{μ(x*(y*z)),κλμ(x*y)}≥κ3λ2min⁡{μ(x*(y*z)),μ(x*y)}.
This completes the proofDenote byFPI(X) the set of all positive implicative filters of a BE-algebra X. Note that a fuzzy subset μ of a BE-algebra X is a fuzzy positive implicative filter of X, if and only if
(23)(∀t∈[0,1])(U(μ;t)∈FPI(X)∪{∅}).
But we know that, for any fuzzy subset μ of a BE-algebra X, there exist λ,κ∈(0,1) and t∈[0,1] such that(1)
μ is a fuzzy positive implicative filter of X with degree (λ,κ),(2)
U(μ;t)∉FPI(X)∪{∅}.Example 28.
Consider aBE-algebra X={1,a,b,c,d} which is given in Example 8 (1). Define a fuzzy subset μ:X→[0,1] by
(24)μ=(1abcd0.40.30.50.30.3).
If t∈(0.4,0.6], then U(μ;t)={1,b} is not a positive implicative filter of X, since b*((a*d)*a)=1,b∈{1,b}, and a∉{1,b}. But μ is a fuzzy positive implicative filter of X with degree (0.4,0.6).Theorem 29.
Letμ be a fuzzy subset of a BE-algebra X. For any t∈[0,1] with t≤max⁡{λ,κ}, if U(μ;t) is an enlarged positive implicative filter of X related to U(μ;t/max⁡{λ,κ}), then μ is a fuzzy positive implicative filter of X with degree (λ,κ).Proof.
Assume thatμ(1)<t≤λμ(x), for some x∈X and t∈(0,λ]. Then μ(x)≥t/λ≥t/max⁡{λ,κ} and so x∈U(μ;t/max⁡{λ,κ}); that is, U(μ;t/max⁡{λ,κ})≠∅. Since U(μ;t) is an enlarged filter of X related to U(μ;t/max⁡{λ,κ}), we have 1∈U(μ;t); that is, μ(1)≥t. This is a contradiction, and thus μ(1)≥λμ(x), for all x∈X.
Now suppose that there exista,b,c∈X such that μ(b)<κmin⁡{μ(a*((b*c)*b)),μ(a)}. If we take t:=κmin⁡{μ(a*((b*c)*b)),μ(a)}, then t∈(0,κ]⊆(0,max⁡{λ,κ}]. Hence, a*((b*c)*b)∈U(μ;t/κ)⊆U(μ;t/max⁡{λ,κ}) and a∈U(μ;t/κ)⊆U(μ;t/max⁡{λ,κ}). It follows from Definition 10 (3) that b∈U(μ;t) so that μ(b)≥t, which is impossible. Therefore,
(25)μ(y)≥κmin⁡{μ(x*((y*z)*y)),μ(x)}
for all x,y,z∈X. Thus, μ is a fuzzy positive implicative filter of X with degree (λ,κ).Corollary 30.
Letμ be a fuzzy subset of a BE-algebra X. For any t∈[0,1] with t≤k/n, if U(μ;t) is an enlarged positive implicative filter of X related to U(μ;(n/k)t), then μ is a fuzzy positive implicative filter of X with degree (k/n,k/n).Theorem 31.
Lett∈[0,1] be such that U(μ;t)(≠∅) is not necessarily a positive implicative filter of a BE-algebra X. If μ is a fuzzy positive implicative filter of X with degree (λ,κ), then U(μ;tmin⁡{λ,κ}) is an enlarged positive implicative filter of X related to U(μ;t).Proof.
Sincetmin⁡{λ,κ}≤t, we have U(μ;t)⊆U(μ;tmin⁡{λ,κ}). Since U(μ;t)≠∅, there exists x∈U(μ;t) and so μ(x)≥t. By (e1), we obtain μ(1)≥λμ(x)≥λt≥tmin⁡{λ,κ}. Therefore, 1∈U(μ;tmin⁡{λ,κ}).
Letx,y,z∈X be such that x*((y*z)*y)∈U(μ;t) and x∈U(μ;t). Then μ(x*((y*z)*y))≥t and μ(x)≥t. It follows from (e4) that
(26)μ(y)≥κmin⁡{μ(x*((y*z)*y)),μ(x)}≥κt≥tmin⁡{λ,κ};
so that y∈U(μ;tmin⁡{λ,κ}). Thus, U(μ;tmin⁡{λ,κ}) is an enlarged positive implicative filter of X related to U(μ;t).



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