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

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 in BE-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

x*x=1 for all x∈X,

x*1=1 for all x∈X,

1*x=x for all x∈X,

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 a subalgebra 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 [<xref ref-type="bibr" rid="B6">1</xref>]).

Let (X;*,1) be a BE-algebra and let F be a nonempty subset of X. Then, F is called a filter of X if

1∈F,

x*y∈F and x∈F imply y∈F,

for all x,y,z∈X.

A nonempty subset F of a BE-algebra X is called an implicative filter of X if it satisfies (F1) and

x*(y*z)∈F and x*y∈F imply x*z∈F, for all x,y,z∈X.

Example 2 (see [<xref ref-type="bibr" rid="B6">1</xref>]).

Let X:={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:

if x≤y,
then z*x≤z*y and y*z≤x*z;

y*z≤(z*x)*(y*x);

y*z≤(x*y)*(x*z).

A BE-algebra (X;*,1) is said to be transitive if it satisfies Proposition 4 (iii).

Definition 5 (see [<xref ref-type="bibr" rid="B7">5</xref>]).

A fuzzy subset μ of a BE-algebra X is called a fuzzy filter of X, if it satisfies, for all x,y∈X,

μ(1)≥μ(x),

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

μ(x*z)≥min{μ(x*(y*z)),μ(x*y)}, for all x,y,z∈X.

Definition 6 (see [<xref ref-type="bibr" rid="B7">5</xref>]).

Let F 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:

F is a subset of G,

1∈G,

(∀x,y∈X)(∀x∈F)(x*y∈F⇒y∈G).

3. Fuzzy Positive Implicative Filters of <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M106"><mml:mi>B</mml:mi><mml:mi>E</mml:mi></mml:math></inline-formula>-Algebras with Degrees in <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M107"><mml:mo mathvariant="bold">(</mml:mo><mml:mn>0,1</mml:mn><mml:mo mathvariant="bold">]</mml:mo></mml:math></inline-formula>

X denotes a BE-algebra unless specified otherwise.

Definition 7.

A nonempty subset F of X is called a positive implicative filter of a BE-algebra X if it satisfies (F1) and

x*((y*z)*y)∈F and x∈F imply y∈F, for all x,y,z∈X.

Note that every positive implicative filter of a BE-algebra X is a filter of X.

Example 8.

(1) Let X:={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 a BE-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) Let X:={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 subset μ of a BE-algebra X is called a fuzzy positive implicative filter of X, if it satisfies (d1) and

μ(y)≥min{μ(x*((y*z)*y)),μ(x)}, for all x,y∈X.

Definition 10.

Let F 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:

F is a subset of G,

1∈G,

(∀x,y,z∈X)(∀x∈F)(x*((y*z)*y)∈F⇒y∈G).

Obviously, every positive implicative filter is an enlarged positive implicative filter of X 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 a BE-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.

Let F 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.

Let G 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 Proposition 12 is not true in general as seen in the following example.

Example 13.

Let X:={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 let λ 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 [<xref ref-type="bibr" rid="B7">5</xref>]).

A fuzzy subset μ of a BE-algebra X is called a fuzzy filter of X with degree (λ,κ), if it satisfies the following:

(∀x∈X)(μ(1)≥λμ(x)),

(∀x,y∈X)(μ(x)≥κmin{μ(y*x),μ(y)}).

A fuzzy subset μ of a BE-algebra X is called a fuzzy implicative filter of X with degree (λ,κ), if it satisfies (e1) and

(∀x,y,z∈X)(μ(x*z)≥κmin{μ(x*(y*z)),μ(x*y)}).

Definition 15.

A fuzzy subset μ of a BE-algebra X is called a fuzzy positive implicative filter of X with degree (λ,κ), if it satisfies (e1) and

(∀x,y,z∈X)(μ(y)≥κmin{μ(x*((y*z)*y)),μ(x)}).

Proposition 16 (see [<xref ref-type="bibr" rid="B7">5</xref>]).

Every fuzzy filter of a BE-algebra X with degree (λ,κ) satisfies the following assertions:

(∀x,y∈X)(μ(x*y)≥λκμ(y));

(∀x,y∈X)(y≤x⇒μ(x)≥λκμ(y));

(∀x,y,z∈X)(x≤y*z⇒μ(z)≥min{κμ(y),λκ2μ(x)}).

Note that if λ≠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-distributive BE-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.

Let X:={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 of X 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.

If μ 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 putting z:=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 Proposition 19 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.

Let μ 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 that μ 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.

Let μ 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.

Let x,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.

Let μ 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 Propositions 20 and 21.

Proposition 23.

Every fuzzy positive implicative filter of a BE-algebra X with degree (λ,κ) satisfies the following assertions:

(∀x,y∈X)(μ(x*y)≥λκμ(y));

(∀x,y∈X)(x≤y⇒μ(y)≥λκμ(x)).

Proof.

It follows from Propositions 16 and 19.

Corollary 24.

Let μ be a fuzzy positive implicative filter of a BE-algebra X with degree (λ,κ). If λ=κ, then

(∀x,y∈X)(μ(x*y)≥λ2μ(y)),

(∀x,y∈X)(x≤y⇒μ(y)≥λ2μ(x)).

Proposition 25.

Let X 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.

Let x,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:

(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.

Let X 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.

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). 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 proof

Denote by FPI(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

μ is a fuzzy positive implicative filter of X with degree (λ,κ),

U(μ;t)∉FPI(X)∪{∅}.

Example 28.

Consider a BE-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.

Let μ 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 that μ(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 exist a,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.

Let μ 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.

Let t∈[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.

Since tmin{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}).

Let x,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).

Conflict of Interests

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

Acknowledgment

The authors are grateful to the referee for their valuable suggestions and help.

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