1 Introduction

In mathematics, an ordered semigroup is a semigroup together with a partial order that is compatible with the semigroup operation. Ordered semigroups have many applications in the theory of sequential machines, formal languages, computer arithmetics, design of fast adders and error-correcting codes. The concept of a fuzzy filter in ordered semigroups was first introduced by Kehayopulu and Tsingelis in [1], where some basic properties of fuzzy filters and prime fuzzy ideals were discussed. A theory of fuzzy generalized sets on ordered semigroups can be developed. Mordeson et al. in [2] presented an up-to-date account of fuzzy sub-semigroups and fuzzy ideals of a semigroup. Murali [3] proposed the definition of a fuzzy point belonging to a fuzzy subset under a natural equivalence on fuzzy subset. The idea of quasi-coincidence of a fuzzy point with a fuzzy set played a vital role in generating different types of fuzzy subgroups. Bhakat and Das [4, 5] gave the concepts of (α,β)-fuzzy subgroups by using the ‘belong to’ () relation and ‘quasi-coincident with’ (q) relation between a fuzzy point and a fuzzy subgroup, and introduced the concept of (,q)-fuzzy subgroup. In [6], Davvaz started the generalized fuzzification in algebra. In [7], Jun et al. initiated the study of (α,β)-fuzzy bi-ideals of an ordered semigroup. In [8], Davvaz and Khan studied (,q)-fuzzy generalized bi-ideals of an ordered semigroup. Shabir et al. [9] studied characterization of regular semigroups by (α,β)-fuzzy ideals. Jun et al. [10] discussed a generalization of (,q)-fuzzy ideals of a BCK/BCI-algebra. Using the idea of a quasi-coincidence of a fuzzy point with a fuzzy set, Jun et al. [11] introduced the concept of (α,β)-intuitionistic fuzzy bi-ideals in an ordered semigroup. They introduced a new sort of intuitionistic fuzzy bi-ideals, called (α,β)-intuitionistic fuzzy bi-ideals, and studied (,q)-intuitionistic fuzzy bi-ideals.

In this paper, we try to have more general form of an (,q)-intuitionistic bi-ideal of an ordered semigroup. We introduce the notion of an (, q k )-intuitionistic bi-ideal of an ordered semigroup, and give examples which are (, q k )-intuitionistic fuzzy bi-ideals but not (,q)-intuitionistic fuzzy bi-ideals. We discuss characterizations of (, q k )-intuitionistic fuzzy bi-ideals in ordered semigroups. We provide a condition for an (, q k )-intuitionistic fuzzy bi-ideal to be an intuitionistic fuzzy bi-ideal. The important achievement of the study with an (, q k )-intuitionistic fuzzy bi-ideal is that the notion of an (,q)-intuitionistic fuzzy bi-ideal is a special case of an (, q k )-intuitionistic fuzzy bi-ideal, and, thus, several results in the paper [11] are the corollaries of our results obtained in this paper.

2 Basic definitions and preliminary results

By an ordered semigroup (or po-semigroup) we mean a structure (S,,), in which the following are satisfied:

(OS1) (S,) is a semigroup,

(OS2) (S,) is a poset,

(OS3) (x,a,bS) (abaxbx, xaxb).

In what follows, xy is simply denoted by xy for all x,yS.

A nonempty subset A of an ordered semigroup S is called a subsemigroup of S if A 2 A. A non-empty subset A of an ordered semigroup S is called a bi-ideal of S if it satisfies

(b1) (bS) (bA) (abbA),

(b2) (a,bS) (a,bAabA),

(b3) ASAA.

An intuitionistic fuzzy set (briefly IFS) A in a non-empty set X is an object having the form A={x, μ A (x), γ A (x)|xX}, where the function μ A :X[0,1] and γ A :X[0,1] denote the degree of membership (namely, μ A (x)) and the degree of non-membership (namely, γ A (x)) for each element xX to the set A, respectively, and 0 μ A (x)+ γ A (x)1 for all xX. For the sake of simplicity, we shall use the symbol A=x, μ A , γ A for the intuitionistic fuzzy set A={x, μ A (x), γ A (x)|xX}.

Let (S,,) be an ordered semigroup and A=x, μ A , γ A be an IFS of S. Then A=x, μ A , γ A is called an intuitionistic fuzzy subsemigroup of S [11] if

(x,yS) ( μ A ( x y ) min { μ A ( x ) , μ A ( y ) }  and  γ A ( x y ) max { γ A ( x ) , γ A ( y ) } ) .

Let (S,,) be an ordered semigroup and A=x, μ A , γ A be an intuitionistic fuzzy subsemigroup of S. Then A=x, μ A , γ A is called an intuitionistic fuzzy bi-ideal of S [11] if

(b4) (x,yS) (xy μ A (x) μ A (y) and γ A (x) γ A (y)),

(b5) (x,yS) ( μ A (xy)min{ μ A (x), μ A (y)} and γ A (xy)max{ γ A (x), γ A (y)}),

(b6) (x,y,zS) ( μ A (xyz)min{ μ A (x), μ A (z)} and γ A (xyz)max{ γ A (x), γ A (z)}).

Let A=x, μ A , γ A be an IFS of S and α(0,1] and β[0,1). Then the sets

U( μ A ;α)= { x S | μ A ( x ) α } andL( γ A ;β)= { x S | γ A ( x ) β }

are called μ A -level and γ A -level cuts of the intuitionistic fuzzy set A=x, μ A , γ A , respectively. For an IFS A=x, μ A , γ A and α(0,1], β[0,1), we define the ( μ A , γ A )-level cut as follows

C ( α , β ) (A)= { x S | μ A ( x ) α  and  γ A ( x ) β } .

Clearly, C ( α , β ) (A)=U( μ A ;α)L( γ A ;β).

Let x be a point of a non-empty set X. If α(0,1] and β[0,1) are two real numbers such that 0α+β1, then the IFS of the form

x ; ( α , β ) =x; x α ,1 x 1 β

is called an intuitionistic fuzzy point (IFP for short) in X, where α (resp. β) is the degree of membership (resp. non-membership) of x;(α,β) and xX is the support of x;(α,β).

Consider an IFP x;(α,β) in S, an IFS A=x, μ A , γ A and α{,q,q}, we define x;(α,β)αA as follows

(b7) x;(α,β)A (resp. x;(α,β)qA) means that μ A (x)α and γ A (x)β (resp. μ A (x)+α>1 and γ A (x)+β<1), and in this case, we say that x;(α,β) belongs to (resp. quasi-coincident with) an IFS A=x, μ A , γ A .

(b8) x;(α,β)qA (resp. x;(α,β)qA) means that x;(α,β)A or x;(α,β)qA (resp. x;(α,β)A and x;(α,β)qA).

By x;(α,β) α ¯ A, we mean that x;(α,β)αA does not hold.

3 (, q k )-Intuitionistic fuzzy bi-ideals

Let k denote an arbitrary element of [0,1) unless specified otherwise. For an IFP x;(α,β) and an IFS A=x, μ A , γ A of X, we say that

(c1) x;(α,β) q k A if μ A (x)+k+α>1 and γ A (x)+k+β<1.

(c2) x;(α,β) q k A if x;(α,β)A or x;(α,β) q k A.

(c3) x;(α,β) α ¯ A if x;(α,β)αA does not hold for α{ q k , q k }.

Theorem 3.1 Let A=x, μ A , γ A be an IFS of an ordered semigroup S. Then the following are equivalent

  1. (1)

    (α( 1 k 2 ,1]) (β[0, 1 k 2 )) ( C ( α , β ) (A) C ( α , β ) (A) is a bi-ideal of S).

  2. (2)

    A=x, μ A , γ A satisfies the following assertions

    (2.1) ( x y μ A ( y ) max { μ A ( x ) , 1 k 2 } and  γ A ( y ) min { γ A ( x ) , 1 k 2 } ) , (2.2) ( min { μ A ( x ) , μ A ( y ) } max { μ A ( x y ) , 1 k 2 } and  max { γ A ( x ) , γ A ( y ) } min { γ A ( x y ) , 1 k 2 } ) , (2.3) ( min { μ A ( x ) , μ A ( z ) } max { μ A ( x y z ) , 1 k 2 }  and max { γ A ( x ) , γ A ( z ) } min { γ A ( x y z ) , 1 k 2 } ) ,

for all x,y,zS.

Proof Assume that C ( α , β ) (A) is a bi-ideal of S for all α( 1 k 2 ,1] and β[0, 1 k 2 ) with C ( α , β ) (A). If there exist a,bS such that condition (2.1) is not valid, that is, there exist a,bS with ab and μ A (b)>max{ μ A (a), 1 k 2 }, γ A (b)<min{ γ A (a), 1 k 2 }. Then μ A (b)( 1 k 2 ,1], γ A (b)[0, 1 k 2 ) and b C ( μ A ( b ) , γ A ( b ) ) (A). But μ A (a)< μ A (b) and γ A (a)> γ A (b) imply that a C ( μ A ( b ) , γ A ( b ) ) (A). This is not possible. Hence (2.1) is valid. Suppose that (2.2) is false, that is,

s:=min { μ A ( a ) , μ A ( c ) } >max { μ A ( a c ) , 1 k 2 }

and

t:=max { γ A ( a ) , γ A ( c ) } <min { γ A ( a c ) , 1 k 2 }

for some a,cS. Then s( 1 k 2 ,1], t[0, 1 k 2 ) and a,c C ( s , t ) (A). But ac C ( s , t ) (A), since μ A (ac)<s and γ A (ac)>t. This is not possible, and so (2.2) is valid. If there exist a,b,cS such that (2.3) is not valid, that is,

s 0 :=min { μ A ( a ) , μ A ( b ) } >max { μ A ( a c b ) , 1 k 2 }

and

t 0 :=max { γ A ( a ) , γ A ( b ) } <min { γ A ( a c b ) , 1 k 2 } .

Then s 0 ( 1 k 2 ,1], t 0 [0, 1 k 2 ) and a,b C ( s 0 , t 0 ) (A). But acb C ( s 0 , t 0 ) (A), since μ A (acb)< s 0 and γ A (acb)> t 0 . This is a contradiction, and hence (2.3) is valid.

Conversely, assume that A=x, μ A , γ A satisfies the three conditions (2.1), (2.2) and (2.3). Suppose that C ( α , β ) (A) for all α( 1 k 2 ,1], and β[0, 1 k 2 ). Let x,yS be such that xy and y C ( α , β ) (A). Then μ A (y)α and γ A (y)β. Using (2.1), we have max{ μ A (x), 1 k 2 } μ A (y)α> 1 k 2 and min{ γ A (x), 1 k 2 } γ A (y)β< 1 k 2 . Hence μ A (x)α and γ A (x)β, i.e., x C ( α , β ) (A). If x,y C ( α , β ) (A), then μ A (x)α, γ A (x)β and μ A (y)α, γ A (y)β. By using (2.2), we have

max { μ A ( x y ) , 1 k 2 } min { μ A ( x ) , μ A ( y ) } α> 1 k 2

and

min { γ A ( x y ) , 1 k 2 } max { γ A ( x ) , γ A ( y ) } β< 1 k 2 ,

so that μ A (xy)α and γ A (xy)β, i.e., xy C ( α , β ) (A). Finally, if x,z C ( α , β ) (A) and yS, then μ A (x)α, γ A (x)β and μ A (z)α, γ A (z)β. By using (2.3), we have

max { μ A ( x y z ) , 1 k 2 } min { μ A ( x ) , μ A ( z ) } α> 1 k 2

and

min { γ A ( x y z ) , 1 k 2 } max { γ A ( x ) , γ A ( z ) } β< 1 k 2 ,

so that μ A (xyz)α and γ A (xyz)β, i.e., xyz C ( α , β ) (A). Therefore, C ( α , β ) (A) is a bi-ideal of S. □

If we take k=0 in Theorem 3.1, then we have the following corollary.

Corollary 3.2 [[11], Theorem 3.1]

Let A=x, μ A , γ A be an IFS of S. Then the following assertions are equivalent

  1. (1)

    (α(0.5,1]) (β[0,0.5)) ( C ( α , β ) (A) C ( α , β ) (A) is a bi-ideal of S).

  2. (2)

    A=x, μ A , γ A satisfies the following conditions

    (2.1) ( x y μ A ( y ) max { μ A ( x ) , 0.5 } and  γ A ( y ) min { γ A ( x ) , 0.5 } ) , (2.2) ( min { μ A ( x ) , μ A ( y ) } max { μ A ( x y ) , 0.5 } and  max { γ A ( x ) , γ A ( y ) } min { γ A ( x y ) , 0.5 } ) , (2.3) ( min { μ A ( x ) , μ A ( z ) } max { μ A ( x y z ) , 0.5 } and  max { γ A ( x ) , γ A ( z ) } min { γ A ( x y z ) , 0.5 } ) ,

for all x,y,zS.

Definition 3.3 An IFS A=x, μ A , γ A in S is called an (, q k )-intuitionistic fuzzy bi-ideal of S if for all x,y,zS, t, t 1 , t 2 (0,1] and s, s 1 , s 2 [0,1) it satisfies the following conditions

(q1) (xy, y;(t,s)Ax;(t,s) q k A),

(q2) (x;( t 1 , s 1 )A and y;( t 2 , s 2 )Axy;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A),

(q3) (x;( t 1 , s 1 )A and z;( t 2 , s 2 )Axyz;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A).

An (, q k )-intuitionistic fuzzy bi-ideal of S with k=0 is an (,q)-intuitionistic fuzzy bi-ideal of S.

Example 3.4 Consider the set S={a,b,c,d,e} with the order relation ace, ade, bd and be and ∗-multiplication table (see Table 1 above).

  1. (1)

    Define an IFS A=x, μ A , γ A by

    μ A :S[0,1] μ A (x)={ 0.40 if  x = a , 0.35 if  x = b , 0.30 if  x = c , 0.50 if  x = d , 0.20 if  x = e

and

γ A :S[0,1] γ A (x)={ 0.40 if  x = a , 0.30 if  x = b , 0.50 if  x = c , 0.40 if  x = d , 0.80 if  x = e .

Then A=x, μ A , γ A is an (, q 0.4 )-intuitionistic fuzzy bi-ideal of S.

  1. (2)

    Let A=x, μ A , γ A be an intuitionistic fuzzy set given by

    μ A :S[0,1] μ A (x)={ 0.80 if  x = a , 0.60 if  x = b , e , 0.30 if  x = c , 0.50 if  x = d

and

γ A :S[0,1] γ A (x)={ 0.20 if  x = a , 0.30 if  x = b , e , 0.60 if  x = c , 0.50 if  x = d .

Then A=x, μ A , γ A is an (, q 0.04 )-intuitionistic fuzzy bi-ideal of S.

Table 1 ∗-multiplication table for S

Theorem 3.5 An IFS A=x, μ A , γ A of S is an (, q k )-intuitionistic fuzzy bi-ideal of S if and only if it satisfies the following conditions

(1) ( x y μ A ( x ) min { μ A ( y ) , 1 k 2 } and  γ A ( x ) max { γ A ( y ) , 1 k 2 } ) , (2) ( μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 1 k 2 } and  γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 1 k 2 } ) , (3) ( μ A ( x y z ) min { μ A ( x ) , μ A ( z ) , 1 k 2 } and  γ A ( x y z ) max { γ A ( x ) , γ A ( z ) , 1 k 2 } ) .

Proof Suppose that A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S. Let x,yS be such that xy. Assume that μ A (y) 1 k 2 and γ A (y) 1 k 2 . If μ A (x)< μ A (y) and γ A (x)> γ A (y), then μ A (x)<t μ A (y) and γ A (x)>s γ A (y) for some t(0, 1 k 2 ) and s( 1 k 2 ,1). It follows y;(t,s)A, but x;(t,s) ¯ A. Since μ A (x)+t<2t<1k and γ A (x)+s>2s>1k, we get x;(t,s) q k ¯ A. Therefore, x;(t,s) q k ¯ A, which is a contradiction. Hence μ A (x) μ A (y) and γ A (x) γ A (y). Now, if μ A (y) 1 k 2 and γ A (x) 1 k 2 , then y;( 1 k 2 , 1 k 2 )A, and so, x;( 1 k 2 , 1 k 2 ) q k A, which implies that x;( 1 k 2 , 1 k 2 )A or x;( 1 k 2 , 1 k 2 ) q k A, that is, μ A (x) 1 k 2 and γ A (x) 1 k 2 or μ A (x)+ 1 k 2 >1 and γ A (x)+ 1 k 2 <1. Hence μ A (x) 1 k 2 and γ A (x) 1 k 2 . Otherwise, μ A (x)+ 1 k 2 < 1 k 2 + 1 k 2 =1 and γ A (x)+ 1 k 2 > 1 k 2 + 1 k 2 =1, a contradiction. Consequently,

μ A (x)min { μ A ( y ) , 1 k 2 }

and

γ A (x)max { γ A ( y ) , 1 k 2 }

for all x,yS with xy. Let x,yS be such that min{ μ A (x), μ A (y)}< 1 k 2 and max{ γ A (x), γ A (y)}> 1 k 2 . We claim that μ A (xy)min{ μ A (x), μ A (y)} and γ A (xy)max{ γ A (x), γ A (y)}. If not, then μ A (xy)<tmin{ μ A (x), μ A (y)} and γ A (xy)>smax{ γ A (x), γ A (y)} for some t(0, 1 k 2 ) and s( 1 k 2 ,1). It follows that x;(t,s)A and y;(t,s)A, but xy;(t,s) ¯ A and μ A (xy)+t<2t<1k and γ A (xy)+s>2s>1k, i.e., xy;(t,s) q k ¯ A. This is a contradiction. Thus, μ A (xy)min{ μ A (x), μ A (y)} and γ A (xy)max{ γ A (x), γ A (y)} for all x,yS with min{ μ A (x), μ A (y)}< 1 k 2 and max{ γ A (x), γ A (y)}> 1 k 2 . If min{ μ A (x), μ A (y)} 1 k 2 and max{ γ A (x), γ A (y)} 1 k 2 , then x;( 1 k 2 , 1 k 2 )A and y;( 1 k 2 , 1 k 2 )A. Using (q2), we have

x y ; ( 1 k 2 , 1 k 2 ) = x y ; min { 1 k 2 , 1 k 2 } , max { 1 k 2 , 1 k 2 } q k A ,

and so, μ A (xy) 1 k 2 and γ A (xy) 1 k 2 or μ A (xy)+ 1 k 2 >1 and γ A (xy)+ 1 k 2 <1. If μ A (xy)< 1 k 2 and γ A (xy)> 1 k 2 , then

μ A (xy)+ 1 k 2 < 1 k 2 + 1 k 2 =1

and

γ A (xy)+ 1 k 2 > 1 k 2 + 1 k 2 =1,

which is impossible. Consequently, μ A (xy)min{ μ A (x), μ A (y), 1 k 2 } and γ A (xy)max{ γ A (x), γ A (y), 1 k 2 } for all x,yS. Let a,b,cS be such that min{ μ A (a), μ A (c)}< 1 k 2 and max{ γ A (a), γ A (c)}> 1 k 2 . We claim that

μ A (abc)min { μ A ( a ) , μ A ( c ) } and γ A (abc)max { γ A ( a ) , γ A ( c ) } .

If not, then μ A (abc)< t 0 min{ μ A (a), μ A (c)} and γ A (abc)> s 0 max{ γ A (a), γ A (c)} for some t 0 (0, 1 k 2 ) and s 0 ( 1 k 2 ,1). It follows that a;( t 0 , s 0 )A and c;( t 0 , s 0 )A, but abc;( t 0 , s 0 ) ¯ A and μ A (abc)+ t 0 <2 t 0 <1k and γ A (abc)+ s 0 >2 s 0 >1k, i.e., abc;( t 0 , s 0 ) q k ¯ A. This is a contradiction. Thus, μ A (abc)min{ μ A (a), μ A (c)} and γ A (abc)max{ γ A (a), γ A (c)} for all a,b,cS with min{ μ A (a), μ A (c)}< 1 k 2 and max{ γ A (a), γ A (c)}> 1 k 2 . If min{ μ A (a), μ A (c)} 1 k 2 and max{ γ A (a), γ A (c)} 1 k 2 , then a;( 1 k 2 , 1 k 2 )A and c;( 1 k 2 , 1 k 2 )A. Using (q3), we have

a b c ; ( 1 k 2 , 1 k 2 ) = a b c ; min { 1 k 2 , 1 k 2 } , max { 1 k 2 , 1 k 2 } q A ,

and so μ A (abc) 1 k 2 and γ A (abc) 1 k 2 or μ A (abc)+ 1 k 2 >1 and γ A (abc)+ 1 k 2 <1. If μ A (abc)< 1 k 2 and γ A (abc)> 1 k 2 , then

μ A (abc)+ 1 k 2 < 1 k 2 + 1 k 2 =1

and

γ A (abc)+ 1 k 2 > 1 k 2 + 1 k 2 =1,

which is impossible. Therefore, μ A (xyz)min{ μ A (x), μ A (z), 1 k 2 } and γ A (xyz)max{ γ A (x), γ A (z), 1 k 2 } for all x,y,zS.

Conversely, let A=x, μ A , γ A be an IFS of S that satisfies the three conditions (1), (2) and (3). Let x,yS, t(0,1] and s[0,1) be such that xy and [y;(t,s)]A. Then μ A (y)t and γ A (y)s, and so,

μ A ( x ) min { μ A ( y ) , 1 k 2 } min { t , 1 k 2 } = { t if  t 1 k 2 , 1 k 2 if  t > 1 k 2

and

γ A ( x ) max { γ A ( y ) , 1 k 2 } max { s , 1 k 2 } = { s if  s 1 k 2 , 1 k 2 if  s < 1 k 2 .

It follows that μ A (x)t and γ A (x)s or μ A (x)+t 1 k 2 +t>1k and γ A (x)+s 1 k 2 +s<1k, i.e., x;(t,s)A or x;(t,s) q k A. Hence, x;(t,s) q k A. Let x,yS, t 1 , t 2 (0,1] and s 1 , s 2 [0,1) be such that x;( t 1 , s 1 )A and y;( t 2 , s 2 )A. Then μ A (x) t 1 , γ A (x) s 1 and μ A (y) t 2 , γ A (y) s 2 . It follows from (2) that

μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 1 k 2 } min { t 1 , t 2 , 1 k 2 } = { min { t 1 , t 2 } if  min { t 1 , t 2 } 1 k 2 , 1 k 2 if  min { t 1 , t 2 } 1 k 2

and

γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 1 k 2 } max { s 1 , s 2 , 1 k 2 } = { max { s 1 , s 2 } if  max { s 1 , s 2 } 1 k 2 , 1 k 2 if  max { s 1 , s 2 } < 1 k 2 .

It follows that xy;min{ t 1 , t 2 },max{ s 1 , s 2 }A or μ A (xy)+min{ t 1 , t 2 } 1 k 2 +min{ t 1 , t 2 }> 1 k 2 + 1 k 2 =1k and γ A (xy)+max{ s 1 , s 2 } 1 k 2 +max{ s 1 , s 2 }< 1 k 2 + 1 k 2 =1k, i.e., xy;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A. Therefore, xy;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A. Let x,y,zS, t 1 , t 2 (0,1] and s 1 , s 2 [0,1) be such that x;( t 1 , s 1 )A and z;( t 2 , s 2 )A. Then μ A (x) t 1 , γ A (x) s 1 and μ A (z) t 2 , γ A (z) s 2 . It follows from (3) that

μ A ( x y z ) min { μ A ( x ) , μ A ( z ) , 1 k 2 } min { t 1 , t 2 , 1 k 2 } = { min { t 1 , t 2 } if  min { t 1 , t 2 } 1 k 2 , 1 k 2 if  min { t 1 , t 2 } 1 k 2

and

γ A ( x y z ) max { γ A ( x ) , γ A ( z ) , 1 k 2 } max { s 1 , s 2 , 1 k 2 } = { max { s 1 , s 2 } if  max { s 1 , s 2 } 1 k 2 , 1 k 2 if  max { s 1 , s 2 } < 1 k 2 .

Thus, we have xyz;min{ t 1 , t 2 },max{ s 1 , s 2 }A or μ A (xyz)+min{ t 1 , t 2 } 1 k 2 +min{ t 1 , t 2 }> 1 k 2 + 1 k 2 =1k and γ A (xyz)+max{ s 1 , s 2 } 1 k 2 +max{ s 1 , s 2 }< 1 k 2 + 1 k 2 =1k, i.e., xyz;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A. Therefore, xyz;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A. Thus, A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S. □

If we take k=0 in Theorem 3.5, then we have the following corollary.

Corollary 3.6 [[11], Theorem 3.5]

An IFS A=x, μ A , γ A of S is an (, q k )-intuitionistic fuzzy bi-ideal of S if and only if it satisfies the conditions

(1) ( x y μ A ( x ) min { μ A ( y ) , 0.5 } and  γ A ( x ) max { γ A ( y ) , 0.5 } ) , (2) ( μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 0.5 } and  γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 0.5 } ) , (3) ( μ A ( x y z ) min { μ A ( x ) , μ A ( z ) , 0.5 } and  γ A ( x y z ) max { γ A ( x ) , γ A ( z ) , 0.5 } ) .

Obviously, every intuitionistic fuzzy bi-ideal is an (,)-intuitionistic fuzzy bi-ideal, and we know that every (,)-intuitionistic fuzzy bi-ideal of S is an (,q)-intuitionistic fuzzy bi-ideal of S, and every (,q)-intuitionistic fuzzy bi-ideal is an (, q k )-intuitionistic fuzzy bi-ideal of S. But the converse may not be true. The following example shows that every (, q k )-intuitionistic fuzzy bi-ideal of S may not be an (,q)-intuitionistic fuzzy bi-ideal nor an intuitionistic fuzzy bi-ideal of S.

Example 3.7 Consider the ordered semigroup of Example 3.4 and define an IFS A=x, μ A , γ A by

μ A :S[0,1] μ A (x)={ 0.80 if  x = a , 0.60 if  x = b , 0.40 if  x = c , 0.30 if  x = d , e

and

γ A :S[0,1] γ A (x)={ 0.20 if  x = a , 0.30 if  x = b , 0.40 if  x = c , 0.70 if  x = d , e .

Then A=x, μ A , γ A is an (, q 0.4 )-intuitionistic fuzzy bi-ideal of S. But

  1. (1)

    A=x, μ A , γ A is not an (,q)-intuitionistic fuzzy bi-ideal of S. Since a;(0.8,0.2)A and b;(0.6,0.3)A but ab;(0.6,0.3) q ¯ A.

  2. (2)

    A=x, μ A , γ A is not an intuitionistic fuzzy bi-ideal of S. Since

    μ A (ab)= μ A (d)=0.30<m { μ A ( a ) = 0.80 , μ A ( b ) = 0.60 }

and

γ A (ab)= γ A (d)=0.70>M { γ A ( a ) = 0.20 , μ A ( b ) = 0.30 } .

In the following, we give a condition for an (, q k )-intuitionistic fuzzy bi-ideal of S to be an ordinary intuitionistic fuzzy bi-ideal of S.

Theorem 3.8 Let A=x, μ A , γ A be an (, q k )-intuitionistic fuzzy bi-ideal of S. If μ A (x) 1 k 2 and γ A (x) 1 k 2 for all xS, then A=x, μ A , γ A is an (,)-intuitionistic fuzzy bi-ideal of S.

Proof The proof is straightforward by Theorem 3.5. □

Corollary 3.9 [[11], Theorem 3.8]

Let A=x, μ A , γ A be an (,q)-intuitionistic fuzzy bi-ideal of S. If μ A (x)0.5 and γ A (x)0.5 for all xS, then A=x, μ A , γ A is an (,)-intuitionistic fuzzy bi-ideal of S.

Proof The proof follows from Theorem 3.8, by taking k=0. □

Theorem 3.10 For an IFS A=x, μ A , γ A of S, the following are equivalent:

  1. (1)

    A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S.

  2. (2)

    (t(0, 1 k 2 ]) (s[ 1 k 2 ,1)) ( C ( α , β ) (A) C ( α , β ) (A)is a bi-ideal ofS) .

Proof Assume that A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S, let t(0, 1 k 2 ] and s[ 1 k 2 ,1) be such that C ( α , β ) (A). Using Theorem 3.5(1), we have

μ A (x)min { μ A ( y ) , 1 k 2 } and γ A (x)max { γ A ( y ) , 1 k 2 }

for any x,yS with xy and x C ( t , s ) (A). It follows that μ A (x)min{t,0.5}=t and γ A (x)max{s,0.5}=s, so that y C ( t , s ) (A). Let x,y C ( α , β ) (A). Then μ A (x)t, γ A (x)s and μ A (y)t, γ A (y)s. Theorem 3.5(2) implies that

μ A (xy)min { μ A ( x ) , μ A ( y ) , 1 k 2 } min { t , 1 k 2 } =t

and

γ A (xy)max { γ A ( x ) , γ A ( y ) , 1 k 2 } max { s , 1 k 2 } =s.

Thus, xy C ( α , β ) (A). Now let x,z C ( α , β ) (A). Then μ A (x)t, γ A (x)s and μ A (z)t, γ A (z)s. Theorem 3.5(3) induces that

μ A (xyz)min { μ A ( x ) , μ A ( z ) , 1 k 2 } min { t , 1 k 2 } =t

and

γ A (xyz)max { γ A ( x ) , γ A ( z ) , 1 k 2 } max { s , 1 k 2 } =s.

Thus, xyz C ( α , β ) (A), therefore, C ( α , β ) (A) is a bi-ideal of S.

Conversely, let A=x, μ A , γ A be an IFS of S such that C ( α , β ) (A) is non-empty and is a bi-ideal of S for all t(0, 1 k 2 ] and s[ 1 k 2 ,1). If there exist a,bS with ab and b C ( α , β ) (A) such that μ A (a)<min{ μ A (b), 1 k 2 } and γ A (a)>max{ γ A (b), 1 k 2 }, then μ A (a)< t a min{ μ A (b), 1 k 2 } and γ A (a)> s a max{ γ A (b), 1 k 2 } for some t a (0, 1 k 2 ] and s a [ 1 k 2 ,1). Then a C ( t a , s a ) (A), a contradiction. Therefore, μ A (x)min{ μ A (y), 1 k 2 } and γ A (x)>max{ γ A (y), 1 k 2 } for all x,yS with xy. Assume that there exist a,bS such that

μ A (ab)<min { μ A ( a ) , μ A ( b ) , 1 k 2 }

and γ A (ab)>max{ γ A (a), γ A (b), 1 k 2 }. Then

μ A (ab)< t 0 min { μ A ( a ) , μ A ( b ) , 1 k 2 }

and

γ A (ab)> s 0 max { γ A ( a ) , γ A ( b ) , 1 k 2 }

for some t 0 (0, 1 k 2 ] and s 0 [ 1 k 2 ,1). It follows that a C ( t 0 , s 0 ) (A) and b C ( t 0 , s 0 ) (A), but ab C ( t 0 , s 0 ) (A). This is a contradiction. Hence

μ A (xy)min { μ A ( x ) , μ A ( y ) , 1 k 2 }

and

γ A (xy)max { γ A ( x ) , γ A ( y ) , 1 k 2 }

for all x,yS. Suppose that

μ A (abc)<min { μ A ( a ) , μ A ( c ) , 1 k 2 }

and

γ A (abc)>max { γ A ( a ) , γ A ( c ) , 1 k 2 }

for some a,b,cS. Then there exist t 1 (0, 1 k 2 ] and s 1 [ 1 k 2 ,1) such that

μ A (abc)< t 1 min { μ A ( a ) , μ A ( c ) , 1 k 2 }

and

γ A (abc)> s 1 max { γ A ( a ) , γ A ( c ) , 1 k 2 } .

Then a C ( t 1 , s 1 ) (A) and c C ( t 1 , s 1 ) (A), but abc C ( t 1 , s 1 ) (A). This is impossible, and hence μ A (xyz)<min{ μ A (x), μ A (z), 1 k 2 } and

γ A (xyz)>max { γ A ( x ) , γ A ( z ) , 1 k 2 }

for all x,y,zS. Therefore, A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S. □

By taking k=0 in Theorem 3.10, we get the following corollary.

Corollary 3.11 [[11], Theorem 3.10]

For an IFS A=x, μ A , γ A of an ordered semigroup (S,,), the following are equivalent

  1. (1)

    A=x, μ A , γ A is an (,q)-intuitionistic fuzzy bi-ideal of S.

  2. (2)

    (t(0,0.5]) (s[0.5,1)) ( C ( α , β ) (A) C ( α , β ) (A) is a bi-ideal of S) .

For an IFP x;(α,β) of S and an IFS A=x, μ A , γ A of S, we say that

(c4) x;(α,β) q ̲ A if μ A (x)+α1 and γ A (x)+β1,

(c5) x;(α,β) q k ̲ A if μ A (x)+α+k1 and γ A (x)+β+k1.

We denote by Q ( α , β ) k (A) (resp. Q k ̲ ( α , β ) (A)) the set {xS|x;(α,β) q k A} (resp. {xS|x;(α,β) q k ̲ A}), and [ A ] ( t , s ) k :={xS|[x;(t,s)] q k A}. It is obvious that [ A ] ( t , s ) k = C ( t , s ) (A) Q ( t , s ) k (A).

Proposition 3.12 If A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S, then

( t ( 1 k 2 , 1 ] ) ( s [ 0 , 1 k 2 ) ) ( Q ( t , s ) k ( A ) Q ( t , s ) k ( A ) is a bi-ideal of S ) .

Proof Assume that A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S. Let t[ 1 k 2 ,1] and s[0, 1 k 2 ] be such that Q ( t , s ) k (A). Let y Q ( t , s ) k (A) and xS be such that xy . Then μ A (y)+t+k>1 and γ A (y)+s+k<1. By means of Theorem 3.5(1), we have

μ A ( x ) min { μ A ( y ) , 1 k 2 } = { 1 k 2 if  μ A ( y ) 1 k 2 , μ A ( y ) if  μ A ( y ) < 1 k 2 > 1 t k

and

γ A ( x ) max { γ A ( y ) , 1 k 2 } = { 1 k 2 if  γ A ( y ) < 1 k 2 , γ A ( y ) if  γ A ( y ) 1 k 2 < 1 s k .

It follows that x Q ( t , s ) (A). Let x,y Q ( t , s ) (A). Then μ A (x)+t>1k and γ A (x)+s<1k, μ A (y)+t>1k and γ A (y)+s<1k. Using (2) of Theorem 3.5, we have that

μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 1 k 2 } = { 1 k 2 if  min { μ A ( x ) , μ A ( y ) } 1 k 2 , min { μ A ( x ) , μ A ( y ) } if  min { μ A ( x ) , μ A ( y ) } < 1 k 2 > 1 t k ,

and

γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 1 k 2 } = { 1 k 2 if  max { γ A ( x ) , γ A ( y ) } < 1 k 2 , max { γ A ( x ) , γ A ( y ) } if  max { γ A ( x ) , γ A ( y ) } 1 k 2 < 1 s k .

Thus, xy Q ( t , s ) (A). Let x,z Q ( t , s ) (A) and yS. Then μ A (x)+t>1k and γ A (x)+s<1k, μ A (z)+t>1k and γ A (z)+s<1k. Using (3) of Theorem 3.5, we have that

μ A ( x y z ) min { μ A ( x ) , μ A ( z ) , 1 k 2 } = { 1 k 2 if  min { μ A ( x ) , μ A ( z ) } 1 k 2 , min { μ A ( x ) , μ A ( z ) } if  min { μ A ( x ) , μ A ( z ) } < 1 k 2 > 1 t k

and

γ A ( x y z ) max { γ A ( x ) , γ A ( z ) , 1 k 2 } = { 1 k 2 if  max { γ A ( x ) , γ A ( z ) } < 1 k 2 , max { γ A ( x ) , γ A ( z ) } if  max { γ A ( x ) , γ A ( z ) } 1 k 2 < 1 s .

Hence xyz Q ( t , s ) k (A). Therefore, Q ( t , s ) k (A) is a bi-ideal of S. □

Theorem 3.13 For any IFS A=x, μ A , γ A of S, the following are equivalent

  1. (1)

    A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S.

  2. (2)

    (t(0,1]) (s[0,1)) ( [ A ] ( t , s ) k [ A ] ( t , s ) k is a bi-ideal of S).

We call [ A ] ( t , s ) k an (q)-level bi-ideal of A=x, μ A , γ A .

Proof Assume that A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S, and let t(0,1] and s[0,1) be such that [ A ] ( t , s ) k . Let y [ A ] ( t , s ) k and xS be such that xy. Then y C ( t , s ) (A) or y Q ( t , s ) k (A), i.e., μ A (y)t and γ A (y)s or μ A (y)+t>1k and γ A (y)+s<1k. Using Theorem 3.5(1), we get

μ A (x)min { μ A ( y ) , 1 k 2 } and γ A (x)max { γ A ( y ) , 1 k 2 } .
(3.1)

We consider two cases μ A (y) 1 k 2 , γ A (y) 1 k 2 and μ A (y)> 1 k 2 , γ A (y)< 1 k 2 . The first case implies from (3.1) that μ A (x) μ A (y) and γ A (x) γ A (y). Thus, if μ A (y)t and γ A (y)s, then μ A (x)t and γ A (x)s, and so, x C ( t , s ) (A) [ A ] ( t , s ) k . If μ A (y)+t>1k and γ A (y)+s<1k, then μ A (x)+t μ A (y)+t>1k and γ A (x)+s γ A (y)+s<1k, which implies that [x;(t,s)] q k A, i.e., x Q ( t , s ) k (A) [ A ] ( t , s ) k . Combining the second case and (3.1), we have μ A (x) 1 k 2 and γ A (x) 1 k 2 . If t 1 k 2 and s 1 k 2 , then μ A (x)t and γ A (x)s, and hence x C ( t , s ) (A) [ A ] ( t , s ) k . If t> 1 k 2 and s< 1 k 2 , then μ A (x)+t> 1 k 2 + 1 k 2 =1k and γ A (x)+s< 1 k 2 + 1 k 2 =1k, which implies that x Q ( t , s ) k (A) [ A ] ( t , s ) k . Therefore, [ A ] ( t , s ) k satisfies the condition (b1). Let x,y [ A ] ( t , s ) k . Then x C ( t , s ) (A) or x;(t,s) q k A and y C ( t , s ) (A) or y;(t,s) q k A, that is, μ A (x)t, γ A (x)s or μ A (x)+t+k>1, γ A (x)+s+k<1 and μ A (y)t, γ A (y)s or μ A (y)+t+k>1, γ A (y)+s+k<1. We consider the following four cases

  1. (i)

    μ A (x)t, γ A (x)s and μ A (y)t, γ A (y)s,

  2. (ii)

    μ A (x)t, γ A (x)s and μ A (y)+t+k>1, γ A (y)+s+k<1,

  3. (iii)

    μ A (x)+t+k>1, γ A (x)+s+k<1 and μ A (y)t, γ A (y)s,

  4. (iv)

    μ A (x)+t+k>1, γ A (x)+s+k<1 and μ A (y)+t+k>1, γ A (y)+s+k<1.

For the case (i), Theorem 3.5(2) implies that

μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 1 k 2 } min { t , 1 k 2 } = { 1 k 2 if  t > 1 k 2 , t if  t 1 k 2

and

γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 1 k 2 } max { s , 1 k 2 } = { 1 k 2 if  s < 1 k 2 , s if  s 1 k 2 .

Then xy C ( t , s ) (A) or μ A (xy)+t+k> 1 k 2 + 1 k 2 +k=1 and γ A (xy)+s+k< 1 k 2 + 1 k 2 +k=1, that is, xy Q ( t , s ) k (A). Hence xy C ( t , s ) (A) Q ( t , s ) k (A)= [ A ] ( t , s ) k . For the second case, assume that t> 1 k 2 and s< 1 k 2 , then 1tk1t< 1 k 2 and 1sk1s 1 k 2 . Hence

μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 1 k 2 } = { min { μ A ( y ) , 1 k 2 } > 1 t k if  min { μ A ( y ) , 1 k 2 } μ A ( x ) , μ A ( x ) t if  min { μ A ( y ) , 1 k 2 } > μ A ( x )

and

γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 1 k 2 } = { max { γ A ( y ) , 1 k 2 } < 1 s k if  max { γ A ( y ) , 1 k 2 } γ A ( x ) , γ A ( x ) s if  max { γ A ( y ) , 1 k 2 } < γ A ( x ) .

Thus xy C ( t , s ) (A) Q ( t , s ) k (A)= [ A ] ( t , s ) k . Suppose that t 1 k 2 and s 1 k 2 . Then

μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 1 k 2 } = { min { μ A ( x ) , 1 k 2 } t if  min { μ A ( x ) , 1 k 2 } μ A ( y ) , μ A ( y ) > 1 t k if  min { μ A ( x ) , 1 k 2 } > μ A ( y ) ,

and

γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 1 k 2 } = { max { γ A ( x ) , 1 k 2 } s if  max { γ A ( x ) , 1 k 2 } γ A ( y ) , γ A ( y ) < 1 s k if  max { γ A ( x ) , 1 k 2 } < γ A ( y ) .

Thus xy C ( t , s ) (A) Q ( t , s ) k (A)= [ A ] ( t , s ) k . We have a similar result for the case (iii). For the final case, if t> 1 k 2 and s< 1 k 2 , then 1tk< 1 k 2 and 1sk> 1 k 2 . Hence

μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 1 k 2 } = 1 k 2 > 1 t k whenever  min { μ A ( x ) , μ A ( y ) } 1 k 2

and

μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 1 k 2 } = min { μ A ( x ) , μ A ( y ) } > 1 s k whenever  min { μ A ( x ) , μ A ( y ) } 1 k 2

and

γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 1 k 2 } = 1 k 2 < 1 s k whenever  max { γ A ( x ) , γ A ( y ) } 1 k 2

and

γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 1 k 2 } = max { γ A ( x ) , γ A ( y ) } < 1 s k , whenever  max { γ A ( x ) , γ A ( y ) } 1 k 2 .

Thus, xy Q ( t , s ) k (A) [ A ] ( t , s ) k . If t 1 k 2 and s 1 k 2 , then

μ A ( x y ) min { μ A ( x ) , μ A ( y ) , 1 k 2 } = { 1 k 2 t if  min { μ A ( x ) , μ A ( y ) } 1 k 2 , min { μ A ( x ) , μ A ( y ) } > 1 t k if  min { μ A ( x ) , μ A ( y ) } < 1 k 2 ,

and

γ A ( x y ) max { γ A ( x ) , γ A ( y ) , 1 k 2 } = { 1 k 2 s if  max { γ A ( x ) , γ A ( y ) } 1 k 2 , max { γ A ( x ) , γ A ( y ) } < 1 s k if  max { γ A ( x ) , γ A ( y ) } > 1 k 2 ,

which implies that xy Q ( t , s ) k (A) [ A ] ( t , s ) k . Let x,z [ A ] ( t , s ) k . Then x C ( t , s ) (A) or x;(t,s) q k A and z C ( t , s ) (A) or z;(t,s) q k A, that is, μ A (x)t, γ A (x)s or μ A (x)+t+k>1, γ A (x)+s+k<1 and μ A (z)t, γ A (z)s or μ A (z)+t+k>1, γ A (z)+s+k<1. We consider the following four cases

  1. (i)

    μ A (x)t, γ A (x)s and μ A (z)t, γ A (z)s,

  2. (ii)

    μ A (x)t, γ A (x)s and μ A (z)+t>1k, γ A (z)+s<1k,

  3. (iii)

    μ A (x)+t>1k, γ A (x)+s<1k and μ A (z)t, γ A (z)s,

  4. (iv)

    μ A (x)+t>1k, γ A (x)+s<1k and μ A (z)+t>1k, γ A (z)+s<1k.

For the case (i), Theorem 3.5(3) implies that

μ A (xyz)min { μ A ( x ) , μ A ( z ) , 1 k 2 } min { t , 1 k 2 } ={ 1 k 2 if  t > 1 k 2 , t if  t 1 k 2

and

γ A (xyz)max { γ A ( x ) , γ A ( z ) , 1 k 2 } max { s , 1 k 2 } ={ 1 k 2 if  s < 1 k 2 , s if  s 1 k 2 .

Then xyz C ( t , s ) (A) or μ A (xyz)+t+k> 1 k 2 + 1 k 2 +k=1 and γ A (xyz)+s+k< 1 k 2 + 1 k 2 +k=1, that is, xyz Q ( t , s ) k (A). Hence xyz C ( t , s ) (A) Q ( t , s ) k (A)= [ A ] ( t , s ) k . For the second case, assume that t> 1 k 2 and s< 1 k 2 , then 1tk1t< 1 k 2 and 1sk1s 1 k 2 . Hence

μ A ( x y z ) min { μ A ( x ) , μ A ( z ) , 1 k 2 } = { min { μ A ( x ) , 1 k 2 } > 1 t k if  min { μ A ( z ) , 1 k 2 } μ A ( x ) , μ A ( x ) t if  min { μ A ( z ) , 1 k 2 } > μ A ( x )

and

γ A ( x y z ) max { γ A ( x ) , γ A ( z ) , 1 k 2 } = { max { γ A ( z ) , 1 k 2 } < 1 s k if  max { γ A ( z ) , 1 k 2 } γ A ( x ) , γ A ( x ) s if  max { γ A ( z ) , 1 k 2 } < γ A ( x ) .

Thus, xyz C ( t , s ) (A) Q ( t , s ) k (A)= [ A ] ( t , s ) k . Suppose that t 1 k 2 and s 1 k 2 . Then

μ A ( x y z ) min { μ A ( x ) , μ A ( z ) , 1 k 2 } = { min { μ A ( x ) , 1 k 2 } t if  min { μ A ( x ) , 1 k 2 } μ A ( z ) , μ A ( z ) > 1 t k if  min { μ A ( x ) , 1 k 2 } > μ A ( z ) ,

and

γ A ( x y z ) max { γ A ( x ) , γ A ( z ) , 1 k 2 } = { max { γ A ( x ) , 1 k 2 } s if  max { γ A ( x ) , 1 k 2 } γ A ( z ) , γ A ( y ) < 1 s k if  max { γ A ( x ) , 1 k 2 } < γ A ( z ) .

Thus, xyz C ( t , s ) (A) Q ( t , s ) k (A)= [ A ] ( t , s ) k . We have a similar result for the case (iii). For the final case, if t> 1 k 2 and s< 1 k 2 , then 1tk< 1 k 2 and 1sk> 1 k 2 . Hence

μ A ( x y z ) min { μ A ( x ) , μ A ( z ) , 1 k 2 } = 1 k 2 > 1 t k whenever  min { μ A ( x ) , μ A ( z ) } 1 k 2

and

μ A ( x y z ) min { μ A ( x ) , μ A ( z ) , 1 k 2 } = min { μ A ( x ) , μ A ( z ) } > 1 s k whenever  min { μ A ( x ) , μ A ( z ) } 1 k 2

and

γ A ( x y z ) max { γ A ( x ) , γ A ( z ) , 1 k 2 } = 1 k 2 < 1 s k whenever  max { γ A ( x ) , γ A ( z ) } 1 k 2

and

γ A ( x y z ) max { γ A ( x ) , γ A ( z ) , 1 k 2 } = max { γ A ( x ) , γ A ( z ) } < 1 s k whenever  max { γ A ( x ) , γ A ( z ) } 1 k 2 .

Thus, xyz Q ( t , s ) k (A) [ A ] ( t , s ) k . If t 1 k 2 and s 1 k 2 , then

μ A ( x y z ) min { μ A ( x ) , μ A ( z ) , 1 k 2 } = { 1 k 2 t if  min { μ A ( x ) , μ A ( z ) } 1 k 2 , min { μ A ( x ) , μ A ( z ) } > 1 t k if  min { μ A ( x ) , μ A ( z ) } < 1 k 2 ,

and

γ A ( x y z ) max { γ A ( x ) , γ A ( z ) , 1 k 2 } = { 1 k 2 s if  max { γ A ( x ) , γ A ( z ) } 1 k 2 , max { γ A ( x ) , γ A ( z ) } < 1 s k if  max { γ A ( x ) , γ A ( z ) } > 1 k 2 ,

which implies that xyz Q ( t , s ) k (A) [ A ] ( t , s ) k . Therefore, [ A ] ( t , s ) k is a bi-ideal of S.

Conversely, suppose that (2) is valid. If there exist a,bS such that ab and

μ A (a)<min { μ A ( b ) , 1 k 2 } and γ A (a)>max { γ A ( b ) , 1 k 2 } .

Then μ A (a)< t a min{ μ A (b), 1 k 2 } and γ A (a)> s a max{ γ A (b), 1 k 2 } for some t a (0,1] and s a [0,1). It follows that b C ( t a , s a ) (A) [ A ] ( t a , s a ) k but a C ( t a , s a ) (A). Also we have μ A (a)+ t a <2 t a 1k and γ A (a)+ s a >2 s a 1k and so a;( t a , s a ) q k ¯ A, i.e., b Q ( t a , s a ) k (A). Therefore, a [ A ] ( t a , s a ) , a contradiction. Hence μ A (x)min{ μ A (y), 1 k 2 } and γ A (x)max{ γ A (y), 1 k 2 } for all x,yS with xy. Suppose that there exist a,bS such that

μ A (ab)<min { μ A ( a ) , μ A ( b ) , 1 k 2 }

and

γ A (ab)>max { γ A ( a ) , γ A ( b ) , 1 k 2 } .

Then

μ A (ab)<tmin { μ A ( a ) , μ A ( b ) , 1 k 2 }

and

γ A (ab)>smax { γ A ( a ) , γ A ( b ) , 1 k 2 }

for t(0,1] and s[0,1). It follows that a C ( t , s ) (A) [ A ] ( t , s ) k and b C ( t , s ) (A) [ A ] ( t , s ) k , so from (b2) ab [ A ] ( t , s ) k . Thus, μ A (ab)t, γ A (ab)s or μ A (ab)+t+k>1, γ A (ab)+s+k<1, a contradiction. Therefore, μ A (xy)min{ μ A (x), μ A (y), 1 k 2 } and γ A (xy)max{ γ A (x), γ A (y), 1 k 2 } for all x,yS. Assume that there exist a,b,cS such that

μ A (abc)<min { μ A ( a ) , μ A ( c ) , 1 k 2 }

and

γ A (abc)>max { γ A ( a ) , γ A ( c ) , 1 k 2 } .

Then μ A (abc)< t 0 min{ μ A (a), μ A (c), 1 k 2 } and γ A (abc)> s 0 max{ γ A (a), γ A (c), 1 k 2 } for t 0 (0,1] and s 0 [0,1). It follows that a C ( t 0 , s 0 ) (A) [ A ] ( t 0 , s 0 ) k and c C ( t 0 , s 0 ) (A) [ A ] ( t 0 , s 0 ) k so from (b2) abc [ A ] ( t 0 , s 0 ) . Thus, μ A (ab) t 0 , γ A (ab) s 0 or μ A (ab)+ t 0 +k>1, γ A (ab)+ s 0 +k<1, a contradiction. Therefore, μ A (xyz)min{ μ A (x), μ A (z), 1 k 2 } and γ A (xyz)max{ γ A (x), γ A (z), 1 k 2 } for all x,y,zS. Thus, A=x, μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S. □

Theorem 3.14 Let { A i |iΛ} be a family of (, q k )-intuitionistic fuzzy bi-ideals of S. Then A= i Λ A i is an (, q k )-intuitionistic fuzzy bi-ideal of S, where i Λ A i =x, i Λ μ A i , i Λ γ A i and

i Λ μ A i ( x ) = inf { μ A i ( x ) | i Λ and x S } , i Λ γ A i ( x ) = sup { γ A i ( x ) | i Λ and x S } .

Proof Let x,yS with xy, t(0,1] and s[0,1) be such that y;(t,s)A. Assume that x;(t,s) q k ¯ A. Then μ A (x)<t, γ A (x)>s and μ A (x)+t+k1, γ A (x)+s+k1, which imply that

μ A (x)< 1 k 2 and γ A (x)> 1 k 2 .

Let Ω 1 :={iΛ| μ A i (x)t, γ A i (x)s} and Ω 2 :={iΛ|x;(t,s) q k A i  and  μ A i (x)<t, γ A i (x)>s}.

Then Λ= Ω 1 Ω 2 and Ω 1 Ω 2 =. If Ω 2 =, then μ A i (x)t, γ A i (x)s for all iΛ, and so, μ A (x)t, γ A (x)s, which is a contradiction. Hence Ω 2 , and so, μ A i (x)+t+k>1, γ A i (x)+s+k<1 and μ A i (x)<t, γ A i (x)>s for every iΛ. It follows that t> 1 k 2 and s< 1 k 2 , so that μ A i (x) μ A (x)t> 1 k 2 and γ A i (x) γ A (x)s< 1 k 2 for all iΛ. Now, suppose that t x := μ A i (x)< 1 k 2 and s x := γ A i (x)> 1 k 2 for some iΛ. Let t x (0, 1 k 2 ) and s x ( 1 k 2 ,1) be such that t x < t x and s x < s x . Then μ A i (y)> 1 k 2 > t x and γ A i (y)< 1 k 2 < s x , i.e., y;( t x , s x ) A i . But μ A i (x)= t x < t x , γ A i (x)= s x > s x and μ A i (x)+ t x +k<1, γ A i (x)+ s x +k>1, that is, x;( t x , s x ) q k ¯ A i . This is a contradiction, and so, μ A i (x) 1 k 2 and γ A i (x) 1 k 2 for all iΛ. Thus, μ A (x) 1 k 2 and γ A (x) 1 k 2 , which is impossible. Therefore, y;(t,s) q k A.

Let x,yS, t 1 , t 2 (0,1] and s 1 , s 2 [0,1) be such that x;( t 1 , s 1 )A and y;( t 2 , s 2 )A. Assume that xy;min{ t 1 , t 2 },max{ s 1 , s 2 } q k ¯ A. Then

μ A (xy)<min{ t 1 , t 2 }, γ A (xy)>max{ s 1 , s 2 }

and

μ A (xy)+min{ t 1 , t 2 }<1k, γ A (xy)+max{ s 1 , s 2 }>1k.

It follows that μ A (xy)< 1 k 2 and γ A (xy)> 1 k 2 . Let Ω 3 :={iΛ| μ A i (xy)min{ t 1 , t 2 } and γ A i (xy)max{ s 1 , s 2 }} and Ω 4 :={iΛ|xy;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A i  and  μ A i (xy)<min{ t 1 , t 2 } and  γ A i (xy)>max{ s 1 , s 2 }}. Then Ω 3 Ω 4 =Λ and Ω 3 Ω 4 =. If Ω 4 =, then μ A i (xy)min{ t 1 , t 2 } and γ A i (xy)max{ s 1 , s 2 } for all iΛ, and so, μ A (xy)min{ t 1 , t 2 } and γ A (xy)max{ s 1 , s 2 }, which is a contradiction. Hence Ω 4 and xy;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A i , i.e., μ A i (xy)+min{ t 1 , t 2 }>1k, γ A i (xy)+max{ s 1 , s 2 }<1k. It follows that min{ t 1 , t 2 }> 1 k 2 and max{ s 1 , s 2 }< 1 k 2 , so that μ A i (x) μ A (x) t 1 min{ t 1 , t 2 }> 1 k 2 and γ A i (x) γ A (x) s 1 max{ s 1 , s 2 } for all iΛ. By a similar way, we have μ A i (y) μ A (y) t 1 min{ t 1 , t 2 }> 1 k 2 and γ A i (y) γ A (y) s 1 max{ s 1 , s 2 } for all iΛ. Now, suppose that t:= μ A i (xy)< 1 k 2 and s:= γ A i (xy)> 1 k 2 for some iΛ. Let t (0, 1 k 2 ) and s ( 1 k 2 ,1) be such that t< t and s> s . Then μ A i (x)> 1 k 2 > t , γ A i (x)< 1 k 2 < s and μ A i (y)> 1 k 2 > t , γ A i (y)< 1 k 2 < s , i.e., x;( t , s )A and y;( t , s )A. But μ A i (xy)=t< t , γ A i (xy)=s> s and μ A i (xy)+ t +k<1, γ A i (xy)+ s +k>1, that is, xy;( t , s ) ¯ q k ¯ A i . This is a contradiction. Thus, μ A i (xy) 1 k 2 and γ A i (xy) 1 k 2 for all iΛ. Therefore, μ A (xy) 1 k 2 and γ A (xy) 1 k 2 , which is invalid. Consequently, xy;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A. Finally, suppose that x,y,zS, t 1 , t 2 (0,1] and s 1 , s 2 [0,1) be such that x;( t 1 , s 1 )A and z;( t 2 , s 2 )A. Assume that xyz;min{ t 1 , t 2 },max{ s 1 , s 2 } q k ¯ A. Then

μ A (xyz)<min{ t 1 , t 2 }, γ A (xyz)>max{ s 1 , s 2 }

and

μ A (xyz)+min{ t 1 , t 2 }<1k, γ A (xyz)+max{ s 1 , s 2 }>1k.

It follows that μ A (xyz)< 1 k 2 and γ A (xyz)> 1 k 2 . Let

Ω 5 := { i Λ | μ A i ( x y z ) min { t 1 , t 2 }  and  γ A i ( x y z ) max { s 1 , s 2 } }

and

Ω 6 : = { i Λ | x y z ; min { t 1 , t 2 } , max { s 1 , s 2 } q k A i and  μ A i ( x y z ) < min { t 1 , t 2 }  and  γ A i ( x y z ) > max { s 1 , s 2 } } .

Then Ω 5 Ω 6 =Λ and Ω 5 Ω 6 =. If Ω 6 =, then μ A i (xyz)min{ t 1 , t 2 } and γ A i (xyz)max{ s 1 , s 2 } for all iΛ, and so μ A (xyz)min{ t 1 , t 2 } and γ A (xyz)max{ s 1 , s 2 } which is a contradiction. Hence Ω 6 and

x y z ; min { t 1 , t 2 } , max { s 1 , s 2 } q k A i ,

i.e.,

μ A i (xyz)+min{ t 1 , t 2 }>1k, γ A i (xyz)+max{ s 1 , s 2 }<1k.

It follows that min{ t 1 , t 2 }> 1 k 2 and max{ s 1 , s 2 }< 1 k 2 , so that

μ A i (x) μ A (x) t 1 min{ t 1 , t 2 }> 1 k 2

and

γ A i (x) γ A (x) s 1 max{ s 1 , s 2 }

for all iΛ. Similarly, we have

μ A i (z) μ A (z) t 1 min{ t 1 , t 2 }> 1 k 2

and

γ A i (z) γ A (z) s 1 max{ s 1 , s 2 }

for all iΛ. Now, suppose that t:= μ A i (xyz)< 1 k 2 and s:= γ A i (xyz)> 1 k 2 for some iΛ. Let t (0, 1 k 2 ) and s ( 1 k 2 ,1) be such that t< t and s> s . Then μ A i (x)> 1 k 2 > t , γ A i (x)< 1 k 2 < s and μ A i (y)> 1 k 2 > t , γ A i (y)< 1 k 2 < s , i.e., x;( t , s )A and y;( t , s )A. But

μ A i (xyz)=t< t , γ A i (xyz)=s> s

and

μ A i (xyz)+ t <1, γ A i (xyz)+ s >1,

that is, xyz;( t , s ) ¯ q k ¯ A i . This is a contradiction. Thus, μ A i (xyz) 1 k 2 and γ A i (xyz) 1 k 2 for all iΛ. Therefore, μ A (xyz) 1 k 2 and γ A (xyz) 1 k 2 , which is invalid. Thus, xyz;min{ t 1 , t 2 },max{ s 1 , s 2 } q k A. Therefore, i Λ A i is an (, q k )-intuitionistic fuzzy bi-ideal of S. □

The following example shows that the union of two (, q k )-intuitionistic fuzzy bi-ideals of S may not be an (, q k )-intuitionistic fuzzy bi-ideal of S.

Example 3.15 Consider the ordered semigroup of Example 3.4 with the ∗-multiplication Table 1 and the IFS of Examples 3.4 and 3.7, then a;(0.7,0.3)AB and b;(0.5,0.4)AB, but abmin{0.7,0.5},max{0.3,0.4}=d;(0.5,0.4) q k ¯ AB.

Definition 3.16 An IFS A=x, μ A , γ A of S is called an ( ¯ , ¯ q k ¯ )-intuitionistic fuzzy bi-ideal of S if for all x,y,zS, t, t 1 , t 2 (0,1] and s, s 1 , s 2 [0,1), it satisfies the following conditions

(q4) (x;(t,s) ¯ Ay;(t,s) ¯ q k ¯ A with xy),

(q5) (xy;min{ t 1 , t 2 },max{ s 1 , s 2 } ¯ Ax;( t 1 , s 1 ) ¯ q k ¯ A or y;( t 2 , s 2 ) ¯ q k ¯ A),

(q6) (xyz;min{ t 1 , t 2 },max{ s 1 , s 2 } ¯ Ax;( t 1 , s 1 ) ¯ q k ¯ A or z;( t 2 , s 2 ) ¯ q k ¯ A).

Let A=x, μ A , γ A be an ( ¯ , ¯ q k ¯ )-intuitionistic fuzzy bi-ideal of an ordered semigroup S. Suppose that there exist a,bS with ab such that

μ A (b)>max { μ A ( a ) , 1 k 2 } and γ A (b)<min { γ A ( a ) , 1 k 2 } .

Then μ A (b)t>max{ μ A (a), 1 k 2 } and γ A (b)s<min{ γ A (a), 1 k 2 } for some t( 1 k 2 ,1] and s[0, 1 k 2 ). It follows that a;(t,s) ¯ A, b;(t,s)A and μ A (b)+t2t>1k, γ A (b)+s2s<1k, i.e., b;(t,s) q k A. This is a contradiction, and so the following inequalities hold.

(e1) ( x y μ A ( b ) max { μ A ( a ) , 1 k 2 }  and  γ A ( b ) min { γ A ( a ) , 1 k 2 } ) .

Suppose that max{ μ A (ab), 1 k 2 }<min{ μ A (a), μ A (b)} and min{ γ A (ab), 1 k 2 }>max{ γ A (a), γ A (b)} for some a,bS. Then max{ μ A (ab), 1 k 2 }<tmin{ μ A (a), μ A (b)} and min{ γ A (ab), 1 k 2 }>smax{ γ A (a), γ A (b)}. Thus, a;(t,s) ¯ A, b;(t,s) ¯ A, μ A (a)+t2t>1k, γ A (a)+s2s<1k, i.e., a;(t,s) q k A and μ A (b)+t2t>1k, γ A (b)+s2s<1k, i.e., b;(t,s) q k A. This is impossible, and hence A=x, μ A , γ A satisfies the following assertion

(e2) max{ μ A (xy), 1 k 2 }<min{ μ A (x), μ A (y)} and min{ γ A (xy), 1 k 2 }>max{ γ A (x), γ A (y)} for all x,yS.

Now, assume that

max { μ A ( a b c ) , 1 k 2 } <min { μ A ( a ) , μ A ( c ) }

and

min { γ A ( a b c ) , 1 k 2 } >max { γ A ( a ) , γ A ( c ) }

for some a,b,cS. Then

max { μ A ( a b c ) , 1 k 2 } <tmin { μ A ( a ) , μ A ( c ) }

and

min { γ A ( a b c ) , 1 k 2 } >smax { γ A ( a ) , γ A ( c ) } .

Thus, a;(t,s) ¯ A,c;(t,s) ¯ A, μ A (a)+t2t>1k, γ A (a)+s2s<1k, i.e., a;(t,s) q k A and μ A (c)+t2t>1k, γ A (c)+s2s<1k, i.e., c;(t,s) q k A. This is a contradiction, and hence we have the following assertion

(e3) max{ μ A (xyz), 1 k 2 }<min{ μ A (x), μ A (z)} and min{ γ A (xyz), 1 k 2 }>max{ γ A (x), γ A (z)} for all x,y,zS.

Let A=x, μ A , γ A be an IFS of S satisfying the three conditions (e1), (e2) and (e3). Let t( 1 k 2 ,1] and s[0, 1 k 2 ) be such that C ( t , s ) (A). Then there exist b C ( t , s ) (A) and Sab, by using (e1), we get

1 k 2 <t μ A (b)max { μ A ( a ) , 1 k 2 } = μ A (a)

and

1 k 2 >s γ A (b)min { γ A ( a ) , 1 k 2 } = γ A (a).

Hence a C ( t , s ) (A). Let a,bS be such that a C ( t , s ) (A) and b C ( t , s ) (A). Then μ A (a)t, γ A (a)s and μ A (b)t, γ A (b)s. Using (e2), we get

max { μ A ( a b ) , 1 k 2 } min { μ A ( a ) , μ A ( b ) } t> 1 k 2

and

min { γ A ( a b ) , 1 k 2 } max { γ A ( a ) , γ A ( b ) } s< 1 k 2 ,

which implies that μ A (ab)=max{ μ A (ab), 1 k 2 }t and γ A (ab)=min{ γ A (ab), 1 k 2 }s. Thus, ab C ( t , s ) (A). Now, suppose that a,c C ( t , s ) (A) and bS. Then μ A (a)t, γ A (a)s and μ A (c)t, γ A (c)s. Using (e3), we get

max { μ A ( a b c ) , 1 k 2 } min { μ A ( a ) , μ A ( c ) } t> 1 k 2

and

min { γ A ( a b c ) , 1 k 2 } max { γ A ( a ) , γ A ( c ) } s< 1 k 2 ,

which implies that μ A (abc)=max{ μ A (abc), 1 k 2 }t and γ A (abc)=min{ γ A (abc), 1 k 2 }s. Thus, abc C ( t , s ) (A). Consequently, C ( t , s ) (A) is a bi-ideal of S. Therefore, we conclude that if an IFS A=x, μ A , γ A of S satisfies the three conditions (e1), (e2) and (e3), then the following assertion is valid

(e4) (t(0.5,1]) (s[0, 1 k 2 )) ( C ( t , s ) (A) C ( t , s ) (A) is a bi-ideal of S).

Now, let A=x, μ A , γ A be an IFS of S satisfying (e4). Let a,bS with ab and t(0,1] and s[0,1) be such that b;(t,s) ¯ q k ¯ ¯ A. Then b;(t,s)A and b;(t,s) q k A. Hence b C ( t , s ) (A) and C ( t , s ) (A). Thus, by (e4), a C ( t , s ) (A) and so μ A (a)t, γ A (a)s, i.e., a;(t,s)A. This shows that (q4) is valid. Let a,bS, t 1 , t 2 (0,1] and s 1 , s 2 [0,1) be such that a;( t 1 , s 1 ) ¯ q k ¯ ¯ A and b;( t 2 , s 2 ) ¯ q k ¯ ¯ A. Then a;( t 1 , s 1 )A, a;( t 1 , s 1 ) q k A and b;( t 2 , s 2 )A, b;( t 2 , s 2 ) q k A, which implies that a C ( t , s ) (A) C ( min { t 1 , t 2 } , max { s 1 , s 2 } ) (A) and b C ( t , s ) (A) C ( min { t 1 , t 2 } , max { s 1 , s 2 } ) (A). Since C ( min { t 1 , t 2 } , max { s 1 , s 2 } ) (A) is bi-ideal of S by (e4), it follows by (b2) that ab C ( min { t 1 , t 2 } , max { s 1 , s 2 } ) (A), that is, μ A (ab)min{ t 1 , t 2 }, and γ A (ab)max{ s 1 , s 2 }, so that ab;min{ t 1 , t 2 },max{ s 1 , s 2 }A. Hence (q5) is valid. Finally, let a,b,cS, t 1 , t 2 (0,1] and s 1 , s 2 [0,1) be such that a;( t 1 , s 1 ) ¯ q k ¯ ¯ A and c;( t 2 , s 2 ) ¯ q k ¯ ¯ A. Then a;( t 1 , s 1 )A, a;( t 1 , s 1 ) q k A and c;( t 2 , s 2 )A, c;( t 2 , s 2 ) q k A, which implies that a C ( t , s ) (A) C ( min { t 1 , t 2 } , max { s 1 , s 2 } ) (A) and c C ( t , s ) (A) C ( min { t 1 , t 2 } , max { s 1 , s 2 } ) (A). Since C ( min { t 1 , t 2 } , max { s 1 , s 2 } ) (A) is bi-ideal of S by (e4), it follows by (b3) that abc C ( min { t 1 , t 2 } , max { s 1 , s 2 } ) (A), that is, μ A (abc)min{ t 1 , t 2 }, and γ A (abc)max{ s 1 , s 2 }, so that abc;min{ t 1 , t 2 },max{ s 1 , s 2 }A. Hence (q6) is valid.

Therefore, as a concluding remark, we have the following theorem.

Theorem 3.17 For an IFS A=x, μ A , γ A of S, the following are equivalent

  1. (1)

    A=x, μ A , γ A is an ( ¯ , ¯ q k ¯ )-intuitionistic fuzzy bi-ideal of S.

  2. (2)

    A=x, μ A , γ A satisfies the condition (e4).

  3. (3)

    A=x, μ A , γ A satisfies the three conditions (e1), (e2) and (e3).

For an IFS A=x; μ A , γ A of S, we consider the following sets

Γ 1 : = { t ( 0 , 1 ] | U ( μ A ; t ) U ( μ A ; t )  is a bi-ideal of  S } , Γ 2 : = { s [ 0 , 1 ) | L ( μ A ; t ) L ( μ A ; t )  is a bi-ideal of  S } .

Then

  1. (1)

    If Γ 1 =(0,1] and Γ 2 =[0,1), then A=x; μ A , γ A is an intuitionistic fuzzy bi-ideal of S.

  2. (2)

    If Γ 1 =(0,0.5] and Γ 2 =[0.5,1), then A=x; μ A , γ A is an (,q)-intuitionistic fuzzy bi-ideal of S.

  3. (3)

    If Γ 1 =(0, 1 k 2 ] and Γ 2 =[ 1 k 2 ,1), then A=x; μ A , γ A is an (, q k )-intuitionistic fuzzy bi-ideal of S.

  4. (4)

    If Γ 1 =( 1 k 2 ,1] and Γ 2 =[0, 1 k 2 ), then A=x; μ A , γ A is an ( ¯ , ¯ q k ¯ )-intuitionistic fuzzy bi-ideal of S.