The concept of intuitionistic fuzzy ideal of an intuitionistic fuzzy lattice is introduced, and its certain characterizations are provided. We defined the quotient (or residual) of ideals of an intuitionistic fuzzy sublattice and studied their properties.

1. Introduction

The concept of intuitionistic fuzzy sets was introduced by Atanassov [1, 2] as a generalization of that of fuzzy sets and it is a very effective tool to study the case of vagueness. Further many researches applied this notion in various branches of mathematics especially in algebra and defined intuitionistic fuzzy subgroups (IFG), intuitionistic fuzzy subrings (IFR), and intuitionistic fuzzy sublattice (IFL), and so forth. In the last five years there are so many articles appeared in this direction. Kim [3], Kim and Jun [4], Kim and Lee [5], introduced different types of IFI’s in Semigroups. Torkzadeh and Zahedi [6] defined intuitionistic fuzzy commutative hyper K-ideals, Akram and Dudek [7] defined intuitionistic fuzzy Lie ideals of Lie algebras, and Hur et al. [8] introduced intuitionistic fuzzy prime ideals of a Ring.

The concept of ideal of a fuzzy subring was introduced by Mordeson and Malik in [9]. After that N Ajmal and A.S Prajapathi introduced the concept of residual of ideals of an L-Ring in [10]. Motivated by this, in this paper we first defined the intuitionistic fuzzy ideal of an IFL and certain characterizations are given. Lastly we defined quotients (residuals) of ideals of an intuitionistic fuzzy sublattice and studied their properties.

2. Preliminaries

We recall the following definitions and results which will be used in the sequel. Throughout this paper L stands for a lattice (L,∨,∧) with zero element “0” and unit element “1”.

Definition 1 (see [<xref ref-type="bibr" rid="B5">1</xref>]).

Let X be a nonempty set. An intuitionistic fuzzy set [IFS] A of X is an object of the following form A={〈x,μA(x),νA(x)〉∣x∈X}, where μA:X→[01] and νA:X→[01] define the degree of membership and the degree of non membership of the element x∈X, respectively, and ∀x∈X, 0≤μA(x)+νA(x)≤1.

The set of all IFS’s on X is denoted by IFS (X).

Definition 2 (see [<xref ref-type="bibr" rid="B5">1</xref>]).

If A={〈x,μA(x),νA(x)〉∣x∈S} and B={〈x,μB(x),νB(x)〉∣x∈S} are any two IFS of X then

Definition 3 (see [<xref ref-type="bibr" rid="B16">11</xref>]).

Let L be a lattice and A={〈x,μA(x),νA(x)〉∣x∈L} be an IFS of L. Then A is called an intuitionistic fuzzy sublattice [IFL] of L if the following conditions are satisfied.

μA(x∨y)≥min{μA(x),μA(y)};

μA(x∧y)≥min{μA(x),μA(y)};

νA(x∨y)≤max{νA(x),νA(y)};

νA(x∧y)≤max{νA(x),νA(y)},∀x,y∈L.

The set of all intuitionist fuzzy sublattices (IFL’s) of L is denoted as IFL (L).Definition 4 (see [<xref ref-type="bibr" rid="B16">11</xref>]).

An IFS A of L is called an intuitionistic fuzzy ideal (IFI) of L if the following conditions are satisfied.

μA(x∨y)≥min{μA(x),μA(y)};

μA(x∧y)≥max{μA(x),μA(y)};

νA(x∨y)≤max{νA(x),νA(y)};

νA(x∧y)≤min{νA(x),νA(y)},∀x,y∈L.

The set of all IFI’s of L is denoted as IFI (L).

Definition 5 (see [<xref ref-type="bibr" rid="B7">12</xref>]).

Let A, B∈ IFS (L). Then we define an IFS (L),

A+B={〈z,μA+B(z),νA+B(z)〉∣z∈L}, where
μA+B(z)=supz=x∨y{min{μA(x),μB(y)}},νA+B(z)=infz=x∨y{max{νA(x),νB(y)}}.

AB={〈z,μAB(z),νAB(z)〉∣z∈L}, where
μAB(z)=supz=x∧y{min{μA(x),μB(y)}}νAB(z)=infz=x∧y{max{νA(x),νB(y)}}.

Lemma 1 (see [<xref ref-type="bibr" rid="B15">13</xref>]).

Let A, B, and C be IFS (L), then the following assertions hold.

AB=BA, A+B=B+A, A·B=B·A.

AB⊆A·B.

C(A+B)⊆CA+CB.

(C+B)A⊆CA+BA.

(A∩B)C⊆AC∩BC.

A⊆B⇒AC⊆BC and A·C⊆B·C.

Lemma 2 (see [<xref ref-type="bibr" rid="B15">13</xref>]).

Let A be an IFL of L, then

A+A=A.

AA=A.

3. Ideal of an Intuitionistic Fuzzy Lattice

In this section we define the ideal of an IFL, and give some characterization of these ideals in terms of operations on IFS (L). We also used ∨ and ∧ to represent maximum and minimum, respectively, which is clear from the context.

Definition 6.

Let A be an IFL of L and B an IFS of L with B⊆A. Then B is called an intuitionistic fuzzy ideal (IFI) of A if the following conditions are satisfied.

μB(x∨y)≥μB(x)∧μB(y).

μB(x∧y)≥[μA(x)∧μB(y)]∨[μB(x)∧μA(y)].

νB(x∨y)≤νB(x)∨νB(y).

νB(x∧y)≤[νA(x)∨νB(y)]∧[νB(x)∨νA(y)]∀x,y∈L.

If B IFI of A, then we write B⊲AExample 1.

Consider the lattice L={1,2,5,10} under divisibility.

Let A={〈x,μA(x),νA(x)〉∣x∈L} be an IFL of L defined by 〈1,.5,.1〉, 〈2,.4,.5〉, 〈5,.4,.3〉, 〈10,.7,.3〉 and B={〈x,μB(x),νB(x)〉∣x∈L} be an IFS of L given by 〈1,.5,.3〉, 〈2,.4,.5〉, 〈5,.3,.4〉, 〈10,.3,.4〉. Clearly B⊲A.

Definition 7.

Let A be an IFL and B is also an IFL with B⊆A. Then B is called an intuitionistic fuzzy sublattice of A.

Lemma 3.

The intersection of two IFI’s of A is again an IFI of A.

Proof.

Let B, C be IFI’s of A. Then we can prove that B∩Cis also an IFI of A. Since B⊆A and C⊆A, we have B∩C⊆A Also
μB∩C(x∨y)=min{μB(x∨y),μC(x∨y)}≥min{μB(x)∧μB(y),μC(x)∧μC(y)},sinceBandCareIFI’sofA.≥min{μB(x)∧μC(x),μB(y)∧μC(y)}≥min{μB∩C(x),μB∩C(y)}≥μB∩C(x)∧μB∩C(y),μB∩C(x∧y)=min{μB(x∧y),μC(x∧y)}≥min{[μB(x)∧μA(y)]∨[μB(y)∧μA(x)],[μC(x)∧μA(y)]∨[μA(x)∧μC(y)]}sinceBandCareIFI’sofA.≥min{μB(x),μC(x)}∧μA(y)∨min{μB(y),μC(y)}∧μA(x)≥[μB∩C(x)∧μA(y)]∨[μB∩C(y)∧μA(x)].
Also
νB∩C(x∨y)=max{νB(x∨y),νC(x∨y)}≤max{νB(x)∨νB(y),νC(x)∨νC(y)},sinceBandCareIFI’sofA.≤max{νB(x)∨νC(x),νB(y)∨νC(y)}≤max{νB∩C(x),νB∩C(y)}≤νB∩C(x)∨νB∩C(y),νB∩C(x∧y)=max{νB(x∧y),νC(x∧y)}≤max{[νB(x)∨νA(y)]∧[νB(y)∨νA(x)],[νC(x)∨νA(y)]∧[νA(x)∨νC(y)]},sinceBandCareIFI’sofA.≤max{νB(x),νC(x)}∨νA(y)∧max{νB(y),νC(y)}∨νA(x)≤[νB∩C(x)∨νA(y)]∧[νB∩C(y)∨νA(x)].
Hence, B∩C is an IFI of A.

Theorem 1.

Let A an IFL and B an IFS of L with B⊆A. Then B is an IFI of A if and only if

μB(x∨y)≥μB(x)∧μB(y),

νB(x∨y)≤νB(x)∨νB(y),

AB⊆B.

Proof.

Suppose that conditions (1), (2), and (3) hold. Then we prove that B is an IFI of A.

We have
μB(x∧y)≥μAB(x∧y),sinceB⊇AB=supx∧y=xi∧yi(μA(xi)∧μB(yi))≥μA(x)∧μB(y).
Similarly
μB(x∧y)≥μB(x)∧μA(y),sinceB⊇AB=BA.
Hence
μB(x∧y)≥[μA(x)∧μB(y)]∨[μB(x)∧μA(y)].
Also
νB(x∧y)≤νAB(x∧y),sinceB⊇AB=infx∧y=xi∧yi(νA(xi)∨νB(yi))≤νA(x)∨νB(y).
Similarly
νB(x∧y)≤νB(x)∨νA(y),sinceB⊇AB=BA.
Hence
νB(x∧y)≤[νA(x)∨νB(y)]∧[νB(x)∨νA(y)].
So from (1), (2), (a), and (b) B is an IFI of A.

Conversely suppose B is an IFI of A. Then obviously conditions (1) and (2) holds. Also we have
μB(x∧y)≥μA(x)∧μB(y),νB(x∧y)≤νA(x)∨νB(y),∀x,y∈L.

So ∀z∈L with z=x∧yμB(z)≥⋁z=x∧y[μA(x)∧μB(y)]=μAB(z),νB(z)≤⋀z=x∧y[νA(x)∨νB(y)]=νAB(z).
Hence AB⊆B.

Theorem 2.

Let A be an IFL of L and B an IFS with B⊆A. Then B is an IFI of A if and only if

μB(x∨y)≥μB(x)∧μB(y),

νB(x∨y)≤νB(x)∨νB(y),

A·B⊆B.

Proof.

Suppose conditions (1), (2), and (3) holds. We prove B is an IFI of A.

We have
μB(x∧y)≥μA⋅B(x∧y),sinceB⊇A⋅B=supx∧y=∨i=1n(xi∧yi)[⋀i=1n(μA(xi)∧μB(yi))]≥μA(x)∧μB(y).
Similarly, we can obtain
μB(x∧y)≥μB(x)∧μA(y),sinceB⊇A⋅B=B⋅A.
Hence
μB(x∧y)≥[μA(x)∧μB(y)]∨[μB(x)∧μA(y)].
Also
νB(x∧y)≤νA⋅B(x∧y),sinceB⊇A⋅B=infx∧y=⋁i=1n(xi∧yi)[⋁i=1n(νA(xi)∨νB(yi))]≤νA(x)∨νB(y).
Similarly
νB(x∧y)≤νB(x)∨νA(y),sinceB⊇A⋅B=B⋅A.
Hence
νB(x∧y)≤[νA(x)∨νB(y)]∧[νB(x)∨νA(y)].
So from (1), (2), (a.1), and (b.1) B is an IFI of A.

Conversely suppose that B is an IFI of A. Then obviously conditions (1) and (2) hold.

Let z∈L and z=⋁i=1n(xi∧yi), where xi∈A,yi∈B.

We have
μB(z)=μB[⋁i=1n(xi∧yi)]≥⋀i=1nμB(xi∧yi)≥⋀i=1n{μA(xi)∧μB(yi)},sinceBIFIofA.
Thus
μB(z)≥⋁[⋀i=1n{μA(xi)∧μB(yi)}]=μA⋅B(z).
Also
νB(z)=νB[⋁i=1n(xi∧yi)]≤⋁i=1nνB(xi∧yi)≤⋁i=1n{νA(xi)∨νB(yi)},sinceBIFIofA.
Thus
νB(z)≤∧[⋁i=1n{νA(xi)∨νB(yi)}]=νA⋅B(z).
Hence A·B⊆B.

Theorem 3.

Let A be an IFL of L and B, C are IFI’s of A. Then B+C is an IFI of A.

Proof.

We have
μB+C(x∨y)≥μB+C(x)∧μB+C(y),νB+C(x∨y)≤νB+C(x)∨νB+C(y)
(by [11, Theorem 5.2]).

And
A(B+C)⊆AB+AC⊆B+C,(B+C)A⊆BA+CA⊆B+C.
(by Lemma 1 and Theorem 1).

Hence B+C is an IFI of A.

4. Quotient of Ideals

Here first we define the residual of ideals of an IFL and prove that the residual of ideals is again an IFI of the IFL. Moreover we establish that it is the largest ideal with respect to some property on the operation ·.

Definition 8.

Let A be an IFL of L and B, C be IFI’s of A. Then the quotient (residual) of B by C denoted as B/C is defined by
B/C=∪{D/D⊲A,DC⊆B}.

Theorem 4.

Let A be an IFL of L and B,C are IFI’s of A. Then the quotient B/C is an IFI of A. Also B⊆B/C⊆A.

Proof.

Let η={D/D⊲A,DC⊆B}. Suppose D, D′∈η. Then D and D′ are IFI’s of A such that DC⊆B and D′C⊆B. Then by Theorem 3D+D′ is an IFI of A. So by Lemmas 1 and 2(D+D′)C⊆DC+D′C⊆B+B=B. Thus D+D′∈η. Now
μB/C(x)∧μB/C(y)=[⋁D∈ημD(x)]∧[⋁D′∈ημD′(y)]=⋁{μD(x)∧μD′(y)/D,D′∈η}≤⋁{μD+D′(x∨y)/D,D′∈η}≤μB/C(x∨y),sinceD+D′∈η.
That is,
μB/C(x∨y)≥μB/C(x)∧μB/C(y).
Also
μB/C(x∧y)=⋁D∈ημD(x∧y)≥⋁D∈η{μD(x)∧μA(y)},sinceD⊲A=[⋁D∈ημD(x)]∧μA(y)=μB/C(x)∧μA(y).
Similarly
μB/C(x∧y)≥μB/C(y)∧μA(x).
Thus
μB/C(x∧y)≥[μB/C(x)∧μA(y)]∨[μB/C(y)∧μA(x)].
Now
νB/C(x)∨νB/C(y)=[⋀D∈ηνD(x)]∨[⋀D′∈ηνD′(y)]=⋀{νD(x)∨νD′(y)/D,D′∈η}≥⋀{νD+D′(x∨y)/D,D′∈η}≥νB/C(x∨y),sinceD+D′∈η.
That is
νB/C(x∨y)≤νB/C(x)∨νB/C(y).
Also
νB/C(x∧y)=⋀D∈ηνD(x∧y)≤⋀D∈η{νD(x)∨νA(y)},sinceD⊲A=[⋀D∈ηνD(x)]∨νA(y)=νB/C(x)∨νA(y).
Similarly
νB/C(x∧y)≤νB/C(y)∨νA(x).
Thus
νB/C(x∧y)≤[νB/C(x)∨νA(y)]∧[νB/C(y)∨νA(x)].
From (24), (27), (29), and (32) B/C is an IFI of A.

Clearly B/C⊆A.

Since B is an IFI of A, BA⊆B (by Theorem 1).

Since C⊆A, by Lemma 1BC⊆BA⊆B. Hence B∈η. So B⊆B/C.

Thus we haveB⊆B/C⊆A.

Theorem 5.

Let A be an IFL and B, C be IFI’s of A. Then B/C is the largest IFI of A with the property (B/C)·C⊆B.

Proof.

Let η={D/D⊲AandDC⊆B}. We have B/C=⋃D∈ηD Let x∈L such that x=⋁i=1n(ai∧bi).

Then
μB(ai∧bi)≥μDC(ai∧bi)≥μD(ai)∧μC(bi),∀D∈η.
So
μB(ai∧bi)≥⋁D∈η[μD(ai)∧μC(bi)]=[⋁D∈ημD(ai)]∧μC(bi)=μB/C(ai)∧μC(bi).
Hence
μB(x)=μB(⋁i=1n(ai∧bi))≥⋀i=1nμB(ai∧bi),sinceBisanIFIofA≥⋀i=1n[μB/C(ai)∧μC(bi)].
Consequently
μB(x)≥⋁{⋀i=1n[μB/C(ai)∧μC(bi)]/x=⋁i=1n(ai∧bi)}=μ(B/C)⋅C(x).
Also
νB(ai∧bi)≤νDC(ai∧bi)≤νD(ai)∨νC(bi),∀D∈η.
So
νB(ai∧bi)≤⋀D∈η[νD(ai)∨νC(bi)]=[⋀D∈ηνD(ai)]∨νC(bi)=νB/C(ai)∧νC(bi).
Hence
νB(x)=νB(⋁i=1n(ai∧bi))≤⋁i=1nνB(ai∧bi),sinceBisanIFIofA≤⋁i=1n[νB/C(ai)∧νC(bi)].
Consequently
νB(x)≤⋀{⋁i=1n[νB/C(ai)∧νC(bi)]/x=⋁i=1n(ai∧bi)}=ν(B/C)⋅C(x).
Thus from (37) and (41) (B/C)·C⊆B.

If D is an ideal of A such that D·C⊆B then DC⊆D·C⊆B. So D∈η. Hence D⊆B/C. Thus B/C is the largest IFI of A such that (B/C)·C⊆B.

Theorem 6.

Let A be an IFL and B, C, D be IFI’s of A. Then the following holds.

If B⊆C then B/D⊆C/D and D/C⊆D/B.

If B⊆C then C/B=A.

B/B=A.

Proof.

(1) Let B⊆C. Write η={E/E⊲AandED⊆B} and ξ={E/E⊲AandED⊆C}. If E∈η then E⊲A and ED⊆B⊆C. Thus E∈ξ and hence η⊆ξ. So B/D=⋃E∈ηE⊆⋃E∈ξE=C/D.

Similarly, let η1={E/E⊲AandEC⊆D} and ξ1={E/E⊲AandEB⊆D}. If E∈η1 then EC⊆D. But B⊆C. So EB⊆EC⊆D. Thus E∈ξ1 and hence η1⊆ξ1. So D/C=⋃E∈η1E⊆⋃E∈ξ1E=D/B.

(2) Let η={E/E⊲AandEB⊆C}. Since B⊲A, we have AB⊆B⊆C, and A⊲A. Thus A∈η and hence A⊆⋃E∈ηE=C/B⊆A, since C/B is an IFI of A. Therefore C/B=A.

(3) We have B⊆B. So from (2) B/B=A.

Corollary 1.

Let A be an IFL of L and B, and C be IFI’s of A. Then

(B/C)/B=A,

(B/B)/C=A,

B/(B∩C)=A.

Proof.

(1) Since B⊆B/C, by Theorem 6 (2), (B/C)/B=A.

(2) By Theorem 6 (3) B/B=A. Since C⊆A=B/B by Theorem 6 (2), (B/B)/C=A.

(3) Since B⊲A and C⊲A. So B∩C⊲A and B∩C⊆B. Hence by Theorem 6 (2), B/(B∩C)=A.

Theorem 7.

Let A be an IFL of L and Bii=1,2……m, C, are IFI’s of A. Then
(⋂i=1mBi)/C=⋂i=1m(Bi/C).

Proof.

Since ⋂i=1mBi⊆Bi by Theorem 6 (1) (⋂i=1mBi)/C⊆Bi/C,∀i.

Hence
(⋂i=1mBi)/C⊆⋂i=1m(Bi/C).

Let
η1={E/E⊲A,EC⊆B1},η2={E/E⊲A,EC⊆B2},η3={E/E⊲A,EC⊆B1∩B2}.
Then ∀x∈LμB1/C∩B2/C(x)=μB1/C(x)∧μB2/C(x)=(⋁E∈η1μE(x))∧(⋁E′∈η2μE′(x))=∨{[μE(x)∧μE′(x)]/E∈η1,E′∈η2}.
Similarly
νB1/C∩B2/C(x)=νB1/C(x)∨νB2/C(x)=(⋀E∈η1νE(x))∨(⋀E′∈η2νE′(x))=∧{[νE(x)∨νE′(x)]/E∈η1,E′∈η2}.
Now let E∈η1 and E′∈η2. Then EC⊆B1 and E′C⊆B2. Also E∩E′ IFI of A.

So that
(E∩E′)C⊆EC∩E′C⊆B1∩B2.
Thus E∩E′∈η3. So η1∩η2⊆η3.

Hence
(B1∩B2)/C=⋃E∈η3E⊇⋃E∈η1,E′∈η2(E∩E′).
So
μB1∩B2/C(x)≥⋁μE∩E′(x)=⋁[μE(x)∧μE′(x)]=μB1/C∩B2/C(x),from(a.2),νB1∩B2/C(x)≤⋀νE∩E′(x)=⋀[νE(x)∨νE′(x)]=νB1/C∩B2/C(x),from(b.2).
Hence
B1∩B2/C⊇B1/C∩B2/C.
From (43) and (48)(B1∩B2)/C=B1/C∩B2/C. This completes the proof.

Next, we denote the set of all IFI’s {Bi}i=1,2…,m of an IFL A that satisfies the property μBi(0)=μBj(0)andνBi(0)=νBj(0)∀i,j by IFI (A*). Then we have the following results.

Lemma 4.

Let A be an IFL of L and B, C∈ IFI (A*). Then

B⊆B+C and C⊆B+C.

B/C=B/B+C.

B+C/B=A and B+C/B∩C=A.

Proof.

(1) We have
μB+C(x)=⋁x=y∨z(μB(y)∧μC(z))≥μB(x)∧μC(0)asx=x∨0=μB(x)∧μB(0)sinceμB(0)=μC(0)=μB(x)sinceμB(0)≥μB(x),νB+C(x)=⋀x=y∨z(νB(y)∨νC(z))≤νB(x)∨νC(0)asx=x∨0=νB(x)∨νB(0)sinceνB(0)=νC(0)=νB(x)sinceνB(0)≤νB(x).
So B⊆B+C. Similarly ⊆B+C.

(2) We have B+C⊲A (by Theorem 3) and C⊆B+C (by (1)). So by Theorem 6 (1),
B/B+C⊆B/C.

Write η={E/E⊲AandEC⊆B} and ξ={E/E⊲AandE(B+C)⊆B}.

Let E∈η, then E⊲A and E⊆A. So EB⊆AB (by Lemma 1). But AB⊆B, since B⊲A. Hence EB⊆B and also EC⊆B. So By Lemma 1, E(B+C)⊆EB+EC⊆B+B=B. Therefore E∈ξ. So η⊆ξ. Thus
B/C=⋃E∈ηE⊆⋃E∈ξE=B/B+C.
From (50) and (51) B/C=B/B+C.

(3) We have B+C⊲A and B⊆B+C. So by Theorem 6 (2), B+C/B=A.

Also we have B∩C⊲A and B∩C⊆B+C. Hence by Theorem 6 (2) B+C/B∩C=A.

Theorem 8.

Let A be an IFL of L and {Bi}i=1,2…m∈ IFI (A*) and C any IFI of A. Then C/∑i=1mBi=⋂i=1m(C/Bi).

Proof.

We have B1+B2⊲A and B1⊆B1+B2, B2⊆B1+B2.

So by Theorem 6 (1), C/B1+B2⊆C/B1 and C/B1+B2⊆C/B2.

Therefore
C/B1+B2⊆C/B1⋂C/B2.
Let
η1={E/E⊲A,EB1⊆C},η2={E/E⊲A,EB2⊆C},η3={E/E⊲A,E(B1+B2)⊆C}.
Then ∀x∈LμC/B1∩C/B2(x)=μC/B1(x)∧μC/B2(x)=(⋁E∈η1μE(x))∧(⋁E′∈η2μE′(x))=⋁{[μE(x)∧μE′(x)]/E∈η1,E′∈η2}.
Similarly
νC/B1∩C/B2(x)=νC/B1(x)∨νC/B2(x)=(⋀E∈η1νE(x))∨(⋀E′∈η2νE′(x))=∧{[νE(x)∨νE′(x)]/E∈η1,E′∈η2}.
Now let E∈η1 and E′∈η2. Then EB1⊆C and E′B2⊆C. Also E∩E′ IFI of A, so that(E∩E′)(B1+B2)⊆(E∩E′)B1+(E∩E′)B2⊆EB1+E′B2⊆C+C=C.
So E∩E′∈η3. Hence η1∩η2⊆η3. Thus C/B1+B2=⋃E∈η3E⊇⋃E∈η1,E′∈η2(E∩E′).

So
μC/B1+B2(x)≥∨μE∩E′(x)=∨[μE(x)∧μE′(x)]=μC/B1∩C/B2(x)from(a.3),νC/B1+B2(x)≤∧νE∩E′(x)=∧[νE(x)∨νE′(x)]=νC/B1∩C/B2(x)from(b.3).
Therefore
C/B1+B2⊇C/B1⋂C/B2.
From (52) and (56) C/B1+B2=C/B1∩C/B2, hence the result.

AtanassovK. T.Intutionistic fuzzy setsAtanassovK. T.New operations defined over the intuitionistic fuzzy setsKimK. H.On intuitionistic Q-fuzzy semiprime ideals in semigroupsKimK. H.JunY. B.Intuitionistic fuzzy interior ideals of semigroupsKimK. H.LeeJ. G.On intuitionistic fuzzy Bi-ideals of semigroupsTorkzadehL.ZahediM. M.Intuitionistic fuzzy commutative hyper K-idealsAkramM.DudekW.Interval-valued intuitionistic fuzzy Lie ideals of Lie algebrasHurK.JangS. Y.KangH. W.Intuitionistic fuzzy ideals of a ringMordesonJ. N.MalikD. S.PrajapatiA. S.Residual of ideals of an L-ringThomasK. V.NairL. S.Intuitionistic fuzzy sublattices and idealsFuzzy Information and Engineering. In pressAtanassovL.On intuitionistic fuzzy versions of L. Zadeh's extension principleThomasK. V.NairL. S.Operations on intuitionistic fuzzy ideals of a latticeInternational Journal of Fuzzy Mathematics. In press