# Quotient of Ideals of an Intuitionistic Fuzzy Lattice

Authors: K. V. Thomas; Latha S.  Nair
Journal: Advances in Fuzzy Systems
 (2010)

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## Abstract

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.


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## Body

## 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 paperL stands for a lattice (L,∨,∧) with zero element “0” and unit element “1”.Definition 1 (see [1]).
LetX 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 onX is denoted by IFS (X).Definition 2 (see [1]).
IfA={〈x,μA(x),νA(x)〉∣x∈S} and B={〈x,μB(x),νB(x)〉∣x∈S} are any two IFS of X  then (i)
A⊆B⇔μA(x)≤μB and νA(x)≥νB(x)∀x∈X;(ii)
A=B⇔μA(x)=μB(x) and νA(x)=νB(x);(iii)
A-={〈x,νA(x),μA(x)〉∣x∈X};(iv)
[A]={〈x,μA(x),μAc(x)〉∣x∈X}, where μAc(x)=1-μA(x);(v)
〈A〉={〈x,νAc(x),νA(x)〉∣x∈X}, where νAc(x)=1-νA(x);(vi)
A∩B=  {〈x,min{μA(x),μB(x)},max{νA(x),νB(x)}〉∣x∈X}= {〈x,μA∩B(x),νA∩B(x)〉∣x∈X};(vii)
A∪B=  {〈x,max{μA(x),μB(x)},min{νA(x),νB(x)}〉∣x∈X}=  {〈x,μA∪B(x),νA∪B(x)〉∣x∈X}.Definition 3 (see [11]).
LetL 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.(i)
μA(x∨y)≥min{μA(x),μA(y)};(ii)
μA(x∧y)≥min{μA(x),μA(y)};(iii)
νA(x∨y)≤max{νA(x),νA(y)};(iv)
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 [11]).
An IFSA of L is called an intuitionistic fuzzy ideal (IFI) of L if the following conditions are satisfied. (i)
μA(x∨y)≥min{μA(x),μA(y)};(ii)
μA(x∧y)≥max{μA(x),μA(y)};(iii)
νA(x∨y)≤max{νA(x),νA(y)};(iv)
νA(x∧y)≤min{νA(x),νA(y)},∀x,y∈L.
The set of all IFI’s ofL is denoted as IFI (L).Definition 5 (see [12]).
LetA, B∈ IFS (L). Then we define an IFS (L),(i)
A+B={〈z,μA+B(z),νA+B(z)〉∣z∈L}, where
(1)μA+B(z)=sup⁡z=x∨y{min⁡{μA(x),μB(y)}},νA+B(z)=inf⁡z=x∨y{max⁡{νA(x),νB(y)}}.(ii)
AB={〈z,μAB(z),νAB(z)〉∣z∈L}, where
(2)μAB(z)=sup⁡z=x∧y{min⁡{μA(x),μB(y)}}νAB(z)=inf⁡z=x∧y{max⁡{νA(x),νB(y)}}.(iii)
A·B=  {〈z,μA·B(z),νA·B(z)〉∣z∈L}where
(3)μA⋅B(z)=sup⁡z=∨i=1n(xi∧yi)(min⁡i{min⁡⁡{μA(xi),μB(yi)}}),νA⋅B(z)=inf⁡z=∨i=1n(xi∧yi)(max⁡i{max⁡⁡{νA(xi),νB(yi)}})Lemma 1 (see [13]).
LetA, B, and C be IFS (L), then the following assertions hold.(1)
AB=BA, A+B=B+A, A·B=B·A.(2)
AB⊆A·B.(3)
C(A+B)⊆CA+CB.(4)
(C+B)A⊆CA+BA.(5)
(A∩B)C⊆AC∩BC.(6)
A⊆B⇒AC⊆BC and A·C⊆B·C.Lemma 2 (see [13]).
Let A be an IFL of L, then(1)
A+A=A.(2)
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.
LetA be an IFL of L and B an IFS of L with B⊆A. Then B is called an intuitionistic fuzzy ideal (IFI) ofA if the following conditions are satisfied.(i)
μB(x∨y)≥μB(x)∧μB(y).(ii)
μB(x∧y)≥[μA(x)∧μB(y)]∨[μB(x)∧μA(y)].(iii)
νB(x∨y)≤νB(x)∨νB(y).(iv)
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 latticeL={1,2,5,10} under divisibility.
LetA={〈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.
LetA be an IFL and B is also an IFL with B⊆A. Then B is called an intuitionistic fuzzy sublattice ofA.Lemma 3.
The intersection of two IFI’s ofA is again an IFI of A.Proof.
LetB, 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
(4)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
(5)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.
LetA an IFL and B an IFS of L with B⊆A. Then B is an IFI of A if and only if(1)
μB(x∨y)≥μB(x)∧μB(y),(2)
νB(x∨y)≤νB(x)∨νB(y),(3)
AB⊆B.Proof.
Suppose that conditions (1), (2), and (3) hold. Then we prove thatB is an IFI of A.
We have(6)μB(x∧y)≥μAB(x∧y),sinceB⊇AB=sup⁡x∧y=xi∧yi(μA(xi)∧μB(yi))≥μA(x)∧μB(y).
Similarly
(7)μB(x∧y)≥μB(x)∧μA(y),sinceB⊇AB=BA.
Hence
(a)μB(x∧y)≥[μA(x)∧μB(y)]∨[μB(x)∧μA(y)].
Also
(8)νB(x∧y)≤νAB(x∧y),sinceB⊇AB=inf⁡x∧y=xi∧yi(νA(xi)∨νB(yi))≤νA(x)∨νB(y).
Similarly
(9)νB(x∧y)≤νB(x)∨νA(y),sinceB⊇AB=BA.
Hence
(b)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 supposeB is an IFI of A. Then obviously conditions (1) and (2) holds. Also we have
(10)μB(x∧y)≥μA(x)∧μB(y),νB(x∧y)≤νA(x)∨νB(y),∀x,y∈L.
So∀z∈L with z=x∧y(11)μB(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.
LetA be an IFL of L and B an IFS with B⊆A. Then B is an IFI of A if and only if(1)
μB(x∨y)≥μB(x)∧μB(y),(2)
νB(x∨y)≤νB(x)∨νB(y),(3)
A·B⊆B.Proof.
Suppose conditions (1), (2), and (3) holds. We proveB is an IFI of A.
We have(12)μB(x∧y)≥μA⋅B(x∧y),sinceB⊇A⋅B=sup⁡x∧y=∨i=1n(xi∧yi)[⋀i=1n(μA(xi)∧μB(yi))]≥μA(x)∧μB(y).
Similarly, we can obtain
(13)μB(x∧y)≥μB(x)∧μA(y),sinceB⊇A⋅B=B⋅A.
Hence
(a.1)μB(x∧y)≥[μA(x)∧μB(y)]∨[μB(x)∧μA(y)].
Also
(14)νB(x∧y)≤νA⋅B(x∧y),sinceB⊇A⋅B=inf⁡x∧y=⋁i=1n(xi∧yi)[⋁i=1n(νA(xi)∨νB(yi))]≤νA(x)∨νB(y).
Similarly
(15)νB(x∧y)≤νB(x)∨νA(y),sinceB⊇A⋅B=B⋅A.
Hence
(b.1)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 thatB is an IFI of A. Then obviously conditions (1) and (2) hold.
Letz∈L and z=⋁i=1n(xi∧yi), where xi∈A,yi∈B.
We have(16)μB(z)=μB[⋁i=1n(xi∧yi)]≥⋀i=1nμB(xi∧yi)≥⋀i=1n{μA(xi)∧μB(yi)},sinceBIFIofA.
Thus
(17)μB(z)≥⋁[⋀i=1n{μA(xi)∧μB(yi)}]=μA⋅B(z).
Also
(18)νB(z)=νB[⋁i=1n(xi∧yi)]≤⋁i=1nνB(xi∧yi)≤⋁i=1n{νA(xi)∨νB(yi)},sinceBIFIofA.
Thus
(19)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. ThenB+C is an IFI of A.Proof.
We have(20)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(21)A(B+C)⊆AB+AC⊆B+C,(B+C)A⊆BA+CA⊆B+C.
(by Lemma 1 and Theorem 1).
HenceB+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.
LetA be an IFL of L and B, C be IFI’s of A. Then the quotient  (residual) ofB by C denoted as B/C is defined by
(22)B/C=∪{D/D⊲A,DC⊆B}.Theorem 4.
LetA 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.
Letη={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 3  D+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
(23)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,
(24)μB/C(x∨y)≥μB/C(x)∧μB/C(y).
Also
(25)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
(26)μB/C(x∧y)≥μB/C(y)∧μA(x).
Thus
(27)μB/C(x∧y)≥[μB/C(x)∧μA(y)]∨[μB/C(y)∧μA(x)].
Now
(28)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
(29)νB/C(x∨y)≤νB/C(x)∨νB/C(y).
Also
(30)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
(31)νB/C(x∧y)≤νB/C(y)∨νA(x).
Thus
(32)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.
ClearlyB/C⊆A.
SinceB is an IFI of A, BA⊆B (by Theorem 1).
SinceC⊆A, by Lemma 1  BC⊆BA⊆B. Hence B∈η. So B⊆B/C.
Thus we have(33)B⊆B/C⊆A.Theorem 5.
LetA 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.
Letη={D/D⊲AandDC⊆B}. We have B/C=⋃D∈ηD Let x∈L such that x=⋁i=1n(ai∧bi).
Then(34)μB(ai∧bi)≥μDC(ai∧bi)≥μD(ai)∧μC(bi),∀D∈η.
So
(35)μB(ai∧bi)≥⋁D∈η[μD(ai)∧μC(bi)]=[⋁D∈ημD(ai)]∧μC(bi)=μB/C(ai)∧μC(bi).
Hence
(36)μB(x)=μB(⋁i=1n(ai∧bi))≥⋀i=1nμB(ai∧bi),sinceBisanIFIofA≥⋀i=1n[μB/C(ai)∧μC(bi)].
Consequently
(37)μB(x)≥⋁{⋀i=1n[μB/C(ai)∧μC(bi)]/x=⋁i=1n(ai∧bi)}=μ(B/C)⋅C(x).
Also
(38)νB(ai∧bi)≤νDC(ai∧bi)≤νD(ai)∨νC(bi),∀D∈η.
So
(39)νB(ai∧bi)≤⋀D∈η[νD(ai)∨νC(bi)]=[⋀D∈ηνD(ai)]∨νC(bi)=νB/C(ai)∧νC(bi).
Hence
(40)νB(x)=νB(⋁i=1n(ai∧bi))≤⋁i=1nνB(ai∧bi),sinceBisanIFIofA≤⋁i=1n[νB/C(ai)∧νC(bi)].
Consequently
(41)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.
IfD 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.
LetA be an IFL and B, C, D be IFI’s of A. Then the following holds.(1)
IfB⊆C then B/D⊆C/D and D/C⊆D/B.(2)
IfB⊆C then C/B=A.(3)
B/B=A.Proof.
(1) LetB⊆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, letη1={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) Letη={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 haveB⊆B. So from (2) B/B=A.Corollary 1.
LetA be an IFL of L and B, and C be IFI’s of A. Then(1)
(B/C)/B=A,(2)
(B/B)/C=A,(3)
B/(B∩C)=A.Proof.
(1) SinceB⊆B/C, by Theorem 6 (2), (B/C)/B=A.
(2) By Theorem6 (3) B/B=A. Since C⊆A=B/B by Theorem 6 (2), (B/B)/C=A.
(3) SinceB⊲A and C⊲A. So B∩C⊲A and B∩C⊆B. Hence by Theorem 6 (2), B/(B∩C)=A.Theorem 7.
LetA be an IFL of L and Bii=1,2……m, C, are  IFI’s of A. Then
(42)(⋂i=1mBi)/C=⋂i=1m(Bi/C).Proof.
Since⋂i=1mBi⊆Bi  by Theorem 6 (1) (⋂i=1mBi)/C⊆Bi/C,∀i.
Hence(43)(⋂i=1mBi)/C⊆⋂i=1m(Bi/C).
Let(44)η1={E/E⊲A,EC⊆B1},η2={E/E⊲A,EC⊆B2},η3={E/E⊲A,EC⊆B1∩B2}.
Then ∀x∈L(a.2)μB1/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
(b.2)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(45)(E∩E′)C⊆EC∩E′C⊆B1∩B2.
Thus E∩E′∈η3. So η1∩η2⊆η3.
Hence(46)(B1∩B2)/C=⋃E∈η3E⊇⋃E∈η1,E′∈η2(E∩E′).
So
(47)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
(48)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.
LetA be an IFL of L and B, C∈ IFI (A*). Then(1)
B⊆B+C and C⊆B+C.(2)
B/C=B/B+C.(3)
B+C/B=A and B+C/B∩C=A.Proof.
(1) We have(49)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 haveB+C⊲A (by Theorem 3) and C⊆B+C (by (1)). So by Theorem 6 (1),
(50)B/B+C⊆B/C.
Writeη={E/E⊲AandEC⊆B} and ξ={E/E⊲AandE(B+C)⊆B}.
LetE∈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
(51)B/C=⋃E∈ηE⊆⋃E∈ξE=B/B+C.
From (50) and (51) B/C=B/B+C.
(3) We haveB+C⊲A and B⊆B+C. So by Theorem 6 (2), B+C/B=A.
Also we haveB∩C⊲A and B∩C⊆B+C. Hence by Theorem 6 (2) B+C/B∩C=A.Theorem 8.
LetA 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 haveB1+B2⊲A and B1⊆B1+B2, B2⊆B1+B2.
So by Theorem6 (1), C/B1+B2⊆C/B1 and C/B1+B2⊆C/B2.
Therefore(52)C/B1+B2⊆C/B1⋂C/B2.
Let
(53)η1={E/E⊲A,EB1⊆C},η2={E/E⊲A,EB2⊆C},η3={E/E⊲A,E(B1+B2)⊆C}.
Then ∀x∈L(a.3)μC/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
(b.3)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(54)(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(55)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
(56)C/B1+B2⊇C/B1⋂C/B2.
From (52) and (56) C/B1+B2=C/B1∩C/B2, hence the result.



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*Source: 781672-2011-01-02.xml*