Advances in Fuzzy Systems
Volume 2010 (2010), Article ID 781672, 8 pages
doi:10.1155/2010/781672
Research Article

Quotient of Ideals of an Intuitionistic Fuzzy Lattice

K. V. Thomas1 and Latha S. Nair2

1Bharata Mata College, Mahatma Gandhi University, Kochi-Kerala, Thrikkakara 682 021, India
2Mar Athanasius College, Mahatma Gandhi University, Kochi, Kothamangalam, Kerala 686 666, India

Received 22 September 2010; Revised 10 November 2010; Accepted 2 December 2010

Academic Editor: José Luis Verdegay

Copyright © 2010 K. V. Thomas and Latha S. Nair. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

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

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

Definition 2 (see [1]). If 𝐴 = { ⟨ 𝑥 , 𝜇 𝐴 ( 𝑥 ) , 𝜈 𝐴 ( 𝑥 ) ⟩ ∣ 𝑥 ∈ 𝑆 } and 𝐵 = { ⟨ 𝑥 , 𝜇 𝐵 ( 𝑥 ) , 𝜈 𝐵 ( 𝑥 ) ⟩ ∣ 𝑥 ∈ 𝑆 } are any two IFS of 𝑋   then (i) 𝐴 ⊆ 𝐵 ⇔ 𝜇 𝐴 ( 𝑥 ) ≤ 𝜇 𝐵 and 𝜈 𝐴 ( 𝑥 ) ≥ 𝜈 𝐵 ( 𝑥 ) ∀ 𝑥 ∈ 𝑋 ;(ii) 𝐴 = 𝐵 ⇔ 𝜇 𝐴 ( 𝑥 ) = 𝜇 𝐵 ( 𝑥 ) and 𝜈 𝐴 ( 𝑥 ) = 𝜈 𝐵 ( 𝑥 ) ;(iii) − 𝐴 = { ⟨ 𝑥 , 𝜈 𝐴 ( 𝑥 ) , 𝜇 𝐴 ( 𝑥 ) ⟩ ∣ 𝑥 ∈ 𝑋 } ;(iv) [ 𝐴 ] = { ⟨ 𝑥 , 𝜇 𝐴 ( 𝑥 ) , 𝜇 𝐴 𝑐 ( 𝑥 ) ⟩ ∣ 𝑥 ∈ 𝑋 } , where 𝜇 𝐴 𝑐 ( 𝑥 ) = 1 − 𝜇 𝐴 ( 𝑥 ) ;(v) ⟨ 𝐴 ⟩ = { ⟨ 𝑥 , 𝜈 𝑐 𝐴 ( 𝑥 ) , 𝜈 𝐴 ( 𝑥 ) ⟩ ∣ 𝑥 ∈ 𝑋 } , where 𝜈 𝐴 𝑐 ( 𝑥 ) = 1 − 𝜈 𝐴 ( 𝑥 ) ;(vi) 𝐴 ∩ 𝐵 =    { ⟨ 𝑥 , m i n { 𝜇 𝐴 ( 𝑥 ) , 𝜇 𝐵 ( 𝑥 ) } , m a x { 𝜈 𝐴 ( 𝑥 ) , 𝜈 𝐵 ( 𝑥 ) } ⟩ ∣ 𝑥 ∈ 𝑋 } =   { ⟨ 𝑥 , 𝜇 𝐴 ∩ 𝐵 ( 𝑥 ) , 𝜈 𝐴 ∩ 𝐵 ( 𝑥 ) ⟩ ∣ 𝑥 ∈ 𝑋 } ; (vii) 𝐴 ∪ 𝐵 =    { ⟨ 𝑥 , m a x { 𝜇 𝐴 ( 𝑥 ) , 𝜇 𝐵 ( 𝑥 ) } , m i n { 𝜈 𝐴 ( 𝑥 ) , 𝜈 𝐵 ( 𝑥 ) } ⟩ ∣ 𝑥 ∈ 𝑋 } =    { ⟨ 𝑥 , 𝜇 𝐴 ∪ 𝐵 ( 𝑥 ) , 𝜈 𝐴 ∪ 𝐵 ( 𝑥 ) ⟩ ∣ 𝑥 ∈ 𝑋 } .

Definition 3 (see [11]). Let 𝐿 be a lattice and 𝐴 = { ⟨ 𝑥 , 𝜇 𝐴 ( 𝑥 ) , 𝜈 𝐴 ( 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) 𝜇 𝐴 ( 𝑥 ∨ 𝑦 ) ≥ m i n { 𝜇 𝐴 ( 𝑥 ) , 𝜇 𝐴 ( 𝑦 ) } ; (ii) 𝜇 𝐴 ( 𝑥 ∧ 𝑦 ) ≥ m i n { 𝜇 𝐴 ( 𝑥 ) , 𝜇 𝐴 ( 𝑦 ) } ;(iii) 𝜈 𝐴 ( 𝑥 ∨ 𝑦 ) ≤ m a x { 𝜈 𝐴 ( 𝑥 ) , 𝜈 𝐴 ( 𝑦 ) } ;(iv) 𝜈 𝐴 ( 𝑥 ∧ 𝑦 ) ≤ m a x { 𝜈 𝐴 ( 𝑥 ) , 𝜈 𝐴 ( 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 IFS 𝐴 of 𝐿 is called an intuitionistic fuzzy ideal (IFI) of 𝐿 if the following conditions are satisfied. (i) 𝜇 𝐴 ( 𝑥 ∨ 𝑦 ) ≥ m i n { 𝜇 𝐴 ( 𝑥 ) , 𝜇 𝐴 ( 𝑦 ) } ; (ii) 𝜇 𝐴 ( 𝑥 ∧ 𝑦 ) ≥ m a x { 𝜇 𝐴 ( 𝑥 ) , 𝜇 𝐴 ( 𝑦 ) } ;(iii) 𝜈 𝐴 ( 𝑥 ∨ 𝑦 ) ≤ m a x { 𝜈 𝐴 ( 𝑥 ) , 𝜈 𝐴 ( 𝑦 ) } ;(iv) 𝜈 𝐴 ( 𝑥 ∧ 𝑦 ) ≤ m i n { 𝜈 𝐴 ( 𝑥 ) , 𝜈 𝐴 ( 𝑦 ) } , ∀ 𝑥 , 𝑦 ∈ 𝐿 .
The set of all IFI’s of 𝐿 is denoted as IFI ( 𝐿 ).

Definition 5 (see [12]). Let 𝐴 , 𝐵 ∈ IFS ( 𝐿 ). Then we define an IFS ( 𝐿 ),(i) 𝐴 + 𝐵 = { ⟨ 𝑧 , 𝜇 𝐴 + 𝐵 ( 𝑧 ) , 𝜈 𝐴 + 𝐵 ( 𝑧 ) ⟩ ∣ 𝑧 ∈ 𝐿 } , where 𝜇 𝐴 + 𝐵 ( 𝑧 ) = s u p 𝑧 = 𝑥 ∨ 𝑦 𝜇 m i n 𝐴 ( 𝑥 ) , 𝜇 𝐵 , 𝜈 ( 𝑦 ) 𝐴 + 𝐵 ( 𝑧 ) = i n f 𝑧 = 𝑥 ∨ 𝑦 𝜈 m a x 𝐴 ( 𝑥 ) , 𝜈 𝐵 . ( 𝑦 ) ( 1 ) (ii) 𝐴 𝐵 = { ⟨ 𝑧 , 𝜇 𝐴 𝐵 ( 𝑧 ) , 𝜈 𝐴 𝐵 ( 𝑧 ) ⟩ ∣ 𝑧 ∈ 𝐿 } , where 𝜇 𝐴 𝐵 ( 𝑧 ) = s u p 𝑧 = 𝑥 ∧ 𝑦 𝜇 m i n 𝐴 ( 𝑥 ) , 𝜇 𝐵 𝜈 ( 𝑦 ) 𝐴 𝐵 ( 𝑧 ) = i n f 𝑧 = 𝑥 ∧ 𝑦 𝜈 m a x 𝐴 ( 𝑥 ) , 𝜈 𝐵 . ( 𝑦 ) ( 2 ) (iii) 𝐴 ⋅ 𝐵 =    { ⟨ 𝑧 , 𝜇 𝐴 ⋅ 𝐵 ( 𝑧 ) , 𝜈 𝐴 ⋅ 𝐵 ( 𝑧 ) ⟩ ∣ 𝑧 ∈ 𝐿 } where 𝜇 𝐴 ⋅ 𝐵 ( 𝑧 ) = s u p 𝑧 = ∨ 𝑛 𝑖 = 1 ( 𝑥 𝑖 ∧ 𝑦 𝑖 ) m i n 𝑖 𝜇 m i n 𝐴 𝑥 𝑖 , 𝜇 𝐵 𝑦 𝑖 , 𝜈 𝐴 ⋅ 𝐵 ( 𝑧 ) = i n f 𝑧 = ∨ 𝑛 𝑖 = 1 ( 𝑥 𝑖 ∧ 𝑦 𝑖 ) m a x 𝑖 𝜈 m a x 𝐴 𝑥 𝑖 , 𝜈 𝐵 𝑦 𝑖 ( 3 )

Lemma 1 (see [13]). Let 𝐴 , 𝐵 , and 𝐶 be IFS ( 𝐿 ), then the following assertions hold. (1) 𝐴 𝐵 = 𝐵 𝐴 , 𝐴 + 𝐵 = 𝐵 + 𝐴 , 𝐴 ⋅ 𝐵 = 𝐵 ⋅ 𝐴 .(2) 𝐴 𝐵 ⊆ 𝐴 ⋅ 𝐵 .(3) 𝐶 ( 𝐴 + 𝐵 ) ⊆ 𝐶 𝐴 + 𝐶 𝐵 .(4) ( 𝐶 + 𝐵 ) 𝐴 ⊆ 𝐶 𝐴 + 𝐵 𝐴 .(5) ( 𝐴 ∩ 𝐵 ) 𝐶 ⊆ 𝐴 𝐶 ∩ 𝐵 𝐶 .(6) 𝐴 ⊆ 𝐵 ⇒ 𝐴 𝐶 ⊆ 𝐵 𝐶 and 𝐴 ⋅ 𝐶 ⊆ 𝐵 ⋅ 𝐶 .

Lemma 2 (see [13]). Let A be an IFL of L, then (1) 𝐴 + 𝐴 = 𝐴 .(2) 𝐴 𝐴 = 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.(i) 𝜇 𝐵 ( 𝑥 ∨ 𝑦 ) ≥ 𝜇 𝐵 ( 𝑥 ) ∧ 𝜇 𝐵 ( 𝑦 ) .(ii) 𝜇 𝐵 ( 𝑥 ∧ 𝑦 ) ≥ [ 𝜇 𝐴 ( 𝑥 ) ∧ 𝜇 𝐵 ( 𝑦 ) ] ∨ [ 𝜇 𝐵 ( 𝑥 ) ∧ 𝜇 𝐴 ( 𝑦 ) ] .(iii) 𝜈 𝐵 ( 𝑥 ∨ 𝑦 ) ≤ 𝜈 𝐵 ( 𝑥 ) ∨ 𝜈 𝐵 ( 𝑦 ) .(iv) 𝜈 𝐵 ( 𝑥 ∧ 𝑦 ) ≤ [ 𝜈 𝐴 ( 𝑥 ) ∨ 𝜈 𝐵 ( 𝑦 ) ] ∧ [ 𝜈 𝐵 ( 𝑥 ) ∨ 𝜈 𝐴 ( 𝑦 ) ] ∀ 𝑥 , 𝑦 ∈ 𝐿 .If 𝐵 IFI of 𝐴 , then we write 𝐵 ⊲ 𝐴

Example 1. Consider the lattice 𝐿 = { 1 , 2 , 5 , 1 0 } under divisibility.
Let 𝐴 = { ⟨ 𝑥 , 𝜇 𝐴 ( 𝑥 ) , 𝜈 𝐴 ( 𝑥 ) ⟩ ∣ 𝑥 ∈ 𝐿 } be an IFL of 𝐿 defined by ⟨ 1 , . 5 , . 1 ⟩ , ⟨ 2 , . 4 , . 5 ⟩ , ⟨ 5 , . 4 , . 3 ⟩ , ⟨ 1 0 , . 7 , . 3 ⟩ and 𝐵 = { ⟨ 𝑥 , 𝜇 𝐵 ( 𝑥 ) , 𝜈 𝐵 ( 𝑥 ) ⟩ ∣ 𝑥 ∈ 𝐿 } be an IFS of L given by ⟨ 1 , . 5 , . 3 ⟩ , ⟨ 2 , . 4 , . 5 ⟩ , ⟨ 5 , . 3 , . 4 ⟩ , ⟨ 1 0 , . 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 ∩ 𝐶 is also an IFI of 𝐴 . Since 𝐵 ⊆ 𝐴 and 𝐶 ⊆ 𝐴 , we have 𝐵 ∩ 𝐶 ⊆ 𝐴 Also 𝜇 𝐵 ∩ 𝐶 𝜇 ( 𝑥 ∨ 𝑦 ) = m i n 𝐵 ( 𝑥 ∨ 𝑦 ) , 𝜇 𝐶 𝜇 ( 𝑥 ∨ 𝑦 ) ≥ m i n 𝐵 ( 𝑥 ) ∧ 𝜇 𝐵 ( 𝑦 ) , 𝜇 𝐶 ( 𝑥 ) ∧ 𝜇 𝐶 ( , 𝑦 ) s i n c e 𝐵 a n d 𝐶 a r e I F I 𝜇 s o f 𝐴 . ≥ m i n 𝐵 ( 𝑥 ) ∧ 𝜇 𝐶 ( 𝑥 ) , 𝜇 𝐵 ( 𝑦 ) ∧ 𝜇 𝐶 𝜇 ( 𝑦 ) ≥ m i n 𝐵 ∩ 𝐶 ( 𝑥 ) , 𝜇 𝐵 ∩ 𝐶 ( 𝑦 ) ≥ 𝜇 𝐵 ∩ 𝐶 ( 𝑥 ) ∧ 𝜇 𝐵 ∩ 𝐶 𝜇 ( 𝑦 ) , 𝐵 ∩ 𝐶 𝜇 ( 𝑥 ∧ 𝑦 ) = m i n 𝐵 ( 𝑥 ∧ 𝑦 ) , 𝜇 𝐶 𝜇 ( 𝑥 ∧ 𝑦 ) ≥ m i n 𝐵 ( 𝑥 ) ∧ 𝜇 𝐴 ∨ 𝜇 ( 𝑦 ) 𝐵 ( 𝑦 ) ∧ 𝜇 𝐴 , 𝜇 ( 𝑥 ) 𝐶 ( 𝑥 ) ∧ 𝜇 𝐴 ∨ 𝜇 ( 𝑦 ) 𝐴 ( 𝑥 ) ∧ 𝜇 𝐶 ( 𝑦 ) s i n c e 𝐵 a n d 𝐶 a r e I F I 𝜇 s o f 𝐴 . ≥ m i n 𝐵 ( 𝑥 ) , 𝜇 𝐶 ( 𝑥 ) ∧ 𝜇 𝐴 𝜇 ( 𝑦 ) ∨ m i n 𝐵 ( 𝑦 ) , 𝜇 𝐶 ( 𝑦 ) ∧ 𝜇 𝐴 ≥ 𝜇 ( 𝑥 ) 𝐵 ∩ 𝐶 ( 𝑥 ) ∧ 𝜇 𝐴 ∨ 𝜇 ( 𝑦 ) 𝐵 ∩ 𝐶 ( 𝑦 ) ∧ 𝜇 𝐴 . ( 𝑥 ) ( 4 ) Also 𝜈 𝐵 ∩ 𝐶 𝜈 ( 𝑥 ∨ 𝑦 ) = m a x 𝐵 ( 𝑥 ∨ 𝑦 ) , 𝜈 𝐶 𝜈 ( 𝑥 ∨ 𝑦 ) ≤ m a x 𝐵 ( 𝑥 ) ∨ 𝜈 𝐵 ( 𝑦 ) , 𝜈 𝐶 ( 𝑥 ) ∨ 𝜈 𝐶 ( , 𝑦 ) s i n c e 𝐵 a n d 𝐶 a r e I F I 𝜈 s o f 𝐴 . ≤ m a x 𝐵 ( 𝑥 ) ∨ 𝜈 𝐶 ( 𝑥 ) , 𝜈 𝐵 ( 𝑦 ) ∨ 𝜈 𝐶 𝜈 ( 𝑦 ) ≤ m a x 𝐵 ∩ 𝐶 ( 𝑥 ) , 𝜈 𝐵 ∩ 𝐶 ( 𝑦 ) ≤ 𝜈 𝐵 ∩ 𝐶 ( 𝑥 ) ∨ 𝜈 𝐵 ∩ 𝐶 𝜈 ( 𝑦 ) , 𝐵 ∩ 𝐶 𝜈 ( 𝑥 ∧ 𝑦 ) = m a x 𝐵 ( 𝑥 ∧ 𝑦 ) , 𝜈 𝐶 𝜈 ( 𝑥 ∧ 𝑦 ) ≤ m a x 𝐵 ( 𝑥 ) ∨ 𝜈 𝐴 ∧ 𝜈 ( 𝑦 ) 𝐵 ( 𝑦 ) ∨ 𝜈 𝐴 , 𝜈 ( 𝑥 ) 𝐶 ( 𝑥 ) ∨ 𝜈 𝐴 ∧ 𝜈 ( 𝑦 ) 𝐴 ( 𝑥 ) ∨ 𝜈 𝐶 , ( 𝑦 ) s i n c e 𝐵 a n d 𝐶 a r e I F I 𝜈 s o f 𝐴 . ≤ m a x 𝐵 ( 𝑥 ) , 𝜈 𝐶 ( 𝑥 ) ∨ 𝜈 𝐴 𝜈 ( 𝑦 ) ∧ m a x 𝐵 ( 𝑦 ) , 𝜈 𝐶 ( 𝑦 ) ∨ 𝜈 𝐴 ≤ 𝜈 ( 𝑥 ) 𝐵 ∩ 𝐶 ( 𝑥 ) ∨ 𝜈 𝐴 ∧ 𝜈 ( 𝑦 ) 𝐵 ∩ 𝐶 ( 𝑦 ) ∨ 𝜈 𝐴 . ( 𝑥 ) ( 5 ) 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 (1) 𝜇 𝐵 ( 𝑥 ∨ 𝑦 ) ≥ 𝜇 𝐵 ( 𝑥 ) ∧ 𝜇 𝐵 ( 𝑦 ) ,(2) 𝜈 𝐵 ( 𝑥 ∨ 𝑦 ) ≤ 𝜈 𝐵 ( 𝑥 ) ∨ 𝜈 𝐵 ( 𝑦 ) ,(3) 𝐴 𝐵 ⊆ 𝐵 .

Proof. Suppose that conditions (1), (2), and (3) hold. Then we prove that 𝐵 is an IFI of 𝐴 .
We have 𝜇 𝐵 ( 𝑥 ∧ 𝑦 ) ≥ 𝜇 𝐴 𝐵 ( 𝑥 ∧ 𝑦 ) , s i n c e 𝐵 ⊇ 𝐴 𝐵 = s u p 𝑥 ∧ 𝑦 = 𝑥 𝑖 ∧ 𝑦 𝑖 𝜇 𝐴 𝑥 𝑖 ∧ 𝜇 𝐵 𝑦 𝑖 ≥ 𝜇 𝐴 ( 𝑥 ) ∧ 𝜇 𝐵 ( 𝑦 ) . ( 6 ) Similarly 𝜇 𝐵 ( 𝑥 ∧ 𝑦 ) ≥ 𝜇 𝐵 ( 𝑥 ) ∧ 𝜇 𝐴 ( 𝑦 ) , s i n c e 𝐵 ⊇ 𝐴 𝐵 = 𝐵 𝐴 . ( 7 ) Hence 𝜇 𝐵 𝜇 ( 𝑥 ∧ 𝑦 ) ≥ 𝐴 ( 𝑥 ) ∧ 𝜇 𝐵 ∨ 𝜇 ( 𝑦 ) 𝐵 ( 𝑥 ) ∧ 𝜇 𝐴 ( 𝑦 ) . ( a ) Also 𝜈 𝐵 ( 𝑥 ∧ 𝑦 ) ≤ 𝜈 𝐴 𝐵 ( 𝑥 ∧ 𝑦 ) , s i n c e 𝐵 ⊇ 𝐴 𝐵 = i n f 𝑥 ∧ 𝑦 = 𝑥 𝑖 ∧ 𝑦 𝑖 𝜈 𝐴 𝑥 𝑖 ∨ 𝜈 𝐵 𝑦 𝑖 ≤ 𝜈 𝐴 ( 𝑥 ) ∨ 𝜈 𝐵 ( 𝑦 ) . ( 8 ) Similarly 𝜈 𝐵 ( 𝑥 ∧ 𝑦 ) ≤ 𝜈 𝐵 ( 𝑥 ) ∨ 𝜈 𝐴 ( 𝑦 ) , s i n c e 𝐵 ⊇ 𝐴 𝐵 = 𝐵 𝐴 . ( 9 ) Hence 𝜈 𝐵 𝜈 ( 𝑥 ∧ 𝑦 ) ≤ 𝐴 ( 𝑥 ) ∨ 𝜈 𝐵 ∧ 𝜈 ( 𝑦 ) 𝐵 ( 𝑥 ) ∨ 𝜈 𝐴 ( 𝑦 ) . ( b ) 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 𝜇 𝐵 ( 𝑥 ∧ 𝑦 ) ≥ 𝜇 𝐴 ( 𝑥 ) ∧ 𝜇 𝐵 𝜈 ( 𝑦 ) , 𝐵 ( 𝑥 ∧ 𝑦 ) ≤ 𝜈 𝐴 ( 𝑥 ) ∨ 𝜈 𝐵 ( 𝑦 ) , ∀ 𝑥 , 𝑦 ∈ 𝐿 . ( 1 0 )
So ∀ 𝑧 ∈ 𝐿 with 𝑧 = 𝑥 ∧ 𝑦 𝜇 𝐵 ( 𝑧 ) ≥ 𝑧 = 𝑥 ∧ 𝑦 𝜇 𝐴 ( 𝑥 ) ∧ 𝜇 𝐵 ( 𝑦 ) = 𝜇 𝐴 𝐵 𝜈 ( 𝑧 ) , 𝐵 ( 𝑧 ) ≤ 𝑧 = 𝑥 ∧ 𝑦 𝜈 𝐴 ( 𝑥 ) ∨ 𝜈 𝐵 ( 𝑦 ) = 𝜈 𝐴 𝐵 ( 𝑧 ) . ( 1 1 ) Hence 𝐴 𝐵 ⊆ 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 (1) 𝜇 𝐵 ( 𝑥 ∨ 𝑦 ) ≥ 𝜇 𝐵 ( 𝑥 ) ∧ 𝜇 𝐵 ( 𝑦 ) ,(2) 𝜈 𝐵 ( 𝑥 ∨ 𝑦 ) ≤ 𝜈 𝐵 ( 𝑥 ) ∨ 𝜈 𝐵 ( 𝑦 ) ,(3) 𝐴 ⋅ 𝐵 ⊆ 𝐵 .

Proof. Suppose conditions (1), (2), and (3) holds. We prove 𝐵 is an IFI of 𝐴 .
We have 𝜇 𝐵 ( 𝑥 ∧ 𝑦 ) ≥ 𝜇 𝐴 ⋅ 𝐵 ( 𝑥 ∧ 𝑦 ) , s i n c e 𝐵 ⊇ 𝐴 ⋅ 𝐵 = s u p 𝑥 ∧ 𝑦 = ∨ 𝑛 𝑖 = 1 ( 𝑥 𝑖 ∧ 𝑦 𝑖 ) 𝑛 𝑖 = 1 𝜇 𝐴 𝑥 𝑖 ∧ 𝜇 𝐵 𝑦 𝑖 ≥ 𝜇 𝐴 ( 𝑥 ) ∧ 𝜇 𝐵 ( 𝑦 ) . ( 1 2 ) Similarly, we can obtain 𝜇 𝐵 ( 𝑥 ∧ 𝑦 ) ≥ 𝜇 𝐵 ( 𝑥 ) ∧ 𝜇 𝐴 ( 𝑦 ) , s i n c e 𝐵 ⊇ 𝐴 ⋅ 𝐵 = 𝐵 ⋅ 𝐴 . ( 1 3 ) Hence 𝜇 𝐵 𝜇 ( 𝑥 ∧ 𝑦 ) ≥ 𝐴 ( 𝑥 ) ∧ 𝜇 𝐵 ∨ 𝜇 ( 𝑦 ) 𝐵 ( 𝑥 ) ∧ 𝜇 𝐴 ( 𝑦 ) . ( a . 1 ) Also 𝜈 𝐵 ( 𝑥 ∧ 𝑦 ) ≤ 𝜈 𝐴 ⋅ 𝐵 ( 𝑥 ∧ 𝑦 ) , s i n c e 𝐵 ⊇ 𝐴 ⋅ 𝐵 = i n f ⋁ 𝑥 ∧ 𝑦 = 𝑛 𝑖 = 1 𝑥 𝑖 ∧ 𝑦 𝑖 𝑛 𝑖 = 1 𝜈 𝐴 𝑥 𝑖 ∨ 𝜈 𝐵 𝑦 𝑖 ≤ 𝜈 𝐴 ( 𝑥 ) ∨ 𝜈 𝐵 ( 𝑦 ) . ( 1 4 ) Similarly 𝜈 𝐵 ( 𝑥 ∧ 𝑦 ) ≤ 𝜈 𝐵 ( 𝑥 ) ∨ 𝜈 𝐴 ( 𝑦 ) , s i n c e 𝐵 ⊇ 𝐴 ⋅ 𝐵 = 𝐵 ⋅ 𝐴 . ( 1 5 ) Hence 𝜈 𝐵 𝜈 ( 𝑥 ∧ 𝑦 ) ≤ 𝐴 ( 𝑥 ) ∨ 𝜈 𝐵 ∧ 𝜈 ( 𝑦 ) 𝐵 ( 𝑥 ) ∨ 𝜈 𝐴 ( 𝑦 ) . ( b . 1 ) 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 = 𝑛 𝑖 = 1 ( 𝑥 𝑖 ∧ 𝑦 𝑖 ) , where 𝑥 𝑖 ∈ 𝐴 , 𝑦 𝑖 ∈ 𝐵 .
We have 𝜇 𝐵 ( 𝑧 ) = 𝜇 𝐵 𝑛 𝑖 = 1 𝑥 𝑖 ∧ 𝑦 𝑖 ≥ 𝑛 𝑖 = 1 𝜇 𝐵 𝑥 𝑖 ∧ 𝑦 𝑖 ≥ 𝑛 𝑖 = 1 𝜇 𝐴 𝑥 𝑖 ∧ 𝜇 𝐵 𝑦 𝑖 , s i n c e 𝐵 I F I o f 𝐴 . ( 1 6 ) Thus 𝜇 𝐵 ( 𝑧 ) ≥ 𝑛 𝑖 = 1 𝜇 𝐴 𝑥 𝑖 ∧ 𝜇 𝐵 𝑦 𝑖 = 𝜇 𝐴 ⋅ 𝐵 ( 𝑧 ) . ( 1 7 ) Also 𝜈 𝐵 ( 𝑧 ) = 𝜈 𝐵 𝑛 𝑖 = 1 𝑥 𝑖 ∧ 𝑦 𝑖 ≤ 𝑛 𝑖 = 1 𝜈 𝐵 𝑥 𝑖 ∧ 𝑦 𝑖 ≤ 𝑛 𝑖 = 1 𝜈 𝐴 𝑥 𝑖 ∨ 𝜈 𝐵 𝑦 𝑖 , s i n c e 𝐵 I F I o f 𝐴 . ( 1 8 ) Thus 𝜈 𝐵 ( 𝑧 ) ≤ ∧ 𝑛 𝑖 = 1 𝜈 𝐴 𝑥 𝑖 ∨ 𝜈 𝐵 𝑦 𝑖 = 𝜈 𝐴 ⋅ 𝐵 ( 𝑧 ) . ( 1 9 ) 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 𝜇 𝐵 + 𝐶 ( 𝑥 ∨ 𝑦 ) ≥ 𝜇 𝐵 + 𝐶 ( 𝑥 ) ∧ 𝜇 𝐵 + 𝐶 𝜈 ( 𝑦 ) , 𝐵 + 𝐶 ( 𝑥 ∨ 𝑦 ) ≤ 𝜈 𝐵 + 𝐶 ( 𝑥 ) ∨ 𝜈 𝐵 + 𝐶 ( 𝑦 ) ( 2 0 ) (by [11, Theorem  5.2]).
And 𝐴 ( 𝐵 + 𝐶 ) ⊆ 𝐴 𝐵 + 𝐴 𝐶 ⊆ 𝐵 + 𝐶 , ( 𝐵 + 𝐶 ) 𝐴 ⊆ 𝐵 𝐴 + 𝐶 𝐴 ⊆ 𝐵 + 𝐶 . ( 2 1 ) (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 , 𝐷 𝐶 ⊆ 𝐵 } . ( 2 2 )

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 , 𝐷 𝐶 ⊆ 𝐵 } . Suppose 𝐷 , 𝐷 ∈ 𝜂 .  Then 𝐷 and 𝐷 are IFI’s of 𝐴 such that 𝐷 𝐶 ⊆ 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 ⊆ 𝐷 𝐶 + 𝐷 𝐶 ⊆ 𝐵 + 𝐵 = 𝐵 . Thus 𝐷 + 𝐷 ∈ 𝜂 . Now 𝜇 𝐵 / 𝐶 ( 𝑥 ) ∧ 𝜇 𝐵 / 𝐶 ( 𝑦 ) = 𝐷 ∈ 𝜂 𝜇 𝐷 ∧ ⎡ ⎢ ⎢ ⎣ ( 𝑥 ) 𝐷 ′ ∈ 𝜂 𝜇 𝐷 ′ ⎤ ⎥ ⎥ ⎦ = 𝜇 ( 𝑦 ) 𝐷 ( 𝑥 ) ∧ 𝜇 𝐷 ′ ( 𝑦 ) / 𝐷 , 𝐷 ≤ 𝜇 ∈ 𝜂 𝐷 + 𝐷 ′ ( 𝑥 ∨ 𝑦 ) / 𝐷 , 𝐷 ∈ 𝜂 ≤ 𝜇 𝐵 / 𝐶 ( 𝑥 ∨ 𝑦 ) , s i n c e D + 𝐷 ∈ 𝜂 . ( 2 3 ) That is, 𝜇 𝐵 / 𝐶 ( 𝑥 ∨ 𝑦 ) ≥ 𝜇 𝐵 / 𝐶 ( 𝑥 ) ∧ 𝜇 𝐵 / 𝐶 ( 𝑦 ) . ( 2 4 ) Also 𝜇 𝐵 / 𝐶 ( 𝑥 ∧ 𝑦 ) = 𝐷 ∈ 𝜂 𝜇 𝐷 ≥ ( 𝑥 ∧ 𝑦 ) 𝐷 ∈ 𝜂 𝜇 𝐷 ( 𝑥 ) ∧ 𝜇 𝐴 = ( 𝑦 ) , s i n c e 𝐷 ⊲ 𝐴 𝐷 ∈ 𝜂 𝜇 𝐷 ( 𝑥 ) ∧ 𝜇 𝐴 ( 𝑦 ) = 𝜇 𝐵 / 𝐶 ( 𝑥 ) ∧ 𝜇 𝐴 ( 𝑦 ) . ( 2 5 ) Similarly 𝜇 𝐵 / 𝐶 ( 𝑥 ∧ 𝑦 ) ≥ 𝜇 𝐵 / 𝐶 ( 𝑦 ) ∧ 𝜇 𝐴 ( 𝑥 ) . ( 2 6 ) Thus 𝜇 𝐵 / 𝐶 𝜇 ( 𝑥 ∧ 𝑦 ) ≥ 𝐵 / 𝐶 ( 𝑥 ) ∧ 𝜇 𝐴 ∨ 𝜇 ( 𝑦 ) 𝐵 / 𝐶 ( 𝑦 ) ∧ 𝜇 𝐴 ( 𝑥 ) . ( 2 7 ) Now 𝜈 𝐵 / 𝐶 ( 𝑥 ) ∨ 𝜈 𝐵 / 𝐶 ( 𝑦 ) = 𝐷 ∈ 𝜂 𝜈 𝐷 ∨ ⎡ ⎢ ⎢ ⎣ ( 𝑥 ) 𝐷 ′ ∈ 𝜂 𝜈 𝐷 ′ ⎤ ⎥ ⎥ ⎦ = 𝜈 ( 𝑦 ) 𝐷 ( 𝑥 ) ∨ 𝜈 𝐷 ′ ( 𝑦 ) / 𝐷 , 𝐷 ≥ 𝜈 ∈ 𝜂 𝐷 + 𝐷 ′ ( 𝑥 ∨ 𝑦 ) / 𝐷 , 𝐷 ∈ 𝜂 ≥ 𝜈 𝐵 / 𝐶 ( 𝑥 ∨ 𝑦 ) , s i n c e 𝐷 + 𝐷 ∈ 𝜂 . ( 2 8 ) That is 𝜈 𝐵 / 𝐶 ( 𝑥 ∨ 𝑦 ) ≤ 𝜈 𝐵 / 𝐶 ( 𝑥 ) ∨ 𝜈 𝐵 / 𝐶 ( 𝑦 ) . ( 2 9 ) Also 𝜈 𝐵 / 𝐶 ( 𝑥 ∧ 𝑦 ) = 𝐷 ∈ 𝜂 𝜈 𝐷 ≤ ( 𝑥 ∧ 𝑦 ) 𝐷 ∈ 𝜂 𝜈 𝐷 ( 𝑥 ) ∨ 𝜈 𝐴 = ( 𝑦 ) , s i n c e 𝐷 ⊲ 𝐴 𝐷 ∈ 𝜂 𝜈 𝐷 ( 𝑥 ) ∨ 𝜈 𝐴 ( 𝑦 ) = 𝜈 𝐵 / 𝐶 ( 𝑥 ) ∨ 𝜈 𝐴 ( 𝑦 ) . ( 3 0 ) Similarly 𝜈 𝐵 / 𝐶 ( 𝑥 ∧ 𝑦 ) ≤ 𝜈 𝐵 / 𝐶 ( 𝑦 ) ∨ 𝜈 𝐴 ( 𝑥 ) . ( 3 1 ) Thus 𝜈 𝐵 / 𝐶 𝜈 ( 𝑥 ∧ 𝑦 ) ≤ 𝐵 / 𝐶 ( 𝑥 ) ∨ 𝜈 𝐴 ∧ 𝜈 ( 𝑦 ) 𝐵 / 𝐶 ( 𝑦 ) ∨ 𝜈 𝐴 ( 𝑥 ) . ( 3 2 ) From (24), (27), (29), and (32) 𝐵 / 𝐶 is an IFI of 𝐴 .
Clearly 𝐵 / 𝐶 ⊆ 𝐴 .
Since 𝐵 is an IFI of 𝐴 , 𝐵 𝐴 ⊆ 𝐵 (by Theorem 1).
Since 𝐶 ⊆ 𝐴 , by Lemma 1   𝐵 𝐶 ⊆ 𝐵 𝐴 ⊆ 𝐵 . Hence 𝐵 ∈ 𝜂 . So 𝐵 ⊆ 𝐵 / 𝐶 .
Thus we have 𝐵 ⊆ 𝐵 / 𝐶 ⊆ 𝐴 . ( 3 3 )

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 ⊲ 𝐴 a n d 𝐷 𝐶 ⊆ 𝐵 } . We have ⋃ 𝐵 / 𝐶 = 𝐷 ∈ 𝜂 𝐷 Let 𝑥 ∈ 𝐿 such that ⋁ 𝑥 = 𝑛 𝑖 = 1 ( 𝑎 𝑖 ∧ 𝑏 𝑖 ) .
Then 𝜇 𝐵 𝑎 𝑖 ∧ 𝑏 𝑖 ≥ 𝜇 𝐷 𝐶 𝑎 𝑖 ∧ 𝑏 𝑖 ≥ 𝜇 𝐷 𝑎 𝑖 ∧ 𝜇 𝐶 𝑏 𝑖 , ∀ 𝐷 ∈ 𝜂 . ( 3 4 ) So 𝜇 𝐵 𝑎 𝑖 ∧ 𝑏 𝑖 ≥ 𝐷 ∈ 𝜂 𝜇 𝐷 𝑎 𝑖 ∧ 𝜇 𝐶 𝑏 𝑖 = 𝐷 ∈ 𝜂 𝜇 𝐷 𝑎 𝑖 ∧ 𝜇 𝐶 𝑏 𝑖 = 𝜇 𝐵 / 𝐶 𝑎 𝑖 ∧ 𝜇 𝐶 𝑏 𝑖 . ( 3 5 ) Hence 𝜇 𝐵 ( 𝑥 ) = 𝜇 𝐵 𝑛 𝑖 = 1 𝑎 𝑖 ∧ 𝑏 𝑖 ≥ 𝑛 𝑖 = 1 𝜇 𝐵 𝑎 𝑖 ∧ 𝑏 𝑖 , ≥ s i n c e 𝐵 i s a n I F I o f 𝐴 𝑛 𝑖 = 1 𝜇 𝐵 / 𝐶 𝑎 𝑖 ∧ 𝜇 𝐶 𝑏 𝑖 . ( 3 6 ) Consequently 𝜇 𝐵 ( 𝑥 ) ≥ 𝑛 𝑖 = 1 𝜇 𝐵 / 𝐶 𝑎 𝑖 ∧ 𝜇 𝐶 𝑏 𝑖 / 𝑥 = 𝑛 𝑖 = 1 𝑎 𝑖 ∧ 𝑏 𝑖 = 𝜇 ( 𝐵 / 𝐶 ) ⋅ 𝐶 ( 𝑥 ) . ( 3 7 ) Also 𝜈 𝐵 𝑎 𝑖 ∧ 𝑏 𝑖 ≤ 𝜈 𝐷 𝐶 𝑎 𝑖 ∧ 𝑏 𝑖 ≤ 𝜈 𝐷 𝑎 𝑖 ∨ 𝜈 𝐶 𝑏 𝑖 , ∀ 𝐷 ∈ 𝜂 . ( 3 8 ) So 𝜈 𝐵 𝑎 𝑖 ∧ 𝑏 𝑖 ≤ 𝐷 ∈ 𝜂 𝜈 𝐷 𝑎 𝑖 ∨ 𝜈 𝐶 𝑏 𝑖 = 𝐷 ∈ 𝜂 𝜈 𝐷 𝑎 𝑖 ∨ 𝜈 𝐶 𝑏 𝑖 = 𝜈 𝐵 / 𝐶 𝑎 𝑖 ∧ 𝜈 𝐶 𝑏 𝑖 . ( 3 9 ) Hence 𝜈 𝐵 ( 𝑥 ) = 𝜈 𝐵 𝑛 𝑖 = 1 𝑎 𝑖 ∧ 𝑏 𝑖 ≤ 𝑛 𝑖 = 1 𝜈 𝐵 𝑎 𝑖 ∧ 𝑏 𝑖 , ≤ s i n c e 𝐵 i s a n I F I o f 𝐴 𝑛 𝑖 = 1 𝜈 𝐵 / 𝐶 𝑎 𝑖 ∧ 𝜈 𝐶 𝑏 𝑖 . ( 4 0 ) Consequently 𝜈 𝐵 ( 𝑥 ) ≤ 𝑛 𝑖 = 1 𝜈 𝐵 / 𝐶 𝑎 𝑖 ∧ 𝜈 𝐶 𝑏 𝑖 / 𝑥 = 𝑛 𝑖 = 1 𝑎 𝑖 ∧ 𝑏 𝑖 = 𝜈 ( 𝐵 / 𝐶 ) ⋅ 𝐶 ( 𝑥 ) . ( 4 1 ) Thus from (37) and (41) ( 𝐵 / 𝐶 ) ⋅ 𝐶 ⊆ 𝐵 .
If 𝐷 is an ideal of 𝐴 such that 𝐷 ⋅ 𝐶 ⊆ 𝐵 then 𝐷 𝐶 ⊆ 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. (1)If 𝐵 ⊆ 𝐶 then 𝐵 / 𝐷 ⊆ 𝐶 / 𝐷 and 𝐷 / 𝐶 ⊆ 𝐷 / 𝐵 .(2)If 𝐵 ⊆ 𝐶 then 𝐶 / 𝐵 = 𝐴 .(3) 𝐵 / 𝐵 = 𝐴 .

Proof. (1) Let 𝐵 ⊆ 𝐶 . Write 𝜂 = { 𝐸 / 𝐸 ⊲ 𝐴 a n d 𝐸 𝐷 ⊆ 𝐵 } and 𝜉 = { 𝐸 / 𝐸 ⊲ 𝐴 a n d 𝐸 𝐷 ⊆ 𝐶 } . If 𝐸 ∈ 𝜂 then 𝐸 ⊲ 𝐴 and 𝐸 𝐷 ⊆ 𝐵 ⊆ 𝐶 . Thus 𝐸 ∈ 𝜉 and hence 𝜂 ⊆ 𝜉 . So ⋃ 𝐵 / 𝐷 = 𝐸 ∈ 𝜂 ⋃ 𝐸 ⊆ 𝐸 ∈ 𝜉 𝐸 = 𝐶 / 𝐷 .
Similarly, let 𝜂 1 = { 𝐸 / 𝐸 ⊲ 𝐴 a n d 𝐸 𝐶 ⊆ 𝐷 } and 𝜉 1 = { 𝐸 / 𝐸 ⊲ 𝐴 a n d 𝐸 𝐵 ⊆ 𝐷 } . If 𝐸 ∈ 𝜂 1 then 𝐸 𝐶 ⊆ 𝐷 . But 𝐵 ⊆ 𝐶 . So 𝐸 𝐵 ⊆ 𝐸 𝐶 ⊆ 𝐷 . Thus 𝐸 ∈ 𝜉 1 and hence 𝜂 1 ⊆ 𝜉 1 . So ⋃ 𝐷 / 𝐶 = 𝐸 ∈ 𝜂 1 ⋃ 𝐸 ⊆ 𝐸 ∈ 𝜉 1 𝐸 = 𝐷 / 𝐵 .
(2) Let 𝜂 = { 𝐸 / 𝐸 ⊲ 𝐴 a n d 𝐸 𝐵 ⊆ 𝐶 } . Since 𝐵 ⊲ 𝐴 , we have 𝐴 𝐵 ⊆ 𝐵 ⊆ 𝐶 , and 𝐴 ⊲ 𝐴 . Thus 𝐴 ∈ 𝜂 and hence ⋃ 𝐴 ⊆ 𝐸 ∈ 𝜂 𝐸 = 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 (1) ( 𝐵 / 𝐶 ) / 𝐵 = 𝐴 ,(2) ( 𝐵 / 𝐵 ) / 𝐶 = 𝐴 ,(3) 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 𝐵 𝑖 𝑖 = 1 , 2 … … 𝑚 , 𝐶 , are IFI’s of 𝐴 . Then 𝑚 𝑖 = 1 𝐵 𝑖 / 𝐶 = 𝑚 𝑖 = 1 𝐵 𝑖 / 𝐶 . ( 4 2 )

Proof. Since ⋂ 𝑚 𝑖 = 1 𝐵 𝑖 ⊆ 𝐵 𝑖   by Theorem 6 (1) ( ⋂ 𝑚 𝑖 = 1 𝐵 𝑖 ) / 𝐶 ⊆ 𝐵 𝑖 / 𝐶 , ∀ 𝑖 .
Hence 𝑚 𝑖 = 1 𝐵 𝑖 / 𝐶 ⊆ 𝑚 𝑖 = 1 𝐵 𝑖 / 𝐶 . ( 4 3 )
Let 𝜂 1 = 𝐸 / 𝐸 ⊲ 𝐴 , 𝐸 𝐶 ⊆ 𝐵 1 , 𝜂 2 = 𝐸 / 𝐸 ⊲ 𝐴 , 𝐸 𝐶 ⊆ 𝐵 2 , 𝜂 3 = 𝐸 / 𝐸 ⊲ A , 𝐸 𝐶 ⊆ 𝐵 1 ∩ 𝐵 2 . ( 4 4 ) Then ∀ 𝑥 ∈ 𝐿 𝜇 𝐵 1 / 𝐶 ∩ 𝐵 2 / 𝐶 ( 𝑥 ) = 𝜇 𝐵 1 / 𝐶 ( 𝑥 ) ∧ 𝜇 𝐵 2 / 𝐶 = ( 𝑥 ) 𝐸 ∈ 𝜂 1 𝜇 𝐸 ∧ ⎛ ⎜ ⎜ ⎝ ( 𝑥 ) 𝐸 ′ ∈ 𝜂 2 𝜇 𝐸 ′ ⎞ ⎟ ⎟ ⎠ 𝜇 ( 𝑥 ) = ∨ 𝐸 ( 𝑥 ) ∧ 𝜇 𝐸 ′ ( 𝑥 ) / 𝐸 ∈ 𝜂 1 , 𝐸 ∈ 𝜂 2 . ( a . 2 ) Similarly 𝜈 𝐵 1 / 𝐶 ∩ 𝐵 2 / 𝐶 ( 𝑥 ) = 𝜈 𝐵 1 / 𝐶 ( 𝑥 ) ∨ 𝜈 𝐵 2 / 𝐶 = ( 𝑥 ) 𝐸 ∈ 𝜂 1 𝜈 𝐸 ∨ ⎛ ⎜ ⎜ ⎝ ( 𝑥 ) 𝐸 ′ ∈ 𝜂 2 𝜈 𝐸 ′ ⎞ ⎟ ⎟ ⎠ 𝜈 ( 𝑥 ) = ∧ 𝐸 ( 𝑥 ) ∨ 𝜈 𝐸 ′ ( 𝑥 ) / 𝐸 ∈ 𝜂 1 , 𝐸 ∈ 𝜂 2 . ( b . 2 ) Now let 𝐸 ∈ 𝜂 1 and 𝐸 ∈ 𝜂 2 . Then 𝐸 𝐶 ⊆ 𝐵 1 and 𝐸 𝐶 ⊆ 𝐵 2 . Also 𝐸 ∩ 𝐸 IFI of 𝐴 .
So that 𝐸 ∩ 𝐸 𝐶 ⊆ 𝐸 𝐶 ∩ 𝐸 𝐶 ⊆ 𝐵 1 ∩ 𝐵 2 . ( 4 5 ) Thus 𝐸 ∩ 𝐸 ′ ∈ 𝜂 3 . So 𝜂 1 ∩ 𝜂 2 ⊆ 𝜂 3 .
Hence 𝐵 1 ∩ 𝐵 2 / 𝐶 = 𝐸 ∈ 𝜂 3 𝐸 ⊇ 𝐸 ∈ 𝜂 1 , 𝐸 ′ ∈ 𝜂 2 𝐸 ∩ 𝐸 . ( 4 6 ) So 𝜇 𝐵 1 ∩ 𝐵 2 / 𝐶 𝜇 ( 𝑥 ) ≥ 𝐸 ∩ 𝐸 ′ = 𝜇 ( 𝑥 ) 𝐸 ( 𝑥 ) ∧ 𝜇 𝐸 ′ ( 𝑥 ) = 𝜇 𝐵 1 / 𝐶 ∩ 𝐵 2 / 𝐶 𝜈 ( 𝑥 ) , f r o m ( a . 2 ) , 𝐵 1 ∩ 𝐵 2 / 𝐶 𝜈 ( 𝑥 ) ≤ 𝐸 ∩ 𝐸 ′ = 𝜈 ( 𝑥 ) 𝐸 ( 𝑥 ) ∨ 𝜈 𝐸 ′ ( 𝑥 ) = 𝜈 𝐵 1 / 𝐶 ∩ 𝐵 2 / 𝐶 ( 𝑥 ) , f r o m ( b . 2 ) . ( 4 7 ) Hence 𝐵 1 ∩ 𝐵 2 / 𝐶 ⊇ 𝐵 1 / 𝐶 ∩ 𝐵 2 / 𝐶 . ( 4 8 ) From (43) and (48) ( 𝐵 1 ∩ 𝐵 2 ) / 𝐶 = 𝐵 1 / 𝐶 ∩ 𝐵 2 / 𝐶 . This completes the proof.

Next, we denote the set of all IFI’s { 𝐵 𝑖 }    𝑖 = 1 , 2 … , 𝑚 of an IFL 𝐴 that satisfies the property 𝜇 𝐵 𝑖 ( 0 ) = 𝜇 𝐵 𝑗 ( 0 ) a n d 𝜈 𝐵 𝑖 ( 0 ) = 𝜈 𝐵 𝑗 ( 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 (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 𝜇 𝐵 + 𝐶 ( 𝑥 ) = 𝑥 = 𝑦 ∨ 𝑧 𝜇 𝐵 ( 𝑦 ) ∧ 𝜇 𝐶 ( 𝑧 ) ≥ 𝜇 𝐵 ( 𝑥 ) ∧ 𝜇 𝐶 ( 0 ) a s 𝑥 = 𝑥 ∨ 0 = 𝜇 𝐵 ( 𝑥 ) ∧ 𝜇 𝐵 ( 0 ) s i n c e 𝜇 𝐵 ( 0 ) = 𝜇 𝐶 ( 0 ) = 𝜇 𝐵 ( 𝑥 ) s i n c e 𝜇 𝐵 ( 0 ) ≥ 𝜇 𝐵 𝜈 ( 𝑥 ) , 𝐵 + 𝐶 ( 𝑥 ) = 𝑥 = 𝑦 ∨ 𝑧 𝜈 𝐵 ( 𝑦 ) ∨ 𝜈 𝐶 ( 𝑧 ) ≤ 𝜈 𝐵 ( 𝑥 ) ∨ 𝜈 𝐶 ( 0 ) a s 𝑥 = 𝑥 ∨ 0 = 𝜈 𝐵 ( 𝑥 ) ∨ 𝜈 𝐵 ( 0 ) s i n c e 𝜈 𝐵 ( 0 ) = 𝜈 𝐶 ( 0 ) = 𝜈 𝐵 ( 𝑥 ) s i n c e 𝜈 𝐵 ( 0 ) ≤ 𝜈 𝐵 ( 𝑥 ) . ( 4 9 ) 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 . ( 5 0 )
Write 𝜂 = { 𝐸 / 𝐸 ⊲ 𝐴 a n d 𝐸 𝐶 ⊆ 𝐵 } and 𝜉 = { 𝐸 / 𝐸 ⊲ 𝐴 a n d 𝐸 ( 𝐵 + 𝐶 ) ⊆ 𝐵 } .
Let 𝐸 ∈ 𝜂 , then 𝐸 ⊲ 𝐴 and 𝐸 ⊆ 𝐴 . So 𝐸 𝐵 ⊆ 𝐴 𝐵 (by Lemma 1). But 𝐴 𝐵 ⊆ 𝐵 , since 𝐵 ⊲ 𝐴 . Hence 𝐸 𝐵 ⊆ 𝐵 and also 𝐸 𝐶 ⊆ 𝐵 . So By Lemma 1, 𝐸 ( 𝐵 + 𝐶 ) ⊆ 𝐸 𝐵 + 𝐸 𝐶 ⊆ 𝐵 + 𝐵 = 𝐵 . Therefore 𝐸 ∈ 𝜉 . So 𝜂 ⊆ 𝜉 . Thus 𝐵 / 𝐶 = 𝐸 ∈ 𝜂 𝐸 ⊆ 𝐸 ∈ 𝜉 𝐸 = 𝐵 / 𝐵 + 𝐶 . ( 5 1 ) 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 { 𝐵 𝑖 } 𝑖 = 1 , 2 … 𝑚 ∈ IFI ( 𝐴 *) and C any IFI of 𝐴 . Then ∑ 𝐶 / 𝑚 𝑖 = 1 𝐵 𝑖 = ⋂ 𝑚 𝑖 = 1 ( 𝐶 / 𝐵 𝑖 ) .

Proof. We have 𝐵 1 + 𝐵 2 ⊲ 𝐴 and 𝐵 1 ⊆ 𝐵 1 + 𝐵 2 , 𝐵 2 ⊆ 𝐵 1 + 𝐵 2 .
So by Theorem 6 (1), 𝐶 / 𝐵 1 + 𝐵 2 ⊆ 𝐶 / 𝐵 1 and 𝐶 / 𝐵 1 + 𝐵 2 ⊆ 𝐶 / 𝐵 2 .
Therefore 𝐶 / 𝐵 1 + 𝐵 2 ⊆ 𝐶 / 𝐵 1 𝐶 / 𝐵 2 . ( 5 2 ) Let 𝜂 1 = 𝐸 / 𝐸 ⊲ 𝐴 , 𝐸 𝐵 1 , 𝜂 ⊆ 𝐶 2 = 𝐸 / 𝐸 ⊲ 𝐴 , 𝐸 𝐵 2 , 𝜂 ⊆ 𝐶 3 = 𝐵 𝐸 / 𝐸 ⊲ 𝐴 , 𝐸 1 + 𝐵 2 . ⊆ 𝐶 ( 5 3 ) Then ∀ 𝑥 ∈ 𝐿 𝜇 𝐶 / 𝐵 1 ∩ 𝐶 / 𝐵 2 ( 𝑥 ) = 𝜇 𝐶 / 𝐵 1 ( 𝑥 ) ∧ 𝜇 𝐶 / 𝐵 2 = ( 𝑥 ) 𝐸 ∈ 𝜂 1 𝜇 𝐸 ∧ ⎛ ⎜ ⎜ ⎝ ( 𝑥 ) 𝐸 ′ ∈ 𝜂 2 𝜇 𝐸 ′ ⎞ ⎟ ⎟ ⎠ = 𝜇 ( 𝑥 ) 𝐸 ( 𝑥 ) ∧ 𝜇 𝐸 ′ ( 𝑥 ) / 𝐸 ∈ 𝜂 1 , 𝐸 ∈ 𝜂 2 . ( a . 3 ) Similarly 𝜈 𝐶 / 𝐵 1 ∩ 𝐶 / 𝐵 2 ( 𝑥 ) = 𝜈 𝐶 / 𝐵 1 ( 𝑥 ) ∨ 𝜈 𝐶 / 𝐵 2 = ( 𝑥 ) 𝐸 ∈ 𝜂 1 𝜈 𝐸 ∨ ⎛ ⎜ ⎜ ⎝ ( 𝑥 ) 𝐸 ′ ∈ 𝜂 2 𝜈 𝐸 ′ ⎞ ⎟ ⎟ ⎠ 𝜈 ( 𝑥 ) = ∧ 𝐸 ( 𝑥 ) ∨ 𝜈 𝐸 ′ ( 𝑥 ) / 𝐸 ∈ 𝜂 1 , 𝐸 ∈ 𝜂 2 . ( b . 3 ) Now let 𝐸 ∈ 𝜂 1 and 𝐸 ∈ 𝜂 2 . Then 𝐸 𝐵 1 ⊆ 𝐶 and 𝐸 𝐵 2 ⊆ 𝐶 . Also 𝐸 ∩ 𝐸 IFI of 𝐴 , so that E ∩ 𝐸 B 1 + B 2 ⊆ E ∩ 𝐸 B 1 + E ∩ 𝐸 B 2 ⊆ E B 1 + 𝐸 B 2 ⊆ C + C = C . ( 5 4 ) So 𝐸 ∩ 𝐸 ′ ∈ 𝜂 3 . Hence 𝜂 1 ∩ 𝜂 2 ⊆ 𝜂 3 . Thus 𝐶 / 𝐵 1 + 𝐵 2 = ⋃ 𝐸 ∈ 𝜂 3 ⋃ 𝐸 ⊇ 𝐸 ∈ 𝜂 1 , 𝐸 ′ ∈ 𝜂 2 ( 𝐸 ∩ 𝐸 ) .
So 𝜇 𝐶 / 𝐵 1 + 𝐵 2 ( 𝑥 ) ≥ ∨ 𝜇 𝐸 ∩ 𝐸 ′ 𝜇 ( 𝑥 ) = ∨ 𝐸 ( 𝑥 ) ∧ 𝜇 𝐸 ′ ( 𝑥 ) = 𝜇 𝐶 / 𝐵 1 ∩ 𝐶 / 𝐵 2 𝜈 ( 𝑥 ) f r o m ( a . 3 ) , 𝐶 / 𝐵 1 + 𝐵 2 ( 𝑥 ) ≤ ∧ 𝜈 𝐸 ∩ 𝐸 ′ 𝜈 ( 𝑥 ) = ∧ 𝐸 ( 𝑥 ) ∨ 𝜈 𝐸 ′ ( 𝑥 ) = 𝜈 𝐶 / 𝐵 1 ∩ 𝐶 / 𝐵 2 ( 𝑥 ) f r o m ( b . 3 ) . ( 5 5 ) Therefore 𝐶 / 𝐵 1 + 𝐵 2 ⊇ 𝐶 / 𝐵 1 𝐶 / 𝐵 2 . ( 5 6 ) From (52) and (56) 𝐶 / 𝐵 1 + 𝐵 2 = 𝐶 / 𝐵 1 ∩ 𝐶 / 𝐵 2 , hence the result.

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