Int. J. Anal. Appl. (2022), 20:45 Bipolar Fuzzy Sublattices and Ideals U. Venkata Kalyani1,∗, T. Eswarlal1, J. KaviKumar2, A. Iampan3 1Department of Engineering Mathematics, College of Engineering, Koneru Lakshmaiah Education Foundation, Vaddeswaram, AP, India 2Department of Mathematics and Statistics, Faculty of Applied Sciences and Technology, Universiti Tun Hussein Onn Malaysia Campus, Pagoh 84600, Johor, Malaysia 3Department of Mathematics, School of Science, University of Phayao, Mae Ka, Mueang, Phayao 56000, Thailand ∗Corresponding author: u.v.kalyani@gmail.com Abstract. In this article, we introduce and study the theory of bipolar fuzzy sublattices (BFLs) and bipolar fuzzy ideals (BFIs) of a lattice, and some interesting properties of these BFLs and BFIs are established. Moreover, we study the properties of BFIs under lattice homomorphisms and also an application of BFLs. 1. Introduction Zadeh [4] introduced the concept of fuzzy sets (FSs) in 1965, and it has become a thriving area of research in a variety of fields. Following that, several researchers applied this concept to various algebraic structures. The fuzzy set theory has various expansions, such as vague sets (VSs), interval- valued fuzzy sets (IVFSs), intuitionistic fuzzy sets (IFS) and so on. The IFS was introduced by Atanassov [2] in 1986 as a generalization of the FS. In both the FS and IFS, the membership value range is in [0,1]. Later, Ajmal and Thomas [5] specifically applied the concept of FSs in lattice theory and developed the theory of fuzzy sublattices (FSLs). Thereafter, Thomas and Nair [3] introduced the concept of intuitionistic fuzzy sublattices (IFSLs) in 2011. Characterization of intuitionistic fuzzy ideals and filters based on lattice operations were studied by Milles [11] in 2017. Later, rough vague lattices were studied by Rao [13] in 2019. Vague lattices were introduced by Rao [12] in 2020. In Received: Jul. 29, 2022. 2010 Mathematics Subject Classification. 06D50, 06D72, 03E72. Key words and phrases. bipolar fuzzy set; bipolar fuzzy sublattice; bipolar fuzzy ideal; homomorphism. https://doi.org/10.28924/2291-8639-20-2022-45 ISSN: 2291-8639 © 2022 the author(s). https://doi.org/10.28924/2291-8639-20-2022-45 2 Int. J. Anal. Appl. (2022), 20:45 2020, Milles [8, 9] researched on the principal intuitionistic fuzzy ideals and filters on a lattice and the lattice of intuitionistic fuzzy topologies generated by intuitionistic fuzzy relations. Zhang [10] studied intuitionistic fuzzy filters on residuated lattices. Nowadays, bipolarity is playing a vital role in many areas. This has become a thriving area of research in many fields like artificial intelligence (AI), machine learning (ML) etc. Lee [1] introduced the concept of bipolar fuzzy sets (BFSs) in 2000, with membership values ranging from [-1,1]. Eswarlal and Kalyani [6,7] investigated bipolar vague cosets, homomorphism, and anti homomorphism in bipolar vague normal groups (BVNGs), and used bipolar vague sets (BVSs) to solve MCDM problems. In this paper, we introduce the concepts of BFLs and BFIs of a lattice. Some interesting character- izations and properties of these BFLs and BFIs are established. In addition, we study the properties of BFIs under lattice homomorphisms and also an application of BFLs. 2. Preliminaries Throughout this paper, unless otherwise stated, L always represents a lattice (L,∨,∧) and L1 represents a lattice (L1,∨,∧). Here, we will review a few standard definitions that are relevant to this work. Definition 2.1. [4] A mapping δ : Z → [0, 1] is represented as a fuzzy set (FS) in a non-empty set Z. Definition 2.2. [3] Let δ be a FS in L. Then δ is called a fuzzy sublattice (FL) of L if for all T ,k ∈ L, (i) δ(T ∨k) ≥ min{δ(T ),δ(k)}, (ii) δ(T ∧k) ≥ min{δ(T ),δ(k)}. Definition 2.3. [1] Suppose X is a universal set. A bipolar fuzzy set (BFS) Bδ in X is an object having the form Bδ = {< T ,BPδ (T ),B N δ (T ) >| T ∈ X} determined by a positive and a negative membership function, respectively, where BPδ : X → [0, 1] and B N δ : X → [−1, 0]. For convenience, the BFS Bδ is denoted by Bδ = (BPδ ,B N δ ). Definition 2.4. [1] Let Bδ and Bω be BFSs in a non-empty set X. (i) Bδ is a subset of Bω, denoted by Bδ ⊆ Bω, if for each T ∈ X, BPδ (T ) ≤ B P ω (T ) and BNδ (T ) ≥ BNω (T ). (ii) The complement of Bδ, denoted by Bcδ = ((B c δ ) P , (Bcδ ) N), is a BFS in X defined as: for each T ∈ X, Bcδ (T ) = (1−B P δ (T ),−1−B N δ (T )), i.e., (B c δ ) P (T ) = 1−BPδ (T ) and (B c δ ) N(T ) = −1−BNδ (T ). (iii) The intersection of Bδ and Bω, denoted by Bδ ∩ Bω, is a BFS in X defined as: for each T ∈ X, (Bδ ∩Bω)(T ) = (BPδ (T ) ∧B P ω (T ),BNδ (T ) ∨B N ω (T )). (iv) The union of Bδ and Bω, denoted by Bδ ∪Bω, is a BFS in X defined as: for each T ∈ X, (Bδ ∪ Bω)(T ) = (BPδ (T ) ∨B P ω (T ),BNδ (T ) ∧B N ω (T )). Int. J. Anal. Appl. (2022), 20:45 3 3. Bipolar fuzzy sublattices and ideals In this section, we introduce and study BFLs and BFIs and their characterizations. Theorem 3.1. Let Bδ = (BPδ ,B N δ ) be a BFS in L. Then for all T ,k ∈ L, the following conditions are equivalent: (i) T ≤ k ⇒ (BPδ (T ) ≥ B P δ (k),B N δ (T ) ≤ B N δ (k)), (ii) BPδ (T ∧k) ≥ max{B P δ (T ),B P δ (k)},B N δ (T ∧k) ≤ min{B N δ (T ),B N δ (k)}, (iii) BPδ (T ∨k) ≤ min{B P δ (T ),B P δ (k)},B N δ (T ∨k) ≥ max{B N δ (T ),B N δ (k)}. Proof. For any T ,k ∈ L, we have T ∧k ≤T and T ∧k ≤ k. Then from (i), we have BPδ (T ∧k) ≥ B P δ (T ),B N δ (T ∧k) ≤ B N δ (T ), BPδ (T ∧k) ≥ B P δ (k), and B N δ (T ∧k) ≤ B N δ (k). Thus BPδ (T ∧k) ≥ max{B P δ (T ),B P δ (k)} and B N δ (T ∧k) ≤ min{B N δ (T ),B N δ (k)}. Now for any T ,k ∈ L, we have T ≤T ∨k and k ≤T ∨k, using (i) we have BPδ (T ) ≥ B P δ (T ∨ k),B P δ (k) ≥ B P δ (T ∨ k),B N δ (T ) ≤ B N δ (T ∨ k), and B N δ (k) ≤ BNδ (T ∨k). Thus BPδ (T ∨k) ≤ min{B P δ (T ),B P δ (k)} and B N δ (T ∨k) ≥ max{B N δ (T ),B N δ (k)}. Hence, (ii) and (iii) are valid. Suppose that (ii) is true. Let T ,k ∈ L be such that T ≤ k. Then T ∧ k = T ⇒ BPδ (T ) = B P δ (T ∧ k) ≥ max{B P δ (T ),B P δ (k)} and B N δ (T ) = B N δ (T ∧ k) ≤ min{BNδ (T ),B N δ (k)}. Thus BPδ (T ) ≥ B P δ (k) and B N δ (T ) ≤ B N δ (k). Finally, suppose (iii) holds. Let T ,k ∈ L be such that T ≤ k. Then T ∨ k = k ⇒ BPδ (k) = B P δ (T ∧ k) ≤ min{B P δ (T ),B P δ (k)} and B N δ (T ) = B N δ (T ∨ k) ≥ max{BNδ (T ),B N δ (k)}. Thus BPδ (T ) ≥ B P δ (k) and B N δ (T ) ≤ B N δ (k). Hence, the proof is completed. � Similar to Theorem 3.1, we get the following theorem. Theorem 3.2. Let Bδ = (BPδ ,B N δ ) be a BFS in L. Then for all T ,k ∈ L, the following conditions are equivalent: (i) T ≤ k ⇒ BPδ (T ) ≤ B P δ (k),B N δ (T ) ≥ B N δ (k), (ii) BPδ (T ∧k) ≤ min{B P δ (T ),B P δ (k)},B N δ (T ∧k) ≥ max{B N δ (T ),B N δ (k)}, (iii) BPδ (T ∨k) ≥ max{B P δ (T ),B P δ (k)},B N δ (T ∨k) ≤ min{B N δ (T ),B N δ (k)}. Definition 3.1. Let Bδ = (BPδ ,B N δ ) be a BFS in L. Then Bδ is called a bipolar fuzzy sublattice (BFL) of L if the following conditions are satisfied for all T ,k ∈ L, (i) BPδ (T ∨k) ≥ min{B P δ (T ),B P δ (k)}, 4 Int. J. Anal. Appl. (2022), 20:45 (ii) BPδ (T ∧k) ≥ min{B P δ (T ),B P δ (k)}, (iii) BNδ (T ∨k) ≤ max{B N δ (T ),B N δ (k)}, (iv) BNδ (T ∧k) ≤ max{B N δ (T ),B N δ (k)}. Example 3.1. Consider the lattice L of "divisors of 10". Then L = {1, 2, 5, 10}. Let Bδ = (BPδ ,B N δ ) be given by < 1, 0.6,−0.4 >,< 2, 0.1,−0.5 >,< 5, 0.3,−0.4 >,< 10, 0.2,−0.7 > . We can routinely prove that Bδ is a BFL of L. Definition 3.2. Let Bδ = (BPδ ,B N δ ) be a BFS in L. Then Bδ is called a bipolar fuzzy ideal (BFI) of L if the following conditions are satisfied for all T ,k ∈ L, (i) BPδ (T ∨ k) ≥ min{B P δ (T ),B P δ (k)}, (ii) B P δ (T ∧ k) ≥ max{B P δ (T ),B P δ (k)}, (iii) B N δ (T ∨ k) ≤ max{BNδ (T ),B N δ (k)}, (iv) B N δ (T ∧k) ≤ min{B N δ (T ),B N δ (k)}. Example 3.2. Consider the lattice L in Example 3.1. Let Bδ = (BPδ ,B N δ ) be given by < 1, 0.7,−0.5 >,< 2, 0.5,−0.3 >,< 5, 0.6,−0.5 >,< 10, 0.4,−0.2 > . We can routinely prove that Bδ is a BFI of L. Theorem 3.3. If J and M are two BFLs (BFIs) of a lattice L, then J ∩M is a BFL (BFI) of L. Proof. Let J = (δPJ ,δ N J ) and M = (δ P M,δ N M) be two BFLs of L. Now, δPJ∩M(T ∨k) = min{δ P J (T ∨k),δ P M(T ∨k)} ≥ min{min{δPJ (T ),δ P M(k)}, min{δ P J (T ),δ P M(k)} = min{min{δPJ (T ),δ P M(T )}, min{δ P J (k),δ P M(k)} = min{δPJ∩M(T ),δ P J∩M(k)}. Thus δPJ∩M(T ∨k) ≥ min{δ P J∩M(T ),δ P J∩M(k)} for all T ,k ∈ L. Similarly, we get δPJ∩M(T ∧k) ≥ min{δ P J∩M(T ),δ P J∩M(k)} for all T ,k ∈ L. Now, δNJ∩M(T ∨k) = max{δ N J (T ∨k),δ N M(T ∨k)} ≤ max{max{δNJ (T ),δ N M(k)}, max{δ N J (T ),δ N M(k)} = max{max{δNJ (T ),δ N M(T )}, max{δ N J (k),δ N M(k)} = max{δNJ∩M(T ),δ N J∩M(k)}. Int. J. Anal. Appl. (2022), 20:45 5 Thus δNJ∩M(T ∨k) ≤ max{δ N J∩M(T ),δ N J∩M(k)} for all T ,k ∈ L. Similarly, we get δNJ∩M(T ∧k) ≤ max{δ N J∩M(T ),δ N J∩M(k)} for all T ,k ∈ L. Hence, J ∩M is a BFL of L. Similarly, we can prove that J ∩M is a BFI of L if J and M are BFIs of L. � Example 3.3. Consider the lattice L given in Example 3.1. If J = {< 1, 0.7,−0.3 >,< 2, 0.4,−0.4 >,< 5, 0.1,−0.3 >,< 10, 0.2,−0.5 >} and M = {< 1, 0.6,−0.4 >,< 2, 0.1,−0.5 >,< 5, 0.3,−0.4 >,< 10, 0.2,−0.7 >} are two BFLs of L, then J ∩M = {< 1, 0.6,−0.3 >,< 2, 0.1,−0.4 >,< 5, 0.1,−0.3 >,< 10, 0.2,−0.5 >} is a BFL of L. Example 3.4. Consider the lattice L given in Example 3.1. If J = {< 1, 0.7,−0.5 >,< 2, 0.5,−0.6 >,< 5, 0.6,−0.5 >,< 10, 0.5,−0.6 >} and M = {< 1, 0.7,−0.3 >,< 2, 0.4,−0.2 >,< 5, 0.2,−0.1 >,< 10, 0.2,−0.1 >} are two BFIs of L, then J ∩M = {< 1, 0.7,−0.3 >,< 2, 0.4,−0.2 >,< 5, 0.2,−0.1 >,< 10, 0.2,−0.1 >} is a BFI of L. Remark 3.1. The union of two BFLs of a lattice L need not be a BFL. Consider the lattice L given in Example 3.1. If J = {< 1, 0.7,−0.3 >,< 2, 0.4,−0.4 >,< 5, 0.1,−0.3 >,< 10, 0.2,−0.5 >} and M = {< 1, 0.6,−0.4 >,< 2, 0.1,−0.5 >,< 5, 0.3,−0.4 >,< 10, 0.2,−0.7 >} are two BFLs of L, then J ∪M = {< 1, 0.7,−0.4 >,< 2, 0.4,−0.5 >,< 5, 0.3,−0.4 >,< 10, 0.2,−0.7 >}. Here, δPJ∪M(2∨5) = δ P J∪M(10) = 0.2 � 0.3 = min{0.4, 0.3} = min{δ P J∪M(2),δ P J∪M(5)}. Hence, J∪M is not a BFL of L. 6 Int. J. Anal. Appl. (2022), 20:45 Remark 3.2. Every BFI of L is a BFL, but the converse need not be true. Consider the lattice L given in Example 3.1. Then the BFI given by J = {< 1, 0.7,−0.5 >,< 2, 0.5,−0.6 >,< 5, 0.6,−0.5 >,< 10, 0.5,−0.6 >} is a BFL of L. But the BFL given by M = {< 1, 0.5,−0.3 >,< 2, 0.4,−0.4 >,< 5, 0.4,−0.3 >,< 10, 0.7,−0.5 >} is not a BFI of L as δPM(2 ∧ 10) = 0.4 � 0.7 = max{δ P M(2),δ P B(10)}. Remark 3.3. The union of two BFIs of a lattice L need not be a BFI. Consider the lattice L given in Example 3.1. If J = {< 1, 0.7,−0.5 >,< 2, 0.5,−0.6 >,< 5, 0.6,−0.5 >,< 10, 0.5,−0.6 >} and M = {< 1, 0.7,−0.3 >,< 2, 0.4,−0.2 >,< 5, 0.2,−0.1 >,< 10, 0.2,−0.1 >} are two BFIs of L, then J ∪M = {< 1, 0.7,−0.5 >,< 2, 0.5,−0.6 >,< 5, 0.6,−0.5 >,< 10, 0.5,−0.6 >}. Here, δNJ∪M(2 ∧ 5) = δ N J∪M(1) = −0.5 −0.6 = {−0.6,−0.5} = min{δ N J∪M(2),δ N J∪M(5)}. Hence, J ∪M is not a BFI of L. Theorem 3.4. Let Bδ = (BPδ ,B N δ ) be a BFL of L. Then for all T ,k ∈ L, the following four statements hold: (i) BPδ (T ∧k) ≥ max{B P δ (T ),B P δ (k)}⇔ (T ≤ k ⇒ B P δ (T ) ≥ B P δ (k)), (ii) BPδ (T ∨k) ≥ max{B P δ (T ),B P δ (k)}⇔ (T ≤ k ⇒ B P δ (T ) ≤ B P δ (k)), (iii) BNδ (T ∧k) ≤ min{B N δ (T ),B N δ (k)}⇔ (T ≤ k ⇒ B N δ (T ) ≤ B N δ (k)), (iv) BNδ (T ∨k) ≤ min{B N δ (T ),B N δ (k)}⇔ (T ≤ k ⇒ B N δ (T ) ≥ B N δ (k)). Proof. Let T ,k ∈ L. (i) Suppose BPδ (T ∧k) ≥ max{B P δ (T ),B P δ (k)}. If T ≤ k, then T ∧k = T . Since B P δ (T ∧k) ≥ max{BPδ (T ),B P δ (k)}, we have B P δ (T ) = B P δ (T ∧ k) ≥ max{B P δ (T ),B P δ (k)}. Hence, B P δ (T ) ≥ BPδ (k). Conversely, suppose (T ≤ k ⇒ BPδ (T ) ≥ B P δ (k)). Then B P δ (T ∧k) ≥ B P δ (T ) and B P δ (T ∧k) ≥ BPδ (k). Hence, B P δ (T ∧k) ≥ max{B P δ (T ),B P δ (k)}. (ii) Suppose BPδ (T ∨k) ≥ max{B P δ (T ),B P δ (k)}. If T ≤ k, then T ∨k = k. Since B P δ (T ∨k) ≥ max{BPδ (T ),B P δ (k)}, we have B P δ (k) = B P δ (T ∨ k) ≥ max{B P δ (T ),B P δ (k)}. Hence, B P δ (T ) ≤ BPδ (k). Conversely, suppose (T ≤ k ⇒ BPδ (T ) ≤ B P δ (k)). Then B P δ (T ) ≤ B P δ (T ∨ k) and B P δ (k) ≤ BPδ (T ∨k). Hence, B P δ (T ∨k) ≥ max{B P δ (T ),B P δ (k)}. Int. J. Anal. Appl. (2022), 20:45 7 (iii) Suppose BNδ (T ∧k) ≤ min{B N δ (T ),B N δ (k)}. If T ≤ k, then T ∧k = T . Since B N δ (T ∧k) ≤ min{BNδ (T ),B N δ (k)}, we have B N δ (T ) = B N δ (T ∧ k) ≤ min{B N δ (T ),B N δ (k)}. Hence, B N δ (T ) ≤ BNδ (k). Conversely, suppose (T ≤ k ⇒ BNδ (T ) ≤ B N δ (k)). Then B N δ (T ∧k) ≤ B N δ (T ) and (B N δ (T ∧k) ≤ BNδ (k). Hence, B N δ (T ∧k) ≤ min{B N δ (T ),B N δ (k)}. (iv) Suppose BNδ (T ∨k) ≤ min{B N δ (T ),B N δ (k)}. If T ≤ k, then T ∨k = k. Since (B N δ (T ∨k) ≤ min{BNδ (T ),B N δ (k)), we have B N δ (k) = B N δ (T∨k) ≤ min{B N δ (T ),B N δ (k)}. Hence, B N δ (T ) ≥ B N δ (k). Conversely, suppose (T ≤ k ⇒ BNδ (T ) ≥ B N δ (k)). Then B N δ (T ) ≥ (B N δ (T ∨ k) and (B N δ (k) ≥ BNδ (T ∨k) ≥ B N δ (k). Hence, B N δ (T ∨k) ≥ min{B N δ (T ),B N δ (k)}. � Theorem 3.5. Let Bη = (BPη ,B N η ) be a BFL of L. Then Bη is a BFI of L if and only if the following two conditions are satisfied for all T ,k ∈ L, (i) BPη (T ∨k) = min{BPη (T ),BPη (k)}, (ii) BNη (T ∨k) = max{BNη (T ),BNη (k)}. Proof. Suppose that Bη is a BFI of L. Let T ,k ∈ L. Then BPη (T ∨k) ≥ min{BPη (T ),BPη (k)} and BNη (T ∨k) ≤ max{BNη (T ),BNη (k)}. Since T ≤T ∨k and k ≤T ∨k, then by Theorem 3.4, we have BPη (T ) ≥ BPη (T ∨ k) and BPη (k) ≥ BPη (T ∨ k). Hence, min{BPη (T ),BPη (k)} ≥ BPη (T ∨ k). Thus BPη (T ∨k) = min{BPη (T ),BPη (k)}. Now, since T ≤T ∨k and k ≤T ∨k, then by Theorem 3.4, we have BNη (T ) ≤ BNη (T ∨ k) and BNη (k) ≤ BNη (T ∨ k). Hence, max{BNη (T ),BNη (k)} ≤ BNη (T ∨ k). Thus BNη (T ∨k) = max{BNη (T ),BNη (k)}. Conversely, suppose that BPη (T∨k) = min{BPη (T ),BPη (k)} and BNη (T∨k) = max{BNη (T ),BNη (k)} for any T ,k ∈ L. Then it is clear that BPη (T ∨ k) ≥ min{BPη (T ),BPη (k)} and BNη (T ∨ k) ≤ max{BNη (T ),BNη (k)} for any T ,k ∈ L. Next, we shall show that BPη (T ∧k) ≥ max{BPη (T ),BPη (k)} and BNη (T ∧k) ≤ min{BNη (T ),BNη (k)} for any T ,k ∈ L. Let T ,k ∈ L. Since T ∨ (T ∧k) = T and k ∨ (T ∧k) = k, we have BPη (T ∨ (T ∧k)) = BPη (T ) and BPη (k ∨ (T ∧k)) = BPη (k). Thus min{BPη (T ),BPη (T ∧k)} = BPη (T ) and min{BPη (k),BPη (T ∧ k)} = BPη (k), hence, BPη (T ∧ k) ≥ BPη (T ) and BPη (T ∧ k) ≥ BPη (k). Therefore, BPη (T ∧ k) ≥ max{BPη (T ),BPη (k)} for any T ,k ∈ L. Let T ,k ∈ L. Since T ∨ (T ∧k) = T and k ∨ (T ∧k) = k, we have BNη (T ∨ (T ∧k)) = BNη (T ) and BNη (k ∨ (T ∧k)) = BNη (k). Thus max{BNη (T ),BNη (T ∧k)} = BNη (T ) and max{BNη (k),BNη (T ∧ k)} = BNη (k). Hence, BNη (T ∧ k) ≤ BNη (T ) and BNη (T ∧ k) ≥ BNη (k). Therefore, BNη (T ∧ k) ≤ min{BNη (T ),BNη (k)} for any T ,k ∈ L. Hence, Bη is a BFI of L. � 8 Int. J. Anal. Appl. (2022), 20:45 4. Bipolar fuzzy ideals under lattice homomorphisms Definition 4.1. Let θ : L → L1 be a mapping and Bδ = (BPδ ,B N δ ) be a BFS in L. Then the image θ(Bδ) is defined as θ(Bδ) = {< k,θ(BPδ )(k),θ(B N δ )(k) >| k ∈ L 1}, θ(BPδ )(k) = { sup{BPδ (T ) | T ∈ θ −1(k)} if θ−1(k) 6= ∅, 0 if otherwise and θ(BNδ )(k) = { inf{BNδ (T ) | T ∈ θ −1(k)} if θ−1(k) 6= ∅, 0 if otherwise. Similarly, if Bη = (BPη ,B N η ) be a BFS in L 1, then θ−1(Bη) = {< T ,θ−1(BPη (T )),θ−1(BNη (T )) >| t ∈ L}, where θ−1(BPη (T )) = BPη (θ(T )) and θ−1(BNη (T )) = BNη (θ(T )). Theorem 4.1. Let θ : L → L1 be an epimorphism. If Bδ is a BFI of L, then θ(Bδ) is a BFI of L1. Proof. Let Bδ = (BPδ ,B N δ ) be a BFI of L. Let s,w ∈ L 1. Then θ(BPδ )(s ∨w) = sup{B P δ (T ) : T ∈ θ −1(s ∨w)} ≥ sup{BPδ (h∨k) | h ∈ θ −1(s),k ∈ θ−1(w)} ≥ sup{min{BPδ (h),B P δ (k)} | h ∈ θ −1(s),k ∈ θ−1(w)} = min{sup{BPδ (h) | h ∈ θ −1(s)}, sup{BPδ (k) | k ∈ θ −1(w)}} = min{θ(BPδ )(s),θ(B P δ )(w)}, θ(BPδ )(s ∧w) = sup{B P δ (T ) | T ∈ θ −1(s ∧w)} ≥ sup{BPδ (h∧k) | h ∈ θ −1(s),k ∈ θ−1(w)} ≥ sup{max{BPδ (h),B P δ (k)} | h ∈ θ −1(s),k ∈ θ−1(w)} = max{sup{BPδ (h) | h ∈ θ −1(s)}, sup{BPδ (k) | k ∈ θ −1(w)}} = max{θ(BPδ )(s),θ(B P δ )(w)}, θ(BNδ )(s ∨w) = inf{B N δ (T ) | T ∈ θ −1(s ∨w)} ≤ inf{BNδ (h∨k) | h ∈ θ −1(s),k ∈ θ−1(w)} ≤ inf{max{BNδ (h),B N δ (k)} | h ∈ θ −1(s),k ∈ θ−1(w)} = max{inf{BNδ (h) | h ∈ θ −1(s)}, inf{BNδ (k) | k ∈ θ −1(w)}} = max{θ(BNδ )(s),θ(B N δ )(w)}, Int. J. Anal. Appl. (2022), 20:45 9 and θ(BNδ )(s ∧w) = inf{B N δ (T ) | T ∈ θ −1(s ∧w)} ≤ inf{BNδ (h∧k) | h ∈ θ −1(s),k ∈ θ−1(w)} ≤ inf{min{BNδ (h),B N δ (k)} | h ∈ θ −1(s),k ∈ θ−1(w)} = min{inf{BNδ (h) | h ∈ θ −1(s)}, inf{BNδ (k) | k ∈ θ −1(w)}} = min{θ(BNδ )(s),θ(B N δ )(w)}. Hence, θ(Bδ) is a BFI of L1. � . Theorem 4.2. Let θ : L → L1 be a homomorphism. If Bη is a BFI of L1, then θ−1(Bη) is a BFI of L. Proof. Let Bη = (BPη ,B N η ) be a BFI of L 1. Let T ,k ∈ L. Then θ−1(BPη )(T ∨k) = B P η (θ(T ∨k)) = BPη {(θ(T ) ∨θ(k)} ≥ min{BPη (θ(T )),B P η (θ(k))} = min{θ−1(BPη )(T ),θ −1(BPη )(k)}, θ−1(BPη )(T ∧k) = B P η (θ(T ∧k)) = BPη {(θ(T ) ∧θ(k)} ≥ max{BPη (θ(T )),B P η (θ(k))} = max{θ−1(BPη )(T ),θ −1(BPη )(k)}, θ−1(BNη )(T ∨k) = B N η (θ(T ∨k)) = BNη {(θ(T ) ∨θ(k)}} ≤ max{BNη (θ(T )),B N η (θ(k)) = max{θ−1(BNη )(T ),θ −1(BNη )(k)}, and θ−1(BNη )(T ∧k) = B N η (θ(T ∧k)) = BNη {(θ(T ) ∧θ(k)} ≤ min{BNη (θ(T )),B N η (θ(k))} = min{θ−1(BNη )(T ),θ −1(BNη )(k)}. 10 Int. J. Anal. Appl. (2022), 20:45 Hence, θ−1(Bη) is a BFI of L. � Theorem 4.3. Let θ : L → L1 be a homomorphism and let µ and η be BFLs of L and L1, respectively. Then (i)θ(µ) is a BFL of L1, (ii) θ−1(η) is a BFL L. Proof. The proof is omitted since it follows the same proof of Theorems 4.1 and 4.2. � Theorem 4.4. If θ : L → L1 is an surjection and Bη,Bδ are BFSs of L and L1, respectively, then (i) θ[θ−1(Bδ)] = Bδ, (ii) Bη ⊆ θ−1[θ(Bη)]. Proof. (i) Let α ∈ L1. Then θ[θ−1(BPδ )](α) = sup{θ −1(BPδ )(γ) | γ ∈ θ −1(α)} = sup{BPδ (θ(γ)) | γ ∈ L,θ(γ) = α} = BPδ (α) because θ is onto, for every α ∈ L 1, there exists γ in L such that θ(γ) = α. Similarly, θ[θ−1(BNδ )](α) = B N δ (α). Hence, θ[θ −1(Bδ)] = Bδ. (ii) Let γ ∈ L. Then θ−1[θ(BPη )](γ) = θ(BPη )(θ(γ)) = sup{BPη (γ) | γ ∈ θ−1[θ(γ)]} ≥ BPη (γ) and θ−1[θ(BNη )](γ) = θ(B N η )(θ(γ)) = inf{BNη (γ) | γ ∈ θ−1[θ(γ)]} ≤ BNη (γ). Hence, Bη ⊆ θ−1[θ(Bη)]. � Definition 4.2. Let f : L → L1 be a surjection and Bδ = (BPδ ,B N δ ) be a BFS in L. Then Bδ is said to be f -invariant if for any w,s ∈ L1 such that f (w) = f (s) implies BPδ (w) = B P δ (s) and BNδ (w) = B N δ (s). From Theorem 4.4 and Definition 4.2, we have the following theorem. Theorem 4.5. Let f : L → L1 be a surjection and Bδ = (BPδ ,B N δ ) be a BFS in L. If a BFS Bδ is f -invariant, then f−1(f (Bδ)) = Bδ. Theorem 4.6. Let f : L → L1 be a surjection, Bδ and Bη be BFSs of L, and B1δ and B 1 η be BFSs of L1. Then (i) Bδ ⊆ Bη ⇒ f (Bδ) ⊆ f (Bη), (ii) B1δ ⊆ B 1 η ⇒ f−1(B1δ ) ⊆ f −1(B1η). Proof. Let Bδ = (BPδ ,B N δ ) and Bη = (B P η ,B N η ) be BFSs in L such that Bδ ⊆ Bη. Then BPδ ≤ B P η and B N δ ≥ B N η . Also, f (Bδ) = {< t,f (BPδ )(t), f (B N δ )(t) >| t ∈ L 1} and f (Bη) = {< t,f (BPη )(t), f (BNη )(t) >| t ∈ L1}. Now, for any t ∈ L, we have f (BPδ )(t) = sup{(B P δ (k) | k ∈ f−1(t))} ≤ sup{(BPη (k) | k ∈ f−1(t))} = f (BPη )(t) and f (BNδ )(t) = inf{(B N δ (k) | k ∈ f −1(t))} ≤ sup{(BNη (k) | k ∈ f−1(t))} = f (BNη )(t). Hence, f (Bδ) ⊆ f (Bη). Similarly, we can prove that B1δ ⊆ B 1 η ⇒ f−1(B1δ ) ⊆ f −1(B1η). � Theorem 4.7. If f : L → L1 is an epimorphism, then there is one to one order preserving correspon- dence between the BFIs of L1 and those of L which are f -invariant. Int. J. Anal. Appl. (2022), 20:45 11 Proof. Let B(L1) denote the set of all BFIs of L1 and B(L) denote the set of all BFIs of L which are f -invariant. Define ς : B(L) → B(L1) and Ψ : B(L1) → B(L) such that ς(Bδ) = f (Bδ) and Ψ(B1δ ) = f −1(B1δ ). By Theorems 4.1 and 4.2, we have ς and Ψ are well-defined. Also by Theorems 4.4 and 4.5, we have ς and Ψ are the inverse to each other which gives that the one- to-one correspondence. Also by Theorem 4.6, we get Bδ ⊆ Bη ⇒ f (Bδ) ⊆ f (Bη). Hence, the correspondence is order preserving. � 5. An application of bipolar fuzzy sublattices The single pattern: the one-minute microwave [15]. The one-minute microwave is a simple system with the following requirements: 1. There is a single button available for the user. 2. If the door is closed and the button is pushed, the oven will be energized for one minute. 3. If the button is pushed while the oven is energized, the cooking time is increased by one minute. 4. If the door is open, pushing the button has no effect. 5. The oven has a light that is turned on when the door is open, and also when the oven is cooking. Otherwise, the light is off. 6. Opening the door stops the cooking and clears the timer (i.e., the remaining cooking time is set to zero). 7. When the cooking is complete (oven times out) a beeper sounds and the light is turned off. Here in this application, we consider the one-minute microwave as a lattice L = {button, timer, oven-door} with the operations ON and OFF. The final output will be cooking the food or not cooking the food. When we consider the button there may be two cases that is the button may be pressed or unpressed. When we consider the timer the cases will be the timer may be initiated or uninitiated. When we consider the oven-door then there may be a chance that the door is closed or open. To check whether it forms a bipolar fuzzy lattice first let us know what the possible cases arise. The cases will be like, oven-door closed and button pressed, oven-door closed and button unpressed, button pressed and timer initiated, timer initiated and oven-door opened, timer initiated and oven-door closed, oven-door closed and button unpressed etc. We shall represent B for the button, T for timer, D for oven-door, and join operator as ‘ON’ and meet operator as ‘OFF’. If we take the operator ‘ON’ between B and T then it is considered as a button pressed and timer initiated. Then cooking will be done. So, the important thing here in this case is pushing the button. So B ∨T = B. If we take the operator ‘OFF’ between B and T then it is considered the button is unpressed. In this case, there is no question about whether the timer is initiated or not. So B∧T = T. Similarly, if we take the operator ‘ON’ between B and D then it is considered the button is pressed and the door is closed. Then cooking will be done. So, B ∨D = B. If we take the operator ‘OFF’ 12 Int. J. Anal. Appl. (2022), 20:45 between B and D then it is considered the button is unpressed. In this case, there is no question about whether the timer is initiated or not. So, B ∧D = D. Similarly, if we take the operator ‘ON’ between T and D then it is considered the timer is initiated and oven-door is closed. Then cooking will be done. So, T ∨D = T . If we take the operator ‘OFF’ between T and D then it is considered the timer is uninitiated and door is open. So, T ∧D = B (here the minimum considered to be B, because the possibility that the timer is uninitiated is that the button is unpressed). Let us consider a BFS in L, as Bδ = {(B, 0.5,−0.5), (T, 0.4,−0.6)(D, 0.3,−0.5)}. 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Fuzzy Syst. 16 (2019), 75-88. https://doi.org/10.22111/ijfs.2019.4485. https://digitalcommons.pvamu.edu/aam/vol15/iss2/13 https://doi.org/10.1007/s00500-018-3647-2 https://doi.org/10.5899/2017/jfsva-00399 https://doi.org/10.1016/j.fss.2016.11.014 https://doi.org/10.22111/ijfs.2019.4485 1. Introduction 2. Preliminaries 3. Bipolar fuzzy sublattices and ideals 4. Bipolar fuzzy ideals under lattice homomorphisms 5. An application of bipolar fuzzy sublattices 6. Conclusion and future work References