# On Rough Hyperideals in Hyperlattices

Authors: Pengfei He; Xiaolong Xin; Jianming Zhan
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We introduce and study rough hyperideals in hyperlattices. First, we give some interesting examples of hyperlattices and introduce hyperideals of hyperlattices. Then, applying the notion of rough sets to hyperlattices, we introduce rough hyperideals in hyperlattices, which are extended notions of hyperideals of hyperlattices. In addition, we consider rough hyperideals in Cartesian products and quotients of hyperlattices. Finally, we investigate some properties about homomorphic images of rough hyperideals in hyperlattices.


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## Body

## 1. Introduction

In applied mathematics, we encounter many examples of mathematical objects that can be added to each other and multiplied by scalar numbers. First of all, the real numbers themselves are such objects. Other examples are real-valued functions, the complex numbers, infinite series, vectors inn-dimensional space, and vector valued functions. Sometimes the sum of two elements is not an element. There are many examples in chemistry where the sum of two elements is a set of elements. In this case we have a hyperstructure. The concept of hyperstructure was introduced in 1934 by a French mathematician, Marty [1]. Algebraic hyperstructures are suitable generalizations of classical algebraic structures. In a classical algebraic structure, the composition of two elements is an element, while in an algebraic hyperstructure, the composition of two elements is a set. Since then, there appeared many components of hyperalgebras such as hypergroups in [2] and hyperrings in [3]. Moreover, Konstantinidou and Mittas introduced the concept of hyperlattices in [4] and superlattices in [5]; also see [6–8]. In particular, Rasouli and Davvaz further studied the theory of hyperlattices and obtained some interesting results [9, 10], which enriched the theory of hyperlattices.Recently, a number of different hyperstructures are widely studied from the theoretical point of view and for their applications to many subjects of pure and applied mathematics by many mathematicians. Also, a recent book [11] contains a wealth of applications on geometry, binary relations, lattices, fuzzy sets and rough sets, automata, combinatorics, codes, artificial intelligence, and probabilistic. Another book [12] is devoted especially to the study of hyperring theory, written by Davvaz and Leoreanu-Fotea. Several kinds of hyperrings are introduced and analyzed. The volume ends with an outline of applications in chemistry and physics, analyzing several special kinds of hyperstructures: e-hyperstructures and transposition hypergroups. The theory of suitable modified hyperstructures can serve as a mathematical background in the field of quantum communication systems.The theory of rough sets was introduced by Pawlak [13] to deal with uncertain knowledge in information systems. It is an expanding research area which stimulates explorations on both real-world applications and on the theory itself. Rough set theory is an extension of set theory, in which a subset of a universe is described by a pair of ordinary sets called the lower and upper approximations. It is a natural question to ask what happens if we substitute an algebraic system with the universe set. Some authors studied algebraic properties of rough sets. Since Biswas and Nanda [14] applied the notion of rough sets to algebra and introduced the notion of rough subgroups, Davvaz et al. have been engaged in extending concepts and methods of rough set theory to various algebraic structures [15–21]. With the development of the hyperstructure theory, Leoreanu-Fotea et al. attached importance to the connections among rough sets, fuzzy sets, and algebraic hyperstructures in [22–30]. These not only enriched the theory of rough sets but also provided new ideas in the study of pure algebra and algebraic hyperstructures.The combination of rough set theory and algebraic systems may provide more new interesting research topics, which have drawn attention of many mathematicians and computer scientists. One can introduce roughness into an algebraic system and investigate algebraic properties of various rough objects. In this paper, in order to broaden application fields of the theory of rough sets and hyperstructures, we introduce the rough set theory into hyperlattices. We introduce rough hyperideals in hyperlattices, which are extended notions of hyperideals in hyperlattices. And we study some properties about rough hyperideals in hyperlattices.


## 2.   Hyperideals in Hyperlattices

In this section, we recall the notion of hyperlattices and give several new examples of it. Moreover, we will introduce hyperideals in hyperlattices and discuss some basic properties of them, which will be used in the following paragraphs.LetL be a nonempty set, and let P*(L) be the set of all nonempty subsets of L. A hyperoperation on L is a map ∘:L×L→P*(L), which associates a nonempty subset a∘b with any pair (a,b) of elements of L×L. The couple (L,∘) is called a hypergroupoid.IfA and B are nonempty subsets of L, for all a,b,x∈L, we denote(1)
x∘A={x}∘A=⋃a∈Ax∘a, A∘x=A∘{x}=⋃a∈Aa∘x;(2)
A∘B=⋃a∈A,b∈Ba∘b.In what follows, let us see what a hyperlattice is.There are several kinds of hyperlattices that can be defined on a nonempty set; see [4–8]. Throughout the paper, we shall consider one of general types of hyperlattices [8]; also see [4].Definition 1 (see [8]).
LetL be a nonempty set endowed with two hyperoperations ⊗ and ⊕. The triple (L,⊗,⊕) is called a hyperlattice if the following conditions hold: for all a,b,c∈L, (1)
a∈a⊗a, a∈a⊕a;(2)
a⊗b=b⊗a, a⊕b=b⊕a;(3)
(a⊗b)⊗c=a⊗(b⊗c), (a⊕b)⊕c=a⊕(b⊕c);(4)
a∈a⊗(a⊕b), a∈a⊕(a⊗b).Let(L,∧,∨) be a lattice. Define hyperoperations “⊗” and “⊕” on L as follows: for all a,b∈L, a⊗b={a∧b}, a⊕b={a∨b}, then (L,⊗,⊕) is a hyperlattice. From this, we can see that hyperlattices are suitable generalizations of lattices.Now, we give some new examples of hyperlattices. From these examples, we can see that hyperlattices are connected to several domains of mathematics.Example 2.
Let(L,≤) be a partially ordered set. Define the following hyperoperations on L: for all a,b∈L, a⊗b={x∈L∣x≤a,x≤b}, a⊕b={x∈L∣a≤x,b≤x}. Then (L,⊗,⊕) is a hyperlattice.Example 3.
LetSub(V) be the set of all subspaces of n-dimensional vectors space V. Define hyperoperations ⊗ and ⊕ on Sub(V) as follows: for all V1,V2∈Sub(V), V1⊗V2=Sub(V1∩V2), V1⊕V2=Sub(V1+V2), where V1+V2 represents the sum space of V1 and V2. One can check that (Sub(V),⊗,⊕) is a hyperlattice.Example 4.
Let(L,∧,∨) be a lattice. Define the following hyperoperations on L: for all a,b∈L, a⊗b={x∈L∣a∨x=b∨x=a∨b}, a⊕b={x∈L∣x≤a∧b}. Then (L,⊗,⊕) is a hyperlattice.Example 5.
Let(L,∧,∨) be a lattice. We define two hyperoperations on L: for all a,b∈L, a⊗b={x∈L∣a∨b≤x}, a⊕b={x∈L∣a∧x=b∧x=a∧b}. Then (L,⊗,⊕) is also a hyperlattice.Example 6.
LetN+ be the set of all positive integers. We define hyperoperations ⊗ and ⊕ on N+ as follows: for all a,b∈N+, a⊗b={x∈N+∣x∣(a,b)}, a⊕b={x∈N+∣x∣[a,b]}, in which a∣b represents that a divides b, (a,b)=gcd{a,b} and [a,b]=lcm{a,b}. We can check that (N+,⊗,⊕) is a hyperlattice.Definition 7 (see [8]).
Let(L,⊗,⊕) be a hyperlattice. A nonempty subset A of L is called a subhyperlattice of L if (A,⊗,⊕) is itself a hyperlattice.It is easy to see that a nonempty subsetA of (L,⊗,⊕) is a subhyperlattice of L if and only if A holds: for all a,b∈A, a⊗b∈P*(A), a⊕b∈P*(A). That is to say, A is a subhyperlattice of (L,⊗,⊕) if and only if A⊗A⊆A, A⊕A⊆A.Example 8.
LetL be a nonempty set. Define hyperoperations on L as follows: for all a,b∈L, a⊗b={a,b}, a⊕b={a,b}. Then (L,⊗,⊕) is a hyperlattice. Each nonempty subset of L is a subhyperlattice of (L,⊗,⊕).Example 9 (see [8]).
LetL={a,b,c,d}, and let hyperoperations ⊗ and ⊕ on L be defined as follows:Then(L,⊗,⊕) is a hyperlattice, in which {a,b} and {c,d} are subhyperlattices.In what follows, we introduce hyperideals of a hyperlattice.Definition 10.
Let(L,⊗,⊕) be a hyperlattice, and let A be a nonempty subset of L.(1)
A is called a ⊕-hyperideal of L if for all a,b∈A and x∈L,
(i)
a⊗b⊆A,(ii)
a⊕x⊆A.(2)
A is called a ⊗-hyperideal of L if for all a,b∈A and x∈L,
(i)
a⊕b⊆A,(ii)
a⊗x⊆A.Obviously, a subhyperlatticeA of (L,⊗,⊕) is a ⊗-hyperideal of L if and only if A⊗L⊆A. Similarly, a subhyperlattice A of (L,⊗,⊕) is a ⊕-hyperideal of L if and only if A⊕L⊆A.Now, we present some examples of⊕-hyperideals and ⊗-hyperideals of hyperlattices.Example 11.
Let(L,∧,∨) be a lattice. Define the hyperoperations ⊗ and ⊕ on L as follows: for all a,b∈L, a⊗b={a∧b}, a⊕b={a∨b}, then (L,⊗,⊕) is a hyperlattice. Every ideal and filter of the lattice (L,∧,∨) are ⊗-hyperideal and ⊕-hyperideal of the hyperlattice (L,⊗,⊕), respectively.From the previous example, we can see that⊗-hyperideals and ⊕-hyperideals of hyperlattices are suitable generalizations of ideals and filters of lattices, respectively.Example 12.
Let(L,⊗,⊕) be a hyperlattice in Example 4. For any element a of the lattice L, denote the principal ideal generated by a of the lattice (L,∧,∨) by I(a), which means that I(a)={x∈L∣x≤a}, then it is easy to check that I(a) is a ⊕-hyperideal of the hyperlattice (L,⊗,⊕).Example 13.
Let(L,⊗,⊕) be a hyperlattice in Example 5. For any element a of the lattice L, denote the principal filter generated by a of the lattice (L,∧,∨) by F(a), which means that F(a)={x∈L∣x≥a}, then F(a) is a ⊗-hyperideal of the hyperlattice (L,⊗,⊕).Example 14.
Let(L,⊗,⊕) be the hyperlattice in Example 9. One can check that {a,b} is a ⊗-hyperideal, but not a ⊕-hyperideal of L and {c,d} is a ⊕-hyperideal, but not a ⊗-hyperideal of L.Next, we discuss some basic properties of hyperideals, which will be used in the following paragraphs.Proposition 15.
Let(L,⊗,⊕) be a hyperlattice, and let A be a nonempty subset of L. Then the following conditions are equivalent.(1)
Ais a⊗-hyperideal of(L,⊗,⊕).(2)
a⊕b⊆Aanda⊗x⊆Afor alla,b∈Aandx∈L.(3)
A⊕A⊆AandA⊗L⊆A.
Similarly, the following conditions are equivalent.(1)
Ais a⊕-hyperideal of(L,⊗,⊕).(2)
a⊗b⊆Aanda⊕x⊆Afor alla,b∈Aandx∈L.(3)
A⊗A⊆AandA⊕L⊆A.Proof.
It is obvious.Let(L1,⊗1,⊕1) and (L2,⊗2,⊕2) be two hyperlattices. Define hyperoperations on the Cartesian product L1×L2 as follows: for all (x1,y1),(x2,y2)∈L1×L2, (x1,y1)⊗(x2,y2)={(x,y)∣x∈x1⊗1x2,y∈y1⊗2y2}, (x1,y1)⊕(x2,y2)={(x,y)∣x∈x1⊕1x2,y∈y1⊕2y2}. One can check that (L1×L2,⊗,⊕) is a hyperlattice, which is called the Cartesian product hyperlattice of L1 and L2.Proposition 16.
LetA and B be two nonempty subsets of (L1,⊗1,⊕1) and (L2,⊗2,⊕2), respectively.(1)
IfAandBare subhyperlattices ofL1andL2, respectively, thenA×Bis a subhyperlattice of(L1×L2,⊗,⊕).(2)
IfAandBare⊗1-hyperideals (⊕1-hyperideals) and⊗2-hyperideals (⊕2-hyperideals) ofL1andL2, respectively, thenA×Bis a⊗-hyperideal (⊕-hyperideal) of(L1×L2,⊗,⊕).Proof.
The proof is straightforward.Let(L1,⊗1,⊕1) and (L2,⊗2,⊕2) be two hyperlattices. A map f:L1→L2 is called a weak homomorphism if f(a⊗1b)⊆f(a)⊗2f(b) and f(a⊕1b)⊆f(a)⊕2f(b) for all a,b∈L1. In particular, if f(a⊗1b)=f(a)⊗2f(b) and f(a⊕1b)=f(a)⊕2f(b), then f is called a homomorphism.If such a homomorphismf is surjective, injective, or bijective, then f is called an epimorphism, a monomorphism, or an isomorphism from (L1,⊗1,⊕1) to (L2,⊗2,⊕2), respectively.Proposition 17.
Letf be a surjective homomorphism from a hyperlattice (L1,⊗1,⊕1) to a hyperlattice (L2,⊗2,⊕2). (1)
IfAis a⊗1-hyperideal (⊕1-hyperideal) ofL1, thenf(A)is a⊗2-hyperideal (⊕2-hyperideal) ofL2.(2)
IfBis a⊗2-hyperideal (⊕2-hyperideal) ofL2, thenf-1(B)is a⊗1-hyperideal (⊕1-hyperideal) ofL1.Proof.
(
1
) Assume that A is a ⊗1-hyperideal of L1. For all a′,b′∈f(A), there exist a,b∈A such that f(a)=a′, f(b)=b′. Then a′⊕2b′=f(a)⊕2f(b)=f(a⊕1b)={f(z)∣z∈a⊕1b}. It follows from a⊕1b∈P*(A) that a′⊕2b′∈P*(f(A)). Now, let x′∈L2; notice that f is surjective, then there exists x∈L1 such that f(x)=x′. Hence, a′⊗2x′=f(a)⊕2f(x)=f(a⊕1x)={f(z)∣z∈a⊗1x}. By a⊗1x⊆A, we have a′⊗2x′∈P*(f(A)). Therefore, f(A) is a ⊗2-hyperideal of L2.
(
2
) Suppose that B is a ⊗2-hyperideal of L2. For all a,b∈f-1(B), then f(a⊕1b)=f(a)⊕2f(b)⊆B. It follows that a⊕1b⊆f-1(B). On the other hand, let x∈L1, then f(a⊗1x)=f(a)⊗2f(x)⊆B; that is, a⊗1x⊆f-1(B). Therefore, f-1(B) is a ⊗1-hyperideal of L1.


## 3. Rough Hyperideals in Hyperlattices

In this section, we introduce the notion of rough hyperideals in hyperlattices and discuss some properties of them.Given a hyperlatticeL, by P*(L) we will denote the set of all nonempty subsets of L. If θ is an equivalence relation on L, then, for every a∈L, [a]θ stands for the equivalence class of a with the represent θ. For any nonempty subset A of L, we denote [A]θ={[a]θ∣a∈A}.For anyA,B∈P*(L), we denote Aθ-B if the following conditions hold: (1)
for alla∈A, ∃b∈B such that aθb;(2)
for alld∈B, ∃c∈A such that cθd.Now, we can introduce the notion of hypercongruences on hyperlattices in the following manner.Definition 18.
Let(L,⊗,⊕) be a hyperlattice. An equivalence relation θ on L is called a hypercongruence on L if for all a,a′,b,b′∈L, the following implication holds: aθa′ and bθb′ imply (a⊗b)θ-(a′⊗b′) and (a⊕b)θ-(a′⊕b′).Obviously, an equivalence relationθ on (L,⊗,⊕) is a hypercongruence if and only if for all a,b,x∈L, we have that aθb implies (a⊗x)θ-(b⊗x) and (a⊕x)θ-(b⊕x).Lemma 19.
Let(L,⊗,⊕) be a hyperlattice, and let θ be a hypercongruence on L. For all a,b∈L, then [a]θ⊗[b]θ⊆[a⊗b]θ, [a]θ⊕[b]θ⊆[a⊕b]θ.Proof.
Suppose thatx∈[a]θ⊗[b]θ, then there exist x1∈[a]θ and x2∈[b]θ such that x∈x1⊗x2. Since aθx1, bθx2, by Definition 18, we have (a⊗b)θ-(x1⊗x2). So x∈x1⊗x2 implies that there exists y∈a⊗b such that xθy. Therefore, we have x∈[a⊗b]θ, which implies [a]θ⊗[b]θ⊆[a⊗b]θ. Similarly, we can prove that [a]θ⊕[b]θ⊆[a⊕b]θ.A hypercongruence relationθ on (L,⊗,⊕) is called ⊗-complete if [a]θ⊗[b]θ=[a⊗b]θ for all a,b∈L. Similarly, θ is called ⊕-complete if [a]θ⊕[b]θ=[a⊕b]θ for all a,b∈L. We call θ  complete if it is both ⊗-complete and ⊕-complete.Now, we briefly recall the rough set theory in Pawlak's sense. Letθ be an equivalence relation on L, and let A be a nonempty subset of L. Then, the sets θ¯(A)={x∈L∣[x]θ∩A≠ϕ} and θ_(A)={x∈L∣[x]θ⊆A} are called, respectively, the upper and lower approximations of A with respect to θ. θ(A)=(θ_(A),θ¯(A)) is called a rough set with respect to θ.Proposition 20.
Letθ be a hypercongruence on a hyperlattice (L,⊗,⊕). If A,B are two nonempty subsets of L, then(1)
θ¯(A)⊗θ¯(B)⊆θ¯(A⊗B). In particular, ifθis⊗-complete, thenθ¯(A)⊗θ¯(B)=θ¯(A⊗B),(2)
θ¯(A)⊕θ¯(B)⊆θ¯(A⊕B). In particular, ifθis⊕-complete, thenθ¯(A)⊕θ¯(B)=θ¯(A⊕B).Proof.
(
1
) Suppose that x∈θ¯(A)⊗θ¯(B). There exist x1∈θ¯(A) and x2∈θ¯(B) such that x∈x1⊗x2. It follows that there exist a,b∈L such that a∈[x1]θ∩A and b∈[x2]θ∩B. Since θ is a hypercongruence on L, we have a⊗b⊆[x1]θ⊗[x2]θ⊆[x1⊗x2]θ by Lemma 19. On the other hand, since a⊗b⊆A⊗B, we obtain a⊗b⊆[x1⊗x2]θ∩(A⊗B), which implies x∈x1⊗x2⊆θ¯(A⊗B). Therefore, θ¯(A)⊗θ¯(B)⊆θ¯(A⊗B).
Ifθ is ⊗-complete, let x∈θ¯(A⊗B), then [x]θ∩(A⊗B)≠ϕ. Therefore, there exists y∈[x]θ∩(A⊗B), and so for some a∈A and b∈B, we have y∈a⊗b. Since θ is ⊗-complete, we can obtain x∈[y]θ⊆[a⊗b]θ=[a]θ⊗[b]θ. Thus, there exist x1∈[a]θ and x2∈[b]θ such that x∈x1⊗x2. It follows that a∈[x1]θ∩A and b∈[x2]θ∩B. Hence, x1∈θ¯(A) and x2∈θ¯(B), and we have x∈x1⊗x2⊆θ¯(A)⊗θ¯(B), which implies θ¯(A⊗B)⊆θ¯(A)⊗θ¯(B). Therefore, θ¯(A)⊗θ¯(B)=θ¯(A⊗B).
(
2
) The proof is similar to that of (1).Proposition 21.
Letθ be a hypercongruence on a hyperlattice (L,⊗,⊕), and A,B are two nonempty subsets of L. (1)
IfAandBare two⊗-hyperideals ofL, thenθ¯(A)⊗θ¯(B)=θ¯(A⊗B).(2)
IfAandBare two⊕-hyperideals ofL, thenθ¯(A)⊕θ¯(B)=θ¯(A⊕B).Proof.
(
1
) Let x∈θ¯(A⊗B), then there exist a∈A and b∈B such that [x]θ∩(a⊗b)≠ϕ, which implies that there exists t∈a⊗b such that xθt. Since A is a ⊗-hyperideal of L, we have a⊗b⊆A. It follows that t∈A. Hence, we obtain that [x]θ∩A=[t]θ∩A≠ϕ, which implies x∈θ¯(A). In a similar way, we have x∈θ¯(B). Thus, x∈x⊗x⊆θ¯(A)⊗θ¯(B). Combining Proposition 20, we have θ¯(A)⊗θ¯(B)=θ¯(A⊗B).
(
2
) The proof is similar to that of (1).Proposition 22.
Letθ be a hypercongruence relation on a hyperlattice (L,⊗,⊕), and let A, B be two nonempty subsets of L.(1)
Ifθis⊗-complete, thenθ_(A)⊗θ_(B)⊆θ_(A⊗B).(2)
Ifθis⊕-complete, thenθ_(A)⊕θ_(B)⊆θ_(A⊕B).Proof.
(
1
) Let x∈θ_(A)⊗θ_(B), then there exist x1∈θ_(A) and x2∈θ_(B) such that x∈x1⊗x2. It follows that [x1]θ⊆A and [x2]θ⊆B. Since θ is ⊗-complete, we have [x1⊗x2]θ=[x1]θ⊗[x2]θ⊆A⊗B, which implies x∈x1⊗x2⊆θ_(A⊗B). Therefore, θ_(A)⊗θ_(B)⊆θ_(A⊗B).
(
2
) Similar to the proof of (1).The following example shows that the converses of Proposition22 do not hold in general.Example 23.
LetL={0,a,b,c,1} be a lattice (L,∧,∨), where the partial order relation on X is defined as shown in Figure 1. For all x,y∈L, x⊗y={x∧y}, x⊕y={x∨y}, then (L,⊗,⊕) is a hyperlattice. Let θ be a complete hypercongruence relation on the hyperlattice L with the following equivalence classes: [1]θ={1},[c]θ={a,c},[b]θ={b,0}. If A={a}, B={b,c,0}, then θ_(A)=ϕ and θ_(B)={b,0}. We have θ_(A⊕B)={a,c}, θ_(A)⊕θ_(B)=ϕ. So, θ_(A⊕B)⊈θ_(A)⊕θ_(B). If the partial order is reverse, then θ_(A⊗B)={a,c}, θ_(A)⊗θ_(B)=ϕ. Therefore, θ_(A⊗B)⊈θ_(A)⊗θ_(B).Figure 1
The lattice in Example23.Proposition 24.
Letθ be a hypercongruence relation on a hyperlattice (L,⊗,⊕). If A and B are ⊗-hyperideals (⊕-hyperideals) of L, then θ¯(A∩B)=θ¯(A)∩θ¯(B).Proof.
Letx∈θ¯(A)∩θ¯(B), we have [x]θ∩A≠ϕ and [x]θ∩B≠ϕ. Then, there exist x1∈A and x2∈B such that x1θx and x2θx. It follows from θ which is a hypercongruence relation that x1⊗x2θ¯x⊗x, which implies that there exists t∈x1⊗x2 such that tθx. Since A and B are ⊗-hyperideals of L, we have x1⊗x2⊆A∩B. So, t∈A∩B. It follows that [x]θ∩(A∩B)=[t]θ∩(A∩B)≠ϕ, which implies x∈θ¯(A∩B). Hence, θ¯(A)∩θ¯(B)⊆θ¯(A∩B). On the other hand, it is clear that θ¯(A∩B)⊆θ¯(A)∩θ¯(B). Therefore, θ¯(A∩B)=θ¯(A)∩θ¯(B). In a similar way, if A and B are ⊕-hyperideals of L, we can also obtain θ¯(A∩B)=θ¯(A)∩θ¯(B).Up to now, we have studied some properties of the lower and upper approximations in hyperlattices. Next, we will introduce and investigate a new algebraic structure called rough hyperideals in hyperlattices. Let us begin with introducing the following definitions.Definition 25.
Letθ be a hypercongruence on a hyperlattice (L,⊗,⊕), and let A be a nonempty subset of L. A is called a lower (an upper) rough subhyperlattice of L if θ_(A) (θ¯(A)) is a subhyperlattice of L. A is called a rough subhyperlattice of L if A is both a lower rough subhyperlattice and an upper rough subhyperlattice of L.
Similarly,A is called a lower (an upper) rough ⊗-hyperideal of L if θ_(A) (θ¯(A)) is a ⊗-hyperideal of L. And we call A a rough ⊗-hyperideal of L if A is both a lower rough ⊗-hyperideal and an upper rough ⊗-hyperideal of L. In a similar way, a rough ⊕-hyperideal of L can be defined.Example 26.
LetL={a,b,c} and hyperoperations ⊗ and ⊕ on L defined as follows:Then,(L,⊗,⊕) is a hyperlattice. Let θ be a complete hypercongruence relation on the hyperlattice L with the following equivalence classes: [a]θ={a,b}, [c]θ={c}. Now for A={b,c}, θ_(A)={c} and θ¯(A)=L. It is clear that {c} and L are ⊕-hyperideals, so A is a rough ⊕-hyperideal of L.Example 27.
LetL={a,b,c,d} be the hyperlattice in Example 9. Let θ be a hypercongruence relation on the hyperlattice L with the following equivalence classes: [a]θ={a,b}, [c]θ={c,d}. Considering A={a,b,c}, we can obtain that  θ_(A)={a,b}  and  θ¯(A)=L. Notice that {a,b} and L are ⊗-hyperideals, so A is a rough ⊗-hyperideal of L. If A={b,c,d}, we have that θ_(A)={c,d} and θ¯(A)=L. From Example 14, we obtain that {c,d} and L are ⊕-hyperideals, so A is a rough ⊕-hyperideal of L.Theorem 28.
Letθ be a hypercongruence on a hyperlattice (L,⊗,⊕), and let A be a nonempty subset of L.(1)
IfA is a subhyperlattice of L, then A is an upper rough subhyperlattice of L.(2)
IfA is a ⊗-hyperideal (⊕-hyperideal) of L, then A is an upper rough ⊗-hyperideal (⊕-hyperideal) of L.Proof.
(
1
) Suppose that a,b∈θ¯(A), then [a]θ∩A≠ϕ and [b]θ∩A≠ϕ. It follows that there exist x1∈[a]θ∩A and x2∈[b]θ∩A. Since A is a subhyperlattice of L, we have x1⊗x2⊆A. Also, by Lemma 19, we can obtain x∈x1⊗x2⊆[a]θ⊗[b]θ⊆[a⊗b]θ. Hence, [a⊗b]θ∩A≠ϕ, which implies that a⊗b⊆θ¯(A). In a similar way, we have a⊕b⊆θ¯(A). Therefore, θ¯(A) is a subhyperlattice of L; that is, A is an upper rough subhyperlattice of L.
(
2
) Let A be a ⊗-hyperideal of L; then A is a subhyperlattice of L. By (1), we have θ¯(A)⊕θ¯(A)⊆θ¯(A). On the other hand, by Proposition 20, we have θ¯(A)⊗L=θ¯(A)⊗θ¯(L)⊆θ¯(A⊗L)⊆θ¯(A). Thus, θ¯(A) is a ⊗-hyperideal of L. Therefore, A is an upper rough ⊗-hyperideal of L. The other case can be proved in a similar way.Theorem 29.
LetA be a nonempty subset of (L,⊗,⊕), and let θ be a complete hypercongruence relation on L such that θ_(A)≠ϕ. (1)
IfA is a subhyperlattice of L, then A is a lower rough subhyperlattice of L.(2)
IfA is a ⊗-hyperideal (⊕-hyperideal) of L, then A is a lower rough ⊗-hyperideal (⊕-hyperideal) of L.Proof.
(
1
) Let A be a subhyperlattice of L. Since θ_(A)≠ϕ, it follows from Proposition 22 that θ_(A)⊗θ_(A)⊆θ_(A⊗A)⊆θ_(A) and θ_(A)⊕θ_(A)⊆θ_(A⊕A)⊆θ_(A). Therefore, θ_(A) is a subhyperlattice of L; that is, A is a lower rough subhyperlattice of L.
(
2
) Assume that A is a ⊗-hyperideal of L; then A is a subhyperlattice of L. Notice that θ is complete; by the statement of (1), we obtain θ_(A)⊕θ_(A)⊆θ_(A). On the other hand, by Proposition 22, we have θ_(A)⊗L=θ_(A)⊗θ_(L)⊆θ_(A⊗L)⊆θ_(A). Thus, θ_(A) is a ⊗-hyperideal of L. Therefore, A is a lower rough ⊗-hyperideal of L. In a similar way, we can prove that A is a lower rough ⊕-hyperideal of L.By the two theorems above, we have immediately the following corollary.Corollary 30.
LetA be a nonempty subset of (L,⊗,⊕), and let θ be a complete hypercongruence relation on L such that θ_(A)≠ϕ. (1)
IfA is a subhyperlattice of L, then A is a rough subhyperlattice of L.(2)
IfA is a ⊗-hyperideal (⊕-hyperideal) of L, then A is a rough ⊗-hyperideal (⊕-hyperideal) of L.The above corollary shows that under some conditions⊗-hyperideals (⊕-hyperideals) are rough ⊗-hyperideals (⊕-hyperideals) in hyperlattices. The following example shows that the converse of this result does not hold in general.Example 31.
In Example26, A={b,c} is a rough ⊕-hyperideal of (L,⊗,⊕), but A is not a ⊕-hyperideal of L.Example 32.
In Example27, A={a,b,c} is a rough ⊗-hyperideal of (L,⊗,⊕), but A is not a ⊗-hyperideal of L.Based on the discussion above, we obtain that rough hyperideals are extended notions of hyperideals in hyperlattices.


## 4. Rough Hyperideals in the Product Hyperlattices and Quotient Hyperlattices

In this section, we consider rough hyperideals in Cartesian products and quotients of hyperlattices. Let us begin with introducing the following proposition.Letθ1 and θ2 be two hypercongruence relations on (L1,⊗1,⊕1) and (L2,⊗2,⊕2), respectively. Define a relation θ* on L1×L2 as follows: for all (x1,y1),(x2,y2)∈L1×L2, (x1,y1)θ*(x2,y2)⇔x1θ1x2 and y1θ2y2. It is easy to check that θ* is a hypercongruence on the product hyperlattice (L1×L2,⊗,⊕). Then we can obtain the following proposition.Proposition 33.
LetA and B be two nonempty subsets of (L1,⊗1,⊕1) and (L2,⊗2,⊕2), respectively. Then,(1)
θ*¯(A×B)=θ1¯(A)×θ2¯(B).(2)
θ*_(A×B)=θ1_(A)×θ2_(B).Proof.
(
1
) For all (x,y)∈L1×L2, (x,y)∈θ*¯(A×B)⇔[(x,y)]θ*∩(A×B)≠ϕ⇔∃(x′,y′)∈A×B such that (x′,y′)θ*(x,y)⇔∃x′∈A,y′∈B,x′θ1x,y′θ2y⇔[x]θ1∩A≠ϕ and [y]θ2∩B≠ϕ⇔x∈θ1¯(A), y∈θ2¯(B)⇔(x,y)∈θ1¯(A)×θ2¯(B). It follows that θ*¯(A×B)=θ1¯(A)×θ2¯(B).
(
2
) For all (x,y)∈L1×L2, (x,y)∈θ*_(A×B)⇔[(x,y)]θ*⊆(A×B)⇔forall(x′,y′)θ*(x,y), (x′,y′)∈A×B⇔ for all x′θ1x,y′θ2y,x′∈A,y′∈B⇔[x]θ1⊆A and [y]θ2⊆B⇔x∈θ1_(A),y∈θ2_(B)⇔(x,y)∈θ1_(A)×θ2_(B). We conclude that θ*_(A×B)=θ1_(A)×θ2_(B).Theorem 34.
Letθ1 and θ2 be hypercongruence relations on (L1,⊗1,⊕1) and (L2,⊗2,⊕2), respectively. If A and B are two nonempty subsets of L1 and L2, respectively, then(1)
A×B is a rough subhyperlattice of  (L1×L2,⊗,⊕) if and only if A and B are rough subhyperlattices of L1 and L2, respectively.(2)
A×B is a rough ⊗-hyperideal (⊕-hyperideal) of (L1×L2,⊗,⊕) if and only if A and B are rough ⊗1-hyperideals (⊕1-hyperideals) and rough ⊗2-hyperideals (⊕2-hyperideals) of L1 and L2, respectively.Proof.
(
1
)
⇒ Assume that A×B is an upper rough subhyperlattice of (L1×L2,⊗,⊕). Let x1,x2∈θ1¯(A),y1,y2∈θ2¯(B), it follows from Proposition 33 that (x1,y1),(x2,y2)∈θ1¯(A)×θ2¯(B)=θ*¯(A×B). Since θ*¯(A×B) is a subhyperlattice of L1×L2, we have (x1,y1)⊗(x2,y2)={(x,y)∣x∈x1⊗1x2,y∈y1⊗2y2}⊆θ*¯(A×B)=θ1¯(A)×θ2¯(B), (x1,y1)⊕(x2,y2)={(x,y)∣x∈x1⊕1x2,y∈y1⊕2y2}⊆θ*¯(A×B)=θ1¯(A)×θ2¯(B). It follows that x1⊗1x2⊆θ1¯(A), x1⊕1x2⊆θ1¯(A), y1⊗2y2⊆θ2¯(B), y1⊕2y2⊆θ2¯(B), which implies that θ1¯(A) and θ2¯(B) are subhyperlattices of L1 and L2, respectively. Therefore, A and B are upper rough subhyperlattices of L1 and L2, respectively. The case of the lower approximation can be seen in a similar way.
⇐ This follows from Propositions 16 and 33.
(
2
)
⇒ Assume that A×B is an upper rough ⊗-hyperideal of (L1×L2,⊗,⊕). Let x1,x2∈θ1¯(A),y1,y2∈θ2¯(B); it follows from (1) that x1⊕1x2⊆θ1¯(A), y1⊕2y2⊆θ2¯(B). Now, for all a1∈L1,b1∈L2, (x1,y1)⊗(a1,b1)={(x,y)∣x∈x1⊗1a1,y∈y1⊗2b1}⊆θ*¯(A×B)=θ1¯(A)×θ2¯(B) since θ*¯(A×B) is a ⊗-hyperideal of L1×L2. Hence, x1⊗1a1⊆θ1¯(A), y1⊗2b1∈θ2¯(B), which implies that A and B are ⊗1-hyperideals and ⊗2-hyperideals of L1 and L2, respectively. Similarly, the case of the lower approximation can be proved.
⇐ This follows from Propositions 16 and 33.Letθ be a hypercongruence relation on (L,⊗,⊕). For all [a]θ,[b]θ∈L/θ, we define [a]θ⊗′[b]θ={[z]θ∣z∈a⊗b} and [a]θ⊕′[b]θ={[z]θ∣z∈a⊕b}. Then, one can check (L/θ,⊗′,⊕′) is a hyperlattice, which is called the quotient hyperlattice of L with respect to θ.WhenL is finite, L/θ is smaller than L, and its structure is usually less complicated than that of L. At the same time, L/θ simulates L in many ways. In fact, we may think of a quotient hyperlattice of L as a less complicated approximation of L.The lower and upper approximations can be presented in an equivalent form as shown bellow.Letθ be a hypercongruence relation on (L,⊗,⊕), and let A be a nonempty subset of L. Denote θ¯(A)/θ={[x]θ∈L/θ∣[x]θ∩A≠ϕ}, θ_(A)/θ={[x]θ∈L/θ∣[x]θ⊆A}.Theorem 35.
Letθ be a hypercongruence relation on (L,⊗,⊕), and let A be a nonempty subset of L. Then, (1)
θ¯(A) is a subhyperlattice of (L,⊗,⊕) if and only if θ¯(A)/θ is a subhyperlattice of (L/θ,⊗′,⊕′).(2)
θ¯(A) is a ⊗-hyperideal (⊕-hyperideal) of (L,⊗,⊕) if and only if θ¯(A)/θ is a ⊗′-hyperideal (⊕′-hyperideal) of (L/θ,⊗′,⊕′).Proof.
(
1
)
⇒ Let [a]θ,[b]θ∈θ¯(A)/θ; then [a]θ∩A≠ϕ and [b]θ∩A≠ϕ. This implies that a,b∈θ¯(A). Since θ¯(A) is a subhyperlattice of L, we have a⊗b⊆θ¯(A), a⊕b⊆θ¯(A). Then, for all x∈a⊗b, [x]θ∩A≠ϕ. Thus, [a]θ⊗′[b]θ={[x]θ∣x∈a⊗b}⊆θ¯(A)/θ. Similarly, [a]θ⊕′[b]θ⊆θ¯(A)/θ. Therefore, θ¯(A)/θ is a subhyperlattice of L/θ.
⇐ Assume that θ¯(A)/θ is a subhyperlattice of L/θ. Let a,b∈θ¯(A); then [a]θ∩A≠ϕ and [b]θ∩A≠ϕ. This implies that [a]θ,[b]θ∈θ¯(A)/θ. Since θ¯(A)/θ is a subhyperlattice of L/θ, we conclude that [a]θ⊗′[b]θ={[x]θ∣x∈a⊗b}⊆θ¯(A)/θ, which implies that for all x∈a⊗b, [x]θ∩A≠ϕ; that is, x∈θ¯(A). Hence a⊗b⊆θ¯(A). In a similar way, we have a⊕b⊆θ¯(A). Therefore, θ¯(A) is a subhyperlattice of L.
(
2
)
⇒ Assume that θ¯(A) is a ⊗-hyperideal of L. Let [a]θ,[b]θ∈θ¯(A)/θ; it follows from the necessity of (1) that [a]θ⊕′[b]θ⊆θ¯(A)/θ. Now, for every [a]θ∈θ¯(A)/θ and [x]θ∈L/θ, then [a]θ∩A≠ϕ. This implies that a∈θ¯(A). Since θ¯(A) is a ⊗-hyperideal of L, we have a⊗x⊆θ¯(A). It follows that [a⊗x]θ∩A≠ϕ. Then, for every t∈a⊗x, [t]θ∩A≠ϕ, which implies that [a]θ⊗′[x]θ={[t]θ∣t∈a⊗x}⊆θ¯(A)/θ. Therefore, θ¯(A)/θ is a ⊗′-hyperideal of L/θ. The other case can be seen in a similar way.
⇐ Suppose that θ¯(A)/θ is a ⊗′-hyperideal of L/θ. Let a,b∈θ¯(A); it follows from the sufficiency of (1) that a⊕b⊆θ¯(A). Now, let x∈L; then [a]θ∈θ¯(A)/θ and [x]θ∈L/θ. Hence, we conclude that [a]θ⊗′[x]θ={[t]θ∣t∈a⊗x}⊆θ¯(A)/θ, which implies that for all t∈a⊗x, [t]θ∩A≠ϕ; that is, t∈θ¯(A). Hence a⊗x⊆θ¯(A). Therefore, θ¯(A) is a ⊗-hyperideal of L. In a similar way, the other case can be seen.Combining Theorems28 and 35, we have the following corollary.Corollary 36.
Letθ be a hypercongruence relation on (L,⊗,⊕), and let A be a nonempty subset of L.(1)
IfA is a subhyperlattice of L, then θ¯(A)/θ is a subhyperlattice of (L/θ,⊗′,⊕′).(2)
IfA is a ⊗-hyperideal (⊕-hyperideal) of L, then θ¯(A)/θ is a ⊗′-hyperideal (⊕′-hyperideal) of (L/θ,⊗′,⊕′).Theorem 37.
LetA be a nonempty subset of (L,⊗,⊕), and let θ be a hypercongruence relation on L such that θ_(A)≠ϕ. (1)
θ_(A) is a subhyperlattice of L if and only if θ_(A)/θ is a subhyperlattice of (L/θ,⊗′,⊕′).(2)
θ_(A) is a ⊗-hyperideal (⊕-hyperideal) of L if and only if θ_(A)/θ is a ⊗′-hyperideal (⊕′-hyperideal) of (L/θ,⊗′,⊕′).Proof.
(
1
)
⇒ Let [a]θ,[b]θ∈θ_(A)/θ, then [a]θ⊆A and [b]θ⊆A. That is, a∈θ_(A) and b∈θ_(A). Since θ_(A) is a subhyperlattice of L, we have a⊗b⊆θ_(A), a⊕b⊆θ_(A). It follows that [a⊗b]θ⊆A, [a⊕b]θ⊆A. Thus, [a]θ⊗′[b]θ⊆θ_(A)/θ, [a]θ⊕′[b]θ⊆θ_(A)/θ. Therefore, θ_(A)/θ is a subhyperlattice of L/θ.
⇐ Assume that θ_(A)/θ is a subhyperlattice of L/θ. Let a,b∈θ_(A); then [a]θ⊆A and [b]θ⊆A. This implies that [a]θ,[b]θ∈θ_(A)/θ. Since θ_(A)/θ is a subhyperlattice of L/θ, we infer that [a]θ⊗′[b]θ={[x]θ∣x∈a⊗b}⊆θ_(A)/θ, which implies that for all x∈a⊗b, [x]θ⊆A; that is, x∈θ_(A). It follows that a⊗b⊆θ_(A). Likewise, we have a⊕b⊆θ_(A). Therefore, θ_(A) is a subhyperlattice of L.
(
2
)
⇒ Assume that A is a ⊗-hyperideal of L. Let [a]θ,[b]θ∈θ_(A)/θ; it follows from the necessity of (1) that [a]θ⊕′[b]θ⊆θ_(A)/θ. Now, for every [a]θ∈θ_(A)/θ and [x]θ∈L/θ, then [a]θ⊆A. This implies that a∈θ_(A). Since θ_(A) is a ⊗-hyperideal of L, we have a⊗x⊆θ_(A). Then, for all t∈a⊗x, we have t∈θ_(A), which implies that [t]θ⊆A. Hence, [t]θ∈θ_(A)/θ. On the other hand, from t∈a⊗x, we have [t]θ∈[a]θ⊗′[x]θ. It follows that [a]θ⊗′[x]θ⊆θ_(A)/θ. Therefore, θ_(A)/θ is a ⊗′-hyperideal of L/θ. The other case can be seen in a similar way.
⇐ Assume that θ_(A)/θ is a ⊗′-hyperideal of L/θ. Let a,b∈θ_(A); it follows from the sufficiency of (1) that a⊕b⊆θ_(A). Now, let x∈L; then [a]θ∈θ_(A)/θ and [x]θ∈L/θ. Hence, we conclude that [a]θ⊗′[x]θ={[t]θ∣t∈a⊗x}⊆θ_(A)/θ, which implies that for all t∈a⊗x, [t]θ⊆A; that is, t∈θ_(A). Hence a⊗x⊆θ_(A). Therefore, θ_(A) is a ⊗-hyperideal of L. In a similar way, the other case can be seen.Combining Theorems29 and 37, we have the following corollary.Corollary 38.
LetA be a nonempty subset of (L,⊗,⊕), and let θ be a complete hypercongruence relation on L such that θ_(A)≠ϕ.(1)
IfA is a subhyperlattice of L, then θ_(A)/θ is a subhyperlattice of (L/θ,⊗′,⊕′).(2)
IfA is a ⊗-hyperideal (⊕-hyperideal) of L, then θ_(A)/θ is a ⊗′-hyperideal (⊕′-hyperideal) of (L/θ,⊗′,⊕′).


## 5. Homomorphic Images of Rough Hyperideals

In this section, we will discuss relations between the upper (lower) rough hyperideals of hyperlattices and the upper (lower) approximations of their homomorphic images. Finally, combining results in the previous sections, we obtain the corresponding relationships between rough hyperideals of quotient hyperlattices of two homomorphic hyperlattices.Lemma 39.
Let(L1,⊗1,⊕1) and (L2,⊗2,⊕2) be two hyperlattices, and let f:L1→L2 be a homomorphism from L1 to L2. Then θ=kerf={(a,b)∈L1×L1∣f(a)=f(b)} is a hypercongruence on (L1,⊗1,⊕1), which is called the kernel of f.Proof.
Clearly,θ=kerf is an equivalence relation on L1.
For alla,a′,b,b′∈L1, let aθa′ and bθb′; then f(a)=f(a′) and f(b)=f(b′). Let x∈a⊗1b; then f(x)∈f(a⊗1b)=f(a)⊗2f(b)=f(a′)⊗2f(b′)=f(a′⊗1b′), which implies that there exists y∈a′⊗1b′ such that f(x)=f(y). That is, there exists y∈a′⊗1b′ such that xθy. Conversely, for any s∈a′⊗1b′, there also exists t∈a⊗1b such that sθt. It follows that (a⊗1b)θ-(a′⊗1b′). Similarly, we can prove that (a⊕1b)θ-(a′⊕1b′). Therefore, θ=kerf is a hypercongruence on (L1,⊗1,⊕1).Theorem 40.
Letf be a homomorphism from the hyperlattice (L1,⊗1,⊕1) to the hyperlattice (L2,⊗2,⊕2) and θ=kerf. If A is a nonempty subset of L1, then(1)
f(θ¯(A))=f(A).(2)
iff is one to one, f(θ_(A))=f(A).Proof.
(
1
) Since A⊆θ¯(A), it follows that f(A)⊆f(θ¯(A)). Conversely, let y∈f(θ¯(A)); there exists x∈θ¯(A) such that f(x)=y, so we have [x]θ∩A≠ϕ. Thus, there exists a∈[x]θ∩A. So, aθx; that is, f(a)=f(x)=y∈f(A). Thus, f(θ¯(A))⊆f(A). Therefore, f(θ¯(A))=f(A).
(
2
) It is obvious that f(θ_(A))⊆f(A). Let y∈f(A); there exists x∈A such that f(x)=y. If a∈[x]θ, then f(a)=f(x). Since f is one to one, we have a=x∈A. So [x]θ⊆A, which implies x∈θ_(A). Thus, y=f(x)∈f(θ_(A)). Therefore, we obtain f(θ_(A))=f(A).In order to discuss some relations between the upper (lower) rough hyperideals of hyperlattices and the upper (lower) approximations of their homomorphic images, we give the following lemma.Lemma 41.
Letf be a surjective homomorphism from a hyperlattice (L1,⊗1,⊕1) to a hyperlattice (L2,⊗2,⊕2), and let θ2 be a hypercongruence relation on L2. If A is a nonempty subset of L1, then (1)
θ1={(a,b)∈L1×L1∣(f(a),f(b))∈θ2} is a hypercongruence relation on a hyperlattice L1.(2)
f(θ1¯(A))=θ2¯(f(A)).(3)
f(θ1_(A))⊆θ2_(f(A)). Furthermore, if f is injective, then f(θ1_(A))=θ2_(f(A)).Proof.
(
1
) It is clear that θ1 is an equivalence relation. Now, let x∈L1; if aθ1b, then f(a)θ2f(b). Since θ2 is a hypercongruence relation on L2, we have that (f(a)⊗2f(x))θ2¯(f(b)⊗2f(x)) and (f(a)⊕2f(x))θ2¯(f(b)⊕2f(x)). It follows that f(a⊗1x)θ2¯f(b⊗1x), f(a⊕1x)θ2¯f(b⊕1x). By the definition of θ1, we easily get that (a⊗1x)θ1¯(b⊗1x), (a⊕1x)θ1¯(b⊕1x). Therefore, θ1 is a hypercongruence relation on a hyperlattice L1.
(
2
) For any y∈f(θ1¯(A)), then there exists x∈θ1¯(A) such that f(x)=y; hence, [x]θ1∩A≠ϕ. This means that there exists a∈[x]θ1∩A such that f(a)∈f(A) and f(a)θ2f(x). It follows that [f(x)]θ2∩f(A)≠ϕ. Then y=f(x)∈θ2¯(f(A)). Therefore, we conclude that f(θ1¯(A))⊆θ2¯(f(A)). Conversely, let y∈θ2¯(f(A)); then there exists x∈L1 such that f(x)=y. Hence [f(x)]θ2∩f(A)≠ϕ. So, there exists a∈A such that f(a)∈f(A) and f(a)∈[f(x)]θ2. Now, by definition of θ1, we have a∈A, a∈[x]θ1. Thus [x]θ1∩A≠ϕ, which implies x∈θ1¯(A). So, y=f(x)∈f(θ1¯(A)). This means that θ2¯(f(A))⊆f(θ1¯(A)). Therefore, f(θ1¯(A))=θ2¯(f(A)).
(
3
) If y∈f(θ1_(A)), then there exists x∈θ1_(A) such that f(x)=y, so we have [x]θ1⊆A. Now, let b∈[y]θ2; then there exists a∈L1 such that f(a)=b; then f(a)∈[f(x)]θ2. Therefore, a∈[x]θ1⊆A, which implies b=f(a)∈f(A). Thus [y]θ2⊆f(A); that is, y∈θ2_(f(A)). Hence, f(θ1_(A))⊆θ2_(f(A)). Now, let f is injective. For any y∈θ2_(f(A)), then there exists x∈L1 such that f(x)=y and [f(x)]θ2⊆f(A). Let a∈[x]θ1, then f(a)∈[f(x)]θ2⊆f(A), and so a∈A. Thus [x]θ1⊆A, which implies x∈θ1_(A). Then y=f(x)∈f(θ1_(A)), and so θ2_(f(A))⊆f(θ1_(A)). From the above, we conclude that f(θ1_(A))=θ2_(f(A)).Now, we arrive at one of our main theorems.Theorem 42.
Letf be a surjective homomorphism from a hyperlattice (L1,⊗1,⊕1) to a hyperlattice (L2,⊗2,⊕2), let θ2 be a hypercongruence relation on L2, and θ1 the hypercongruence on L1 defined in Lemma 41. If A is a nonempty subset of L1, then (1)
θ1¯(A) is a subhyperlattice of L1 if and only if θ2¯(f(A)) is a subhyperlattice of L2.(2)
θ1¯(A) is a ⊗1-hyperideal (⊕1-hyperideal) of L1 if and only if θ2¯(f(A)) is a ⊗2-hyperideal (⊕2-hyperideal) of L2.Proof.
(
1
)
⇒ Let x,y∈θ2¯(f(A)). By Lemma 41, we have that x,y∈f(θ1¯(A))=θ2¯(f(A)). So there exist a,b∈θ1¯(A) such that x=f(a),y=f(b). Then x⊗2y=f(a)⊗2f(b)=f(a⊗1b),x⊕2y=f(a)⊕2f(b)=f(a⊕1b). Since θ1¯(A) is a subhyperlattice of L1, we have that a⊗1b⊆θ1¯(A), a⊕1b⊆θ1¯(A). It follows that x⊗2y⊆f(θ1¯(A))=θ2¯(f(A)), x⊕2y⊆f(θ1¯(A))=θ2¯(f(A)). Therefore, θ2¯(f(A)) is a subhyperlattice of L2.
⇐ Assume that x,y∈θ1¯(A); then f(x),f(y)∈f(θ1¯(A)). By Lemma 41, we have that f(x),f(y)∈f(θ1¯(A))=θ2¯(f(A)). Since θ2¯(f(A)) is a subhyperlattice of L2, f(x)⊗2f(y)⊆θ2¯(f(A)), f(x)⊕2f(y)⊆θ2¯(f(A)). It follows that f(x⊗1y)⊆f(θ1¯(A)), f(x⊕1y)⊆f(θ1¯(A)). Then for all m∈x⊗1y,n∈x⊕1y, there exist m′,n′∈θ1¯(A) such that f(m)=f(m′), f(n)=f(n′). So we have that [m′]θ1∩A≠ϕ, [n′]θ1∩A≠ϕ and m∈[m′]θ1, n∈[n′]θ1. This implies [m]θ1∩A≠ϕ, [n]θ1∩A≠ϕ, and so m∈θ1¯(A), n∈θ1¯(A). Hence, x⊗1y⊆θ1¯(A), x⊕1y⊆θ1¯(A). Therefore, θ1¯(A) is a subhyperlattice of L1.
(
2
)
⇒ Assume that θ1¯(A) is a ⊗1-hyperideal of L1; let x,y∈θ2¯(f(A)). From the proof of necessity of (1), we obtain x⊕2y⊆θ2¯(f(A)). On the other hand, it follows from x∈θ2¯(f(A))=f(θ1¯(A)) that there exists s∈θ1¯(A) such that x=f(s). Now, let a=f(t)∈L2; then a⊗2x=f(t)⊗2f(s)=f(t⊗1s). Since θ1¯(A) is a ⊗1-hyperideal of L1, we have t⊗1s⊆θ1¯(A). Hence a⊗2x=f(t⊗1s)⊆f(θ1¯(A))=θ2¯(f(A)). Therefore, θ2¯(f(A)) is a ⊗2-hyperideal of L2.
⇐ Let x,y∈θ1¯(A). From the proof of sufficiency of (1), we have x⊕1y⊆θ1¯(A). From x∈θ1¯(A), then f(x)∈f(θ1¯(A)). Now, let a∈L1; then f(a)∈L2. Since θ2¯(f(A)) is a ⊗2-hyperideal of L2, f(a)⊗2f(x)=f(a⊗1x)⊆θ2¯(f(A))=f(θ1¯(A)). Then for all m∈a⊗1x, there exists m′∈θ1¯(A) such that f(m)=f(m′). So we have that [m′]θ1∩A≠ϕ and m∈[m′]θ1. This implies [m]θ1∩A≠ϕ, and so m∈θ1¯(A). Hence, a⊗1x⊆θ1¯(A). Therefore, θ1¯(A) is a ⊗2-hyperideal of L1.
The other cases can be seen in a similar way.Theorem 43.
Letf be an isomorphism from a hyperlattice (L1,⊗1,⊕1) to a hyperlattice (L2,⊗2,⊕2), let θ2 be a hypercongruence relation on L2, and θ1 the hypercongruence on L1 defined in Lemma 41. If A is a nonempty subset of L1, then (1)
θ1_(A) is a subhyperlattice of L1 if and only if  θ2_(f(A)) is a subhyperlattice of L2.(2)
θ1_(A) is a ⊗1-hyperideal (⊕1-hyperideal) of L1 if and only if   θ2_(f(A)) is a ⊗2-hyperideal (⊕2-hyperideal) of L2.Proof.
Sincef is one to one, we have f(θ1_(A))=θ2_(f(A)) by Lemma 41. Therefore, the proof is similar to that of Theorem 42.Combining Theorems42 and 43, we have immediately the following corollary.Corollary 44.
Letf be an isomorphism from a hyperlattice (L1,⊗1,⊕1) to a hyperlattice (L2,⊗2,⊕2), let θ2 be a hypercongruence relation on L2, and θ1 the hypercongruence on L1 defined in Lemma 41. If A is a nonempty subset of L1, then (1)
A is a rough subhyperlattice of L1 if and only if f(A) is a rough subhyperlattice of L2.(2)
A is a rough ⊗1-hyperideal (⊕1-hyperideal) of L1 if and only if f(A) is a rough ⊗2-hyperideal (⊕2-hyperideal) of L2.In what follows, we obtain the corresponding relationships between rough hyperideals of quotient hyperlattices of two homomorphic hyperlattices.Combining Theorems35 and 42, we conclude the following theorem.Theorem 45.
Letf be a surjective homomorphism from a hyperlattice (L1,⊗1,⊕1) to a hyperlattice (L2,⊗2,⊕2), let θ2 be a hypercongruence relation on L2, and θ1 the hypercongruence on L1 defined in Lemma 41. If A is a nonempty subset of L1, then (1)
θ1¯(A)/θ1 is a subhyperlattice of L1/θ1 if and only if θ2¯(f(A))/θ2 is a subhyperlattice of L2/θ2.(2)
θ1¯(A)/θ1 is a ⊗1-hyperideal (⊕1-hyperideal) of L1 if and only if θ2¯(f(A))/θ2 is a ⊗2-hyperideal (⊕2-hyperideal) of L2/θ2.Combining Theorems37 and 43, we can obtain the following results.Theorem 46.
Letf be an isomorphism from a hyperlattice (L1,⊗1,⊕1) to a hyperlattice (L2,⊗2,⊕2), let θ2 be a hypercongruence relation on L2, and θ1 the hypercongruence on L1 defined in Lemma 41. If A is a nonempty subset of L1, then (1)
θ1_(A)/θ1 is a subhyperlattice of L1/θ1 if and only if θ2_(f(A))/θ2 is a subhyperlattice of L2/θ2.(2)
θ1_(A)/θ1 is a ⊗1-hyperideal (⊕1-hyperideal) of L1/θ1 if and only if θ2_(f(A))/θ2 is a ⊗2-hyperideal (⊕2-hyperideal) of L2/θ2.


## 6. Conclusions

In the present paper, we have combined rough set theory and hyperlattices. We introduce rough hyperideals in hyperlattices, which are extended notions of hyperideals of hyperlattices. We conclude the corresponding relationships between rough hyperideals of hyperlattices and that of quotient hyperlattices. Also, we have the relations between rough hyperideals of two homomorphic hyperlattices. Based on this, we obtain the corresponding relationships between rough hyperideals of quotient hyperlattices of two homomorphic hyperlattices. We hope that our work can broaden application fields of the theory of rough sets and hyperlattices.



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