Rough sets based on fuzzy ideals in distributive lattices

Definition 2.1

[19] Let L be a lattice and ∅ ≠ I ⊆ L. Then I is called an ideal of L if it satisfies the following conditions: for any x, y ∈ L,

1. x ∈ I and y ∈ I imply x ∨ y ∈ I;

2. x ∈ L and x ≤ y imply x ∈ I.

Let A, B be subsets of L, we define the join and meet as follows:

Let I, J be ideals of L, then I ∨ J is an ideal of L [18].

Definition 2.2

[19] Let L be a lattice. A relation R is called an equivalence relation on L if for all a, b, c ∈ L,

1. Reflexive: (a, a) ∈ R;

2. Symmetry: (a, b) ∈ R implies (b, a) ∈ R;

3. Transitivity: (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.

An equivalence relation R is called a congruence relation on L, if for all a, b, c, d ∈ L, (a, b) ∈ R, (c, d) ∈ R, then (a ∨ c, b ∨ d ) ∈ R and (a ∧ c, b ∧ d) ∈ R.

Definition 2.3

[20] Let μ be a fuzzy set of a lattice L. Then μ is called a fuzzy sublattice of L if μ(x ∧ y) ∧ μ(x ∨ y) ≥ μ(x) ∧ μ(y), for all x, y ∈ L.

Let μ be a fuzzy sublattice of L. Then μ is a fuzzy ideal of L, if μ(x ∨ y) = μ(x) ∧ μ(y) for all x, y ∈ L.

Proposition 2.4

[20] Let μ be a fuzzy sublattice of a lattice L. Then μ is a fuzzy ideal of L if and only if x ≤ y implies that μ(x) ≥ μ(y), for all x, y ∈ L.

Proposition 2.5

[21] Let μ be a fuzzy set of a lattice L. Then μ is a fuzzy ideal of L if and only if any one of the following sets of conditions is satisfied: for all x, y ∈ L,

1. μ(0) = 1 and μ(x ∨ y) = μ(x) ∧ μ(y);

2. μ(0) = 1, μ(x ∨ y) ≥ μ(x) ∧ μ(y) and μ(x ∧ y) ≥ μ(x) ∨ μ(y).

Let μ be a fuzzy subset of a lattice L and t ∈ [0, 1]. Then the set μ_t = {x ∈ L∣μ(x) ≥ t} is called a t-level subset of μ.

Remark 2.6

A fuzzy set μ is a fuzzy ideal of a lattice L if and only if every subset μ_t is an ideal of L for all t ∈ [0, 1].

Definition 2.7

[1] Let R be an equivalence relation on the universe U and (U, R) be a Pawlak approximation space. A subset X ⊆ U is called definable if R_*X = R^*X; otherwise, X is said to be a rough set, where two operators are defined as:

Definition 2.8

[1] Let X and Y be two non-empty sets and B ⊆ Y. Let T : X → 𝒫^∗(Y) be a set-valued mapping, where 𝒫^∗(Y) denotes the family of all non-empty subsets of Y. The lower and upper approximations T(B) and T(B) are defined by

respectively. If T(B) ≠ T(B), then the pair (T(B), T(B)) is said to be a generalized rough set.

Definition 3.1

Let μ be a fuzzy ideal of L. For each t ∈ [0, 1], the set

is called a t-level relation of μ.

Example 3.2

Let L = {0, a, b, c, 1}. We define the binary relation ≤ in the following Hasse diagram. It is easy to check that L is a distributive lattice. Let μ=10+0.8a+0.6b+0.4c+01. Then it is clear that μ is a fuzzy ideal of L. Choose t = 0.9, then we have U(μ, 0.9) = {(0, 0), (a, a), (b, b), (c, c), (1, 1)}. Thus U(μ, 0.9) is called a 0.9-level relation of μ.

Now we prove that U(μ, t) is a congruence relation on L.

Lemma 3.3

Let μ be a fuzzy ideal of L and t ∈ [0, 1]. Then U(μ, t) is a congruence relation on L.

Proof

It is easy to see that μ(0) = 1 and for any x ∈ L, ⋁a∨x=a∨x μ(a) = ⋁ μ(a) ≥ μ(0) = 1 ≥ t. From Definition 3.1, we get that (x, x) ∈ U(μ, t), i.e., U(μ, t) is reflexive. Obviously, U(μ, t) is symmetric. Let (x, y) ∈ U(μ, t) and (y, z) ∈ U(μ, t). Then we have

and so (⋁a∨x=a∨yμ(a))∧(⋁b∨y=b∨zμ(b))≥t. Since μ is a fuzzy ideal of L, we obtain that

For a ∨ x = a ∨ y, b ∨ y = b ∨ z, we have a ∨ b ∨ x = a ∨ b ∨ y, a ∨ b ∨ y = a ∨ b ∨ z. Thus a ∨ b ∨ x = a ∨ b ∨ z, i.e., c ∨ x = c ∨ z, where c = a ∨ b ∈ L. It follows that

and so ⋁c∨x=c∨z μ(c) ≥ t. According to Definition 3.1, we get that (x, z) ∈ U(μ, t). Therefore, U(μ, t) is an equivalence relation on L. Now we show that U(μ, t) is a congruence relation on L. Let (x, y) ∈ U(μ, t) and (u, v) ∈ U(μ, t). Then

and so

Further, we have

For a ∨ x = a ∨ y, b ∨ u = b ∨ v, we have a ∨ b ∨ (x ∨ u) = a ∨ b ∨ (y ∨ v), i.e., c ∨ (x ∨ u) = c ∨ (y ∨ v), where c = a ∨ b ∈ L. Hence,

Consequently, ⋁c∨(x∨u)=c∨(y∨v) μ(c) ≥ t, which implies that (x ∨ u, y ∨ v) ∈ U(μ, t).

Further, let (x₁, y₁) ∈ U(μ, t) and (x₂, y₂) ∈ U(μ, t). Then

So

For b ∨ x₁ = b ∨ y₁ and c ∨ x₂ = c ∨ y₂, we have

On the other hand, since L is a distributive lattice, we have

Since (b ∨ x₁) ∧ c = (c ∨ y₁) ∧ c and (c ∨ x₂) ∧ b = (c ∨ y₂) ∧ b, we have

Notice that μ is a fuzzy ideal of L, we get that

It follows from b ∧ c ≤ b, x₁ ∧ c ≤ c, x₂ ∧ b ≤ b that

Thus

and therefore (x₁ ∧ x₂, y₁ ∧ y₂) ∈ U(μ, t). According to the above discussing, we get that U(μ, t) is a congruence relation on L.□

Remark 3.4

In Lemma 3.3, we say x is congruent to y mod μ, written x ≡_t y (mod μ) if

It follows from Definition 3.1 and Lemma 3.3 that we can get many useful properties of these congruence relations. We denote by [x]_(μ,t) the equivalence class of U(μ, t) containing x of L.

Lemma 3.5

Let μ be a fuzzy ideal of L and t ∈ [0, 1]. Then for all x, y ∈ L,

1. [x]_(μ,t) ∨ [y]_(μ,t) ⊆ [x ∨ y]_(μ,t);

2. [x]_(μ,t) ∧ [y]_(μ,t) ⊆ [x ∧ y]_(μ,t).

Proof

The proof is easy, and we omit the details.□

Let μ be a fuzzy ideal of L and t ∈ [0, 1]. Then U(μ, t) is a congruence relation on L. Thus, when U = L and R is the above equivalence relation (congruence relation), then we use (L, μ, t) instead of approximation space (U, R).

Definition 3.6

Let μ be a fuzzy ideal of L, t ∈ [0, 1] and ∅ ⊊ X ⊆ L. Then

and

are called the lower approximation and the upper approximation of the set X with respect to μ and t, respectively. It is easy to know that U(μ, t)(X) ⊆ X ⊆ U(μ, t)(X).

Lemma 3.7

Let μ and ν be two fuzzy ideals of L such that μ ⊆ ν and t ∈ [0, 1]. Then [x]_(μ,t) ⊆ [x]_(ν,t) for all x ∈ L.

Proof

Let a ∈ [x]_(μ,t). Then we have (a, x) ∈ U(μ, t), i.e., ⋁b∨a=b∨x μ(b) ≥ t. Since μ ⊆ ν, we have μ(b) ≤ ν(b). Thus ⋁b∨a=b∨x ν(b) ≥ ⋁b∨a=b∨x μ(b) ≥ t, which implies that (a, x) ∈ U(ν, t), i.e., a ∈ [x]_(ν,t). Therefore, [x]_(μ,t) ⊆ [x]_(ν,t).□

From Lemma 3.7, we get the the following conclusion easily.

Lemma 3.8

Let μ and ν be two fuzzy ideals of L such that μ ⊆ ν, t ∈ [0, 1] and ∅ ⊊ X ⊆ L. Then

1. U(ν, t)(X) ⊆ U(μ, t)(X);

2. U(μ, t)(X) ⊆ U(ν, t)(X);

3. U(μ, t)(X) ∪ U(ν, t)(X) ⊆ U(μ ∩ ν, t)(X);

4. U(μ ∩ ν, t)(X) ⊆ U(μ, t)(X) ∩ U(ν, t)(X).

The following example shows that the containedness in (3) and (4) of Lemma 3.8 need not be an equality.

Example 3.9

Consider the lattice L in Example 3.2, let μ=10+0.6a+0.8b+0.4c+01,ν=10+0.8a+0.5b+0.3c+01. Then it is clear that μ and ν are fuzzy ideals of L. Choose t = 0.8, then we have

Thus

If X = {0, c}, then

Therefore U(μ ∩ ν, t)(X) ⫋ U(μ, t)(X) ∩ U(ν, t)(X). Further, if X = {c, 1}, then

Hence U(μ, t)(X) ∪ U(ν, t)(X) ⫋U(μ ∩ ν, t)(X).

The following definition is from Zadeh’s expansion principle.

Definition 3.10

Let μ and ν be two fuzzy sets over L. Define μ ∨ ν over L as follows:

for all x ∈ L.

Now we investigate the operations of lower approximations and upper approximations of the set X with respect to μ and t, respectively.

Proposition 3.11

Let μ and ν be two fuzzy ideals of L, t ∈ [0, 1] and ∅ ⊊ X ⊆ L. Then

1. U(μ ∨ ν, t)(X) ⊆ U(μ, t)(X) ∩ U(ν, t)(X);

2. U(μ ∨ ν, t)(X) ⊇ U(μ, t)(X) ∪ U(ν, t)(X).

Proof

Since L is a distributive lattice, we have that μ ∨ ν is a fuzzy ideal of L. Let x ∈ L. Then (μ ∨ ν)(x) = ⋁x=a∨b (μ(a) ∧ ν(b)) ≥ μ(x) ∧ ν(0). Notice that ν is a fuzzy ideal of L, we obtain that ν(0) = 1. It follows that

and so μ ⊆ μ ∨ ν. In a similar way, we have ν ⊆ μ ∨ ν. According to Lemma 3.8, we get that U(μ ∨ ν, t)(X) ⊆ U(μ, t)(X) ∩ U(ν, t)(X) and U(μ ∨ ν, t)(X) ⊇ U(μ, t)(X) ∪ U(ν, t)(X).□

Proposition 3.12

Let μ and ν be two fuzzy ideals of L, t ∈ [0, 1] and ∅ ⊊ X ⊆ L. Then

1. U(μ, t) ∩ U(ν, t) is a congruence relation on L;

2. U(μ, t) ∩ U(ν, t)(X) ⊇ U(μ, t)(X) ∪ U(ν, t)(X);

3. U(μ, t) ∩ U(ν, t)(X) ⊆ U(μ, t)(X) ∩ U(ν, t)(X).

Proof

It is straightforward.□

Theorem 3.13

Let μ and ν be two fuzzy ideals of L, t ∈ [0, 1] and ∅ ⊊ X ⊆ L. Then

1. U(μ ∩ ν, t)(X) = U(μ, t) ∩ U(ν, t)(X);

2. U(μ ∩ ν, t)(X) = U(μ, t) ∩ U(ν, t)(X).

Proof

1. We first show that U(μ, t) ∩ U(ν, t)(X) ⊆ U(μ ∩ ν, t)(X). Let x ∈ U(μ, t) ∩ U(ν, t) and y ∈ [x]_(μ∩ν,t). Then (x, y) ∈ U(μ ∩ ν, t),

   ⋁a∨y=a∨x(μ∩ν)(a)≥t,i.e.,⋁a∨y=a∨x(μ(a)∧ν(a))≥t.

   Thus,

   ⋁a∨y=a∨xμ(a)≥tand⋁a∨y=a∨xν(a)≥t.

   Hence, y ∈ [x]_(μ,t) and y ∈ [x]_(ν,t). So y ∈ [x]_{(μ,t)∩(ν,t)}, and therefore y ∈ X, which implies that x ∈ U(μ ∩ ν, t)(X). Therefore, U(μ, t) ∩ U(ν, t)(X) ⊆ U(μ ∩ ν, t)(X).

   Next we show that U(μ ∩ ν, t)(X) ⊆ U(μ, t) ∩ U(ν, t)(X). Let x ∈ U(μ ∩ ν, t)(X) and x′ ∈ [x]_{(μ,t)∩(ν,t)}. Then x′ ∈ [x]_(μ,t) and x′ ∈ [x]_(ν,t), i.e.,

   ⋁a∨x′=a∨xμ(a)≥tand⋁b∨x′=b∨xν(b)≥t.

   For a ∨ x′ = a ∨ x and b ∨ x′ = b ∨ x, we have

   (a∨x′)∧(b∨x′)=(a∨x)∧(b∨x).

   Since L is a distributive lattice and μ and ν are fuzzy ideals of L, we have

   x′∨(a∧b)=x∨(a∧b)andμ(a∧b)≥μ(a),ν(a∧b)≥ν(b),

   i.e.,

   t≤(⋁a∨x′=a∨xμ(a))∧(⋁b∨x′=b∨xν(b))≤⋁x′∨(a∧b)=x∨(a∧b)(μ(a)∧ν(b))≤⋁x′∨(a∧b)=x∨(a∧b)(μ(a∧b)∧ν(a∧b))=⋁x′∨(a∧b)=x∨(a∧b)(μ∩ν)(a∧b).

   Thus x′ ∈ [x]_(μ∩ν,t), then x′ ∈ X, which implies that x ∈ U(μ, t) ∩ U(ν, t)(X). Thus

   U(μ,t)∩U(ν,t)_(X)⊆U(μ∩ν,t)_(X).

   Therefore, U(μ ∩ ν, t)(X) = U(μ, t) ∩ U(ν, t)(X).

2. Let x ∈ U(μ ∩ ν, t)(X). Then there exists x′ ∈ [x]_{(μ,t)∩(ν,t)} ∩ X, i.e., x′ ∈ X and (x, x′) ∈ U(μ ∩ ν, t), so

   ⋁a∨y=a∨x(μ∩ν)(a)≥t,i.e.,⋁a∨y=a∨x(μ(a)∧ν(a))≥t.

   Thus,

   ⋁a∨x=a∨x′μ(a)≥tand⋁a∨x=a∨x′ν(a)≥t.

   Hence, x′ ∈ [x]_(μ,t) and x′ ∈ [x]_(ν,t), which implies that x ∈ U(μ, t) ∩ U(ν, t)(X). So

   U(μ∩ν,t)¯(X)⊆U(μ,t)∩U(ν,t)¯(X).

   In a similar way, we have U(μ ∩ ν, t)(X) ⊇ U(μ, t) ∩ U(ν, t)(X). Therefore, U(μ ∩ ν, t)(X) = U(μ, t) ∩ U(ν, t)(X).□

Theorem 3.14

Let μ be a fuzzy ideal of L and t ∈ [0, 1]. Then

Proof

It is easy to know that U(μ, t)(μ_t) ⊆ μ_t ⊆ U(μ, t)(μ_t). Now we show that U(μ, t)(μ_t) ⊆ μ_t ⊆ U(μ, t)(μ_t). Let x ∈ U(μ, t)(μ_t). Then [x]_(μ,t) ∩ μ_t ≠ ∅, which means that there exists y ∈ μ_t and y ∈ [x]_(μ,t), i.e., μ(y) ≥ t and ⋁a∨y=a∨x μ(a) ≥ t. So there exists a ∈ L such that μ(a) ≥ t satisfying a ∨ y = a ∨ x. Then we have a ∈ μ_t. Since μ is a fuzzy ideal of L, we have μ_t is an ideal of L and a ∨ y ∈ μ_t. Thus a ∨ x ∈ μ_t. Since x ≤ a ∨ x, we have x ∈ μ_t, which implies that U(μ, t)(μ_t) ⊆ μ_t. Therefore, U(μ, t)(μ_t) = μ_t. Further, let x ∈ μ_t and y ∈ [x]_(μ,t). Then (x, y) ∈ U(μ, t), i.e., ⋁b∨x=b∨y μ(b) ≥ t. So there exists b ∈ L such that μ(b) ≥ t satisfying b ∨ y = b ∨ x. Then we have b ∈ μ_t and b ∨ y ∈ μ_t. Since y ≤ a ∨ y, we have y ∈ μ_t. So [x]_(μ,t) ⊆ μ_t, which implies that x ∈ U(μ, t)(μ_t). Hence μ_t ⊆ U(μ, t)(μ_t). From the above, U(μ, t)(μ_t) = μ_t = U(μ, t)(μ_t).□

Theorem 3.15

Let μ and ν be two fuzzy ideals of L and t ∈ [0, 1]. Then μ_t = U(μ, t)(μ ∩ ν)_t.

Proof

It is easy to know that (μ ∩ ν)_t = μ_t ∩ ν_t. Now we show that μ_t = U(μ, t)(μ_t ∩ ν_t). Let x ∈ μ_t, y ∈ ν_t. Then μ(x) ≥ t. Since μ is a fuzzy ideal of L, we have μ_t is an ideal of L. Further, x ∧ y ≤ x and x ∧ y ≤ y, we have x ∧ y ∈ μ_t and x ∧ y ∈ ν_t, i.e., x ∧ y ∈ μ_t ∩ ν_t. Since x ∨ x = x ∨ (x ∧ y), we have ⋁a∨x=a∨(x∧y) μ(a) ≥ μ(x) ≥ t, which implies that x ∧ y ∈ [x]_(μ,t). Thus [x]_(μ,t) ∩ (μ_t ∩ ν_t) ≠ ∅. So x ∈ U(μ, t)(μ_t ∩ ν_t), that is, μ_t ⊆ U(μ, t)(μ_t ∩ ν_t). On the other hand, it is easy to see that U(μ, t)(μ_t ∩ ν_t) ⊆ U(μ, t)(μ_t). Moreover, it follows from Theorem 3.14 that U(μ, t)(μ_t) = μ_t. So U(μ, t)(μ_t ∩ ν_t) ⊆ μ_t. Therefore, μ_t = U(μ, t)(μ_t ∩ ν_t), i.e., μ_t = U(μ, t)(μ ∩ ν)_t.□

Corollary 3.16

Let μ and ν be two fuzzy ideals of L and t ∈ [0, 1]. Then ν_t ⊆ U(μ, t)(μ_t ∨ ν_t).

Proof

Since μ and ν are two fuzzy ideals of L, we have μ_t and ν_t are ideals of L. Further, since L is a distributive lattice, we have μ_t ∨ ν_t is an ideal of L. Let x ∈ μ_t and y ∈ ν_t. Then x ∨ y ∈ μ_t ∨ ν_t. On the other hand, ⋁b∨x=b∨(x∨y) μ(b) ≥ μ(x) ≥ t, which implies that x ∨ y ∈ [x]_(μ,t). Thus [x]_(μ,t) ∩ (μ_t ∨ ν_t) ≠ ∅. So y ∈ U(μ, t)(μ_t ∨ ν_t). Therefore, ν_t ⊆ U(μ, t)(μ_t ∨ ν_t).□

In the following discussion, we denote by ↓ a = {x ∈ L∣x ≤ a} for a ∈ L.

Theorem 3.17

Let μ be a fuzzy ideal of L, t ∈ [0, 1]. Then

1. U(μ, t)(↓ a) = μ_t for each a ∈ μ_t;

2. ⋃a∈μt U(μ, t)(↓ a) ⊆ μ_t.

Proof

1. Since μ is a fuzzy ideal of L, we have μ_t is an ideal of L. It follows from the definition of ↓ a that ↓ a is an ideal and ↓ a ⊆ μ_t for each a ∈ μ_t. It follows from the Theorem 3.15 that U(μ, t)(↓ a) = μ_t.

2. Let a ∈ μ_t. Then ↓ a ⊆ μ_t. It is easy to see that U(μ, t)(↓ a) ⊆ U(μ, t)(μ_t). Follows from Theorem 3.14, we obtain that U(μ, t)(μ_t) = μ_t. Thus U(μ, t)(↓ a) ⊆ μ_t. Therefore, ⋃a∈μt U(μ, t)(↓ a) ⊆ μ_t.□

Theorem 3.18

Let μ and ν be two fuzzy ideals of L and t ∈ [0, 1]. Then the followings are equivalent:

1. μ ⊆ ν;

2. ν_t = U(μ, t)(ν_t);

3. ν_t = U(μ, t)(ν_t).

Proof

(1) ⇒ (2) Let μ ⊆ ν and x ∈ U(μ, t)(ν_t). Then [x]_(μ,t) ∩ ν_t ≠ ∅. This means that there exists a ∈ ν_t such that a ∈ [x]_(μ,t), i.e.,

Since μ ⊆ ν, we have

So there exists b ∈ L such that ν(b) ≥ t satisfying b ∨ a = b ∨ x, i.e., b ∈ ν_t. So b ∨ a = b ∨ x ∈ ν_t. Since x ≤ b ∨ x, we have x ∈ ν_t. Hence, U(μ, t)(ν_t) ⊆ ν_t. On the other hand, it is easy to see that ν_t ⊆ U(μ, t)(ν_t). Therefore, ν_t = U(μ, t)(ν_t).

(2) ⇒ (1) If ν_t = U(μ, t)(ν_t), it follows from Theorem 3.14 and Theorem 3.15 that μ_t = U(μ, t)(μ_t ∩ ν_t) ⊆ U(μ, t)(ν_t) = ν_t. Therefore, μ ⊆ ν.

(2) ⇒ (3) Let ν_t = U(μ, t)(ν_t), x ∈ ν_t and a ∈ [x]_(μ,t). Assume that a ∉ ν_t, then a ∉ U(μ, t)(ν_t). Thus, [x]_(μ,t) ∩ ν_t = ∅, this implies that a ∉ U(μ, t)(ν_t) = ν_t, which contradicts with x ∈ ν_t. Thus a ∈ ν_t. Hence, [x]_(μ,t) ⊆ ν_t, this means that x ∈ U(μ, t)(ν_t). Thus ν_t ∈ U(μ, t)(ν_t). On the other hand, it is easy to see that U(μ, t)(ν_t) ⊆ ν_t. Therefore, ν_t = U(μ, t)(ν_t).

(3) ⇒ (2) Assume that ν_t = U(μ, t)(ν_t). Let x ∈ U(μ, t)(ν_t). Then [x]_(μ,t) ∩ ν_t ≠ ∅, which means that there exists a ∈ ν_t such that a ∈ [x]_(μ,t). Since ν_t = U(μ, t)(ν_t), we have [x]_(μ,t) = [a]_(μ,t) ⊆ ν_t, so x ∈ U(μ, t)(ν_t) = ν_t, i.e., U(μ, t)(ν_t) ⊆ ν_t. On the other hand, it is easy to see that ν_t ⊆ U(μ, t)(ν_t). Therefore, ν_t = U(μ, t)(ν_t).□

Theorem 3.19

Let μ, ν and ω be fuzzy ideals of L such that μ ⊆ ω and t ∈ [0, 1]. Then

Proof

Since μ ⊆ ω, we have μ_t ⊆ ω_t. It follows from Theorem 3.14 that U(μ, t)(ω_t) = ω_t. So U(ν, t)(ω_t) = U(ν, t)(U(μ, t)(ω_t)). Next we show that U(μ, t)(U(ν, t)(ω_t)) = U(ν, t)(ω_t). First of all, we prove that U(ν, t)(ω_t) is an ideal of L. Since ω is a fuzzy ideal of L, we have ω_t is an ideal of L. On the other hand, it is easy to see that a ∨ b ∈ U(ν, t)(ω_t) for all a, b ∈ U(ν, t)(ω_t). Let c ∈ L, d ∈ U(ν, t)(ω_t) and c ≤ d. Then there exists e ∈ [d]_(ν,t) ∩ ω_t. Now let f ∈ [c]_(ν,t). Then e ∧ f ∈ [d]_(ν,t) ∧ [c]_(ν,t) ⊆ [c ∧ d]_(ν,t) = [c]_(ν,t). Since e ∧ f ≤ e, we have e ∧ f ∈ ω_t. Thus [c]_(ν,t) ∩ A ≠ ∅, this means that c ∈ U(ν, t)(ω_t). Thus U(ν, t)(ω_t) is an ideal of L. Further, μ_t ⊆ ω_t ⊆ U(ν, t)(ω_t). It follows from Theorem 3.14 that U(μ, t)(U(ν, t)(ω_t)) = U(ν, t)(ω_t).□

Theorem 3.20

Let μ, ν and ω be fuzzy ideals of L such that μ ⊆ ω and t ∈ [0, 1]. Then

Proof

Let x ∈ U(μ, t)(ω_t) ∩ U(ν, t)(ω_t). Since μ and ω are two fuzzy ideals of L and μ ⊆ ω, we have μ_t ⊆ ω_t. It follows from Theorem 3.14 that x ∈ ω_t ∩ U(ν, t)(ω_t) = ω_t ⊆ U(ν, t) ∩ U(ν, t)(ω_t). So U(μ, t)(ω_t) ∩ U(ν, t)(ω_t) ⊆ U(μ, t) ∩ U(ν, t)(ω_t). It follows from Proposition 3.12 that U(μ, t) ∩ U(ν, t)(ω_t) = U(μ, t)(ω_t) ∩ U(ν, t)(ω_t).□

Theorem 3.21

Let μ and ν be two fuzzy ideals of L such that μ ⊆ ν and t ∈ [0, 1]. If ∅ ⊊ A ⊆ L, then

Proof

It is easy to see that U(μ, t)(ν_t ∩ A) ⊆ U(μ, t)(ν_t) ∩ U(μ, t)(A). Now we show that U(μ, t)(ν_t) ∩ U(μ, t)(A) ⊆ U(μ, t)(ν_t ∩ A). Let x ∈ U(μ, t)(ν_t) ∩ U(μ, t)(A). Since ν is a fuzzy ideal of L, we have ν_t is an ideal of L. It follows from Theorem 3.14 that x ∈ ν_t ∩ U(μ, t)(A). Thus x ∈ ν_t and x ∈ U(μ, t)(A), i.e., [x]_(μ,t) ∩ A ≠ ∅. Thus there exists a ∈ A such that a ∈ [x]_(μ,t), which implies that ⋁b∨a=b∨x μ(b) ≥ t. This means that there exists b ∈ L such that μ(b) ≥ t satisfying b ∨ a = b ∨ x, i.e., b ∈ μ_t. Since μ ⊆ ν, we have μ_t ⊆ ν_t. Thus b ∈ ν_t and b ∨ a = b ∨ x ∈ ν_t. Since a ≤ b ∨ a, we have a ∈ ν_t. So a ∈ A ∩ ν_t, it follows that x ∈ U(μ, t)(ν_t ∩ A). And therefore U(μ, t)(ν_t ∩ A) = U(μ, t)(ν_t) ∩ U(μ, t)(A).□

Theorem 3.22

Let μ be a fuzzy ideal of L and t ∈ [0, 1]. If A, B are ideals of L and μ_t ⊆ A ∪ B, then

1. U(μ, t)(A) ∨ U(μ, t)(B) = U(μ, t)(A ∨ B);

2. U(μ, t)(A) ∨ U(μ, t)(B) ⊆ U(μ, t)(A ∨ B).

Proof

1. Let x ∈ U(μ, t)(A) ∨ U(μ, t)(B). Then there exist y ∈ U(μ, t)(A) and z ∈ U(μ, t)(B) such that x = y ∨ z, i.e., [y]_(μ,t) ∩ A ≠ ∅ and [z]_(μ,t) ∩ B ≠ ∅, which means that there exist a ∈ A and b ∈ B such that a ∈ [y]_(μ,t) and b ∈ [z]_(μ,t), i.e.,

   ⋁y′∨a=y′∨yμ(y′)≥t,⋁z′∨b=z′∨zμ(z′)≥t.

   For y′ ∨ a = y′ ∨ y, z′ ∨ b = z′ ∨ z, we have (y′ ∨ z′) ∨ (a ∨ b) = (y′ ∨ z′) ∨ (y ∨ z) = (y′ ∨ z′) ∨ x. Thus

   t≤(⋁y′∨a=y′∨yμ(y′))∧(⋁z′∨b=z′∨zμ(z′))≤⋁(y′∨z′)∨(a∨b)=(y′∨z′)∨(y∨z)(μ(y′)∧μ(z′))≤⋁(y′∨z′)∨(a∨b)=(y′∨z′)∨xμ(y′∨z′).

   So a ∨ b ∈ [x]_(μ,t). Thus [x]_(μ,t) ∧ (A ∨ B) ≠ ∅, i.e., x ∈ U(μ, t)(A ∨ B). Therefore, U(μ, t)(A) ∨ U(μ, t)(B) ⊆ U(μ, t)(A ∨ B). Next we show that U(μ, t)(A ∨ B) ⊆ U(μ, t)(B). Since A and B are ideals of L and L is a distributive lattice, we have A ∨ B is also an ideal of L. Since μ_t ⊆ A ∪ B, we have μ_t ⊆ A ∪ B ⊆ A ∨ B. According to Theorem 3.14, we get that U(μ, t)(A ∨ B) = A ∨ B ⊆ U(μ, t)(A) ∨ U(μ, t)(B). Therefore, U(μ, t)(A) ∨ U(μ, t)(B) = U(μ, t)(A ∨ B).

2. It follows from Theorem 3.15 that U(μ, t)(A ∨ B) = A ∨ B. Since U(μ, t)(A) ∨ U(μ, t)(B) ⊆ A ∨ B, we have U(μ, t)(A) ∨ U(μ, t)(B) ⊆ U(μ, t)(A ∨ B).□

   Let μ and ν be two fuzzy ideals of L and t ∈ [0, 1]. The composition of U(μ, t) and U(ν, t) is defined as follows:

   U(μ,t)∗U(ν,t)={(x,y)∈L×L|∃z∈Lsuchthat(x,z)∈U(μ,t)and(z,y)∈U(ν,t)}

   It is not difficult to check that U(μ, t)∗ U(ν, t) is a congruence relation on L if and only if U(μ, t)∗ U(ν, t) = U(ν, t)∗ U(μ, t).

Theorem 3.23

Let μ and ν be two fuzzy ideals of L, t ∈ [0, 1] and U(μ, t)∗ U(ν, t) = U(ν, t)∗ U(μ, t).

1. If A is a non-empty subset of L, then U(μ, t)∗ U(ν, t)(A) ⊆ U(μ, t)(A) ∩ U(μ, t)(A).

2. If A is a sublattice of L, then U(μ, t)(A) ∩ U(μ, t)(A) ⊆ U(μ, t)∗ U(ν, t)(A).

Proof

1. Let x ∈ U(μ, t)∗ U(ν, t)(A) and a ∈ [x]_(μ,t). Since x ∈ [x]_(ν,t), we have a ∈ [x]_{(μ, t)∗(ν, t)}. Thus a ∈ A. So x ∈ U(μ, t)(A). In a similar way, we have x ∈ U(ν, t)(A). Therefore,

   U(μ,t)∗U(ν,t)_(A)⊆U(μ,t)_(A)∩U(μ,t)_(A).

2. Let x ∈ U(μ, t)(A) ∩ U(μ, t)(A). Then there exist y, z ∈ A such that y ∈ [x]_(μ,t) and z ∈ [x]_(ν,t), i.e.,

   ⋁a∨y=a∨xμ(a)≥t,⋁b∨z=b∨xν(a)≥t.

   For a ∨ y = a ∨ x, b ∨ z = b ∨ x, we have (z ∨ y) ∨ a = (z ∨ x) ∨ a, (z ∨ x) ∨ b = x ∨ b. Hence

   ⋁(z∨y)∨a=(z∨x)∨aμ(a)≥⋁a∨y=a∨xμ(a)≥t,

   and

   ⋁(z∨x)∨b=x∨bν(a)≥⋁z∨b=x∨bν(b)≥t.

   Thus (z ∨ y) ∈ [z ∨ x]_(μ,t), (z ∨ x) ∈ [x]_(ν,t), i.e., (z ∨ y) ∈ [x]_{(μ, t)∗(ν, t)}. Since A is a sublattice of L, we have z ∨ y ∈ A. Thus z ∨ y ∈ [x]_{(μ, t)∗(ν, t)} ∩ A, i.e., x ∈ U(μ, t)∗ U(ν, t)(A). Therefore, U(μ, t)(A) ∩ U(μ, t)(A) ⊆ U(μ, t)∗ U(ν, t)(A).□

   The following example shows that the containedness in Theorem 3.22 (2) and Theorem 3.23 need not be an equality.

Example 3.24

Consider the lattice in Example 3.2. Let μ=10+0.8a+0.6b+0.4c+01andν=10+0.7a+0.8b+0.3c+01. Then it is clear that μ and ν are fuzzy ideals of L. Choose t = 0.8, then μ_t = {0, a} and ν_t = {0, b}. Now let A = {a, b}, B = {0, b}. Then we have μ_t ⊆ A ∪ B and A ∨ B = {a, b, c}. Thus

Therefore,

Let A = {a, b, c}. Then U(μ, t)(A) = {b, c}, U(ν, t)(A) = {a, c}, and

Therefore, U(μ, t)∗ U(ν, t)(A) ⊆ U(μ, t)(A) ∩ U(μ, t)(A).

Let A = {a, c} be a sublattice of L. Then U(μ, t)(A) = {0, a, b, c}, U(ν, t)(A) = {a, c}, and

Therefore, U(μ, t)(A) ∩ U(μ, t)(A) ⊆ U(μ, t)∗ U(ν, t)(A).

Theorem 3.25

Let μ and ν be two fuzzy ideals of L, t ∈ [0, 1], U(μ, t)∗ U(ν, t) = U(ν, t)∗ U(μ, t) and A be an ideal of L.

1. If μ_t ⊆ A, then U(μ, t)∗ U(ν, t)(A) = U(μ, t)(A) ∩ U(ν, t)(A).

2. If μ_t, ν_t ⊆ A, then U(μ, t)∗ U(ν, t)(A) = U(μ, t)(A) ∩ U(ν, t)(A).

Proof

1. Let x ∈ U(μ, t)(A) ∩ U(ν, t)(A) and x′ ∈ [x]_{(μ, t)∗ (ν, t)}. Then there exists y ∈ L such that x′ ∈ [y]_(μ,t) and y ∈ [x]_(ν,t). So ⋁x′∨d=y∨d μ(d) ≥ t and y ∈ A, which means that there exists d ∈ L such that μ(d) ≥ t satisfying x′ ∨ d = y ∨ d. Thus d ∈ μ_t. Since A is an ideal of L and μ_t ⊆ A, we get that y ∨ d ∈ A. Further, since x′ ∨ d = y ∨ d ≥ x′, we have x′ ∈ A. So x ∈ U(μ, t)∗ U(ν, t)(A). Therefore, U(μ, t)(A) ∩ U(ν, t)(A) ⊆ U(μ, t)∗ U(ν, t)(A). On the other hand, it follows from Theorem 3.23 that U(μ, t)∗ U(ν, t)(A) = U(μ, t)(A) ∩ U(ν, t)(A).

2. Let x ∈ U(μ, t)∗ U(ν, t)(A). Then there exist x′ ∈ A and y ∈ L such that x′ ∈ [y]_(μ,t) and y ∈ [x]_(ν,t). So y ∈ U(μ, t)(A). Since A is an ideal of L and μ_t ⊆ A, it follows from Theorem 3.15 that U(μ, t)(A) = A. So y ∈ A. Thus x ∈ U(ν, t)(A). Since U(μ, t)∗ U(ν, t) = U(ν, t)∗ U(μ, t), we have x ∈ U(μ, t)(A). Therefore, U(μ, t)∗ U(ν, t)(A) ⊆ U(μ, t)(A) ∩ U(ν, t)(A). From Theorem 3.23, we get that U(μ, t)∗ U(ν, t)(A) = U(μ, t)(A) ∩ U(ν, t)(A).□