Generalized granular variable precision approximation operators 1

Abstract

In this paper, as an extension of variable precision (θ, σ)-fuzzy rough sets, we introduce the notion of the generalized granular variable precision $({\underline{R}}^{β}, {\bar{R}}^{β})$ -fuzzy rough sets determined by fuzzy granules with R ∈ L^X×Y on a residuated and co-residuated lattice with an involution N. We investigate the properties of ${\underline{R}}^{β}$ and ${\bar{R}}^{β}$ with e_{L
^Y} and d_{L
^Y} are (L, ⊙)-fuzzy preorder and (L, ⊕)-fuzzy co-preorder, respectively. Moreover, we investigate Alexander topologies and Alexander co-topologies induced by ${\underline{R}}^{⊤}$ and ${\bar{R}}^{⊥}$ , respectively. For an L-fuzzy relation R ∈ L^X×X and ${\underline{R}}^{β}, {\bar{R}}^{β}$ for each β ∈ L, we obtain Alexandrov L-fuzzy pretopologies and Alexandrov L-fuzzy precotopologies on X, respectively. We give their examples.

Keywords

Residuated(co-residuated) lattices generalized granular variable precision approximation operators fuzzy preorders (co-preorders)

1 Introduction

Pawlak [16, 17] introduced the rough set theory as a formal tool to deal with imprecision and uncertainty in the data analysis. It is based on equivalence relation and crisp sets. Information granules have been studied in rough set theory which is considered as one of granular computing, and applied in formal concepts and decision-making activities [1–3 , 34].

As an extension of rough sets, Ziarko [34] introduced the variable precision rough set which cannot only solve problems with uncertain data, but also relax the strict definition of the rough set. He studied the relative error limit of the partition blocks with the inclusion order A ⊂ _βB iff e (A, B) ≤ β. As an extension of Ziarko’s rough sets, Dai et al. [3] introduced the generalized variable precision rough set $({\underline{apr}}_{R}^{β} (A), {\bar{apr}}_{R}^{β})$ defined by, for x ∈ R^-1 [y] iff (x, y) ∈ R iff y ∈ R [x], $\begin{matrix} {\underline{apr}}_{R}^{β} (A) = \cup {R [x] ∣ e (R^{- 1} [y], A) \leq β}, \\ {\bar{apr}}_{R}^{β} (A) = \cup {R [x] ∣ e (R^{- 1} [y], A) < 1 - β} . \end{matrix}$

Dudois and Prade [4] introduced the notions of fuzzy rough sets as fuzziness of concepts and vagueness of information in decision making. As a new approach to develop fuzzy rough sets, Chen et al. [2] introduced the granular (θ, σ)-fuzzy rough set defined as, for all B ∈ [0, 1] ^Y and a left-continuous t-norm ⊙, $\begin{matrix} {\underline{R}}_{θ} (B) (y) & = & ⋁ {R (x, y) ⊙ λ ∣ R (x, y) ⊙ λ \leq B (y)} \\ {\bar{R}}_{σ} (B) (y) & = & ⋀ {(1 - R (x, y)) \oplus (1 - λ) ∣ \\ B (y) & \leq & (1 - R (x, y)) \oplus (1 - λ)} \end{matrix}$ where x ⊕ y = 1 - ((1 - x) ⊙ (1 - y)). Moreover, they combined granular computing with the granular (θ, σ)-fuzzy rough sets. Yao et al. [31] studied variable precision $({\underline{R}}_{θ}^{β}, {\bar{R}}_{σ}^{β})$ -fuzzy rough sets defined as two maps ${\underline{R}}_{θ}^{β}, {\bar{R}}_{σ}^{β} : [0, 1]^{Y} \to [0, 1]^{Y}$ as $\begin{matrix} {\underline{R}}_{θ}^{β} (B) (y) = ⋁ {R (x, y) ⊙ λ ∣ \frac{| B_{y} |}{| Y |} \geq β} \\ {\bar{R}}_{σ}^{β} (B) (y) = ⋀ {(1 - R (x, y)) \oplus (1 - λ) ∣ \frac{| B^{y} |}{| Y |} \geq β} \end{matrix}$ where B_y = {y ∈ Y ∣ R (x, y) ⊙ λ ≤ B (y)} and B^y = {y ∈ Y ∣ B (y) ≤ (1 - R (x, y)) ⊕ (1 - λ)}. Moreover, the relative error limit of the partition blocks was investigated.

On the other hand, Ward [28] introduced a complete residuated lattice which is an algebraic structure for many valued logic. It is an important mathematical tool for algebraic structure of fuzzy contexts [1, 19]. By using the concepts of lower and upper approximation operators, information systems and decision rules are investigated in complete residuated lattices [1 , 24]. Qiao et al. [19] introduced granular variable precision L-fuzzy rough set based on residuated and co-residuated lattices as a generalization of left-continuous t-norms and right-continuous t-norms on [0, 1].

In this paper, as an extension of variable precision (θ, σ)-fuzzy rough sets in Chen et al. [2], Qiao et al. [19] and Yao et al. [31], we introduce the notion of the generalized granular variable precision $({\underline{R}}^{β}, {\bar{R}}^{β})$ -fuzzy rough sets determined by fuzzy granules where R ∈ L^X×Y, e_{L
^Y} and d_{L
^Y} are (L, ⊙)-fuzzy preorder and (L, ⊕)-fuzzy co-preorder, respectively. For an (L, ⊙)-fuzzy preorder R ∈ L^X×X, we obtain an Alexandrov topology and Alexandrov co-topology on X as follows: $\begin{matrix} τ_{R} & = & {A \in L^{X} ∣ A = {\underline{R}}^{⊤} (A)}, \\ κ_{R} & = & {A \in L^{X} ∣ A = {\bar{R}}^{⊥} (A)} . \end{matrix}$

For an L-fuzzy relation R ∈ L^X×X and ${\underline{R}}^{β}, {\bar{R}}^{β}$ for each β ∈ L, we obtain an Alexandrov L-fuzzy pretopology and Alexandrov L-fuzzy precotopology on X as follows: $\begin{matrix} T_{{\underline{R}}^{β}} (A) & = & ⋀_{x \in X} (A (x) \to {\underline{R}}^{β} (A) (x)), \\ F_{{\bar{R}}_{β}} (A) & = & ⋁_{x \in X} (A (x) \Rightarrow {\bar{R}}_{β} (A) (x)) . \end{matrix}$

2 Preliminaries

Definition 2.1. [1 , 28] A structure (L, ∨, ∧, ⊙, →, ⊥, ⊤) is called a complete residuated lattice iff it satisfies the following properties:

(L, ∨, ∧, ⊥, ⊤) is a complete lattice where ⊥ is the bottom element and ⊤ is the top element;

(L, ⊙, ⊤) is a commutative monoid;

for all x, y, z ∈ L, x ≤ y → z iff x ⊙ y ≤ z.

Definition 2.2. [5–7 , 19] A structure (L, ∨, ∧, ⊕, ⇒, ⊥, ⊤) is called a complete co-residuated lattice iff it satisfies the following properties:

(L, ∨, ∧, ⊥, ⊤) is a complete lattice where ⊥ is the bottom element and ⊤ is the top element;

(L, ⊕, ⊥) is a commutative monoid;

for all x, y, z ∈ L, y ⇒ z ≤ x iff z ≤ x ⊕ y.

A decreasing map N : L → L is called an involutive negation if N (⊥) = ⊤, N (⊤) = ⊥, N (N (a)) = a for all a ∈ L.

For α ∈ L, A ∈ L^X, we denote (α → A), (α ⊙ A), α ∈ L^X as (α → A) (x) = α → A (x), (α ⊙ A) (x) = α ⊙ A (x), α (x) = α.

In this paper, unless otherwise specified, we assume that (L, ∨, ∧, ⊙, ⊕, →, ⇒, N, ⊥, ⊤) is a complete residuated and co-residuated lattice with an involutive negation N such that N (x ⊙ y) = N (x) ⊕ N (y).

Lemma 2.3. [1 , 19] For each x, y, z, x_i, y_i ∈ L, the following properties hold.

⊤ → x = x and ⊥ ⇒ x = x.

x ≤ y → x and y ⇒ x ≤ x.

x→ y = ⊤ iff x ≤ y iff y⇒ x = ⊥.

x → (y → z) = y → (x → z), x ⇒ (y ⇒ z) = y ⇒ (x ⇒ z) and x ⊙ y → x ⊙ z ≥ y → z.

(x → y) ⊙ (y → z) ≤ (x → z) and (x ⇒ y) ⊕ (y ⇒ z) ≥ (x ⇒ z).

x ⊙ (y → z) ≤ y → x ⊙ z and y ⇒ x ⊕ z ≤ x ⊕ (y ⇒ z).

y ⊙ (y → z) ≤ z and y ⊕ (y ⇒ z) ≥ z.

x → (y → z) = (x ⊙ y) → z and x ⇒ (y ⇒ z) = (x ⊕ y) ⇒ z.

x → (⋀ _i∈Γy_i) = ⋀ _i∈Γ (x → y_i) and x ⇒ (⋁ _i∈Γy_i) = ⋁ _i∈Γ (x ⇒ y_i).

(⋁ _i∈Γx_i) → y = ⋀ _i∈Γ (x_i → y) and (⋀ _i∈Γx_i) ⇒ y = ⋁ _i∈Γ (x_i ⇒ y).

x ⊙ (⋀ _i∈Γy_i) ≤ ⋀ _i∈Γ (x ⊙ y_i) and x ⊕ (⋁ _i∈Γy_i) ≥ ⋁ _i∈Γ (x ⊕ y_i)

⋁_i∈Γx_i → ⋁ _i∈Γy_i ≥ ⋀ _i∈Γ (x_i → y_i) and ⋀_i∈Γx_i → ⋀ _i∈Γy_i ≥ ⋀ _i∈Γ (x_i → y_i).

⋁_i∈Γx_i ⇒ ⋁ _i∈Γy_i ≤ ⋀ _i∈Γ (x_i ⇒ y_i) and ⋀_i∈Γx_i ⇒ ⋀ _i∈Γy_i ≤ ⋀ _i∈Γ (x_i ⇒ y_i).

x → y = ⋀ _z∈L (z → x) → (z → y) and x → y = ⋀ _z∈L (y → z) → (x → z).

x ⇒ y = ⋁ _z∈L (z ⇒ x) ⇒ (z ⇒ y) and x ⇒ y = ⋁ _z∈L (y ⇒ z) ⇒ (x ⇒ z).

Definition 2.4. [1, 19] Let (L, ⊙, →, ⊥, ⊤) be a complete residuated lattice. Let X be a set. A map e_X : X × X → L is called:

reflexive if e_X (x, x) =⊤ for all x ∈ X,

⊙-transitive if e_X (x, y) ⊙ e_X (y, z) ≤ e_X (x, z), for all x, y, z ∈ X.

If e_X satisfies (E1) and (E2), e_X is called an (L, ⊙)-fuzzy preorder.

Definition 2.5. Let (L, ⊕, ⇒, N, ⊥, ⊤) be a complete co-residuated lattice. Let X be a set. A map d_X : X × X → L is called:

co-reflexive if d_X (x, x) =⊥ for all x ∈ X,

⊕-co-transitive if d_X (x, y) ⊕ d_X (y, z) ≥ d_X (x, z), for all x, y, z ∈ X.

If d_X satisfies (D1) and (D2), d_X is an (L, ⊕)-fuzzy co-preorder.

Example 2.6. (1) Define a map e_L : L × L → L as e_L (x, y) = x → y. By Lemma 2.3(3,5), e_L is an (L, ⊙)-fuzzy preorder.

(2) Define a map d_L : L × L → L as d_L (x, y) = x ⇒ y. By Lemma 2.3(3,5), d_L is an (L, ⊕)-fuzzy co-preorder.

(3) Define a map e_{L
^X} : L^X × L^X → L as e_{L
^X} (A, B) = ⋀ _x∈X (A (x) → B (x)). Then e_{L
^X} is an (L, ⊙)-fuzzy preorder from Lemma 2.3 (5).

(4) Define a map d_{L
^X} : L^X × L^X → L as d_{L
^X} (A, B) = ⋁ _x∈X (A (x) ⇒ B (x)). Then d_{L
^X} is an (L, ⊕)-fuzzy co-preorder from Lemma 2.3 (5).

Definition 2.7. [2] Let R ∈ L^X×Y be an L-fuzzy relation. Then, for all x ∈ X and λ ∈ L, two L-fuzzy granules are defined as follows $[x_{λ}]_{R}^{⊙} (y) = R (x, y) ⊙ λ$ and $[x_{λ}]_{R}^{\oplus} (y) = R^{N} (x, y) \oplus λ$ , for all y ∈ X.

3 Generalized granular variable precision approximation operators

Definition 3.1. Let R ∈ L^X×Y be an L-fuzzy relation and β ∈ L. Two maps ${\underline{R}}^{β}, {\bar{R}}^{β} : L^{Y} \to L^{Y}$ are defined as follows: for all B ∈ L^Y, $\begin{matrix} {\underline{R}}^{β} (B) (y) = ⋁ {[x_{λ}]_{R}^{⊙} (y) ∣ e_{L^{Y}} ([x_{λ}]_{R}^{⊙}, B) \geq β} \\ {\bar{R}}^{β} (B) (y) = ⋀ {[x_{λ}]_{R}^{\oplus} (y) ∣ d_{L^{Y}} ([x_{λ}]_{R}^{\oplus}, B) \leq β} . \end{matrix}$

Then ${\underline{R}}^{β}$ and ${\bar{R}}^{β}$ are called the generalized granular variable precision lower approximation operator on Y and the generalized granular variable precision upper approximation operator on Y, respectively.

The pair $({\underline{R}}^{β}, {\bar{R}}^{β})$ is called the generalized granular variable precision $({\underline{R}}^{β}, {\bar{R}}^{β})$ -fuzzy rough sets determined by fuzzy granules where e_{L
^Y} and d_{L
^Y} are (L, ⊙)-fuzzy preorder and (L, ⊕)-fuzzy co-preorder, respectively.

Remark 3.2. (1) Let ⊙ be a left continuous t-norm, β =⊤ and x ⊕ y = 1 - ((1 - x) ⊙ (1 - y)), since $e_{L^{Y}} ([x_{λ}]_{R}^{⊙}, B) = ⊤$ , by Lemma 2.3(3), then $[x_{λ}]_{R}^{⊙} \leq B$ . Hence ${\underline{R}}^{⊤} (B) (y) = ⋁ {[x_{λ}]_{R}^{⊙} (y) ∣ [x_{λ}]_{R}^{⊙} \leq B} .$

Put β =⊥. Since $d_{L^{Y}} ([x_{λ}]_{R}^{⊙}, B) = ⊥$ , by Lemma 2.3(3), then $B \leq [x_{λ}]_{R}^{\oplus}$ . Hence ${\bar{R}}^{⊥} (B) (y) = ⋀ {[x_{λ}]_{R}^{\oplus} (y) ∣ B \leq [x_{λ}]_{R}^{\oplus}} .$

Then ${\underline{R}}^{⊤}$ and ${\bar{R}}^{⊥}$ coincide with the definition in [2].

(2) Define e_P(Y), d_P(Y) : P (Y) × P (Y) → [0, 1] as $\begin{matrix} e_{P (Y)} (A, B) = 1 - | \frac{n (A) - n (B)}{n (Y)} |, \\ d_{P (Y)} (A, B) = | \frac{n (A) - n (B)}{n (Y)} | . \end{matrix}$ where n (A) is the number of elements in A. Let (L = [0, 1], ⊙, →, ⊕, ⇒, N) be a complete residuated and co-residuated lattice with an involutive negation defined by N (x) =1 - x and $\begin{matrix} x ⊙ y = (x + y - 1) \lor 0, x \to y = (1 - x + y) \land 1, \\ x \oplus y = (x + y) \land 1, x \Rightarrow y = (y - x) \lor 0 . \end{matrix}$

We easily show that e_P(Y) and d_P(Y) are (L, ⊙)-fuzzy preorder and (L, ⊕)-fuzzy co-preorder. Two maps ${\underline{R}}^{β}, {\bar{R}}^{β} : L^{Y} \to L^{Y}$ are defined as follows: for all B ∈ L^Y, $\begin{matrix} {\underline{R}}^{β} (B) (y) = ⋁ {[x_{λ}]_{R}^{⊙} (y) ∣ e_{P (Y)} (B_{y}, Y) \geq β} \\ {\bar{R}}^{β} (B) (y) = ⋀ {[x_{λ}]_{R}^{\oplus} (y) ∣ d_{P (Y)} (B^{y}, Y) \leq 1 - β} . \end{matrix}$ where $B_{y} = {y \in Y ∣ [x_{λ}]_{R}^{⊙} (y) \leq B (y)}$ and $B^{y} = {y \in Y ∣ B (y) \leq [x_{λ}]_{R}^{\oplus} (y)} .$

Then ${\underline{R}}^{β}$ and ${\bar{R}}^{β}$ coincide with the definition in [19, 31].

Theorem 3.3. Let R ∈ L^X×Y be an L-fuzzy relation and β ∈ L. For each B ∈ L^Y, y ∈ Y, $\begin{matrix} {\underline{R}}^{β} (B) (y) & = & ⋁_{x \in X} [x_{B_{β} (x)}]_{R}^{⊙} (y) \\ = & ⋁_{x \in X} (R (x, y) ⊙ B_{β} (x)) \end{matrix}$ where B_β (x) = ⋀ _y∈Y (R (x, y) ⊙ β → B (y)).

Proof. Since $e_{L^{Y}} ([x_{λ}]_{R}^{⊙}, B) = ⋀_{y \in Y} ([x_{λ}]_{R}^{⊙} (y) \to B (y)) \geq β$ , we have $β \leq [x_{λ}]_{R}^{⊙} (y) \to B (y) iff β \leq R (x, y) ⊙ λ \to B (y) iff R (x, y) ⊙ λ \leq β \to B (y) iff λ \leq R (x, y) \to (β \to B (y)) iff λ \leq (R (x, y) ⊙ β) \to B (y) .$ Put B_β (x) = ⋀ _y∈Y (R (x, y) ⊙ β → B (y)). Thus, $[x_{λ}]_{R}^{⊙} (y) = R (x, y) ⊙ λ \leq R (x, y) ⊙ B_{β} (x) = [x_{B_{β} (x)}]_{R}^{⊙} (y)$ . Hence ${\underline{R}}^{β} (B) (y) \leq ⋁_{x \in X} [x_{B_{β} (x)}]_{R}^{⊙} (y) .$

Conversely, $\begin{matrix} e_{L^{Y}} ([x_{B_{β} (x)}]_{R}^{⊙}, B) = ⋀_{y \in Y} ([x_{B_{β} (x)}]_{R}^{⊙} (y) \to B (y)) \\ = ⋀_{y \in Y} (⋁_{x \in X} (R (x, y) ⊙ B_{β} (x)) \to B (y)) \\ = ⋀_{y \in Y} (⋁_{x \in X} (R (x, y) ⊙ ⋀_{y \in Y} (R (x, y) ⊙ β \to B (y))) \\ \to B (y)) \\ \geq ⋀_{y \in Y} (⋁_{x \in X} (R (x, y) ⊙ (R (x, y) \to (β \to B (y))) \\ \to B (y)) (by Lemma 2.3 (7)) \\ \geq ⋀_{y \in Y} ((β \to B (y)) \to B (y)) \geq β . \end{matrix}$

Hence ${\underline{R}}^{β} (B) (y) \geq ⋁_{x \in X} [x_{B_{β} (x)}]_{R}^{⊙} (y) .$

Theorem 3.4. Let R ∈ L^X×Y be an L-fuzzy relation and β ∈ L. For each B ∈ L^Y, we have $\begin{matrix} {\bar{R}}^{β} (B) (y) = ⋀_{x \in X} [x_{B_{β} (x)}]_{R}^{\oplus} (y) \\ = ⋀_{x \in X} (R^{N} (x, y) \oplus B^{β} (x)) \end{matrix}$ where B^β (x) = ⋁ _y∈Y (R^N (x, y) ⊕ β ⇒ B (y)).

Proof. Since $d_{L^{Y}} ([x_{λ}]_{R}^{\oplus}, B) = ⋁_{y \in Y} ([x_{λ}]_{R}^{\oplus} (y) \Rightarrow B (y)) \leq β$ , we have $\begin{matrix} [x_{λ}]_{R}^{\oplus} (y) \Rightarrow B (y) \leq β \\ iff R^{N} (x, y) \oplus λ \Rightarrow B (y) \leq β \\ iff B (y) \leq R^{N} (x, y) \oplus λ \oplus β \\ iff λ \geq (R^{N} (x, y) \oplus β) \Rightarrow B (y) . \end{matrix}$

Put B^β (x) = ⋁ _y∈Y (R^N (x, y) ⊕ β ⇒ B (y)). Thus, $[x_{λ}]_{R}^{\oplus} (y) \geq [x_{B^{β} (x)}]_{R}^{\oplus} (y)$ . Hence ${\bar{R}}^{β} (B) (y) \geq ⋀_{x \in X} [x_{B_{β} (x)}]_{R}^{\oplus} (y) .$

Conversely, $\begin{matrix} d_{L^{Y}} ([x_{B_{β} (x)}]_{R}^{\oplus}, B) = ⋁_{y \in Y} ([x_{B_{β} (x)}]_{R}^{\oplus} (y) \Rightarrow B (y)) \\ = ⋁_{y \in Y} (⋀_{x \in X} (R^{N} (x, y) \oplus B^{β} (x)) \Rightarrow B (y)) \\ = ⋁_{y \in Y} (⋀_{x \in X} (R^{N} (x, y) \oplus ⋁_{y \in Y} (R^{N} (x, y) \oplus β \\ \Rightarrow B (y))) \Rightarrow B (y)) \\ \leq ⋁_{y \in Y} (⋀_{x \in X} (R^{N} (x, y) \oplus (R^{N} (x, y) \Rightarrow (β \Rightarrow B (y))) \\ \Rightarrow B (y)) (by Lemma 2.3 (7)) \\ \leq ⋀_{y \in Y} ((β \Rightarrow B (y)) \Rightarrow B (y)) \leq β . \end{matrix}$

Hence ${\bar{R}}^{β} (B) (y) \leq ⋁_{x \in X} [x_{B^{β} (x)}]_{R}^{\oplus} (y) .$

Theorem 3.5. Let R ∈ L^X×Y be an L-fuzzy relation and β ∈ L, B, B₁, B₂ ∈ L^Y.

${\underline{R}}^{β} (B) \leq β \to B .$

${\underline{R}}^{β} (B) \leq β \to {\underline{R}}^{⊤} (B) .$

If β₁ ≤ β₂, then ${\underline{R}}^{β_{2}} (B) \leq {\underline{R}}^{β_{1}} (B) .$

If B₁ ≤ B₂, then ${\underline{R}}^{β} (B_{1}) \leq {\underline{R}}^{β} (B_{2}) .$

${\underline{R}}^{β} (α \to B) = {\underline{R}}^{α ⊙ β} (B) \leq α \to {\underline{R}}^{β} (B) .$

${\underline{R}}^{β} (α ⊙ B) \geq α ⊙ {\underline{R}}^{β} (B) .$

${\underline{R}}^{⊤} (⊥_{Y}) = ⊥_{Y}$ and ${\underline{R}}^{⊤} (⊤_{Y}) (y) = ⋁_{x \in X} R (x, y) .$

If for each y ∈ Y, there exists x ∈ X such that R (x, y) =⊤, then ${\underline{R}}^{⊤} (⊤_{Y}) = ⊤_{Y} .$

Proof. (1) It follows from: $\begin{matrix} {\underline{R}}^{β} (B) (y) = ⋁_{x \in X} (R (x, y) ⊙ (⋀_{y \in Y} (R (x, y) ⊙ β \\ \to B (y)))) \\ \leq ⋁_{x \in X} (R (x, y) ⊙ (R (x, y) \to (β \to B (y)))) \\ \leq β \to B (y) (by Lemma 2.3 (7)) . \end{matrix}$

(2) Since a ⊙ (b → c) ≤ b → (a ⊙ c), we have $\begin{matrix} {\underline{R}}^{β} (B) (y) = ⋁_{x \in X} (R (x, y) ⊙ (⋀_{y \in Y} (R (x, y) ⊙ β \\ \to B (y)))) \\ = ⋁_{x \in X} (R (x, y) ⊙ (β \to ⋀_{y \in Y} (R (x, y) \to B (y)))) \\ \leq β \to ⋁_{x \in X} (R (x, y) ⊙ ⋀_{y \in Y} (R (x, y) \to B (y)))) \\ = β \to {\underline{R}}^{⊤} (B) . \end{matrix}$

(3) and (4) are easily proved.

(5) For each B ∈ L^Y and y ∈ Y, we have $\begin{matrix} {\underline{R}}^{β} (α \to B) (y) = ⋁_{x \in X} (R (x, y) ⊙ \\ (⋀_{y \in Y} (R (x, y) ⊙ β \to (α \to B) (y)))) \\ = ⋁_{x \in X} (R (x, y) ⊙ (⋀_{y \in Y} (R (x, y) ⊙ β ⊙ α \\ \to B) (y)))) (by Lemma 2.3 (6)) \\ = {\underline{R}}^{α ⊙ β} (B) \\ \leq α \to ⋁_{x \in X} (R (x, y) ⊙ (⋀_{y \in Y} (R (x, y) ⊙ β \\ \to B) (y)))) \\ = α \to {\underline{R}}^{β} (B) (by Lemma 2.3 (6)) . \end{matrix}$

(6) For each B ∈ L^Y and y ∈ Y, we have $\begin{matrix} {\underline{R}}^{β} (α ⊙ B) (y) = ⋁_{x \in X} (R (x, y) ⊙ \\ (⋀_{y \in Y} (R (x, y) ⊙ β \to (α ⊙ B) (y)))) \\ \geq ⋁_{x \in X} (R (x, y) ⊙ (⋀_{y \in Y} (α ⊙ (R (x, y) ⊙ β \\ \to B (y))))) \\ \geq ⋁_{x \in X} (R (x, y) ⊙ α ⊙ (⋀_{y \in Y} (R (x, y) ⊙ β \to B (y)))) \\ = ⊙ α ⋁_{x \in X} (R (x, y) ⊙ (⋀_{y \in Y} (R (x, y) ⊙ β \to B (y)))) \\ = α ⊙ {\underline{R}}^{β} (B) . \end{matrix}$

(7) By (1), put β =⊤, ${\underline{R}}^{⊤} (⊥_{Y}) = ⊥_{Y}$ .

Since (⊤ _X) _β = ⊤ _Y, ${\underline{R}}^{⊤} (⊥_{Y}) (y) = ⋁_{x \in X} R (x, y) ⊙ (⊤_{X})_{β} (x) = ⋁_{x \in X} R (x, y) .$

(8) If for each y ∈ Y, there exists x ∈ X such that R (x, y) =⊤, then ${\underline{R}}^{⊤} (⊤_{Y}) (y) = ⋁_{x \in X} R (x, y) = ⊤ .$

Theorem 3.6. Let R ∈ L^X×Y be an L-fuzzy relation and β ∈ L.

N (a → b) = N (a) ⇒ N (b), for each a, b ∈ L.

N (B_β) = (N (B)) ^N(β) for each B ∈ L^Y.

${\bar{R}}^{N (β)} (N (B)) = N ({\underline{R}}^{β} (B))$ , for each B ∈ L^Y.

Proof. (1) It follows from: $\begin{matrix} N (b) \leq N (a) \oplus (N (a) \Rightarrow N (b)) \\ iff b \geq a ⊙ N (N (a) \Rightarrow N (b)) \\ iff a \to b \geq N (N (a) \Rightarrow N (b)) \\ iff N (a \to b) \leq N (a) \Rightarrow N (b) . \\ b \geq a ⊙ (a \to b) \\ iff N (b) \leq N (a) \oplus N (a \to b) \\ iff N (a) \Rightarrow N (b) \leq N (a \to b) . \end{matrix}$

(2) For each B ∈ L^Y and x ∈ X, we have $\begin{matrix} (N (B))^{N (β)} (x) \\ = ⋁_{y \in Y} (R^{N} (x, y) \oplus N (β) \Rightarrow N (B) (y)) \\ = N (⋀_{y \in Y} (R (x, y) ⊙ β \to B (y)) = N (B_{β} (x)) . \end{matrix}$

(3) For each B ∈ L^Y and y ∈ Y, we have $\begin{matrix} {\bar{R}}^{N (β)} (N (B)) (y) = ⋀_{x \in X} (R^{N} (x, y) \\ \oplus (⋁_{y \in Y} (R^{N} (x, y) \oplus N (β) \Rightarrow N (B) (y)))) \\ = ⋀_{x \in X} (R^{N} (x, y) \oplus (N (B))^{N (β)} (x)) \\ = N (⋁_{x \in X} (R (x, y) ⊙ B_{β} (x))) \\ = N ({\underline{R}}^{β} (B)) . \end{matrix}$

Example 3.7. Let (L = [0, 1], ∧, ∨, N) be a complete lattice with an involutive negation N : L → L defined by N (x) =1 - x. We obtain →, ⇒ : L → L defined by $\begin{matrix} x \to y = ⋁ {z \in L ∣ x \land z \leq y}, \\ x \Rightarrow y = ⋀ {z \in L ∣ x \lor z \geq y} . \end{matrix}$

Then we obtain: $x \to y = {\begin{matrix} 1, & if x \leq y, \\ y, & if x ≰ y, \end{matrix} x \Rightarrow y = {\begin{matrix} 0, & if x \geq y, \\ y, & if x ≰ y . \end{matrix}$

Hence (L = [0, 1], ⊙ = ∧, →, ⊕ = ∨, ⇒, N) is a complete residuated and co-residuated lattice. $N (x \to y) = N (x) \Rightarrow N (y) = {\begin{matrix} 0, & if x \leq y, \\ N (y), & if x ≰ y . \end{matrix}$

Let R ∈ [0, 1] ^X×Y be a [0, 1]-fuzzy relation and β ∈ [0, 1]. Then the followings hold: $\begin{matrix} (N (B))^{N (β)} (x) = (1 - B)^{1 - β} (x) \\ = ⋁_{y \in Y} ((1 - R (x, y)) \lor (1 - β) \Rightarrow (1 - B) (y)) \\ = 1 - (⋀_{y \in Y} (R (x, y) \land β \to B (y)) = 1 - B_{β} (x), \\ {\bar{R}}^{N (β)} (N (B)) (y) = {\bar{R}}^{1 - β} (1 - B) (y) \\ = 1 - (⋁_{x \in X} (R (x, y) \land B_{β} (x))) = 1 - {\underline{R}}^{β} (B) . \end{matrix}$

Example 3.8. Let (P (X) = {A ∣ A ⊂ X}, ∩, ∪, N) be a complete lattice with an involutive negation N : P (X) → P (X) defined by N (A) = A^c. We obtain →, ⇒ : P (X) → P (X) defined by $\begin{matrix} A \to B = ⋃ {C \in P (X) ∣ A \cap C \subset B} = A^{c} \cup B, \\ A \Rightarrow B = ⋂ {C \in P (X) ∣ A \cup C \supset B} = A^{c} \cap B . \end{matrix}$

Hence (P (X), ⊙ = ∩, →, ⊕ = ∪, ⇒, N) be a complete residuated and co-residuated lattice. $N (A \to B) = A \cap B^{c} = N (A) \Rightarrow N (B) .$

For R ⊂ X × Y and D ∈ P (Y), two maps ${\underline{R}}^{D}, {\bar{R}}^{D} : P (Y) \to P (Y)$ for all B ∈ P (Y), $\begin{matrix} {\underline{R}}^{D} (B) = ⋃ {R [x] ∣ e_{P (Y)} (R [x], B) \supset D} \\ {\bar{R}}^{D} (B) = ⋂ {R^{c} [x] ∣ d_{P (Y)} (R^{c} [x], B) \subset D} \end{matrix}$ where R [x] = {y ∈ X ∣ (x, y) ∈ R} and e_P(Y), d_P(Y) : P (Y) × P (Y) → P (Y) as $e_{P (Y)} (A, B) = A \to B, d_{P (Y)} (A, B) = A \Rightarrow B .$

Detailed results follow in [14].

Theorem 3.9. Let R ∈ L^X×Y be an L-fuzzy relation and β ∈ L, B, B₁, B₂ ∈ L^Y.

${\bar{R}}^{β} (B) \geq β \Rightarrow B .$

${\bar{R}}^{β} (B) \geq β \Rightarrow {\bar{R}}^{⊥} (B) .$

If β₁ ≤ β₂, then ${\bar{R}}^{β_{2}} (B) \leq {\bar{R}}^{β_{1}} (B) .$

If B₁ ≤ B₂, then ${\bar{R}}^{β} (B_{1}) \leq {\bar{R}}^{β} (B_{2}) .$

${\bar{R}}^{β} (α \Rightarrow B) = {\bar{R}}^{α \oplus β} (B) \geq α \Rightarrow {\bar{R}}^{β} (B) .$

${\bar{R}}^{β} (α \oplus B) \leq α \oplus {\bar{R}}^{β} (B) .$

${\bar{R}}^{⊥} (⊤_{Y}) = ⊤_{Y}$ and ${\bar{R}}^{⊥} (⊥_{Y}) (y) = ⋀_{x \in X} R^{N} (x, y) .$

If for each y ∈ Y, there exists x ∈ X such that R (x, y) =⊤, then ${\bar{R}}^{⊥} (⊥_{Y}) = ⊥_{Y} .$

Proof. (1) For each B ∈ L^Y and y ∈ Y, we have

$\begin{matrix} {\bar{R}}^{β} (B) (y) = ⋀_{x \in X} (R^{N} (x, y) \\ \oplus (⋁_{y \in Y} (R^{N} (x, y) \oplus β \Rightarrow B (y)))) \\ \geq ⋀_{x \in X} (R^{N} (x, y) \oplus (R^{N} (x, y) \Rightarrow (β \Rightarrow B (y)))) \\ \geq β \Rightarrow B (y) (by Lemma 2.3 (7)) . \end{matrix}$

(2) Since a ⊕ (b ⇒ c) ≥ b ⇒ a ⊕ c, for each B ∈ L^Y and y ∈ Y, $\begin{matrix} {\bar{R}}^{β} (B) (y) = ⋀_{x \in X} (R^{N} (x, y) \\ \oplus (⋁_{y \in Y} (R^{N} (x, y) \oplus β \Rightarrow B (y)))) \\ = ⋀_{x \in X} (R^{N} (x, y) \oplus (β \Rightarrow ⋁_{y \in Y} (R^{N} (x, y) \Rightarrow B (y)))) \\ \geq ⋀_{x \in X} (β \Rightarrow R^{N} (x, y) \oplus ⋁_{y \in Y} (R^{N} (x, y) \Rightarrow B (y)))) \\ \geq β \Rightarrow ⋀_{x \in X} (R^{N} (x, y) \oplus ⋁_{y \in Y} (R (x, y) \to B (y)))) \\ = β \Rightarrow {\bar{R}}^{⊥} (B) (y) . \end{matrix}$

(3) and (4) are easily proved.

(5) For each B ∈ L^Y and y ∈ Y, $\begin{matrix} {\bar{R}}^{β} (α \Rightarrow B) (y) = ⋀_{x \in X} (R^{N} (x, y) \oplus \\ (⋁_{y \in Y} (R^{N} (x, y) \oplus β \Rightarrow (α \Rightarrow B) (y)))) \\ = ⋀_{x \in X} (R^{N} (x, y) \oplus (⋁_{y \in Y} (R^{N} (x, y) \oplus β \oplus α \\ \Rightarrow B) (y)))) \\ = {\bar{R}}^{α ⊙ β} (B) \\ = ⋀_{x \in X} (R^{N} (x, y) \oplus (α \Rightarrow (⋁_{y \in Y} (R^{N} (x, y) \oplus β \oplus α \\ \Rightarrow B) (y)))) \\ \geq ⋀_{x \in X} (α \Rightarrow (R^{N} (x, y) \oplus (⋁_{y \in Y} (R^{N} (x, y) \oplus β \\ \Rightarrow B) (y)))) \\ \geq α \Rightarrow ⋀_{x \in X} ((R^{N} (x, y) \oplus (⋁_{y \in Y} (R^{N} (x, y) \oplus β \\ \Rightarrow B) (y)))) \\ = α \Rightarrow {\bar{R}}^{β} (B) (y) . \end{matrix}$

(6) Since b ⊕ c ≤ b ⊕ (a ⊕ (a ⇒ c)), we have a ⇒ b ⊕ c ≤ b ⊕ (a ⇒ c). For each B ∈ L^Y and y ∈ Y, $\begin{matrix} {\bar{R}}^{β} (α \oplus B) (y) = ⋀_{x \in X} (R^{N} (x, y) \oplus \\ (⋁_{y \in Y} (R^{N} (x, y) \oplus β \Rightarrow (α \oplus B) (y)))) \\ \leq ⋀_{x \in X} (R^{N} (x, y) \oplus (⋁_{y \in Y} (α \oplus (R^{N} (x, y) \\ \oplus β \Rightarrow B (y)))) \\ \leq ⋀_{x \in X} (R^{N} (x, y) \oplus (α \oplus (⋁_{y \in Y} (R^{N} (x, y) \\ \oplus β \Rightarrow B (y)))) \\ \leq α \oplus ⋀_{x \in X} (R^{N} (x, y) \oplus \oplus (⋁_{y \in Y} (R^{N} (x, y) \\ \oplus β \Rightarrow B (y)))) \\ = α \oplus {\bar{R}}^{β} (B) (y) . \end{matrix}$

(7) By (1), put β =⊥, ${\bar{R}}^{⊥} (⊤_{Y}) \geq ⊥ \Rightarrow ⊤_{Y} = ⊤_{Y}$ .

Since (⊥ _Y) ^⊥ (x) = ⋁ _y∈Y (R^N (x, y) ⇒ ⊥ _Y (x)) = N (⋀ _y∈Y (R (x, y) → ⊤ _Y (x))) = ⊥, ${\bar{R}}^{⊥} (⊥_{Y}) (y) = ⋀_{x \in X} (R^{N} (x, y) \oplus (⊥_{Y})^{⊥} (x)) = ⋀_{x \in X} R^{N} (x, y) .$

(8) If for each y ∈ Y, there exists x ∈ X such that R (x, y) =⊤, then ${\bar{R}}^{⊥} (⊥_{Y}) (y) = ⋀_{x \in X} R^{N} (x, y) = ⊥ .$

Example 3.10. Let (L = [0, 1], ⊙, →, ⊕, ⇒, N) be a complete residuated and co-residuated lattice with an involutive negation defined by $\begin{matrix} x ⊙ y = (x + y - 1) \lor 0, \\ x \to y = (1 - x + y) \land 1, \\ x \oplus y = (x + y) \land 1, \\ x \Rightarrow y = (y - x) \lor 0, N (x) = 1 - x . \end{matrix}$

Let (X = {x₁, x₂, x₃}, Y = {y₁, y₂, y₃, y₄}, R) be an information system where X is a set of object, Y is a set of attributes and R, R^N ∈ [0, 1] ^X×Y as $\begin{matrix} R & = & (\begin{matrix} 1 & 0.6 & 1 & 0.7 \\ 0.4 & 0.8 & 1 & 0.5 \\ 0.8 & 0.7 & 0.6 & 0.4 \end{matrix}) \\ R^{N} & = & (\begin{matrix} 0 & 0.4 & 0 & 0.3 \\ 0.6 & 0.2 & 0 & 0.5 \\ 0.2 & 0.3 & 0.4 & 0.6 \end{matrix}) \end{matrix}$

For B = (0.3, 0.1, 0.2, 0.5) and β = 0.8, we obtain B_0.8 = (0.4, 0.4, 0.6) and ${\underline{R}}^{0.8} (B) = (0.4, 0.3, 0.4, 0.1)$ .

By(1), since N (B) = (0.7, 0.9, 0.8, 0.5) and N (β) =0.2, we obtain N (B) ^N(0.8) = (0.6, 0.6, 0.4) = N (B_0.8). and ${\bar{R}}^{N (0.8)} (N (B)) = (0.6, 0.7, 0.6, 0.9) = N ({\underline{R}}^{0.8} (B)) .$

For B = (0.9, 0.7, 1, 0.6) and β = 0.3, we obtain B^0.3 = (0.7, 0.7, 0.4) and ${\bar{R}}^{0.3} (B) = (0.9, 1, 1, 1)$ .

Since it does not satisfy the condition of Theorems 3.5(8) and 3.9(8), ${\underline{R}}^{1} (1_{Y}) = (1, 0.8, 1, 0.7) \neq 1_{Y}$ and ${\bar{R}}^{0} (0_{Y}) = (0, 0.2, 0, 0.3) \neq 0_{Y}$ .

Theorem 3.11. Let R ∈ L^X×X be a reflexive L-fuzzy relation and β ∈ L.

${\underline{R}}^{β} (B) \geq B_{β}$ and B_β ≤ β → B.

${\underline{R}}^{⊤} (B) \geq B_{⊤} = ⋀_{y \in Y} (R (x, y) \to B (y))$ and B_⊤ ≤ B.

${\bar{R}}^{β} (B) \leq B^{β}$ and B^β ≥ β ⇒ B.

${\bar{R}}^{⊥} (B) \leq B^{⊥} = ⋁_{y \in Y} (R^{N} (x, y) \Rightarrow B (y))$ and B_⊥ ≥ B.

If R (x, y) ⊙ β ≤ B (x) → B (y), then $B (y) \leq ⋁_{x \in X} (R (x, y) ⊙ B (x)) \leq {\underline{R}}^{β} (B) (y) .$

If R (x, y) ≤ B (x) → B (y), then $B (y) = ⋁_{x \in X} (R (x, y) ⊙ B (x)) = {\underline{R}}^{⊤} (B) (y) .$

If R^N (x, y) ⊕ β ≥ B (x) ⇒ B (y), then ${\bar{R}}^{β} (B) (y) \leq ⋀_{x \in X} (R^{N} (x, y) \oplus B (x)) \leq B (y) .$

If R^N (x, y) ≥ B (x) ⇒ B (y), then $B (y) = ⋀_{x \in X} (R^{N} (x, y) \oplus B (x)) = {\bar{R}}^{⊥} (B) (y) .$

Proof. (1) For each B ∈ L^Y and y ∈ Y, $\begin{matrix} {\underline{R}}^{β} (B) (y) & = & ⋁_{x \in X} (R (x, y) \\ ⊙ (⋀_{z \in Y} (R (x, z) ⊙ β \to B (z)))) \\ \geq & R (y, y) ⊙ (⋀_{z \in Y} (R (y, z) ⊙ β \to B (z)))) \\ = & B_{β} (y), \\ B_{β} (y) & = & ⋀_{z \in Y} (R (y, z) ⊙ β \to B (z)))) \\ \leq & R (y, y) ⊙ β \to B (y) = β \to B (y) . \end{matrix}$

(2) By (1), put β =⊤. It is trivial.

(3) For each B ∈ L^Y and y ∈ Y, $\begin{matrix} {\bar{R}}^{β} (B) (y) & = & ⋀_{x \in X} (R^{N} (x, y) \\ \oplus (⋁_{y \in Y} (R^{N} (x, y) \oplus β \Rightarrow B (y)))) \\ \leq & R^{N} (y, y) \oplus (⋁_{z \in Y} (R^{N} (y, z) \oplus β \Rightarrow B (z))) \\ = & B^{β} (y) \\ B^{β} (y) & = & ⋁_{z \in Y} (R^{N} (y, z) \oplus β \Rightarrow B (z)))) \\ \geq & R^{N} (y, y) \oplus β \Rightarrow B (y) = β \Rightarrow B (y) . \end{matrix}$

(4) It follows from, by (1), β =⊥.

(5) Since B (x) ≤ R (x, y) ⊙ β → B (y), $\begin{matrix} {\underline{R}}^{β} (B) (y) = ⋁_{x \in X} (R (x, y) \\ ⊙ (⋀_{y \in Y} (R (x, y) ⊙ β \to B (y)))) \\ \geq ⋁_{x \in X} (R (x, y) ⊙ B (x)) \\ \geq R (y, y) ⊙ B (y) = B (y) . \end{matrix}$

(6) By (5) and Theorem 3.5(1), put β =⊤. Then $B (y) \leq ⋁_{x \in X} (R (x, y) ⊙ B (x)) \leq {\underline{R}}^{⊤} (B) (y) \leq B (y) .$

(7) Since B (x) ≥ R^N (x, y) ⊕ β ⇒ B (y), $\begin{matrix} {\underline{R}}^{β} (B) (y) = ⋀_{x \in X} (R^{N} (x, y) \\ \oplus (⋁_{y \in Y} (R^{N} (x, y) \oplus β \Rightarrow B (y)))) \\ \leq ⋀_{x \in X} (R^{N} (x, y) \oplus B (x)) \\ \leq R^{N} (y, y) \oplus B (y) = B (y) . \end{matrix}$

(8) By (7) and Theorem 3.5(1), put β =⊥. Then $B (y) \geq ⋀_{x \in X} (R^{N} (x, y) \oplus B (x)) \leq {\bar{R}}^{⊥} (B) (y) \geq B (y) .$

Theorem 3.12. Let R ∈ L^X×X be an ⊙-transitive L-fuzzy relation and β ∈ L.

R^N is an ⊕-co-transitive L-fuzzy relation.

${\underline{R}}^{β} (B) \leq B_{β}$ and ${\bar{R}}^{β} (B) \geq B^{β}$ .

If p = ⋀ _x∈YR (x, x), then ${\underline{R}}^{β} (B) \geq p ⊙ B_{β}$ and ${\bar{R}}^{β} (B) \leq N (p) \oplus B^{β}$ .

If p = ⋀ _x∈YR (x, x), then $p ⊙ {\underline{R}}^{α ⊙ β} (B) \leq {\underline{R}}^{α} ({\underline{R}}^{β} (B))$ .

If p = ⋀ _x∈YR (x, x), then ${\bar{R}}^{α} ({\bar{R}}^{β} (B)) \leq N (p) \oplus {\bar{R}}^{α \oplus β} (B)$ .

Proof. (1) It follows from R^N (x, y) ⊕ R^N (y, z) = N (R (x, y) ⊙ R (y, z)) ≥ N (R (x, z)) = R^N (x, z).

(2) For each B ∈ L^Y and y ∈ Y, we have $\begin{matrix} {\underline{R}}^{β} (B) (y) = ⋁_{x \in X} (R (x, y) \\ ⊙ (⋀_{z \in Y} (R (x, z) ⊙ β \to B (z)))) \\ \leq ⋁_{x \in X} (R (x, y) ⊙ (⋀_{z \in Y} (R (x, y) ⊙ R (y, z) ⊙ β \\ \to B (z)))) \\ = ⋁_{x \in X} (R (x, y) ⊙ (R (x, y) \to ⋀_{z \in Y} (R (y, z) ⊙ β \\ \to B (z)))) \\ \leq ⋀_{z \in Y} (R (y, z) ⊙ β \to B (z)) = B_{β} (y), \\ {\bar{R}}^{β} (B) (y) = ⋀_{x \in X} (R^{N} (x, y) \\ \oplus (⋁_{z \in Y} (R^{N} (x, z) \oplus β \Rightarrow B (z)))) \\ \geq ⋀_{x \in X} (R^{N} (x, y) \oplus (⋁_{z \in Y} (R^{N} (x, y) \oplus R^{N} (y, z) \oplus β \\ \Rightarrow B (z)))) \\ = ⋀_{x \in X} (R^{N} (x, y) \oplus (R^{N} (x, y) \\ \Rightarrow ⋁_{z \in Y} (R^{N} (y, z) \oplus β \Rightarrow B (z)))) \\ \geq ⋁_{z \in Y} (R^{N} (y, z) \oplus β \Rightarrow B (z)) = B^{β} (y) . \end{matrix}$

(3) For each B ∈ L^Y and z ∈ Y, $\begin{matrix} {\underline{R}}^{β} (B) (z) = ⋁_{x \in X} (R (x, y) ⊙ B_{β} (x)) \\ \geq R (z, z) ⊙ B_{β} (z) \\ \geq (⋀_{z \in X} R (z, z)) ⊙ B_{β} (z) = p ⊙ B_{β} (z), \\ {\bar{R}}^{β} (B) (z) = ⋀_{x \in X} (R^{N} (x, y) \oplus B^{β} (x)) \\ \leq R^{N} (z, z) \oplus B^{β} (z) \\ \leq N (⋀_{z \in X} R (z, z)) \oplus B^{β} (z) = N (p) \oplus B^{β} (z) . \end{matrix}$

(4) Since b → a ⊙ c ≥ a ⊙ (b → c), we have $\begin{matrix} ({\underline{R}}^{β} (B))_{α} (y) = ⋀_{z \in X} (R (y, z) ⊙ α \to {\underline{R}}^{β} (B) (z)) \\ \geq ⋀_{z \in X} (R (x, z) ⊙ α \to p ⊙ B_{β} (z)))) (by (3)) \\ \geq ⋀_{z \in X} (p ⊙ (R (y, z) ⊙ α \to B_{β} (z))) \\ \geq p ⊙ (⋀_{z \in X} (R (y, z) ⊙ α \to B_{β} (z)))) \\ = p ⊙ (⋀_{z \in Y} (R (y, z) ⊙ α \\ \to ⋀_{w \in X} (R (z, w) ⊙ β \to B (w)))) \\ = p ⊙ (⋀_{z, w \in Y} (R (y, z) ⊙ α ⊙ R (z, w) ⊙ β \\ \to B (w))) \\ \geq p ⊙ (⋀_{w \in X} (R (y, w) ⊙ α ⊙ β \to B (w))) \\ = p ⊙ B_{α ⊙ β} (y), \\ {\underline{R}}^{α} ({\underline{R}}^{β} (B)) (y) = ⋁_{x \in X} (R (x, y) ⊙ ({\underline{R}}^{β} (B))_{α} (x)) \\ \geq ⋁_{x \in X} (R (x, y) ⊙ (p ⊙ B_{α ⊙ β} (x))) \\ = p ⊙ {\underline{R}}^{α ⊙ β} (B) (y) . \end{matrix}$

(5) Since b ⇒ a ⊕ c ≤ a ⊕ (b ⇒ c), we have $\begin{matrix} ({\bar{R}}^{β} (B))^{α} (y) = ⋁_{z \in X} (R^{N} (y, z) \oplus α \Rightarrow {\bar{R}}^{β} (B) (z)) \\ \leq ⋁_{z \in X} (R^{N} (x, z) \oplus α \Rightarrow N (p) \oplus B^{β} (z)))) (by (3)) \\ \leq ⋁_{z \in X} (N (p) \oplus (R^{N} (y, z) \oplus α \Rightarrow B^{β} (z))) \\ = N (p) \oplus (⋁_{z \in X} (R^{N} (y, z) \oplus α \Rightarrow B^{β} (z)))) \\ = N (p) \oplus (⋁_{z \in Y} (R^{N} (y, z) \oplus α \\ \Rightarrow ⋁_{w \in X} (R^{N} (z, w) \oplus β \Rightarrow B (w)))) \\ = N (p) \oplus (⋁_{z, w \in Y} (R^{N} (y, z) \oplus α \oplus R^{N} (z, w) \oplus β \\ \Rightarrow B (w))) \\ \leq N (p) \oplus (⋁_{w \in X} (R^{N} (y, w) \oplus α \oplus β \Rightarrow B (w))) \\ = N (p) \oplus B^{α \oplus β} (y), \\ {\bar{R}}^{α} ({\bar{R}}^{β} (B)) (y) = ⋁_{x \in X} (R (x, y) \oplus ({\bar{R}}^{β} (B))_{α} (x)) \\ \leq ⋁_{x \in X} (R (x, y) \oplus (N (p) \oplus B_{α \oplus β} (x))) \\ = N (p) \oplus {\bar{R}}^{α \oplus β} (B) (y) . \end{matrix}$

Lemma 3.13. Let (L, ⊕, ⇒, ⊥, ⊤) be a complete co-residuated lattice.

If x₁ ≤ x₂, then x₁ ⊕ y ≤ x₂ ⊕ y, x₁ ⇒ y ≤ x₂ ⇒ y and y ⇒ x₁ ≤ y ⇒ x₂.

x ⊕ ⋀ _i∈Γy_i = ⋀ _i∈Γ (x ⊕ y_i).

x ⊕ y ⇒ x ⊕ z ≤ y ⇒ z and (x ⇒ y) ⇒ (x ⇒ z) ≤ y ⇒ z.

Proof. (1) Let x₁ ≤ x₂ be given. Since y ⇒ x₁ ⊕ y ≤ x₁ ≤ x₂, we have x₁ ⊕ y ≤ x₂ ⊕ y. By Lemma 2.3(7), since y ≤ x₁ ⊕ (x₁ ⇒ y) ≤ x₂ ⊕ (x₁ ⇒ y), we have x₁ ⇒ y ≤ x₂ ⇒ y.

(2) By (1), x ⊕ ⋀ y_i ≤ ⋀ (x ⊕ y_i). Since x ⇒ ⋀ (x ⊕ y_i) ≤ y_i, then x ⇒ ⋀ (x ⊕ y_i) ≤ ⋀ y_i. Hence x ⊕ ⋀ y_i ≥ ⋀ (x ⊕ y_i).

(3) Since z ≤ y ⊕ (y ⇒ z), by (1), x ⊕ z ≤ x ⊕ y ⊕ (y ⇒ z). Thus, x ⊕ y ⇒ x ⊕ z ≤ y ⇒ z.

Theorem 3.14. Let R ∈ L^X×X be an L-fuzzy relation and p = ⋀ _x∈YR (x, x).

If R (x, y) ⊙ β ≤ B (x) → B (y), then ${\underline{R}}^{α} (p ⊙ B) \leq {\underline{R}}^{α} ({\underline{R}}^{β} (B))$ .

If R^N (x, y) ⊕ β ≥ B (x) ⇒ B (y), then ${\bar{R}}^{α} (N (p) \oplus B) (y) \geq {\bar{R}}^{α} ({\bar{R}}^{β} (B)) .$

Proof. (1) We have $(p ⊙ B)_{α} \leq ({\underline{R}}^{β} (B))_{α}$ from: $\begin{matrix} (p ⊙ B)_{α} (x) \to ({\underline{R}}^{β} (B))_{α} (x) \\ = ⋀_{z \in X} (R (x, z) ⊙ α \to (p ⊙ B) (z)) \\ \to ⋀_{z \in X} (R (x, z) ⊙ α \to {\underline{R}}^{β} (B) (z)) \\ \geq ⋀_{z \in X} ((R (x, z) ⊙ α \to (p ⊙ B) (z)) \\ \to (R (x, z) ⊙ α \to {\underline{R}}^{β} (B) (z))) \\ \geq ⋀_{z \in X} ((p ⊙ B) (z) \to {\underline{R}}^{β} (B) (z)) \\ \geq ⋀_{z \in X} ((p ⊙ B) (z) \to (P ⊙ B_{β} (z))) \\ \geq ⋀_{z \in X} (B (z) \to B_{β} (z)) \\ \geq ⋀_{z \in X} (B (z) \to ⋀_{y \in X} (R (z, y) ⊙ β \to B (y)) \\ = ⋀_{z, y \in X} (R (z, y) ⊙ β \to (B (z) \to B (y))) = ⊤ . \end{matrix}$

Hence ${\underline{R}}^{α} (p ⊙ B) \leq {\underline{R}}^{α} ({\underline{R}}^{β} (B))$ .

(2) We have $(N (p) \oplus B)^{α} \geq ({\bar{R}}^{β} (B))^{α}$ from $\begin{matrix} (N (p) \oplus B)^{α} (x) \Rightarrow ({\bar{R}}^{β} (B))^{α} (x) \\ = ⋁_{z \in X} (R^{N} (x, z) \oplus α \Rightarrow (N (p) \oplus B) (z)) \\ \Rightarrow ⋁_{z \in X} (R^{N} (x, z) \oplus α \Rightarrow {\bar{R}}^{β} (B) (z)) \\ \leq ⋁_{z \in X} ((R^{N} (x, z) \oplus α \Rightarrow (N (p) \oplus B) (z)) \\ \Rightarrow (R^{N} (x, z) \oplus α \Rightarrow {\bar{R}}^{β} (B) (z))) \\ (by Lemma 3.13 (3)) \\ \leq ⋁_{z \in X} ((N (p) \oplus B) (z) \Rightarrow {\bar{R}}^{β} (B) (z)) \\ \leq ⋁_{z \in X} ((N (p) \oplus B) (z) \Rightarrow (N (p) \oplus B^{β} (z))) \\ \leq ⋁_{z \in X} (B (z) \Rightarrow B^{β} (z)) \\ \leq ⋁_{z \in X} (B (z) \Rightarrow ⋁_{y \in X} (R^{N} (z, y) \oplus β \Rightarrow B (y))) \\ = ⋁_{z, y \in X} (R^{N} (z, y) \oplus β \Rightarrow (B (z) \Rightarrow B (y))) = ⊥ . \end{matrix}$

Hence ${\bar{R}}^{α} (N (p) \oplus B) \geq {\bar{R}}^{α} ({\bar{R}}^{β} (B))$ .

Theorem 3.15. Let R ∈ L^X×X be an (L, ⊙)-fuzzy preorder and β ∈ L.

R^N is an (L, ⊕)-fuzzy co-preorder.

${\underline{R}}^{β} (B) = B_{β}$ and ${\bar{R}}^{β} (B) = B^{β}$ .

${\underline{R}}^{α ⊙ β} (B) \leq {\underline{R}}^{α} ({\underline{R}}^{β} (B))$ .

${\underline{R}}^{β} (B) = {\underline{R}}^{⊤} ({\underline{R}}^{β} (B))$ . Moreover, ${\underline{R}}^{⊤} (B) = {\underline{R}}^{⊤} ({\underline{R}}^{⊤} (B))$ .

${\bar{R}}^{α} ({\bar{R}}^{β} (B)) \leq {\bar{R}}^{α \oplus β} (B)$ .

${\bar{R}}^{⊥} ({\bar{R}}^{β} (B)) = {\bar{R}}^{β} (B)$ . Moreover, ${\bar{R}}^{⊥} (B) = {\bar{R}}^{⊥} ({\bar{R}}^{⊥} (B))$ .

If R (x, y) ⊙ β ≤ B (x) → B (y), then ${\underline{R}}^{α} (B) \leq {\underline{R}}^{α} ({\underline{R}}^{β} (B)) .$

If R^N (x, y) ⊕ β ≥ B (x) ⇒ B (y), then ${\bar{R}}^{β} (B) \geq {\underline{R}}^{α} ({\underline{R}}^{β} (B)) .$

R (x, y) ≤ B (x) → B (y) iff $B = {\underline{R}}^{⊤} (B) = B_{⊤} .$

R^N (x, y) ≥ B (x) ⇒ B (y) iff $B = {\bar{R}}^{⊥} (B) = B^{⊥} .$

Proof. (1) It follows from R^N (x, x) = N (R (x, x)) =⊥ and Theorem 3.12(1).

(2) It follows from Theorems 3.11(1,3) and 3.12(2).

(3) and (5) Since p =⋀ _x∈YR (x, x) = ⊤, by Theorem 3.12(4,5).

(4) By (3), put α =⊤. By Theorem 3.5(1), ${\underline{R}}^{β} (B) \leq {\underline{R}}^{⊤} ({\underline{R}}^{β} (B)) \leq {\underline{R}}^{β} (B)$ .

(6) By (5), put α =⊥. By Theorem 3.9(1), ${\bar{R}}^{β} (B) \leq {\bar{R}}^{⊤} ({\bar{R}}^{β} (B)) \leq {\bar{R}}^{β} (B)$ .

(7) and (8) follow from p =⋀ _x∈YR (x, x) = ⊤, N (p) =⊥ and Theorem 3.14(1,2).

(9) Since R (x, y) ≤ B (x) → B (y), by Theorem 3.11(5), $B \leq {\underline{R}}^{⊤} (B) = B_{⊤} \leq B .$

Since B (x) = B_⊤ (x) = ⋀ _y∈X (R (x, y) → B (y)), B (x) ⊙ R (x, y) ≤ B (y).

(10) Since R^N (x, y) ≥ B (x) ⇒ B (y), by Theorem 3.11(7), $B \geq {\bar{R}}^{⊤} (B) = B^{⊥} \geq B .$

Since B (x) = B^⊥ (x) = ⋁ _y∈X (R^N (x, y) ⇒ B (y)), B (x) ⊕ R^N (x, y) ≥ B (y).

Definition 3.16. [8 , 26] A subset τ ⊂ L^X is called an Alexandrov topology on X iff it satisfies the following conditions:

⊥_X, ⊤ _X ∈ τ.

If A_i ∈ τ for all i ∈ I, then ⋀_i∈IA_i, ⋁ _i∈I ∈ τ.

If A ∈ τ and α ∈ L, then α ⊙ A, α → A ∈ τ.

A subset κ ⊂ L^X is called an Alexandrov co-topology on X iff it satisfies the following conditions:

⊥_X, ⊤ _X ∈ κ.

If A_i ∈ κ for all i ∈ I, then ⋀_i∈IA_i, ⋁ _i∈I ∈ κ.

If A ∈ κ and α ∈ L, then α ⊕ A, α ⇒ A ∈ κ.

Definition 3.17. A map $T : L^{X} \to L$ is called an Alexandrov L-fuzzy pretopology on X if it satisfies the following conditions:

$T (⊥_{X}) = ⊤$ ,

$T (⋁_{i \in Γ} A_{i}) \geq ⋀_{i} T (A_{i})$ for all {A_i} _i∈Γ ⊆ L^X.

$T (α ⊙ A) \geq T (A)$ for all A ∈ L^X and α ∈ L.

A map $F : L^{X} \to L$ is called an Alexandrov L-fuzzy precotopology on X if it satisfies the following conditions:

$F (⊤_{X}) = ⊤$ ,

$F (⋀_{i \in Γ} A_{i}) \geq ⋀_{i} F (A_{i})$ for all {A_i} _i∈Γ ⊆ L^X.

$F (α \Rightarrow A) \geq F (A)$ for all A ∈ L^X and α ∈ L.

Theorem 3.18. Let R ∈ L^X×X be an (L, ⊙)-fuzzy preorder.

$τ_{R} = {A \in L^{X} ∣ A = {\underline{R}}^{⊤} (A)} = {{\underline{R}}^{⊤} (A) ∣ A \in L^{X}}$ is an Alexandrov topology on X.

$κ_{R} = {A \in L^{X} ∣ A = {\bar{R}}^{⊥} (A)} = {{\bar{R}}^{⊥} (A) ∣ A \in L^{X}}$ is an Alexandrov co-topology on X.

A ∈ τ_R iff N (A) ∈ κ_R.

Proof. (1) From Theorem 3.15(4), since ${\underline{R}}^{⊤} (B) = {\underline{R}}^{⊤} ({\underline{R}}^{⊤} (B))$ , we have ${A \in L^{X} ∣ A = {\underline{R}}^{⊤} (A)} = {{\underline{R}}^{⊤} (A) ∣ A \in L^{X}} .$

(AT1) By Theorem 3.5(1), since ${\underline{R}}^{⊤} (⊥_{X}) \leq ⊥_{X}$ , ⊥_X ∈ τ_R. By Theorem 3.11(2), since ${\underline{R}}^{⊤} (⊤_{X}) \geq ⊤_{X}$ , ⊤_X ∈ τ_R.

(AT2) If $A_{i} = {\underline{R}}^{⊤} (A_{i})$ for i ∈ Γ, by Theorem 3.15(2), $\begin{matrix} {\underline{R}}^{⊤} (⋀_{i \in Γ} A_{i}) (y) = (⋀_{i \in Γ} A_{i})_{⊤} (y) \\ = ⋀_{x \in X} (R (x, y) \to ⋀_{i \in Γ} A_{i} (x)) \\ = ⋀_{i \in Γ} ⋀_{x \in X} (R (x, y) \to A_{i} (x)) \\ = ⋀_{i \in Γ} {\underline{R}}^{⊤} (A_{i}) (y) = ⋀_{i \in Γ} A_{i} . \end{matrix}$

Hence ⋀_i∈IA_i ∈ τ_R.

We have $⋁_{i \in Γ} A_{i} = ⋁_{i \in Γ} {\underline{R}}^{⊤} (A_{i}) \leq {\underline{R}}^{⊤} (⋁_{i \in Γ} A_{i}) \leq ⋁_{i \in Γ} A_{i}$ from Theorem 3.5(1). Hence ⋁_i∈ΓA_i ∈ τ_R.

(AT3) Let $A = {\underline{R}}^{⊤} (A)$ . By Theorem 3.5(6), ${\underline{R}}^{⊤} (α ⊙ A) \geq α ⊙ {\underline{R}}^{⊤} (A) = α ⊙ A .$

By (2), ${\underline{R}}^{β} (α ⊙ A) = (α ⊙ A)_{⊤} \leq α ⊙ A .$ Hence $α ⊙ A = {\underline{R}}^{⊤} (α ⊙ A)$ . Moreover, $\begin{matrix} {\underline{R}}^{⊤} (α \to A) (y) = (α \to A)_{⊤} (y) \\ = ⋀_{x \in X} (R (x, y) \to (α \to A (x)) \\ = α \to ⋀_{x \in X} (R (x, y) \to A (x)) \\ = α \to {\underline{R}}^{⊤} (A) (y) = α \to A . \end{matrix}$

Hence α ⊙ A, α → A ∈ τ_R.

(2) From Theorem 3.15(6), since ${\bar{R}}^{⊥} ({\bar{R}}^{⊥} (B)) = {\bar{R}}^{⊥} (B)$ , ${A \in L^{X} ∣ A = {\bar{R}}^{⊥} (A)} = {{\bar{R}}^{⊥} (A) ∣ A \in L^{X}}$ .

(AF1) By Theorem 3.11(4), since ${\bar{R}}^{⊥} (⊥_{X}) (x) \leq ⊥_{X}^{⊥} (x) = ⋁_{y \in Y} (R^{N} (x, y) \Rightarrow ⊥_{X} (y)) = ⊥_{X} (x),$ ⊥_X ∈ κ_R. By Theorem 3.9(1), since ${\bar{R}}^{⊥} (⊤_{X}) \geq ⊥ \Rightarrow ⊤_{X} = ⊤_{X}$ , ⊤_X ∈ κ_R.

(AF2) If $A_{i} = {\bar{R}}^{⊥} (A_{i})$ for i ∈ Γ, by Theorem 3.15(2), $\begin{matrix} {\bar{R}}^{⊥} (⋁_{i \in Γ} A_{i}) (y) = (⋁_{i \in Γ} A_{i})^{⊥} (y) \\ = ⋁_{x \in X} (R (x, y) \Rightarrow ⋁_{i \in Γ} A_{i} (x)) \\ = ⋁_{i \in Γ} ⋁_{x \in X} (R (x, y) \Rightarrow A_{i} (x)) \\ = ⋁_{i \in Γ} {\bar{R}}^{⊥} (A_{i}) (y) = ⋁_{i \in Γ} A_{i} . \end{matrix}$

Hence ⋁_i∈IA_i ∈ κ_R. We have $⋀_{i \in Γ} A_{i} = ⋀_{i \in Γ} {\bar{R}}^{⊥} (A_{i}) \geq {\bar{R}}^{⊥} (⋀_{i \in Γ} A_{i}) \geq ⊥ \Rightarrow ⋀_{i \in Γ} A_{i} = ⋀_{i \in Γ} A_{i}$ from Theorem 3.9(1). Hence ⋀_i∈γA_i ∈ κ_R.

(AF3) Let $A = {\bar{R}}^{⊥} (A)$ . By Theorem 3.9(6), ${\bar{R}}^{⊥} (α \oplus A) \leq α \oplus {\bar{R}}^{⊥} (A) = α \oplus A .$ By (2), ${\bar{R}}^{⊥} (α \oplus A) = (α \oplus A)^{⊥} \geq α \oplus A .$ Hence $α \oplus A = {\bar{R}}^{⊥} (α \oplus A)$ . Moreover, $\begin{matrix} {\bar{R}}^{⊥} (α \Rightarrow A) (y) = (α \Rightarrow A)^{⊥} (y) \\ = ⋁_{x \in X} (R^{N} (x, y) \Rightarrow (α \Rightarrow B (x)) \\ = α \Rightarrow ⋁_{x \in X} (R^{N} (x, y) \Rightarrow A (x)) \\ = α \Rightarrow {\bar{R}}^{⊥} (A) (y) = α \Rightarrow A . \end{matrix}$

Hence α ⊕ A, α ⇒ A ∈ τ_R.

(3) It follows from A ∈ τ_R; i.e. $A = {\underline{R}}^{⊤} (A)$ iff $N (A) = N ({\underline{R}}^{⊤} (A)) = {\bar{R}}^{⊥} (N (A))$ iff $N (A) = {\bar{R}}^{⊥} (N (A))$ iff N (A) ∈ κ_R.

Theorem 3.19. Let R ∈ L^X×X be an L-fuzzy relation and β ∈ L. Define two maps $T_{{\underline{R}}^{β}}, F_{{\bar{R}}_{β}} : L^{X} \to L$ as $\begin{matrix} T_{{\underline{R}}^{β}} (A) = ⋀_{x \in X} (A (x) \to {\underline{R}}^{β} (A) (x)), \\ F_{{\bar{R}}_{β}} (A) = ⋁_{x \in X} (A (x) \Rightarrow {\bar{R}}_{β} (A) (x)) . \end{matrix}$

Then the following properties hold.

$T_{{\underline{R}}^{β}}$ is an Alexandrov L-fuzzy pretopology.

$F_{{\bar{R}}_{β}}$ is an Alexandrov L-fuzzy precotopology.

For A ∈ L^X, $T_{{\underline{R}}^{N (β)}} (A) = N (F_{{\bar{R}}_{β}} (N (A)))$ .

Proof. {(1) (PT1) It follows from ${\underline{R}}^{β} (⊥_{X}) = ⊥_{X}$ .

(PT2) By Lemma 2.3(12),we have $\begin{matrix} T_{{\underline{R}}^{β}} (⋁_{i} A_{i}) = ⋀_{x \in X} (⋁_{i} A_{i}) (x) \to {\underline{R}}^{β} (⋁_{i} A_{i}) (x)) \\ \geq ⋀_{x \in X} (⋁_{i} A_{i}) (x) \to ⋁_{i} {\underline{R}}^{β} (A_{i}) (x)) \\ \geq ⋀_{i} ⋀_{x \in X} (A_{i} (x) \to {\underline{R}}^{β} (A_{i}) (x)) = ⋀_{i} T_{{\underline{R}}^{β}} (A_{i}) . \end{matrix}$

(PT3) By Lemma 2.3 (4) and Theorem 3.5(6), $\begin{matrix} T_{{\underline{R}}^{β}} (α ⊙ A) \\ = ⋀_{x \in X} (α ⊙ A (x) \to {\underline{R}}^{β} (α ⊙ A) (x)) \\ \geq ⋀_{x \in X} (α ⊙ A (x) \to α ⊙ {\underline{R}}^{β} (A) (x)) \\ \geq ⋀_{x \in X} (A (x) \to {\underline{R}}^{β} (A) (x)) = T_{{\underline{R}}^{β}} (A) . \end{matrix}$

Thus, $T_{{\underline{R}}^{β}}$ is an Alexandrov L-fuzzy pretopology.

(2) (CF1) It is easily proved from ${\bar{R}}_{β} (⊤_{X}) = ⊤_{X}$ .

(CF2) By Lemma 2.3(13), we have $\begin{matrix} F_{{\bar{R}}_{β}} (⋀_{i} A_{i}) = ⋁_{x \in X} (⋀_{i} A_{i} (x) \Rightarrow {\bar{R}}_{β} (⋀_{i} A_{i}) (x)) \\ \leq ⋁_{x \in X} (⋀_{i} A_{i} (x) \Rightarrow ⋀_{i} {\bar{R}}_{β} (A_{i}) (x)) \\ \leq ⋁_{x \in X} (⋀_{i} (A_{i} (x) \Rightarrow {\bar{R}}_{β} (A_{i}) (x))) \\ \leq ⋀_{i} ⋁_{x \in X} (A_{i} (x) \Rightarrow {\bar{R}}_{β} (A_{i}) (x))) = ⋀_{i} F_{{\bar{R}}_{β}} (A_{i}) . \end{matrix}$

(CF3) By Lemma 3.13(3) and Theorem 3.9(6), $\begin{matrix} F_{{\bar{R}}_{β}} (α \oplus A) = ⋁_{x \in X} ((α \oplus A) (x) \Rightarrow {\bar{R}}_{β} (α \oplus A) (x)) \\ \leq ⋁_{x \in X} ((α \oplus A) (x) \Rightarrow (α \oplus {\bar{R}}_{β} (A) (x))) \\ \leq ⋁_{x \in X} (A (x) \Rightarrow {\bar{R}}_{β} (A) (x)) = F_{{\bar{R}}_{β}} (A) . \end{matrix}$

Thus $F_{{\bar{R}}_{β}}$ is an Alexandrov L-fuzzy precotopology.

(3) For each A ∈ L^X, $\begin{matrix} N (F_{{\bar{R}}_{β}} (N (A))) \\ = N (⋁_{x \in X} (N (A) (x) \Rightarrow {\bar{R}}_{β} (N (A)) (x))) \\ = ⋀_{x \in X} N (N (A) (x) \Rightarrow {\bar{R}}_{β} (N (A)) (x)) \\ = ⋀_{x \in X} (A (x) \to N ({\bar{R}}_{β} (N (A))) (x)) \\ = ⋀_{x \in X} (A (x) \to {\underline{R}}^{N (β)} (A) (x)) = T_{{\underline{R}}^{N (β)}} (A) . \end{matrix}$

Example 3.20. Let (L = {0, a, b, 1}, ⊙ = ∧, ⊕ = ∨, N) be a complete residuated and co-residuated lattice with an involutive negation N : L → L as N (0) =1, N (1) =0, N (a) = b, N (b) = a where 0 < a, b < 1 and a and b are incomparable. We obtain →, ⇒ : L → L defined by x → y = ⋁ {z ∈ L ∣ x ∧ z ≤ y}, x ⇒ y = ⋀ {z ∈ L ∣ x ∨ z ≥ y} where $\begin{matrix} x \to y = {\begin{matrix} 1, & if x = 0 or y = 1, or x = y \\ b, & if x = a, y = 0, \\ b, & if x = a, y = b, \\ a, & if x = b, y = 0, \\ a, & if x = b, y = a, \\ y, & if x = 1, \end{matrix} \\ x \Rightarrow y = {\begin{matrix} 0, & if x = 1 or y = 0, or x = y \\ b, & if x = a, y = b, \\ b, & if x = a, y = 1, \\ a, & if x = b, y = a, \\ a, & if x = b, y = 1, \\ y, & if x = 0, \end{matrix} \end{matrix}$

Define X = {x₁, x₂, x₃} and R, R^N ∈ L^X×X as $R = (\begin{matrix} 1 & a & b \\ 0 & 1 & 0 \\ a & a & 1 \end{matrix}) R^{N} = (\begin{matrix} 0 & b & a \\ 1 & 0 & 1 \\ b & b & 0 \end{matrix})$

Then R is an (L, ∧)-fuzzy preorder on X. Moreover, R^N is an (L, ∨)-fuzzy copreorder on X.

(1) For B = (b, a, 1), we obtain $\begin{matrix} B_{a} = (b, 1, b) = {\underline{R}}^{a} (B), B_{b} = (1, a, 1) = {\underline{R}}^{b} (B) \\ B_{1} = (b, a, b) = {\underline{R}}^{1} (B), B_{0} = (1, 1, 1) = {\underline{R}}^{0} (B) \\ {\underline{R}}^{1} ({\underline{R}}^{a} (B)) = {\underline{R}}^{a} (B), {\underline{R}}^{1} ({\underline{R}}^{b} (B)) = {\underline{R}}^{b} (B) \\ {\underline{R}}^{a} ({\underline{R}}^{a} (B)) = {\underline{R}}^{a} (B), {\underline{R}}^{b} ({\underline{R}}^{b} (B)) = {\underline{R}}^{b} (B), \\ (1, 1, 1) = {\underline{R}}^{a} ({\underline{R}}^{b} (B)) = {\underline{R}}^{a \land b} (B) \geq {\underline{R}}^{a} (B) \\ (1, 1, 1) = {\underline{R}}^{b} ({\underline{R}}^{a} (B)) = {\underline{R}}^{a \land b} (B) \geq {\underline{R}}^{b} (B) . \end{matrix}$

Since N (B) = (a, b, 0) and ${\bar{R}}^{b} (N (B)) = (a, 0, a)$ , we have $T_{{\underline{R}}^{a}} (B) = N (F_{{\bar{R}}^{N (a)}} (N (B))) = b .$

(2) By(1), since N (B) = (a, b, 0) and N (b) = a, we obtain N (B) ^N(b) = (0, b, 0) = N (B_b) and ${\bar{R}}^{N (b)} (N (B)) = (0, b, 0) = N ({\underline{R}}^{b} (B)) .$

(3) For B = (0, a, b), we obtain $\begin{matrix} B^{a} = {\bar{R}}^{a} (B), B^{b} = (a, a, a) = {\bar{R}}^{b} (B) \\ B^{1} = (0, 0, 0) = {\bar{R}}^{1} (B), B^{0} = (1, a, b) = {\bar{R}}^{0} (B) \\ {\bar{R}}^{1} ({\bar{R}}^{a} (B)) = {\bar{R}}^{a} (B), {\bar{R}}^{1} ({\bar{R}}^{b} (B)) = {\bar{R}}^{b} (B) \\ {\bar{R}}^{a} ({\bar{R}}^{a} (B)) = {\bar{R}}^{a} (B), {\bar{R}}^{b} ({\bar{R}}^{b} (B)) = {\bar{R}}^{b} (B) \\ (0, 0, 0) = {\bar{R}}^{a} ({\bar{R}}^{b} (B)) = {\bar{R}}^{a \lor b} (B) \leq {\bar{R}}^{a} (B) \\ (0, 0, 0) = {\bar{R}}^{b} ({\bar{R}}^{a} (B)) = {\bar{R}}^{a \lor b} (B) \leq {\bar{R}}^{b} (B) . \end{matrix}$

From Theorem 3.18, we can obtain (4-5).

(4) $τ_{R} = {{\underline{R}}^{⊤} (A) ∣ A \in L^{X}}$ is an Alexandrov topology on X where ${\underline{R}}^{⊤} (A) = (\begin{matrix} A (x_{1}) \land (a \to A (x_{2})) \land (b \to A (x_{3})) \\ A (x_{2}) \\ (a \to A (x_{1}) \land (a \to A (x_{2}) \land A (x_{3}) \end{matrix})$

(5) $κ_{R} = {{\bar{R}}^{⊥} (A) ∣ A \in L^{X}}$ is an Alexandrov co-topology on X where ${\bar{R}}^{⊥} (A) = (\begin{matrix} A (x_{1}) \lor (b \Rightarrow A (x_{2})) \lor (a \Rightarrow A (x_{3})) \\ A (x_{2}) \\ (b \Rightarrow A (x_{1})) \lor (b \Rightarrow A (x_{2})) \lor A (x_{3}) \end{matrix})$

Then ${\underline{R}}^{⊤} (A) = N ({\bar{R}}^{⊤} (N (A)))$ .

From Theorem 3.19, we can obtain (4-5).

(6) For B = (b, a, 1), by (1), ${\underline{R}}^{a} (B) = B_{a} = (b, 1, b)$ . Since N (B) = (a, b, 0) and ${\bar{R}}^{b} (N (B)) = (a, 0, a)$ , we have $T_{{\underline{R}}^{a}} (B) = N (F_{{\bar{R}}^{N (a)}} (N (B))) = b .$

(7) For B = (0, a, b), by (3), ${\bar{R}}^{a} (B) = (b, 0, b)$ . Since N (B) = (1, b, a) and ${\underline{R}}^{b} (N (B)) = (a, 1, a)$ , we have $T_{{\underline{R}}^{N (a)}} (N (B)) = N (F_{{\bar{R}}^{a}} (B)) = a .$

4 Conclusion

In this paper, as an extension of variable precision (θ, σ)-fuzzy rough sets in Chen et al. [2], Qiao et al. [19] and Yao et al. [31], we study the notion of the generalized granular variable precision $({\underline{R}}^{β}, {\bar{R}}^{β})$ -fuzzy rough sets determined by fuzzy granules where R ∈ L^X×Y, e_{L
^Y} and d_{L
^Y} are (L, ⊙)-fuzzy preorder and (L, ⊕)-fuzzy co-preorder, respectively. For an (L, ⊙)-fuzzy preorder R ∈ L^X×X and ${\underline{R}}^{⊤}, {\bar{R}}^{⊥}$ , we obtain Alexandrov topologies and Alexandrov cotopologies on X, respectively. For an L-fuzzy relation R ∈ L^X×X and ${\underline{R}}^{β}, {\bar{R}}^{β}$ for each β ∈ L, we obtain Alexandrov L-fuzzy pretopologies and Alexandrov L-fuzzy precotopologies on X, respectively.

In the future, the generalized granular variable precision $({\underline{R}}^{β}, {\bar{R}}^{β})$ -fuzzy rough sets need to be applied formal concept analysis and decision-making systems.

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