On maximal filters in triangle algebras

Abstract

In this paper, we define two types of maximal filters in triangle algebras. The concepts of the radical of a triangle algebra and radical of an interval valued residuated lattice-filter (IVRL-filter) in triangle algebras are introduced. Some characterizations of the radical of an IVRL-filter are given. The notion of IVRL-extended semi maximal filter in triangle algebras also introduced. We investigate the relations between this notion and other types of IVRL-filters in triangle algebras. Special elements such as dense elements and double complemented elements in triangle algebras are defined. Properties of the collections of these elements as well as properties of IVRL-filters in triangle algebras are studies.

Keywords

Triangle algebra IVRL-filter maximal radical dense element double complemented element

1 Introduction

Van Gass et al. [11] introduced the notion of triangle algebras as a variety of residuated lattices equipped with approximation operators ν and μ together with a third angular point u different from 0 and 1. In that paper (Theorem 26) they showed that these algebras serve as an equational representation of interval-valued residuated lattices (IVRLs). Based on the definition and properties of triangle algebras, the authors defined triangle logic TL and showed that this logic is sound and complete with respect to the variety of triangle algebras.

The same authors in [10] introduced the notion of some filters such as Boolean and prime filters in triangle algebras and examined their mutual dependencies and connections.

This paper organized as follows: in the next section, we give some definitions, lemmas and theorems that are needed in the sequel. In Section 3, we define two types of maximal filters in triangle algebras and show that under some conditions they are equivalent. In Sections 4 and 5, the notion of radical in triangle algebras and radical of an IVRL filter are introduced. Next, the notion of IVRL-extended semi maximal filter is defined and the relationship between this and other types of IVRL filters is studied. By some examples we show that these notions are different. All these connections are given in a diagram of Fig. 1. In Section 6, special sets are defined and some results are obtained.

2 Preliminaries

2.1 Residuated lattices

Definition 2.1. [5] A residuated lattice is an algebra $L = (L, \lor, \land, *, \to, 0, 1)$ with four binary operations and two constant 0,1 such that:

(L, ∨ , ∧ , 0, 1) is a bounded lattice with 0 as smallest and 1 as greatest element,

∗ is commutative and associative, with 1 as neutral element, and

x ∗ y ≤ z if and only if x ≤ y → z, for all x, y and z in L (residuation principle).

The ordering ≤ and negation ¬ in a residuated lattice $L = (L, \lor, \land, *, \to, 0, 1)$ are defined as follow, for all x and y in L: x ≤ y if and only if x ∧ y = x (or equivalently, if and only if x ∨ y = y; or, also equivalently, if and only if x → y = 1) and ¬x = x → 0.

Lemma 2.2. [5, 9] Let (L, ∨ , ∧ , ∗ , → , 0, 1) be a residuated lattice. Then the following properties are valid, for all x, y and z in L:

x ∗ y ≤ x ∧ y, x ≤ x → y,

x ∗ (x → y) ≤ x ∧ y (in particular x ∗ ¬ = 0),

x ∨ y ≤ (x → y) → y (in particular x ≤ ¬¬ x), ¬¬ ¬ x = ¬ x,

x ≤ y if and only if x → y = 1, and x ∗ y = 0 if and only if x ≤ ¬ y,

(x → y) ∗ z ≤ x → (y ∗ z) ,

x → (y ∧ z) = (x → y) ∧ (x → z),

(x ∨ y) → z = (x → z) ∧ (y → z) (in particular ¬ (x ∨ y) = ¬ x ∧ ¬ y),

(x ∗ y) → z = x → (y → z) (in particular ¬ (x ∗ y) = x → ¬ y),

x ∗ (y ∨ z) = (x ∗ y) ∨ (x ∗ z),

(y → z) ≤ (x → y) → (x → z),

x → (y → z) = y → (x → z),

x → y ≤ (y → z) → (x → z),

x → ¬ y = y → ¬ x = ¬¬ x → ¬ y = ¬ (x ∗ y).

Definition 2.3. [5] A filter of a residuated lattice $L = (L, \lor, \land, *, \to, 0, 1)$ is a non-empty subset F of L satisfying:

for all x, y in L: if x ∈ F and x ≤ y, then y ∈ F,

for all x and y in F: x ∗ y ∈ F(i.e. F is closedunder ∗).

Definition 2.4. [5 , 9] Let F be a filter of a residuated lattice L. Then

F is said to be proper if F ≠ L. A proper filter F of L is said to be a maximal filter if and only if it is not included in any other proper filter of L.

A Boolean filter of L is a filter F satisfying:

for all x in L: x ∨ ¬ x ∈ F.

A Boolean filter of the second kind of L is a filter F satisfying:

for all x in L: x ∈ F or ¬x ∈ F (or both).

A prime filter of L is a filter F satisfying:

for all x and y in L: x → y ∈ F or x → y ∈ F (or both).

A prime filter of the second kind is a filter F satisfying:

for all x and y in L: if x ∨ y ∈ F, then x ∈ F or y ∈ F (or both).

An implicative filter of $L$ is a subset F of L such that

1 ∈ F and for all x, y and z in L: if (x ∗ y) → z ∈ F and x → y ∈ F, then x → z ∈ F.

A positive implicative filter of $L$ is a subset F of L such that

(x → y) → x ∈ F implies x ∈ F.

An obstinate filter of L is a subset F of L such that

for all x, y in L: if x, y ∉ F imply x → y ∈ F and y → x ∈ F.

An alternative definition for a filter F (see e.g. [9]) of a residuated lattice $L = (L, \lor, \land, *, \to, 0, 1)$ is the following:

1 ∈ F,

for all x and y in L: if x ∈ F and x → y ∈ F, then y ∈ F.

Definition 2.5. [5] The order of x ∈ L, denoted by ord (x), is the smallest $n \in ℕ$ such that xⁿ = 0 . If there is no such n, then ord (x) = ∞ .

2.2 Interval-valued residuated lattice and triangle algebras

Definition 2.6. [2] Given a lattice $A = (A, \lor, \land)$ its triangularization $𝕋 (A)$ is the structure $𝕋 (A) = (Int (A), \lor, \land)$ defined by

$Int (A) = {[x_{1}, x_{2}] : (x_{1}, x_{2}) \in A^{2}$ and x₁ ≤ x₂},

[x₁, x₂] ∧ [y₁, y₂] = [x₁ ∧ y₁, x₂ ∧ y₂],

[x₁, x₂] ∨ [y₁, y₂] = [x₁ ∨ y₁, x₂ ∨ y₂].

The set $D_{A} = {[x, x] : x \in A}$ is called the diagonal of $𝕋 (A)$ .

Definition 2.7. [11] An interval-valued residuated lattice (IVRL) is a residuated lattice $(Int (A), \lor, \land, ⊙, \to_{⊙}, [0, 0], [1, 1])$ on the triangularization $𝕋 (A)$ of a bounded lattice $A$ , in which the diagonal $D_{A}$ is closed under ⊙ and →_⊙, i.e. $[x, x] ⊙ [y, y] \in D_{A}$ and $[x, x] \to_{⊙} [y, y] \in D_{A}$ for all x, y in A.

Definition 2.8. [11] A triangle algebra is a structure $A = (A, \lor, \land, *, \to, ν, μ, 0, u, 1)$ in which (A, ∨ , ∧ , ∗ , → , 0, 1) is a residuated lattice, ν and μ are unary operators on A, u a constant, and satisfying the following conditions:

(T . 7) ν (x → y) ≤ νx → νy,

(T . 8) (νx ↔ νy) ∗ (μx ↔ μy) ≤ (x ↔ y) ,

(T . 9) νx → νy ≤ ν (νx → νy) .

Definition 2.9. [10] Let $A = (A, \lor, \land, *, \to, ν, μ, 0, u, 1)$ be a triangle algebra. An element x in A is called exact if νx = x. The set of exact elements of $A$ is denoted by $E (A)$ .

Theorem 2.10. [10] In a triangle algebra $A = (A, \lor, \land, *, \to, ν, μ, 0, u, 1)$ , the implication → and the product ∗ are completely determined by More specifically:

ν (x → y) = (νx → νy) ∧ (μx → μy) ,

μ (x → y) = (μx → (μ (u ∗ u) → μy)) ∧ (νx → μy) ,

ν (x ∗ y) = νx ∗ νy,

μ (x ∗ y) = (νx ∗ μy) ∨ (μx ∗ νy) ∨ (μx ∗ μy ∗ μ (u ∗ u)) .

Definition 2.11. [10] An IVRL-filter (IF) of triangle algebra $A = (A, \lor, \land, *, \to, ν, μ, 0, u, 1)$ is a non-empty subset F of A satisfying:

if x ∈ F, y ∈ A and x ≤ y, then y ∈ F,

if x, y ∈ F, then x ∗ y ∈ F,

if x ∈ F, then νx ∈ F.

For all x, y ∈ A, we write x ∼ _Fy if and only if x → y and y → x are both in F.

Van Gass et al. suggested two different ways to define specific kinds of IVRL-filters of triangle algebras. The first is to impose a property on a filter of the subalgebra of exact elements and extend this filter to the whole triangle algebra, using (F . 3). We call these IVRL-extended filters. For example, An IVRL-extended prime filter of triangle algebra $A = (A, \lor, \land, *, \to, ν, μ, 0, u, 1)$ is a subset F of A such that $F \cap E (A)$ is a prime filter of $E (A)$ and x ∈ F if and only if $ν x \in F \cap E (A) .$ The second way is to impose a property on the whole IVRL-filter. For example, a prime IVRL-filter of a triangle algebra $A = (A, \lor, \land, *, \to, ν, μ, 0, u, 1)$ is an IVRL-filter of A such that F is a prime filter of (A, ∨ , ∧ , ∗ , → , 0, 1) [10].

3 Maximal and prime filters in triangle algebras

Hereafter, (A, ∨ , ∧ , ∗ , → , ν, μ, 0, u, 1) or A is a triangle algebra unless otherwise specified.

Definition 3.1. A proper filter F of A is said to be a maximal IVRL-filter if and only if it is not included in any other proper filter of A and the condition of (F . 3) meet. Also, a proper IVRL-filter F of A is said to be an IVRL-extended maximal filter if and only if it is not included in any other proper IVRL-filter of A.

By the definition of generated IVRL-filter in a triangle algebra [12], if F and G are IVRL-filters of A, then

[F ∪ G) = {a ∈ A : νa ≥ f ∗ g for some f ∈ F, g ∈ G}.

Theorem 3.2. M is an IVRL-extended maximal filter of A if and only if for all x ∈ A, x ∉ M there exist m ∈ M, n ≥ 1 such that m ∗ νxⁿ = 0.

Proof. Let M be an IVRL-extended maximal filter of A and x ∉ M . Then by Proposition 4.2 [12] we have $\begin{matrix} [M \cup {x}) & = & {t \in A : ν t \geq a * x^{n}, \\ for all a \in M, n \geq 1} = A . \end{matrix}$

Hence 0 ∈ [M ∪ {x}). So there exist n ≥ 1 and m ∈ M such that m ∗ νxⁿ ≤ m ∗ xⁿ ≤ 0. Thus m ∗ νxⁿ = 0. So ¬ (νxⁿ) = m ∈ M.

Conversely, assume that there is a proper IVRL-filter F′ such that M ⊆ F′ . Then there exists x ∈ F′ such that x ∉ M . By hypothesis there exist m ∈ M and n ≥ 1 such that m ∗ νxⁿ = 0 . But x, m ∈ F′; hence 0 ∉ F′, that is a contradiction.□

Proposition 3.3. Let F ⊆ A . Then F is a maximal IVRL-filter of A if and only if F is an IVRL-extended maximal filter of A and ¬ (μ (u ∗ u)) ⁿ ∈ F.

Proof. IVRL-filter F of A is a maximal IVRL-filter if and only if for all x ∈ A if x ∉ F, there exists $n \in ℕ$ such that ν (¬ xⁿ) ∈ F. By (T . 7), we have ν (¬ xⁿ) ≤ νxⁿ → ν0, so F is an IVRL-extended maximal filter of A. By Theorem 2.10, ν (¬ xⁿ) = (νxⁿ → ν0) ∧ (μxⁿ → μ0) = ¬ (μxⁿ) = ¬ ((νx ∗ μ (xⁿ)) ∨ (μ x ∗ νxⁿ) ∨ ((μx ∗ μ (x^n-1)) ∗ μ (u ∗ u))). Taking x = u,we get that ¬ (μ (u ∗ u)) ⁿ ∈ F.

Conversely, let F be an IVRL-extended maximal filter of A and ¬μ (u ∗ u) ∈ F. Then νxⁿ ∗ ¬ μ (u ∗ u) ∈ F . By Lemma 2.2, we have $\begin{matrix} ν x^{n} * \neg μ (u * u) \leq \neg μ (u * u) \\ \leq (μ x * μ (x^{n - 1})) \to \neg μ (u * u) \\ = μ x \to (μ (x^{n - 1}) \to \neg μ (u * u)) \\ = μ x \to ((\neg μ (x^{n - 1})) \land (μ (x^{n - 1}) \to \neg μ (u * u))) \\ = (μ x \to \neg μ (x^{n - 1})) \land (μ x \to (μ (x^{n - 1}) \\ \to \neg μ (u * u))) \\ = (μ x \to \neg μ (x^{n - 1})) \land ((μ x * μ (x^{n - 1})) \\ \to \neg μ (u * u)) \\ = (μ x \to (\neg μ (x^{n - 1}) \land \neg ν x^{n - 1})) \\ \land ((μ x * μ (x^{n - 1})) \to \neg μ (u * u)) \\ = (μ x \to \neg μ (x^{n - 1})) \land (μ x \to \neg ν x^{n - 1})) \\ \land ((μ x * μ (x^{n - 1})) \to \neg μ (u * u)) \\ \leq (ν x \to \neg μ (x^{n - 1})) \land (μ x \to \neg ν x^{n - 1})) \\ \land ((μ x * μ (x^{n - 1})) \to \neg μ (u * u)) \\ = \neg (ν x * μ (x^{n - 1})) \land \neg (μ x * ν x^{n - 1}) \\ \land \neg ((μ x * μ (x^{n - 1})) * μ (u * u)) \\ = \neg ((ν x * μ (x^{n - 1})) \lor (μ x * ν x^{n - 1}) \\ \lor ((μ x * μ (x^{n - 1})) * μ (u * u))) \\ = \neg μ (x^{n}) \\ = ν (\neg x^{n}) . □ \end{matrix}$

Van. Gass et al. [10] showed that every IVRL-extended prime filter is an IVRL-extended prime filter of the second kind but the converse is true if (x → y) ⊔ (y → x) =1 (prelinearity), for all x, y ∈ A.

By Lemma 2.2, we have

Proposition 3.4. Let F be an IVRL-filter of A. If a, b ∈ A such that a ∨ b ∈ F, then F (a) ∩ F (b) = F, where F (a) = [F ∪ {a}).

Corollary 3.5. Let F, F₁, F₂ be IVRL-filters of A. Then the following are equivalent:

If F = F₁ ∩ F₂, then F = F₁ or F = F₂,

F is an IVRL-extended prime filter of the second kind.

Theorem 3.6. Let F be an IVRL-filter of A. Then the following statements hold:

Any IVRL-extended prime filter of the second kind of A is contained in a unique IVRL-extended maximal filter of A.

If a ∈ A \ F, then there is an IVRL-extended prime filter of the second kind P of A such that F ⊆ P and a ∉ P.

Proof. (i) Let M be an IVRL-extended maximal filter of A. By Corollary 3.5, if F₁, F₂ are IVRL-filters of A such that M = F₁ ∩ F₂. By the maximality of M we deduce that M = F₁ or M = F₂ . Hence M is an IVRL-extended prime filter of the second kind of A.

(ii) Let F^′′ = {F′ : F ⊆ F′, F′ ∩ [a) = ∅}. Since the condition of Zorn^,s lemma hold, we deduce that F^′′ has a maximal element P. Suppose that P is not an IVRL- extended prime filter of the second kind. Hence there are a, b ∈ A, such that νa ∨ νb ∈ P, but νa ∉ P, νb ∉ P. By the maximality of P, we have P (a) , P (b) ∉ F^′′ . Hence P (a)∩ [a) ≠ ∅ and P (b)∩ [a) ≠ ∅. So, there are x ∈ P (a) ∩ [a), y ∈ P (b) ∩ [a) , such that νx ≥ f ∗ a^m, νy ≥ g ∗ bⁿ, where $f, g \in P, m, n \in ℕ$ . Then νx ∨ νy ≥ (f ∗ a^m) ∨ (g ∗ bⁿ) ≥ (f ∨ g) ∗ (g ∨ a^m) ∗ (f ∨ bⁿ) ∗ (a^m ∨ bⁿ) . Since f ∨ g, g ∨ a^m, f ∨ bⁿ ∈ P, νx ∨ νy ∈ P, but νx ∨ νy ∈ [a) . Thus P ∩ [a) ≠ ∅ , that is a contradiction. Therefore, P is an IVRL-extended prime filter of the second kind.□

Corollary 3.7. Every proper IVRL-filter of A is the intersection of those IVRL-extended prime filter of the second kind which contains F.

4 Radical of a triangle algebra

Here, we introduce the notion of radical of a triangle algebra and study it in detail.

Definition 4.1. The intersection of all IVRL-extended maximal filter of A is called the radical of A and is denoted by Rad (A).

It is clear that Rad (A) ≠ A since IVRL-extended maximal filters of A are proper. Since 1 ∈ F, Rad (A)≠ ∅, where F is an IVRL-extended maximal filter of A.

Theorem 4.2. Let A be a triangle algebra. Then $Rad (A) \supseteq {a \in A : ν a \geq \neg (ν a^{n}) for any n \in ℕ} .$

Proof. Let a ∉ Rad (A). Then there exists an IVRL-extended maximal filter M of A such that F ⊆ M an a ∉ M. Since M is an IVRL-extended maximal filter of A, there exists $n \in ℕ$ such that ¬ (νaⁿ) ∈ M. We have ¬ (νaⁿ) → νa = 1 ∈ M, and so νa ∈ M. Then νaⁿ ∈ M and 0 = ¬ (νaⁿ) ∗ νaⁿ ∈ M, that is a contradiction. Therefore a ∈ Rad (F).□

Definition 4.3. A triangle algebra A is called an MTL-algebra if (a → b) ∨ (b → a) =1 (prelinearity), for all a, b ∈ A.

Theorem 4.4. Let A be an MTL- triangle algebra. Then $Rad (A) = {a \in A : ν a \geq \neg (ν a^{n}) for any n \in ℕ} .$

Proof. By Theorem 4.2, we have to show Rad (A) ⊆ {a ∈ A : νa ≥ ¬ (νaⁿ) for any $n \in ℕ}$ . Let a ∈ Rad (A) and there exists $k \in ℕ$ such that νa≱ ¬ (νa^k). Then by Corollary 3.7, there exists an IVRL-extended prime filter P of A such that ¬ (νa^k) → νa ∉ P . Since P is an IVRL-extended prime filter of A, we obtain νa → ¬ (νa^k) ∈ P. By Theorem 3.6 (i), there exists a unique IVRL-extended maximal filter M of A such that P ⊆ M . So νa → ¬ (νaⁿ) ∈ M . If a ∈ M, then aⁿ ∈ M, for all $n \in ℕ$ ; hence a^k ∈ M in particular νa^k ∈ M. In addition, we have νa → ¬ (νaⁿ) ∈ M . Thus ¬ (νa^k) ∈ M and so 0 = νa^k ∗ ¬ (νa^k) ∈ M, which is a contradiction. Hence a ∉ M. We have F ⊆ P ⊆ M and a ∉ M. Thus a ∉ Rad (F), which is a contradiction. Therefore, νa → ¬ (νa^k) ∈ F, for all $k \in ℕ$ .□

Example 4.5. Let A = {[0, 0] , [0, a] , [0, b] , [a, a] , [b, b] , [0, 1] , [a, 1] , [b, 1] , [1, 1]}. Define ⊙ and ⇒ as follow:

⊙	a	b	1
0	0	0	0
a	a	0	a
b	0	b	b
1	a	b	1

⇒	0	a	b	1
0	1	1	1	1
a	b	1	b	1
b	a	a	1	1
1	0	a	b	1

and we define ν,μ, ∗ and → one as follow:

ν [x₁, x₂] = [x₁, x₁], μ [x₁, x₂] = [x₂, x₂],[x₁, x₂] ∗ [y₁, y₂] = [x₁ ⊙ y₁, x₂ ⊙ y₂],[x₁, x₂] → [y₁, y₂] = [(x₁ ⇒ y₁) ⊓ (x₂ ⇒ y₂) , x₂ ⇒ y₂].

Then (A, ∨ , ∧ , ∗ , → , ν, μ, [0, 0] , [0, 1] , [1, 1]) is an MTL-triangle algebra with [0, 0] as smallest and [1, 1] as greatest element.

It is clear that F₁ = {[b, b] , [b, 1] , [1, 1]} and F₂ = {[a, a] , [a, 1] , [1, 1]} are IVRL-extended maximal filters of A. Thus Rad (A) = {[1, 1]}.

5 Radicals of IVRL-filters in triangle algebras

Here, we introduce the notion of radical of an IVRL-filter in triangle algebras and study it in detail.

According to Proposition 3.3, we define radical of IVRL-extended maximal filters of trianglealgebras.

Definition 5.1. Let F be a proper IVRL-filter of triangle algebra A. The intersection of all IVRL-extended maximal filters of A that contain F is called the radical of F and is denoted by Rad (F).

It is clear that Rad (F) is an IVRL-filter and F ⊆ Rad (F).

In the following example we show that if F ≠ {1}, then Rad (F) does not always coincide with Rad ({1}).

Example 5.2. In Example 4.5, it is clear that F = {[a, a] , [a, 1] , [1, 1]} is an IVRL-filter. We obtain that Rad (F) = {[a, a] , [a, 1] , [1, 1]} = F ≠ {[1, 1]} = Rad ({[1, 1]}).

By Theorem 3.6, we have the following lemma.

Lemma 5.3. (1) If M is an IVRL-extended maximal filter of triangle algebra A, then Rad (M) = M.

(2) If P is an IVRL-extended prime filter of the second kind, then there exists an unique IVRL-extended maximal filter M of triangle algebra A such that P ⊆ M . Therefore Rad (P) = M.

In the following theorem we characterize Rad (F), where F is a proper IVRL-filter of trianglealgebra A.

Theorem 5.4. Let F be an IVRL-filter of A. Then $Rad (F) \supseteq {a \in A : \neg (ν a^{n}) \to ν a \in F, for all n \in ℕ} .$

From now A is an MTL-triangle algebra unless otherwise specified.

Theorem 5.5. Let F be an IVRL-filter of A. Then $Rad (F) = {a \in A : \neg (ν a^{n}) \to ν a \in F, for all n \in ℕ} .$

Theorem 5.6. Let F be an IVRL-filter of A. Then a ∈ Rad (F) implies that ¬νa → νaⁿ ∈ F, for all $n \in ℕ$ .

Proof. Let a ∈ Rad (F). Since Rad (F) is an IVRL-filter of A, νaⁿ ∈ Rad (F), for all $n \in ℕ$ . By Theorem 5.5, ¬ ((νaⁿ) ^m) → νaⁿ ∈ F, for all $m \in ℕ$ . We have νa^nm ⩽ νa and by Lemma 2.2, ¬ ((νaⁿ) ^m) → νaⁿ ≤ ¬ νa → νaⁿ. So, ¬νa → νaⁿ ∈ F, for all $n \in ℕ$ .□

Theorem 5.7. Let F and G be IVRL-filters of A and a, b ∈ A. Then the following conditionshold:

If F ⊆ G, then Rad (F) ⊆ Rad (G).

Rad (Rad (F)) = Rad (F).

If a is a nilpotent element of A, then a ∉ Rad (F). Also, Rad (F) ⊆ {a ∈ A : ord (a) = ∞}, where F is proper IVRL-filter of A.

If a, b ∈ Rad (F), then ¬νa → νb ∈ F.

If a, b ∈ Rad (F), then ¬ (¬ νa ∗ ¬ νb) ∈ F.

If Rad (F) = A \ {0}, where F is proper IVRL-filter of A, then ¬a = 0, for all a ∈ A \ {0}.

If F is a linear triangle algebra and a ∈ Rad (F), then (¬ (νaⁿ) ∗ ¬ (νaⁿ)) → νa = 1, for all $n \in ℕ$ , where F is proper IVRL-filter of A.

If for every a ∈ F there exists $k \in ℕ$ such that a^k ∈ G, then Rad (F) ⊆ Rad (G).

[Rad (F) ∪ Rad (G)) ⊆ Rad ([F ∪ G)).

Proof. (1) It is clear.

(2) By (1), we have Rad (F) ⊆ Rad (Rad (F)) . We show that Rad (Rad (F)) ⊆ Rad (F). Let x ∈ Rad (Rad (F)). Then x ∈ M, for all IVRL-extended maximal filter M of A containing Rad (F). Let M₀ be an arbitrary IVRL-extended maximal filter of A containing F. Then M₀ = Rad (M₀) ⊇ Rad (F) and so x ∈ M₀ . Thus x ∈ Rad (F) , that is Rad (Rad (F)) ⊆ Rad (F). Therefore Rad (Rad (F)) = Rad (F).

(3) Let a be a nilpotent element of A and a ∈ Rad (F). Thus there exists $m \in ℕ$ such that a^m = 0. We obtain that a^m ∈ Rad (F) . So 0 ∈ Rad (F) , which is a contradiction. Thus a ∉ Rad (F).

(4) Let a, b ∈ Rad (F) . Then a ∗ b ∈ Rad (F) and ¬ (ν (a ∗ b)) → ν (a ∗ b) ∈ F. We have a ∗ b ⩽ a, thus ¬ (ν (a ∗ b)) → ν (a ∗ b) ⩽ ¬ νa → ν (a ∗ b) . Hence ¬νa → ν (a ∗ b) ∈ F . Since a ∗ b ⩽ b, a ∗ b → b = 1 ∈ F. So (¬ νa → ν (a ∗ b)) ∗ (ν (a ∗ b) → νb) ∈ F. By Lemma 2.2, we obtain ¬νa → νb ∈ F .

(5) Let a, b ∈ Rad (F) . By (4), we have ¬νa → νb ∈ F . Since νb ⩽ ¬¬ νb, ¬νa → νb ⩽ ¬ νa → ¬¬ νb, thus ¬νa → ¬¬ νb ∈ F. By Lemma 2.2, ¬ (¬ νa ∗ ¬ νb) ∈ F .

(6) Let Rad (F) = A \ {0} and a ∈ A such that ¬a ≠ 0 . By hypothesis, we have ¬a, a ∈ Rad (F) , hence 0 ∈ Rad (F) , which is acontradiction.

(7) Let a ∈ Rad (F) . Consider two elements ¬ (νaⁿ) → νa and ¬ (νaⁿ) in A for fixed $n \in ℕ$ . By hypothesis, we have ¬ (νaⁿ) → νa ≤ ¬ (νaⁿ) or ¬ (νaⁿ) ≤ ¬ (νaⁿ) → νa . Let ¬ (νaⁿ) → νa ≤ ¬ (νaⁿ). Since a ∈ Rad (F), ¬ (νaⁿ) → νa ∈ F . Thus ¬ (νaⁿ) ∈ F and so νa ∈ F. Hence νaⁿ ∈ F, for all $n \in ℕ,$ so 0 = νaⁿ ∗ ¬ (νaⁿ), which is a contradiction. Thus, ¬ (νaⁿ) ≤ ¬ (νaⁿ) → νa . Then (¬ (νaⁿ) → ¬ (νaⁿ)) → νa = 1, for all $n \in ℕ .$ So (¬ (νaⁿ) → νa) → νa = 1, for all $n \in ℕ .$

(8) Let a ∈ F . Assume there exists $k \in ℕ$ such that a^k ∈ G . we have a^k ≤ a and thus a ∈ G . So by (1) we have Rad (F) ⊆ Rad (G).

(9) Since F, G ⊆ [F ∪ G) , Rad (F) , Rad (G) ⊆ Rad ([F ∪ G)), and so Rad (F) ∪ Rad (G) ⊆ Rad ([F ∪ G)) .□

From parts (1) , (2) of Theorem 5.7, radical of an IVRL-filter is closure operator on Fil (A).

The following example shows that the converse of Theorem 5.7 (3),(4) and (5) may not hold.

Example 5.8. Consider Example 4.5. It is clear that F = {[b, b] , [b, 1] , [1, 1]} is an IVRL-filter and Rad (F) = {[b, b] , [b, 1] , [1, 1]}.

{a ∈ A : ord (a) = ∞} = {[0, a] , [0, b] , [0, 1] ,

[a, a] , [b, b] , [a, 1] , [b, 1] , [1, 1]} ⊈ Rad (F) .

¬ν [a, a] → ν [b, b] = [b, b] → [b, b] = [1, 1] ∈ F, while [a, a] ∉ F. So the converse of Theorem 5.7 (4) is not true in general.

¬ (¬ ν [a, a] ∗ ¬ ν [b, b]) = ¬ ([b, b] ∗ [a, a]) = ¬ [0, 0] = [1, 1] ∈ F, while [a, a] ∉ Rad (F). Hence the converse of Theorem 5.7 (5) is not true in general.

In the following example we show that the converse of Theorem 5.7 (8) is not true.

Example 5.9. [11] Let $L^{I} = (L^{I}, \lor, \land, *, \to, 0, 1)$ be a residuated lattice in which: $x * y = \min (x, y) and x \to y = {\begin{matrix} 1 & x \leq y \\ y & y < x \end{matrix} .$

[x₁, x₂] ⊙ [y₁, y₂] = [x₁ ∗ y₁, x₂ ∗ y₂] , [x₁, x₂] ⇒ [y₁, y₂] = [(x₁ → y₁) ∧ (x₂ → y₂) , x₂ → y₂].

Then the structure $L^{I} = (L^{I}, \lor, \land, ⊙, \Rightarrow, [0, 0], [1, 1])$ is a residuated lattice, too. Consider $ν [x_{1}, x_{2}] = [x_{1}, x_{1}], μ [x_{1}, x_{2}] = [x_{2}, x_{2}], u = [0, 1]$ then $(Int (L), \lor, \land, ⊙, \Rightarrow, ν, μ, [0, 0], [0, 1], [1, 1])$ is an MTL-triangle algebra. Let $F = {[x_{1}, x_{2}] \in Int (L) : x_{1}, x_{2} \geq 0.8}$ and G = {[1, 1]}. Then $Rad (F) = Rad (G) = Int (L) \ [0, 0]$ , but for every a ∈ F there is not $k \in ℕ$ such that a^k ∈ G .

Now, we investigate the relationships between the notions of various types of IVRL-filters and radicals of IVRL-filters in MTL-triangle algebras.

Lemma 5.10. Let F and G be IVRL-filters of A. Then $Rad ([F \cup G)) \subseteq Rad ([Rad (F) \cup Rad (G))) .$

Proof. Since F ⊆ Rad (F) and G ⊆ Rad (G), we have F ∪ G ⊆ Rad (F) ∪ Rad (G), therefore [F ∪ G) ⊆ [Rad (F) ∪ Rad (G)). Thus, Rad ([F ∪ G)) ⊆ Rad ([Rad (F) ∪ Rad (G))).□

In the following example we show that the converse of above lemma and the converse of Theorem 5.7 (9) are not true.

Example 5.11. Let L = {0, a, b, c, d, e, f, g, 1}, where 0 < a < b < c < d < f < g < 1, 0 < a < b < c < d < e < g < 1. Define ⊙ and ⇒ as follow:

⊙	a	b	c	d	e	f	g	1
0	0	0	0	0	0	0	0	0
a	0	0	a	a	a	a	a	a
b	0	b	b	b	b	b	b	b
c	a	b	c	c	c	c	c	c
d	a	b	c	c	c	c	c	d
e	a	b	c	c	e	c	e	e
f	a	b	c	c	c	f	f	f
g	a	b	c	c	e	f	g	g
1	a	b	c	d	e	f	g	1

⇒	0	a	b	c	d	e	f	g	1
0	1	1	1	1	1	1	1	1	1
a	b	1	1	1	1	1	1	1	1
b	a	b	1	1	1	1	1	1	1
c	0	a	b	1	1	1	1	1	1
d	0	a	b	g	1	1	1	1	1
e	0	a	b	f	f	1	f	1	1
f	0	a	b	e	e	e	1	1	1
g	0	a	b	d	d	e	f	1	1
1	0	a	b	c	d	e	f	g	1

And we define ν, μ, ∗ , → as follow

ν [x₁, x₂] = [x₁, x₁], μ [x₁, x₂] = [x₂, x₂], [x₁, x₂] ∗ [y₁, y₂] = [x₁ ⊙ y₁, x₂ ⊙ y₂], [x₁, x₂] → [y₁, y₂] = [(x₁ ⇒ y₁) ⊓ (x₂ ⇒ y₂) , x₂ ⇒ y₂]. A = {[x, y] : x, y ∈ L, x ≤ y}, so A has 44 elements. Thus (A, ∨ , ∧ , ∗ , → , ν, μ, [0, 0] , [0, 1] , [1, 1]) is an MTL-triangle algebra. It is clear that F = A \ {[0, 0] , [0, a] , . . . [0, 1] , [a, a] , . . . , [a, 1]} and G = {[f, f] , [f, 1] , [f, g] , [g, g] , [g, 1] , [1, 1]} are IVRL-filters of A. We have Rad (F) = Rad (G) = F. Thus

[Rad (F) ∪ Rad (G)) = A, so Rad ([Rad (F) ∪ Rad (G))) = A . [F ∪ G) = A \ {[0, 0] , [0, a] , . . . [0, 1] ,

[a, a] , . . . , [a, 1]} . So, Rad ([F ∪ G)) = A \ {[0, 0] , [0, a] , . . . [0, 1]}. Hence, Rad ([Rad (F) ∪ Rad (G))) ⊈ Rad ([F ∪ G)).

[Rad (F) ∪ Rad (G)) = A and Rad ([F ∪ G)) = A \ {[0, 0] , [0, a] , . . . [0, 1]}. Hence the equality of Theorem 5.7 (9) is not true in general.

It is clear that F ⊆ Rad (F). So we have the following proposition.

Proposition 5.12. Let F be an IVRL-extended (prime, obstinate, fantastic) Boolean filter of A. Then so is Rad (F).

Proposition 5.13. Let {F_i} _i∈I be a family of IVRL-filters of A. Then Rad (∩ _i∈IF_i) = ∩ _i∈IRad (F_i).

Proof. We have ∩_i∈IF_i ⊆ F_i ⊆ Rad (F_i) for all i ∈ I . By Theorem 5.7 (1) and (2), we have Rad (∩ _i∈IF_i) ⊆ Rad (F_i), for all i ∈ I . Thus Rad (∩ _i∈IF_i) ⊆ ∩ _i∈IRad (F_i).

Conversely, let a ∈ ∩ _i∈IRad (F_i) . Then a ∈ Rad (F_i) and so ¬ (νaⁿ) → νa ∈ F_i for all i ∈ I, $n \in ℕ$ . Hence ¬ (νaⁿ) → νa ∈ ∩ _i∈IF_i for all $n \in ℕ,$ so a ∈ Rad (∩ _i∈IF_i) . This implies that Rad (∩ _i∈IF_i) = ∩ _i∈IRad (F_i).□

Proposition 5.14. If F and G are IVRL-filters of A, then [a] _Rad(F∩G) ⊆ [a] _Rad(F) ∩ [a] _Rad(G).

Proof. We have Rad (F ∩ G) ⊆ Rad (F) ∩ Rad (G). Then $\begin{matrix} [a]_{Rad (F \cap G)} = {y \in A : a \equiv_{Rad (F \cap G)} b} \\ = {b \in A : a \to b \in Rad (F \cap G), b \\ \to a \in Rad (F \cap G)} \\ \subseteq {b \in A : a \to b \in Rad (F) \cap Rad (G), \\ b \to a \in Rad (F) \cap Rad (G)} \\ = {b \in A : a \to b \in Rad (F), \\ b \to a \in Rad (F)} \cap {b \in A : a \to b \in Rad (G), \\ b \to a \in Rad (G)} . \end{matrix}$

So [a] _Rad(F∩G) ⊆ [a] _Rad(F) ∩ [a] _Rad(G).□

Definition 5.15. Let A and B be two triangle algebras. It is easy to see that A × B with the corresponding operations defined pointwise is a triangle algebra

(a, b) ∧ (c, d) = (a ∧ c, b ∧ d) , (a, b) ∨ (c, d) = (a ∨ c, b ∨ d) , (a, b) → (c, d) = (a → c, b → d) , (a, b) ∗ (c, d) = (a ∗ c, b ∗ d) , (a, b) ≤ (c, d) = (a ≤ c, b ≤ d) , ν (a, b) = (νa, νb) , μ (a, b) = (μa, μb) .

We can also show that if F and G are IVRL-filters of A and B respectively, then F × G is an IVRL- filter of A × B.

Proposition 5.16. If F and G are IVRL-filters of A, then Rad (F × G) = Rad (F) × Rad (G).

Proof. Rad (F × G) = {(a, b) ∈ A × B : (¬ (ν (a, b) ⁿ))

→ν (a, b) ∈ F × G} = {(a, b) ∈ A × B : (¬ ν (aⁿ) , ¬ ν (bⁿ)) → (νa, νb) ∈ F × G} = {(a, b) ∈ A × B : (¬ ν (aⁿ) → νa, ¬ ν (bⁿ) → νb) ∈ F × G} = Rad (F) × Rad (G) . □

Corollary 5.17. Let {F_i} _i∈I be a family of IVRL-filters of A. Then $Rad (\prod_{i \in I} F_{i}) = \prod_{i \in I} Rad (F_{i})$ .

Definition 5.18. Let $A = (A, \lor, \land, *, \to, ν, μ, 0, u, 1)$ and $B = (B, ⊔, ⊓, ⊙, \Rightarrow, \bar{ν}, \bar{μ}, 0, \bar{u}, 1)$ be any two triangle algebras. If the mapping h : A ⟶ B satisfies for a, b ∈ A

$h (a \lor b) = h (a) ⊔ h (b), h (a \land b) = h (a) ⊓ h (b), h (a * b) = h (a) ⊙ h (b), h (a \to b) = h (a) \Rightarrow h (b), h (ν a) = \bar{ν} h (a), h (μ a) = \bar{μ} h (a) .$

Then h is called a homomorphism.

Proposition 5.19. Let F and G be IVRL-filters of A. Then (A × B)/Rad (F × G) ≅ A/Rad (F) × B/Rad (G).

Proof. Consider the natural homomorphisms

α : A → A/Rad (F) andδ : B → B/Rad (G). That α (a) = [a] , β (b) = [b]. We define Φ : A × B → A/Rad (F) × B/Rad (G) by Φ (a, b) = ([a] , [b]) for all (a, b) ∈ A × B . Then it is clear that Φ is a well defined onto homomorphism. Moreover,

$\begin{matrix} ker (Φ) = {(a, b) \in A \times B : Φ (a, b) = ([1], [1])} \\ = {(a, b) \in A \times B : ([a], [b]) = ([1], [1])} \\ = {(a, b) \in A \times B : [a] = [1], [b] = [1]} \\ = {(a, b) \in A \times B : a \in Rad (F), b \in Rad (G)} \\ = Rad (F) \times Rad (G) \\ = Rad (F \times G) . \end{matrix}$

So we have A × B/Rad (F × G) ≅ A/Rad (F) × B/Rad (G).□

If G ⊆ F, we can define IVRL-filters F/G of A/G by F/G = {[a] : a ∈ F}.

Proposition 5.20. If F and G are IVRL-filter of A and G ⊆ F, then Rad (F/G) = Rad (F)/G.

Proof. $\begin{matrix} Rad (F / G) & = & {[a] \in A / G : \neg ν ([a]^{n}) \to ν [a] \in F / G} \\ = & {[a] \in A / G : \neg ν (a^{n}) \to ν a \in F} \\ = & {[a] \in A / G : a \in Rad (F)} \\ = & Rad (F) / G . □ \end{matrix}$

Proposition 5.21. Let f : A → B be an MTL-triangle algebra monomorphism. Then Rad (f (F)) = f (Rad (F)) and Rad (f^-1 (F)) = f^-1 (Rad (F)).

Proof. Let a ∈ f (Rad (F)). f^-1 (a) ∈ Rad (F) if and only if ¬ν ((f^-1 (a)) ⁿ) → ν (f^-1 (a)) ∈ F if and only if f^-1 (¬ ν (aⁿ) → νa) ∈ F if and only if ¬ν (aⁿ) → νa ∈ f (F) if and only if a ∈ Rad (f (F)). The proof of the other equality is similar.□

Proposition 5.22. Let f : A → B be a homomorphism of MTL-triangle algebras. Then Rad (ker(f)) = f^-1 (Rad ({1})), where F is an IVRL-filter of A.

Proof. It is clear that ker(f) is an IVRL-filter of A. By Theorem 5.5, a ∈ Rad (ker(f)) iff ¬ (νaⁿ) → νa ∈ ker(f), for all $n \in ℕ$ iff f (¬ (νaⁿ) → νa) =1, for all $n \in ℕ$ iff ¬ (ν (f (aⁿ))) → ν (f (a)) =1,for all $n \in ℕ$ iff f (a) ∈ Rad ({1}) iff a ∈ f^-1 (Rad ({1})) . □

Definition 5.23. [5] An element a ∈ A is called complemented if there is an element b ∈ A such that a ∨ b = 1 and a ∧ b = 0. If such element b exists it is called complement of a. The set of all complemented elements in A is denoted by B (L).

Corollary 5.24. We see that for a ∈ A, if a ∈ B (A), then νa ∈ B (A). By (T . 7) and (T . 4), we have 1 = νa ∨ ν ¬ a ≤ νa ∨ ¬ νa. Thus νa ∨ ¬ νa = 1, and so νa ∈ B (A).

Lemma 5.25. For every a ∈ A, if a ∈ B (A), then ¬a → a = a.

Proof. If a ∈ B (A) , then a ∨ ¬ a = 1. Since 1 = a ∨ ¬ a ≤ [(a → ¬ a) → ¬ a] ∨ [(¬ a → a) → a], by Lemma 2.2, we deduce that (a → ¬ a) → ¬ a = (¬ a → a) → a = 1 . Hence a → ¬ a ≤ ¬ a and ¬a → a ≤ a. By Lemma 2.2, we have a → ¬ a = ¬ a and ¬a → a = a.□

Theorem 5.26. Let F be an IVRL-filter of A. Then Rad (F) ∩ B (A) ⊆ F.

Proof. Let a ∈ Rad (F) ∩ B (A) ⊆ F . Then a ∈ Rad (F) and a ∈ B (A). So ¬ (νaⁿ) → νa ∈ F, for all $n \in ℕ$ . By Corollary 5.24 and Lemma 5.25 ¬νa → νa = νa. So, a ∈ F. Hence, Rad (F) ∩ B (A) ⊆ F.□

The following example shows that equality of Theorem 5.26 may not always hold.

Example 5.27. Consider Example 4.5. It is clear that F = {[a, a] , [a, 1] , [1, 1]} is an IVRL-filter. We have B (A) = {[b, b] , [a, a] , [0, 0] , [1, 1]} and Rad (F) = {[a, a] , [a, 1] , [1, 1]}, and so Rad (F) ∩ B (A) ≠ F.

Theorem 5.28. Let F be an IVRL-filter of A. Then the following statements hold:

Rad ([1]/F) = Rad (F)/F, where [1]/F is an IVRL-filter in the quotient algebra A/F.

If Rad (F) ⊆ B (A), then B (A/Rad (F)) = B (A)/Rad (F).

If a ∈ A is a nilpotent element, then a/Rad (F) ∈ A/Rad (F) is a nilpotent element.

Proof. (1) $\begin{matrix} Rad ([1] / F) & = & ⋂_{N \in Max (A), F \subseteq N} (N / F) \\ = & (⋂_{N \in Max (A), F \subseteq N} N) / F \\ = & Rad (F) / F . \end{matrix}$

(2) Let Rad (F) ⊆ B (A). Then $\begin{matrix} B (A) / Rad (F) & = & {e / Rad (F) : e \in B (A)} \\ = & {e / Rad (F) : e \lor \neg e = 1} \\ = & {e / Rad (F) : e / Rad (F) \\ \lor \neg (e / Rad (F)) = 1 / Rad (F)} \\ = & B (A / Rad (F)) . \end{matrix}$

(3) Let a ∈ A be a nilpotent element. Then there exists $n \in ℕ$ such that aⁿ = 0. Thus 0/Rad (F) = aⁿ/Rad (f) = (a/Rad (F)) ⁿ. So a/Rad (F) ∈ A/Rad (F) is a nilpotent element.□

In the following example we show that the converse of Theorem 5.28 (3) may not hold.

Example 5.29. Consider Example 4.5. It is clear that F = {[b, b] , [b, 1] , [1, 1]} is an IVRL- filter of A and Rad (F) = {[b, b] , [b, 1] , [1, 1]}. We obtain that {[0, 0] , [0, a] , [a, a]} = [[0, a]] _Rad(F) = [[0, 0]] _Rad(F). So [[0, a]] _Rad(F) ∈ A/Rad (F) is a nilpotent element, while [0, a] is not a nilpotent element of A.

Lemma 5.30. Rad ({1}) = {1} if and only if for every a ∈ A, νa = inf {¬ (νaⁿ) → νa : for all $n \in ℕ}$ .

Proof. Let Rad ({1}) = {1}. Then νa ≤ ¬ (νaⁿ) → νa, for all $n \in ℕ$ . Hence νa is a lower bound for {¬ (νaⁿ) → νa, for all $n \in ℕ}$ . We show that νa is the largest lower bound for {¬ (νaⁿ) → νa, for all $n \in ℕ}$ . Let b be a lower bound for {¬ (νaⁿ) → νa, for all $n \in ℕ}$ . We must show that b ≤ νa. For all $n \in ℕ$ , we have b ≤ ¬ (νaⁿ) → νa then b ∗ ¬ (νaⁿ) ≤ νa. We know that νa ≤ b → νa, for all $n \in ℕ$ . Then νaⁿ ≤ (b → νa) ⁿ and ¬ ((b → νa) ⁿ) ≤ ¬ (νaⁿ). So b ∗ ¬ ((b → νa) ⁿ) ≤ b ∗ ¬ (νaⁿ) ≤ νa thus ¬ ((b → νa) ⁿ) ≤ b → νa and so ¬ ((b → νa) ⁿ) → (b → νa) =1. Hence b → νa ∈ Rad ({1}) = {1} and b ≤ νa .

Conversely, let νa = inf {¬ (νaⁿ) → νa : for all $\in ℕ}$ , for every a ∈ A. We must show that Rad ({1}) ⊆ {1}. Let a ∈ Rad ({1}). Then ¬ (νaⁿ) → νa = 1, for all $n \in ℕ$ , by hypothesis we get that a = 1. Hence Rad ({1}) = {1}. □

In the following definition we consider the closed set of closure operator Rad.

Definition 5.31. Let F be a proper IVRL-filter of A. If Rad (F) = F, then F is called an IVRL-extended semi maximal filter (ISF).

According to Theorem 5.5, we can represent an IVRL-extended semi maximal filter F of A by $\begin{matrix} F & = & Rad (F) \\ = & {a \in A : \neg (ν a^{n}) \to ν a \in F, for all n \in ℕ} . \end{matrix}$

Proposition 5.32. For every IVRL-filter F of A, If F is an IVRL-extended positive implicative (IPIF)(maximal (IMF), obstinate (IOF), Boolean (IBF)) filter of A, then F is an IVRL-extended semi maximal filter of A.

Proof. Let F be an IVRL-extended positive implicative filter of A. We must show that Rad (F) ⊆ F. Let a ∈ Rad (F). Then ¬ (νaⁿ) → νa ∈ F, for all $n \in ℕ$ . So (νa → 0) → νa = ¬ νa → νa ∈ F. By Definition 2.4, νa ∈ F and so a ∈ F . Thus Rad (F) = F; hence F is an IVRL-extended semi maximal filter of A.□

Theorem 5.33. Let F be an IVRL-filter of A. Then F is an IVRL-extended obstinate filter if and only if F is an IVRL-extended Boolean filter of the second kind.

Proof. Assume that F is an IVRL-extended obstinate filter and νx ∉ F thus, x ∉ F. By the hypothesis ¬νx ∈ F.

Conversely, let νx ∉ F. So, x ∉ F. To prove that F is an IVRL-extended obstinate filter we only need to show that ¬νx ∈ F. Since F is an IVRL-extended Boolean filter of the second kind, ¬νx ∈ F. Hence F is an IVRL-extended obstinate filter of A.□

Remark 5.34. By Figure 9 in [10], every IVRL-extended Boolean filter of the second kind of A (IVRL-extended obstinate filter) is IVRL-extended Boolean filter of A.

Theorem 5.35. Let F be an IVRL-filter of A. Then F is an IVRL-extended obstinate filter of A if and only if F is an IVRL-extended maximal filter and IVRL-extended Boolean filter of A.

Proof. Suppose that F is an IVRL-extended obstinate filter which is not IVRL-extended maximal filter. So there exists a proper IVRL-filter G strictly greatest than F. Let a ∈ G \ F . Then (¬ νa) ⁿ ∈ F, for some n ≥ 1 . We know that (¬ νa) ⁿ ≤ ¬ νa. By definition of IVRL-filter, it follows that ¬νa ∈ F, so ¬νa ∈ G. Thus νa, ¬ νa ∈ G, so νa ∗ ¬ νa = 0 ∈ G, which is a contradiction to the fact that G is proper. By Remark 5.34, every IVRL-extended obstinate filter of A is an IVRL-extended Boolean filter of A.

Conversely, since F is an IVRL-extended Boolean filter, νa ∨ ¬ νa ∈ F, for all a ∈ A . Since F is an IVRL-extended maximal filter, F is an IVRL-extended prime filter of the second kind, so νa ∈ F or ¬νa ∈ F. Hence F is an IVRL-extended obstinate filter of A.□

Theorem 5.36. Let F be an IVRL-filter of A. Then the following conditions are equivalent:

F is an IVRL-extended maximal filter and IVRL-extended implicative filter,

F is an IVRL-extended maximal filter and IVRL-extended positive implicative filter,

F is an IVRL-extended obstinate filter.

Proof. (i ⇒ ii) By Theorem 4.4 [12].

(ii ⇒ iii) Let b ∉ F . Firstly, we show that A_b = {k ∈ A : νb → νk ∈ F} is an IVRL-filter. Since νb → 1 =1 ∈ F, 1 ∈ A_b. If a, a → t ∈ A_b, then by (T . 7), νb → νa, νb → (νa → νt) ∈ F. Since F is an IVRL-extended positive implicative filter, νb → νt ∈ F, so t ∈ A_b. Since ννk = νk, if k ∈ A_y, then νk ∈ A_b. If a ∈ F, since νa ≤ νb → νa, νb → νa ∈ F. Thus a ∈ A_b, hence F ⊆ A_b ⊆ A . Since b ∉ F and νb → νb = 1 ∈ F, we have b ∈ A_y . It follows that F ⊈ A_b . Since F is an IVRL-extended maximal filter and b ∉ F, we get A_b = A . Then a ∈ A_b, for each a ∈ A and so νb → νa ∈ F . Similarly, if a, b ∈ A such that a ∉ F, then νa → νb ∈ F.

(iii ⇒ i) Assume on the contrary that F is not an IVRL-extended positive implicative filter. So there exist a, b ∈ A such that (νa → νb) → νa ∈ F and νa ∉ F . We consider two cases: νb ∈ F or νb ∉ F. If νb ∈ F, then νa → νb ∈ F and so νa ∈ F. If νb ∉ F by hypothesis, νa → νb ∈ F and νa ∈ F . In every cases we get a contradiction. So F is an IVRL-extended implicative filter. According to Theorem 5.35, F is an IVRL-extended maximal filter. □

The following examples determonstrate the relationship between other IVRL-filters and IVRL-extended semi maximal filter.

Example 5.37. (a) Consider Example 4.5. It is clear that F = {[1, 1]} is IVRL-extended semi maximal filter of A. However, F is not an IVRL-extended prime filter, since [a, a] → [b, b] , [b, b] → [a, a] ∉ F and is not an IVRL-extended obstinate filter. Since [a, 1] ∉ F, ¬ (ν [a, 1]) = [b, b] ∉ F, F is not an IVRL-extended maximal filter. As ν [a, 1] ⊔ ν [b, 1] = [1, 1] ∈ F but ν [a, 1] , ν [b, 1] ∉ F, F is not an IVRL-extended prime filter of the second kind.

(b) Consider Example 5.9. It is clear that F = {[1, 1]} is an IVRL-extended prime filter of $Int (L)$ . So F is an IVRL-extended prime filter of the second kind. We obtain that $Rad (F) = Int (L) \ [0, 0]$ ; hence Rad (F) ≠ F . This implies that F is not an IVRL-extended semi maximal filter.

(c) Let A = {[0, 0] , [0, v] , [v, v] , [v, a] , [v, b] , [v, 1] , [0, a] , [0, b] , [a, a] , [b, b] , [0, 1] , [a, 1] , [b, 1] , [1, 1]}. we define ⊙ and ⇒ as follow:

⊙	v	a	b	1	⇒	0	v	a	b	1
0	0	0	0	0		0	1	1	1	1	1
v	0	v	v	v		v	v	1	1	1	1
a	v	a	v	a		a	0	b	1	b	1
b	v	v	b	b		b	0	a	a	1	1
1	v	a	b	1		1	0	v	a	b	1

And we define ∗, → , ν, μ as follow:

ν [x₁, x₂] = [x₁, x₁], μ [x₁, x₂] = [x₂, x₂], [x₁, x₂] ∗ [y₁, y₂] = [x₁ ⊙ y₁, x₂ ⊙ y₂], [x₁, x₂] → [y₁, y₂] = [(x₁ ⇒ y₁) ⊓ (x₂ ⇒ y₂) , x₂ ⇒ y₂].

Then (A, ∨ , ∧ , ∗ , → , ν, μ, [0, 0] , [0, 1] , [1, 1]) is an MTL-triangle algebra. It is clear that F = {[1, 1]} is an IVRL-extended fantastic filter of A, but $\begin{matrix} Rad (F) & = & {[v, v], [n, a], [v, b], [v, 1], [a, a], [b, b], \\ [a, 1], [b, 1], [1, 1]} \\ \neq & F . \end{matrix}$

So F is not an IVRL-extended semi maximal filter of A.

(d) Consider Example 5.9. It is clear that F = [1, 1] is an IVRL-extended implicative filter. We get that Rad (F) = A \ [0, 0] ≠ F, hence F is not an IVRL-extended semi maximal filter of A.

By Proposition 5.32, Theorem 5.35, Theorem 5.36 and Example 5.37, we determine the relations between IVRL-extended semi maximal filter and other types IVRL-filters that we introduced them as follow:

Corollary 5.38. {F_i} _i∈I is a family of IVRL-extended semi maximal filter of A if and only if ∩_i∈IF_i is an IVRL-extended semi maximal filter of A.

Proof. Let {F_i} _i∈I be a family of IVRL-extended semi maximal filters of A. So Rad (F_i) = F_i, for all i ∈ I, by Proposition 5.13, we have Rad (∩ _i∈IF_i) = ∩ _i∈IRad (F_i) = ∩ _i∈IF_i.

Conversely, assume that a ∈ Rad (F_i). So ¬ (νaⁿ) → νa ∈ F_i for all i ∈ I, $n \in ℕ$ . Hence for all $n \in ℕ,$ ¬ (νaⁿ) → νa ∈ ∩ _i∈IF_i, that is a ∈ Rad (∩ _i∈IF_i) = ∩ _i∈IF_i . So a ∈ ∩ _i∈IF_i for all i ∈ I . Thus a ∈ F_i and Rad (F_i) ⊆ F_i. It is clear that F_i ⊆ Rad (F_i) and so Rad (F_i) = F_i. This implies that {F_i} _i∈I is a family of IVRL-extended semi maximal filter of A.□

Proposition 5.39. Let F be an IVRL-filter of A. Then F is an IVRL-extended semi maximal filters of A if and only if {1}/F is an IVRL-extended semi maximal filter of A/F.

Proof. Assume that F is an IVRL-extended semi maximal filter of A, so Rad (F) = F. By Theorem 5.28(1), Rad (F) = Rad (F)/F = F/F = {1}/F. So {1}/F is an IVRL-extended semi maximal filter of A/F.

Conversely, let {1}/F be an IVRL-extended semi maximal filter of A/F. Then Rad ({1/F}) = {1}/F. By Theorem 5.28(1), Rad (F)/F = {1}/F. We must show that Rad (F) ⊆ F . Let x ∈ Rad (F). Then x/F ∈ Rad (F)/F = {1}/F, and x/F = 1/F that is x ∈ F, hence Rad (F) ⊆ F. Thus F is an IVRL-extended semi maximal filter of A.□

6 Dense and double complemented elements of an IVRL-filter

We study the set of dense and double complemented elements of an IVRL-filter. Next, we consider the relationships between these sets and radical of an IVRL-filter.

Definition 6.1. Dense elements of a triangle algebra A is defined as D_s (A) = {a ∈ A : ¬ νa = 0}. Its restriction to an IVRL-filter F is defined as D_s (F) = {a ∈ F : ¬ νa = 0}. Also, the set of double complements of F define as D (F) = {a ∈ A : ¬¬ νa ∈ F}.

Remark 6.2. Let F be an IVRL-filter of triangle algebra A. Then D_s (F) = D_s (A) ∩ F. It is clear that D_s (F) is an IVRL-filter of A. Also, we have D_s (F) ⊆ Rad (F) , D_s (A) ∪ F ⊆ D (F).

Example 6.3. Consider Example 4.5. It is clear that F = {[a, a] , [a, 1] , [1, 1]} is an IVRL-filter of A. We have D_s (F) = {[1, 1]}, D (F) = {[a, a] , [a, 1] , [1, 1]}.

Proposition 6.4. Let F be an IVRL-filter of triangle algebra A. Then Rad (F/D_s (F)) = Rad (F)/D_s (F) .

Proof. $\begin{matrix} Rad (F / D_{s} (F)) \\ = ⋂_{N \in Max (A), D_{s} (F) \subseteq F \subseteq N} (N / D_{s} (F)) \\ = (⋂_{N \in Max (A), D_{s} (F) \subseteq F \subseteq N} N) / D_{s} (F) \\ = Rad (F) / D_{s} (F) . □ \end{matrix}$

Theorem 6.5. Let F be an IVRL-extended (positive implicative, maximal, obstinate, Boolean) filter of A. Then Rad (F) ⊆ D (F) .

In the following example we show that Rad (F) does not always coincide with D (F).

Example 6.6. Consider Example 5.37 (c). But we define ⊙ and ⇒ as follow:

It is clear that F = {[1, 1]} is an IVRL-filter. We have Rad (F) = {[v, v] , [v, a] , [v, b] , [v, 1] , [a, a] , [b, b] , [a, 1] , [b, 1] , [1, 1]} and D (F) = {[a, a] , [b, b] , [a, 1] , [b, 1] , [1, 1]}. So Rad (F) ⊈ D (F).

Proposition 6.7. Idempotent elements of D (F) are a subset of Rad (F).

Proof. Let a be an idempotent element of D (F). Then ¬¬ νa ∈ F . Since 0 ≤ a and ν, → are increasing, ¬νa → 0 ≤ ¬ νa → νa . That is ¬¬ νa ≤ ¬ νa → νa, so ¬νa → νa ∈ F. Because ¬νa → νa = ¬ (νaⁿ) → νa for all $n \in ℕ$ , we have a ∈ Rad (F).□

In the following example we show that D (F) does not always coincide with Rad (F).

Example 6.8. Consider Example 5.11. It is clear that F = {[1, 1]} is an IVRL-filter and A is not a Gödel triangle algebra, so D (F) = A \ {[0, 0] , [0, a] , . . . , [0, 1]} and Rad (F) = A \ {[0, 0] , [0, a] , . . . , [0, 1] , [a, a] , . . . , [a, 1]}. Hence, D (F) ⊈ Rad (F).

In the following example we show that D (F) does not always coincide with D_s (F).

Example 6.9. Consider Example 5.37 (c). It is clear that F = {[a, a] , [a, 1] , [1, 1]} is an IVRL-filter of A. So D (F) = {[v, v] , [v, a] , [v, b] , [v, 1] , [a, a] , [a, 1] , [b, b] , [b, 1] , [1, 1]} and D_s (F) = {[a, a] , [a, 1] , [1, 1]}. Thus D (F) ⊈ D_s (F).

Proposition 6.10. Let A be a triangle algebra. Then D_s (A) = D ({1}).

7 Conclusion

We introduced the notion of the maximal filters of triangle algebras. We defined radicals of triangle algebras and radical of an IVRL-filter F in triangle algebras. Also, we presented a characterization and many important properties of Rad (F). In addition, we defined IVRL-extended semi maximal filters and compared this filter with other types of filters. So, Fig. 1 gives a schematic summary of this relations. Also, we showed that if F is an IVRL-extended (positive implicative, fantastic, obstinate, Boolean) filter of A, then so is Rad (F). Next, we introduced the special sets of dense elements and double complemented elements of F. Finally, we determined the relationship between these sets and Rad (F).

In our future work, we will continue our study on algebraic structures specially triangle algebras and like to use these results to find some classification for this structure.

Acknowledgments

We wish to thank the reviewers for excellent suggestions that have been incorporated into the paper.

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⊙	a	b	c	d	e	f	g	1
0	0	0	0	0	0	0	0	0
a	0	0	a	a	a	a	a	a
b	0	b	b	b	b	b	b	b
c	a	b	c	c	c	c	c	c
d	a	b	c	c	c	c	c	d
e	a	b	c	c	e	c	e	e
f	a	b	c	c	c	f	f	f
g	a	b	c	c	e	f	g	g
1	a	b	c	d	e	f	g	1

⇒	0	a	b	c	d	e	f	g	1
0	1	1	1	1	1	1	1	1	1
a	b	1	1	1	1	1	1	1	1
b	a	b	1	1	1	1	1	1	1
c	0	a	b	1	1	1	1	1	1
d	0	a	b	g	1	1	1	1	1
e	0	a	b	f	f	1	f	1	1
f	0	a	b	e	e	e	1	1	1
g	0	a	b	d	d	e	f	1	1
1	0	a	b	c	d	e	f	g	1

⊙	v	a	b	1	⇒	0	v	a	b	1
0	0	0	0	0		0	1	1	1	1	1
v	0	v	v	v		v	v	1	1	1	1
a	v	a	v	a		a	0	b	1	b	1
b	v	v	b	b		b	0	a	a	1	1
1	v	a	b	1		1	0	v	a	b	1

⊙	a	b	c	d	e	f	g	1
0	0	0	0	0	0	0	0	0
a	0	0	a	a	a	a	a	a
b	0	b	b	b	b	b	b	b
c	a	b	c	c	c	c	c	c
d	a	b	c	c	c	c	c	d
e	a	b	c	c	e	c	e	e
f	a	b	c	c	c	f	f	f
g	a	b	c	c	e	f	g	g
1	a	b	c	d	e	f	g	1

⇒	0	a	b	c	d	e	f	g	1
0	1	1	1	1	1	1	1	1	1
a	b	1	1	1	1	1	1	1	1
b	a	b	1	1	1	1	1	1	1
c	0	a	b	1	1	1	1	1	1
d	0	a	b	g	1	1	1	1	1
e	0	a	b	f	f	1	f	1	1
f	0	a	b	e	e	e	1	1	1
g	0	a	b	d	d	e	f	1	1
1	0	a	b	c	d	e	f	g	1

⊙	v	a	b	1	⇒	0	v	a	b	1
0	0	0	0	0		0	1	1	1	1	1
v	0	v	v	v		v	v	1	1	1	1
a	v	a	v	a		a	0	b	1	b	1
b	v	v	b	b		b	0	a	a	1	1
1	v	a	b	1		1	0	v	a	b	1

⊙	a	b	c	d	e	f	g	1
0	0	0	0	0	0	0	0	0
a	0	0	a	a	a	a	a	a
b	0	b	b	b	b	b	b	b
c	a	b	c	c	c	c	c	c
d	a	b	c	c	c	c	c	d
e	a	b	c	c	e	c	e	e
f	a	b	c	c	c	f	f	f
g	a	b	c	c	e	f	g	g
1	a	b	c	d	e	f	g	1

⇒	0	a	b	c	d	e	f	g	1
0	1	1	1	1	1	1	1	1	1
a	b	1	1	1	1	1	1	1	1
b	a	b	1	1	1	1	1	1	1
c	0	a	b	1	1	1	1	1	1
d	0	a	b	g	1	1	1	1	1
e	0	a	b	f	f	1	f	1	1
f	0	a	b	e	e	e	1	1	1
g	0	a	b	d	d	e	f	1	1
1	0	a	b	c	d	e	f	g	1

⊙	v	a	b	1	⇒	0	v	a	b	1
0	0	0	0	0		0	1	1	1	1	1
v	0	v	v	v		v	v	1	1	1	1
a	v	a	v	a		a	0	b	1	b	1
b	v	v	b	b		b	0	a	a	1	1
1	v	a	b	1		1	0	v	a	b	1