Some results in hoop algebras

1 Introduction

Hoop algebras are naturally ordered commutative residuated integral monoids, introduced by B. Bosbach [8, 9] then study by J.R. Büchi and T.M. Owens [10], a paper never published. In the last years, hoops theory was enriched with deep structure theorems (see [1, 13]). Many of these results have a strong impact with fuzzy logic. Particularly, from the structure theorem of finite basic hoops ([1], Corollary 2.10) one obtains an elegant short proof of the completeness theorem for propositional basic logic (see [1], Theorem 3.8), introduced by Hájek in [14]. The algebraic structures corresponding to Hájek’s propositional (fuzzy) basic logic, BL-algebras, are particular cases of hoops. Kondo [16], considered fundamental properties of some types of filters (implicative, positive implicative and fantastic filters) of hoops and proved that for any hoop A and filter F of A, (a) F is an implicative filter if and only if A/F is a relatively pseudo-complemented semilattice, that is, Brouwerian semilattice; (b) F is a positive implicative filter if and only if A/F is a {∧ , ∨ , → , 1}-reduct of Heyting algebra; (c) F is a fantastic filter if and only if A/F is a Wajsberg hoop. Moreover he proved that, for any filter of a hoop, it is a positive implicative filter if and only if it is an implicative and fantastic filter. R.A. Borzooei and M. Aaly Kologani investigate the relation between these filter [6].

In this paper, we introduce the notion of obstinate filter in hoop algebras and we prove some results which determine the relationship between this notion and other types of filters in hoops. We prove that, if F is an obstinate filter of hoop A, then A/F is a local hoop and Boolean algebra. Finally, we define the notion of locally finite hoop and we investigate the obstinate filters in these structures.

2 Preliminaries

In this section, we recollect some definitions and results which will be used not cite them every time they are used.

Definition 2.1. [1] A hoop algebra or hoop is an algebra (A, ⊙ , → , 1) of type (2, 2, 0) such that for all x, y, z ∈ A:

(A, ⊙ , 1) is a commutative monoid,

On hoop A, we define x ≤ y if and only if x → y = 1. It is easy to see that ≤ is a partial order relation on A. A hoop A is bounded if there is an element 0 ∈ A such that 0 ≤ x, for all x ∈ A. We let x⁰ = 1, x ⁿ  = x^n-1 ⊙ x, for any n ∈ ℕ. Let A be a bounded hoop. We define a negation “^-” on A by, x^- = x → 0, for all x ∈ A. If (x^-) ^- = x, for all x ∈ A, then the bounded hoop A is said to have the double negation property, or (DNP), for short. The order of a ∈ A, a ≠ 1, in symbols ord (a) is the smallest n ∈ ℕ such that a ⁿ  = 0. If no such n exists, then ord (a) =∞.

Proposition 2.2. [8, 9] In any hoop (A, ⊙ , → , 1) the following properties hold, for all x, y, z, ∈ A:

x ⊙ y ≤ x, y and x ⁿ  ≤ x, for any n ∈ ℕ,

Proposition 2.3. [1] Let A be a bounded hoop and for any x, y ∈ A, we define, x ⊔ y = ((x → y) → y) ∧ ((y → x) → x). Then the following conditions are equivalent:

⊔ is associative,

Definition 2.4. [6] A hoop A is called a ⊔-hoop, if ⊔ is a join operation on A.

Proposition 2.5. [13] In any ⊔-hoop A, if x ⊔ y = 1, then x ∧ y = x ⊙ y.

Definition 2.6. [13] Let A be a hoop. A non-empty subset F of A is called a filter of A if,

x ∈ F and x ≤ y, y ∈ A, then y ∈ F,

Clearly, 1 ∈ F, for all filter of A. A filter F of A is called proper filter if F ≠ A. It can be easily to see that, if A is a bounded hoop, then a filter is proper if and only if it is not containing 0.

Proposition 2.7. [6] Let F be a non-empty subset of hoop A. Then F is a filter of A if and only if x ∈ F and x → y ∈ F, then y ∈ F, for any x, y ∈ A.

Definition 2.8. [6] A proper filter F of a ⊔-hoop A is called prime filter of A if x ⊔ y ∈ F implies x ∈ F or y ∈ F, for any x, y ∈ A.

A maximal filter is a proper filter M of hoop A such that is not included in any other proper filter.

Definition 2.9. [7, 16] Let F be a subset of A such that 1 ∈ F. Then, for any x, y, z ∈ A, F is called:

a positive implicative filter of A, if x → (y → z) ∈ F and x → y ∈ F, then x → z ∈ F,

Theorem 2.10. [16] Every implicative filter of A is a fantastic filter.

Definition 2.11. [13] Let A and B be two bounded hoops. A map f : A → B is called a hoop homomorphism if and only if for all x, y ∈ A, f (0) =0,  f (1) =1, f (x ⊙ y) = f (x) ⊙ f (y) and f (x → y) = f (x) → f (y).

The set of all hoop homomorphism from A to B is shown by Hom (A, B).

Definition 2.12. [13] A simple hoop is a hoop which has just two trivial filters.

Note. From now one, in this paper, we let A be a bounded hoop, unless otherwise state.

3 Obstinate filters

In this section, we define the notion of obstinate filter in bounded hoops and we study some properties of them.

Definition 3.1. Let F be a proper filter of A. Then F is called an obstinate filter of A if, for any x, y ∉ F, x → y ∈ F and y → x ∈ F.

Example 3.2. Let A = {0, a, b, c, d, 1} be a set, where 0 < b < a < 1, 0 < d < a < 1 and 0 < d < c < 1. Define operations ⊙ and → on A as follows,

Then (A, ⊙ , → , 0, 1) is a bonded hoop. By routine calculation, we can see that filter F = {1, a, b} is an obstinate filter.

Proposition 3.3. Let F be a proper filter of A. Then the following statements are equivalent:

F is an obstinate filter of A,

Proof. (i) ⇒ (ii) Suppose that F is an obstinate filter and x ∉ F. Since F is a proper filter and A is bounded, 0 ∉ F. Moreover, since F is an obstinate filter, 1 = 0 → x ∈ F and x^- = x → 0 ∈ F. Then (x^-)  ⁿ  ∈ F, for all n ∈ ℕ.

(ii) ⇒ (i) Let x, y ∉ F. Then by assumption, (x^-)  ⁿ  ∈ F and (y^-)  ^m  ∈ F, for some n, m ≥ 1. Since F is a filter, by Proposition 2.2(i) and (F1), (x^-)  ⁿ  ≤ x^- and (y^-)  ^m  ≤ y^-, and so x^- ∈ F and y^- ∈ F. Also, by Proposition 2.2(v), x^- ≤ x → y and y^- ≤ y → x, for all x, y ∈ A. Then by (F1), x → y ∈ F and y → x ∈ F. Thus, F is an obstinate filter of A.

(i) ⇒ (iii) Suppose that x ∉ F. Since F is an obstinate filter, by (ii), (x^-)  ⁿ  ∈ F, for some n ≥ 1. Since (x^-)  ⁿ  ≤ x^- and F is filter, by (F1), x^- ∈ F.

(iii) ⇒ (i) Let x, y ∉ F. Then by assumption, x^-, y^- ∈ F. Thus, by Proposition 2.2(v), x^- ≤ x → y and y^- ≤ y → x. Since F is a filter, by (F1), x → y ∈ F and y → x ∈ F. Hence, F is an obstinate filter of A.

Let F be a filter of A. Define x ≡  _F y if and only if x → y ∈ F and y → x ∈ F, for any x, y ∈ A. Then ≡ _F is a congruence relation on A. The set of all congruence classes is denoted by A/F, it means A/F = {[x] ∣ x ∈ A}, where [x] = {y ∈ A ∣ x ≡  _F y}. Define operations ⊙ and → on A/F in this way, [x] ⊙ [y] = [x ⊙ y] ,   [x] → [y] = [x → y]. Therefore, (A/F, ⊙ , → , [1] , [0]) is a bounded hoop with respect to F and [x] ≤ [y] if and only if x → y ∈ F.

Theorem 3.4. Let F be an obstinate filter of A. Then A/F is a Boolean algebra.

Proof. Let x ∈ A. Since F is an obstinate filter, by Proposition 3.3 (iii), x ∈ F or x^- ∈ F. Thus, [x] = [1] or [x^-] = [1]. Hence, [x] =1 or [x^--] = [0]. Since [x] ≤ [x^--], we have [x] = [0]. Therefore, A/F is a Boolean algebra. □

Proposition 3.5. Let A, B be bounded hoops and f ∈ Hom (A, B). If G is an obstinate filter of B, then f^-1 (G) is an obstinate filter of A.

Proof. Let G be an obstinate filter of B. It is clear that f^-1 (G) is a proper filter of A. Suppose that x ∈ A such that x ∉ f^-1 (G). Then f (x) ∉ G. Since G is an obstinate filter, by Proposition 3.3 (iii), (f (x)) ^- ∈ G, and so f (x^-) ∈ G. Thus, x^- ∈ f^-1 (G). Hence, by Proposition 3.3, f^-1 (G) is an obstinate filter of A. □

Note. Let B be a bounded hoop and f ∈ Hom (A, B). Then ker(f) = {x ∈ A ∣ f (x) =1} is a filter of A.

Theorem 3.6. If F is an obstinate filter of A and for any x ∈ A, x² = x, then there exists f ∈ Hom (A, A) such that ker(f) = F.

Proof. Let a ∈ A. We define a map f on A as follows, f ( x ) = { 1 if x ∈ F a if x ∉ F

Now, we prove that f is a hoop homomorphism of A. Let x, y be two arbitrary elements of A. We consider the following cases:

Case 1. Let x, y ∈ F. Since F is a filter, by (F2), x ⊙ y ∈ F. Then by definition of f, f (x) =1 = f (y) and f (x ⊙ y) =1. Thus, f (x) ⊙ f (y) = f (x ⊙ y) =1. Since y ∈ F and F is a filter, by Proposition 2.2(iii) and (F1), y ≤ x → y, and so x → y ∈ F. Thus, f (x → y) =1. Then, 1 = f (x → y) = f (x) → f (y). Hence, f is a hoop homomorphism.

Case 2. Let x, y ∉ F. Then x ⊙ y ∉ F. By definition of f, f (x) = f (y) = f (x ⊙ y) = a. Hence, f (x ⊙ y) = a = a ⊙ a = f (x) ⊙ f (y). Also, since F is an obstinate filter and x, y ∉ F, x → y ∈ F, and so f (x → y) =1. Then, f (x → y) =1 = a → a = f (x) → f (y). Hence, f is a hoop homomorphism.

Case 3. Let x ∉ F and y ∈ F. Then by Proposition 2.2(i) and (F1), x ⊙ y ∉ F. Then by definition of f, f (x) = a,  f (y) =1 and f (x ⊙ y) = a. Thus, f (x) ⊙ f (y) = a ⊙ 1 = a = f (x ⊙ y). Moreover, since y ∈ F and F is a filter, by Proposition 2.2(iii) and (F1), y ≤ x → y, and so x → y ∈ F. Thus, f (x → y) =1. Then, f (x) → f (y) = a → 1 =1 = f (x → y). Hence, f is a hoop homomorphism.

Case 4. Let x ∈ F and y ∉ F. Similar to Case 3, f (x ⊙ y) = f (x) ⊙ f (y). If x → y ∈ F, then by (F2), x ⊙ (x → y) ∈ F. By Proposition 2.2 (i) and (HP4) and (F1), y ∈ F, which is a contradiction. Thus, x → y ∉ F. So, by definition of f, f (x) =1,  f (y) = a and f (x → y) = a. Then, f (x) → f (y) =1 → a = a = f (x → y). Hence, f is a hoop homomorphism. □

4 Relation between obstinate filters and other filters in hoops

In this section, we investigate the relationship between obstinate filters and other filters in hoops.

Theorem 4.1. Every obstinate filter of A is a maximal filter of A.

Proof. Let F be an obstinate filter of A which is not maximal. Then there exists a proper filter G of A such that F ⊆ G. Let a ∈ G ∖ F. Since F is an obstinate filter, by Proposition 3.3 (iii), a^- ∈ F. Since a^- ∈ G and a ∈ G, by Proposition 2.2(iv), a ⊙ a^- = 0 ∈ G, which is a contradiction. Hence, F is a maximal filter. □

The next example shows that the converse of Theorem 4.1, is not true, in general.

Example 4.2. Let A = {0, a, b, c, d, 1} be a set. Define operations ⊙ and → on A as follows,

Then (A, ⊙ , → , 0, 1) is a bounded hoop. It is clear that F = {1, a} is a maximal filter but it is not an obstinate filter. Because b, 0 ∉ F and b → 0 = d ∉ F.

Corollary 4.3. Every obstinate filter of a bounded ⊔-hoop A is a prime filter of A.

Proof. Since any maximal filter of a bounded hoop A is a prime filter, so by Theorem 4.1, the proof is clear. □

Proposition 4.4. Let A be a hoop with condition x² = x and F is a maximal filter. Then F be an obstinate filter.

Proof. Let F be a maximal filter of A and x ∉ F. So for some n ∈ ℕ , ( x n ) - ∈ F. Hence by hypothesis, x^- ∈ F. Thus by Proposition 3.3, F be an obstinate filter. □

Let A be a bounded hoop. Define D _s  (A) = {x ∈ A ∣ x^- = 0} (See [15]). In the following example, we show that D _s  (A) need not be an obstinate filter.

Example 4.5. According to Example 3.2, F = {a, b, 1} is an obstinate filter. It is easy to see that D _s  (A) = {1}, and so D _s  (A) ⊂ F ⊂ A. Then D _s  (A) is not a maximal filter and by Theorem 4.1, D _s  (A) is not an obstinate filter.

Proposition 4.6. Let F be an obstinate filter of bounded ⊔-hoop A. Then, for any x, y ∉ F, there exists z ∉ F such that x ≤ z and y ≤ z.

Proof. Let F be an obstinate filter of A. Then by Corollary 4.3, F is a prime filter. If x ⊔ y ∈ F, since F is a prime, then x ∈ F or y ∈ F, which is a contradiction. So x ⊔ y ∉ F and x, y ≤ x ⊔ y. □

Proposition 4.7. Let A be a bounded hoop such that D _s  (A) = A ∖ {0}. Then any proper implicative filter of A is an obstinate filter.

Proof. Suppose that F is a proper implicative filter of A. Since F is a proper filter, x^- ∉ F. Also by [[6], Corollary 4.5], x^-- → x ∈ F, and so 1 → x = x ∈ F. Then by Proposition 3.3, F is an obstinate filter. □

Theorem 4.8. Let F be a proper filter of A. Then for any x, y ∈ A, the following conditions are equivalent:

F is a maximal and implicative filter,

Proof. (i) ⇒ (ii) By [[16], Proposition 3], the proof is clear.

(ii) ⇒ (iii) Let y ∉ F. We prove that, A _y  = {u ∈ A ∣ y → u ∈ F} is a filter of A. It is clear that 1 ∈ A _y . If x, x → t ∈ A _y , then y → x, y → (x → t) ∈ F. Since F is a positive implicative filter, we have y → t ∈ F, and so t ∈ A _y . Thus, by Proposition 2.7, A _y is a filter. Let x ∈ F. Since F is a filter, by Proposition 2.2(iii) and (F1), y → x ∈ F, and so x ∈ A _y . Hence, F ⊆ A _y  ⊆ A. Since y → y = 1 ∈ F and y ∉ F, we have y ∈ A _y . Then F ⊂ A _y . Moreover, since F is a maximal filter and y ∉ F, we get that A _y  = A. Then, for any x ∈ A, x ∈ A _y , and so y → x ∈ F. By the similarly way if x ∉ F, then x → y ∈ F. Hence, F is an obstinate filter.

(iii) ⇒ (i) Suppose that F is not an implicative filter. Then there exist x, y ∈ A such that (x → y) → x ∈ F and x ∉ F. If y ∈ F, then By Proposition 2.2 (iii) and (F1), x → y ∈ F. Now, since (x → y) → x, x → y ∈ F and F is a filter, by Proposition 2.7, x ∈ F, which is a contradiction. If y ∉ F, since x ∉ F and F is an obstinate filter, then x → y ∈ F. Thus, by Proposition 2.7, x ∈ F, which is a contradiction. Hence, F is an implicative filter. On the other hand, by Theorem 4.1, F is a maximal filter. □

Proposition 4.9. Let F be a positive implicative filter of bounded hoop A with condition x² = 0 for x ≠ 1. Then F is a maximal filter of A.

Proof. If F is a positive implicative filter and x ∉ F, then by [[6], Theorem 4.9], x → x² ∈ F for some x ∈ A. So by condition x^- ∈ F. Therefore by Theorem 4.1, F is a maximal filter of A. □

Proposition 4.10. Let F be a prime filter of bounded ⊔-hoop A with condition x ⊔ x^- = 1. Then F is a maximal filter of A.

Proof. If F is a prime filter and x ⊔ x^- = 1, then x ∨ x^- ∈ F. Let x ∉ F, so x^- ∈ F. Therefore, F is a maximal filter of A. □

Theorem 4.11. (i) Every obstinate filter of A is a fantastic and perfect filter,

(ii) If D _s  (A) = A ∖ {0}, then every fantastic filter is an obstinate filter,

(iii) If A has (DNP) such that for any x ∈ A, x² = x, then every perfect filter is an obstinate filter.

Proof. (i) By Theorems 2.10 and 4.8, every obstinate filter of A is a fantastic filter. Now, suppose that F is an obstinate filter of A and for some n ∈ ℕ , ( x n ) - ∈ F. Then by Proposition 3.3, x ⁿ  ∉ F and so x ∉ F. Thus, by Proposition 3.3 (iii), x^- ∈ F, and so (x^-)  ^m  ∈ F. Therefore, ((x^-)  ^m ) ^- ∉ F, for any m ∈ ℕ. Conversely, suppose that for any m ∈ ℕ, ((x^-)  ^m ) ^- ∉ F. Then by Proposition 3.3, (x^-)  ^m  ∈ F. Since x ^m  ≤ x, by Proposition 2.2(vi) and Proposition 2.2(i), (x^-)  ^m  ≤ x^- ≤ (x ^m ) ^-. Since F is a filter, by (F1), (x ^m ) ^- ∈ F. Hence, F is a perfect filter.

(ii) Let F be a proper fantastic filter of A. Then 0 ∉ F, and so x^- ∉ F, for any 0 ≠ x ∈ A. Since 0 → x = 1 ∈ F and F is a fantastic filter, ((x → 0) →0) → x = 1 → x = x ∈ F. Then by Proposition 3.3, F is an obstinate filter.

(iii) Let F be a perfect filter of A and x ∉ F. Since A has (DNP) and x² = x, for any x ∈ A, we have ((x^-)  ^m ) ^- = (x^-) ^- = x ∉ F. Hence, for any m ∈ ℕ, ((x^-)  ^m ) ^- ∉ F. Since F is a perfect filter, there exists n ∈ ℕ such that (x ⁿ ) ^- = x^- ∈ F. Then by Proposition 3.3, F is an obstinate filter. □

Example 4.12. (i) According to Example 4.2, it is easy to see that F = {1, a} is a fantastic filter but is not an obstinate filter. Because, 0, a ∉ F, but a → 0 = a ∉ F.

(ii) Let A = {0, a, b, c, 1} be a set. Define operations ⊙ and → on A as follows:

Then (A, ⊙ , → , 0, 1) is a bounded hoop. By routine calculation, we can see that F = {1, c} is a perfect filter but it is not an obstinate filter. Because a, 0 ∉ F but a → 0 =0 ∉ F.

Proposition 4.13. Let F be an obstinate filter of A. Then the following conditions hold,

for all 0, 1 ≠ x ∈ A,  x → x^- ∈ F or x^- → x ∈ F,

Proof. (i) Let x ∈ F. Then by Proposition 2.2(iii), x ≤ x^- → x. Since F is a filter, by (F1), x^- → x ∈ F. If x ∉ F, then by Proposition 3.3, x^- ∈ F. By Proposition 2.2(iii), x^- ≤ x → x^-, and so x → x^- ∈ F.

(ii) Let x ∈ F. Then by Proposition 2.2(iii), x ≤ x^-- → x. Since F is a filter, by (F1), x^-- → x ∈ F. If x ∉ F, by Proposition 3.3, x^- ∈ F. By Proposition 2.2(v), x^- ≤ x^-- → x, and so by (F1), x^-- → x ∈ F.

(iii) Let x ∉ F and y ≤ x. Then by Proposition 2.2(vi), x^- ≤ y^-. Since F is an obstinate filter, by Proposition 3.3, x^- ∈ F and by (F1), y^- ∈ F.

(iv) Let x ∉ F. Since F is an obstinate filter, by Proposition 3.3, x^- ∈ F. By Proposition 2.2(iii) and (F1), y → x^- ∈ F. Also, by Proposition 2.2(vii), x → y^- ∈ F and (x ⊙ y) ^- ∈ F.

(v) By [[6], Theorem 4.9] and Theorem 4.8, the proof is clear. □

In the following example, we show that the converse of Proposition 4.13, is not true, in general.

Example 4.14. (1) Let A = {0, a, b, 1} be a chain. Define operations ⊙ and → on A as follows:

Then (A, ⊙ , → , 0, 1) is a bounded hoop. By routine calculation, we can see that F = {1, b} is a filter. So for any 0, 1 ≠ x ∈ A, x → x^- and x^- → x ∈ F, but F is not an obstinate filter.

(2) In Example 3.2, F = {1, c} is not an obstinate filter, but for any x ∈ A, x^-- → x = 1 ∈ F. So, (ii) is not true.

(3) In Example 4.12(ii), for any x ∈ A, x → x² = 1, but filter F = {1} is not an obstinate filter. Because, a, a^- ∉ F.

Theorem 4.15. Let A be a bounded hoop. If A is a simple and has an obstinate filter, then A = {0, 1}.

Proof. Let A = {0, 1}. We see that F = {1} is an obstinate filter. Now, let A be a simple hoop which is not the two-element chain. Then A has at least one element x which is different from 0 and 1. Let F be an obstinate filter of A. Then F is a proper filter of A. Since A is a simple hoop, we have F = {1}, and so x ∉ F. Since F is an obstinate filter, by Theorem 4.13, x^- ∈ F = {1}. Thus, x → 0 =1. It means that x ≤ 0, which is a contradiction. Therefore, A has no obstinate filter. □

Theorem 4.16. Suppose that F and G are two proper filters of A such that F ⊆ G. If F is an obstinate filter, then G is an obstinate filter, too.

Proof. Let F be an obstinate filter and F ⊆ G. Then by Theorem 4.1, F is a maximal filter. Since G is a proper filter, we get that F = G. Hence, G is an obstinate filter. □

Example 4.17. In Example 3.2, {1} ⊆ {1, a, b} such that {1, a, b} is an obstinate filter. But {1} is not an obstinate filter. Then the converse of Theorem 4.16 is not true, in general.

Corollary 4.18. {1} is an obstinate filter of A if and only if every filter F of A is an obstinate filter.

Proof. By Theorem 4.16, the proof is clear. □

Theorem 4.19. Let F be a proper filter of A. Then F is an obstinate filter if and only if every filter of the quotient hoop A/F is an obstinate filter.

Proof. (⇒) Suppose that F is an obstinate filter of A. We prove that {} is an obstinate filter of A/F. Let x ∈ A such that [x] ∉ {[1]} i.e., [x] ≠ [1]. Then x ∉ F. Since F is an obstinate filter, by Proposition 3.3, x^- ∈ F, and so [x^-] = [1] i.e., [x^-] ∈ {[1]}. Then, by Proposition 3.3, {[1]} is an obstinate filter of A/F. Hence, by Corollary 4.18, every filter of A/F is an obstinate filter.

(⇐) Suppose that every filter of A/F is an obstinate filter. Then {[1]} = F is an obstinate filter. □

In Fig. 1, we show the relationship between obstinate filter and other types of filters of hoop algebras.

5 Obstinate filters in local and locally finite hoops

In this section, we introduce the notion of locally finite hoop and investigate some properties of obstinate filters.

Definition 5.1. A bounded hoop is called local if ord (x)< ∞ or ord (x^-)< ∞, for all x ∈ A.

Example 5.2. The hoop of Example 4.14(1) is a local and the hoop of Example 3.2 is not a local hoop.

Theorem 5.3. A is a local if and only if it has exactly one maximal filter.

Proposition 5.4. Let A, B be bounded hoops and f ∈ Hom (A, B). Then the following conditions hold:

if f ∈ Hom (A, B) is onto and A is a local hoop, then B is a local hoop.

Proof. (i) Let x ∈ B. Since f is onto, there exists y ∈ A such that f (y) = x. Since A is a local hoop, then ord (y)< ∞ or ord (y^-)< ∞. If ord (y)< ∞, then there exists n ∈ ℕ such that y ⁿ  = 0. So 0 = f (y ⁿ ) = (f (y))  ⁿ  = x ⁿ . Hence, ord (x)< ∞. By the similar way, if ord (y^-)< ∞, then ord (x^-)< ∞. Hence, B is a local hoop.

(ii) Let x ∈ A. Then for some y ∈ B,  f (x) = y. Since B is a local hoop, thus ord (y)< ∞ or ord (y^-)< ∞. If ord (y)< ∞, then there exists n ∈ ℕ such that y ⁿ  = 0. So 0 = y ⁿ  = (f (x))  ⁿ  = f (x ⁿ ). Since f is injective, then x ⁿ  = 0. Hence, ord (x)< ∞. By the similar way, if ord (y^-)< ∞, then ord (x^-)< ∞. Hence, A is a local hoop. □

Corollary 5.5. Let B be a bounded hoop and f ∈ Hom (A, B) be onto. If A is a local, then B has exactly one maximal filter.

Proof. By Proposition 5.4, and Theorem 5.3, the proof is clear. □

Note. By Theorem 4.1, if {1} is an obstinate filter of A, then {1} is a maximal filter of A. Thus, {1} is the only maximal filter of A and {1} is the only obstinate filter. Moreover, {1} is the only proper filter of A. Then A is a simple hoop.

Proposition 5.6. Any simple hoop is a local hoop.

Proof. Let A be a simple hoop. Then has exactly one maximal filter. So by Theorems 5.3, A is a local hoop. □

We show that in the following example the converse of above proposition is not true, in general.

Example 5.7. Let A be a hoop in Example 4.14 (i). It is clear that A is a local hoop and is not a simple hoop.

Proposition 5.8. Let D _s  (A) = A ∖ {0}. Then A is a local hoop.

Proof. If D _s  (A) = A ∖ {0}, then for any x ∈ A ∖ {0}, x^- = 0. So ord (x^-)< ∞. Therefore, A is a local hoop. □

Theorem 5.9. (i) If {1} is an obstinate filter of A, then A is a local hoop,

(ii) If F is an obstinate filter of A, then A/F is a local hoop.

Proof. (i) Let {1} be an obstinate filter of A. Then by Theorem 4.1, {1} is a maximal filter. Then by Theorem 5.3, A is a local hoop.

(ii) Let F be an obstinate filter of A. Then, by Proposition 3.3, x ∈ F or x^- ∈ F. If x ∈ F, then [x^-] = [0] and ord ([x^-])< ∞. If x^- ∈ F, then [x] = [0] and ord ([x])< ∞. Hence A/F is a local hoop. □

We show that in the following example, the converse of above theorem is not true in general.

Example 5.10. The hoop of Example 4.14(1) is a local hoop, but filter F = {1} is not obstinate. Because a, a^- ∉ F. Also for F = {1, b}, A/F is a local hoop but F is not an obstinate filter.

Proposition 5.11. Let A be a local hoop with (DNP) and x² = x, for all x ∈ A. Then any proper filter of A is an obstinate filter.

Proof. Let F be a proper filter of A and x ∈ A. Since A is local, there exsits n ∈ ℕ such that x ⁿ  = 0 or (x^-)  ⁿ  = 0. If x ⁿ  = 0, since x² = x, then x = 0, and so x^- = 1 ∈ F. If (x^-)  ⁿ  = 0, then x^- = 0. Since A has (DNP), we have x = 1 ∈ F. Hence, by Proposition 3.3, F is an obstinate filter. □

Proposition 5.12. If any proper filter of A is an obstinate filter, then A is a local hoop.

Proof. If any proper filter of A is an obstinate filter, then F = {1} is an obstinate filter. So by Theorem 5.9(i), A is a local hoop. □

Proposition 5.13. Let A be a local ⊔-hoop.

If for any x, y ∈ A, x ⊔ y = 1, then x = 1 or y = 1,

Proof. (i) Let x ≠ 1 and y ≠ 1. By Proposition 2.5, x ⊙ y = x ∧ y. Since A is a local hoop, ord (x)< ∞ or ord (x^-)< ∞. If ord (x)< ∞, then there exists n ∈ ℕ such that x ⁿ  = 0. Hence, x ⁿ  = x^n-1 ⊙ x = x^n-1 ∧ x = x^n-1 = 0, which is a contradiction. The proof of other case is similar.

(ii) Let F be a proper filter and for some proper filter M of A, F ⊂ M. Then there exists x ∈ M ∖ F. Since A is local, then ord (x)< ∞ or ord (x^-)< ∞. Thus, there exists n ∈ ℕ such that x ⁿ  = 0 or (x^-)  ⁿ  = 0. If x ⁿ  = 0, since M is a filter, then 0 ∈ M, and so M = A. If (x^-)  ⁿ  = 0, then by assumption, 1 = x ⊔ (x^-)  ⁿ  = x ⊔ 0 = x. Since F is a filter, then x ∈ F, which is a contradiction. Hence, F = M. Therefore, F is a maximal filter.

(iii) By (ii), F = {1} is the only proper maximal filter if A. So A is a simple hoop. □

Definition 5.14. Let A be a bounded hoop. A is called locally finite if for any 1 ≠ x ∈ A, ord (x)< ∞.

Note. Every locally finite hoop is a local but the converse may not be true.

Example 5.15. (i) Let A = {0, a, b, c, 1} be a chain. Define operations ⊙ and → on A as follows:

Then (A, ⊙ , → , 0, 1) is a bounded hoop. We can see that, ord (0) =1,  ord (a) =2,  ord (b) =2 and ord (c) =3. Then A is locally finite.

(ii) In Example 4.14(1), A is a local hoop. But it is not locally finite. Because, ord (b) =∞.

Proposition 5.16. A is a locally finite if and only if every filter of A is a trivial filter.

Proof. Let F be a filter of locally finite A. If 1 ≠ x ∈ F, then [x) is a filter. Since x ≠ 1, we have ord (x)< ∞. Hence, 0 ∈ [x) ⊆ F, and so F = A. Therefore, F is a trivial filter.

Conversely, let every filter of A be a trivial filter. Let x ∈ A, so [x) = {1} or [x) = A. If [x) = {1}, then x = 1. If [x) = A, then for some n ∈ ℕ , x n = 0. So ord (x)< ∞. Therefore, A is a locally finite hoop. □

Corollary 5.17. If A is a locally finite hoop, then it has not any obstinate filter.

Proof. Since obstinate filter is a proper filter, then by Proposition 5.16, A has not any obstinate filter. □

Proposition 5.18. (i) Let F be an obstinate filter of A. Then A/F is a locally finite. Moreover, if A is bounded, then A/F = {[0], [1]}.

(ii) If A/F = {[0] , [1]}, then F is an obstinate filter of A.

Proof. (i) Let F be an obstinate filter. Then by Theorem 4.1, F is a maximal filter. So, A/F is a locally finite. Now, let [x] ∈ A/F. Then x ∈ A ∖ F. Since F is an obstinate filter, by Proposition 3.3 (iii), x^- ∈ F. Then [x^-] = [1]. Since [x] ≤ [x^--] = [0], we get that [x] = [0].

(ii) Let x ∉ F. So x/F ≠ [1] and x/F = [0]. Then x^- ∈ F. Hence by Proposition 3.3, F is an obstinate filter. □

Proposition 5.19. Let A, B be a bounded hoop and f ∈ Hom (A, B). Then the following conditions hold:

if f ∈ Hom (A, B) is onto and A is a locally finite hoop, then B is a locally finite hoop.

Proof. (i) Let 1 ≠ y ∈ B. Since f is onto, there exists x ∈ A such that f (x) = y. If x = 1, then y = f (x) =1. which is a contradiction. Since A is a locally finite hoop, then ord (x)< ∞ and there exists n ∈ ℕ such that x ⁿ  = 0. So 0 = f (x ⁿ ) = (f (x))  ⁿ  = y ⁿ . Hence, ord (y)< ∞. Therefore, B is a locally finite hoop.

(ii) Let 1 ≠ x ∈ A. Then for any y ∈ B,  f (x) = y. Since B is a locally finite hoop, then ord (y)< ∞ and there exists n ∈ ℕ such that y ⁿ  = 0. So 0 = y ⁿ  = (f (x))  ⁿ  = f (x ⁿ ). Since f is injective, then x ⁿ  = 0. Hence, ord (x)< ∞. Therefore, A is a locally finite hoop. □

6 Conclusion and future research

In this paper, we have introduced the notion of obstinate filters in hoop. We have established properties of obstinate filters in hoop and proved the relationships between obstinate filters and other types of filters in hoops. We have shown that, if F is an obstinate filter of a hoop A, then A/F is boolean algebra. Also, we introduced the concepts of local, locally finite and quasi local hoops and we study the quotient of them.

Some important issues for future research are trying to define the notion of n-fold obstinate filter and fuzzy obstinate filter BL-algebra and try to define other types of filters in hoop.