On co-annihilators in hoops

1 Introduction

It is well known that various logical algebras have been proposed and researched as the semantical systems of non-classical logical systems. Among these logical algebras, residuated lattices were introduced by Ward and Dilworth in 1939 to constitute the semantics of Höhle Monoidal Logic which are the basis for the majority of formal fuzzy logic. Apart from their logical interest, residuated lattices have interesting algebraic properties and include two important classes of algebras: BL-algebras and MV-algebras. In order to study the basic logic framework of fuzzy set system, based on continuous triangle module and under the theoretical framework of residuated lattices theory, Hájek [21], proposed a new fuzzy logic system BL-system and the corresponding logical algebraic system BL-algebra. MV-algebras were introduced by Chang [16], in order to give an algebraic proof of the completeness theorem of Lukasiewice system of many valued logic. The notion of filters has been introduced in many algebraic structures such as lattices, rings, MV-algebras. Filters theory is a very effective tool for studying various algebraic and logical systems. In the theory of MV-algebras the notion of deductive systems and ideals are the dual notions, in BL-algebra, with the lack of a suitable algebraic addition, the focus is shifted to deductive systems also called filters. Turunen [30], defined co-annihilator of a non-empty set X of L and proved some of its properties on BL-algebras. They got A^⊤ as a prime filter if and only if A is linear and A ≠ {1}. Also, in [25], B. A. Laurentiu Leustean introduced the notion of the co-annihilator relative to F on pseudo-BL-algebras, which is a generalization of the co-annihilator, and they also extended some results obtained in [18]. Moreover, in [26], B. L. Meng et al. defined the generalized co-annihilator of BL-algebras as a generalization of co-annihilator on BL-algebras.

Hoops are naturally ordered commutative residuated integral monoids, introduced by B. Bosbach in [12, 13] then study by J. R. Büchi and T. M. Owens in [14], a paper never published. In the last years, hoops theory was enriched with deep structure theorems (see [1, 7–13]). Many of these results have a strong impact with fuzzy logic. Particularly, from the structure theorem of finite basic hoops(2[Corollary 2.10]10) one obtains an elegant short proof of the completeness theorem for propositional basic logic (see [2, Theorem 3.8]10), introduced by Hájek in [21]. The algebraic structures corresponding to Hájek’s propositional (fuzzy) basic logic, BL-algebras, are particular cases of hoops. The main example of BL-algebras in interval [0,1] endowed with the structure induced by a t-norm. MV-algebras, product algebras and Gödel algebras are the most known classes of BL-algebras. Recent investigations are concerned with non-commutative generalizations for these structures.

Now in this paper, we introduce the notion of co-annihilator in hoops and investigate some related properties of them.

2 Preliminaries

In this section, we gather some basic notions relevant to hoop which will need in the next sections.

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

(HP1) (A, ⊙ , 1) is a commutative monoid,

(HP2) x → x = 1,

(HP3) (x ⊙ y) → z = x → (y → z), (HP4) x ⊙ (x → y) = y ⊙ (y → x).

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(See [2]).

The following proposition provides some properties of hoop.

Proposition 2.1. [12, 13] Let (A, ⊙ , → , 1) be a hoop. Then the following conditions hold, for all x, y, z, a ∈ A:

(i) (A, ≤) is a ∧-semilattice with x ∧ y = x ⊙ (x → y),

(ii) x ⊙ y ≤ z if and only if x ≤ y → z,

(iii) x ⊙ y ≤ x, y,

(iv) x ≤ y → x,

(v) x → 1 =1,

(vi) 1 → x = x,

(vii) x ⊙ (x → y) ≤ y,

(viii) x ≤ y implies x ⊙ a ≤ y ⊙ a,

(ix) x ≤ y implies z → x ≤ z → y, (x) x ≤ y implies y → z ≤ x → z,

(xi) x → (y ∧ z) = (x → y) ∧ (x → z).

Proposition 2.2. [20] Let A be a bounded hoop. Then the following conditions hold, for all x, y ∈ A:

(i) x ≤ y″,

(ii) x ⊙ x′ = 0 and x″′ = x′,

(iii) x ≤ x′ → y,

(iv) if x = x″′, then x → y = y′ → x′.

Proposition 2.3. [20] Let A be a hoop and for any x, y ∈ A, we define, x ⊔ y = ((x → y) → y) ∧ ((y → x) → x). Then the following conditions are equivalent:

(i) ⊔ is an associative operation on A,

(ii) x ≤ y implies x ⊔ z ≤ y ⊔ z, for all x, y, z ∈ A,

(iii) x ⊔ (y ∧ z) ≤ (x ⊔ y) ∧ (x ⊔ z), for all x, y, z ∈ A,

(iv) ⊔ is the join operation on A.

A hoop A is called a ⊔-hoop, if ⊔ is a join operation on A.

Proposition 2.4. [5] Let A be a ⊔-hoop. Then, for all x, y, z ∈ A and n ∈ ℕ ,

(i) (x ⊔ y) → z = (x → z) ∧ (y → z),

(ii) x ⊙ (y ⊔ z) = (x ⊙ y) ⊔ (x ⊙ z),

(iii) (x ⊔ y)  ⁿ  → z = ⋀ {(x₁ ⊙ . . . ⊙ x _n ) → z ∣ x _i  ∈ {x, y}},

(iv) (A, ⊔ , ∧) is a distributive lattice.

Let A be a hoop. A non-empty subset F of A is called a filter of A if it satisfies,

(F1) x, y ∈ F implies x ⊙ y ∈ F;

(F2) x ∈ F and x ≤ y imply y ∈ F, for any x, y ∈ A. We use F ( A ) to the denote the set of all filters of A. Clearly, 1 ∈ F, for all F ∈ F ( A ) . A filter F of A is called a proper filter if F ≠ A. It can be easily seen that, if A is a bounded hoop, then a filter is proper if and only if it is not containing 0. If A is a ⊔-hoop and F is a proper filter of A, then F is called a prime filter of A if x ⊔ y ∈ F, then x ∈ F or y ∈ F, for some x, y ∈ A(See [5, 11]).

If X ⊆ A, we denote by 〈X〉 the filter generated by X in A. A description of 〈X〉 is easily obtained.

Proposition 2.5. [20] Let A be a hoop, X ⊆ A and F be a filter of A. Then 〈 X 〉 = { a ∈ A ∣ x 1 ⊙ x 2 ⊙ . . . ⊙ x n ≤ a , for some n ∈ ℕ and x 1 , . . . , x n ∈ X } = { a ∈ A ∣ x 1 → ( x 2 → ( . . . → ( x n → a ) . . . ) ) = 1 , for some n ∈ ℕ and x 1 , . . . , x n ∈ X } In particular, for any element x ∈ A, we have 〈 x 〉 = { a ∈ A ∣ x n ≤ a , for some n ∈ ℕ } = { a ∈ A ∣ x n → a = 1 , for some n ∈ ℕ } Let F be a filter of hoop A and x ∈ A. Then 〈 F ∪ { x } 〉 = { a ∈ A ∣ f ⊙ x n ≤ a , ∃ n ∈ ℕ , f ∈ F } = { a ∈ A ∣ x n → a ∈ F , ∃ n ∈ ℕ }

Let X be a non-empty subset of bounded hoop A with (DNP). Then X^⊥ = {a ∈ A ∣ a ∧ x = 0,  any  x ∈ X} is called an annihilator of X.

Definition 2.6. [22, 23] Consider X as a non-empty set. A mapping ϕ : P ( X ) → P ( X ) is called a closure map on X, if the following holds, for all Y , Z ∈ P ( X ) :

(C1) Y ⊆ ϕ (Y),

(C2) ϕ² (Y) = ϕ (Y),

(C3) Y ⊆ Z implies ϕ (Y) ⊆ ϕ (Z).

Theorem 2.7. [5] Let (L, ∧ , ∨) be a lattice and f : L → L be a closure map. Then Imf is a lattice in which the lattice operations are given by inf {a, b} = a ∧ b and sup {a, b} = f (a ∨ b).

Definition 2.8. [1] Let A and B be two hoops. Then h : A → B is called a homomorphism if, for all a, b ∈ A, h ( 1 A ) = 1 B , h ( a ⊙ A b ) = h ( a ) ⊙ B h ( b ) , h ( a → A b ) = h ( a ) → B h ( b ) If A and B are two bounded hoops, then h (0 _A ) =0 _B . Also, if a hoop homomorphism h is a surjection, then it is called a hoop epimorphism. A one-to-one and onto hoop homomorphism h is called a hoop isomorphism from A to B, in this case, A and B are called to be isomorphism and denoted by A ≅ B.

Definition 2.9. [6] A Heyting algebra is an algebra (A, ∨ , ∧ , → , 1) of type (2, 2, 2, 0), where (A, ∨ , ∧ , 1) is a lattice with the greatest element 1 and, for any x, y, z ∈ A, x ≤ y → z if and only if x ∧ y ≤ z

Notation. From now on, in this paper, (A, ⊙ , → , 1) or simply A is a ⊔-hoop, unless otherwise state.

3 Co-annihilators in hoops

In this section, we introduce the notion of co-annihilator set in a hoop and investigate some related properties of them. We show that ( F ( A ) , ⊆ ) is a pseudo-complemented lattice, and for any F ∈ F ( A ) , it’s pseudo-complement is the co-annihilator of F. Then we prove that the set of all co-annihilators of A is a Boolean algebra.

Definition 3.1. Let X be a nonempty subset of A. Then the set X ⊤ = { a ∈ A ∣ a ⊔ x = 1 , for all x ∈ X } is called a co-annihilator of X.

Example 3.2. Let A = {0, a, b, c, d, 1} where 0 ≤ a ≤ b ≤ 1, 0 ≤ a ≤ d ≤ 1 and 0 ≤ c ≤ d ≤ 1. Suppose that operations → and ⊙ on A are defined by the Cayley tables as follows:

Then by routine calculations we can see that (A, → , ⊙ , 0, 1) is a bounded ⊔-hoop. Now, if X₁ = {b, 1}, X₂ = {d, 1} and X₃ = {a, c, d, 1}, then it is easy to check that X 1 ⊤ = { c , d , 1 } , X 2 ⊤ = { b , 1 } and X 3 ⊤ = { 1 } .

Lemma 3.3. For any x, y, z ∈ A, ( x ⊔ z ) ⊙ ( y ⊔ z ) ≤ ( x ⊙ y ) ⊔ z

Proof. Let x, y, z ∈ A. Then; [ ( x ⊔ z ) ⊙ ( y ⊔ z ) ] → [ ( x ⊙ y ) ⊔ z ] = ( x ⊔ z ) → [ ( y ⊔ z ) → ( ( x ⊙ y ) ⊔ z ) ] , by Proposition 2.4 ( i ) = ( x ⊔ z ) → [ ( y → ( ( x ⊙ y ) ⊔ z ) ) ∧ ( z → ( ( x ⊙ y ) ⊔ z ) ) ] = ( x ⊔ z ) → ( y → ( ( x ⊙ y ) ⊔ z ) ) , by ( HP 3 ) = y → ( ( x ⊔ z ) → ( ( x ⊙ y ) ⊔ z ) ) , by Proposition 2.4 ( i ) = y → [ ( x → ( ( x ⊙ y ) ⊔ z ) ) ∧ ( z → ( ( x ⊙ y ) ⊔ z ) ) ] = y → ( x → ( ( x ⊙ y ) ⊔ z ) ) = ( y ⊙ x ) → ( ( x ⊙ y ) ⊔ z ) = 1 Hence, (x ⊔ z) ⊙ (y ⊔ z) ≤ (x ⊙ y) ⊔ z. □

Proposition 3.4. Let ∅ ≠ X ⊆ A. Then X^⊤ is a filter of A.

Proof. Let ∅ ≠ X ⊆ A. Since 1 ∈ A, so for any x ∈ X, x ⊔ 1 =1 and then 1 ∈ X^⊤. Suppose a, b ∈ X^⊤. Then, for any x ∈ X, a ⊔ x = b ⊔ x = 1. Thus, by Lemma 3.3, 1 = (a ⊔ x) ⊙ (b ⊔ x) ≤ (a ⊙ b) ⊔ x, and so (a ⊙ b) ⊔ x = 1. Hence, a ⊙ b ∈ X^⊤. Moreover, let a ≤ b, for a ∈ X^⊤ and b ∈ A. Then, for any x ∈ X, a ⊔ x = 1. Since a ⊔ x ≤ b ⊔ x, we have b ⊔ x = 1, thus, b ∈ X^⊤. Therefore, X ⊤ ∈ F ( A ) . □

Proposition 3.5. Let A be a bounded ⊔-hoop with (DNP). Then, for any ∅ ≠ X ⊆ A, (X^⊤) ′ = (X′) ^⊥, where X′ = {x′ ∣ x ∈ X}.

Proof. Let a ∈ (X^⊤) ′. Then, there exists b ∈ X^⊤ such that a = b′. Since b ∈ X^⊤, for any x ∈ X, b ⊔ x = 1. Then by Proposition 2.4(i), 0 = (b ⊔ x) ′ = b′ ∧ x′ = a ∧ x′. So, for any x′ ∈ X′, a ∧ x′ = 0. Thus, a ∈ (X′) ^⊥. Hence, (X^⊤) ′ ⊆ (X′) ^⊥. Conversely, let a ∈ (X′) ^⊥. Then, for any x′ ∈ X′, a ∧ x′ = 0. Thus, for any x″ ∈ X″, a′ ⊔ x″ = 1. Since A has (DNP), for any x ∈ X, a′ ⊔ x = 1. Then, a′ ∈ X^⊤, and so a ∈ (X^⊤) ′. Hence, (X′) ^⊥ ⊆ (X^⊤) ′. Therefore, (X^⊤) ′ = (X′) ^⊥. □ In the following example, we show that the condition (DNP) is necessary in Proposition 3.5.

Example 3.6. Let A = {0, a, b, c, d, 1} where 0 ≤ a ≤ c ≤ 1, 0 ≤ b ≤ d ≤ 1 and 0 ≤ b ≤ c ≤ 1. Suppose that operations → and ⊙ on A are defined by the Cayley tables as follows:

Then by routine calculations we can see that (A, → , ⊙ , 0, 1) is a bounded ⊔-hoop that has not (DNP). Now, if X = {b, d}. Then X′ = {a}, then it is easy to see that (X^⊤) ′ = {0} and (X′) ^⊥ = {0, b, d}, and so (X^⊤) ′ ≠ (X′) ^⊥.

Proposition 3.7. The following statements hold, for any a, b ∈ A:

(i) {0} ^⊤ = A^⊤ = {1} and {1} ^⊤ = A,

(ii) if a ≤ b, then {a} ^⊤ ⊆ {b} ^⊤,

(iii) {a} ^⊤ ∩ {b} ^⊤ = {a ∧ b} ^⊤,

(iv) {a} ^⊤ ∪ {b} ^⊤ ⊆ {a ⊔ b} ^⊤,

(v) if x ∈ {a} ^⊤ and y ∈ {b} ^⊤, then x ⊔ y ∈ {a ∧ b} ^⊤ and x ∧ y, x ⊙ y, x ⊔ y ∈ {a ⊔ b} ^⊤,

(vi) if x ∈ {a} ^⊤ and y ∈ {a ∧ b} ^⊤ or y ∈ {a ⊙ b} ^⊤, then x ∧ y, x ⊙ y, x ⊔ y ∈ {a ⊔ b} ^⊤.

(vii) if f : A → B is a ⊔-hoop isomorphism, then f ({x} ^⊤) = {f (x)} ^⊤, for any x ∈ A.

Proof. (i) According to the definition of co-annihilator, we have { 0 } ⊤ = { a ∈ A ∣ a ⊔ 0 = 1 } = { a ∈ A ∣ a = 1 } = { 1 } Also, A ⊤ = { a ∈ A ∣ a ⊔ x = 1 , for any x ∈ A } = { a ∈ A ∣ a ⊔ a = 1 } = { 1 } Moreover, { 1 } ⊤ = { a ∈ A ∣ a ⊔ 1 = 1 } = A

(ii) Let x ∈ {a} ^⊤. Then a ⊔ x = 1. Since a ≤ b and a ⊔ x ≤ b ⊔ x, we have b ⊔ x = 1. Hence, x ∈ {b} ^⊤.

(iii) Since a ∧ b ≤ a, b, by (ii), {a ∧ b} ^⊤ ⊆ {a} ^⊤, {b} ^⊤, then {a ∧ b} ^⊤ ⊆ {a} ^⊤ ∩ {b} ^⊤. Let x ∈ {a} ^⊤ ∩ {b} ^⊤. Then a ⊔ x = b ⊔ x = 1. Since A is a ⊔-hoop, by Proposition 2.4(iv), A is a distributive lattice, thus, 1 = (x ⊔ a) ∧ (x ⊔ b) = x ⊔ (a ∧ b). Hence, x ∈ {a ∧ b} ^⊤. Therefore, {a} ^⊤ ∩ {b} ^⊤ = {a ∧ b} ^⊤.

(iv) Since a, b ≤ a ⊔ b, by (ii), {a} ^⊤, {b} ^⊤ ⊆ {a ⊔ b} ^⊤. So, it is clear that {a} ^⊤ ∪ {b} ^⊤ ⊆ {a ⊔ b} ^⊤.

(v) Let x ∈ {a} ^⊤ and y ∈ {b} ^⊤. Then by Proposition 2.4(iv), A is a distributive lattice, thus, ( x ⊔ y ) ⊔ ( a ∧ b ) = [ ( x ⊔ y ) ⊔ a ] ∧ [ ( x ⊔ y ) ⊔ b ] = [ ( x ⊔ a ) ⊔ y ] ∧ [ x ⊔ ( y ⊔ b ) ] = 1 So, x ⊔ y ∈ {a ∧ b} ^⊤. Also, since A is a distributive lattice, we have, ( x ∧ y ) ⊔ ( a ⊔ b ) = [ ( x ∧ y ) ⊔ a ] ⊔ b = [ ( x ⊔ a ) ∧ ( a ⊔ y ) ] ⊔ b = ( a ⊔ y ) ⊔ b = a ⊔ ( y ⊔ b ) = 1

By Lemma 3.3, we get that, ( x ⊙ y ) ⊔ ( a ⊔ b ) = [ ( x ⊙ y ) ⊔ a ] ⊔ b ≥ [ ( x ⊔ a ) ⊙ ( a ⊔ y ) ] ⊔ b = ( a ⊔ y ) ⊔ b = a ⊔ ( y ⊔ b ) = 1 Moreover, (x ⊔ y) ⊔ (a ⊔ b) = (x ⊔ a) ⊔ (y ⊔ b) =1. Therefore, x ∧ y, x ⊙ y, x ⊔ y ∈ {a ⊔ b} ^⊤.

(vi) Since a ∧ b, a ⊙ b ≤ b, by (ii), {a ⊙ b} ^⊤ ⊆ {b} ^⊤ and {a ∧ b} ^⊤ ⊆ {b} ^⊤. Then by (vi), the proof is clear.

(vii) Let x ∈ A and a ∈ f ({x} ^⊤). Then there exists b ∈ {x} ^⊤ such that f (b) = a. Since b ∈ {x} ^⊤ and f is a ⊔-hoop homomorphism, b ⊔ x = 1 and a ⊔ f (x) = f (b) ⊔ f (x) = f (b ⊔ x) =1. Then a ∈ {f (x)} ^⊤. Conversely, suppose a ∈ {f (x)} ^⊤. Since f is an isomorphism, there is b ∈ A such that a = f (b) and f (b) ⊔ f (x) =1, then f (b ⊔ x) =1. So, b ⊔ x = 1. Thus, b ∈ {x} ^⊤. Hence, a = f (b) ∈ f ({x} ^⊤). Therefore, f ({x} ^⊤) = {f (x)} ^⊤. □

In the following example, we show that the converse of Proposition 3.7(iv), may not be true, in general.

Example 3.8. Let A be a hoop as Example 3.6. It is easy to see that {c} ^⊤ = {1, d} and {d} ^⊤ = {1, a, c}. Then we consequence that A = {1} ^⊤ = {c ⊔ d} ^⊤ ⊈ {c} ^⊤ ∪ {d} ^⊤ = {1, a, c, d}.

Proposition 3.9. Let ∅ ≠ X ⊆ A. Then X ⊤ = { a ∈ A ∣ a → x = x and x → a = a , for all x ∈ X }

Proof. Let B = {a ∈ A ∣ a → x = x and x → a = a,  forall x ∈ X} and a ∈ X^⊤. Then, for any x ∈ X, a ⊔ x = 1. By Proposition 2.3, 1 = a ⊔ x = ( ( a → x ) → x ) ∧ ( ( x → a ) → a ) ≤ ( a → x ) → x and ( x → a ) → a Thus, (x → a) → a = 1, and so x → a ≤ a. Moreover, by Proposition 2.1(iv), a ≤ x → a. Hence, x → a = a. By the similar way, we consequence that a → x = x, and so a ∈ B. Conversely, suppose a ∈ B. Then, for any x ∈ X, a → x = x and x → a = a. By Proposition 2.3, a ⊔ x = ((a → x) → x) ∧ ((x → a) → a) =1. Hence, a ∈ X^⊤. Therefore, X^⊤ = {a ∈ A ∣ a → x = x and x → a = a,  forall x ∈ X}. □

Notation. For any ∅ ≠ X ⊆ A, we denoted by X^∗ = {a ∈ A ∣ a → x = x,  forall x ∈ X}.

Proposition 3.10. Let ∅ ≠ X ⊆ A. Then X ∗ ∈ F ( A ) .

Proof. By Proposition 2.1(vi), 1 → x = x, for any x ∈ X. Then 1∈ X^∗ ≠ ∅. Let a ≤ b and a ∈ X^∗, for b ∈ A. Since a ∈ X^∗, for any x ∈ X, a → x = x and by Proposition 2.1(iv) and (x), x ≤ b → x ≤ a → x = x. Thus, for any x ∈ X, b → x = x, and so b ∈ X^∗. Moreover, if a, b ∈ X^∗, then, for any x ∈ X, a → x = b → x = x. Thus, by (HP3), (a ⊙ b) → x = a → (b → x) = a → x = x. So, a ⊙ b ∈ X^∗. Hence, X ∗ ∈ F ( A ) . □

Proposition 3.11. Let ∅ ≠ X ⊆ A. Then X^⊤ ⊆ X^∗.

Proof. Let a ∈ X^⊤. Then, for any x ∈ X, a ⊔ x = 1. By Propositions 2.1(iv) and 2.4(i), we have x = 1 → x = ( a ⊔ x ) → x = ( a → x ) ∧ ( x → x ) Thus, a → x = x, and so x ∈ X^∗. Hence, X^⊤ ⊆ X^∗. □

In the following example we show that the converse of Proposition 3.11, may not be true, in general.

Example 3.12. Let A = {0, a, b, 1} be a chain. Suppose the operations → and ⊙ on A are defined by the Cayley tables as follows:

Then routine calculations show that A with these operations is a bounded ⊔-hoop. Let X = {a}. Then X^⊤ = {1} and X^∗ = {1, b}. So, it is clear that X^∗ ⊈ X^⊤.

Proposition 3.13. If A is a bounded ⊔-hoop with (DNP), then X^∗ = X^⊤, for any ∅ ≠ X ⊆ A.

Proof. Let A be a bounded ⊔-hoop with (DNP) and x, y ∈ A. Since y″ = y, we have x → y = x → y″, then by (HP3), x → y″ = x → (y′ → 0) = y′ → (x → 0), and so x → y = y′ → x′. Now, we prove that (x → y) → y = (y → x) → x. Since y″ = y and x → y = y′ → x′, we have (x → y) → y = (y′ → x′) → y″. Thus, by (HP3), ( y ′ → x ′ ) → y ″ = ( y ′ → x ′ ) → ( y ′ → 0 ) = ( y ′ ⊙ ( y ′ → x ′ ) ) → 0

By Proposition 2.1(i), (y′ ⊙ (y′ → x′)) →0 = (y′ ∧ x′) ′. By the similar way, we can prove that (y → x) → x = (y′ ∧ x′) ′. Hence, (x → y) → y = (y → x) → x, and so x ⊔ y = (y → x) → x. Let a ∈ X^∗. Then, for any x ∈ X, a ⊔ x = (a → x) → x = x → x = 1. Thus, X^∗ ⊆ X^⊤, and by Proposition 3.11, X^∗ = X^⊤, for any ∅ ≠ X ⊆ A.□

Proposition 3.14. For any ∅ ≠ X ⊆ A, 〈X〉 ∩ X^⊤ = {1}.

Proof. Since 〈X〉 is a generated filter of A and by Proposition 3.4, X ⊤ ∈ F ( A ) , it is clear that {1} ⊆ 〈X〉 ∩ X^⊤. Let a ∈ 〈X〉 ∩ X^⊤. Since a ∈ 〈X〉, by Proposition 2.5, there exist n ∈ ℕ and x₁, x₂, . . . , x _n  ∈ X such that x₁ ⊙ x₂ ⊙ . . . ⊙ x _n  ≤ a. Moreover, a ∈ X^⊤, then, for any x ∈ X, a ⊔ x = 1, so, for any 1 ≤ i ≤ n, a ⊔ x _i  = 1. By Lemma 3.3, we have, 1 = ( x 1 ⊔ a ) ⊙ ( x 2 ⊔ a ) ⊙ . . . ⊙ ( x n ⊔ a ) ≤ ( x 1 ⊙ x 2 ⊙ . . . ⊙ x n ) ⊔ a ≤ a ⊔ a = a

Then a = 1. Hence, 〈X〉 ∩ X^⊤ ⊆ {1}. Therefore, 〈X〉 ∩ X^⊤ = {1}.□

Proposition 3.15. Let ∅ ≠ X, Y ⊆ A. Then the following statements hold:

(i) X∩ X^∗ = ∅ or X ∩ X^∗ = {1},

(ii) if X ⊆ Y, then Y^∗ ⊆ X^∗.

Proof. (i) If 1 ∈ X, by Proposition 3.10, X ∗ ∈ F ( A ) , then 1 ∈ X ∩ X^∗. Let x ∈ X ∩ X^∗. Since x ∈ X^∗, x = x → x = 1, thus, x = 1, and so X ∩ X^∗ = {1}. Now, if 1 ∉ X, then, for any x ∈ X ∩ X^∗, x ≠ 1. Moreover, since x ∈ X^∗, 1 = x → x = x, so, x = 1, which is a contradiction. Hence, X∩ X^∗ = ∅.

(ii) Let a ∈ Y^∗. Then, for any y ∈ Y, a → y = y. Since X ⊆ Y, for any x ∈ X, we have x ∈ Y, then a → x = x. Thus, a ∈ X^∗. Hence, Y^∗ ⊆ X^∗. □

Corollary 3.16. Let ∅ ≠ X ⊆ A. Then 〈X〉 ∩ X^∗ = {1}.

Proof. By Proposition 3.10, X ∗ ∈ F ( A ) , then by Proposition 3.15(i), the proof is clear. □

Proposition 3.17. If F ∈ F ( A ) and F is linear, then F^⊤ is a prime filter of A.

Proof. Let F ∈ F ( A ) . Then by Proposition 3.4, F ⊤ ∈ F ( A ) . If x ∈ F, then x = x ⊔ x ≠ 1, thus, x ∉ F^⊤. Hence, F^⊤ is a proper filter of A. Suppose a ⊔ b ∈ F^⊤ and a, b ∉ F^⊤. Then there exist x₁, x₂ ∈ F such that a ⊔ x₁ ≠ 1 and b ⊔ x₂ ≠ 1. Let x = x₁ ∧ x₂. Since x₁, x₂ ∈ F, F ∈ F ( A ) and by Proposition 2.1(iii), x₁ ⊙ x₂ ≤ x₁ ∧ x₂ = x, x ∈ F. Also, by Proposition 2.4(iv), A is a distributive lattice, then a ⊔ x = a ⊔ ( x 1 ∧ x 2 ) = ( a ⊔ x 1 ) ∧ ( a ⊔ x 2 ) ≠ 1 Similarly, b ⊔ x ≠ 1. Moreover, F is linear, then without the lost of generality, let a ⊔ x ≤ b ⊔ x. Thus, 1 = ( a ⊔ b ) ⊔ x = b ⊔ ( a ⊔ x ) ≤ b ⊔ ( b ⊔ x ) = b ⊔ x and so, b ⊔ x = 1, which is a contradiction. Hence, a ∈ F^⊤ or b ∈ F^⊤. Therefore, F^⊤ is a prime filter of A. □

Proposition 3.18. Let ∅ ≠ X, Y ⊆ A. Then the following statements hold:

(i) if X ⊆ Y, then Y^⊤ ⊆ X^⊤,

(ii) X^⊤ = ⋂ _x∈X {x} ^⊤,

(iii) X ⊆ X^⊤⊤,

(iv) X^⊤ = X^⊤⊤⊤,

(v) X^⊤ = 〈X〉^⊤,

(vi) X^⊤ ∩ X^⊤⊤ = {1},

(vii) (X ∪ Y) ^⊤ = X^⊤ ∩ Y^⊤,

(viii) X^⊤ ∪ Y^⊤ ⊆ (X ∩ Y) ^⊤.

Proof. (i) Let a ∈ Y^⊤. Then, for any y ∈ Y, a ⊔ y = 1. Since X ⊆ Y, for any x ∈ X, x ∈ Y, then a ⊔ x = 1. Hence, a ∈ X^⊤.

(ii) Let a ∈ X^⊤. Then, for any x ∈ X, a ⊔ x = 1. Thus, for any x ∈ X, a ∈ {x} ^⊤. Hence, a ∈ ⋂ _x∈X {x} ^⊤. The proof of other side is similar.

(iii) By definition of co-annihilator, we have X ⊤ ⊤ = { a ∈ A ∣ a ⊔ x = 1 , for any x ∈ X ⊤ } So, for all x ∈ X^⊤, if b ∈ X, then b ⊔ x = 1, hence, b ∈ X^⊤⊤.

(iv) By (iii), X ⊆ X^⊤⊤. Then by (i), X^⊤⊤⊤ ⊆ X^⊤. Now, let Y = X^⊤. Then by (iii), Y ⊆ Y^⊤⊤. Thus, X^⊤ ⊆ X^⊤⊤⊤. Hence, X^⊤ = X^⊤⊤⊤.

(v) Since 〈 X 〉 ∈ F ( A ) , we have X ⊆ 〈X〉. Then by (i), 〈X〉^⊤ ⊆ X^⊤. Let a ∈ X^⊤. Then, for any x ∈ X, a ⊔ x = 1. Also, for any z ∈ 〈X〉, by Proposition 2.5, there exist n ∈ ℕ and x₁, x₂, . . . , x _n  ∈ X such that x₁ ⊙ x₂ ⊙ . . . ⊙ x _n  ≤ z. Then by Lemma 3.3, 1 = ( x 1 ⊔ a ) ⊙ ( x 2 ⊔ a ) ⊙ . . . ⊙ ( x n ⊔ a ) ≤ ( x 1 ⊙ x 2 ⊙ . . . ⊙ x n ) ⊔ a ≤ z ⊔ a Thus, z ⊔ a = 1, and so a ∈ 〈X〉^⊤. Therefore, X^⊤ = 〈X〉^⊤.

(vi) By Proposition 3.4, X ⊤ ∈ F ( A ) . Then it is clear that {1} ⊆ X^⊤ ∩ X^⊤⊤. Let a ∈ X^⊤ ∩ X^⊤⊤. Then by definition of co-annihilator, we have a = a ⊔ a = 1. Hence, X^⊤ ∩ X^⊤⊤ = {1}.

(vii) Since X, Y ⊆ X ∪ Y, by (i) we have, (X ∪ Y) ^⊤ ⊆ X^⊤, Y^⊤, and so (X ∪ Y) ^⊤ ⊆ X^⊤ ∩ Y^⊤. Now, let a ∈ X^⊤ ∩ Y^⊤. Then, for any x ∈ X and y ∈ Y, a ⊔ x = a ⊔ y = 1. Suppose α ∈ X ∪ Y. Then α ∈ X or α ∈ Y. So, it is clear that a ⊔ α = 1. Hence, a ∈ (X ∪ Y) ^⊤. Therefore, (X ∪ Y) ^⊤ = X^⊤ ∩ Y^⊤.

(viii) Since X ∩ Y ⊆ X, Y, by (i), it is clear that X^⊤ ∪ Y^⊤ ⊆ (X ∩ Y) ^⊤. □

In the following example we show that the converse of Proposition 3.18(viii), may not be true, in general.

Example 3.19. Let A be a hoop as Example 3.6. Let X = {1, c} and Y = {1, d}. Then by routine calculation, we get that X^⊤ = {1, d} and Y^⊤ = {1, a, c}. Then we consequence that A = {1} ^⊤ = (X ∩ Y) ^⊤ ⊈ X^⊤ ∪ Y^⊤ = {1, a, c, d}.

Notation. By routine calculations, we can see that all items in Proposition 3.18 hold for X^∗.

Proposition 3.20. Let X and Y be two non-empty subsets of A such that a ∈ X^⊤ and b ∈ Y^⊤, for any a, b ∈ A. Then a ⊔ b ∈ (X ∪ Y) ^⊤ and a ∧ b ∈ (X ∩ Y) ^⊤.

Proof. Let a ∈ X^⊤ and b ∈ Y^⊤. Then, for any x ∈ X and y ∈ Y, a ⊔ x = b ⊔ y = 1. By Proposition 2.4(iv), A is a distributive lattice, thus, for any α ∈ X ∩ Y, ( a ∧ b ) ⊔ α = ( a ⊔ α ) ∧ ( b ⊔ α ) = 1 So, a ∧ b ∈ (X ∩ Y) ^⊤. Moreover, for any x ∈ X ∪ Y, ( a ⊔ b ) ⊔ x = a ⊔ ( b ⊔ x ) = ( a ⊔ x ) ⊔ b = 1 Hence, a ⊔ b ∈ (X ∪ Y) ^⊤. □

Proposition 3.21. Let X and Y be two non-empty subsets of A. Then X^⊤ ∩ Y^⊤ = {1} if and only if X^⊤ ⊆ Y^⊤⊤ and Y^⊤ ⊆ X^⊤⊤.

Proof. Let a ∈ X^⊤ and b ∈ Y^⊤. Then by Proposition 3.4, X ⊤ , Y ⊤ ∈ F ( A ) . Thus, a ⊔ b ∈ X^⊤ ∩ Y^⊤ = {1}, so a ⊔ b = 1. Hence, X^⊤ ⊆ Y^⊤⊤. By the similar way, we have Y^⊤ ⊆ X^⊤⊤. Conversely, suppose X^⊤ ⊆ Y^⊤⊤. Then X^⊤ ∩ Y^⊤ ⊆ Y^⊤⊤ ∩ Y^⊤. By Proposition 3.18(vi), X^⊤ ∩ Y^⊤ = {1}. □

Proposition 3.22. Let F , G ∈ F ( A ) . Then the following statements hold:

(i) F ∩ G = {1} if and only if F ⊆ G^∗,

(ii) F ⊆ F^∗∗.

Proof. (i) Let F , G ∈ F ( A ) such that F ∩ G = {1}. Then by Proposition 2.1(iv) and (vii), for any y ∈ G, x, y ≤ (x → y) → y. Since F , G ∈ F ( A ) , (x → y) → y ∈ F ∩ G = {1}. Then x → y ≤ y. Thus, by Proposition 2.1(iv), x → y = y, and so x ∈ G^∗. Conversely, since F ⊆ G^∗, we have F ∩ G ⊆ G ∩ G^∗. Then by Proposition 3.15, F ∩ G = {1} or F∩ G = ∅. Since F , G ∈ F ( A ) , it is clear that F ∩ G = {1}.

(ii) Let F ∈ F ( A ) . Then by Proposition 3.10, F ∗ ∈ F ( A ) , so by Proposition 3.15, F ∩ F^∗ = {1}. Thus, by (i), F ⊆ F^∗∗. □

Proposition 3.23. Let F be a non-empty subset of A. Then there exists ∅ ≠ X ⊆ A such that F = X^⊤ if and only if F is an upset, F^⊤⊤ = F, F ∩ X^⊤ = {1} and X^⊤ ∩ F^⊤ = {1}.

Proof. (⇒) Suppose there exists ∅ ≠ X ⊆ A such that F = X^⊤. Then by Proposition 3.18(iv), F^⊤⊤ = X^⊤⊤⊤ = X^⊤ = F. Also, by Proposition 3.18(vi), we have F ∩ X^⊤⊤ = X^⊤ ∩ X^⊤⊤ = {1} and X^⊤ ∩ F^⊤ = X^⊤ ∩ X^⊤⊤ = {1}. Moreover, by Proposition 3.4, F = X ⊤ ∈ F ( A ) , then it is clear that F is an upset. (⇐) Let a ∈ F. Then, for any x ∈ X, a ≤ a ⊔ x. Since F is an upset, a ⊔ x ∈ F, also X ⊤ ⊤ ∈ F ( A ) , then a ⊔ x ∈ F ∩ X^⊤⊤ = {1}. Thus, by Proposition 3.18(vi), a ∈ X^⊤⊤⊤ = X^⊤, and so F ⊆ X^⊤. Moreover, since F^⊤ ∩ X^⊤ = {1}, by Proposition 3.21, we have X^⊤ ⊆ F^⊤⊤ = F. Hence, X^⊤ ⊆ F. Therefore, X^⊤ = F. □

Proposition 3.24. Let A be a bounded ⊔-hoop, F be a linear filter of A, x ∈ F such that x ≠ 1 and x′ ⊔ x = 1. Then x is the smallest element of F.

Proof. For proving that x is the smallest element of F, it is enough to show that, for any a ∈ F, x ≤ a. For this, since x′ ⊔ x = 1, by Propositions 2.4(i), 0 = (x′ ⊔ x) ′ = x″ ∧ x′. Also, by Proposition 2.2(i), x ≤ x″, then x ∧ x′ ≤ x″ ∧ x′ = 0, so x ∧ x′ = 0. Let a ∈ F. Since A is a bounded ⊔-hoop, by Propositions 2.4(iv), A is a distributive lattice, then a = a ⊔ 0 = a ⊔ ( x ∧ x ′ ) = ( a ⊔ x ) ∧ ( a ⊔ x ′ ) Moreover, F is linear, by Proposition 3.17, F^⊤ is a prime filter of A. Then x ⊔ x′ = 1 ∈ F^⊤, so x ∈ F^⊤ or x′ ∈ F^⊤. If x ∈ F^⊤, then, for any x ∈ F, x = x ⊔ x = 1, and so x = 1, which is a contradiction. Thus, x′ ∈ F^⊤. So, for any a ∈ F, a ⊔ x′ = 1. Hence, a = a ⊔ x, and so x ≤ a. Therefore, x is the smallest element of F. □

In the following example we show that there is a bounded ⊔-hoop with properties in Proposition 3.24.

Example 3.25. Let A be a hoop as Example 3.6 and F = {1, d}. Then it is clear that A is a bounded ⊔-hoop, F is a linear filter of A and there exists an element 1 ≠ x ∈ F such that x′ ⊔ x = 1.

Definition 3.26. [37] In a lattice L with bottom element 0, an element x ∈ L is said to has a pseudo-complement element if there exists a greatest element x^∗ ∈ L, disjoint from x, with the property that x ∧ x^∗ = 0. More formally, x^∗ = max {y ∈ L ∣ x ∧ y = 0}. The lattice L itself is called a pseudo-complemented lattice if every element of L has a pseudo-complement element.

Theorem 3.27. Let A be a bounded ⊔-hoop. Then ( F ( A ) , ⊆ , ∧ , ∨ , { 1 } , A ) is a bounded distributive lattice, where F ∧ G = F ∩ G and F ∨ G = 〈F ∪ G〉, for any F , G ∈ F ( A ) .

Proof. By Proposition 2.5, it is clear that F ∧ G , F ∨ G ∈ F ( A ) , for any F , G ∈ F ( A ) . So, it is enough to prove that ( F ( A ) , ⊆ , ∧ , ∨ ) is a distributive lattice. For this, we prove that F ∧ (G ∨ H) = (F ∧ G) ∨ (F ∧ H), for any F , G , H ∈ F ( A ) . Since G, H ⊆ 〈G ∪ H〉 = G ∨ H, we have F ∩ G, F ∩ H ⊆ F ∧ (G ∨ H). Then ( F ∧ G ) ∨ ( F ∧ H ) = 〈 ( F ∩ G ) ∪ ( F ∩ H ) 〉 ⊆ F ∧ ( G ∨ H ) Conversely, let x ∈ F ∧ (G ∨ H). Then x ∈ F and x ∈ G ∨ H = 〈G ∪ H〉. By Proposition 2.5, there exist n ∈ ℕ and a _i  ∈ G ∪ H, for 1 ≤ i ≤ n, such that a₁ ⊙ a₂ ⊙ . . . ⊙ a_n-1 ⊙ a _n  ≤ x. Since a _i , x ≤ x ⊔ a _i , for 1 ≤ i ≤ n, we get that a _i  ⊔ x ∈ F ∩ G or a _i  ⊔ x ∈ F ∩ H. Then by Lemma 3.3, we have x = x ⊔ x ≥ x ⊔ ( a 1 ⊙ a 2 ⊙ . . . ⊙ a n - 1 ⊙ a n ) ≥ ( x ⊔ a 1 ) ⊙ ( x ⊔ a 2 ) ⊙ . . . ⊙ ( x ⊔ a n ) ∈ ( F ∩ G ) ∪ ( F ∩ H )

Thus, x ∈ 〈 (F ∩ G) ∪ (F ∩ H) 〉 = (F ∧ G) ∨ (F ∧ H). Hence, F ∧ ( G ∨ H ) = ( F ∧ G ) ∨ ( F ∧ H )

Therefore, ( F ( A ) , ⊆ , ∧ , ∨ , { 1 } , A ) is a bounded distributive lattice. □

Theorem 3.28. Let A be a bounded ⊔-hoop. Then ( F ( A ) , ⊆ , ∧ , ∨ , { 1 } , A ) is a pseudo-complemented lattice and for any F ∈ F ( A ) , F^⊤ is pseudo-complement of F.

Proof. By Propositions 3.4 and 3.14, it is clear that, for any F ∈ F ( A ) , F ∩ F^⊤ = {1}. Suppose G ∈ F ( A ) such that G ∩ F = {1}. We prove that G ⊆ F^⊤. Let a ∈ G. Then, for any x ∈ F, x, a ≤ x ⊔ a. Since F , G ∈ F ( A ) , a ⊔ x ∈ F ∩ G = {1}. Then a ⊔ x = 1, and so a ∈ F^⊤. Hence, F^⊤ is the largest filter of A such that G ∩ F = {1}. Therefore, F^⊤ is a pseudo-complement of F. □

Theorem 3.29. Let A be a bounded ⊔-hoop. Then ( F ( A ) , ⊆ , ∧ , ∨ , { 1 } , A ) is a pseudo-complemented lattice and for any F ∈ F ( A ) , F^∗ is pseudo-complement of F.

Proof. By Proposition 3.22, it is clear that, for any F ∈ F ( A ) , F ∩ F^∗ = {1}. Suppose G ∈ F ( A ) such that G ∩ F = {1}. We prove that G ⊆ F^∗. Let a ∈ G. Then, for any x ∈ F, by Proposition 2.1(vii) and (iv), x, a ≤ (x → a) → a. Since F , G ∈ F ( A ) , (x → a) → a ∈ F ∩ G = {1}. Then (x → a) → a = 1, and so x → a ≤ a. Also, by Proposition 2.1(iv), a ≤ x → a, so, x → a = a. Thus, a ∈ F^∗. Hence, F^∗ is the largest filter of A such that G ∩ F = {1}. Therefore, F^∗ is a pseudo-complement of F. □

Notation. By Theorems 3.28 and 3.29, F ( A ) form two pseudo-complemented lattices (with ∗ and ⊤). By Proposition 3.13, if A has (DNP), then these two pseudo-complemented lattices are the same.

Lemma 3.30. Let a, b ∈ A. Then 〈a〉 ∨ 〈b〉 = 〈a ⊙ b〉.

Proof. Since a ⊙ b ≤ a ∧ b ≤ a, b, then it is clear that 〈a〉, 〈b〉 ⊆ 〈a ∧ b〉 ⊆ 〈a ⊙ b〉. Hence, 〈a〉 ∨ 〈b〉 ⊆ 〈a ⊙ b〉. For the converse inclusions, let x ∈ 〈a ⊙ b〉. Then there exists n ∈ ℕ such that (a ⊙ b)  ⁿ  ≤ x, and so a ⁿ  ⊙ b ⁿ  ≤ x. Since 〈a〉 ∨ 〈b〉 = 〈 {a} ∪ {b} 〉 = 〈 {a, b} 〉, a ⁿ  ∈ 〈a〉 and b ⁿ  ∈ 〈b〉, by Proposition 2.4(iii), x ≥ a ⁿ  ⊙ b ⁿ  ∈ 〈a〉 ∨ 〈b〉. Hence, 〈a ⊙ b〉 ⊆ 〈a〉 ∨ 〈b〉. Therefore, 〈a〉 ∨ 〈b〉 = 〈a ⊙ b〉. □

Notation. Let F 1 , F 2 ∈ F ( A ) . Then we consider F₁ → F₂ by {a ∈ A ∣ F₁ ∩ 〈a〉 ⊆ F₂}.

Theorem 3.31. Let F 1 , F 2 , F 3 ∈ F ( A ) . Then the following statements hold:

(i) F 1 → F 2 ∈ F ( A ) ,

(ii) F₁ → F₂ = {a ∈ A ∣ a ⊔ b ∈ F₂,  forall b ∈ F₁},

(iii) F₁ ∩ F₂ ⊆ F₃ if and only if F₁ ⊆ F₂ → F₃,

(iv) F ( A ) is a Heyting algebra.

Proof. (i) Since 〈1〉 = {1} and {1} ∩ F₁ = {1} ⊆ F₂, we consequence that 1 ∈ F₁ → F₂. Now, let x, y ∈ A such that x ≤ y and x ∈ F₁ → F₂, that is 〈x〉 ∩ F₁ ⊆ F₂. Since 〈y〉 ⊆ 〈x〉, we have 〈y〉 ∩ F₁ ⊆ 〈x〉 ∩ F₁ ⊆ F₂. Thus, 〈y〉 ∩ F₁ ⊆ F₂, and so y ∈ F₁ → F₂. Now, let x, y ∈ F₁ → F₂. Then 〈x〉 ∩ F₁ ⊆ F₂ and 〈y〉 ∩ F₁ ⊆ F₂. Thus, by definition of ∨ and Lemma 3.30, F 2 = F 2 ∨ F 2 = ( 〈 x 〉 ∩ F 1 ) ∨ ( 〈 y 〉 ∩ F 1 ) = 〈 ( 〈 x 〉 ∩ F 1 ) ∪ ( 〈 y 〉 ∩ F 1 ) 〉 = 〈 F 1 ∩ ( 〈 x 〉 ∪ 〈 y 〉 ) 〉 = 〈 F 1 ∩ ( 〈 x 〉 ∨ 〈 y 〉 ) 〉 = 〈 F 1 ∩ 〈 x ⊙ y 〉 〉 = F 1 ∩ 〈 x ⊙ y 〉 Hence, a ⊙ b ∈ F₁ → F₂. Therefore, F 1 → F 2 ∈ F ( A ) .

(ii) Let B = {a ∈ A ∣ a ⊔ b ∈ F₂,  forall b ∈ F₁}, x ∈ B and z ∈ 〈x〉 ∩ F₁. Since z ∈ 〈x〉, there exists n ∈ ℕ such that x ⁿ  ≤ z and z ∈ F₁. Since x ∈ B, for any y ∈ F₁, x ⊔ y ∈ F₂, and so, x ⊔ z ∈ F₂. By x ⁿ  ≤ z, we get z = x ⁿ  ⊔ z ≥ (x ⊔ z)  ⁿ and this implies that, z ∈ F₂. Hence, F₁ ∩ 〈x〉 ⊆ F₂, and so z ∈ F₁ → F₂. Therefore, B ⊆ F₁ → F₂. Conversely, suppose x ∈ F₁ → F₂. Then F₁ ∩ 〈x〉 ⊆ F₂. If y ∈ F₁, then x, y ≤ x ⊔ y, and so, x ⊔ y ∈ F₁ ∩ 〈x〉 ⊆ F₂. Thus, x ⊔ y ∈ F₂. Hence, x ∈ B, and so B ⊆ F₁ → F₂. Therefore, F₁ → F₂ = {a ∈ A ∣ a ⊔ b ∈ F₂,  forall b ∈ F₁}.

(iii) Suppose F₁ ∩ F₂ ⊆ F₃ and x ∈ F₁. Then 〈x〉 ⊆ F₁. Thus, 〈x〉 ∩ F₂ ⊆ F₁ ∩ F₂ ⊆ F₃. Hence, x ∈ F₂ → F₃. Conversely, let F₁ ⊆ F₂ → F₃ and x ∈ F₁ ∩ F₂. Since x ∈ F₁, we have x ∈ F₂ → F₃. Then 〈x〉 ∩ F₂ ⊆ F₃. Thus, x ∈ F₃. Hence, F₁ ∩ F₂ ⊆ F₃.

(iv) By (iii) and Theorem 3.27, the proof is clear. □

Theorem 3.32. ( F ( A ) , ∩ , → , { 1 } , A ) is a bounded hoop.

Proof. First of all, we notice that the operation ∩, is the product operation of F ( A ) . Hence, it is clear that ( F ( A ) , ∩ , A ) is a commutative monoid, so (HP1) holds. Also, by Theorem 3.31(i), for any F ∈ F ( A ) , F → F = {a ∈ A ∣ F ∩ 〈a〉 ⊆ F} = A, and so (HP2) holds. For proving (HP4), first of all we show that, for any F 1 , F 2 ∈ F ( A ) , F₁ ∧ F₂ = F₁ ∩ (F₁ → F₂). For this, let x ∈ F₁ ∧ F₂. Then by Theorem 3.27, x ∈ F₁ ∩ F₂. Thus, F₁ ∩ 〈x〉 ⊆ F₁ ∩ F₂ ⊆ F₂. Hence, by Theorem 3.31(i), x ∈ F₁ → F₂, and so x ∈ F₁ ∩ (F₁ → F₂). Conversely, if x ∈ F₁ ∩ (F₁ → F₂), then x ∈ F₁ ∩ 〈x〉 ⊆ F₂, and so x ∈ F₂. Hence, x ∈ F₁ ∩ F₂ = F₁ ∧ F₂. Therefore, F₁ ∧ F₂ = F₁ ∩ (F₁ → F₂). Moreover, by Theorem 3.27, it is clear that in this structure the operations product and meet are the same. Thus, F₁ ∩ (F₁ → F₂) = F₁ ∧ F₂ = F₂ ∧ F₁ = F₂ ∩ (F₂ → F₁). Hence, (HP4) holds. Finally, we prove that F₁ → (F₂ → F₃) = (F₁ ∩ F₂) → F₃, for any F 1 , F 2 , F 3 ∈ F ( A ) . Let x ∈ (F₁ ∩ F₂) → F₃. Then by Theorem 3.31(i), (F₁ ∩ F₂) ∩ 〈x〉 ⊆ F₃. Thus, F₂ ∩ (F₁ ∩ 〈x〉) ⊆ F₃, and so F₁ ∩ 〈x〉 ⊆ F₂ → F₃. Hence, x ∈ F₁ → (F₂ → F₃). So, (F₁ ∩ F₂) → F₃ ⊆ F₁ → (F₂ → F₃). Conversely, suppose x ∈ F₁ → (F₂ → F₃). Then by Theorem 3.31(i), F₁ ∩ 〈x〉 ⊆ F₂ → F₃. Thus, F₂ ∩ 〈F₁ ∩ 〈x〉〉 ⊆ F₃, and so (F₁ ∩ F₂) ∩ 〈x〉 ⊆ F₃. Hence, x ∈ (F₁ ∩ F₂) → F₃. So, F₁ → (F₂ → F₃) ⊆ (F₁ ∩ F₂) → F₃. Then F₁ → (F₂ → F₃) = (F₁ ∩ F₂) → F₃, and so (HP3) holds. Therefore, ( F ( A ) , ∩ , → , { 1 } , A ) is a bounded hoop. □

Note. Let C - Ann (A) = {X^⊤ ∣ ∅ ≠ X ⊆ A} be the set of all co-annihilators of A. By Proposition 3.18(v), X^⊤ = 〈X〉^⊤, we get that C - Ann ( A ) = { F ⊤ ∣ F ∈ F ( A ) } . Hence, C - Ann (A) is the set of all pseudo-complements of the pseudo-complement lattice F ( A ) .

Proposition 3.33. The following statements hold, for any ∅ ≠ X ⊆ A and F , G ∈ F ( A ) :

(i) {1} , A ∈ C - Ann (A),

(ii) F ∈ C - Ann (A) if and only if F^⊤⊤ = F,

(iii) a map ⊤⊤ : X → X^⊤⊤ is a closure map,

(iv) F ∩ (F ∩ G) ^⊤ = F ∩ G^⊤,

(v) (F ∩ G) ^⊤⊤ = F^⊤⊤ ∩ G^⊤⊤,

(vi) (F ∨ G) ^⊤ = F^⊤ ∩ G^⊤,

(vii) if F, G ∈ C - Ann (A), then F  ⋁ _C-Ann(A) G = (F^⊤ ∩ G^⊤) ^⊤.

Proof. (i) By Proposition 3.7(i), {1} ^⊤ = A and A^⊤ = {1}. So {1} , A ∈ C - Ann (A).

(ii) Let F ∈ C - Ann (A). Then there exists ∅ ≠ X ⊆ A such that F = X^⊤. Thus, by Proposition 3.18(iv), F^⊤⊤ = X^⊤⊤⊤ = X^⊤ = F. Conversely, if F^⊤⊤ = F, then X = F^⊤, and so X^⊤ ∈ C - Ann (A). Hence, F = F^⊤⊤ ∈ C - Ann (A).

(iii) According to Definition 2.6, by Proposition 3.18(i), (iii) and (iv), the proof is clear.

(iv) Since F ∩ G ⊆ G, by Proposition 3.18(i), G^⊤ ⊆ (F ∩ G) ^⊤. Then F ∩ G^⊤ ⊆ F ∩ (F ∩ G) ^⊤. On the other side, by Proposition 3.14, (F ∩ G) ∩ (F ∩ G) ^⊤ = {1}. Then by Proposition 3.21, F ∩ (F ∩ G) ^⊤ ⊆ G^⊤. Thus, F ∩ (F ∩ G) ^⊤ ⊆ F ∩ G^⊤. Hence, F ∩ (F ∩ G) ^⊤ = F ∩ G^⊤.

(v) Since F ∩ G ⊆ F, G, by Proposition 3.18(i), (F ∩ G) ^⊤⊤ ⊆ F^⊤⊤ ∩ G^⊤⊤. On the other side, by Proposition 3.14, (F ∩ G) ∩ (F ∩ G) ^⊤ = {1}. Then by Proposition 3.21, F ∩ (F ∩ G) ^⊤ ⊆ G^⊤. Also, by Proposition 3.18(iv), G^⊤ = G^⊤⊤⊤. Then we have F ∩ (F ∩ G) ^⊤ ⊆ G^⊤⊤⊤, and so F ∩ G^⊤⊤ ∩ (F ∩ G) ^⊤ = {1}. By the similar way, we can see that G^⊤⊤ ∩ (F ∩ G) ^⊤ ⊆ F^⊤⊤⊤, and so F^⊤⊤ ∩ G^⊤⊤ ∩ (F ∩ G) ^⊤ = {1}. Thus, F^⊤⊤ ∩ G^⊤⊤ ⊆ (F ∩ G) ^⊤⊤. Therefore, (F ∩ G) ^⊤⊤ = F^⊤⊤ ∩ G^⊤⊤.

(vi) Since F, G ⊆ F ∨ G, by Proposition 3.18(i), (F ∨ G) ^⊤ ⊆ F^⊤ ∩ G^⊤. Now, by Proposition 3.18(iv) and (v), (F ∨ G) ^⊤ ⊆ F^⊤⊤⊤ ∩ G^⊤⊤⊤ = (F^⊤ ∩ G^⊤) ^⊤⊤. Also, by Proposition 3.18(iii), F ⊆ F^⊤⊤ ⊆ (F^⊤ ∩ G^⊤) ^⊤. By the similar way, G ⊆ (F^⊤ ∩ G^⊤) ^⊤, and so F ∨ G ⊆ (F^⊤ ∩ G^⊤) ^⊤. Hence, (F^⊤ ∩ G^⊤) ^⊤⊤ ⊆ (F ∨ G) ^⊤ ⊆ (F^⊤ ∩ G^⊤) ^⊤⊤, and so, (F ∨ G) ^⊤ = (F^⊤ ∩ G^⊤) ^⊤⊤. Then by (v), ( F ∨ G ) ⊤ = ( F ⊤ ∩ G ⊤ ) ⊤ ⊤ = F ⊤ ⊤ ⊤ ∩ G ⊤ ⊤ ⊤ = F ⊤ ∩ G ⊤

(vii) By Theorem 3.27 and the proof of (vii), F  ⋁ _C-Ann(A)G = (F ∨ G) ^⊤⊤ = (F^⊤ ∩ G^⊤) ^⊤. □

Theorem 3.34. (C - Ann (A) , ∧ , ∨ , ⊤ , {1} , A) is a Boolean algebra.

Proof. By Proposition 3.33, {1} , A ∈ C - Ann (A). Then C - Ann (A)≠ ∅ and it is bounded. By Proposition 3.33(vi) and (viii), for any F, G ∈ C - Ann (A), F ∧ _C-Ann(A)G = F ∩ G and F ∨ _C-Ann(A)G = (F^⊤ ∩ G^⊤) ^⊤. Thus, (C - Ann (A) , ∧ , ∨ , ⊤ , {1} , A) is a bounded lattice. Now, we prove that it is a distributive lattice. For this, let F, G, H ∈ C - Ann (A). Then H ∧ (F ∨ G) = H ∩ (F ∨ G). Suppose K = (H ∩ F) ∨ (H ∩ G). By Proposition 3.18(iii), H ∩ F ⊆ K ⊆ K^⊤⊤, then H ∩ F ∩ K^⊤ = {1}, and so H ∩ K^⊤ ⊆ F^⊤. By the similar way, H ∩ K^⊤ ⊆ G^⊤. Thus, H ∩ K^⊤ ⊆ F^⊤ ∩ G^⊤. By Proposition 3.18(iii), H ∩ K^⊤ ⊆ F^⊤ ∩ G^⊤ ⊆ (F^⊤ ∩ G^⊤) ^⊤⊤, and so H ∩ K^⊤ ∩ (F^⊤ ∩ G^⊤) ^⊤ = {1}. Then by Proposition 3.33(ii), H ∩ (F^⊤ ∩ G^⊤) ^⊤ ⊆ K^⊤⊤ = K. Hence, H ∧ ( F ∨ G ) = H ∩ ( F ⊤ ∩ G ⊤ ) ⊤ ⊆ K = ( H ∩ F ) ∨ ( H ∩ G ) = ( H ∧ F ) ∨ ( H ∧ G )

Also, since H ∩ F ⊆ H, F and H ∩ G ⊆ H, G, by Proposition 3.18(i), F^⊤, H^⊤ ⊆ (F ∩ H) ^⊤ and G^⊤, H^⊤ ⊆ (G ∩ H) ^⊤. Then F^⊤ ∩ G^⊤ ⊆ (H ∩ F) ^⊤ ∩ (H ∩ G) ^⊤, and so [(H ∩ F) ^⊤ ∩ (H ∩ G) ^⊤] ^⊤ ⊆ (F^⊤ ∩ G^⊤) ^⊤. By Proposition 3.33(viii), (H ∧ F) ∨ (H ∧ G) ⊆ F ∨ G. Moreover, H ∩ F ⊆ H and H ∩ G ⊆ H, and so by Proposition 3.18(i), H^⊤ ⊆ (F ∩ H) ^⊤ ∩ (G ∩ H) ^⊤. Then [(H ∩ F) ^⊤ ∩ (H ∩ G) ^⊤] ^⊤ ⊆ H^⊤⊤ = H. Hence, ( H ∧ F ) ∨ ( H ∧ G ) = ( H ∩ F ) ∨ ( H ∩ G ) = [ ( H ∩ F ) ⊤ ∩ ( H ∩ G ) ⊤ ] ⊤ ⊆ H ∩ ( F ⊤ ∩ G ⊤ ) ⊤ ⊆ H ∧ ( F ∨ G )

Thus, (H ∧ F) ∨ (H ∧ G) = H ∧ (F ∨ G) and so C - Ann (A) is a distributive lattice. Also, for any F ∈ C - Ann (A), F ∩ F^⊤ = {1} and F ∨ F^⊤ = (F^⊤ ∩ F^⊤⊤) ^⊤ = ({1}) ^⊤ = A. Then the complement of F ∈ C - Ann (A) is F^⊤. Therefore, (C - Ann (A) , ∧ , ∨ , ⊤ , {1} , A) is a Boolean algebra. □

4 α-filter of hoops

In this section, we introduce an extension of a filter, which leads to a useful characterization of α-filters on hoops. For instance, we obtain a series of characterizations of α-filters. In addition, we show that there are no non-trivial α-filters on hoop-chains. That implies the structure of all α-filters contains only trivial α-filters on hoops. On hoops, we prove that the set of all α-filters is a pseudo-complemented lattice. Moreover, the structure of all α-filters can form a Boolean algebra under certain conditions.

Definition 4.1. A filter F of A is called an α-filter of A if, a^⊤⊤ ⊆ F, for any a ∈ F. We note by F α ( A ) the set of all α-filters of A.

Note. For any bounded ⊔-hoop A, it is easy to see that {1} and A are α-filters of A, which are called trivial α-filters of A.

Example 4.2. Let A be a hoop as Example 3.8. By routine calculation, we can see that F = {1, d} is an α-filter of A.

The largest element of A which is not 1, is called a co-atom. It means that if a ∈ A is a co-atom and a ≤ x, then x = 1 or x = a.

Proposition 4.3. If A contains only one co-atom, then A has no non-trivial α-filter.

Proof. Let the element a ∈ A be the unique co-atom of A. Then, for any 1 ≠ x ∈ A, x ≤ a. Suppose F is a non-trivial α-filter of A. Then, there exists 1 ≠ y ∈ A such that y ≤ a. Thus, x ⊔ y ≤ a ⊔ a = a ≠ 1. Hence, y^⊤ = {1} and so, by Proposition 3.7(i), y^⊤⊤ = A ⊈ F, which is a contradiction. Therefore, A has no non-trivial α-filter. □

Proposition 4.4. If A is a linear hoop, then A has no non-trivial α-filter.

Proof. Suppose A has non-trivial α-filter such as F. Then, for 1 ≠ x ∈ F and any y ∈ A, since A is linear, we have x ≤ y or y ≤ x. Without the lost of generality, let x ≤ y. Then x ⊔ y = y ≠ 1. Thus, x^⊤ = {1}, and so by Proposition 3.7(i), x^⊤⊤ = A ⊈ F, which is a contradiction. Therefore, A has no non-trivial α-filter. □

In the following example, we show that the converse of Proposition 4.4, may not be true, in general.

Example 4.5. Let A = {0, a, b, c, d, 1} where 0 ≤ a ≤ c ≤ 1 and 0 ≤ b ≤ c ≤ 1. Suppose operations → and ⊙ are defined on A by the Cayley tables as follows:

Then routine calculations show that (A, → , ⊙ , 0, 1) is a bounded ⊔-hoop. It is clear that A has no non-trivial α-filter and A is not linear.

Definition 4.6. For any non-empty subset X of A, define E (X) the smallest α-filter of A containing X. We can get that E (X) = E (〈X〉). Now, we give an extension of a filter which leads to a useful characterization of α-filter.

Theorem 4.7. Let F ∈ F ( A ) . Then E ( F ) = { a ∈ A ∣ ∃ f ∈ F such that f ⊤ ⊆ a ⊤ }

Proof. Let B = {a ∈ A ∣   ∃ f ∈ F suchthat f^⊤ ⊆ a^⊤}. It is enough to prove that B ∈ F α ( A ) and it is the smallest α-filter containing F. For this, since A is a ⊔-hoop and F ∈ F ( A ) , 1 ∈ A and 1 ∈ F. By Proposition 3.7(i), {1} ^⊤ = A. Thus, 1 ∈ B. Let a ≤ b and a ∈ B, for any b ∈ A. Since a ∈ B, there exists f ∈ F such that f^⊤ ⊆ a^⊤. By Proposition 3.7(ii), a^⊤ ⊆ b^⊤, and so, f^⊤ ⊆ a^⊤ ⊆ b^⊤. Thus, b ∈ B. Now, suppose a, b ∈ B. Then there exist f, g ∈ F such that f^⊤ ⊆ a^⊤ and g^⊤ ⊆ b^⊤. Since F ∈ F ( A ) , f ⊙ g ∈ F. Let x ∈ (f ⊙ g) ^⊤. Then x ⊔ (f ⊙ g) =1. By Proposition 2.1(iii), f ⊙ g ≤ f, g. Then 1 = x ⊔ (f ⊙ g) ≤ x ⊔ f and 1 = x ⊔ (f ⊙ g) ≤ x ⊔ g. So x ⊔ f = 1 and x ⊔ g = 1. Thus, x ∈ f^⊤ and x ∈ g^⊤. Since f^⊤ ⊆ a^⊤ and g^⊤ ⊆ b^⊤, we have x ⊔ a = x ⊔ b = 1. Then by Lemma 3.3, x ⊔ ( a ⊙ b ) ≥ ( x ⊔ a ) ⊙ ( x ⊔ b ) = 1 Thus, x ⊔ (a ⊙ b) =1. Hence, (f ⊙ g) ^⊤ ⊆ (a ⊙ b) ^⊤, and so a ⊙ b ∈ B. Hence, B ∈ F ( A ) . Now, let a ∈ B and x ∈ a^⊤⊤. By Proposition 3.18(iv), a^⊤ = a^⊤⊤⊤ ⊆ x^⊤. Since a ∈ B, there exists f ∈ F such that f^⊤ ⊆ a^⊤. Then f^⊤ ⊆ a^⊤ ⊆ x^⊤. Thus, x ∈ B. Hence, a^⊤⊤ ⊆ B, and so B ∈ F α ( A ) . Suppose C is an α-filter of A containing F. Then, for any a ∈ B, there is f ∈ F such that f^⊤ ⊆ a^⊤. By Propositions 3.18(ii) and 3.33(ii), a ∈ 〈 a 〉 ⊆ 〈 a 〉 ⊤ ⊤ = a ⊤ ⊤ ⊆ f ⊤ ⊤ ⊆ C Hence, B ⊆ C. Therefore, E (F) = {a ∈ A ∣   ∃ f ∈ F suchthat f^⊤ ⊆ a^⊤}. □

Corollary 4.8. Let F ∈ F ( A ) . Then E ( F ) = { a ∈ A ∣ ∃ f ∈ F such that a ⊤ ⊤ ⊆ f ⊤ ⊤ }

Proof. Let B = {a ∈ A ∣   ∃ f ∈ F suchthat a^⊤⊤ ⊆ f^⊤⊤} and a ∈ E (F). Then there exists f ∈ F such that f^⊤ ⊆ a^⊤. By Proposition 3.18(i), a^⊤⊤ ⊆ f^⊤⊤. Then a ∈ B. Conversely, suppose a ∈ B. Then there exists f ∈ F such that a^⊤⊤ ⊆ f^⊤⊤. By Proposition 3.18(i) and (iv), f^⊤ = f^⊤⊤⊤ ⊆ a^⊤⊤⊤ = a^⊤. Hence, a ∈ E (F). Therefore, E (F) = {a ∈ A ∣   ∃ f ∈ F suchthat a^⊤⊤ ⊆ f^⊤⊤}. □

Notation. By Proposition 3.7(i), it is clear that {1} ^⊤⊤ = {1}. Then {1} is an α-filter of A. So, we can consequence that ⋂ { F ∣ F ∈ F α ( A ) } = { 1 } .

Example 4.9. Let A be a hoop as Example 3.6 and F = {1, b, d}. Then E (F) = {0, b, d, 1}.

Proposition 4.10. Let F ∈ F ( A ) . Then the following statements are equivalent:

(i) F is an α-filter of A,

(ii) E (F) = F,

(iii) for any x, y ∈ A, if x^⊤ = y^⊤ and x ∈ F, then y ∈ F,

(iv) F = ⋃ _x∈F {x} ^⊤⊤.

Proof. (i) ⇒ (ii) According to definition of E (F), it is clear that F ⊆ E (F). Let a ∈ E (F). Then by Theorem 4.7, there exists f ∈ F such that f^⊤ ⊆ a^⊤. Thus, by Proposition 3.18(i), a^⊤⊤ ⊆ f^⊤⊤. Let x ∈ a^⊤⊤. Since F is an α-filter of A and a^⊤⊤ ⊆ f^⊤⊤, x ∈ F, and so E (F) ⊆ F. Hence, E (F) = F. (ii) ⇒ (iii) Let x ∈ F. By (ii), F = E (F), then x ∈ E (F). So, there exists f ∈ F such that f^⊤ ⊆ x^⊤. Since x^⊤ = y^⊤, we have f^⊤ ⊆ y^⊤. Then y ∈ E (F) = F. Hence, y ∈ F. (iii) ⇒ (i) Let a ∈ F. Then by Proposition 3.18(iv), a^⊤ = a^⊤⊤⊤. Thus, a^⊤ = (a^⊤⊤) ^⊤ and a ∈ F, then by (iii), a^⊤⊤ ⊆ F. Hence, F ∈ F α ( A ) . (i) ⇒ (iv) Since F ∈ F α ( A ) , for any x ∈ F, x^⊤⊤ ⊆ F. Then ⋃_x∈F {x} ^⊤⊤ ⊆ F. Now, let a ∈ F. Then by Proposition 3.18(iii), {a} ⊆ {a} ^⊤⊤ ⊆ ⋃ _x∈F {x} ^⊤⊤. Hence, F ⊆ ⋃ _x∈F {x} ^⊤⊤. Therefore, F = ⋃ _x∈F {x} ^⊤⊤. (iv) ⇒ (i) Let a ∈ F. Since F = ⋃ _x∈F {x} ^⊤⊤, there exists x ∈ F such that a ∈ {x} ^⊤⊤. Then by Proposition 3.18(iii), {a} ⊆ {a} ^⊤⊤. Thus, {a} ^⊤⊤ ⊆ ⋃ _x∈F {x} ^⊤⊤ = F. Hence, F ∈ F α ( A ) . □

Proposition 4.11. Let F 1 , F 2 ∈ F ( A ) . Then E (F₁) = E (F₂) if and only if F₁ ⊆ E (F₂) and F₂ ⊆ E (F₁).

Proof. Let F 1 , F 2 ∈ F ( A ) , E (F₁) = E (F₂). Since F₁ ⊆ E (F₁) and E (F₁) = E (F₂), it is clear that F₁ ⊆ E (F₂). By the similar way, F₂ ⊆ E (F₁). Conversely, let F₁ ⊆ E (F₂) and a ∈ E (F₁). Then there exists f ∈ F₁ such that f^⊤ ⊆ a^⊤. Since F₁ ⊆ E (F₂), we get that f ∈ E (F₂). Thus, there is g ∈ F₂ such that g^⊤ ⊆ f^⊤, and so g^⊤ ⊆ a^⊤. Hence, a ∈ E (F₂), and so, E (F₁) ⊆ E (F₂). By the similar way, we can see that E (F₂) ⊆ E (F₁). Therefore, E (F₁) = E (F₂). □

Proposition 4.12. Let A be a bounded hoop-chain. Then E (F) = {1} or E (F) = A, for any F ∈ F ( A ) .

Proof. Let F ∈ F ( A ) . If F = {1}, then by Proposition 3.7(i), we get that E ( F ) = { a ∈ A ∣ f ⊤ ⊆ a ⊤ , for some f ∈ F } = { a ∈ A ∣ { 1 } ⊤ ⊆ a ⊤ } = { a ∈ A ∣ A ⊆ a ⊤ } = { 1 } Now, if F ≠ {1}, then there exists 1 ≠ a ∈ F. Since A is chain, for any b ∈ A, a ⊔ b = a ≠ 1 or a ⊔ b = b ≠ 1. Thus, a^⊤ = {1}, and so by Proposition 3.7(i), we get that         E (F) = {a ∈ A ∣ f^⊤ ⊆ a^⊤,  forsome f ∈ F} = {a ∈ A ∣ f^⊤ ⊆ {1} ,  forsome f ∈ F} = A □

Lemma 4.13. Let X and Y be two non-empty subsets of ⊔-hoops A and B, respectively. Then X^⊤ × Y^⊤ = (X × Y) ^⊤.

Proof. We have ( X × Y ) ⊤ = { ( a , b ) ∈ A × B ∣ ( a , b ) ⊔ ( x , y ) = ( 1 , 1 ) , for all ( x , y ) ∈ X × Y } = { ( a , b ) ∈ A × B ∣ ( a ⊔ x , b ⊔ y ) = ( 1 , 1 ) , for all ( x , y ) ∈ X × Y } = { ( a , b ) ∈ A × B ∣ a ⊔ x = 1 and b ⊔ y = 1 , for all ( x , y ) ∈ X × Y } = { a ∈ A ∣ a ⊔ x = 1 , for all x ∈ X } × { b ∈ B ∣ b ⊔ y = 1 , for all y ∈ Y } = X ⊤ × Y ⊤ □