Semi maximal filter in Hilbert algebra

Abstract

In this paper, the radical of an implicative filter in Hilbert algebras is studied and some theorems are given on its properties and we show that radical form a closure operator. Also, the radical of the set of all regular elements is characterized. Finally, the notion of semi maximal filter is introduced as a closed set of that closure operator and studied it in detail.

Keywords

(semisimple,local)Hilbert algebra (with supremum)radical of an implicative filter semi maximal filter

1 Introduction

The variety of Hilbert algebras is an important tool for investigations in intuitionistic logic and other non-classical logics. Hilbert algebras represent the algebraic counterpart of the implicative fragment of Intuitionistic Propositional Logic. Hilbert algebra was studied by A. Diego [8], D. Buşneag [1 –3], D. Buşneag and M. Ghita [4], S. A. Celani and D. Montangie [6], S. A. Celani [5], S. M. Hong and Y. B. Jun [10] and Figalloet al. [9].

Hilbert algebras with infimum are the algebraic counterpart of a propositional calculus weaker than the {→ , ∧}-fragment of the intuitionistic propositional calculus. The associated order of Hilbert algebras with infimum is a meet- semilattice. Also Hilbert algebras with supremum are Hilbert algebras which the associated order is a join- semilattice. This class of algebras is a particular class of BCK-algebras with lattice operations and was studied by Idziak in [11]. In this work, we use algebraic techniques for study Hilbert algebra and we get some results in this structure. The structure of the paper is as follows: in Section 2, some definitions and properties of Hilbert algebras are recalled. In Section 3, some characterizations are given for semi maximal filter and the radical of an implicative filter in Hilbert algebras. We prove that if F and G are implicative filters of H and K respectively, then (H × K)/Rad (F × G) ≅ H/Rad (F) × K/Rad (G). We also characterize the radical of bounded Hilbert algebra of R (H) and prove that, Rad (R (H)) = Rad (H) ∩ R (H). Furthermore we have proved some theorems which determine the relationship between semi maximal filter and other types of filters in Hilbert algebra. In Section 4, we have summarized the results of this paper, and the previous results in this field, and given the relationships between all types of filters in a Hilbert algebra.

2 Preliminaries

We recall some basic definitions and results that are necessary in the sequel.

A Hilbert algebra [8] is an algebra (H, → , 1) of type (2, 0) such that the following axioms hold, for all x, y, z ∈ H:

x → (y → x) =1;

(x → (y → z)) → ((x → y) → (x → z)) =1;

if x → y = y → x = 1, then x = y.

For a Hilbert algebra H, (H, ≤) is a poset by defining an order relation ≤ such that x ≤ y if and onlyif x → y = 1(called the natural order on H), with respect to this order, 1 is the greatest element of H. If H has a smallest element 0, we say that H is bounded, in this case for x ∈ H we have x^* = x → 0.

The following proposition includes some properties of Hilbert algebra [1 , 8].

Proposition 2.1. [4] If H is a bounded Hilbert algebra and x, y ∈ H, then

0^* = 1, 1^* = 0;

x → y^* = y → x^*;

x → x^* = x^*, x^* → x = x^**, x ≤ x^**, x ≤ x^* → y;

x → y ≤ y^* → x^*;

If x ≤ y, then y^* ≤ x^*;

x^*** = x^*;

(x → y) ^** = x → y^** = x^** → y^**;

(y → x) ^* ≤ x → y.

Definition 2.2. [4] A Hilbert algebra (H, → , 1) is called a Hilbert algebra with supremum when the underlying structure (H, ≤) with the natural order induced by → is a join-semilattice.

A subset D of Hilbert algebra H is called a deductive system (or implicative filter or simply filter) of H if: (i)1 ∈ D, (ii) If x ∈ D and x → y ∈ D, then y ∈ D, for all x, y ∈ H. Let D be a deductive system of Hilbert algebra. If x ≤ y and x ∈ D, then y ∈ D. We denote Ds (H)= {D: D is a deductive system of H}. If H is bounded, then deductive system D is proper if and only if 0 ∉ D. We note that, in some papers the deductive system is called implicative filter as well, so we use the term, implicative filter, in this paper. Also, we denote P (H) = {x ∈ H : X > 0}. An implicative filter F of a bounded Hilbert algebra with supremum H is called a Boolean filter of:

first kind (BF1) if x^** → x ∈ F, for all x ∈ H.

second kind (BF2) if x ∨ x^* ∈ F, for all x ∈ H.

We denote, F (H) = {x ∈ H : x^** > x^*}. Also an implicative filter F of a Hilbert algebra with supremum H is called a prime filter of:

first kind (PF1), if x ∨ y ∈ F implies x ∈ F or y ∈ F, for all x, y ∈ H.

second kind (PF2), if x → y ∈ F or y → x ∈ F, for all x, y ∈ H.

third kind (PF3), if (x → y) ∨ (y → x) ∈ F, for all x, y ∈ H.

Let F be an implicative filter of a bounded Hilbert algebra H. Define: x ≡ _Fy if and only if x → y ∈ F and y → x ∈ F. Then ≡_F is a congruence relation on H. The set of all congruence classes is denoted by H/F, i.e, H/F : = {[x] : x ∈ H}, where [x] = {y ∈ H|x ≡ _Fy}.Define → on H/F as follows: [x] → [y] = [x → y], and 1 = 1/F = F. Therefore (H/F, → , [1] , [0]) is a bounded Hilbert algebra with respect to F and the order relation on H/F is given by [x] ≤ [y] if and only if x → y ∈ F. Clearly, [x] = [1] if and only if x ∈ F.

Definition 2.3. [15] A non-empty subset F of Hilbert algebra H is called

A positive implicative filter of H, if 1 ∈ F and x → ((y → z) → y) ∈ F and x ∈ F imply y ∈ F, for every x, y, z ∈ H.

A fantastic filter of H, if 1 ∈ F and z → (y → x) ∈ F and z ∈ F imply ((x → y) → y) → x ∈ F, for every x, y, z ∈ H.

Theorem 2.4. [4] For a bounded Hilbert algebra H, the following assertions are equivalent:

x^** = x, for every x ∈ H;

H is a Boolean algebra relative to naturalordering, where x ∧ y = (x → y^*) ^*, x ∨ y = x^* → y.

Definition 2.5. [1] A proper implicative filter M of a Hilbert algebra H is called maximal if it is not properly contained in any other proper implicative filter of H.

For any Hilbert algebra H, Max (H) denotes the set of all maximal implicative filters of H.

Theorem 2.6. [1] An implicative filter M of a bounded Hilbert algebra H is maximal if and only if for all x ∈ H, if x ∉ M then x^* ∈ M.

Definition 2.7. [10] Hilbert algebra H is semisimple if the intersection of all maximal implicative filters of H is {1}.

Definition 2.8. [1] The intersection of all maximal implicative filters of Hilbert algebra H is called the radical of H and it is denoted by Rad (H).

Definition 2.9. [1] Let H be a Hilbert algebra. We say that an element x ∈ H is regular if for every y ∈ H we have (x → y) → x = x. We denote by R (H) the set of all regular elements of H. We say that an element x ∈ H is dense if for every r ∈ R (H) we have x → r = r. We denote the set of all dense elements of H by D (H).

Proposition 2.10. [1] If H is a bounded Hilbert algebra, then

D (H) = {x ∈ H : x^* = 0};

R (H) = {x ∈ H : x^** = x}.

Proposition 2.11. [1] If H is a bounded Hilbert algebra, then D (H) = ∩ _M∈Max(H)M.

A Hilbert algebra is said to be local if and only if it has exactly one maximal filter [7].

Proposition 2.12. [16] H is a local Hilbert algebra if and only if D (H) = H \ {0}.

Theorem 2.13. [16] The following statements hold:

R (H) = {x^* : x ∈ H} = {x^** : x ∈ H};

(R (H) , → , 0, 1) is a bounded Hilbert algebra.

For subsets A and B of a Hilbert algebra H, define [12] $A \to B : = {x \to y ∣ x \in A, y \in B}$

We use the notation A → b (resp. a → B) instead of A → {b} (resp. {a} → B). Note that A → B = ∪ _a∈A (a → B) = ∪ _b∈B (A → b).

Lemma 2.14. [12] For any subsets A, B and E of a Hilbert algebra H, we have

A ⊆ B ⇒ A → E ⊆ B → E, E → A ⊆ E → B.

(A ∩ B) → E ⊆ (A → E) ∩ (B → E).

E → (A ∩ B) ⊆ (E → A) ∩ (E → B).

(A ∪ B) → E = (A → E) ∪ (B → E).

E → (A ∪ B) = (E → A) ∪ (E → B).

Lemma 2.15. [1] If M ∈ Ds (H), then M ∈ Max (H) if and only if M ∩ R (H) is maximal in R (H).

Theorem 2.16. [1] If H is a bounded Hilbert algebra and there is a deductive system D ≠ H, then there is a maximal deductive system M of H such that D ⊆ M.

Theorem 2.17. [10] Every bounded Hilbert algebra with at least two elements contains at least one maximal deductive system.

3 Semi maximal filter

In this section, we introduce the notion of the semi maximal filter and we prove some theorems which determine the relationship between semi maximal filters and other types of filters in Hilbert algebras.

Definition 3.1. Let F be a proper implicative filter of H. The intersection of all maximal implicative filters of H which contain F is called the radical of F and it is denoted by Rad (F).

In view of Theorem 2.16, for a proper implicative filter F of bounded Hilbert algebra H, Rad (F) always exists.

Note that H is semisimple if and only if Rad (H) = {1}.

Example 3.2.

Let H = {0, a, b, c, 1}. With 0 < a, b < c < 1, and a, b are incomparable. We define $\begin{matrix} \begin{matrix} \to & 0 & a & b & c & 1 \\ 0 & 1 & 1 & 1 & 1 & 1 \\ a & b & 1 & b & 1 & 1 \\ b & a & a & 1 & 1 & 1 \\ c & 0 & a & b & 1 & 1 \\ 1 & 0 & a & b & c & 1 \end{matrix} \end{matrix}$ Then (H, → , 0, 1) is a bounded Hilbert algebra. The implicative filters of H are {1}, {c, 1}, {a, c, 1}, {b, c, 1} and {0, a, b, c, 1}. The implicative filters {a, c, 1}, {b, c, 1} are maximal implicative filters, so Rad (H) = {c, 1}. Thus H is not semisimple. Rad ({1})= Rad ({c, 1})= {c, 1}, Rad ({a, c, 1})= {a, c, 1} and Rad ({b, c, 1})= {b, c, 1}.

Let H = {0, a, b, 1}. With → defined by: $\begin{matrix} \begin{matrix} \to & 0 & a & b & 1 \\ 0 & 1 & 1 & 1 & 1 \\ a & b & 1 & b & 1 \\ b & a & a & 1 & 1 \\ 1 & 0 & a & b & 1 \end{matrix} \end{matrix}$ Then H is a bounded Hilbert algebra with supremum. The proper implicative filters are {1}, {a, 1}, {b, 1}. The implicative filters {a, 1}, {b, 1} are maximal implicative filters. Rad (H) = {1}= Rad ({1}), Rad ({a, 1})= {a, 1} and Rad ({b, 1})= {b, 1}.

Let H = {0, a, b, c, 1}. We define $\begin{matrix} \begin{matrix} \to & 0 & a & b & c & 1 \\ 0 & 1 & 1 & 1 & 1 & 1 \\ a & 0 & 1 & b & b & 1 \\ b & 0 & a & 1 & a & 1 \\ c & 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & a & b & c & 1 \end{matrix} \end{matrix}$ (H, → , 0, 1) is a bounded Hilbert algebra with supremum. Rad (H)= Rad ({1})= Rad ({a, 1})= Rad ({b, 1}) = {c, a, b, 1}.

Define $x \to y = {\begin{matrix} 1 & if x \leq y \\ y & if x > y \end{matrix}$ Then H = ([0, 1] , → , 0, 1) is a Hilbert algebra. F = [a, 1] where 0 < a ≤ 1 is a proper implicative filter, then Rad ([a, 1]) = (0, 1].

Let H = {0, a, b, c, d, e, f, g, 1}, with 0 < a < b <e < 1, 0 < a < d < e < 1, 0 < a < d < g <1, 0 < c < d < e < 1, 0 < c < d < g < 1, 0 < c < f < g < 1 but elements {a, c} , {b, d}, {d, f} , {e, g} and {b, f} are pairwise incomparable. Define $\begin{matrix} \begin{matrix} \to & 0 & a & b & c & d & e & f & g & 1 \\ 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ a & f & 1 & 1 & f & 1 & 1 & f & 1 & 1 \\ b & f & g & 1 & f & g & 1 & f & g & 1 \\ c & b & b & b & 1 & 1 & 1 & 1 & 1 & 1 \\ d & 0 & b & b & f & 1 & 1 & f & 1 & 1 \\ e & 0 & a & b & f & g & 1 & f & g & 1 \\ f & b & b & b & e & e & e & 1 & 1 & 1 \\ g & 0 & b & b & c & e & e & f & 1 & 1 \\ 1 & 0 & a & b & c & d & e & f & g & 1 \end{matrix} \end{matrix}$ H is a Hilbert algebra, The implicative filters of H are {1}, {e, 1}, {g, 1}, {f, g, 1}, {b, e, 1}, {d, e, g, 1}, {a, b, d, e, g, 1}, {c, d, e, f, g, 1} and H. The implicative filters {a, b, d, e, g, 1} and {c, d, e, f, g, 1} are maximal implicative filters,so Rad (H) = {d, e, g, 1}. Thus H is not semi-simple Hilbert algebra. Rad ({1}) = Rad ({e, 1}) =Rad ({g, 1}) =Rad ({f, g, 1})=Rad ({b, e, 1})= R ({d, e, g, 1})= {d, e, g, 1} and Rad ({a, b, d, e, g, 1})= {a, b, d, e, g, 1} and Rad ({c, f, d, e, g, 1})= {c, f, d, e, g, 1}.

The following theorem includes some properties of radicals in Hilbert algebra.

Theorem 3.3. Let F, G and K be proper implicative filters of H and a, b ∈ H. Then the following statements hold:

Rad (F) is an implicative filter;

F ⊆ Rad (F);

Rad (H) ⊆ Rad (F);

If M is a maximal implicative filter, then Rad (M) = M;

If F is an implicative filter of a bounded H, then there exists a maximal filter M of H such that F ⊆ M,

Rad (H) = D (H).

F ⊆ G, then Rad (F) ⊆ Rad (G);

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

If 〈F ∪ G〉 is a proper implicative filter then Rad (F) ∪ Rad (G) ⊆ Rad (〈F ∪ G〉);

F → Rad (G) = Rad (F) → Rad (G);

Rad (G) → F ⊆ Rad (G) → Rad (F);

Rad (F → G) = Rad (F → Rad (G)).

Rad (F ∩ G) → K = (Rad (F) → K) ∩ (Rad (G) → K);

K → Rad (F ∩ G) = (K → Rad (F)) ∩ (K → Rad (G));

Rad (F ∪ G) → K = (Rad (F) → K) ∪ (Rad (G) → K);

K → Rad (F ∪ G) = (K → Rad (F)) ∪ (K → Rad (G));

Rad (〈F ∪ G〉) ⊆ Rad (〈Rad (F) ∪ Rad (G) 〉).

Proof.

Assume x ∈ Rad (F). Then x ∈ ∩ M where F ⊆ M ∈ Max (H). Thus x ∈ ∩ M where G ⊆ M ∈ Max (H). Therefore x ∈ Rad (G).

By (vii) we have Rad (F) ⊆ Rad (Rad (F)), since F ⊆ Rad (F). It is sufficient to show that Rad (Rad (F)) ⊆ Rad (F). Let x ∈ Rad (Rad (F)). Then x ∈ M, for all maximal filters M of H contains Rad (F). Let M₀ be an arbitrary maximal filter of H contains F. Then M₀ = Rad (M₀) ⊇ Rad (F), hence x ∈ M₀. Thus x ∈ Rad (F). Therefore Rad (Rad (F)) = Rad (F).

Since F, G ⊆ 〈F ∪ G〉, thus by (vii), we have Rad (F) , Rad (G) ⊆ Rad (〈F ∪ G〉). Therefore Rad (F) ∪ Rad (G) ⊆ Rad (〈F ∪ G〉).

By Lemma 2.14, we have F → Rad (G) ⊆ Rad (F) → Rad (G). Also, by ([12], Theorem 3.8 and Proposition 3.16) and Theorem 3.3(i),Rad (G) = F → Rad (G) ⊆ Rad (F) → Rad (G) = Rad (G), thus F → Rad (G) = Rad (F) → Rad (G).

By Lemma 2.14 and (vii), we have Rad (F → G) ⊆ Rad (F → Rad (G)). Also, by ([12], Theorem 3.8 and Proposition 3.16) and (viii), Rad (G) = Rad (F → G) ⊆ Rad (F → Rad (G)) = Rad (G), thus Rad (F → G) = Rad (F → Rad (G)).

By (vii) and Lemma 2.14, we have Rad (F ∩ G) → K ⊆ (Rad (F) → K) ∩ (Rad (G) → K). Also, by ([12], Theorem 3.8 andProposition 3.16) Rad (F ∩ G) → K = (Rad (F) → K) ∩ (Rad (G) → K).

See Theorem 3.27.

We have F ⊆ Rad (F) and G ⊆ Rad (G), thus F ∪ G ⊆ Rad (F) ∪ Rad (G), therefore 〈F ∪ G〉 ⊆ 〈Rad (F) ∪ Rad (G) 〉. Then Rad (〈F ∪ G〉) ⊆ Rad (〈Rad (F) ∪ Rad (G) 〉). □

Theorem 3.3 (ii), (vii) and (viii) show that Rad is a closure operator on implicative filters of H.

In the following we show that the converse ofTheorem 3.3 (ii), (iii), (vii) and (xi) is not true.

Example 3.4.

In Example 3.2 (i), if F = {1}, Rad (F) = {c, 1} thus Rad (F) ⊈ F.

In Example 3.2 (i), Rad (H) = {c, 1}, Rad ({a, c, 1})= {a, c, 1}, thus Rad ({a, c, 1} ⊈ Rad (H).

In Example 3.2 (iii), consider F = {a, 1}, G = {b, 1}, Rad (F) = Rad (G) = {c, a, b, 1}, thus Rad (F) ⊆ Rad (G), but F ⊈ G. Therefore the converse of Theorem 3.3 (vii) is not true.

In Example 3.2 (i), Consider F = {1} and G = {a, c, 1}, then Rad (F) = {c, 1} and Rad (G) = {a, c, 1}. Therefore Rad (G) → F = {1} and Rad (G) → Rad (F) = {c, 1}. Thus Rad (G) → F ≠ Rad (G) → Rad (F). So, the converse of Theorem 3.3 (xi) is not true.

The following example shows that the condition “〈F ∪ G〉 is a proper implicative filter” in Theorem 3.3 (ix) is essential.

Example 3.5. In Example 3.2 (ii), consider F = {a, 1} and G = {b, 1}, then F ∪ G = {a, b, 1}. Thus 〈F ∪ G〉 = {0, a, b, 1}, so Rad (〈F ∪ G〉) = Rad (H) = {1}. Also, we have Rad (F) = {a, 1} and Rad (G) = {b, 1}, Thus Rad (F) ∪ Rad (G) = {a, b, 1}. Therefore Rad (F) ∪ Rad (G) ⊈ Rad (〈F ∪ G〉)

Proposition 3.6. Let F be a (Boolean filter, maximal, positive implicative, and fantastic) filter of H. Then Rad (F) is too.

Proof. If F is a Boolean filter, positive implicative filter or fantastic filter, then, by extension theorem Rad (F) is too. Also, if F is a maximal filter of H, then Rad (F) = F. Thus, Rad (F) is a maximal filter.

In the following we show that the converse of above proposition is not true.

Example 3.7. In Example 3.2 (iii),

Rad ({1}) = {c, a, b, 1} is a Boolean filter of the first kind and Boolean filter of the second kind. But {1} is neither Boolean filter of the first kind nor Boolean filter of the second kind, because a^** → a = a ∉ {1} and a ∨ a^* = a ∉ {1}.

Rad ({1}) = {c, a, b, 1} is a maximal filter, but {1} is not maximal filter.

Rad ({1}) = {c, a, b, 1} is a positive implicative filter and fantastic filter, because, it is Boolean filter of the first kind. (See Theorem 6.14 [15]). But {1} is neither positive implicative filter nor fantastic filter.

Proposition 3.8. If F and G are implicative filters of H, then [x] _Rad(F∩G) ⊆ [x] _Rad(F) ∩ [x] _Rad(G).

Proof. Since F ∩ G ⊆ F, G, thus Rad (F ∩ G) ⊆ Rad (F) ∩ Rad (G). Therefore

[x] _Rad(F∩G) = {y ∈ H : x ≡ _Rad(F∩G)y} = {y ∈ H : x → y, y → x ∈ Rad (F ∩ G)} ⊆ {y ∈ H : x → y, y → x ∈ Rad (F) ∩ Rad (G)} = {y ∈ H : x → y, y → x ∈ Rad (F)} ∩ {y ∈ H : x → y, y → x ∈ Rad (G)} .

Hence [x] _Rad(F∩G) ⊆ [x] _Rad(F) ∩ [x] _Rad(G). □

Definition 3.9. Let F and G be implicative filters of Hilbert algebras H and K. Then F × G = {x ∈ H × K : x = (f, g)} is an implicative filter of H × K, where

(a, b) → (c, d) = (a → c, b → d),

(a, b) ≤ (c, d) = (a ≤ c, b ≤ d).

for all a, c ∈ H and b, d ∈ K.

Proposition 3.10. If F and G are implicative filters of H, then Rad (F × G) = Rad (F) × Rad (G) .

Proof. (x, y) ∈ Rad (F × G) if and only if (x, y) ∈ M × N, where maximal filter M × N contains F × G if and only if x ∈ M and y ∈ N, where M and N be arbitrary maximal filters of H contains F and G if and only if (x, y) ∈ Rad (F) × Rad (G). □

Corollary 3.11. Let {F_i} _i∈I be a family of implicative filters of H. Then $Rad (\prod_{i \in I} F_{i}) = \prod_{i \in I} Rad (F_{i})$ .

Proposition 3.12. If F and G are implicative filters of H and K, then (H × K)/Rad (F × G) ≅ H/Rad (F) × K/Rad (G).

Proof. Consider the natural homomorphisms $\prod_{H} : H \to H / Rad (F)$ , for all h ∈ H, $\prod_{H} (h) = [h]$ and $\prod_{K} : K \to K / Rad (G)$ , for all k ∈ K, $\prod_{K} (k) = [k]$ Define φ : H × K → H/Rad (F) × K/Rad (G), by φ (h, k) = ([h] , [k]) for all (h, k) ∈ H × K. Then it is clear that φ is a well defined surjective homomorphism. Moreover, $\begin{matrix} \ker φ & = & {(h, k) \in H \times K : φ (h, k) = ([1], [1])} \\ = & {(h, k) \in H \times K : ([h], [k]) = ([1], [1])} \\ = & {(h, k) \in H \times K : [h] = [1], [k]) = [1]} \\ = & {(h, k) \in H \times K : h \in Rad (F), k \in Rad (G)} \\ = & H / Rad (F) \times K / Rad (G) \\ = & Rad (F \times G) . \end{matrix}$

Therefore, we have (H × K)/Rad (F × G) ≅ H/Rad (F) × K/Rad (G). □

Now, we characterize radical of bounded Hilbert algebra of R (H).

Proposition 3.13. Rad (R (H)) = Rad (H) ∩ R (H).

Proof. It is clear that Rad (R (H)) ⊆ R (H). It is sufficient to show that Rad (R (H)) ⊆ Rad (H). By contrary, suppose that there is an x^* ∈ Rad (R (H)) such that x^* ∉ Rad (H). Then there is M ∈ Max (H) such that x^* ∉ M. Thus x^** ∈ M. Because x^** = (x^**) ^**, therefore x^** ∈ M ∩ R (H), which is maximal in R (H), by Lemma 2.15. Since x^* ∈ Rad (R (H)) ⊆ M ∩ R (H), thus x^* and x^** = x^* → 0 ∈ M ∩ R (H), therefore 0inM ∩ R (H), which is a contradiction. Thus, Rad (R (H)) ⊆ Rad (H) ∩ R (H). On the other hand, we have R (H) ⊆ Rad (R (H)), thus Rad (H) ∩ R (H) ⊆ Rad (R (H)). Therefore Rad (R (H)) = Rad (H) ∩ R (H). □

Proposition 3.14. If x ∈ Rad (H), then x^** ∈ Rad (R (H)).

Proof. Let x ∈ Rad (H), but x^** ∉ Rad (R (H)). Thus there is M ∈ Max (R (H)) such that x^** ∉ M. Therefore x^*** ∈ M. Since (x^**) ^* = [(x^**) ^*] ^**, hence $(x^{* *})^{*} \in \bar{M}$ , which is maximal in H.

Because $x \in Rad (H) \subseteq \bar{M}$ we obtain that $x^{* *} \in \bar{M}$ , thus $x^{* *}, (x^{* *})^{*} = x^{* *} \to 0 \in \bar{M}$ . so $0 \in \bar{M}$ , a contradiction since $\bar{M}$ is proper. Thus, if x ∈ Rad (H), then x^** ∈ Rad (R (H)).

Open problem. Does the converse of Proposition 3.14 hold in every bounded Hilbert algebra ?

Proposition 3.15. If x ∈ H, then $x^{*} \in Rad (H) if and only if x^{*} \in Rad (R (H)) .$

Proof. Suppose that x^* ∈ Rad (H). By Proposition 3.14, (x^*) ^** = x^* ∈ Rad (R (H)).

Conversely, let x^* ∈ Rad (R (H)). From Proposition 3.13, x^* ∈ Rad (H). □

Definition 3.16. A proper implicative filter F of H is called semi maximal filter if Rad (F) = F.

Example 3.17. In Example 3.2 (i), F = {1, c} is a proper implicative filter which Rad (F) = F. Thus F is a semi maximal filter. Also F = {1} is a proper implicative filter, but Rad (F) ≠ F. Thus F is not a semi maximal filter.

The semi maximal filters of H are closed subsets of closure operator Rad.

Proposition 3.18. Any maximal filter of H is a semi maximal filter.

Proof. Let M be a maximal filter of H. Then Rad (M) = M. Thus M is a semi maximal filter of H. □

Corollary 3.19. Let H be a local Hilbert algebra with supremum. Then every Boolean filter of second kind of H is a semi maximal filter of H.

Proposition 3.20. Let H be a local Hilbert algebra. If the proper implicative filter F is a semi maximal filter of H, then x^* = 0, for all x ∈ F.

In the following we show that the converse of above proposition is not true.

Example 3.21. In Example 3.2 (iii), F = {a, 1}, a^* = 0, 1^* = 0, but D (F) = {a, b, c, 1}. Thus F is not a semi maximal filter of H.

The following example shows that the condition "local " in above theorem is essential.

Example 3.22. In Example 3.2 (i), H = {0, a, b, c, 1}, is a bounded Hilbert algebra, but is not local. Also F = {a, c, 1} is a semi maximal filter, but a^* ≠ 0.

Theorem 3.23. Let F be a semi maximal filter of H. Then F = {x ∈ H : x^* → x ∈ F}.

Proof. Let A = {x ∈ H : x^* → x ∈ F}. Clearly, F ⊆ A. Let x ∈ A. Thus x^* → x ∈ F. Also, let M be an arbitrary maximal filter of H that contains F. If x ∉ M, then x^* ∈ M, thus x^*, x^* → x ∈ M, so x ∈ M, which is a contradiction. Therefore x ∈ M. Thus x ∈ Rad (F), so x ∈ F. Hence A = F. □

Corollary 3.24. If F is a semi maximal filter of H, then F = {x ∈ H : x^** ∈ F}.

In the following we show that the converse of above theorem is not true.

Example 3.25. Let H = {0, a, b, c, 1}. Define

$\begin{matrix} \begin{matrix} \to & 0 & a & b & c & 1 \\ 0 & 1 & 1 & 1 & 1 & 1 \\ a & a & 1 & 1 & 1 & 1 \\ b & a & c & 1 & c & 1 \\ c & a & b & b & 1 & 1 \\ 1 & 0 & a & b & c & 1 \end{matrix} \end{matrix}$

(H, → , 0, 1) is a Hilbert algebra with supremum. Consider F = {1}, then F = {x ∈ H : x^* → x ∈ F} = {x ∈ H : x^** ∈ F}. But F is not a semi maximal filter, because Rad (F) = {a, b, c, 1} ≠ F.

Theorem 3.26. Let F be an implicative filter of H and a^* → a ∈ F, for all a ∈ H. Then a ∈ Rad (F).

Proof. Let a ∉ Rad (F). Thus there is a maximalfilter M which a ∉ M, hence a^* ∈ M. On the other hand a^* → a ∈ F ⊆ M, thus a ∈ M, which is acontradiction. □

Theorem 3.27. If {F_i} _i∈I is a family of implicative filters of H, then

Rad (∩ _i∈IF_i) = ∩ _i∈IRad (F_i);

If F_i is semi maximal filter of H, for all i ∈ I, then ∩_i∈IF_i is a semi maximal filter of H.

Proof. (i) Since ∩_i∈IF_i ⊆ F_i, thus by Theorem 3.3 (vii), Rad (∩ _i∈IF_i) ⊆ Rad (F_i), for all i ∈ I. Thus Rad (∩ _i∈IF_i) ⊆ ∩ _i∈IRad (F_i).

Conversely, if x ∈ ∩ _i∈IRad (F_i), then x ∈ Rad (F_i), for all i ∈ I, thus for all maximal filters M of H contains F_i, for all i ∈ I, we have x ∈ M. Therefore x ∈ M, ∩_i∈IRad (F_i) ⊆ M. Hence x ∈ Rad (∩ _i∈IF_i).

(ii) Let F_i be semi maximal filter of H, for all i ∈ I. Then $\underset{i \in I}{\cap} F_{i} \subseteq Rad (i \in I \cap F_{i}) = \underset{i \in I}{\cap} Rad (F_{i}) = \underset{i \in I}{\cap} F_{i} .$

Thus Rad (∩ _i∈IF_i) = ∩ _i∈IRad (F_i) = ∩ _i∈IF_i. □

In the following we show that the converse of part (ii) of above theorem is not true.

Example 3.28. In Example 3.2 (i), F₁ = {a, c, 1} , F₂ = {b, c, 1} and F₃ = {0, a, b, c, 1} are implicative filters. Then F₁ ∩ F₂ ∩ F₃ = {c, 1} is a semi maximal filter. But F₃ is not a semi maximal filter, because Rad (F₃) = {c, 1} ≠ F₃.

Theorem 3.29. Let F be an implicative filter of H. Then Rad (F) is the smallest semi maximal filter of H that contains F.

Proof. By Theorem 3.3 (viii), Rad (F) is a semi maximal filter of H. Let G be a semi maximal filter of H such that F ⊆ G. Then Rad (F) ⊆ Rad (G) = G. □

Theorem 3.30. Let F be an implicative filter of H. Then

Rad ({1/F}) = Rad (F)/F;

F is a semi maximal filter of H if and only if {1/F} is a semi maximal filter of H/F.

Proof.

By definition, we have $Rad ({1 / F}) = ⋂_{F \subseteq N \in Max (H)} (N / F) = (⋂_{F \subseteq N \in Max (H)} N) / F = Rad (F) / F$ .

Let F be a semi maximal filter of H, i.e. Rad (F) = F. Then by (i), we have Rad ({1/F}) = Rad (F)/F = F/F = {1/F}. Thus {1/F} is a semi maximal filter of H/F.

Conversely, let {1/F} be a semi maximal filter of H/F. Therefore Rad ({1/F}) = {1/F}, thus by (i) Rad (F)/F = {1/F}. It is sufficient to show that, Rad (F) ⊆ F. Let x ∈ Rad (F). Then x/F ∈ Rad (F)/F= {1/F}, and x/F = 1/F, thus x ∈ F, so Rad (F) ⊆ F. Therefore F is a semi maximal filter of H.

Corollary 3.31. Let F be an implicative filter of H. Then F is a semi maximal filter of H if and only if H/F is a semisimple Hilbert algebra.

Remark 3.32. As a consequence of corollary 3.31

by using semi maximal filter we can study semisimple Hilbert algebra.

If H is a semisimple Hilbert algebra we get a characterization of it by elements.

Theorem 3.33. If F (H) (P (H)) is a semi maximal filter of H, then F (H) (P (H)) is a maximal filter.

Proof. Assume F (H) is a semi maximal filter of H, thus Rad (F (H)) = F (H), By contrary, if F (H) is not a maximal filter, thus for some x ∈ H, if x ∉ F (H) then x^* ∉ F (H). Thus x^** ≤ x^* and x^* = x^*** ≤ x^**, therefore x^* = x^**, so x^* = 0 and x^** = 1, which is a contradiction. Hence F (H) is a maximal filter. □

For investigating a Hilbert algebra is semisimple it is sufficient the filter {1} investigate.

Theorem 3.34. {1} is a semi maximal filter of H, if and only if H is semisimple.

Proof. Since Rad ({1}) = Rad (H), thus {1} is a semi maximal filter of H, if and only if H is semisimple. □

Example 3.35. In Example 3.2 (i), F = {1, c} is a semi maximal filter, H/F = {[0] , [a] , [b] , [1]} is not local, since [a] ^* = [a^*] = [b] ≠ [0]. Also G = {b, c, 1} is a semi maximal filter, but H/F = {[0] , [1]} is local. Also K = {1} is not a semi maximal filter, H/K = {[0] , [a] , [b] , [c] , [1]} is not local, since [a] ^* = [a^*] = [b] ≠ [0].

Example 3.36.

In Example 3.2 (i), we have

the semi maximal filter {c, 1} is a BF1, BF2, PF1 and PF2, but it is not a PF3 and maximal filter.

the implicative filter {1} is not semi maximal, also c^** → c = c ∉ {1}, thus {1} is not a BF1, but it is a PF1.

F = {b, c, 1} is a proper implicative filter where 0^* = 1 ∈ F, a^* = b ∈ F, thus F is a maximal filter by Theorem 2.6, thus F is a semi maximal filter and PF1, PF2 and PF3.

In Example 3.2 (ii), {1} is a BF1 and semi maximal filter.

In Hilbert algebra {0, a, 1}, 0 < a < 1, which $x \to y = {\begin{matrix} 1 & if x \leq y \\ y & if x y ∥ \end{matrix}$ F = {a, 1} is a maximal and semi maximal of H = {0, a, 1}.

4 Conclusion and future research

In this paper, we studied the radical of an implicative filter in a Hilbert algebra and investigated some of its properties. This paper contains characterizations for the radical of bounded Hilbert algebra of R (H). The notion of a semi maximal filter in Hilbert algebras is introduced and we have proved theorems which determine the relationship between this notion and other types of filters in Hilbert algebra and by some examples we showed that these notions are different. We studied semisimple Hilbert algebras by semi maximal filters. We showed that Rad is a closure operator on implicative filtersof H.

The results of this paper will be devoted to study of Hilbert algebra, intuitionistic^,s logic which are different extensions of basic logic.

In the following diagram we summarize the results of this paper and the previous results in this field and we give the relationships between all types of filters in a Hilbert algebra. The mark A ⟶ B ( $A \underset{⟶}{a} B$ or $A \underset{a}{⟶} B$ ), means that A implies B(respectively, A implies B with the condition “a”).

Some important issues for future works are:

constructing the related logical properties of this structures,

describing all the semi maximal filters on a finite an or infinite sets,

finding useful results on other structures.

We hope this work would serve as a foundation for further studies on the structure of Hilbert algebras and develop corresponding many-valued logical systems.

Acknowledgments

The authors are extremely grateful to the referees for their valuable comments and helpful suggestions which helped to improve the presentation of this paper.

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