Deductive systems and R-congruences of pre-Hilbert algebras

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1 Introduction

Henkin (1950) introduced the notion of “implicative model”, as a model of positive implicative propositional calculus. Monteiro (1960) has given the name “Hilbert algebras” to the dual algebras of Henkin’s implicative models. Iséki (1966) introduced a new notion called a BCK algebra. It is an algebraic formulation of the BCK-propositional calculus system of Meredith (1962), and generalizes the concept of implicative algebras (see Abbott 1967). To solve some problems on BCK algebras, Komori (1984) introduced BCC algebras. These algebras (also called BIK⁺-algebras) are an algebraic model of the BIK⁺-logic. Then, Buşneag and Rudeanu (2010) introduced the notion of a pre-BCK algebra. A BCK algebra is just a pre-BCK with antisymmetry. Iorgulescu (2016) introduced new generalizations of BCK and Hilbert algebras (RML, pre-BCC, pimpl-RML algebras and many others). Recently, as a generalization of Hilbert algebras, Bandaru et al. (2021) introduced GE algebras (generalized exchange algebras) and Walendziak (2024) introduced pre-Hilbert algebras (the definition of a pre-Hilbert algebra is inspired by Henkin’s Positive Implicative Logic). All of the algebras mentioned above are contained in the class of RML algebras (an RML algebra is an algebra (A, → ,1) of type (2, 0) satisfying the identities: $x \to x = 1 = x \to 1$ and $1 \to x = x$ ).

In this paper, we introduce the notion of a deductive system in a pre-Hilbert algebra and investigate its elementary properties. Deductive systems of algebras of logic are an important algebraic notion. From the logical point of view, deductive systems correspond to those sets of formulas which are closed under the inference rule modus ponens. Note that deductive systems are the same as ideals considered among others in the papers Iséki and Tanaka (1976), Meng (1994) and Jun (2001) on BCK algebras (in these papers, the notation with ∗ and 0 is used). In the theory of Hilbert algebras, the concept of a deductive system was introduced by Diego (1996). This concept was further developed by Buşneag (1985, 1987) and Jun (1966) (see also Hong and Jun 1966).

In particular, we introduce the notion of a disjoint union of pre-Hilbert algebras and describe deductive systems in such algebras. We obtain a characterization of deductive systems and prove that the lattice of deductive systems of a pre-Hilbert algebra is algebraic and distributive. We also introduce and investigate the concept of an R-congruence in pre-Hilbert algebras. We show that the lattice of deductive systems of a pre-Hilbert algebra A is isomorphic to the lattice of R-congruences on A. Furthermore, we construct the quotient algebra A/D of a pre-Hilbert algebra A via a deductive system D of A and obtain the fundamental homomorphism theorem. Finally, we study maximal deductive systems. We give some characterizations of them. We provide the homomorphic properties of maximal deductive systems.

The motivation of this study consists of algebraic and logical arguments. Namely, pre-Hilbert algebras are related to Henkin’s Positive Implicative Logic, they belong to a wide class of algebras of logic. Moreover, the results of the paper may have applications for future studies of some generalizations of Hilbert algebras.

2 Preliminaries and properties of pre-Hilbert algebras

Let A = (A, → ,1) be an algebra of type (2, 0). We define the binary relation $\leq$ on A as follows: for all $x, y \in A,$

We consider the following list of properties (Iorgulescu 2016) that may be satisfied by A:

(An) (Antisymmetry) $(x \leq y and y \leq x) \Rightarrow x = y,$

(B) $y \to z \leq (x \to y) \to (x \to z),$

(BB) $y \to z \leq (z \to x) \to (y \to x),$

(D) $y \leq (y \to x) \to x,$

(Ex) (Exchange) $x \to (y \to z) = y \to (x \to z),$

(K) $x \leq y \to x,$

(L) (Last element) $x \leq 1,$

(M) $1 \to x = x,$

(Re) (Reflexivity) $x \leq x,$

(Tr) (Transitivity) $(x \leq y and y \leq z) \Rightarrow x \leq z,$

(*) y $\leq z \Rightarrow x \to y \leq x \to z,$

(**) $y \leq z \Rightarrow z \to x \leq y \to x,$

(p-1) $x \to (y \to z) \leq (x \to y) \to (x \to z),$

(p-2) $(x \to y) \to (x \to z) \leq x \to (y \to z),$

(pimpl) $x \to (y \to z) = (x \to y) \to (x \to z) .$

Remark 2.1

The properties in the list are the most important properties satisfied by a Hilbert algebra (the properties (An) – (**) are satisfied by a BCK algebra).

Lemma 2.2

Let A = (A, → ,1) be an algebra of type (2, 0). Then the following hold:

(i)(M) + (B) imply (Re), (*), (**);
(ii)(M) + (*) imply (Tr);
(iii)(M) + (**) imply (Tr);
(iv)(M) + (L) + (B) imply (K);
(v)(C) + (An) imply (Ex);
(vi)(p-1) + (p-2) + (An) imply (pimpl);
(vii)(M) + (L) + (B) + (An) + (p-1) imply (Ex).

Proof

(i)–(v) follow from Proposition 2.1 and Theorem 2.7 of Iorgulescu (2016).

(vi) is obvious.

(vii) Let $x, y, z \in A$ . From (p-1) we obtain

(1)

By above (i) and (iv), A satisfies (**) and (K). Applying (K), we get $y \leq x \to y$ and hence, by (**),

(2)

From (M) and (**) we conclude that (Tr) holds in A. Using (Tr), from (1) and (2) we have $x \to (y \to z) \leq y \to (x \to z) .$ Thus A satisfies (C). Using above (v), we obtain (Ex). □

Following Iorgulescu (2016), we say that (A, → ,1) is an RML algebra if it satisfies the axioms (Re), (M), (L). We introduce now the following definition.

Definition 2.3

Let A = (A, → ,1) be an RML algebra. The algebra A is said to be a:

1.pre-BCC algebra if it satisfies (B),
2.BCC algebra if it satisfies (B), (An), that is, it is a pre-BCC algebra with (An),
3.pre-BCK algebra if it satisfies (B), (Ex), that is, it is a pre-BCC algebra with (Ex),
4.BCK algebra if it is a pre-BCK algebra satisfying (An).

Denote by RML, pre-BCC, BCC, pre-BCK and BCK the classes of RML, pre-BCC, BCC, pre-BCK and BCK algebras, respectively. By definitions, we have

It is known that ≤ is an order relation in BCC and BCK algebras. By definition, in RML algebras, $\leq$ is a reflexive relation. By Lemma 2.2 (i), (ii), in pre-BCC and pre-BCK algebras, $\leq$ is reflexive and transitive (i.e., it is a pre-order relation).

Recall that an algebra (A, → ,1) is called a Hilbert algebra if it verifies the axioms (An), (K), (p-1). Diego (1966) proved that the class of all Hilbert algebras is a variety.

In Walendziak (2024), we introduced the following notion:

Definition 2.4

A pre-Hilbert algebra is an algebra (A, → ,1) of type (2, 0) satisfying (M), (K) and (p-1).

Let us denote by pre-H and H the classes of pre-Hilbert and Hilbert algebras, respectively.

Remark 2.5

Since pre-H is defined by identities, we see that pre-H is a variety.

Remark 2.6

Since (An) + (K) + (p-1) imply (M) (see Diego 1966), a Hilbert algebra is in fact a pre-Hilbert algebra satisfying (An).

Remark 2.7

A motivation for the definition of pre-Hilbert algebra is the Positive Implicative Logic given by Henkin (1950). This logic is the part of intuitionistic logic corresponding to formulas in which implication occurs as the only connective. The propositional calculus of the Henkin system of positive logic is specified by the following two axiom schemes:

(H1) $α \to (β \to α)$ ,
(H2)

and the modus ponens inference rule. Conditions (K) and (p-1) of Definition 2.4 are inspired by axioms (H1) and (H2), respectively. Moreover, (M) is inspired by the modus ponens (indeed, from (M) we see that if $x = 1$ and $x \to y = 1$ , then $y = 1$ ).

Theorem 2.8

Pre-Hilbert algebras satisfy (Re), (M), (L), (K), (C), (D), (B), (BB), (Tr), (*), (**), (p-1), (p-2).

Remark 2.9

Pre-Hilbert algebras don’t have to satisfy (An), (Ex), (pimpl); see the example below.

Example 2.10

Consider the set $A = {a, b, c, d, 1}$ and the operation → given by the following table:

We can observe that the properties (M), (K), (p-1) (hence (Re), (L), (B), (BB), (C), (D), (*), (**), (Tr), (p-2)) are satisfied. Then, $(A, \to, 1)$ is a pre-Hilbert algebra. It does neither satisfy (An) for $(x, y) = (b, c),$ nor (Ex) and (pimpl) for $(x, y, z) = (a, d, b) .$

Proposition 2.11

Let A = $(A, \to, 1)$ be an algebra of type (2, 0). The following are equivalent:

(i)A is a pre-Hilbert algebra;
(ii)A satisfies (M), (L), (B) and (p-1);
(iii)A is a pre-BCC algebra satisfying (p-1).

Proof

(i) ⇒ (ii).Follows from Theorem 2.8.
(ii) ⇒ (iii).By Lemma 2.2 (i), (M) + (B) imply (Re). Therefore A is a pre-BCC algebra with (p-1).
(iii) ⇒ (i).By Lemma 2.2 (iv), (M) + (L) + (B) imply (K). Then A satisfies (M), (K) and (p-1). Thus A is a pre-Hilbert algebra. □

Proposition 2.12

Let A = $(A, \to, 1)$ be a pre-Hilbert algebra. Then A induces a pre-order $\leq$ on A, defined by: $x \leq y ⟺ x \to y = 1$ and 1 is the element of A satisfying the following conditions:

(L1) $x \leq 1,$

(L2) $1 \leq x \Rightarrow x = 1.$

Proof

Straightforward. □

Proposition 2.13

([Walendziak 2024], Proposition 3.18) Let A be a non-void set of elements, $\leq$ be a pre-order relation on A and 1 be an element of A satisfying (L1) and (L2). We define the operation → by.

Then $(A, \to, 1)$ is a pre-Hilbert algebra.

Example 2.14

Let $Z$ be the set of integers and let for x, y $\in Z$ the symbol x∣y mean that x divides y. Then the relation ∣ is a pre-order on $Z$ which is not an order (for example, $1 ∣ - 1$ and $- 1 ∣ 1$ but $1 \neq - 1$ ). Moreover, $x ∣ 0$ for each x $\in Z$ and if $0 ∣ x,$ then $x = 0$ . If we define the operation → by.

then $(Z, |, 0)$ is a pre-Hilbert algebra.

Remark 2.15

The class of all pre-Hilbert algebras is a variety. Therefore, if A₁ and A₂ are two pre-Hilbert algebras, then the direct product A = A₁ × A₂ is also a pre-Hilbert algebra.

Let I be be any set and, for each $i \in I,$ let $A_{i} = (A_{i}, \to_{i}, 1)$ be a pre-Hilbert algebra. Suppose that $A_{i} \cap A_{j} = {1}$ for $i \neq j$ ; $i, j \in I .$ Set $A = ⋃_{i \in I} A_{i}$ and define the binary operation → on A via

It is easy to check that A = $(A, \to, 1)$ is a pre-Hilbert algebra. We say that A is the disjoint union of $A_{i}$ , $i \in I$ , and write $A = \sum_{i \in I} {A A}_{i} .$

Remark 2.16

Since a Hilbert algebra is a pre-Hilbert algebra satisfying (An), we have H = pre-H + (An). By Proposition 2.11, pre-H = pre-BCC + (p-1). Therefore H = BCC + (p-1). From Lemma 2.2 (vii) we conclude that BCC + (p-1) = BCC + (Ex) + (p-1) = BCK + (p-1), that is, H = BCK + (p-1).

The interrelationships between the classes of algebras mentioned before are visualized in Fig. 1.

3 Deductive systems of pre-Hilbert algebras

Diego (1966) introduced the notion of a deductive system of a Hilbert algebra. Similarly, if A = $(A, \to, 1)$ is a pre-Hilbert algebra, we say that a subset D of A is a deductive system of A if it satisfies:

(d₁) $1 \in D$ ,

(d₂) for all $x, y \in A$ , if $x \in D$ and $x \to y \in D$ , then $y \in D$ .

By DS(A) we denote the set of all deductive systems of A. It is obvious that $1, A \in D S (A) .$

Example 3.1

Let $A = {a, b, c, 1}$ and → be given by the following table:

We can observe that the properties (M), (K), (p-1) are satisfied. Then, A = $(A, \to, 1)$ is a pre-Hilbert algebra. It is easy to see that $D S (A) = {{1}, {1, a, b}, A} .$

Proposition 3.2

Let D be a deductive system of a pre-Hilbert algebra (A, → ,1). Then, for any $x, y \in A$ , if $x \leq y$ and $x \in D$ , then $y \in D$ .

Proof

Straightforward. □

Remark 3.3

Any deductive system D of a pre-Hilbert algebra (A, → ,1) satisfies the following condition: $x \to 1 = 1 \in D$ for every $x \in D$ (such deductive systems are called closed).

Proposition 3.4

Let D be a nonempty subset of a pre-Hilbert algebra A = $(A, \to, 1)$ . Then D is a deductive system of A if and only if for any $x, y \in D, x \leq y \to z$ implies $z \in D$ .

Proof

Let $D \in D S (A)$ and $x, y \in D .$ Suppose that $x \leq y \to z .$ From Proposition 3.2 we conclude that $y \to z \in D .$ By the definition of a deductive system, $z \in D .$

Conversely, assume that $x \leq y \to z$ implies $z \in D$ for all $x, y \in D .$ Since D is nonempty, let $x \in D .$ By (L), $x \leq x \to 1.$ Hence $1 \in D$ by assumption. Let $x \in D$ and $x \to y \in D .$ From (D) we see that $x \leq (x \to y) \to y .$ By assumption, $y \in D .$ Thus D is a deductive system of A. □

Proposition 3.5

A deductive system of a pre-Hilbert algebra A = $(A, \to, 1)$ is a subalgebra of A.

Proof

Let D be a deductive system of A. Obviously $1 \in D .$ Let $x, y \in D .$ By (K), $y \leq x \to y .$ Since $y \in D,$ from Proposition 3.2 we deduce that $x \to y \in D .$ Therefore, D is a subalgebra of A. □

Remark 3.6

The converse of Proposition 3.5 does not hold. Indeed, the subalgebra ${1, a}$ of the pre-Hilbert algebra A from Example 3.1 is not a deductive system.

Proposition 3.7

For each $i \in I,$ let $D_{i}$ be a deductive system of the pre-Hilbert algebra $A_{i} = (A_{i}, \to_{i}, 1)$ . Then $D = \prod_{i \in I} D_{i}$ is a deductive system of $A = \prod_{i \in I} A_{i} .$

Proof

Straightforward. □

Theorem 3.8

Let ${(A_{i})}_{i \in I}$ be an indexed family of pre-Hilbert algebras and $A = \sum_{i \in I} A_{i} .$ Let $D_{i}$ be a deductive system of $A_{i}$ for $i \in I .$ Then $⋃_{i \in I} D_{i}$ is a deductive system of A. Conversely, every deductive system of A is of this form.

Proof

Let D = $⋃_{i \in I} D_{i}$ . Of course, $1 \in D .$ Let $x \to y \in D$ and $x \in D .$ If $x \in A_{i}$ and $y \in A_{j}$ , where $i \neq j,$ then $y = x \to y \in D .$ Suppose that $x, y \in A_{i} .$ Then $x \to y, x \in D_{i} .$ Since $D_{i}$ is a deductive system of $A_{i}$ , we conclude that $y \in D_{i} .$ Hence $y \in D,$ and consequently, $D \in D S (A) .$

Now let D be a deductive system of A. It is easy to see that $D_{i} = D \cap A_{i} \in D S (A_{i})$ for $i \in I .$ We have D = $D \cap ⋃_{i \in I} A_{i} = ⋃_{i \in I} (D \cap A_{i}) = ⋃_{i \in I} D_{i} .$ □

Let A = $(A, \to_{A}, 1_{A})$ and B = $(B, \to_{B}, 1_{B})$ be pre-Hilbert algebras and let $f : A \to B$ be a homomorphism. The kernel of f is the set

that is, Ker(f) = $f^{\leftarrow} ({1_{B}}),$ where $f^{\leftarrow} (X)$ denote the f-inverse image of $X \subseteq B .$ It is easy to see that the next lemma holds.

Lemma 3.9

Let A = $(A, \to_{A}, 1_{A})$ and B = $(B, \to_{B}, 1_{B})$ be pre-Hilbert algebras, $f : A \to B$ be a homomorphism and let $x, y \in A .$ If $f (x) = f (y),$ then $x \to_{A} y, y \to_{A} x \in Ker (f) .$

Proposition 3.10

Let A and B be pre-Hilbert algebras and let $f : A \to B$ be a homomorphism. If $D \in D S (B),$ then $f^{\leftarrow} (D) \in D S (A) .$

Proof

The proof is straightforward. □

Proposition 3.11

Let A = $(A, \to_{A}, 1_{A})$ and B = $(B, \to_{B}, 1_{B})$ be pre-Hilbert algebras and let $f : A \to B$ be a surjective homomorphism. If D is a deductive system of A including Ker(f), then $f (D) \in D S (B)$ .

Proof

Obviously, $1_{B} \in f (D) .$ Let $x \in f (D), y \in B,$ and let $x \to_{B} y \in f (D) .$ Then there are $a, b \in D$ such that $x = f (a)$ and $x \to_{B} y = f (b) .$ Since f is surjective, $y = f (c)$ for some $c \in A .$ We have $f (b) = f (a) \to_{B} f (c) = f (a \to_{A} c)$ and hence, by Lemma 3.9, $b \to_{A} (a \to_{A} c) \in Ker (f) \subseteq D .$ Since $a, b \in D,$ we conclude that $c \in D .$ Therefore $y = f (c) \in f (D) .$ Consequently, $f (D) \in D S (B) .$ □

Let A be a pre-Hilbert algebra and $X \subseteq A .$ The set

D(X) = $⋂ {D \in D S (A) : D \supseteq X}$ is a deductive system of A, called the deductive system generated by X. For $a \in A,$ we write D(a) instead of D({a}). Define $\land$ and $\lor$ on DS(A) by $D \land E = D \cap E$ and $D \lor E = D D (D \cup E) .$ It is easy to see that under these operations DS(A) is a lattice. Moreover, this lattice is complete, since it is closed under arbitrary intersections.

Proposition 3.12

Let X be a nonempty subset of a pre-Hilbert algebra A = $(A, \to, 1)$ and let $Y = {x \in A : a_{1} \to (\dots \to (a_{n} \to x) \dots) = 1 for some a_{1}, \dots, a_{n} \in X, n \in N} .$ Then D(X) = Y. Moreover, D(∅) = {1}.

Proof

Since A satisfies (L), $1 \in Y$ . Suppose that $x, x \to y \in Y .$ By the definition of Y, there are $a_{1}, \dots, a_{n} \in X$ such that

(3)

and, similarly, there exist $b_{1}, \dots, b_{m} \in X$ such that

(4)

Applying (C), we get

Repeating this, by (*) and (C), we have

Then, applying (4), we have $1 = x \to (b_{1} \to (\dots (b_{m} \to y) \dots)),$ and therefore, $x \leq b_{1} \to (\dots (b_{m} \to y) \dots) .$ Hence, by (*),

Thus $y \in Y .$ Consequently, Y is a deductive system of A. Obviously, $X \subseteq Y .$

To prove that Y is the least deductive system including X, let $D \in D S (A)$ and $D \supseteq X .$ For any $x \in Y,$ there are $a_{1}, \dots, a_{n} \in X$ such that (3) holds. Since $X \subseteq D,$ we have $a_{1}, \dots, a_{n} \in D .$ By definition, from (3) we conclude that $x \in D .$ So $Y \subseteq D .$ Then D(X) = Y.

Moreover, it is easily seen that D(∅) = {1}. □

By Proposition 3.12, the mapping $X \to D D (X)$ is an algebraic closure operator on the power set of A, that is, for every $X \subseteq A,$

D(X) = $⋃$ {D( $X_{0}$ ): $X_{0}$ is a finite subset of X}.

Hence, by Theorem 2.16 of [McKenzie et al. 1987], we get.

Theorem 3.13

For any pre-Hilbert algebra A, (DS(A), ⊆) is an algebraic lattice whose the compact elements are exactly the finitely generated deductive systems.

Proposition 3.14

Let D be a deductive system of a pre-Hilbert algebra A = $(A, \to, 1)$ and $a \in A .$ Then D(D ∪ {a}) = ${x \in A : a \to x \in D} .$

Proof

Set $D_{a} = {x \in A : a \to x \in D} .$ We first show that $D_{a}$ is a deductive system of A. By (L), $1 \in D_{a} .$ Suppose that $x \to y \in D_{a}$ and $x \in D_{a} .$ Then $a \to (x \to y) \in D$ and $a \to x \in D .$ Applying (p-1), we get

By Proposition 3.2, $(a \to x) \to (a \to y) \in D,$ and hence $a \to y \in D,$ since $a \to x \in D$ and D ∈ DS(A). Consequently, $y \in D_{a} .$ Thus $D_{a}$ is a deductive system of A. Let $x \in D .$ By (K), $x \leq a \to x .$ Hence $a \to x \in D,$ that is, $x \in D_{a} .$ Therefore, $D \subseteq D_{a} .$ Moreover, $a \in D_{a} .$ Now let E be another deductive system including the set $D \cup {a} .$ If $x \in D_{a},$ then $a \to x \in D .$ This gives $a \to x \in E,$ and hence $x \in E .$ Then $D_{a} \subseteq E .$ Consequently, $D D (D \cup a) = D_{a} .$ □

Proposition 3.15

Let $A = (A, \to, 1)$ is a pre-Hilbert algebra. For any $a \in A,$

Proof

It is obvious that $D D (a) = D D ({1} \cup {a}) .$ Applying Proposition 3.14 for $D = {1},$ we have $D D (a) = {x \in A : a \to x = 1} = {x \in A : a \leq x} .$ □

For every $a \in A,$ D(a) is the principal deductive system generated by a.

Let $A = (A, \to, 1)$ be a pre-Hilbert algebra and $D, E \in D S (A) .$ Write

Lemma 3.16

If $A = (A, \to, 1)$ is a pre-Hilbert algebra and $D, E \in D S (A),$ then $D * E$ is a deductive system of A.

Proof

It is clear that $1 \in D * E .$ Suppose that $x \in D * E$ and $x \to y \in D * E .$ Let $d \in D .$ Applying (K), we see that $d \leq x \to d,$ and hence $x \to d \in D$ by Proposition 3.2. Since $x \to y \in D * E$ and $x \to d \in D,$ we obtain

(5)

From (K) and (p-1) we conclude that

Then, using (**), we get

(6)

Since E is a deductive system, from (5) and (6) we deduce that

(7)

By (B), $(x \to d) \to d \leq ((y \to d) \to (x \to d)) \to ((y \to d) \to d) .$ Since $(x \to d) \to d \in E,$ we see that

(8)

From (7) and (8) we have $(y \to d) \to d \in E .$ Therefore, $y \in D * E .$ Thus, $D * E$ is a deductive system of A. □

Lemma 3.17

If $A = (A, \to, 1)$ is a pre-Hilbert algebra, D and E are deductive systems of A, then $D * E$ is a relative pseudocomplement of D with respect of E in the lattice DS(A).

Proof

By Lemma 3.16 $, D * E \in D S (A) .$ Let $X \in D S (A) .$ It is sufficient to show that

(9)

Let $X \subseteq D * E$ and $x \in X \cap D .$ Then $x = (x \to x) \to x \in E .$ Hence $X \cap D \subseteq E .$ Conversely, assume $X \cap D \subseteq E$ and let $x \in X .$ Let d be an arbitrary element of D. By (K), $d \leq (x \to d) \to d .$ Since D is a deductive system and d $\in$ D, we get $(x \to d) \to d \in D .$ By (D), $x \leq (x \to d) \to d .$ From this we conclude that $(x \to d) \to d \in X,$ because $x \in X$ and $X \in D S (A) .$ Then $(x \to d) \to d \in X \cap D,$ and hence $(x \to d) \to d \in E .$ Therefore $x \in D * E,$ and consequently $X \subseteq D * E .$ □

From Theorem 3.13 and Lemma 3.17 we obtain

Theorem 3.18

For any pre-Hilbert algebra A, (DS(A), ⊆) is an algebraic distributive lattice.

Example 3.19

Let A be the pre-Hilbert algebra from Example 2.10. It is easy to see that A has the following deductive systems: $D_{1} = {1},$ $D_{2} = {1, a},$ $D_{3} = {1, d},$ $D_{4} = {1, a, d},$ $D_{5} = {1, b, c},$ $D_{6} = {1, a, b, c},$ $D_{7} = {1, b, c, d},$ $D_{8} = A .$ Figure 2 is the Hasse diagram of the lattice DS(A).

Proposition 3.20

If A is a pre-Hilbert algebra, then $(D S (A), *, A)$ is a Hilbert algebra.

Proof

Let D and E be deductive systems of A. Observe that

Indeed, by (9), $D \subseteq E ⟺ A \cap D \subseteq E ⟺ A \subseteq D * E ⟺ D * E = A .$ Obviously, DS(A) satisfies (An). Since $D \cap E \subseteq E,$ we have $E \subseteq D * E$ by (9). Consequently, DS(A) satisfies (K). To prove (p-1), let $D, E, F \in D S (A) .$ First we prove that

(10)

Since $E \subseteq D * E,$ we get $D \cap E \subseteq D \cap (D * E) .$ On the other hand, $D \cap (D * E) \subseteq D$ and $D \cap (D * E) \subseteq E .$ Hence $D \cap (D * E) \subseteq D \cap E .$ Thus (10) holds. Applying (10), we obtain

Hence by (9), $(D * E) \cap (D * (E * F)) \subseteq D * F,$ and consequently $D * (E * F) \subseteq (D * E) * (D * F),$ that is, (p-1) holds in DS(A). Thus (DS(A), $*, A$ ) is a Hilbert algebra. □

4 Deductive systems and R-congruences

For $D \in D S (A),$ we first define

and then define

Therefore

$x \sim_{D} y ⟺ x \to y \in D$ and $y \to x \in D .$

Theorem 4.1

Let $A = (A, \to, 1)$ be a pre-Hilbert algebra and D be a deductive system of A. Then $\sim_{D}$ is a congruence on A.

Proof

By (Re), $x \to x = 1 \in D,$ that is, $x \sim_{D} x$ for any $x \in A .$ This means that $\sim_{D}$ is reflexive. From definition, $\sim_{D}$ is symmetric. To prove that $\sim_{D}$ is transitive, let $x \sim_{D} y$ and $y \sim_{D} z .$ Then $x \to y, y \to x \in D$ and $y \to z, z \to y \in D .$ By (BB), $x \to y \leq (y \to z) \to (x \to z)$ and $z \to y \leq (y \to x) \to (z \to x) .$ Hence, by Proposition 3.2 and the definition of a deductive system, we get $x \to z \in D$ and $z \to x \in D .$ Consequently, $x \sim_{D} z,$ and so $\sim_{D}$ is transitive. Thus $\sim_{D}$ is an equivalence relation on A.

Let $x, y, z \in A$ and suppose that $x \sim_{D} y .$ Then $x \to y \in D$ and $y \to x \in D .$ By (B), $x \to y \leq (z \to x) \to (z \to y) .$ Since $x \to y \in D,$ we obtain $(z \to x) \to (z \to y) \in D,$ that is, $z \to x \leq_{D} z \to y .$ Similarly, $z \to y \leq_{D} z \to x,$ because $y \to x \in D .$ Therefore

(11)

Applying (BB), we have $y \to x \leq (x \to z) \to (y \to z)$ and $x \to y \leq (y \to z) \to (x \to z) .$ Hence $(x \to z) \to (y \to z) \in D$ and $(y \to z) \to (x \to z) \in D .$ Thus

(12)

Thus $\sim_{D}$ is an equivalence relation satisfying (11) and (12). Consequently, $\sim_{D}$ is a congruence on A. □

Let Con(A) denote the set of all congruences on A. For $x \in A$ and $θ \in Con (A),$ we write $x /$ θ for the congruence class containing x, that is, $x / θ = {y \in A : y θ x} .$ We say that $θ$ is an R-congruence on A if it satisfies the following property: for all $x, y \in A,$

(R) $(x \to y) θ 1$ and $(y \to x) θ 1$ imply $x θ y .$

We will denote by ${Con}_{R} (A)$ the set of all R-congruences on A.

Observe that $\sim_{D}$ satisfies (R). Let $(x \to y) \sim_{D} 1$ and $(y \to x) \sim_{D} 1.$ Then $x \to y = 1 \to (x \to y) \in D$ and $y \to x = 1 \to (y \to x) \in D .$ Consequently, $x \sim_{D} y .$ Thus we have

Proposition 4.2

If A is a pre-Hilbert algebra and $D \in D S (A),$ then $\sim_{D} \in {Con}_{R} (A) .$

Example 4.3

Let A be the pre-Hilbert algebra from Example 3.1. Then $ω := {(x, x) : x \in A} \in Con (A),$ but it does not satisfy (R). Indeed, $(a \to b) ω 1$ and $(b \to a) ω 1,$ but a $ω$ b does not hold.

As usual, a deductive system D of a pre-Hilbert algebra A is called the kernel of the congruence $θ$ on A if $D = 1 / θ .$

Proposition 4.4

Every deductive system of a pre-Hilbert algebra $A = (A, \to, 1)$ is the kernel of some R-congruence on A.

Proof

Let $D \in D S (A) .$ By Proposition 4.2, $\sim_{D} \in {Con}_{R} (A) .$ Moreover, we have

Therefore, $D = 1 / \sim_{D} .$ □

Proposition 4.5

Let $A = (A, \to, 1)$ be a pre-Hilbert algebra and $θ \in {Con}_{R} (A) .$ Then $1 / θ \in D S (A)$ and $θ = \sim_{1 / θ} .$

Proof

Observe that $1 / θ$ is a deductive system of A. Obviously, $1 \in 1 / θ .$ Let $x \in 1 / θ$ and $x \to y \in 1 / θ .$ Then $x θ 1$ and $(x \to y) θ 1.$ Hence $(x \to y) θ (1 \to y) .$ Consequently, $y θ 1,$ that is, $y \in 1 / θ .$ Thus $1 / θ \in D S (A) .$

Now observe that $θ = \sim_{1 / θ} .$ Indeed,

since $θ$ satisfies (R). Therefore, $θ = \sim_{1 / θ} .$ □

Theorem 4.6

Let $A = (A, \to, 1)$ be a pre-Hilbert algebra. Then the lattices $(D S (A), \subseteq)$ and $({Con}_{R} (A), \subseteq)$ are isomorphic.

Proof

We consider the function

By Proposition 4.5, $φ$ maps ${Con}_{R} (A)$ into DS(A). Since any deductive system of A is the kernel of some R-congruence on A, we conclude that $φ$ is onto DS(A). Observe that

(13)

for all $θ_{1}, θ_{2} \in {Con}_{R} (A) .$ If $θ_{1} \subseteq θ_{2}$ , then clearly ${1 / θ}_{1} \subseteq 1 / θ_{2}$ , that is, $φ (θ_{1}) \subseteq {φ (θ}_{2})$ . Conversely, assume ${1 / θ}_{1} \subseteq 1 / θ_{2}$ and let $x θ_{1} y .$ Then $(x \to y) θ_{} 1$ and hence $(x \to y) θ_{2} 1.$ Similarly, $(y \to x) θ_{2} 1.$ Since $θ_{2}$ satisfies (R), we have $x θ_{2} y .$ Thus $θ_{} \subseteq θ_{2} .$ Consequently, $φ$ maps ${Con}_{R} (A)$ onto DS(A) and satisfies (13). Therefore $({Con}_{R} (A), \subseteq)$ is isomorphic to $(D S (A), \subseteq) .$ □

Example 4.7

Let A be the pre-Hilbert algebra from Example 2.10. By Theorem 4.6 and Example 3.19, ${Con}_{R} (A)$ is the eight-element Boolean algebra.

Since pre-Hilbert algebras form a variety, from the Homomorphism Theorem for universal algebra we have

Lemma 4.8

Let $A = (A, \to, 1)$ be a pre-Hilbert algebra and $θ \in Con (A) .$ Set $x / θ, \to^{'} y / θ = (x \to y) / θ$ for $x, y \in A .$ Then $A / θ := (A / θ, \to^{'}, 1 / θ)$ is a pre-Hilbert algebra.

Proposition 4.9

Let $θ$ be a congruence on a pre-Hilbert algebra $A = (A, \to, 1)$ . Then the following are equivalent:

(i) $θ$ is an R-congruence on A,

(ii) $A /$ θ is a Hilbert algebra.

Proof

(i) ⇒ (ii). By Lemma 4.8, $A / θ$ is a pre-Hilbert algebra. Let $θ$ satisfy (R) and $x, y \in A .$ Suppose that

(14)

Then $(x \to y) / θ = 1 / θ = (y \to x) / θ .$ Hence $(x \to y) θ 1$ and $(y \to x) θ 1.$ Since satisfies (R), we have $x θ y .$ Therefore, $x / θ = y / θ,$ and consequently, (An) holds in $A / θ .$

(ii) ⇒ (i). Let $x, y \in A .$ Assume that $(x \to y) θ 1$ and $(y \to x) θ 1.$ Hence we obtain (14). Since $A /$ θ satisfies (An), we get $x / θ = y / θ,$ that is, $x θ y .$ Thus $θ$ is an R-congruence on A. □

Let D be a deductive system of a pre-Hilbert algebra $A = (A, \to, 1)$ . For $x \in A,$ we write $x / D = {y \in A : x \sim_{D} y} .$ It is easy to see that $x / D = x / \sim_{D} .$ We note that $x \sim_{D} y$ if and only if $x / D = y / D,$ that is,

In particular,

(15)

By the proof of Proposition 4.4, $D = 1 / \sim_{D}$ , that is, $1 / D = D .$ Then $(A / \sim_{D}, \to^{'}, D)$ denotes the quotient algebra $A / \sim_{D} .$ Instead of $A / \sim_{D}$ and $A / \sim_{D}$ we also write A/D and A/D, respectively.

Now, we define the binary relation $\leq^{'}$ on A/D as follows: for all $x, y \in A,$

(that is, $x / D \leq^{'} y / D ⟺ x \to y \in D$ ).

Remark 4.10

Since $\leq_{D}$ is a transitive relation (this follows from the proof of Theorem 4.1), we conclude that $\leq^{'}$ is well defined.

Proposition 4.11

Let D be a deductive system of a pre-Hilbert algebra $A = (A, \to, 1)$ . Then:

(i) A/D is a Hilbert algebra,

(ii) $x / D \leq^{'} y / D ⟺ x / D \to^{'} y / D = D$ for all $x, y \in A .$

Proof

(i) From Proposition 4.2 we see that $\sim_{D} \in {Con}_{R} (A) .$ Hence, by Proposition 4.9, $A / D = A / \sim_{D}$ is a Hilbert algebra.

(ii) Let $x, y \in A .$ We have

Then, (ii) holds. □

Theorem 4.12

Let A be a pre-Hilbert algebra and B be a Hilbert algebra. Let $f : A \to B$ be a homomorphism from A onto B. Then A/Ker(f) is isomorphic to B.

Proof

Since B as a Hilbert algebra satisfies (An), we have $x \sim_{Ker (f)} y$ if and only if $f (x) = f (y)$ (for all $x, y \in A$ ). Then this theorem follows from the Homomorphism Theorem for universal algebras. □

Proposition 4.13

Let D be a deductive system of a pre-Hilbert algebra $A = (A, \to, 1)$ and let $E \subseteq A / D .$ Then $E \in D S (A / D)$ if and only if $E = D_{0} / D$ (where $D_{0} / D := {x / D : x \in D_{0}}$ ) for some $D_{0} \in D S (A)$ such that $D \subseteq D_{0} .$

Proof

Suppose that $E \in D S (A / D) .$ Let $D_{0} = {x \in A : x / D \in E} .$ Suppose that $x \in D .$ By (15), $x / D = 1 / D \in E .$ Hence $x \in D_{0},$ and we obtain $D \subseteq D_{0} .$ Observe that D₀ is a deductive system of A. Indeed, $1 \in D_{0}$ and let $x \to y, x \in D_{0} .$ Then $(x \to y) / D \in E$ and $x / D \in E .$ Hence $y / D \in E$ and therefore $y \in D_{0} .$ Thus $D_{0} \in D S (A) .$ It is easy to see that $E = D_{0} / D .$

Conversely, let $E = D_{0} / D$ for some $D_{0} \in D S (A)$ such that $D \subseteq D_{0} .$ Of course, $1 / D \in E .$ Let $x / D \to^{'} y / D,$ $x / D \in E .$ Then $x \to y \in D_{0}$ and $x \in D_{0} .$ Since D₀ is a deductive system of A, we see that $y \in D_{0},$ hence that $y / D \in E .$ Consequently, $E \in D S (A / D) .$ □

5 Maximal deductive systems

Definition 5.1

Let A be a pre-Hilbert algebra and let D be a proper deductive system of A (i.e., $D \neq A$ ). Then D is called maximal if whenever E is a deductive system such that $D \subseteq E \subseteq A,$ then either $E = D$ or $E = A .$

The next lemma is obvious and its proof will be omitted.

Lemma 5.2

Every proper deductive system of a pre-Hilbert algebra can be extended to a maximal deductive system.

Theorem 5.3

For each $i \in I,$ let $D_{i}$ be a deductive system of the pre-Hilbert algebra $A_{i} .$ A deductive system $D = \prod_{i \in I} D_{i}$ is maximal in $A = \prod_{i \in I} A_{i}$ if and only if there is an unique index $j \in I$ such that $D_{j}$ is maximal in $A_{j}$ and $D_{i} = A_{i}$ for any $i \neq j .$

Proof

Straightforward. □

The following two theorems give the homomorphic properties of maximal deductive systems.

Theorem 5.4

Let $A = (A, \to_{A}, 1_{A})$ and $B = (B, \to_{B}, 1_{B})$ be pre-Hilbert algebras and let $f : A \to B$ be a surjective homomorphism. If D is a maximal deductive system of A including Ker(f), then $f (D)$ is a maximal deductive system of B.

Proof

By Proposition 3.11, $f (D) \in D S (B) .$ Let $x \in A ∖ D$ and suppose that $f (D) = B$ . Then $f (x) = f (y)$ for some $y \in D .$ Applying Lemma 3.9 we conclude that $y \to_{A} x \in D,$ and hence $x \in D,$ a contradiction. Therefore $f (D) \neq B$ . We take a proper deductive system E of B such that $E \supseteq f (D)$ . From Proposition 3.10 we deduce that $f^{\leftarrow} (E) \in D S (A) .$ It is easy to see that $D \subseteq f^{\leftarrow} (E) \subset A .$ Since D is maximal, $f^{\leftarrow} (E) = D$ . Consequently, $f (D) = f (f^{\leftarrow} (E)) = E$ . Thus $f (D)$ is a maximal deductive system of B. □

Theorem 5.5

Let A and B be pre-Hilbert algebras and let $f : A \to B$ be a surjective homomorphism. If D is a maximal deductive system of B, then $f^{\leftarrow} (D)$ is a maximal deductive system of A.

Proof

From Proposition 3.10 we conclude that $E = f^{\leftarrow} (D) \in D S (A)$ . It is easily seen that $E \neq A .$ By Lemma 5.2 there is a maximal deductive system F of A including E. We have

Since $F \supseteq E \supseteq Ker (f)$ , Theorem 5.4 shows that $f (F)$ is a maximal deductive system of B. Obviously, $f (F) \supseteq f (f^{\leftarrow} (D)) = D$ and hence $f (F) = D$ . Then $F \subseteq f^{\leftarrow} (f (F)) = f^{\leftarrow} (D) = E \subseteq F$ , that is, $f^{\leftarrow} (D) = F$ . Thus $f^{\leftarrow} (D)$ is a maximal deductive system of A. □

Theorem 5.6

For every proper deductive system D of a pre-Hilbert algebra $A = (A, \to, 1)$ , the following conditions are equivalent:

(a) D is a maximal deductive system of A;

(b) $y \to x \in D$ for any $x \in A$ and $y \in A ∖ D$ ;

(d) A/D has exactly two R-congruences.

Proof

(a) ⇒ (b). Let $x \in A$ . Suppose that D is a maximal deductive system of A and let $y \in A ∖ D$ . Then $D D (D \cup {y}) = A$ and hence $x \in D D (D \cup {y})$ . By Proposition 3.14, $y \to x \in D$ .

(b) ⇒ (c). Let D satisfy (b). Let $x \in A$ and $y \in A ∖ D$ . Then $y / D \to^{'} x / D = (y \to x) / D = D$ , since $y \to x \in D$ . By Proposition 3.15, $D D (y / D) = {x / D \in A / D : y / D \leq^{'} x / D} = {x / D \in A / D : y / D \to^{'} x / D = D} .$ Hence $D D (y / D) = A / D$ . Let E be a deductive system of A/D. From Proposition 4.13 we see that $E = D_{0} / D$ for some $D_{0} \in D S (A)$ such that $D \subseteq D_{0}$ . If $D = D_{0}$ , then $E = D / D = {D}$ . Suppose that $D \subset D_{0}$ . Then $E \supseteq D D (y_{0} / D)$ for some $y_{0} \in D_{0} ∖ D$ . Since $D D (y_{0} / D) = A / D$ , we conclude that $E = A / D$ . Thus $A / D = {A / D, {D}} .$

(c) ⇒ (a). Let F be a proper deductive system of A including D. By Proposition 4.13, $G := F / D \in D S (A / D) .$ Observe that $G \neq A / D .$ On the contrary, suppose that $G = A / D .$ Let $a \in A ∖ F$ and assume $a / D \in F / D .$ Then there exists some $f \in F$ with $f / D = a / D .$ Now we have $f \in F$ and $f \to a \in D \subseteq F$ and hence $a \in F$ , a contradiction. Therefore $G \neq A / D$ and by assumption, $G = {D},$ that is, $F = D$ , which proves that D is maximal. □

Proposition 5.7

Let D be a maximal deductive system of a pre-Hilbert algebra $A = (A, \to, 1)$ . Then for any $x, y \in A,$ we have $x \to y \in D$ or $y \to x \in D .$

Proof

Let $x, y \in A .$ If $x \in A ∖ D$ and $y \in A,$ then $x \to y \in D$ by Theorem 5.6. Similarly, $y \to x \in D$ for $x \in A$ and $y \in A ∖ D .$ If $x, y \in D,$ then $x \to y \in D$ and $y \to x \in D$ , since D is a subalgebra of A (see Proposition 3.5). □

Remark 5.8

Let A be the pre-Hilbert algebra from Example 3.1. Then for any $x, y \in A,$ we have $x \to y \in {1}$ or $y \to x \in {1},$ but $D = {1}$ is not a maximal deductive system of A. Therefore the converse of Proposition 5.7 does not hold.

6 Conclusions

We obtained the properties and characterizations of deductive systems in pre-Hilbert algebras. We described deductive systems in direct products and disjoint unions of such algebras. We showed that the lattice of deductive systems of a pre-Hilbert algebra is algebraic and distributive. We introduced the concept of an R-congruence and proved that the lattice of R-congruences on a pre-Hilbert algebra A is isomorphic to the lattice of deductive systems of A. Moreover, we constructed the quotient algebra A/D of a pre-Hilbert algebra A via a deductive system D of A and obtained the fundamental homomorphism theorem. Finally, we studied maximal deductive systems. We gave some characterizations of them. We provided the homomorphic properties of maximal deductive systems.