Int. J. Anal. Appl. (2023), 21:42 On Anti-Q-Fuzzy Deductive Systems of Hilbert Algebras M. Vasuki1, P. Senthil Kumar2, N. Rajesh2,∗ 1Department of Mathematics, Srinivasan College of Arts and Science (affiliated to Bharathidasan University), Perambalur, Tamilnadu, India 2Department of Mathematics, Rajah Serfoji Government College (affiliated to Bharathidasan University), Thanjavur-613005, Tamilnadu, India ∗Corresponding author: nrajesh_topology@yahoo.co.in Abstract. In this paper, the concept of anti-Q-fuzzy deductive systems concepts of Hilbert algebras are introduced and proved some results. Further, we discuss the relation between anti-Q-fuzzy deductive system and level subsets of a Q-fuzzy set. Anti Q-fuzzy deductive system is also applied in the Cartesian product of Hilbert algebras. 1. Introduction The concept of fuzzy sets was proposed by Zadeh [19]. The theory of fuzzy sets has several applica- tions in real-life situations, and many scholars have researched fuzzy set theory. After the introduction of the concept of fuzzy sets, several research studies were conducted on the generalizations of fuzzy sets. The integration between fuzzy sets and some uncertainty approaches such as soft sets and rough sets has been discussed in [1,4,7]. The idea of intuitionistic fuzzy sets suggested by Atanassov [2] is one of the extensions of fuzzy sets with better applicability. Applications of intuitionistic fuzzy sets appear in various fields, including medical diagnosis, optimization problems, and multicriteria decision making [12–14]. The concept of Hilbert algebra was introduced in early 50-ties by L.Henkin and T.Skolem for some investigations of implication in intuicionistic and other non-classical logics. In 60-ties, these algebras were studied especially by A.Horn and A.Diego from algebraic point of view. A.Diego proved (cf. [9] that Hilbert algebras form a variety which is locally finite. Hilbert algebras were Received: Feb 14, 2023. 2020 Mathematics Subject Classification. 20N05, 94D05, 03E72. Key words and phrases. Hilbert algebra; anti-Q-fuzzy deductive system; Q-fuzzy set. https://doi.org/10.28924/2291-8639-21-2023-42 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-42 2 Int. J. Anal. Appl. (2023), 21:42 treated by D.Busneag (cf. [5], [6]) and Y.B.Jun (cf. [15]) and some of their filters forming deduc- tive systems were recognized. W.A.Dudek (cf. [11]) considered the fuzzification of subalgebras and deductive systems in Hilbert algebras. In this paper, the concept of anti-Q-fuzzy deductive systems concepts of Hilbert algebras are introduced and proved some results. Further, we discuss the relation between anti-Q-fuzzy deductive system and level subsets of a Q-fuzzy set. Anti Q-fuzzy deductive system is also applied in the Cartesian product of Hilbert algebras. 2. Preliminaries Definition 2.1. [9] A Hilbert algebra is a triplet H = (H, ·, 1), where H is a nonempty set, · is a binary operation and 1 is fixed element of H such that the following axioms hold for each x,y,z ∈ H. (1) x ∗ (y ∗x) = 1, (2) (x ∗ (y ∗z)) ∗ ((x ∗y) ∗ (x ∗z)) = 1, (3) x ∗y = 1 and y ∗x = 1 imply x = y. The following result was proved in [11]. Lemma 2.1. Let H = (H, ·, 1) be a Hilbert algebra and x,y,z ∈ H. Then (1) x ∗x = 1, (2) 1 ∗x = x, (3) x ∗ 1 = 1, (4) x ∗ (y ∗z) = y ∗ (x ∗z). It is easily checked that in a Hilbert algebra H the relation ≤ y defined by x ≤ y ⇔ x ∗y = 1 is a partial order on H with 1 as the largest element. Definition 2.2. [8] A nonempty subset I of a Hilbert algebra H = (H, ·, 1) is called an ideal of H if (1) 1 ∈ I, (2) x ∗y,q ∈ I for all x ∈ H, y ∈ I, (3) (y2 ∗ (y1 ∗x)) ∗x ∈ I for all x ∈ H, y1,y2 ∈ I. Definition 2.3. [10] A fuzzy set µ in a Hilbert algebra H is said to be a fuzzy ideal of H if the following conditions are hold: (1) µ(1,q) ≥ µ(x,q) for all x ∈ H, (2) µ(x ∗y,q) ≥ µ(y,q) for all x,y ∈ H, (3) µ((y1 ∗ (y2 ∗x,q),q) ∗x) ≥ min{µ(y1,q),µ(y2,q)} for all x,y1,y2 ∈ H. Lemma 2.2. Let µ be a fuzzy set in A. Then the following statements hold: for any x,y ∈ A, (1) 1 − max{µ(x),µ(y)} = min{1 −µ(x), 1 −µ(y)}, (2) 1 − min{µ(x),µ(y)} = max{1 −µ(x), 1 −µ(y)}. Int. J. Anal. Appl. (2023), 21:42 3 Definition 2.4. [16] A Q-fuzzy set in a nonempty set H (or a Q-fuzzy subset of H) is an arbitrary function µ : X×Q → [0, 1], where Q is a nonempty set and [0, 1] is the unit segment of the real line. Definition 2.5. [16] Let µ be a Q-fuzzy set in A. The Q-fuzzy set µ defined by µ(x,q) = 1−µ(x,q) for all x ∈ A and q ∈ Q is called the complement of µ in A. Remark 2.1. For a Q-fuzzy set µ in A, we have µ = µ. Definition 2.6. [17] Let f : A → B be a function and µ be a Q-fuzzy set in B. We define a new Q-fuzzy set in A by µf as µ(f (x),q) for all x ∈ A and q ∈ Q. Definition 2.7. [17] Let f : A → B be a bijection and µf be a Q-fuzzy set in A. We define a new Q-fuzzy set in B by µ as µ(y,q) = µf (x,q), where f (x) = y for all y ∈ B and q ∈ Q. Definition 2.8. [17] Let µ be a Q-fuzzy set in A and δ be a Q-fuzzy set in B. The Cartesian product µ × δ : (A × B) × Q → [0, 1] is defined by (µ × δ)((x,y),q) = max{µ(x,q),δ(y,q)} for all x ∈ A, y ∈ B and q ∈ Q. The dot product µ · δ : (A × B) × Q → [0, 1] is defined by (µ∗δ)((x,y),q) = min{µ(x,q),δ(y,q)} for all x ∈ A, y ∈ B and q ∈ Q. Lemma 2.3. For any a,b ∈ R such that a < b, a < b+a 2 < b. 3. Anti-Q-fuzzy deductive systems Definition 3.1. A Q-fuzzy set µ in a Hilbert algebra H is said to be an anti-Q-fuzzy deductive system of H if the following conditions are hold: (∀x ∈ H) ( µ(1,q) ≤ µ(x,q) ) , (3.1) (∀ x,y ∈ H) ( µ(y,q) ≤ max{µ(x ∗y,q),µ(x,q)} ) . (3.2) A Q-fuzzy set µ in a Hilbert algebra H is called an anti-Q-fuzzy deductive system of H if it is an anti-q-fuzzy deductive system of H for all q ∈ Q. Example 3.1. Let H = {1,x,y,z, 0} be a set with a binary operation · defined by the following Cayley table: · 1 x y z 1 1 x y z x 1 1 y z y 1 x 1 z z 1 1 y 1 Then (H, ·, 1) is a Hilbert algebra. Let Q = {q}. We define a q-fuzzy set µ as follows: µ(1,q) = 0.9,µ(x,q) = 0.3,µ(y,q) = 0.1,µ(z,q) = 0.6. Then µ is a Q-fuzzy ideal of H. 4 Int. J. Anal. Appl. (2023), 21:42 Lemma 3.1. If µ is an anti-Q-fuzzy deductive system of a Hilbert algebra H, then (∀ x,y,z ∈ H) ( z ≤ x ∗y ⇒ µ(y,q) ≤ max{µ(x,q),µ(z,q)} ) . (3.3) Proof. Let x,y,z ∈ X such that z ≤ x ∗y. Then z · (x ∗y) = 1. µ(y,q) ≤ max{µ(x ∗y,q),µ(x,q)} ≤ max{max{µ(z · (x ∗y),q),µ(z,q)},µ(x,q)} = max{max{µ(1,q),µ(z,q)},µ(x,q)} = max{µ(x,q),µ(z,q)}. � Lemma 3.2. If µ is an anti-Q-fuzzy deductive system of a Hilbert algebra H, then (∀ x,y ∈ H) ( x ≤ y ⇒ µ(x,q) ≤ µ(y,q) ) . (3.4) Proof. Let x,y ∈ H be such that x ≤ y. Then x∗y = 1 and so µ(y,q) ≤ max{µ(x∗y,q),µ(x,q)} = max{µA(1,q),µA(x,q)} = µA(x,q). � Theorem 3.1. If µ is an anti-Q-fuzzy deductive system of a Hilbert algebra H, then for any x,a1,a2, ...,an ∈ X, (...((x ∗ a1) ∗ a2) · ...) ∗ an = 1 implies µ(x,q) ≤ max{µ(a1,q),µ(a2,q), . . .µ(an,q)}. Proof. Apply induction on n and apply Lemmas 3.1 and 3.2, the proof is clear. � Theorem 3.2. Every anti-Q-fuzzy deductive system of Hilbert algebra X is an anti-Q-fuzzy subalgebra of X. Proof. Let µ be an anti-Q-fuzzy deductive system of X. Since y ≤ x ∗ y for all x,y ∈ X, from Lemma 3.2 that µ(y,q) ≤ µ(x ∗ y,q) for all q ∈ Q. It follows from that µ(x ∗ y,q) ≤ µ(y,q) ≤ max{µ(x ∗y,q),µ(x,q)}≤ max{µ(x,q),µ(y,q)}. Hence µ is an anti-Q-fuzzy subalgebra of X. � Theorem 3.3. If µ is a Q-fuzzy subalgebra of X such that µ(y,q) ≥ min{µ(x,q),µ(z,q)} for all x,y,z ∈ X satisfying x ∗y ≤ z, then µ is an anti-Q-fuzzy deductive system of X. Proof. Let µ be an anti-Q-fuzzy subalgebra of X. Then by definition µ(1,q) ≤ µ(z,q) for all x ∈ X and q ∈ Q. Since x ≤ (x ∗y) ∗y, then by hypothesis, µ(y,q) ≤ max{µ(x ∗y,q),µ(x,q)}. Hence µ is an anti-Q-fuzzy deductive system of X. � Proposition 3.1. If {µi : i ∈ ∆} is a family of anti-Q-fuzzy deductive systems of a Hilbert algebra H, then ∧ i∈∆ µi is an anti-Q-fuzzy deductive system of H. Int. J. Anal. Appl. (2023), 21:42 5 Proof. Let {µi : i ∈ ∆} be a family of anti-Q-fuzzy deductive systems of a Hilbert algebra H. Let x ∈ H and q ∈ Q, we have ( ∧ i∈∆ µi)(1,q) = sup i∈∆ {µi(1,q)} ≤ sup i∈∆ {µi(x,q)} = ( ∧ i∈∆ µi)(x,q). Let x,y ∈ H and q ∈ Q, we have ( ∧ i∈∆ µi)(y,q) = sup i∈∆ {µi(y,q)} ≤ sup i∈∆ {max{µi(x ∗y,q),µi(x,q)}} = max{sup i∈∆ µi(x ∗y,q), sup i∈∆ µi(x,q)} = max{( ∧ i∈∆ µi)(x ∗y,q), ( ∧ i∈∆ µi)(x,q)}. Hence ∧ i∈∆ µi is an anti-Q-fuzzy deductive system of a Hilbert algebra H. � Proposition 3.2. Every Q-fuzzy ideal of a Hilbert algebra H is an anti-Q-fuzzy deductive system of H. Proof. Let µ be an anti-Q-fuzzy ideal of H. If y1 = x ∗ y, y2 = x, where x,y ∈ H and q ∈ Q, then by (1), (2) of Lemma 2.1, we have µ(y,q) = µ(1 ∗ y,q) = µ(((x ∗ y,q) ∗ (x ∗ y,q)) ∗ y,q) ≤ max{µ(x ∗y,q),µ(x,q)}. Hence µ be an anti-Q-fuzzy deductive system of H. � Lemma 3.3. [18] Let µ be a fuzzy set in A and for any t ∈ [0, 1]. Then the following properties hold: (1) L(µ,t) = U(µ, 1 − t), (2) L−(µ,t) = U+(µ, 1 − t), (3) U(µ,t) = L(µ, 1 − t), (4) U+(µ,t) = L−(µ, 1 − t). Lemma 3.4. [18] Let µ be a Q-fuzzy set in A and for any t ∈ [0, 1] and q ∈ Q. Then the following properties hold: (1) L(µ,t,q) = U(µ, 1 − t,q), (2) L−(µ,t,q) = U+(µ, 1 − t,q), (3) U(µ,t,q) = L(µ, 1 − t,q), (4) U+(µ,t,q) = L−(µ, 1 − t,q). Lemma 3.5. [18] Let µ be a Q-fuzzy set in A and for any t ∈ [0, 1] and q ∈ Q. Then the following properties hold: (1) L(µ,t) = ⋂ q∈Q L(µ,t,q), (2) L−(µ,t) = ⋂ q∈Q L−(µ,t,q), (3) U(µ,t) = ⋂ q∈Q U(µ,t,q), (4) U+(µ,t) = ⋂ q∈Q U+(µ,t,q). 6 Int. J. Anal. Appl. (2023), 21:42 Definition 3.2. Let µ be a q-fuzzy set of a Hilbert algebra H and t ∈ [0, 1]. Then we define the sets U(µ,t) = {x ∈ H : µ(x,q) ≥ t f or all q ∈ Q} and U+(µ,t) = {x ∈ H : µ(x,q) > t f or all q ∈ Q} are called an upper α-level subset and an upper α-strong level subset of µ, respectively. The sets L(µ,t) = {x ∈ H : µ(x,q) ≤ t f or all q ∈ Q} and L−(µ,t) = {x ∈ H : µ(x,q) < t f or all q ∈ Q} are called a lower t-level subset and a lower t-strong level subset of µ, respectively. For any q ∈ Q, the sets U(µ,t,q) = {x ∈ H : µ(x,q) ≥ t} and U+(µ,t,q) = {x ∈ H : µ(x,q) > t} are called a q-upper t-level subset and a q-upper t-strong level subset of µ, respectively. The sets L(µ,t,q) = {x ∈ H : µ(x,q) ≤ t} and L+(µ,t,q) = {x ∈ H : µ(x,q) < t} are called a q-lower t-level subset and a q-lower t-strong level subset of µ, respectively. Theorem 3.4. Let µ be a Q-fuzzy set in H. Then the following statements hold: (1) µ is an anti-Q-fuzzy deductive system of H if and only if for any t ∈ [0, 1] and q ∈ Q, L(µ,t,q) is either empty or a deductive system of H. (2) µ is an anti-Q-fuzzy deductive system of H if and only if for any t ∈ [0, 1] and q ∈ Q, L−(µ,t,q) is either empty or a deductive system of H. (3) µ is an anti-Q-fuzzy deductive system of H if and only if for any t ∈ [0, 1] and q ∈ Q, U(µ,t,q) is either empty or a deductive system of H. (4) µ is an anti-Q-fuzzy deductive system of H if and only if for any t ∈ [0, 1] and q ∈ Q, U+(µ,t,q) is either empty or a deductive system of H. Proof. (1). Assume that µ is an anti-Q-fuzzy deductive system of H. Then µ is an anti-Q-fuzzy deductive system of H for all q ∈ Q. Let q ∈ Q and t ∈ [0, 1] be such that L(µ,t,q) 6= ∅ and let x ∈ H be such that x ∈ L(µ,t,q). Then µ(x,q) ≤ t. Thus µ(1,q) = µ(x ∗ x,q) ≤ µ(x,q). Hence µ(1,q) ≤ µ(x,q) ≤ t, so 1 ∈ L(µ,t,q). Let x,y ∈ H and q ∈ Q be such that x ∈ L(µ,t,q) and x ∗ y ∈ L(µ,t,q). Then µ(x,q) ≤ t and µ(x ∗ y,q) ≤ t. Then we have µ(y,q) ≤ max{µ(x ∗ y,q),µ(x,q)}. Then µ(y,q) ≤ max{µ(x∗y,q),µ(x,q)}≤ t, so y ∈ L(µ,t,q). Hence L(µ,t,q) is a deductive system of H. Conversely, assume that every nonempty set L(µ,t,q) is a deductive system in H. If µ(1,q) ≤ µ(x,q) is not true for all x ∈ H and q ∈ Q. Then there exist x0 ∈ H and q ∈ Q such that µ(1,q) > µ(x0,q). Let t = 1 2 (µ(1,q) + µ(x0,q)). Then t ∈ [0, 1] and by Lemma 2.8, we have µ(1,q) > t > µ(x0,q). Thus x0 ∈ L(µ,t,q), so, L(µ,t,q) 6= ∅. By assumption, we have L(µ,t,q) is a deductive system of H. It follows that 1 ∈ L(µ,t,q), so µ(1,q) ≤ t, which is a contradicition. Hence µ(1,q) ≤ µ(x,q) for all x ∈ H and q ∈ Q. Suppose that µ(y,q) ≤ max{µ(x ∗ y,q),µ(x,q)} is not true for all x,y ∈ H and q ∈ Q. Then there exist u0,v0 ∈ H and q ∈ Q such that µ(v0,q) > max{µ(u0 ∗v0,q),µ(u0,q)}. Let p = 12 (µ(v0,q) + max{µ(u0 ∗v0,q),µ(u0,q)}). Then p ∈ [0, 1] and we have µ(v0,q) > p > max{µ(u0 ∗ v0,q),µ(u0,q)}. Then u0,v0 ∈ L(µ,p,q), so, L(µ,p,q) 6= ∅. By assumption, we have L(µ,p,q) is a deductive system of H. It follows that v0 ∈ L(µ,p,q). Hence µ(v0,q) ≤ p, which is a contradiction. Hence µ(y,q) ≤ max{µ(x ∗ y,q),µ(x,q)} is true for all x,y ∈ H and q ∈ Q. Hence µ is an anti-q-fuzzy deductive system of H for all q ∈ Q. Consequently Int. J. Anal. Appl. (2023), 21:42 7 µ is an anti-Q-fuzzy deductive system of H. (2). Assume that µ is an anti-Q-fuzzy deductive system of H. Then µ is an anti-Q-fuzzy deductive system of H for all q ∈ Q. Let q ∈ Q and t ∈ [0, 1] be such that L−(µ,t,q) 6= ∅ and let x ∈ H be such that x ∈ L−(µ,t,q). Then µ(x,q) < t. Thus µ(1,q) = µ(x ∗ x,q) ≤ µ(x,q). Hence µ(1,q) ≤ µ(x,q) < t, so 1 ∈ L−(µ,t,q). Let x,y ∈ H and q ∈ Q be such that x ∈ L−(µ,t,q) and x ∗ y ∈ L−(µ,t,q). Then µ(x,q) < t and µ(x ∗ y,q) < t. Then we have µ(y,q) ≤ max{µ(x ∗ y,q),µ(x,q)}. Then µ(y,q) ≤ max{µ(x ∗y,q),µ(x,q)} < t, so y ∈ L−(µ,t,q). Hence L−(µ,t,q) is a deductive system of H. Conversely, assume that every nonempty set L−(µ,t,q) is a deductive system in H. If µ(1,q) ≤ µ(x,q) is not true for all x ∈ H and q ∈ Q. Then there exist x0 ∈ H and q ∈ Q such that µ(1,q) > µ(x0,q). Let t′ = 12 (µ(1,q) + µ(x0,q)). Then t ′ ∈ [0, 1] and we have µ(1,q) > t′ > µ(x0,q). Thus x0 ∈ L−(µ,t′,q), so, L−(µ,t′,q) 6= ∅. By assumption, we have L−(µ,t′,q) is an ideal of H. It follows that 1 ∈ L−(µ,t′,q), so µ(1,q) < t′, which is a contradicition. Hence µ(1,q) ≤ µ(x,q) for all x ∈ H and q ∈ Q. Suppose that µ(y,q) ≤ max{µ(x ∗ y,q),µ(x,q)} is not true for all x,y ∈ H and q ∈ Q. Then there exist u0,v0 ∈ H and q ∈ Q such that µ(v0,q) > max{µ(u0∗v0,q),µ(u0,q)}. Let p′ = 12 (µ(v0,q) + max{µ(u0∗v0,q),µ(u0,q)}). Then p ′ ∈ [0, 1] and we have µ(v0,q) > p′ > max{µ(u0∗v0,q),µ(u0,q)}. Then u0,v0 ∈ L−(µ,p′,q), so, L−(µ,p′,q) 6= ∅. By assumption, we have L−(µ,p′,q) is a deductive system of H. It follows that v0 ∈ L−(µ,p′,q). Hence µ(v0,q) < p′, which is a contradiction. Hence µ(y,q) ≤ max{µ(x ∗ y,q),µ(x,q)} is true for all x,y ∈ H and q ∈ Q. Hence µ is an anti-q-fuzzy deductive system of H for all q ∈ Q. Consequently µ is an anti-Q-fuzzy deductive system of H. (3). Assume that µ is an anti-Q-fuzzy deductive system of H. Then µ is an anti-Q-fuzzy deductive system of H for all q ∈ Q. Let q ∈ Q and t ∈ [0, 1] be such that U(µ,t,q) 6= ∅ and let x ∈ U(µ,t,q). Then µ(x,q) ≥ t. Thus µ(1,q) = µ(x∗x,q) ≤ µ(x,q). Then 1−µ(1,q) ≤ 1−µ(x,q), so µ(1,q) ≥ µ(x,q) ≥ t. Hence 1 ∈ U(µ,t,q). Let x,y ∈ H be such that x ∈ U(µ,t,q) and x ∗ y ∈ U(µ,t,q). Then µ(x,q) ≥ t and µ(x ∗y,q) ≥ t. Then we have µ(y,q) ≤ max{µ(x ∗y,q),µ(x,q)}. Then µ(y,q) ≤ max{µ(x ∗y,q),µ(x,q)} 1 −µ(y,q) ≤ max{1 −µ(x ∗y,q), 1 −µ(x,q)} 1 −µ(y,q) ≤ 1 − min{µ(x ∗y,q),µ(x,q)} µ(y,q) ≥ min{µ(x ∗y,q),µ(x,q)}≥ t. Thus y ∈ U(µ,t,q). Hence U(µ,t,q) is a deductive system of H. Conversely, assume that every nonempty set U(µ,t,q) is a deductive system in H. If µ(1,q) ≤ µ(x,q) is not true for all x ∈ H and q ∈ Q. Then there exist x0 ∈ H and q ∈ Q such that µ(1,q) > µ(x0,q). Then µ(1,q) > µ(x0,q) 1 −µ(1,q) > 1 −µ(x0,q) µ(1,q) < µ(x0,q). 8 Int. J. Anal. Appl. (2023), 21:42 Let t = 1 2 (µ(1,q) + µ(x0,q)). Then t ∈ [0, 1] and we have µ(1,q) < t < µ(x0,q). Thus x0 ∈ U(µ,t,q), that is U(µ,t,q) 6= ∅. By assumption, we have U(µ,t,q) is a deductive system of H. It follows that 1 ∈ U(µ,t,q), so µ(1,q) ≥ t, which is a contradiction. Hence µ(1,q) ≤ µ(x,q) for all x ∈ H and q ∈ Q. Suppose that µ(y,q) ≤ max{µ(x∗y,q),µ(x,q)} is not true for all x,y ∈ H. Then there exist u0,v0 ∈ H and q ∈ Q such that µ(v0,q) > max{µ(u0 ∗v0,q),µ(u0,q)}. Then µ(v0,q) > max{µ(u0 ∗v0,q),µ(u0,q)} 1 −µ(v0,q) > max{1 −µ(u0 ∗v0,q), 1 −µ(u0,q)} 1 −µ(v0,q) > 1 − min{µ(u0 ∗v0,q),µ(u0,q)} µ(v0,q) < min{µ(u0 ∗v0,q),µ(u0,q)}. Taking p = 1 2 (µ(v0,q) + min{µ(u0 ∗ v0,q),µ(u0,q)}). Then p ∈ [0, 1] and we have µ(v0,q) < p < min{µ(u0 ∗ v0,q),µ(u0,q)}. Thus µ(u0 ∗ v0,q) > p and µ(u0,q) > p, so u0 ∗ v0,u0,∈ U(µ,p,q), so U(µ,p,q) 6= ∅. By assumption, we have U(µ,p,q) is a deductive system of H. It follows that v0 ∈ U(µ,p,q), so µ(v0,q) ≥ p, a contradiction. Hence µ(y,q) ≤ max{µ(x∗y,q),µ(x,q)} is true for all x,y ∈ H and q ∈ Q. Hence µ is an anti-q-fuzzy deductive system of H for all q ∈ Q. Consequently µ is an anti-Q-fuzzy deductive system of H. (4). Assume that µ is an anti-Q-fuzzy deductive system of H. Then µ is an anti-Q-fuzzy deductive system of H for all q ∈ Q. Let q ∈ Q and t ∈ [0, 1] be such that U+(µ,t,q) 6= ∅ and let x ∈ U+(µ,t,q). Then µ(x,q) > t. Thus µ(1,q) = µ(x∗x,q) ≤ µ(x,q). Then 1−µ(1,q) ≤ 1−µ(x,q), so µ(1,q) ≥ µ(x,q) > t. Hence 1 ∈ U+(µ,t,q). Let x,y ∈ H be such that x ∈ U+(µ,t,q) and x∗y ∈ U+(µ,t,q). Then µ(x∗y,q) > t and µ(x,q) > t. Then we have µ(y,q) ≤ max{µ(x∗y,q),µ(x,q)}. Then µ(y,q) ≤ min{µ(x ∗y,q),µ(x,q)} 1 −µ(y,q) ≤ max{1 −µ(x ∗y,q), 1 −µ(x,q)} 1 −µ(y,q) ≤ 1 − min{µ(x ∗y,q),µ(x,q)} µ(y,q) ≥ min{µ(x ∗y,q),µ(x,q)} > t. Thus y ∈ U+(µ,t,q). Hence U+(µ,t,q) is a deductive system of H. Conversely, assume that every nonempty set U+(µ,t,q) is a deductive system in H. If µ(1,q) ≤ µ(x,q) is not true for all x ∈ H and q ∈ Q. Then there exist x0 ∈ H and q ∈ Q such that µ(1,q) > µ(x0,q). Then µ(1,q) > µ(x0,q) 1 −µ(1,q) > 1 −µ(x0,q) µ(1,q) < µ(x0,q). Let t = 1 2 (µ(1,q) + µ(x0,q)). Then t ∈ [0, 1] and we have µ(1,q) < t < µ(x0,q). Thus x0 ∈ U+(µ,t,q), that is U+(µ,t,q) 6= ∅. By assumption, we have U+(µ,s,q) is a deductive system of H. It follows that 1 ∈ U+(µ,t,q), so µ(1,q) > t, which is a contradiction. Hence µ(1,q) ≤ µ(x,q) for all x ∈ H and q ∈ Q. Suppose that µ(y,q) ≤ max{µ(x ∗ y,q),µ(x,q)} is not true for all x,y ∈ H. Int. J. Anal. Appl. (2023), 21:42 9 Then there exist u0,v0 ∈ H and q ∈ Q such that µ(v0,q) > max{µ(u0 ∗v0,q),µ(u0,q)}. Then µ(v0,q) > max{µ(u0 ∗v0,q),µ(u0,q)} 1 −µ(v0,q) > max{1 −µ(u0 ∗v0,q), 1 −µ(u0,q)} 1 −µ(v0,q) > 1 − min{µ(u0 ∗v0,q),µ(u0,q)} µ(v0,q) < min{µ(u0 ∗v0,q),µ(u0,q)}. Taking p = 1 2 (µ(v0,q) + min{µ(u0 ∗ v0,q),µ(u0,q)}). Then p ∈ [0, 1] and we have µ(v0,q) < p < min{µ(u0 ∗ v0,q),µ(u0,q)}. Thus µ(u0 ∗ v0,q) > p and µ(u0,q) > p, so u0 ∗ v0,u0 ∈ U+(µ,p,q), so U+(µ,p,q) 6= ∅. By assumption, we have U+(µ,p,q) is a deductive system of H. It follows that v0 ∈ U+(µ,p,q), so µ(v0,q) ≥ p, a contradiction. Hence µ(y,q) ≤ max{µ(x ∗y,q),µ(y,q)} is true for all x,y ∈ H and q ∈ Q. Hence µ is an anti-q-fuzzy deductive system of H for all q ∈ Q. Then µ is an anti-Q-fuzzy deductive system of H. � Corollary 3.1. Let µ be a Q-fuzzy set in H. Then the following statements hold: (1) µ is an anti-Q-fuzzy deductive system of H, then for any t ∈ [0, 1], L(µ,t) is either empty or a deductive system of H. (2) µ is an anti-Q-fuzzy deductive system of H, then for any t ∈ [0, 1], L−(µ,t) is either empty or a deductive system of H. (3) µ is an anti-Q-fuzzy deductive system of H, then for any t ∈ [0, 1], U(µ,t) is either empty or a deductive system of H. (4) µ is an anti-Q-fuzzy deductive system of H, for any t ∈ [0, 1], U+(µ,t) is either empty or a deductive system of H. Proof. (1). Assume that µ is an anti-Q-fuzzy deductive system of H. Then we have for any t ∈ [0, 1] and q ∈ Q. Let L(µ,t,q) is either empty or a deductive system of H. Let t ∈ [0, 1]. If L(µ,t,q) = ∅ for some q ∈ Q, it follows that L(µ,t) = ⋂ q∈Q L(µ,t,q). If L(µ,t,q) 6= ∅ for all q ∈ Q, it follows from that L(µ,t,q) is a deductive system of H for all q ∈ Q. Then we have L(µ,t) = ⋂ q∈Q L(µ,t,q) is an ideal of H. (2). Similarly to as in the proof of (1). (3). Assume that µ is an anti-Q-fuzzy deductive system of H. Then we have for any t ∈ [0, 1] and q ∈ Q. Let U(µ,t,q) is either empty or a deductive system of H. Let t ∈ [0, 1]. If U(µ,t,q) = ∅ for some q ∈ Q, it follows that U(µ,t) = ⋂ q∈Q U(µ,t,q). If U(µ,t,q) 6= ∅ for all q ∈ Q, it follows that U(µ,t,q) is a deductive system of H for all q ∈ Q. Then we have U(µ,t) = ⋂ q∈Q U(µ,t,q) is a deductive system of H. (4). Similarly to as in the proof of (3). � Corollary 3.2. Let D be a deductive system of H. Then the following statements hold: 10 Int. J. Anal. Appl. (2023), 21:42 (1) for any k ∈ (0, 1], there exists an anti-Q-fuzzy deductive system µ of H such that L(µ,t) = D for all t < k and L(µ,t) = H for all t ≥ k, (2) for any k ∈ (0, 1], there exists an anti-Q-fuzzy deductive system γ of H such that U(γ,t) = D for all t > k and U(γ,t) = H for all t ≤ k. Proof. (1). Let µ be a Q-fuzzy set in H defined by, for all q ∈ Q µ(x,q) = { 0 if x ∈ D k if x /∈ D. Case 1 : To show that L(µ,t) = D for all t < k, let t ∈ [0, 1] be such that t < k. Let x ∈ L(µ,t). Then µ(x,q) ≤ t < k for all q ∈ Q. Thus µ(x,q) 6= k for all q ∈ Q, so µ(x,q) = 0 for all q ∈ Q. Then x ∈ D, so L(µ,t) ⊆ D. Now, let x ∈ D. Then µ(x,q) = 0 ≤ t for all q ∈ Q. Thus x ∈ L(µ,t), so D ⊆ L(µ,t). Hence L(µ,t) = D for all t < k. Case 2 : To show that L(µ,t) = H for all t ≥ k, let t ∈ [0, 1] be such that t ≥ k. Clearly, L(µ,t) ⊆ H. Let x ∈ H. Then for all q ∈ Q, we define µ(x,q) = { 0 < t if x ∈ D k ≤ t if x /∈ D. Then x ∈ L(µ,t), so H ⊆ L(µ,t). Hence L(µ,t) = H for all t ≥ k. We claim that L(µ,t,q) = L(µ,t,q′) for all q,q′ ∈ Q. For q,q′ ∈ Q, we obtain x ∈ L(µ,t,q) ⇔ µ(x,q) ≤ t ⇔ µ(x,q′) ≤ t ⇔ x ∈ L(µ,t,q′). Hence L(µ,t,q) = L(µ,t,q′) for all q,q′ ∈ Q. Then we have L(µ,t) = ⋂ q∈Q L(µ,t,q). By the claim, we have L(µ,t) = L(µ,t,q) for all q ∈ Q. Since L(µ,t,q) = L(µ,t) = D for all t < k and L(µ,t) = L(µ,t) = H for all t ≥ k, it follows that µ is an anti-Q-fuzzy deductive system of H. (2). Let µ be a Q-fuzzy set in H defined by, for all q ∈ Q µ(x,q) = { 1 if x ∈ D k if x /∈ D. Case 1 : To show that U(µ,t) = D for all t > k, let t ∈ [0, 1] be such that t > k. Let x ∈ U(µ,t). Then µ(x,q) ≥ t > k for all q ∈ Q. Then µ(x,q) 6= k for all q ∈ Q, so µ(x,q) = 1 for all q ∈ Q. Thus x ∈ I, so U(µ,t) ⊆ D. Now, let x ∈ D. Then µ(x,q) = 1 ≥ t for all q ∈ Q. Then x ∈ U(µ,t), so D ⊆ U(µ,t). Hence U(µ,t) = D for all t > k. Case 2 : To show that U(µ,t) = H for all t ≤ k, let t ∈ [0, 1] be such that t ≤ k. Clearly, U(µ,t) ⊆ H. Let x ∈ H. Then for all q ∈ Q, µ(x,q) = { 1 > t if x ∈ D k ≥ t if x /∈ D. Then x ∈ U(µ,t), so H ⊆ U(µ,t). Hence U(µ,t) = H for all t ≤ k. We claim that U(µ,t,q) = U(µ,t,q′) for all q,q′ ∈ Q. For q,q′ ∈ Q, we obtain x ∈ U(µ,t,q) ⇔ µ(x,q) ≥ t ⇔ µ(x,q′) ≥ t ⇔ x ∈ U(µ,t,q′). Hence U(µ,t,q) = U(µ,t,q′) for all q,q′ ∈ Q. Then U(µ,t) = ⋂ q∈Q U(µ,t,q). By the claim, we have U(µ,t) = U(µ,t,q) for all q ∈ Q. Since U(µ,t,q) = U(µ,t) = D for all t > k and U(µ,t) = U(µ,t) = H for all t ≤ k, it follows that µ is an anti-Q-fuzzy ideal of H. Then L(µ,t) = L(µ,t) = D for all t > k and L(µ,t) = L(µ,t) = H for all t ≤ k. Let µ = θ. Then θ is an anti-Q-fuzzy ideal of H such that L(µ,t) = L(µ,t) = D for all t > k and L(µ,t) = L(µ,t) = H for all t ≤ k. � Let (A, ·, 1A) and (B,?, 1B) be Hilbert algebras A mapping f : A → B is called a homomorphism if f (x∗y) = f (x)?f (y) for all x,y ∈ A. Note that if f : X → Y is a homomorphism of Hilbert algebras, then f (1A) = 1B. Let f : X → Y be a homomorphism of Hilbert algebras. Int. J. Anal. Appl. (2023), 21:42 11 Theorem 3.5. Let (A, ·, 1A) and (B,?, 1B) be Hilbert algebras and let f : A → B be a homomorphism. If µ is an anti-q-fuzzy deductive system of B, then µf is also a q-fuzzy deductive system of A. Proof. Assume that µ be an anti-q-fuzzy deductive system of B. Let x ∈ A. Then µf (1A,q) = µ(f (1A),q) = µ(1B,q) ≤ µ(f (x),q) = µf (x,q). Let x,y ∈ A. Then µf (y,q) = µ(f (y),q) ≤ max{µ(f (x∗y),q),µ(f (x),q)} = max{µf (x∗y,q),µf (x,q)}. Hence µf is an anti-q-fuzzy deductive system of A. � Corollary 3.3. Let (A, ·, 1A) and (B,?, 1B) be Hilbert algebras and let f : A → B be a homomorphism. If µ is an anti-Q-fuzzy deductive system of B, then µf is also an anti-Q-fuzzy deductive system of A. Theorem 3.6. Let (A, ·, 1A) and (B,?, 1B) be Hilbert algebras and let f : A → B be a isomorphism. If µf is an anti-q-fuzzy deductive system of A, then µ is also an anti-q-fuzzy deductive system of B. Proof. Assume that µf be an anti-q-fuzzy deductive system of A. Let y ∈ B. Then there exists x ∈ A such that f (x) = y, we have µ(1B,q) = µ(y ? 1B,q) = µ(f (x) ?f (1A),q) = µ(f (x ∗1A),q) = µf (x ∗ 1A,q) = µf (1A,q) ≤ µf (x,q) ≤ µ(f (x),q) = µ(y,q). Let x,y ∈ B. Then there exist a,b ∈ X such that f (a) = x and f (b) = y. It follows that µ(y,q) = µ(f (b),q) = µf (b,q) ≤ max{µf (a∗b,q),µf (a,q)} = max{µ(f (a∗b),q),µ(f (a),q)} = max{µ(f (a)?f (b),q),µ(f (a),q)} = max{µ(x ? y,q),µ(x,q)}. Hence µ is an anti-q-fuzzy deductive system of B. � Corollary 3.4. Let (A, ·, 1A) and (B,?, 1B) be Hilbert algebras and let f : A → B be a isomorphism. If µf is an anti-Q-fuzzy deductive system of A, then µ is also an anti-Q-fuzzy deductive system of B. Lemma 3.6. For any a,b,c,d ∈ R, the following properties hold: (1) max{max{a,b}, max{c,d}} = max{max{a,c}, max{b,d}} (2) min{min{a,b}, min{c,d}} = min{min{a,c}, min{b,d}}. Remark 3.1. Let (A, ·, 1A) and (B,?, 1B) be Hilbert algebras. Then A×B is a Hilbert algebra defined by (x,y) � (u,v) = (x ∗u,y ? v) for every x,y ∈ A and u,v ∈ B, then clearly (A×B,�, (1A, 1B)) is a Hilbert algebra. Theorem 3.7. Let (A, ·, 1A) and (B,?, 1B) be Hilbert algebras. If µ is an anti-q-fuzzy deductive system of A and δ is an anti-q-fuzzy deductive system of B, then µ× δ is an anti-q-fuzzy deductive system of A×B. Proof. Assume that µ is an anti-q-fuzzy deductive system of A and δ is an anti-q-fuzzy deduc- tive system B. Let (x,y) ∈ A × B. Then (µ × δ)((1A, 1B),q) = max{µ(1A,q),δ(1B,q)} ≤ 12 Int. J. Anal. Appl. (2023), 21:42 max{µ(x,q),δ(y,q)}} = (µ×δ)((x,y),q). Let (x1,y1), (x2,y2) ∈ A×B. Then (µ×δ)((x2,y2),q) = max{µ(x2,q),δ(y2,q)} ≤ max{max{µ(x1 ∗x2,q),µ(x1,q)}, max{δ(y1 ? y2,q),δ(y1,q)}} = max{max{µ(x1 ∗x2,q),δ(y1 ? y2,q)}, max{µ(x1,q),δ(y1,q)}} = max{(µ×δ)((x1 ∗x2,y1 ? y2),q), (µ ·δ)((x1,y1),q)} = max{(µ×δ)((x1 ∗y1) � (x2 ? y2),q), (µ ·δ)((x1,y1),q)}. Hence µ×δ is an anti-q-fuzzy deductive system of A×B. � Corollary 3.5. Let (A, ·, 1A) and (B,?, 1B) be Hilbert algebras. If µ is an anti-Q-fuzzy deductive system of A and δ is an anti-Q-fuzzy deductive system of B, then µ×δ is an anti-Q-fuzzy deductive system of A×B. Theorem 3.8. If µ is a Q-fuzzy set of A and δ is a Q-fuzzy set of B such that µ×δ is an anti-q-fuzzy deductive system of A×B, then the following statements hold: (1) either µ(1A,q) ≤ µ(x,q) for all x ∈ A or δ(1B,q) ≤ δ(x,q) for all x ∈ B, (2) if µ(1A,q) ≤ µ(x,q) for all x ∈ A, then either δ(1B,q) ≤ µ(x,q) for all x ∈ A or δ(1B,q) ≤ δ(x,q) for all x ∈ B, (3) if δ(1A,q) ≤ δ(x,q) for all x ∈ B, then either µ(1A,q) ≤ µ(x,q) for all x ∈ A or µ(1A,q) ≤ δ(x,q) for all x ∈ B. Proof. (1). Suppose that there exist x ∈ A and y ∈ B such that µ(1A,q) > µ(x,q) and δ(1B,q) > δ(y,q). Then (µ× δ)((x,y),q) = max{µ(x,q),δ(y,q)} < max{µ(1A,q),δ(1B,q)}} = (µ × δ)((1A, 1B),q), which is a contradiction. Hence µ(1A,q) ≤ µ(x,q) for all x ∈ A or δ(1B,q) ≤ δ(x,q) for all x ∈ B. (2). Assume that µ(1A,q) ≤ µ(x,q) for all x ∈ A. Suppose that there exist x ∈ A and y ∈ B such that µ(1A,q) > µ(x,q) and δ(1B,q) > δ(y,q). Then µ(1A,q) ≤ µ(x,q) < δ(1B,q). Thus (µ×δ)((x,y),q) = max{µ(x,q),δ(y,q)} < max{µ(1A,q),δ(1B,q)}} = δ(1B,q) = max{µ(1A,q),δ(1B,q)} = (µ×δ)((1A, 1B),q), which is a contradiction. Hence δ(1A,q) ≤ µ(x,q) for all x ∈ A or δ(1B,q) ≤ δ(x,q) for all x ∈ B. (3). Assume that δ(1A,q) ≤ δ(x,q) for all x ∈ B. Suppose that there exist x ∈ A and y ∈ B such Int. J. Anal. Appl. (2023), 21:42 13 that µ(1A,q) > µ(x,q) and µ(1A,q) > δ(y,q). Then δ(1B,q) ≤ δ(x,q) < µ(1A,q). Thus (µ×δ)((x,y),q) = max{µ(x,q),δ(y,q)} > max{µ(1A,q),µ(1A,q)}} = µ(1A,q) = max{µ(1A,q),δ(1B,q)} = (µ×δ)((1A, 1B),q), which is a contradiction. Hence µ(1A,q) ≤ µ(x,q) for all x ∈ A or µ(1B,q) ≤ δ(x,q) for all x ∈ B. � Corollary 3.6. If µ is a Q-fuzzy set of A and δ is a Q-fuzzy set of B such that µ×δ is an anti-Q-fuzzy deductive system of A×B, then the following statements hold: (1) for all q ∈ Q, either µ(1A,q) ≤ µ(x,q) for all x ∈ A or δ(1B,q) ≤ δ(x,q) for all x ∈ B, (2) for all q ∈ Q, if µ(1A,q) ≤ µ(x,q) for all x ∈ A, then either δ(1B,q) ≤ µ(x,q) for all x ∈ A or δ(1B,q) ≤ δ(x,q) for all x ∈ B, (3) for all q ∈ Q, if δ(1A,q) ≤ δ(x,q) for all x ∈ B, then either µ(1A,q) ≤ µ(x,q) for all x ∈ A or µ(1A,q) ≤ δ(x,q) for all x ∈ B. Theorem 3.9. Let (A, ·, 1A) and (B,?, 1B) be Hilbert algebras and let µ be a Q-fuzzy set in A and δ be a Q-fuzzy set in B. If µ× δ is an anti-q-fuzzy deductive system of A×B, then either µ is an anti-q-fuzzy deductive system of A or δ is an anti-q-fuzzy deductive system of B. Proof. Assume that µ× δ is an anti-q-fuzzy deductive system of A×B. Suppose that µ is not an anti-q-fuzzy deductive system of A and δ is not an anti-q-fuzzy deductive system of B. Then we have µ(1A,q) ≤ µ(x,q) for all x ∈ A or δ(1B,q) ≤ δ(x,q) for all x ∈ B. Suppose that µ(1A,q) ≤ µ(x,q) for all x ∈ A. Then either δ(1B,q) ≤ µ(x,q) for all x ∈ A or δ(1B,q) ≤ δ(x,q) for all x ∈ B. If δ(1B,q) ≤ µ(x,q) for all x ∈ A, then (µ× δ)((x, 1B),q) = max{µ(x,q),δ(1B,q)} = µ(x,q). We consider, for all x,y ∈ A, µ(y,q) = max{µ(y,q),δ(1B,q)} = (µ×δ)((y, 1B),q) ≤ max{(µ×δ)((x, 1B) � (y, 1B),q), (µ×δ)((x, 1B),q)} = max{(µ×δ)((x ∗y, 1B ? 1B),q), (µ×δ)((x, 1B),q)} = max{(µ×δ)((x ∗y, 1B),q), (µ×δ)((x, 1B),q)} = max{max{µ(x ∗y,q),δ(1B,q)}, max{µ(x,q),δ(1B,q)}} = max{µ(x ∗y,q),µ(x,q)}. Hence µ is an anti-q-fuzzy deductive system of A, which is a contradiction. Suppose that δ(1B,q) ≤ δ(x,q) for all x ∈ B. Then either µ(1A,q) ≤ µ(x,q) for all x ∈ B or µ(1A,q) ≤ δ(x,q) for all x ∈ B. 14 Int. J. Anal. Appl. (2023), 21:42 If µ(1A,q) ≤ δ(x,q) for all x ∈ B, then (µ · δ)((1A,x),q) = max{µ(1A,q),δ(x,q)} = δ(x,q). We consider, for all x,y ∈ B, δ(y,q) = max{µ(1A,q),δ(y,q)} = (µ×δ)((1A,y),q) ≤ max{(µ×δ)((1A,x) � (1A,y),q), (µ×δ)((1A,x),q)} = max{(µ×δ)((1A ∗ 1A,x ∗y),q), (µ×δ)((1A,x),q)} = max{(µ×δ)((1A,x ∗y),q), (µ×δ)((1A,x),q)} = max{min{µ(1A,q),δ(x ∗y,q)}, max{µ(1A,q),δ(x,q)}} = max{δ(x ∗y,q),δ(x,q)}. Hence δ is an anti-q-fuzzy deductive system of B, which is a contradiction. Since µ is not an anti-q-fuzzy deductive system of A and δ is not an anti-q-fuzzy deductive system of B, we have µ(1A,q) ≤ µ(x,q) for all x ∈ A and δ(1B,q) ≤ δ(x,q) for all x ∈ B. Let x1,x2 ∈ A and y1,y2 ∈ B such that µ(x2,q) > max{µ(x1 ∗ x2,q),µ(x1,q)} and δ(y2,q) > max{δ(y1 ? y2,q),δ(y1,q)}, so max{µ(x2,q),δ(y2,q)} > max{max{µ(x1 ∗x2,q),µ(x1,q)}, max{δ(y1 ? y2,q),δ(y1,q)}}. 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