Int. J. Anal. Appl. (2022), 20:25 Interval-Valued Intuitionistic Fuzzy Subalgebras/Ideals of Hilbert Algebras Aiyared Iampan1,∗, V. Vijaya Bharathi2, M. Vanishree3, N. Rajesh3 1Fuzzy Algebras and Decision-Making Problems Research Unit, Department of Mathematics, School of Science, University of Phayao, Mae Ka, Mueang, Phayao 56000, Thailand 2No. 42, Cauvery Nagar East, Thanjavur-613005, Tamilnadu, India 3Department of Mathematics, Rajah Serfoji Government College (affliated to Bharathidasan University), Thanjavur-613005, Tamilnadu, India ∗Corresponding author: aiyared.ia@up.ac.th Abstract. In this paper, the concept of interval-valued intuitionistic fuzzy sets to subalgebras and ideals of Hilbert algebras is introduced. The inverse image of interval-valued intuitionistic fuzzy subalgebras and interval-valued intuitionistic fuzzy ideals of Hilbert algebras is studied and some related properties are investigated. Equivalence relations on interval-valued intuitionistic fuzzy ideals are discussed. 1. Introduction The concept of fuzzy sets was proposed by Zadeh [15]. 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, 3, 4]. 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 multi-criteria decision-making [6–8]. The concept of Hilbert algebras was introduced in early 50-ties by Henkin and Skolem for some investigations of implication in intuitionistic and other non-classical logics. In 60-ties, these algebras were studied especially by Horn and Diego from algebraic point of view. Diego proved Received: Apr. 8, 2022. 2010 Mathematics Subject Classification. 20N05, 94D05, 03E72. Key words and phrases. Hilbert algebra; interval-valued intuitionistic fuzzy subalgebra; interval-valued intuitionistic fuzzy ideal. https://doi.org/10.28924/2291-8639-20-2022-25 ISSN: 2291-8639 © 2022 the author(s). https://doi.org/10.28924/2291-8639-20-2022-25 2 Int. J. Anal. Appl. (2022), 20:25 (cf. [11]) that Hilbert algebras form a variety which is locally finite. Hilbert algebras were treated by Busneag (cf. [9], [10]) and Jun (cf. [13]) and some of their filters forming deductive systems were rec- ognized. Dudek (cf. [12]) considered the fuzzification of subalgebras and deductive systems in Hilbert algebras. In this paper, the concept of interval-valued intuitionistic fuzzy sets to subalgebras and ideals of Hilbert algebras is introduced. The inverse image of interval-valued intuitionistic fuzzy subalgebras and interval-valued intuitionistic fuzzy ideals of Hilbert algebras is studied and some related properties are investigated. Equivalence relations on interval-valued intuitionistic fuzzy ideals are discussed. 2. Preliminaries Before we begin the study, let’s review the definition of Hilbert algebras, which was defined by Diego [11] in 1966. Definition 2.1. A Hilbert algebra is a triplet X = (X, ·, 1), where H is a nonempty set, · is a binary operation, and 1 is a fixed element of X such that the following axioms hold: (1) (∀x,y ∈ X)(x · (y ·x) = 1), (2) (∀x,y,z ∈ X)((x · (y ·z)) · ((x ·y) · (x ·z)) = 1), (3) (∀x,y ∈ X)(x ·y = 1,y ·x = 1 ⇒ x = y). The following result was proved in [12]. Lemma 2.1. Let X = (X, ·, 1) be a Hilbert algebra. Then (1) (∀x ∈ X)(x ·x = 1), (2) (∀x ∈ X)(1 ·x = x), (3) (∀x ∈ X)(x · 1 = 1), (4) (∀x,y,z ∈ X)(x · (y ·z) = y · (x ·z)). In a Hilbert algebra X = (X, ·, 1), the binary relation ≤ is defined by (∀x,y ∈ X)(x ≤ y ⇔ x ·y = 1), which is a partial order on X with 1 as the largest element. Definition 2.2. [14] A nonempty subset I of a Hilbert algebra X = (X, ·, 1) is called an ideal of X if (1) 1 ∈ I, (2) (∀x ∈ X,∀y ∈ I)(x ·y ∈ I), (3) (∀x ∈ X,∀y1,y2 ∈ I)((y1 · (y2 ·x)) ·x ∈ I). A fuzzy set [15] in a nonempty set X is defined to be a function µ : X → [0, 1], where [0, 1] is the unit closed interval of real numbers. Int. J. Anal. Appl. (2022), 20:25 3 Definition 2.3. [12] A fuzzy set µ in a Hilbert algebra X = (X, ·, 1) is said to be a fuzzy subalgebra of X if the following condition holds: (∀x,y ∈ X)(µ(x ·y) ≥ min{µ(x),µ(y)}). Definition 2.4. [5] A fuzzy set µ in a Hilbert algebra X = (X, ·, 1) is said to be a fuzzy ideal of X if the following conditions hold: (1) (∀x ∈ X)(µ(1) ≥ µ(x)), (2) (∀x,y ∈ X)(µ(x ·y) ≥ µ(y)), (3) (∀x,y1,y2 ∈ X)(µ((y1 · (y2 ·x)) ·x) ≥ min{µ(y1),µ(y2)}). Definition 2.5. [2] Let X be a nonempty set. An intuitionistic fuzzy set A in X is defined to be a structure A := {(x,µA(x),γA(x)) | x ∈ X}, (2.1) where µA : X → [0, 1] is the degree of membership of x to [0, 1] and γA : X → [0, 1] is the degree of non-membership of x to [0, 1] such that (∀x ∈ X)(0 ≤ µA(x) + γA(x) ≤ 1), and the intuitionistic fuzzy set A in (2.1) is simply denoted by A = (µA,γA). Let D[0, 1] be the set of all closed subintervals of [0, 1]. Consider two elements D1,D2 ∈ D[0, 1]. If D1 = [a1,b1] and D2 = [a2,b2], then rmin{D1,D2} = [min{a1,a2}, min{b1,b2}] and rmax{D1,D2} = [max{a1,a2}, max{b1,b2}]. If Di = [ai,bi] ∈ D[0, 1] for i = 1, 2, . . ., then we define rsupi{Di} = [sup i {ai}, sup i {bi}]. Similarly, we define rinfi{Di} = [inf i {ai}, inf i {bi}]. Now we call D1 ≥ D2 if and only if a1 ≥ a2 and b1 ≤ b2. Similarly, the relations D1 ≤ D2 and D1 = D2 are defined. Definition 2.6. An interval-valued intuitionistic fuzzy (IVIF) set A in X is an object having the form A = {(x,µA(x),γA(x)) | x ∈ X}, where µA : X → D[0, 1] and γA : X → D[0, 1]. The intervals µA(x) and γA(x) denote the intervals of the degree of belongingness and non-belongingness of the element 4 Int. J. Anal. Appl. (2022), 20:25 x to the set D[0, 1], where µA(x) = [µlA(x),µ u A(x)] and γA(x) = [γ l A(x),γ u A(x)] for all x ∈ X with the following condition: (∀x ∈ X)(0 ≤ µuA(x) + γ u A(x) ≤ 1). For the sake of simplicity, we shall use the symbol A = (µA,γA) for the IVIF set A = {(x,µA(x),γA(x)) | x ∈ X}. Also note that µA(x) = [1 − µuA(x), 1 − µ l A(x)] and γA(x) = [1 −γuA(x), 1 −γ l A(x)] for all x ∈ X, where [µA(x),γA(x)] represents the complement of x in A. 3. IVIF subalgebras of Hilbert algebras Definition 3.1. An IVIF set A = (µA,γA) in a Hilbert algebra X is called an IVIF subalgebra of X if (∀x,y ∈ X) ( µA(x ·y) ≥ rmin{µA(x),µA(y)} γA(x ·y) ≤ rmax{γA(x),γA(y)} ) . (3.1) Example 3.1. Let X = {1,x,y,z, 0} with the following Cayley table: · 1 x y z 0 1 1 x y z 0 x 1 1 y z 0 y 1 x 1 z z z 1 1 y 1 y 0 1 1 1 1 1 Then X is a Hilbert algebra. We define an IVIF set A = (µA,γA) as follows: µA(a) = { [0.5, 0.6] if a ∈{1,x,y,z} [0.1, 0.2] if a = 0 and γA(a) = { [0.3, 0.4] if a ∈{1,x,y,z} [0.4, 0.5] if a = 0. Then A is an IVIF subalgebra of X. Proposition 3.1. Every IVIF subalgebra A = (µA,γA) of a Hilbert algebra X satisfies µA(1) ≥ µA(x) and γA(1) ≤ γA(x) for all x ∈ X, where µA(1) and γA(1) are the upper bound and lower bound of µA(x) and γA(x), respectively. Proof. For any x ∈ X, we have µA(1) = µA(x ·x) ≥ rmin{µA(x),µA(x)} = rmin{[µlA(x),µ u A(x)], [µ l A(x),µ u A(x)]} = [µlA(x),µ u A(x)] = µA(x) Int. J. Anal. Appl. (2022), 20:25 5 and γA(1) = γA(x ·x) ≤ rmax{γA(x),γA(x)} = rmax{[γlA(x),γ u A(x)], [γ l A(x),γ u A(x)]} = [γlA(x),γ u A(x)] = γA(x). � Proposition 3.2. If an IVIF set A = (µA,γA) in a Hilbert algebra X is an IVIF subalgebra, then (∀x ∈ X) ( µA(1 ·x) ≥ µA(x) γA(1 ·x) ≤ γA(x) ) . (3.2) Proof. For any x ∈ X, we have µA(1 ·x) ≥ rmin{µA(1),µA(x)} = rmin{µA(x ·x),µA(x)} ≥ rmin{rmin{µA(x),µA(x)},µA(x)} = µA(x) and γA(1 ·x) ≤ rmax{γA(1),γA(x)} = rmax{γA(x ·x),γA(x)} ≤ rmax{rmax{γA(x),γA(x)},γA(x)} = γA(x). � Theorem 3.1. An IVIF set A = (µA,γA) = ([µlA,µ u A], [γ l A,γ u A]) in a Hilbert algebra X is an IVIF subalgebra of X if and only if µlA,µ u A,γ l A, and γ u A are fuzzy subalgebras of X. Proof. Let µlA and µ u A be fuzzy subalgebras of X and x,y ∈ X. Then µ l A(x ·y) ≥ min{µ l A(x),µ l A(y)} and µuA(x ·y) ≤ min{µ u A(x),µ u A(y)}. Now, µA(x ·y) = [µlA(x ·y),µ u A(x ·y)] ≥ [min{µlA(x),µ l A(y)}, min{µ u A(x),µ u A(y)}] = rmin{[µlA, (x),µ u A(x)], [µ l A(y),µ u A(y)]} = rmin{µA(x),µA(y)}. Again, let γlA and γ u A be fuzzy subalgebras of X and x,y ∈ X. Then γ l A(x · y) ≤ max{γ l A(x),γ l A(y)} and γuA(x ·y) ≤ max{γ u A(x),γ u A(y)}. Now, γA(x ·y) = [γlA(x ·y),γ u A(x ·y)] ≤ [max{γlA(x),γ l A(y)}, max{γ u A(x),γ u A(y)}] = rmax{[γlA, (x),γ u A(x)], [γ l A(y),γ u A(y)]} = rmax{γA(x),γA(y)}. 6 Int. J. Anal. Appl. (2022), 20:25 Hence, A = {[µlA,µ u A], [γ l A,γ u A]} is an IVIF subalgebra of X. Conversely, assume that A is an IVIF subalgebra of X. For any x,y ∈ X, [µlA(x ·y),µ u A(x ·y)] = µA(x ·y) ≥ rmin{µA(x),µA(y)} = rmin{[µlA(x),µ u A(x)], [µ l A(y),µ u A(y)]} = [min{µlA(x),µ l A(y)}, min{µ u A(x),µ u A(y)}] and [γlA(x ·y),γ u A(x ·y)] = γA(x ·y) ≤ rmax{γA(x),γA(y)} = rmax{[γlA(x),γ u A(x)], [γ l A(y),γ u A(y)]} = [max{γlA(x),γ l A(y)}, max{γ u A(x),γ u A(y)}]. Thus µlA(x ·y) ≥ min{µ l A(x),µ l A(y)},µ u A(x ·y) ≥ min{µ u A(x),µ u A(y)},γ l A(x ·y) ≤ max{γ l A(x),γ l A(y)}, and γuA(x ·y) ≤ max{γ u A(x),γ u A(y)}. Therefore, µ l A,µ u A,γ l A, and γ u A are fuzzy subalgebras of X. � Theorem 3.2. If A = (µA,γA) and B = (µB,γB) are two IVIF subalgebras of a Hilbert algebra X, then A∩B = (µA∩B,γA∪B) is an IVIF subalgebra of X. Proof. Let x,y ∈ X. Since A and B are IVIF subalgebras of X, by Theorem 3.1, we have µA∩B(x ·y) = [µlA∩B(x ·y),µ u A∩B(x ·y)] = [min{µlA(x ·y),µ l B(x ·y)}, min{µ u A(x ·y),µ u B(x ·y)}] ≥ [min{µlA∩B(x),µ l A∩B(y)}, min{µ u A∩B(x),µ u A∩B(y)}] = rmin{µA∩B(x),µA∩B(y)} and γA∪B(x ·y) = [γlA∪B(x ·y),γ u A∪B(x ·y)] = [max{γlA(x ·y),γ l B(x ·y)}, max{γ u A(x ·y),γ u B(x ·y)}] ≤ [max{γlA∪B(x),γ l A∪B(y)}, max{γ u A∪B(x),γ u A∪B(y)}] = rmax{γA∪B(x),γA∪B(y)}. Hence, A∩B = (µA∩B,γA∪B) is an IVIF subalgebra of X. � Corollary 3.1. Let {Ai | i = 1, 2, 3, · · ·} be a family of IVIF subalgebras of a Hilbert algebra X. Then ∞ ∩ i=1 Ai is also an IVIF subalgebra of X, where ∞ ∩ i=1 Ai = {(x, rminµAi (x), rmaxγAi (x)) | x ∈ X}. For any elements x and y of a Hilbert algebra X, let n∏ x ·y denotes the expression x ·(· · ·(x ·(x ·y))), where x occurred n times. Theorem 3.3. Let A = (µA,γA) be an IVIF subalgebra of a Hilbert algebra X and let n ∈N. Then (1) µA ( n∏ x ·x ) ≥ µA(x) for any odd number n, (2) γA ( n∏ x ·x ) ≤ γA(x) for any odd number n, Int. J. Anal. Appl. (2022), 20:25 7 (3) µA ( n∏ x ·x ) = µA(x) for any even number n, (4) γA ( n∏ x ·x ) = γA(x) for any even number n. Proof. Let x ∈ X and assume that n is odd. Then n = 2p− 1 for some positive integer p. We prove the theorem by induction. Now, µA(x ·x) = µA(1) ≥ µA(x) and γA(x ·x) = γA(1) ≤ γA(x). Suppose that µA ( 2p−1∏ x ·x ) ≥ µA(x) and γA ( 2p−1∏ x ·x ) ≤ γA(x). Then by assumption, µA ( 2(p+1)−1∏ x ·x ) = µA ( 2p+1∏ x ·x ) = µA ( 2p−1∏ x · (x · (x ·x)) ) = µA ( 2p−1∏ x ·x ) ≥ µA(x) and γA ( 2(p+1)−1∏ x ·x ) = γA ( 2p+1∏ x ·x ) = γA ( 2p−1∏ x · (x · (x ·x)) ) = γA ( 2p−1∏ x ·x ) ≤ γA(x), which proves (1) and (2). Proofs are similar for the cases (3) and (4). � Definition 3.2. Let A = (µA,γA) be an IVIF set defined in a Hilbert algebra X. The IVIF sets ⊕A and ⊗A are defined as ⊕A = {(x,µA(x),µA(x)) | x ∈ X} and ⊗A = {(x,γA(x),γA(x)) | x ∈ X}. Theorem 3.4. If A = (µA,γA) is an IVIF subalgebra of a Hilbert algebra X, then ⊕A and ⊗A both are IVIF subalgebras. Proof. Let x,y ∈ X. Then µA(x ·y) = [1, 1] −µA(x ·y) ≤ [1, 1] − rmin{µA(x),µA(y)} = rmax{1 − µA(x), 1 − µA(y)} = rmax{µA(x),µA(y)}. Hence, ⊕A is an IVIF subalgebra of X. Let x,y ∈ X. Then γA(x · y) = [1, 1] −γA(x · y) ≥ [1, 1] − rmax{γA(x),γA(y)} = rmin{1 −γA(x), 1 −γA(y)} = rmin{γA(x),γA(y)}. Hence, ⊗A is also an IVIF subalgebra of X. � The sets {x ∈ X | µA(x) = µA(1)} and {x ∈ X | γA(x) = γA(1)} are denoted by µ1A and γ 1 A, respectively. These two sets are also subalgebra of a Hilbert algebra X. Theorem 3.5. Let A = (µA,γA) be an IVIF subalgebra of a Hilbert algebra X, then the sets µ1A and γ1A are subalgebras of X. 8 Int. J. Anal. Appl. (2022), 20:25 Proof. Let x,y ∈ µ1A. Then µA(x) = µA(1) = µA(y) and so µA(x ·y) ≤ rmin{µA(x),µA(y)} = µA(1). By using Proposition 3.1, we have µA(x ·y) = µA(1); hence, x ·y ∈ µ1A. Again, let x,y ∈ γ 1 A. Then γA(x) = γA(1) = γA(y) and so, γA(x · y) ≤ rmax{γA(x),γA(y)} = γA(1). Again, by Proposition 3.1, we have γA(x ·y) = γA(1); hence, x ·y ∈ γ1A. Therefore, the sets µ 1 A and γ 1 A are subalgebras of X. � Theorem 3.6. Let B be a nonempty subset of a Hilbert algebra X and A = (µA,γA) be an IVIF set in X defined by µA(x) = { [α1,α2] if x ∈ B [β1,β2] otherwise and γA(x) = { [θ1,θ2] if x ∈ B [δ1,δ2] otherwise for all [α1,α2], [β1,β2], [θ1,θ2], [δ1,δ2] ∈ D[0, 1] with [α1,α2] ≥ [β1,β2] and [θ1,θ2] ≤ [δ1,δ2] and α2 +θ2 ≤ 1 and β2 +δ2 ≤ 1. Then A is an IVIF subalgebra of X if and only if B is a subalgebra of X. Moreover, µ1A = B = γ 1 A. Proof. Let A be an IVIF subalgebra of X. Let x,y ∈ B. Then µA(x ·y) ≥ rmin{µA(x),µA(y)} = rmin{[α1,α2], [α1,α2]} = [α1,α2] and γA(x ·y) ≤ rmax{γA(x),γA(y)} = rmax{[α1,α2], [α1,α2]} = [α1,α2]. So x ·y ∈ B. Hence, B is a subalgebra of X. Conversely, suppose that B is a subalgebra of X. Let x,y ∈ X. Consider two cases: Case (i): If x,y ∈ B, then x ·y ∈ B. Thus µA(x ·y) = [α1,α2] = rmin{µA(x),µA(y)} and γA(x ·y) = [θ1,θ2] = rmax{γA(x),γA(y)}. Case (ii): If x /∈ B or y /∈ B, then µA(x · y) ≥ [β1,β2] = rmin{µA(x),µA(y)} and γA(x · y) ≤ [θ1,θ2] = rmax{γA(x),γA(y)}. Hence, A is an IVIF subalgebra of X. Now, µ1A = {x ∈ X | µA(x) = µA(1)} = {x ∈ X | µA(x) = [α1,α2]} = B and γ1A = {x ∈ X | γA(x) = γA(1)} = {x ∈ X | γA(x) = [θ1,θ2]} = B. � Definition 3.3. Let A = (µA,γA) be an IVIF subalgebra of a Hilbert algebra X. For [s1,s2], [t1,t2] ∈ D[0, 1], the sets U(µA : [s1,s2]) = {x ∈ X | µA(x) ≥ [s1,s2]} is called an upper [s1,s2]-level of A and L(γA : [t1,t2]) = {x ∈ X | γA(x) ≤ [t1,t2]} is called a lower [t1,t2]-level of A. Theorem 3.7. If A = (µA,γA) is an IVIF subalgebra of a Hilbert algebra X, then the upper [s1,s2]-level and lower [t1,t2]-level of A are subalgebras of X. Int. J. Anal. Appl. (2022), 20:25 9 Proof. Let x,y ∈ U(µA : [s1,s2]). Then µA(x) ≤ [s1,s2] and µA(y) ≤ [s1,s2]. It follows that µA(x · y) ≤ rmin{µA(x),µA(y)} ≤ [s1,s2] so that x · y ∈ U(µA : [s1,s2]). Hence, U(µA : [s1,s2]) is a subalgebra of X. Let x,y ∈ L(γA : [t1,t2]). Then γA(x) ≤ [t1,t2] and γA(y) ≤ [t1,t2]. Thus γA(x · y) ≤ rmax{γA(x),γA(y)} ≤ [t1,t2] so that x · y ∈ L(γA : [t1,t2]). Hence, L(γA : [t1,t2]) is a subalgebra of X. � Theorem 3.8. Let A = (µA,γA) be an IVIF set in a Hilbert algebra X such that the sets U(µA : [s1,s2]) and L(γA : [t1,t2]) are subalgebras of X for every [s1,s2], [t1,t2] ∈ D[0, 1]. Then A is an IVIF subalgebra of X. Proof. Let [s1,s2], [t1,t2] ∈ D[0, 1] be such that U(µA : [s1,s2]) and L(γA : [t1,t2]) are subal- gebras of X. In contrary, let x0,y0 ∈ X be such that µA(x0 · y0) < rmin{µA(x0),µA(y0)}. Let µA(x0) = [θ1,θ2],µA(y0) = [θ3,θ4], and µA(x0·y0) = [s1,s2]. Then [s1,s2] < rmin{[θ1,θ2], [θ3,θ4]} = [min{θ1,θ3}, min{θ2,θ4}]. So, s1 < min{θ1,θ3} and s2 < min{θ2,θ4}. Consider, [ρ1,ρ2] = 1 2 [µA(x0 ·y0) + rmin{µA(x0),µA(y0)}] = 1 2 [[s1,s2] + [min{θ1,θ3}, min{θ2,θ4}]] = [ 1 2 (s1 + min{θ1,θ3}), 12 (s2 + min{θ2,θ4})]. Therefore, min{θ1,θ3} > ρ1 = 12 (s1 +min{θ1,θ3}) > s1 and min{θ2,θ4} > ρ2 = 1 2 (s2 +min{θ2,θ4}) > s2. Hence, [min{θ1,θ3}, min{θ2,θ4}] > [ρ1,ρ2] > [s1,s2], so that x0 ·y0 /∈ U(µA : [s1,s2]), which is a contradiction because µA(x0) = [θ1,θ2] ≥ [min{θ1,θ3}, min{θ2,θ4}] > [ρ1,ρ2] and µA(y0) = [θ3,θ4] ≥ [min{θ1,θ3}, min{θ2,θ4}] > [ρ1,ρ2]. This implies that x0 · y0 ∈ U(µA : [s1,s2]). Thus µA(x · y) ≤ rmin{µA(x),µA(y)} for all x,y ∈ X. Again, in contrary, let x0,y0 ∈ X be such that γA(x0 · y0) > rmax{γA(x0),γA(y0)}. Let γA(x0) = [η1,η2],γA(y0) = [η3,η4], and γA(x0 · y0) = [t1,t2]. Then [t1,t2] > rmax{[η1,η2], [η3,η4]} = [max{η1,η3}, max{η2,η4}]. So t1 > max{η1,η3} and t2 > max{η2,η4}. Let us consider, [β1,β2] = 1 2 [γA(x0 ·y0) + rmax{γA(x0),γA(y0)}] = 1 2 [[t1,t2] + [max{η1,η3}, max{η2,η4}] = [ 1 2 (t1 + max{η1,η3}), 12 (t2 + max{η2,η4})]. Therefore, max{η1,η3} < β1 = 12 [(t1 + max{η1,η3})] < t1 and max{η2,η4} < β2 = 1 2 [(t2 + max{η2,η4})] < t2. Hence, [max{η1,η3}, max{η2,η4}] < [β1,β2] < [t1,t2] so that x0 · y0 /∈ L(γA : [t1,t2]), which is a contradiction because γA(x0) = [η1,η2] ≤ [max{η1,η3}, max{η2,η4}] > [β1,β2] 10 Int. J. Anal. Appl. (2022), 20:25 and γA(y0) = [η3,η4] ≥ [max{η1,η3}, max{η2,η4}] > [β1,β2]. This implies that x0 · y0 ∈ L(γA : [t1,t2]). Thus γA(x · y) ≥ rmax{γA(x),γA(y)} for all x,y ∈ X. Therefore, A is an IVIF subalgebra of X. � Theorem 3.9. Any subalgebra of a Hilbert algebra X can be realized as both the upper [s1,s2]-level and lower [t1,t2]-level of some IVIF subalgebra of X. Proof. Let B be an IVIF subalgebra of X and A be an IVIF set on X defined by µA(x) ={ [α1,α2] if x ∈ B [0, 0] otherwise and γA(x) = { [β1,β2] if x ∈ B [1, 1] otherwise for all [α1,α2], [β1,β2] ∈ D[0, 1] and α2 + β2 ≤ 1. We consider the following cases: Case (i) If x,y ∈ B, then µA(x) = [α1,α2],γA(x) = [β1,β2], and µA(y) = [α1,α2], γA(y) = [β1,β2]. Thus µA(x ·y) = [α1,α2] = rmin{[α1,α2], [α1,α2]} = rmin{µA(x),µA(y)} and γA(x ·y) = [β1,β2] = rmax{[β1,β2], [β1,β2]} = rmax{γA(x),γA(y)}. Case (ii) If x ∈ B and y /∈ B, then µA(x) = [α1,α2],γA(x) = [β1,β2], and µA(y) = [0, 0], γA(y) = [1, 1]. Thus µA(x ·y) ≥ [0, 0] = rmin{[α1,α2], [0, 0]} = rmin{µA(x),µA(y)} and γA(x ·y) ≥ [1, 1] = rmax{[β1,β2], [1, 1]} = rmax{γA(x),γA(y)}. Case (iii) If x /∈ B and y ∈ B, then µA(x) = [0, 0],γA(x) = [1, 1],µA(y) = [α1,α2], and γA(y) = [β1,β2]. Thus µA(x ·y) ≥ [0, 0] = rmin{[0, 0], [α1,α2]} = rmin{µA(x),µA(y)} and γA(x ·y) ≤ [1, 1] = rmax{[1, 1], [β1,β2]} = rmax{γA(x),γA(y)}. Case (iv) If x /∈ B and y /∈ B, then µA(x) = [0, 0],γA(x) = [1, 1],µA(y) = [0, 0], and γA(y) = [1, 1]. Thus µA(x ·y) ≤ [0, 0] = rmin{[0, 0], [0, 0]} = rmin{µA(x),µA(y)} and γA(x ·y) ≥ [1, 1] = rmax{[1, 1], [1, 1]} = rmax{γA(x),γA(y)}. Therefore, A is an IVIF subalgebra of X. � Int. J. Anal. Appl. (2022), 20:25 11 Theorem 3.10. Let B be a subset of a Hilbert algebra X and A be an IVIF set in X defined by µA(x) ={ [α1,α2] if x ∈ B [0, 0] otherwise and γA(x) = { [β1,β2] if x ∈ B [1, 1] otherwise for all [α1,α2], [β1,β2] ∈ D[0, 1] and α2 + β2 ≤ 1. If A is realized as a lower level subalgebra and an upper level subalgebra of some IVIF subalgebra of X, then B is a subalgebra of X. Proof. Let A be an IVIF subalgebra of X and x,y ∈ B. Then µA(x) = [α1,α2] = µA(y) and γA(x) = [β1,β2] = γA(y). Thus µA(x ·y) ≤ rmin{µA(x),µA(y)} = rmin{[α1,α2], [α1,α2]} = [α1,α2] and γA(x ·y) ≥ rmax{γA(x),γA(y)} = rmax{[β1,β2], [β1,β2]} = [β1,β2], which imply that x ·y ∈ B. Hence, B is a subalgebra of X. � 4. IVIF ideals of Hilbert algebras Definition 4.1. An IVIF set A = (µA,γA) in a Hilbert algebra X is said to be an IVIF ideal of X if the following conditions are hold: (∀x ∈ X) ( µA(1) ≥ µA(x) γA(1) ≤ γA(x) ) , (4.1) (∀x,y ∈ X) ( µA(x ·y) ≥ µA(y) γA(x ·y) ≤ γA(y) ) , (4.2) (∀x,y1,y2 ∈ X) ( µA((y1 · (y2 ·x)) ·x) ≥ rmin{µA(y1),µA(y2)} γA((y1 · (y2 ·x)) ·x) ≤ rmax{γA(y1),γA(y2)} ) . (4.3) Example 4.1. Let X = {1,x,y,z, 0} with the following Cayley table: · 1 x y z 0 1 1 x y z 0 x 1 1 y z 0 y 1 x 1 z z z 1 1 y 1 y 0 1 1 1 1 1 Then X is a Hilbert algebra. We define an IVIF set A = (µA,γA) as follows: µA(x) = { [0.5, 0.6] if x ∈{1,x,y,z} [0.1, 0.2] if x = 0 and γA(x) = { [0.3, 0.4] if x ∈{1,x,y,z} [0.4, 0.5] if x = 0. 12 Int. J. Anal. Appl. (2022), 20:25 Then A is an IVIF ideal of X. Proposition 4.1. If A = (µA,γA) is an IVIF ideal of a Hilbert algebra X, then (∀x,y ∈ X) ( µA((y ·x) ·x) ≥ µA(y) γA((y ·x) ·x) ≤ γA(y) ) . (4.4) Proof. Putting y1 = y and y2 = 1 in (4.3), we have µA((y ·x) ·x) ≥ rmin{µA(y),µA(1)} = µA(y) and γA((y ·x) ·x) ≤ rmax{γA(y),γA(1)} = γA(y). � Lemma 4.1. If A = (µA,γA) is an IVIF ideal of a Hilbert algebra X, then (∀x,y ∈ X) ( x ≤ y ⇒ { µA(x) ≤ µA(y) γA(x) ≥ γA(y) ) . (4.5) Proof. Let x,y ∈ X be such that x ≤ y. Then x ·y = 1 and so µA(y) = µA(1 ·y) = µA(((x ·y) · (x ·y)) ·y) ≥ rmin{µA(x ·y),µA(x)} ≥ rmin{µA(1),µA(x)} = µA(x). Thus γA(y) = γA(1 ·y) = γA(((x ·y) · (x ·y)) ·y) ≤ rmax{γA(x ·y),γA(x)} ≤ rmax{γA(1),γA(x)} = γA(x). � Theorem 4.1. Every IVIF ideal of a Hilbert algebra X is an IVIF subalgebra of X. Proof. Let A = (µA,γA) be an IVIF ideal of X. Since y ≤ x ·y for all x,y ∈ X, it follows from Lemma 4.1 that µA(y) ≥ µA(x ·y),γA(y) ≤ γA(x ·y). It follows from (4.2) that µA(x ·y) ≥ µA(y) ≥ rmin{µA(x ·y),µA(x)} ≥ rmin{µA(x),µA(y)} Int. J. Anal. Appl. (2022), 20:25 13 and γA(x ·y) ≤ γA(y) ≤ rmax{γA(x ·y),γA(x)} ≤ rmax{γA(x),γA(y)}. Hence, A is an IVIF subalgebra of X. � Theorem 4.2. An IVIF set A = (µA,γA) = {[µlA,µ u A], [γ l A,γ u A]} in a Hilbert algebra X is an IVIF ideal of X if and only if µlA,µ u A,γ l A, and γ u A are fuzzy ideals of X. Proof. Since µlA(1) ≥ µ l A(x),µ u A(1) ≥ µ u A(x),γ l A(1) ≤ γ l A(x), and γ u A(1) ≤ γ u A(x), we have µA(1) ≥ µA(x) and γA(1) ≤ γA(x). Let x,y ∈ X. Then µA(x ·y) = [µlA(x ·y),µ u A(x ·y)] ≥ [µ l A(y),µ u A(y)] = µA(y) and γA(x ·y) = [γlA(x ·y),γ u A(x ·y)] ≤ [γ l A(y),γ u A(y)] = γA(y). Let x,y1,y2 ∈ X. Then µA((y1 · (y2 ·x)) ·x) = [µlA((y1 · (y2 ·x)) ·x),µ u A((y1 · (y2 ·x)) ·x)] ≥ [min{µlA(y1),µ l A(y2)}, min{µ u A(y1),µ u A(y2)}] = rmin{[µlA(y1),µ u A(y1)], [µ l A(y2),µ u A(y2)]} = rmin{µA(y1),µA(y2)} and γA((y1 · (y2 ·x)) ·x) = [γlA((y1 · (y2 ·x)) ·x),γ u A((y1 · (y2 ·x)) ·x)] ≤ [max{γlA(y1),γ l A(y2)}, max{γ u A(y1),γ u A(y2)}] = rmax{[γlA(y1),γ u A(y1)], [γ l A(y2),γ u A(y2)]} = rmax{γA(y1),γA(y2)}. Hence, A = {[µlA,µ u A], [γ l A,γ u A]} is an IVIF ideal of X. Conversely, assume that A is an IVIF ideal of X. Let x ∈ X. Then [µlA(1),µ u A(1)] = µA(1) ≥ µA(x) = [µ l A(x),µ u A(x)]; hence, µ l A(1) ≥ µ l A(x) and γ l A(1) ≤ γ l A(x). Let x,y ∈ X. Then [µ l A(x · y),µuA(x ·y)] = µA(x ·y) ≥ µA(y) = [µ l A(y),µ u A(y)]; hence, µ l A(x ·y) ≥ µ l A(y) and µ u A(x ·y) ≥ µ u A(y). Also, [γlA(x · y),γ u A(x · y)] = γA(x · y) ≤ γA(y) = [γ l A(y),γ u A(y)]; hence, γ l A(x · y) ≤ γ l A(y) and γuA(x ·y) ≤ γ u A(y). Let x,y1,y2 ∈ X. Then µlA((y1 · (y2 ·x)) ·x),µ u A((y1 · (y2 ·x)) ·x)] = µA((y1 · (y2 ·x)) ·x) ≥ rmin{µA(y1),µA(y2)} = rmin{[µlA(y1),µ u A(y1)], [µ l A(y2),µ u A(y2)]} = [min{µlA(y1),µ l A(y2)}, min{µ u A(y1),µ u A(y2)}]. 14 Int. J. Anal. Appl. (2022), 20:25 Hence, µlA((y1 ·(y2 ·x)) ·x) ≥ min{µ l A(y1),µ l A(y2)} and µ u A((y1 ·(y2 ·x)) ·x) ≥ min{µ u A(y1),µ u A(y2)}. Also, [γlA((y1 · (y2 ·x)) ·x),γ u A((y1 · (y2 ·x)) ·x)] = γA((y1 · (y2 ·x)) ·x) ≤ rmax{γA(y1),γA(y2)} = rmax{[γlA(y1),γ u A(y1)], [γ l A(y2),γ u A(y2)]} = [max{γlA(y1),γ l A(y2)}, max{γ u A(y1),γ u A(y2)}]. Hence, γlA((y1 ·(y2 ·x)) ·x) ≤ max{γ l A(y1),γ l A(y2)} and γ u A((y1 ·(y2 ·x)) ·x) ≤ max{γ u A(y1),γ u A(y2)}. Therefore, µlA,µ u A,γ l A, and γ u A are fuzzy ideals of X. � Proposition 4.2. If A = (µA,γA) and B = (µB,γB) are IVIF ideals of a Hilbert algebra X, then A∩B is an IVIF ideal of X. Proof. Let A = (µA,γA) and B = (µB,γB) be IVIF ideals of X. Let x ∈ X. Then µA∩B(1) = [µ l A∩B(1),µ u A∩B(1)] = [min{µlA(1),µ l B(1)}, min{µ u A(1),µ u B(1)}] ≥ [min{µlA(x),µ l B(x)}, min{µ u A(x),µ u B(x)}] = [µlA∩B(x),µ u A∩B(x)] = µA∩B(x) and γA∪B(1) = [γ l A∪B(1),γ u A∪B(1)] = [max{γlA(1),γ l B(1)}, max{γ u A(1),γ u B(1)}] ≤ [max{γlA(x),γ l B(x)}, max{γ u A(x),γ u B(x)}] = [γlA∪B(x),γ u A∪B(x)] = γA∪B(x). Let x,y ∈ X. Then µA∩B(x ·y) = [µlA∩B(x ·y),µ u A∩B(x ·y)] = [min{µlA(x ·y),µ l B(x ·y)}, min{µ u A(x ·y),µ u B(x ·y)}] ≥ [min{µlA(y),µ l B(y)}, min{µ u A(y),µ u B(y)}] = [µlA∩B(y),µ u A∩B(y)] = µA∩B(y) and γA∪B(x ·y) = [γlA∪B(x ·y),γ u A∪B(x ·y)] = [max{γlA(x ·y),γ l B(x ·y)}, max{γ u A(x ·y),γ u B(x ·y)}] ≤ [max{γlA(y),γ l B(y)}, max{γ u A(y),γ u B(y)}] = [γlA∪B(y),γ u A∪B(y)] = γA∪B(y). Int. J. Anal. Appl. (2022), 20:25 15 Let x,y1,y2 ∈ X. Then µA∩B((y1 · (y2 ·x)) ·x) = [µlA∩B((y1 · (y2 ·x)) ·x),µ u A∩B((y1 · (y2 ·x)) ·x)] = [ min{µlA((y1 · (y2 ·x)) ·x),µ l B((y1 · (y2 ·x)) ·x)}, min{µuA((y1 · (y2 ·x)) ·x),µ u B((y1 · (y2 ·x)) ·x)} ] ≥ [ min{min{µlA(y1),µ l A(y2)}, min{µ l B(y1),µ l B(y2)}}, min{min{µuA(y1),µ u A(y2)}, min{µ u B(y1),µ u B(y2)}} ] = [ min{min{µlA(y1),µ l B(y1)}, min{µ l A(y2),µ l B(y2)}}, min{min{µuA(y1),µ u B(y2)}, min{µ u A(y2),µ u B(y2)}} ] = [min{µlA∩B(y1),µ l A∩B(y2)}, min{µ l A∩B(y1),µ l A∩B(y2)}] = rmin{µA∩B(y1),µA∩B(y2)} and γA∪B((y1 · (y2 ·x)) ·x) = [γlA∪B((y1 · (y2 ·x)) ·x),γ u A∪B((y1 · (y2 ·x)) ·x)] = [ max{γlA((y1 · (y2 ·x)) ·x),γ l B((y1 · (y2 ·x)) ·x)}, max{γuA((y1 · (y2 ·x)) ·x),γ u B((y1 · (y2 ·x)) ·x)} ] ≤ [ max{min{γlA(y1),γ l A(y2)}, min{γ l B(y1),γ l B(y2)}}, max{min{γuA(y1),γ u A(y2)}, min{γ u B(y1),γ u B(y2)}} ] = [ max{min{γlA(y1),γ l B(y1)}, min{γ l A(y2),γ l B(y2)}}, max{min{γuA(y1),γ u B(y2)}, min{γ u A(y2),γ u B(y2)}} ] = [max{γlA∩B(y1),γ l A∩B(y2)}, max{γ l A∩B(y1),γ l A∩B(y2)}] = rmax{γA∩B(y1),γA∩B(y2)}. Hence, A∩B is an IVIF ideal of X. � Corollary 4.1. If {Ai = (µAi,γAi ) | i ∈ ∆} is a family of IVIF ideals of a Hilbert algebra X, then ⋂ i∈∆ Ai is an IVIF ideal of X. Corollary 4.2. If A = (µA,γA) is an IVIF ideal of a Hilbert algebra X, then A is also an IVIF ideal of X. Theorem 4.3. If A = (µA,γA) is an IVIF ideal of a Hilbert algebra X, then ⊕A and ⊗A are both IVIF ideals. Proof. Assume that A = (µA,γA) is an IVIF ideal of X. Let x ∈ X. Then µA(1) = 1 − µA(1) ≤ 1 − µA(x) ≤ µA(x). Let x,y ∈ X. Then µA(x · y) = 1 − µA(x · y) ≤ 1 − µA(y) ≤ µA(y). Let 16 Int. J. Anal. Appl. (2022), 20:25 x,y1,y2 ∈ X. Then µA((y1 · (y2 ·x)) ·x) = 1 −µA((y1 · (y2 ·x)) ·x) ≤ 1 − rmin{µA(y1),µA(y2)} = rmax{1 −µA(y1), 1 −µA(y2)} = rmax{µA(y1),µA(y2)}. Hence, ⊕A is an IVIF ideal of X. Let x ∈ X. Then γA(1) = 1 − γA(1) ≥ 1 − γA(x) ≥ γA(x). Let x,y ∈ X. Then γA(x · y) = 1 −γA(x ·y) ≥ 1 −γA(y) ≥ γA(y). Let x,y1,y2 ∈ X. Then γA((y1 · (y2 ·x)) ·x) = 1 −γA((y1 · (y2 ·x)) ·x) ≥ 1 − rmax{γA(y1),γA(y2)} = rmin{1 −γA(y1), 1 −γA(y2)} = rmin{γA(y1),γA(y2)}. Hence, ⊗A is an IVIF ideal of X. � Theorem 4.4. An IVIF set A = (µA,γA) is an IVIF ideal of a Hilbert algebra X if and only if for every [s1,s2], [t1,t2] ∈ D[0, 1], the sets U(µA : [t1,t2]) and L(g, [s1,s2]) are either empty or ideals of X. Proof. Let A = (µA,γA) be an IVIF ideal of X and let [s1,s2], [t1,t2] ∈ D[0, 1] be such that U(µA : [t1,t2]) and L(γA : [s1,s2]) are nonempty sets of X. It is clear that 1 ∈ U(µA : [t1,t2])∩L(γA : [s1,s2]) since µA(1) ≥ [t1,t2] and γA(1) ≤ [s1,s2]. Let x ∈ X and y ∈ U(µA : [t1,t2]). Then µA(y) ≥ [t1,t2]. It follows that µA(x · y) ≥ µA(y) ≥ [t1,t2] so that x · y ∈ U(µA : [t1,t2]). Let x ∈ X and y1,y2 ∈ U(µA : [t1,t2]). Then µA(y1) ≥ [t1,t2] and µA(y2) ≥ [t1,t2]. Hence, µA((y1 · (y2 ·x)) ·x) ≥ min{µA(y1),µA(y2)} ≥ [t1,t2] so that (y1 · (y2 · x)) · x ∈ U(µA : [t1,t2]). Hence, U(µA : [t1,t2]) is an ideal of X. Let x ∈ X and y ∈ L(γA : [s1,s2]). Then γA(y) ≤ [s1,s2]. It follows that γA(x·y) ≤ γA(y) ≤ [s1,s2] so that x·y ∈ L(γA : [s1,s2]). Let x ∈ X and y1,y2 ∈ L(γA : [s1,s2]). Then γA(y1) ≤ [s1,s2] and γA(y2) ≤ [s1,s2]. Hence, γA((y1 · (y2 ·x)) ·x) ≤ max{γA(y1),γA(y2)}≤ [s1,s2] so that (y1 · (y2 ·x)) ·x ∈ L(γA : [s1,s2]). Hence, L(γA : [s1,s2]) is an ideal of X. Assume now that every nonempty sets U(µA : [t1,t2]) and L(γA : [s1,s2]) are ideals of X. If µA(1) ≥ µA(x) is not true for all x ∈ X, then there exists x0 ∈ X such that µA(1) < µA(x0). But in this case for [s1,s2] = 1 2 (µA(1) + µA(x0)). Then x0 ∈ U(µA : [s1,s2]), that is U(µA : [s1,s2]) 6= ∅. Since by the assumption, U(µA : [s1,s2]) is an ideal of X, then µA(1) ≥ [s1,s2], which is impossible. Hence, µA(1) ≥ µA(x). If γA(1) ≤ γA(x) is not true, then there exists y0 ∈ X such that γA(1) < γA(y0). But in this case for [s′0,s ′′ 0 ] = 1 2 (γA(1) + γA(y0)). Then y0 ∈ L(γA : [s′0,s ′′ 0 ]), that is L(γA : [s ′ 0,s ′′ 0 ]) 6= ∅. Since by the assumption, L(γA : [s′0,s ′′ 0 ]) is an ideal of X, then γA(1) ≤ [s ′ 0,s ′′ 0 ], which is impossible. Hence, γA(1) ≤ γA(x). If µA(x · y) ≥ µA(y) is not true for all x,y ∈ X, then there exist x0,y0 ∈ X such that µA(x0 · y0) < µA(y0). Let [t1,t2] = 12 (µA(x0 · y0) + µA(y0)). Then t ∈ [0, 1] and µA(x0 · y0) < t < µA(y0), which prove that y0 ∈ U(µA : t). Since U(µA : t) is an ideal of X, Int. J. Anal. Appl. (2022), 20:25 17 x0 · y0 ∈ U(µA : t). Hence, µA(x0 · y0) ≥ t, a contradiction. Thus µA(x · y) ≥ µA(y) is true for all x,y ∈ X. If γA(x · y) ≤ γA(y) is not true for all x,y ∈ X, then there exist x0,y0 ∈ X such that γA(x0 ·y0) > γA(y0). Let [t′0,t ′′ 0 ] = 1 2 (γA(x0 ·y0) + γA(y0)). Then [t′0,t ′′ 0 ] ∈ D[0, 1] and γA(x0 ·y0) > [t′0,t ′′ 0 ] > γA(y0), which prove that y0 ∈ L(γA : [t ′ 0,t ′′ 0 ]). Since L(γA : [t ′ 0,t ′′ 0 ]) is an ideal of X, x0·y0 ∈ L(γA : [t′0,t ′′ 0 ]). Hence, γA(x0·y0) ≤ [t ′ 0,t ′′ 0 ], a contradiction. Thus γA(x·y) ≤ γA(y) is true for all x,y ∈ X. Suppose that µA((y1·(y2·x))·x) ≥ rmin{µA(y1),µA(y2)} is not true for all x,y1,y2 ∈ X. Then there exist u0,v0,x0 ∈ X such that µA((u0 · (v0 · x0))) · x0) < rmin{µA(u0),µA(v0)}. Taking [p′,p′′] = 1 2 (µA((u0 ·(v0 ·x0)))·x0) + rmin{µA(u0),µA(v0)}). Then µA((u0 ·(v0 ·x0))·x0) < [p′,p′′] < rmin{µA(u0),µA(v0)}, which prove that u0,v0 ∈ U(µA : [p′,p′′]). Since U(µA : p[p′,p′′]) is an ideal of X, (u0·(v0·x0))x0 ∈ U(µA : [p′,p′′]), a contradiction. Thus µA((y1·(y2·x))·x) ≥ rmin{µA(y1),µA(y2)} is true for all x,y1,y2 ∈ X. Suppose that γA((y1·(y2·x))·x) ≤ rmax{γA(y1),γA(y2)} is not true for all x,y1,y2 ∈ X. Then there exist u0,v0,x0 ∈ X such that γA((u0 ·(v0 ·x0)·x0) > rmax{γA(u0),γA(v0)}. Taking [p′0,p ′′ 0 ] = 1 2 (γA((u0 · (v0 ·x0)) ·x0) + rmax{γA(u0),γA(v0)}). Then γA((u0 · (v0 ·x0)) ·x0) > [p′0,p ′′ 0 ] > rmax{γA(u0),γA(v0)}, which prove that u0,v0 ∈ L(γA : [p ′ 0,p ′′ 0 ]). Since L(γA : [p ′ 0,p ′′ 0 ]) is an ideal of X, (u0 · (v0 · x0)) · x0 ∈ L(γA : [p′0,p ′′ 0 ]), a contradiction. Thus γA((y1 · (y2 · x)) · x) ≤ rmax{γA(y1),γA(y2)} is true for all x,y1,y2 ∈ X. Hence, A is an IVIF ideal of X. � 5. Product of IVIF subalgebras/ideals in Hilbert algebras Definition 5.1. Let A = (µA,γA) and B = (µB,γB) be IVIF sets in Hilbert algebras X and Y , respectively. The cartesian product A×B = {((x,y), (µA×µB)(x,y), (γA×γB)(x,y)) | x ∈ X,y ∈ Y} defined by (µA ×µB)(x,y) = rmin{µA(x),µB(y)} and (γA ×γB)(x,y) = rmax{γA(x),γB(y)}, where µA ×µB : X ×Y → D[0, 1] and γA ×γB : X ×Y → D[0, 1] for all x ∈ X and y ∈ Y . Remark 5.1. Let X and Y be Hilbert algebras. We define the binary operation · on X × Y by (x,y) · (u,v) = (x ·u,y ·v) for every (x,y), (u,v) ∈ X ×Y , then clearly (X ×Y, ·, (1, 1)) is a Hilbert algebra. Proposition 5.1. If A = (µA,γA) and B = (µB,γB) are IVIF subalgebras of Hilbert algebras X and Y , respectively, then the cartesian product A×B is also an IVIF subalgebra of X ×Y . 18 Int. J. Anal. Appl. (2022), 20:25 Proof. Let (x1,y1), (x2,y2) ∈ X ×Y . Then (µA ×µB)((x1,y1) · (x2,y2)) = (µA ×µB)((x1 ·x2), (y1 ·y2)) = rmin{µA(x1 ·x2),µB(y1 ·y2)} ≥ rmin{rmin{µA(x1),µA(x2)}, rmin{µB(y1),µB(y2)}} = rmin{rmin{µA(x1),µB(y1)}, rmin{µA(x2),µB(y2)}} = rmin{(µA ×µB)(x1,y1), (µA ×µB)(x2,y2)} and (γA ×γB)((x1,y1) · (x2,y2)) = (γA ×γB)((x1 ·x2), (y1 ·y2)) = rmin{γA(x1 ·x2),γB(y1 ·y2)} ≤ rmin{rmax{γA(x1),γA(x2)}, rmax{γB(y1),γB(y2)}} = rmax{rmin{γA(x1),γB(y1)}, rmin{γA(x2),γB(y2)}} = rmax{(µA ×µB)(x1,y1), (µA ×µB)(x2,y2)}. Hence, A×B is an IVIF subalgebra of X ×Y . � Lemma 5.1. If A = (µA,γA) and B = (µB,γB) are two IVIF subalgebras of Hilbert algebras X and Y , respectively, then ⊕(A×B) = (µA ×µB,µA ×µB) is an IVIF subalgebra of X ×Y . Proof. It is sufficient to prove only the part of µA ×µB. Let (x1,y1), (x2,y2) ∈ X ×Y . Then (µA ×µB)((x1,y1) · (x2,y2)) = (µA ×µB)((x1 ·x2), (y1 ·y2)) = rmax{µA(x1 ·x2),µB(y1 ·y2)} = rmax{1 −µA(x1 ·x2), 1 −µB(y1 ·y2)} ≤ rmax{1 − rmin{µA(x1),µA(x2)}, 1 − rmin{µB(y1),µB(y2)}} = rmax{rmax{1 −µA(x1), 1 −µB(y1)}, rmax{1 −µA(x2), 1 −µB(y2)}} = rmax{rmax{µA(x1),µB(y1)}, rmax{µA(x2),µB(y2)}} = rmax{(µA ×µB)(x1,y1), (µA ×µB)(x2,y2)}. Hence, ⊕(A×B) is an IVIF subalgebra of X ×Y . � Lemma 5.2. If A = (µA,γA) and B = (µB,γB) are two IVIF subalgebras of Hilbert algebras X and Y , respectively, then ⊗(A×B) = (γA ×γB,γA ×γB) is an IVIF subalgebra of X ×Y . Int. J. Anal. Appl. (2022), 20:25 19 Proof. It is sufficient to prove only the part of γA ×γB. Let (x1,y1), (x2,y2) ∈ X ×Y . Then (γA ×γB)((x1,y1) · (x2,y2)) = (γA ×γB)((x1 ·x2), (y1 ·y2)) = rmin{γA(x1 ·x2),γB(y1 ·y2)} = rmin{1 −γA(x1 ·x2), 1 −γB(y1 ·y2)} ≥ rmin{1 − rmax{γA(x1),γA(x2)}, 1 − rmax{γB(y1),γB(y2)}} = rmin{rmin{1 −γA(x1), 1 −γB(y1)}, rmin{1 −γA(x2), 1 −γB(y2)}} = rmin{rmin{γA(x1),γB(y1)}, rmin{γA(x2),γB(y2)}} = rmin{(γA ×γB)(x1,y1), (γA ×γB)(x2,y2)}. Hence, ⊗(A×B) is an IVIF subalgebra of X ×Y . � Theorem 5.1. The IVIF sets A = (µA,γA) and B = (µB,γB) are IVIF subalgebras of Hilbert algebras X and Y , respectively if and only if ⊕(A×B) and ⊗(A×B) are IVIF subalgebras of X ×Y . Proof. It follows from Lemmas 5.1 and 5.2. � Proposition 5.2. If A = (µA,γA) and B = (µB,γB) are two IVIF ideals of Hilbert algebras X and Y , respectively, then the cartesian product A×B is also an IVIF ideal of X ×Y . Proof. Let (x,y) ∈ X ×Y . Then (µA ×µB)(1, 1) = rmin{µA(1),µB(1)} ≥ rmin{µA(x),µB(y)} = (µA ×µB)(x,y) and (γA ×γB)(1, 1) = rmax{γA(1),γB(1)} ≤ rmax{γA(x),γB(y)}} = (γA ×γB)(x,y). Let (x1,x2), (y1,y2) ∈ X ×Y . Then (µA ×µB)((x1,x2) · (y1,y2)) = (µA ×µB)((x1 ·y1), (x2 ·y2)) = rmin{µA(x1 ·y1),µB(x2 ·y2)} ≥ rmin{µA(y1),µB(y2)} = (µA ×µB)(y1,y2) and (γA ×γB)((x1,x2) · (y1,y2)) = (γA ×γB)((x1 ·y1), (x2 ·y2)) = rmax{γA(x1 ·y1),γB(x2 ·y2)} ≤ rmax{γA(y1),γB(y2)} = (γA ×γB)(y1,y2). 20 Int. J. Anal. Appl. (2022), 20:25 Let (x1,y1), (x2,y2), (x3,y3) ∈ X ×Y . Then (µA ×µB)(((x2,y2) · ((x3,y3) · (x1,y1))) · (x1,y1)) = (µA ×µB)(((x2 · (x3 ·x1)) ·x1)), (y2 · (y3 ·y1)) ·y1)) = rmin{µA((x2 · (x3 ·x1)) ·x1),µB((y2 · (y3 ·y1)) ·y1)} ≥ rmin{rmin{µA(x2),µA(x3)}, rmin{µB(y2),µB(y3)}} = rmin{rmin{µA(x2),µB(y2)}, rmin{µA(x3),µB(y3)}} = rmin{(µA ×µB)(x2,y2), (µA ×µB)(x3,y3)} and (γA ×γB)(((x2,y2) · ((x3,y3) · (x1,y1))) · (x1,y1)) = (γA ×γB)(((x2 · (x3 ·x1)) ·x1)), (y2 · (y3 ·y1)) ·y1)) = rmax{γA((x2 · (x3 ·x1)) ·x1),γB((y2 · (y3 ·y1)) ·y1)} ≤ rmax{rmax{γA(x2),γA(x3)}, rmax{γB(y2),γB(y3)}} = rmax{rmax{γA(x2),γB(y2)}, rmax{γA(x3),γB(y3)}} = rmax{(γA ×γB)(x2,y2), (γA ×γB)(x3,y3)}. Hence, A×B is an IVIF ideal of X ×Y . � Lemma 5.3. If A = (µA,γA) and B = (µB,γB) are two IVIF ideals of Hilbert algebras X and Y , respectively, then ⊕(A×B) = (µA ×µB,µA ×µB) is an IVIF ideal of X ×Y . Proof. It is sufficient to prove only the part of µA ×µB. Let (x,y) ∈ X ×Y . Then (µA ×µB)(1, 1) = rmax{µA(1),µB(1)} = rmax{1 −µA(1), 1 −µB(1)} ≤ rmax{1 −µA(x), 1 −µB(y)} = (µA ×µB)(x,y). Let (x1,x2), (y1,y2) ∈ X ×Y . Then (µA ×µB)((x1,x2) · (y1,y2)) = (µA ×µB)((x1 ·y1), (x2 ·y2)) = rmax{µA(x1 ·y1),µB(x2 ·y2)} = rmax{1 −µA(x1 ·y1), 1 −µB(x2 ·y2)} ≤ rmax{1 −µA(y1), 1 −µB(y2)} = rmax{µA(y1),µB(y2)} = (µA ×µB)(y1,y2). Int. J. Anal. Appl. (2022), 20:25 21 Let (x1,y1), (x2,y2), (x3,y3) ∈ X ×Y . Then (µA ×µB)(((x2,y2) · ((x3,y3) · (x1,y1))) · (x1,y1)) = (µA ×µB)(((x2 · (x3 ·x1)) ·x1)), (y2 · (y3 ·y1)) ·y1)) = rmax{µA((x2 · (x3 ·x1)) ·x1),µA((y2 · (y3 ·y1)) ·y1)} = rmax{1 −µA((x2 · (x3 ·x1)) ·x1), 1 −µA((y2 · (y3 ·y1)) ·y1)} ≤ rmax{1 − rmin{µA(x2),µA(x3)}, 1 − rmin{µB(y2),µB(y3)}} = rmax{rmax{1 −µA(x2), 1 −µB(y2)}, rmax{1 −µA(x3), 1 −µB(y3)}} = rmax{rmax{µA(x2),µB(y2)}, rmax{µA(x3),µB(y3)}} = rmax{(µA ×µB)(x2,y2), (µA ×µB)(x3,y3)}. Hence, ⊕(A×B) is an IVIF ideal of X ×Y . � Lemma 5.4. If A = (µA,γA) and B = (µB,γB) are two IVIF ideals of Hilbert algebras X and Y , respectively, then ⊗(A×B) = (γA ×γB,γA ×γB) is an IVIF ideal of X ×Y . Proof. It is sufficient to prove only the part of γA ×γB. Let (x,y) ∈ X ×Y . Then (γA ×γB)(1, 1) = rmin{γA(1),γB(1)} = rmin{1 −γA(1), 1 −γB(1)} ≥ rmin{1 −γA(x), 1 −γB(y)} = (γA ×γB)(x,y). Let (x1,x2), (y1,y2) ∈ X ×Y . Then (γA ×γB)((x1,x2) · (y1,y2)) = (γA ×γB)((x1 ·y1), (x2 ·y2)) = rmin{γA(x1 ·y1),γB(x2 ·y2)} = rmin{1 −γA(x1 ·y1), 1 −γB(x2 ·y2)} ≥ rmin{1 −γA(y1), 1 −γB(y2)} = rmin{γA(y1),γB(y2)} = (γA ×γB)(y1,y2). Let (x1,y1), (x2,y2), (x3,y3) ∈ X ×Y . Then (γA ×γB)(((x2,y2) · ((x3,y3) · (x1,y1))) · (x1,y1)) = (γA ×γB)(((x2 · (x3 ·x1)) ·x1), ((y2 · (y3 ·y1)) ·y1)) = rmin{γA((x2 · (x3 ·x1)) ·x1),γA((y2 · (y3 ·y1)) ·y1)} = rmin{1 −γA((x2 · (x3 ·x1)) ·x1), 1 −γA((y2 · (y3 ·y1)) ·y1)} ≥ rmin{1 − rmax{γA(x2),γA(x3)}, 1 − rmax{γB(y2),γB(y3)}} = rmin{rmin{1 −γA(x2), 1 −γB(y2)}, rmin{1 −γA(x3), 1 −γB(y3)}} = rmin{rmin{γA(x2),γB(y2)}, rmin{γA(x3),γB(y3)}} = rmin{(γA ×γB)(x2,y2), (γA ×γB)(x3,y3)}. Hence, ⊗(A×B) is an IVIF subalgebra of X ×Y . � 22 Int. J. Anal. Appl. (2022), 20:25 Theorem 5.2. The IVIF sets A = (µA,γA) and B = (µB,γB) are IVIF ideals of Hilbert algebras X and Y , respectively if and only if ⊕(A×B) and ⊗(A×B) are IVIF ideals of X ×Y . Proof. It follows from Lemmas 5.3 and 5.4. � Theorem 5.3. Let A = (µA,γA) and B = (µB,γB) be any two IVIF sets in Hilbert algebras X and Y , respectively. If A × B is an IVIF subalgebra of X × Y , then nonempty upper [s1,s2]-level cut U(µA × µB : [s1,s2]) and nonempty lower [t1,t2]-level cut L(γA × γB : [t1,t2]) are subalgebras of X ×Y for all [s1,s2], [t1,t2] ∈ D[0, 1]. Proof. It follows from Theorem 3.7. � Theorem 5.4. Let A = (µA,γA) and B = (µB,γB) be any two IVIF sets in Hilbert algebras X and Y , respectively. If A × B is an IVIF ideal of X × Y , then nonempty upper [s1,s2]-level cut U(µA×µB : [s1,s2]) and nonempty lower [t1,t2]-level cut L(γA×γB : [t1,t2]) are ideals of X×Y for all [s1,s2], [t1,t2] ∈ D[0, 1]. Proof. It follows from Theorem 4.4. � A mapping f : X → Y of Hilbert algebras is called a homomorphism if f (x · y) = f (x) · f (y) for all x,y ∈ X. Note that if f : X → Y is a homomorphism of Hilbert algebras, then f (1) = 1. Let f : X → Y be a homomorphism of Hilbert algebras. For any IVIF set A = (µA,γA) in Y , we define an IVIF set f−1(A) = (µf−1(A),γf−1(A)) in X by µf−1(A)(x) = µA(f (x)) and γf−1(A)(x) = γA(f (x)) ∀x ∈ X. Proposition 5.3. Let f : X → Y be a homomorphism of a Hilbert algebra X into a Hilbert algebra Y and A = (µA,γA) an IVIF subalgebra of Y . Then the inverse image f−1(A) of A is an IVIF subalgebra of X. Proof. Let x,y ∈ X. Then µf−1(A)(x ·y) = µA(f (x ·y)) = µA(f (x) · f (y)) ≥ rmin{µA(f (x)),µA(f (y))} = rmin{µf−1(A)(x),µf−1(A)(y)} and γf−1(A)(x ·y) = γA(f (x ·y)) = γA(f (x) · f (y)) ≤ rmax{γA(f (x)),γA(f (y))} = rmax{γf−1(A)(x),γf−1(A)(y)}. Hence, f−1(A) of A is an IVIF subalgebra of X. � Int. J. Anal. Appl. (2022), 20:25 23 Theorem 5.5. Let f : X → Y be a homomorphism of a Hilbert algebra X into a Hilbert algebra Y and A = (µA,γA) be an IVIF ideal of Y . Then the inverse image f−1(A) = (µf−1(A),γf−1(A)) is an IVIF ideal of X. Proof. Since f is a homomorphism of X into Y , then f (1) = 1 ∈ Y and, by the assumption, µA(f (1)) = µA(1) ≥ µA(y) for every y ∈ Y . In particular, µA(f (1)) ≥ µA(f (x)) for all x ∈ X. Hence, µf−1(A)(1) ≥ µf−1(A)(x) for all x ∈ X. Also, γA(f (1)) = γA(1) ≤ γA(y) for every y ∈ Y . In particular, γB(f (1)) ≤ γB(f (x)) for all x ∈ X. Hence, γf−1(A)(1) ≤ γf−1(A)(x) for all x ∈ X, which proves (4.1). Now let x,y ∈ X. Then, by the assumption, µf−1(A)(x ·y) = µA(f (x ·y)) = µA(f (x) · f (y)) ≥ µA(f (y)) = µf−1(A)(y) and γf−1(A)(x ·y) = γA(f (x ·y)) = γA(f (x) · f (y)) ≤ γA(f (y)) = γf−1(A)(y), which proves (4.2). Let x,y1,y2 ∈ X. Then by assumption, µf−1(A)((y1 · (y2 ·x)) ·x) = µA(f (y1 · (y2 ·x) ·x)) = µA(f (y1) · (f (y2 ·x)) · f (x)) = µA(f (y1 · (y2 ·x)) · f (x)) = µA(f (y1 · (y2 ·x)) ·x) ≥ rmin{µA(f (y1)),µA(f (y2))} = rmin{µf−1(A)(y1),µf−1(A)(y2)} and γf−1(A)((y1 · (y2 ·x)) ·x) = γA(f (y1 · (y2 ·x) ·x)) = γA(f (y1) · (f (y2 ·x)) · f (x)) = γA(f (y1 · (y2 ·x)) · f (x)) = γA(f (y1 · (y2 ·x)) ·x)) ≤ rmax{γA(f (y1)),γA(f (y2))} = rmax{γf−1(A)(y1),γf−1(A)(y2)}, which proves (4.3). Hence, f−1(A) is an IVIF ideal of X. � 6. Equivalence relations on IVIF subalgebras/ideals Let I (H) be the family of all IVIF ideals of a Hilbert algebra X and let t = [t1,t2] ∈ D[0, 1]. Define binary relations Ut and Lt on I (H) as follows: (A,B) ∈ Ut ⇔ U(µA : t) = U(µB : t), (A,B) ∈ Lt ⇔ L(γA : t) = L(γB : t), respectively, for A = (µA,γA) and B = (µB,γB) in I (H). Then clearly Ut and Lt are equivalence relations on I (H). For any A = (µA,γA) ∈ I (H), let [A]Ut (resp., [A]Lt) denote the equivalence class 24 Int. J. Anal. Appl. (2022), 20:25 of A modulo Ut (resp., Lt), and denote by I (H)/Ut (resp., I (H)/Lt) the system of all equivalence classes modulo Ut (resp., Lt); so I (H)/Ut = {[A]Ut | A = (µA,γA) ∈ I (H)}, respectively, I (H)/Lt = {[A]Lt | A = (µA,γA) ∈ I (H)}. Now let I(H) denote the family of all ideals of X and let t = [t1,t2] ∈ D[0, 1]. Define maps ft and gt from I (H) to I(H) ∪{∅} by ft(A) = U(µA : t) and gt(A) = L(γA : t), respectively, for all A = (µA,γA) ∈ I (H). Then ft and gt are clearly well defined. Theorem 6.1. For any t = [t1,t2] ∈ D[0, 1], the maps ft and gt are surjective from I (H) to I(H) ∪{∅}. Proof. Let t = [t1,t2] ∈ D[0, 1]. Note that 0 = (0,1) is in I (H), where 0 and 1 are IVIF sets in X defined by 0(x) = [0, 0] and 1(x) = [1, 1] for all x ∈ X. Obviously, ft(0) = U(0,t) = U([0, 0] : [t1,t2]) = ∅ = L([1, 1] : [t1,t2]) = L(1 : t) = gt(0). Let G( 6= ∅) ∈ I(H). For G = (χG,χG) ∈ I (H), we have ft(G) = U(χG : t) = G and gt(G) = L(χG; t) = G. Hence, ft and gt are surjective. � Theorem 6.2. The quotient sets I (H)/Ut and I (H)/Lt are equipotent to I(H) ∪{∅} for every t = [t1,t2] ∈ D[0, 1]. Proof. For t = [t1,t2] ∈ D[0, 1], let f ∗t (resp., g∗t ) be a map from I (H)/Ut (resp., I (H)/Lt) to I(H) ∪{∅} defined by f ∗t ([A]Ut ) = ft(A) (resp., g∗t ([A]Lt ) = gt(A)) for all A = (µA,γA) ∈ I (H). If U(µA : t) = U(µB : t) and L(γA : t) = L(γB : t) for A = (µA,γA) and B = (µB,γB) ∈ I (H), then (A,B) ∈ Ut and (A,B) ∈ Lt; hence, [A]Ut = [B]Ut and [A]Lt = [B]Lt. Therefore, the maps f ∗t and g∗t are injective. Now let G( 6= ∅) ∈ I(H). For G = (χG,χG) ∈ I (H), we have f ∗t ([G]Ut = ft(G) = U(χG : t) = G, g∗t ([G]Lt = gt(G) = L(χG,t) = G. Finally, for 0 = (0,1) ∈ I (H), we get f ∗t ([0]Ut = ft(0) = U(0,t) = ∅, g∗t ([0]Lt = gt(0) = L(1,t) = ∅. This shows that f ∗t and g ∗ t are surjective. � For any t = [t1,t2] ∈ D[0, 1], we define another relation Rt on I (H) as follows: (A,B) ∈ Rt ⇔ U(µA : t) ∩L(γA : t) = U(µB : t) ∩L(γB : t) for any A = (µA,γA),B = (µB,γB) ∈ I (H). Then the relation Rt is also an equivalence relation on I (H). Int. J. Anal. Appl. (2022), 20:25 25 Theorem 6.3. For any t = [t1,t2] ∈ D[0, 1], the map ϕt : I (H) → I(H) ∪ {∅} is defined by ϕt(A) = ft(A) ∩gt(A) for each A = (µA,γA) ∈ I (H) as surjective. Proof. Let t = [t1,t2] ∈ D[0, 1]. For 0 = (0,1) ∈ I (H), ϕt(0) = ft(0) ∩gt(0) = U(0,t) ∩L(0,t) = ∅. For any H ∈ I (H), there exists H = (χH,χH) ∈ I (H) such that ϕt(H) = ft(H) ∩gt(H) = U(χH : t) ∩L(χH,t) = H. Hence, ϕt is surjective. � Theorem 6.4. For any t = [t1,t2] ∈ D[0, 1], the quotient set I (H)/Rt is equipotent to I(H) ∪{∅}. Proof. Let t = [t1,t2] ∈ D[0, 1] and let ϕ∗t : I (H)/Rt → I(H)∪{∅} be a map defined by ϕ∗t([A]Rt ) = ϕt(A) for all [A]Rt ∈ I (H)/Rt. If ϕ∗t([A]Rt ) = ϕ∗t([B]Rt ) for any [A]Rt, [B]Rt ∈ I (H)/Rt, then ft(A) ∩ gt(A) = ft(B) ∩ gt(B), that is, U(µA : t) ∩ L(γA : t) = U(µB : t) ∩ L(γB : t); hence, (A,B) ∈ Rt. It follows that [A]Rt = [B]Rt so that ϕ∗t is injective. For 0 = (0,1) ∈ I (H), ϕ∗t([0]Rt ) = ϕt(0) = ft(0) ∩gt(0) = U(0,t) ∩L(1,t) = ∅. If H ∈ I (H), then for H = (χH,χH) ∈ I (H), we have ϕ∗t([H]Rt ) = ϕt(H) = ft(H) ∩gt(H) = U(χH : t) ∩L(χH,t) = H. Hence, ϕ∗t is surjective, this completes the proof. � The same type of results are also true for IVIF subalgebras. Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication of this paper. References [1] B. Ahmad, A. Kharal, On Fuzzy Soft Sets, Adv. Fuzzy Syst. 2009 (2009), 586507. https://doi.org/10.1155/ 2009/586507. [2] K.T. Atanassov, Intuitionistic Fuzzy Sets, Fuzzy Sets Syst. 20 (1986), 87–96. https://doi.org/10.1016/ s0165-0114(86)80034-3. [3] M. Atef, M.I. Ali, T.M. 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