EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 4, 2022, 1854-1868 ISSN 1307-5543 – ejpam.com Published by New York Business Global Quotient Pseudo Hyper GR-algebras Ramises G. Manzano, Jr.1, Gaudencio C. Petalcorin, Jr.2,∗ 1 Office of the Mathematics and Statistics Programs, College of Science, University of the Philippines Cebu, Lahug, Cebu City, Philippines 2 Department of Mathematics and Statistics, College of Science, MSU-Iligan Institute of Technology, Tibanga, Iligan City, Philippines Abstract. A pseudo hyper GR-algebra is an algebraic structure involving two distinct hyperop- erations. Properties of this hyper algebra have been studied and given illustrations. This paper focuses on the quotient structure of pseudo hyper GR-algebras. From an equivalence relation on a pseudo hyper GR-algebra H, we can define a congruence relation on H that is used in the con- struction of the quotient structure H/I, where I is the congruence class of 0 under the congruence relation. Moreover, some isomorphism theorems of pseudo hyper GR-algebras are included in this paper. 2020 Mathematics Subject Classifications: 14L17, 20N20, 03G25 Key Words and Phrases: Hyper algebras, quotient hyper algebras 1. Introduction Algebraic hyperstructures were introduced by a French mathematician, Marty [6], in 1934. They represent a natural extension of classical hyperstructures in which the composition of two elements of a given set is a set, instead of an element. Afterwards, this new idea was expanded rapidly and showed itself as a new view of sets. The introduction of hyperstructure theory led to the study of several problems of noncommutative algebra. Algebraic hyperstructure theory has multiple ap- plications to other fields such as: geometry, graphs and hypergraphs, binary relations, lattices, groups, relation algebras, artificial intelligence, probabilities, and so on. In 1966, Y. Imai and K. Iséki [1] initiated the notion ofBCK-algebra as a generalization of the concept of set-theoretic difference and propositional calculi. Furthermore, Y.B. Jun et al. [5] applied hyperstructure theory to BCK-algebras and introduced the notion of hyper BCK-algebras as a generalization of BCK-algebra. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i4.4554 Email addresses: rgmanzano@up.edu.ph (R. Manzano, Jr.), gaudencio.petalcorin@g.msuiit.edu.ph (G. Petalcorin, Jr.) https://www.ejpam.com 1854 © 2022 EJPAM All rights reserved. R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1855 R.A. Indangan and G.C. Petalcorin [2] defined a new class of algebraic hyperstructure called hyper GR-algebra. In this algebra, they presented a helpful understanding on how this hyper algebra differs from the rest. R.G. Manzano and G.C. Petalcorin [4] extended the study hyper GR-algebras by intorducing a new definition involving two hyperoperations. This gives birth to pseudo hyper GR-algebras. 2. Preliminaries Let H be a nonempty set endowed with a hyperoperation “ ∗ ”, that is, “ ∗ ” is a function from H × H to P ∗(H) = P (H) \ {∅}. For two nonempty subsets A and B of H, A ∗ B = ⋃ a∈A,b∈B a ∗ b. We shall use x ∗ y instead of x ∗ {y}, {x} ∗ y or {x} ∗ {y}. When A is a nonempty subset of H and x ∈ H, we agree to write A ∗x instead of A ∗ {x}. Similarly, we write x ∗A for {x} ∗A. In effect, A ∗x = ⋃ a∈A a ∗ x and x ∗A = ⋃ a∈A x ∗ a. A set H endowed with a family Γ of hyperoperations is called a hyperstructure. If Γ is singleton, that is, Γ = {f}, then the hyperstructure is called a hypergroupoid. Definition 2.1. [2] Let H be a nonempty set with “⊛” a hyperoperation on H. Then (H;⊛, 0) is called a hyper GR-algebra if it contains a constant 0 ∈ H and for all x, y, z ∈ H, the following conditions are satisfied: [HGR1] (x⊛ z)⊛ (y ⊛ z) ≪ x⊛ y; [HGR2] (x⊛ y)⊛ z = (x⊛ z)⊛ y; [HGR3] x ≪ x; [HGR4] 0⊛ (0⊛ x) ≪ x, for all x ̸= 0; and [HGR5] (x⊛ y)⊛ z ≪ y ⊛ z. where x ≪ y if and only if 0 ∈ x ⊛ y, and for every A,B ⊆ H, A ≪ B means that for every a ∈ A, there exists b ∈ B such that a ≪ b. Example 2.2. [2] Let H = {0, 1, 2}. Define the operation “⊛” by the Cayley table shown below. ⊛ 0 1 2 0 {0} {0} {0} 1 {0, 1, 2} {0, 1} {0, 1} 2 {0, 2} {0, 1, 2} {0, 2} By routine calculations, (H;⊛, 0) is a hyper GR-algebra. Definition 2.3. [2] A hyper GR-algebra H is faithful if for all A,B ⊆ H, 0 ∈ A ⊛ B implies A ≪ B. R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1856 Definition 2.4. [2] Let H be a hyper GR-algebra and S be a subset of H containing 0. If S is a hyper GR-algebra with respect to the hyperoperation ⊛ on H, then we say that S is a hyper subGR-algebra of H. Theorem 2.5. [2] (Hyper SubGR-algebra Criterion) Let H be a hyper GR-algebra and S be a nonempty subset of H. Then S is a hyper subGR-algebra of H if and only if x⊛ y ⊆ S, for all x, y ∈ S. Definition 2.6. [4] Let H be a nonempty set with “⊛” and “◦” be the two hyperopera- tions on H. Then (H;⊛,⊙, 0) is called a pseudo hyper GR-algebra, if it contains a constant 0 ∈ H and for all x, y, z ∈ H, the following conditions are satisfied: [PHGR1] (x⊙ z)⊙ (y ⊙ z) ≪ x⊙ y and (x⊛ z)⊛ (y ⊛ z) ≪ x⊛ y; [PHGR2] (x⊙ y)⊛ z = (x⊛ z)⊙ y; [PHGR3] 0 ∈ x⊛ x and 0 ∈ x⊙ x; [PHGR4] 0⊙ (0⊛ x) ≪ x, for all x ̸= 0; and [PHGR5] (x⊛ y)⊛ z ≪ y ⊙ z. where x ≪ y if and only if 0 ∈ x ⊙ y and 0 ∈ x ⊛ y, and for every A,B ⊆ H, A ≪ B means that for every a ∈ A, there exists b ∈ B such that a ≪ b. Example 2.7. [4] Let H = {0, 1, 2, 3} and consider the following Cayley tables below. ⊛ 0 1 2 3 0 {0, 1} {0, 1} {0, 1} {0, 1} 1 {0, 1} {0, 1} {0, 1} {0, 1} 2 {0, 2} {0, 1, 2} {0, 2} {0, 1, 2} 3 {0, 1, 2} {0, 3} {0, 1, 3} {0, 3} ⊙ 0 1 2 3 0 {0, 1} {0, 1} {0, 1} {0, 1} 1 {1} {0, 1} {0, 1} {0, 1} 2 {0, 2} {0, 2} {0, 1, 2} {0, 1, 2} 3 {0, 3} {0, 1, 3} {0, 1, 3} {0, 1, 3} By routine calculations, we see that (H;⊛,⊙, 0) is a pseudo hyper GR-algebra. Remark 2.8. [4] In a pseudo hyper GR-algebra H, the following are evident: (i) x ≪ x (ii) (x⊙ y)⊛ z ≪ (x⊛ z)⊙ y R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1857 (iii) (A⊙B)⊛ C = (A⊛ C)⊙B (iv) A ⊆ B implies A ≪ B. Example 2.9. [4] Let H = N ∪ {0} be the set of all nonnegative integers and let the hyperoperations “⊛” and “⊙” be defined on H as follows: x⊛ y = {0, x} and x⊙ y = {0, x, y}. Then H is a pseudo hyper GR-algebra. Remark 2.10. [4] Note that if the two hyperoperations are equal , that is, ⊛ = ⊙, then a pseudo hyper-GR algebra H becomes a hyper GR-algebra. Definition 2.11. [4] Let H be a pseudo hyper GR-algebra and S be a subset of H containing 0. If S itself is a pseudo hyper GR-algebra with respect to the hyperoperations ⊛ and ⊙ on H, then S is called a pseudo hyper subGR-algebra of H. Theorem 2.12. [4] (Pseudo Hyper SubGR-algebra Criterion) Let S be a nonempty subset of a pseudo hyper GR-algebra H. Then S is a pseudo hyper subGR-algebra if and only if both x⊛ y ⊆ S and x⊙ y ⊆ S for all x, y ∈ S. Example 2.13. [4] For any pseudo hyper GR-algebra H, the set S = {0} is a pseudo hyper subGR-algebra of H. 3. Quotient Pseudo Hyper GR-algebras In this section, we construct the structure of the quotient pseudo hyper GR-algebra H/I from a pseudo hyper GR-algebra H via congruence relation. All throughout, we denote a pseudo hyper GR-algebra (H,⊛,⊙, 0) simply by H, unless otherwise stated. Definition 3.1. Let θ be an equivalence relation on a pseudo hyper GR-algebra H and A and B be nonempty subsets of H. (i) AθB if there exist a ∈ A and b ∈ B such that aθb; (ii) Aθ̄B if for every a ∈ A, there exists b ∈ B such that aθb and for every b ∈ B, there exists a ∈ A such that aθb; (iii) θ is called a right ⊛-congruence (resp. right ⊙-congruence) on H if aθb implies (a⊛ u)θ̄(b⊛ u) (resp. (a⊙ u)θ̄(b⊙ u)) for all u in H; (iv) θ is called a left ⊛-congruence (resp. left ⊙-congruence) on H if aθb implies (u ⊛ a)θ̄(u⊛ b) (resp. (u⊙ a)θ̄(u⊙ b)) for all u in H; R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1858 (v) θ is called a ⊛-congruence (resp. ⊙-congruence) on H if it is a right and a left ⊛-congruence (resp. a right and left ⊙-congruence); (vi) θ is called a left congruence on H if it is a left ⊛-congruence and a left ⊙-congruence on H; (vii) θ is called a right congruence on H if it is a right ⊛-congruence and a right ⊙- congruence on H; (viii) θ is called a congruence relation on H if it is a ⊛-congruence and ⊙-congruence on H; and (ix) θ is called a regular congruence relation on H, if θ is a congruence relation on H and for any x, y ∈ H, whenever (x⊛ y)θ{0}, (y ⊛ x)θ{0}, (x⊙ y)θ{0}, and (y ⊙ x)θ{0}, we have xθy. Example 3.2. Let H = {0, 1, 2} and consider the following Cayley tables below. ⊛ 0 1 2 0 {0} {0} {0} 1 {0, 1} {0, 1} {0, 1} 2 {0, 2} {0, 2} {0, 2} ⊙ 0 1 2 0 {0} {0, 1} {0, 2} 1 {0, 1} {0, 1} {0, 1, 2} 2 {0, 2} {0, 1, 2} {0, 2} Then (H;⊛,⊙, 0) is a pseudo hyper GR-algebra. Define θ on H by θ = {(0, 0), (0, 1), (1, 0), (1, 1), (2, 2)}. It can be easily verified that θ is an equivalence relation. We will show that θ is a congruence relation using Definition 3.1. Since xθx for all x ∈ H1, we have (a⊛x)θ̄(a⊛x), (x⊛a)θ̄(x⊛a), (a⊙x)θ̄(a⊙x) and (x⊙a)θ̄(x⊙a). Therefore, the remaining elements of θ that is left for verification are (1, 0) and (0, 1) which can be done simultaneously. Since 1θ0, we will show that (1⊛ a)θ̄(0⊛ a), (a⊛ 1)θ̄(a⊛ 0), (1⊙ a)θ̄(0⊙ a) and (a⊙ 1)θ̄(a⊙ 0) for all a ∈ H. For a = 0, 1⊛ 0 = {0, 1}θ̄{0} = 0⊛ 0 and 0⊛ 1 = {0}θ̄{0} = 0⊛ 0 1⊙ 0 = {0, 1}θ̄{0} = 0⊙ 0 and 0⊙ 1 = {0}θ̄{0} = 0⊙ 0. For a = 1, 1⊛ 1 = {0, 1}θ̄{0} = 0⊛ 1 and 1⊛ 1 = {0, 1}θ̄{0, 1} = 1⊛ 0 1⊙ 1 = {0, 1}θ̄{0, 1} = 0⊙ 1 and 1⊙ 1 = {0, 1}θ̄{0, 1} = 1⊙ 0. For a = 2, 1⊛ 2 = {0, 1}θ̄{0} = 0⊛ 2 and 2⊛ 1 = {0, 2}θ̄{0, 2} = 2⊛ 0 1⊙ 2 = {0, 1, 2}θ̄{0, 2} = 0⊙ 2 and 2⊙ 1 = {0, 1, 2}θ̄{0, 2} = 2⊙ 0. Therefore, θ is a congruence relation. R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1859 Definition 3.3. Let θ be a congruence relation on a pseudo hyperGR-algebraH. The con- gruence class of x, denoted by [x]θ or Ix is given by [x]θ = Ix = {y ∈ H |xθy}. Lemma 3.4. Let θ be an equivalence relation on H and A,B ⊆ H. If Aθ̄B and Bθ̄C, then Aθ̄C. Proof. Let a ∈ A. By Definition 3.1(ii), there exists b ∈ B such that aθb. Also, there exists c ∈ C such that bθc. By transitivity, aθc. Let c ∈ C. Then there exists b ∈ B such that bθc. But there exists a ∈ A such that aθb. So, by transitivity, aθc. Therefore, Aθ̄C. □ Lemma 3.5. Let θ be an equivalence relation on H such that x⊛ 0 = x and x⊙ 0 = x, for all x ∈ H. Then the following hold: (i) If θ is a left ⊛-congruence (left ⊙-congruence) on H, then [0]θ is a pseudo hyper GR-ideal of type 8. (ii) If θ is a left congruence on H, then [0]θ is a pseudo hyper GR-ideal of type 4. Proof. (i) Suppose that θ is a left ⊛-congruence. Let y ∈ [0]θ and x ∈ [0]≪θ⊛,y . Then x⊛ y ≪ [0]θ. Then for all a ∈ x⊛ y, there exists b ∈ [0]θ such that 0 ∈ a⊛ b. Since θ is a left ⊛-congruence, bθ0 implies that (a⊛ b)θ̄(a⊛ 0) and so (a⊛ b)θ̄a. Now, 0 ∈ a⊛ b and (a⊛b)θ̄a would imply that 0θa and so x⊛y ⊆ [0]θ. Since yθ0 and θ is a left ⊛-congruence, (x ⊛ y)θ̄(x ⊛ 0). Thus, for all z ∈ x ⊛ y, we have zθx. Since x ⊛ y ⊆ [0]θ, zθ0. By commutativity and transitivity, zθ0 and zθx means that xθ0. Hence, x ∈ [0]θ. Therefore, [0]θ is a pseudo hyper GR-ideal of type 8. Similarly, it is easy to show that [0]θ is a pseudo hyper GR-algebra of type 8 for the case of left ⊙-congruence. (ii) Suppose that θ is a left congruence on H. Let y ∈ [0]θ and x ∈ [0]≪θ⊛,y . Then x ⊛ y ⊆ [0]θ. This means that x ⊛ y ≪ [0]θ. Then for all a ∈ x ⊛ y, there exists b ∈ [0]θ such that 0 ∈ a ⊛ b. Since θ is a left ⊛-congruence, bθ0 implies that (a⊛ b)θ̄(a⊛ 0) and so (a⊛ b)θ̄a. Now, 0 ∈ a⊛ b and (a⊛ b)θ̄a would imply that 0θa and so x ⊛ y ⊆ [0]θ. Since yθ0 and θ is a left ⊛-congruence, (x ⊛ y)θ̄(x ⊛ 0) or equivalently (x⊛ y)θ̄{x}. Thus, for all z ∈ x⊛ y, we have zθx. Since x⊛ y ⊆ [0]θ, zθ0. Now, zθ0 and zθx mean that xθ0. Hence, x ∈ [0]θ. Similarly, if y ∈ [0]θ and x ∈ [0]≪θ⊙,y , then x ∈ [0]θ. Therefore, [0]θ is a pseudo hyper GR-ideal of type 4. □ The next example will give us the idea on how the quotient structure on a pseudo hyper GR-algebra is constructed. Example 3.6. Consider the pseudo hyper GR-algebra H = {0, 1, 2} in Example 3.2. The relation θ defined on H is a congruence relation as shown. More- over, I = [0]θ = {0, 1} and I2 = {2}. Let H/I denote the set of all congruence classes on H, that is, H/I = {Ix |x ∈ H}. In our case, H/I = {I, I2}. Define hyperoperations R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1860 ⊗ and ⊚ on H/I by Ix ⊗ Iy = {Iz | z ∈ x ⊛ y} and Ix ⊚ Iy = {Iz | z ∈ x ⊙ y} and Ix ≪ Iy ⇐⇒ I0 ∈ Ix ⊛ Iy and I0 ∈ Ix ⊙ Iy. Thus, for our case, we have the following Cayley tables: ⊗ I I2 I {I} {I} I2 {I, I2} {I, I2} ⊚ I I2 I {I} {I, I2} I2 {I, I2} {I, I2} By routine calculations, (H/I;⊗,⊚, I) is a pseudo hyper GR-algebra. Lemma 3.7. Let θ be a congruence relation on a pseudo hyper GR-algebra H such that xθx′ and yθy′. Then (x⊛ y)θ(x′ ⊛ y′) and (x⊙ y)θ(x′ ⊙ y′). Proof. Suppose that θ is a congruence relation such that xθx′ and yθy′. Then Ix = Ix′ and Iy = Iy′ . Let z ∈ x ⊛ y. Then Iz ∈ Ix ⊗ Iy = Ix′ ⊗ Iy′ . Thus, Iz ∈ Ix′ ⊗ Iy′ . This means that z ∈ x′ ⊛ y′. Hence, (x⊛ y)θ(x′ ⊛ y′). Similarly, we can show that (x⊙ y)θ(x′ ⊙ y′). □ We will now show in general that using congruence relation, the quotient structure obtained is a pseudo hyper GR-algebra. Theorem 3.8. Let θ be a congruence relation on a pseudo hyper GR-algebra H such that I = [0]θ and H/I = {Ix |x ∈ H}, where Ix = [x]θ for all x ∈ H. Then H/I with hyperoperations ⊗ and ⊚, and hyperorder ≪ which are defined as follows: Ix ⊗ Iy = {Iz | z ∈ x⊛ y} and Ix ⊚ Iy = {Iz | z ∈ x⊙ y}, and Ix ≪ Iy ⇐⇒ I0 ∈ Ix ⊗ Iy and I0 ∈ Ix ⊚ Iy. is a pseudo hyper GR-algebra which we call the quotient pseudo hyper GR-algebra. Proof. Let us show first that the hyperoperations ⊗ and ⊚ on H/I are well-defined. Suppose that x, y, x′, y′ ∈ H such that Ix = Ix′ and Iy = Iy′ . Let Iz ∈ Ix ⊗ Iy. Then there exists u ∈ x ⊛ y such that Iu = Iz. Since xθx′ and yθy′, and θ is a congruence on H, by Lemma 3.7, (x ⊛ y)θ(x′θy′). Hence, there exists z′ ∈ x′ ⊛ y′ such that uθz′ and thus, Iz′ = Iu. Since Iz′ ∈ Ix′ ⊗ Iy′ and Iz = Iu = Iz′ , we have Iz ∈ Ix′ ⊗ Iy′ . Thus, Ix⊗Iy ⊆ Ix′⊗Iy′ . Similarly, we can show that Ix′⊗Iy′ ⊆ Ix⊗Iy. Hence, Ix⊗Iy = Ix′⊗Iy′ . Therefore, the hyperoperation “⊗” is well-defined. Similarly, we can show that Ix ⊚ Iy = Ix′ ⊚ Iy′ so that “⊚” is also well-defined. Now, since H is a pseudo hyper GR-algebra, 0 ∈ H and so, I0 = [0]θ = I ∈ H/I. Hence, H/I is nonempty and I ∈ H/I. It remains to show that H/I satisfies all the axioms of a pseudo hyper GR-algebra. [PHGR1] Let Iw ∈ (Ix ⊚ Iz) ⊚ (Iy ⊚ Iz), for some Ix, Iy, Iz ∈ H/I. Then there are Iu ∈ Ix⊚ Iz and Iv ∈ Iy ⊚ Iz such that Iw ∈ Iu⊚ Iv. Hence, there are u ′ ∈ x⊙ z, v′ ∈ y⊙ z and w′ ∈ u ⊙ v such that Iu = Iu′ , Iv = Iv′ , and Iw = Iw′ . Hence, uθu′, vθv′ and wθw′. R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1861 Since θ is a congruence relation on H, by Lemma 3.7, (u⊙ v)θ(u′ ⊙ v′). From w′ ∈ u⊙ v, there exists a ∈ u′ ⊙ v′ such that w′θa and so, Iw′ = Ia. Thus, Iw = Iw′ = Ia. By PHGR1 on H, a ∈ u′⊙v′ ⊆ (x⊙z)⊙ (y⊙z) ≪ x⊙y. Hence, there exists b ∈ x⊙y such that a ≪ b, which means that 0 ∈ a ⊙ b and 0 ∈ a ⊛ b. Furthermore, Ib ∈ Ix ⊚ Iy, I0 ∈ Ia ⊚ Ib, andI0 ∈ Ia ⊗ Ib. Since Iw = Iw′ = Ia, we have I0 ∈ Iw ⊚ Ib and I0 ∈ Iw ⊗ Ib which means that Iw ≪ Ib. This implies that (Ix ⊚ Iz)⊚ (Iy ⊚ Iz) ≪ Ix ⊚ Iy. Similarly, we can show also that (Ix ⊗ Iz)⊗ (Iy ⊗ Iz) ≪ Ix ⊗ Iy. Therefore, [PHGR1] holds. [PHGR2] Let Iw ∈ (Ix ⊚ Iy)⊗ Iz. Then there exists Iu ∈ Ix ⊚ Iy such that Iw ∈ Iu ⊗ Iz. Since Iu ∈ Ix ⊚ Iy, there exists u′ ∈ x ⊙ y such that uθu′, that is, Iu = Iu′ . Hence, Iu′ ∈ Ix ⊚ Iy. Since Iw ∈ Iu ⊗ Iz = Iu′ ⊗ Iz, there exists w′ ∈ u′ ⊛ z such that w′θw. Now, w′ ∈ u′ ⊛ z ⊆ (x ⊙ y) ⊛ z = (x ⊛ z) ⊙ y, by PHGR2 on H. Hence, w′ ∈ (x ⊛ z) ⊙ y and u′ ⊛ z ⊆ (x ⊙ y) ⊛ z. This means that there exists b ∈ x ⊛ z such that w′ ∈ b ⊙ y. Since b ∈ x ⊛ z, Ib ∈ Ix ⊗ Iz. Also, Iw′ ∈ Ib ⊚ Iy. Thus, Iw = Iw′ ∈ Ib ⊚ Iy ⊆ (Ix ⊗ Iz) ⊚ Iy. Hence, (Ix ⊚ Iy)⊗ Iz ⊆ (Ix ⊗ Iz)⊚ Iy. For the other set inclusion, let Iw ∈ (Ix ⊗ Iz)⊚ Iy. Then there exists Iu ∈ Ix ⊗ Iz such that Iw ∈ Iu⊚Iy. Since Iu ∈ Ix⊗Iz, there exists u ′ ∈ x⊛z such that uθu′, that is, Iu = Iu′ . Hence, Iu′ ∈ Ix⊗ Iz. Since Iw ∈ Iu⊚ Iy = Iu′ ⊚ Iy, there exists w ′ ∈ u′⊙y such that w′θw. Now, w′ ∈ u′ ⊙ y ⊆ (x⊛ z)⊙ y = (x⊙ y)⊛ z, by PHGR2 on H. Hence, w′ ∈ (x⊙ y)⊛ z and u′⊙y ⊆ (x⊛z)⊙y. This means that there exists b ∈ x⊙y such that w′ ∈ b⊛z. Now, b ∈ x⊙y implies Ib ∈ Ix⊚Iy. Also, Iw′ ∈ Ib⊗Iz. Thus, Iw = Iw′ ∈ Ib⊗Iz ⊆ (Ix⊚Iy)⊗Iz. Hence, (Ix ⊗ Iz)⊚ Iy ⊆ (Ix ⊚ Iy)⊗ Iz. Therefore, (Ix ⊗ Iz)⊚ Iy = (Ix ⊚ Iy)⊗ Iz and [PHGR2] holds. [PHGR3] By PHGR3 of H, 0 ∈ x⊛ x and 0 ∈ x⊙ x which means that I0 ∈ Ix ⊗ Ix and I0 ∈ Ix ⊚ Ix. This means that Ix ≪ Ix. Therefore, [PHGR3] holds. [PGHR4] Let Iw ∈ I0 ⊚ (I0 ⊗ Ix). Then there exists Iu ∈ I0 ⊗ Ix for which Iw ∈ I0 ⊚ Iu. Since Iu ∈ I0⊗Ix, there exists u ′ ∈ 0⊛x such that uθu′ and Iu = Iu′ . Hence, Iu′ ∈ I0⊗Ix. Since Iw ∈ I0 ⊚ Iu = I0 ⊚ Iu′ , there exists w′ ∈ 0⊙ u′ such that wθw′ and Iw = Iw′ . Now, w′ ∈ 0⊙u′ ⊆ 0⊙ (0⊛x) ≪ x, by PHGR4 of H. This means that w′ ≪ x. Thus, Iw′ ≪ Ix. Now, Iw = Iw′ and so, Iw ≪ Ix. Since Iw ∈ I0 ⊚ (I0 ⊗ Ix), we have I0 ⊚ (I0 ⊗ Ix) ≪ Ix. Therefore, [PHGR4] holds. [PHRG5] Let Iw ∈ (Ix ⊗ Iy) ⊗ Iz. Then there is Iu ∈ Ix ⊗ Iy such that Iw ∈ Iu ⊗ Iz. Since Iu ∈ Ix ⊗ Iy, there exists u′ ∈ x ⊛ y such that uθu′ and Iu = Iu′ . Also, since Iw ∈ Iu ⊛ Iz, there exists w′ ∈ u ⊛ z such that Iw = Iw′ . Since θ is a congruence on H and uθu′, by Lemma 3.7, (u ⊛ z)θ(u′ ⊛ z). Then there exists a ∈ u′ ⊛ z such that w′θa. Thus, Iw = Iw′ = Ia. Now, a ∈ u′ ⊛ z ⊆ (x⊛ y)⊛ z ≪ y⊙ z, PHGR5 on H. Hence, there exists b ∈ y ⊙ z such that a ≪ b. This means that 0 ∈ a⊛ b and 0 ∈ a⊙ b. Furthermore, Ib ∈ Iy ⊚ Iz, I0 ∈ Ia ⊚ Ib, and I0 ∈ Ia ⊗ Ib . Since Iw = Iw′ = Ia, we have I0 ∈ Iw ⊚ Ib and I0 ∈ Iw ⊗ Ib and hence, Iw ≪ Ib. Note that Ib ∈ Iy ⊚ Iz. Thus, (Ix ⊗ Iy)⊛ Iz ≪ Iy ⊚ Iz. Hence, [PHGR5] holds. Therefore, (H/I;⊗,⊚, I) is a pseudo hyper GR-algebra. □ Lemma 3.9. Let θ and θ′ be two regular congruences on H such that R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1862 [0]θ = [0]θ′ . Then θ = θ′. Proof. It is enough to show that xθy ⇐⇒ xθ′y. If xθy, then (x ⊛ x)θ̄(x ⊛ y). Since 0 ∈ x ⊛ x and θ is a congruence on H, there exists z ∈ x ⊛ y such that 0θz. Then, z ∈ [0]θ = [0]θ′ and z ∈ [0]′θ, that is, 0θ′z. Hence, {0}θ′(x⊛ y). Similarly, we can also show that {0}θ′(y⊛ x). Thus, xθ′y since θ′ is regular. Following the same argument, xθ′y implies xθy. Therefore, θ = θ′. □ 4. Isomorphism Theorems of Pseudo Hyper GR-algebras This section discusses some hyper isomorphism theorems of pseudo hyper GR-algebras, namely, the first and the third hyper isomorphism theorems. All throughout, H and H ′ are pseudo hyper GR-algebras, unless otherwise stated. Lemma 4.1. Let θ be a regular congruence on a pseudo hyper GR-algebraH and I = [0]θ. Then the map π : H −→ H/I defined by π(x) = Ix, for all x ∈ H, is an epimorphism, called the canonical epimorphism. Proof. Let x, y ∈ H such that x = y. Then π(x) = Ix = [x]θ = [y]θ = Iy = π(y). Hence, π is a well-defined map. Now, observe that π(0) = I0 = I. Next, we will show that π is a homomorphism. Pick x, y ∈ H. Let J ∈ π(x) ⊛ π(y). Then J ∈ Ix ⊛ Iy and so there exists element u ∈ x ⊛ y such that J = Iu = π(u) ∈ π(x⊛y). Thus, π(x)⊛π(y) ⊆ π(x⊛y). Now, let L ∈ π(x⊛y). Then there exists an element v ∈ x ⊛ y such that L = π(v) = Iv. Note that Iv ∈ Ix⊛ Iy = π(x)⊛π(y). Hence, we have L ∈ π(x)⊛π(y) and π(x⊛ y) ⊆ π(x)⊛π(y). Thus, π(x⊛ y) = π(x)⊛π(y). Similarly, for the hyperoperation ⊙, we can show that π(x⊙ y) = π(x)⊙ π(y). Thus, π is a hyper homomorphism. Let Ix ∈ H/I with x ∈ H. Then π(x) = Ix ∈ H/I. Therefore, π is a surjective map and so, an epimorphism. □ Theorem 4.2. (Homomorphism Theorem) Let θ be a regular congruence relation on H and I = [0]θ. If f : H −→ H ′ is a homomorphism of pseudo hyper GR-algebras such that f(x) ≪ f(y) and f(y) ≪ f(x) imply that f(x) = f(y) for all x, y ∈ H, then f̄ : H/I −→ H ′, which is defined by f̄(Ix) = f(x), for all x ∈ H, is a unique homomorphism such that f̄ ◦ π = f , where π denotes the canonical epimorphism and ◦ is the composition map. Moreover, if I = ker f , then f̄ is a monomorphism. Proof. Let θ be a regular congruence relation on H and I = [0]θ. Define f̄ : H/I −→ H ′ by f̄(Ix) = f(x) for all x ∈ H. Let x, y ∈ H such that Ix = Iy and let t ∈ Ix = Iy. Since H/I is a pseudo hyper GR-algebra, by [PHGR3], t ≪ t and so, Ix ≪ Iy. Hence I ∈ Ix ⊛ Iy and I ∈ Ix ⊙ Iy. It follows that there exist z ∈ x⊛ y and z ∈ x⊙ y such that I = Iz. Thus, z ∈ I ⊆ ker f and so f(z) = 0′. Since f is a hyper homomorphism, we have R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1863 0′ = f(z) = f(x⊛y) = f(x)⊛f(y) and 0′ = f(z) = f(x⊙y) = f(x)⊙f(y). it follows that f(x) ≪ f(y). Using the same argument, picking t′ ∈ Iy = Ix will imply that f(y) ≪ f(x). Thus, by the hypothesis f(x) ≪ f(y) and f(y) ≪ f(x) imply that f(x) = f(y). Hence, f̄(Ix) = f̄(Iy) and f̄ is a well-defined map. Let Ix, Iy ∈ H/I. We will show that f̄(Ix ⊙ Iy) = f̄(Ix) ⊙ f̄(Iy). Let w ∈ f̄(Ix ⊙ Iy). Then there exists It ∈ Ix ⊙ Iy such that w = f̄(It) = f(t). Now, It ∈ Ix ⊙ Iy implies that t ∈ x⊙ y and w = f(t) ∈ f(x⊙ y) = f(x)⊙ f(y) = f̄(Ix)⊙ f̄(Iy), Thus, f̄(Ix ⊙ Iy) ⊆ f̄(Ix)⊙ f̄(Iy). Now, let u ∈ f̄(Ix)⊙ f̄(Iy) = f(x)⊙f(y) = f(x⊙y). Then there exists an element v ∈ x⊙ y such that u = f(v). Hence, Iv ∈ Ix⊙ Iy, and we have u = f(v) = f̄(Iv) ∈ f̄(Ix⊙ Iy). Thus, f̄(Ix)⊙ f̄(Iy) ⊆ f̄(Ix ⊙ Iy). Hence, f̄(Ix ⊙ Iy) = f̄(Ix)⊙ f̄(Iy). In a similar manner, we can show that f̄(Ix ⊛ Iy) = f̄(Ix)⊛ f̄(Iy). Hence, f̄ is a homomorphism. Now, dom(f̄ ◦ π) = H = domf and for all x ∈ H, (f̄ ◦ π)(x) = f̄(π(x)) = f̄(Ix) = f(x). Hence, f̄ ◦ π = f . To show the uniqueness of f̄ , we suppose that there is another homomorphism g such that g ◦ π = f . Let x ∈ H. Then g(Ix) = g(π(x)) = f(x) = f̄(π(x)) = f̄(Ix). Now, we will show that if I = ker f , then f̄ is a monomorphism. Suppose that f̄(Ix) = f̄(Iy) with x, y ∈ H. Then f(x) = f(y). Since f is a homomorphism and by [PHGR3], 0H′ = f(0H) ∈ f(x⊛ x) = f(x)⊛ f(x) = f(x)⊛ f(y) = f(x⊛ y). So, there exists u ∈ x⊛ y such that f(u) = 0H′ . Hence, u ∈ ker f = I and so, uθ0. It follows that (x⊛ y)θ{0}. Also, 0H′ = f(0H) ∈ f(x⊛ x) = f(x)⊛ f(x) = f(y)⊛ f(x) = f(y ⊛ x). So, there exists v ∈ y ⊛ x such that f(v) = 0H′ . Then v ∈ ker f = I and so, vθ0. Hence, (y ⊛ x)θ{0}. Similarly, 0H′ = f(0H) ∈ f(x⊙ x) = f(x)⊙ f(x) = f(x)⊙ f(y) = f(x⊙ y). So, there exists u ∈ x ⊙ y such that f(u) = 0H′ . This means that u ∈ ker f = I and so, uθ0 which implies that (x⊙ y)θ{0}. Lastly, 0H′ = f(0H) ∈ f(x⊙ x) = f(x)⊙ f(x) = f(y)⊙ f(x) = f(y ⊙ x). R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1864 Thus, there exists v ∈ y ⊙ x such that f(v) = 0H′ . Moreover, v ∈ ker f = I and vθ0. Thus, (y ⊙ x)θ{0}. Since θ is a regular congruence relation, it follows that xθy. Thus, Ix = Iy. Therefore, f̄ is a one-to-one map. This proves the theorem. □ Before we prove the First Isomorphism Theorem, let us consider the following example which is a specific case of the next theorem. Example 4.3. Consider the pseudo hyper GR-algebra H = {0, 1, 2} in Example 3.6. We can verify that the given congruence relation θ on H is regular. In our case, I = [0]θ = {0, 1} and I2 = {2}. Then H/I = {I, I2} whose Cayley table is shown below ⊛ I I2 I {I} {I} I2 {I, I2} {I, I2} ⊙ I I2 I {I} {I, I2} I2 {I, I2} {I, I2} By routine calculations, H/I is a pseudo hyper GR-algebra. Now, consider the set H ′ = {0, 1} together with the Cayley tables below ⊛ 0 1 0 {0} {0} 1 {0, 1} {0, 1} ⊙ 0 1 0 {0} {0, 1} 1 {0, 1} {0, 1} By routine calculations, H ′ is a pseudo hyper GR-algebra. Define the map f : H −→ H ′ by f(0) = 0 = f(1) and f(2) = 1. Then f is a homomorphism as shown in table below x y x⊛ y f(x⊛ y) f(x) f(y) f(x)⊛ f(y) 0 0 {0} {0} 0 0 {0} 0 1 {0} {0} 0 0 {0} 0 2 {0} {0} 0 1 {0} 1 0 {0, 1} {0} 0 0 {0} 1 1 {0, 1} {0} 0 0 {0} 1 2 {0, 2} {0} 0 1 {0} 2 0 {0, 2} {0, 1} 1 0 {0, 1} 2 1 {0, 2} {0, 1} 1 0 {0, 1} 2 2 {0, 2} {0, 1} 1 1 {0, 1} R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1865 x y x⊙ y f(x⊙ y) f(x) f(y) f(x)⊙ f(y) 0 0 {0} {0} 0 0 {0} 0 1 {0, 1} {0} 0 0 {0} 0 2 {0, 2} {0, 1} 0 1 {0, 1} 1 0 {0, 1} {0} 0 0 {0} 1 1 {0, 1} {0} 0 0 {0} 1 2 {0, 1, 2} {0, 1} 0 1 {0, 1} 2 0 {0, 2} {0, 1} 1 0 {0, 1} 2 1 {0, 1, 2} {0, 1} 1 0 {0, 1} 2 2 {0, 2} {0, 1} 1 1 {0, 1} According how f is being defined, we have ker f = {0, 1} = [0]θ = I and Imf = {0, 1} = H ′. Let us define the map φ : H/I −→ H ′ by φ(Ix) = { 0 if Ix = I 1 otherwise. We can verify that φ is an isomorphism. Moreover, H/ker f ∼= Imf . Corollary 4.4. (First Isomorphism Theorem) Let θ be a regular congruence relation on H and I = [0]θ. If f : H −→ H ′ is a hyper homomorphism of pseudo hyper GR- algebras such that f(x) ≪ f(y) and f(y) ≪ f(x) imply that f(x) = f(y) and ker f = I, then H/ker f ∼= Imf . Proof. By Theorem 4.2, the map f̄ : H/ker f −→ H ′ is a monomorphism and by Remark ?? (ii), f̄ : H/ker f ∼= Im f̄ is an isomorphism. Thus, H/ker f ∼= Im f̄ . Since f̄(Ix) = f(x) for all x ∈ H, Im f̄ = Imf . Hence, the result follows. □ Proposition 4.5. Let K be a pseudo hyper subGR-algebra of H and θ a regular congru- ence on H and I = [0]θ. Define K/I = {Ix ∈ H/I |x ∈ K}. Then K/I is a pseudo hyper subGR-algebra of H/I. Proof. Since K is a pseudo hyper subGR-algebra of H, 0 ∈ K and so, [0]θ = I ∈ K/I which means that K/I is nonempty. Let Ix, Iy ∈ K/I. Then x, y ∈ K. Since K is a pseudo hyper subGR algebra of H, x ⊛ y ⊆ K. Suppose that Iz ∈ Ix ⊛ Iy. Then z ∈ x ⊛ y ⊆ K. Thus, Iz ∈ K/I and Ix ⊛ Iy ⊆ K/I. Similarly, suppose that Iz ∈ Ix ⊙ Iy. Then, z ∈ x ⊙ y ⊆ K, and so, Iz ∈ K/I. Thus, Ix ⊙ Iy ⊆ K/I. Therefore, K/I is a pseudo hyper subGR-algebra of H/I. □ Corollary 4.6. Let θ1 and θ2 be regular congruence relations on H, with J = [0]θ1 and I = [0]θ2 . Define J/I = {Ix ∈ H/I |x ∈ J}. If J is a pseudo hyper subGR-algebra of H, then J/I is a pseudo hyper subGR-algebra of H/I. Proof. By definition of J/I, it is easy to see that J/I ⊆ H/I. Since J is a pseudo hyper subGR-algebra of H, by Proposition 4.5, J/I is pseudo hyper subGR-algebra of H. □ R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1866 Lemma 4.7. Let θ1 and θ2 be regular congruence relations on H, with J = [0]θ1 and I = [0]θ2 . Define J/I = {Ix ∈ H/I |x ∈ J} and a relation θ on H/I by IxθIy if and only if xθ1y, for all Ix, Iy ∈ H/I. Then θ is a regular congruence relation and [I]θ = J/I. Proof. Define a relation θ on H/I by IxθIy if and only if xθ1y, for all Ix, Iy ∈ H/I. We will show that θ is a regular congruence relation on H/I. We will show first that θ is an equivalence relation on H/I. Note that θ1 is an equiv- alence relation on H. Let Ix ∈ H/I. Then x ∈ H and xθ1x on H, that is, IxθIx, which means that reflexivity of θ on H/I holds. Now, let Ix, Iy ∈ H/I such that IxθIy. Then x, y ∈ H and xθ1y on H. Since θ1 is a symmetric relation on H, yθ1x on H and so IyθIx on H/I, that is, θ is a symmetric relation on H/I. Assume that IxθIy and IyθIz, where Ix, Iy, Iz ∈ H/I. Then x, y, z ∈ H and xθ1y and yθ1z on H. Since θ1 is transitive relation on H, we have xθ1z which tells us that IxθIz. Hence, θ is a transitive relation on H/I. Therefore, θ is an equivalence relation. Now, we will show that θ is a congruence relation on H/I. Note that θ1 is a congruence relation on H. Let Ia, Ix, Iy ∈ H/I for some a, x, y ∈ H such that IxθIy on H/I. Note that θ1 is a regular congruence. By definition of θ on H/I, xθ1y on H and Definition 3.1(viii), (a⊛x)θ̄1(a⊛y), (x⊛a)θ̄1(y⊛a), (a⊙x)θ̄1(a⊙y), and (x⊙a)θ̄1(y⊙a). Thus, for each u ∈ a⊛ x, there exists v ∈ a⊛ y such that uθ1v and for each v ∈ a⊛ y, there exists u ∈ a⊛ x such that uθ1v. This means that for each Iu ∈ Ia ⊛ Ix, there exists Iv ∈ Ia ⊛ Iy such that IuθIv and for every Iv ∈ Ia ⊛ Iy, there exists Iu ∈ Ia ⊛ Ix such that IuθIv. Hence, (Ia⊛ Ix)θ̄(Ia⊛ Iy). In a similar manner, we can also show that (Ix⊛ Ia)θ̄(Iy ⊛ Ia). On the other hand, for each r ∈ a ⊙ x, there exists s ∈ a ⊙ y such that rθ1s and for all s ∈ a ⊙ y, there exists r ∈ a ⊙ x such that rθ1s. This means that for each Ir ∈ Ia ⊙ Ix, there exists Is ∈ Ia⊙ Iy such that IrθIs and for every Is ∈ Ia⊛ Iy, there exists Ir ∈ Ia⊛ Ix such that IrθIs. Hence, (Ia ⊙ Ix)θ̄(Ia ⊙ Iy). In a similar manner, we can also show that (Ix ⊙ Ia)θ̄(Iy ⊛ Ia). Therefore, θ is a congruence relation on H/I. Finally, we will show that θ is a regular congruence relation on H/I. Suppose that (Ix⊛ Iy)θ{I}, (Iy ⊛ Ix)θ{I}, (Ix⊙ Iy)θ{I}, and (Iy ⊙ Ix)θ{I}. Then there exist u ∈ x⊛ y, v ∈ y ⊛ x, r ∈ x ⊙ y, and s ∈ y ⊙ x such that Iuθ{I}, Ivθ{I}, Irθ{I}, and Isθ{I}, that is, uθ10, vθ10, rθ10, and sθ10. So we have, (x⊛ y)θ1{0}, (y ⊛ x)θ1{0},(x⊙ y)θ1{0}, and (y ⊙ x)θ1{0}. Since θ1 is a regular congruence relation on H, xθ1y, that is, IxθIy. Therefore, θ is a regular congruence relation on H/I. Now, [I]θ = {Ix ∈ H/I | IxθI} = {Ix ∈ H/I |xθ10} = {Ix ∈ H/I |x ∈ [0]θ1 = J} = {Ix ∈ H/I |x ∈ J} = J/I Theorem 4.8. (Third Isomorphism Theorem) Let f : H −→ H ′ be a homomorphism of pseudo hyper GR-algebras H and H ′ such that f(x) ≪ f(y) and f(y) ≪ f(x) imply that f(x) = f(y). Suppose further that θ1 and θ2 are regular congruence relations on H R. Manzano, Jr., G. Petalcorin, Jr. / Eur. J. Pure Appl. Math, 15 (4) (2022), 1854-1868 1867 with J = [0]θ1 and I = [0]θ2 . Then (H/I)/(J/I) ∼= H/J , where J/I = {Ix ∈ H/I |x ∈ J}. Proof. Let f : H −→ H ′ be a homomorphism of pseudo hyper GR-algebras H and H ′ such that f(x) ≪ f(y) and f(y) ≪ f(x) imply that f(x) = f(y). Suppose further that θ1 and θ2 are regular congruence relations on H with J = [0]θ1 and I = [0]θ2 . Let θ be a regular congruence relation on H/I defined in Lemma 4.7. We define the map φ : H/I → H/J by φ(Ix) = Jx. Suppose that Ix = Iy. Since θ is a reflexive relation on H/I, IxθIy on H/I which means that xθ1y. Note that J = [0]θ1 . Now, if x ∈ J = [0]θ2 , then xθ10 and 0θ1x. Since, xθ1y, by transitivity of θ1 on H, 0θ1y and so yθ10 which will imply that y ∈ [0]θ1 = J . Thus, φ(Ix) = Jx = J = Jy = φ(Iy) and φ is a well-defined map. We will now show that φ is a homomorphism. Note that φ)(I) = J . Let Ix, Iy ∈ H/I. Let Jz ∈ φ(Ix ⊛ Iy). Then, there exists Iu ∈ Ix ⊛ Iy such that φ(Iu) = Jz, that is, Ju = φ(Iu) = Jz. Since Iu ∈ Ix ⊛ Iy, u ∈ x ⊛ y. Hence, Ju ∈ Jx ⊛ Jy. Thus, Jz = Ju ∈ Jx ⊛ Jy = φ(Ix) ⊛ φ(Iy) and so, φ(Ix ⊛ Iy) ⊆ φ(Ix) ⊛ φ(Iy). Next, let Jz′ ∈ φ(Ix) ⊛ φ(Iy) = Jx ⊛ Jy. Then, there exists an element u′ ∈ x ⊛ y such that Jz′ = Ju′ = φ(Iu′). Since u′ ∈ x ⊛ y, Iu′ ∈ Ix ⊛ Iy and Jz′ = φ(Iu′) ∈ φ(Ix ⊛ Iy). So, φ(Ix)⊛ φ(Iy) ⊆ φ(Ix ⊛ Iy). Therefore, φ(Ix ⊛ Iy) = φ(Ix)⊛ φ(Iy). Now, let Jz ∈ φ(Ix ⊙ Iy). Then, there exists an element Iu ∈ Ix ⊙ Iy such that φ(Iu) = Jz, that is, Ju = φ(Iu) = Jz. Since Iu ∈ Ix ⊙ Iy, u ∈ x⊙ y. Hence, Ju ∈ Jx ⊙ Jy. Thus, Jz = Ju ∈ Jx ⊙ Jy = φ(Ix) ⊙ φ(Iy) and so, φ(Ix ⊙ Iy) ⊆ φ(Ix) ⊙ φ(Iy). Next, let Jz′ ∈ φ(Ix) ⊙ φ(Iy) = Jx ⊙ Jy. Then, there exists an element u′ ∈ x ⊙ y such that Jz′ = Ju′ = φ(Iu′). Since u′ ∈ x ⊙ y, Iu′ ∈ Ix ⊙ Iy and so, Jz′ = φ(Iu′) ∈ φ(Ix ⊙ Iy). So, φ(Ix)⊙ φ(Iy) ⊆ (Ix ⊙ Iy). Hence, φ(Ix ⊙ Iy) = φ(Ix)⊙ φ(Iy) and φ is a homomorphism. Now, ker φ = {Ix ∈ H/I |φ(Ix) = [0]θ1} = {Ix ∈ H/I | Jx = [0]θ1 = J} = {Ix ∈ H/I |x ∈ J} = J/I. By Lemma 4.7, ker φ = J/I = [I]θ. Lastly, we will show that φ is onto. Let z ∈ H and Iz ∈ H/I. Thus, there exists an element Iz ∈ H/I such that φ(Iz) = Jz and so φ is onto. Therefore, by the First Isomorphism Theorem, (H/I)/(J/I) ∼= Imφ = H/J . □ Acknowledgements The authors would like to thank the referees who gave their brilliant suggestions in refining this paper before publication to the European Journal of Pure and Applied Math- ematics. REFERENCES 1868 References [1] Imai, Y. and Iseki, K., On Axiom systems of Propositional Calculi XIV, Proc. Japan Academy, 42 (1966), 19-22. [2] Indangan R.A. and Petalcorin G.C., Some Results on Hyper GR-ideals of a Hyper GR-algebra, Journal of Algebra and Applied Mathematics, 14, (2016), 101-119. [3] Indangan R.A., Petalcorin G.C. and Villa, A.A., Some Hyper Homomorphic Properties on Hyper GR-Algebras, Journal of Algebra and Applied Mathematics, 15, (2017), 100- 121. [4] Manzano, Jr., R.G. and Petalcorin, Jr., G.C. 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