108 This work is licensed under a Creative Commons Attribution 4.0 International License. (𝜽𝟏,𝜽𝟐)-Derivation Pair on Rings Mohammed Khalid Shahoodh Ministry of Education, Ramadi Directorate of Education, Anbar-Iraq moha861122@yahoo.com Abstract Ring theory is one of the influential branches of abstract algebra. In this field, many algebraic problems have been considered by mathematical researchers who are working in this field. However, some new concepts have been created and developed to present some algebraic structures with their properties. Rings with derivations have been studied fifty years ago, especially the relationships between the derivations and the structure of a ring. By using the notatin of derivation, many results have been obtained in the literature with different types of derivations. In this paper, the concept of the derivation theory of a ring has been considered. This study presented the definition of (𝜃1,𝜃2)-derivation pair and Jordan (𝜃1, 𝜃2)-derivation pair on an associative ring Γ, and the relation between them. Furthermore, we study the concept of prime rings under this notion by introducing some of its properties where 𝜃1 and 𝜃2 are two mappings of Γ into itself. Keywords: Ring Theory, Derivation theory, Prime ring, Derivation pair, Semiprime ring. 1. Introduction The study of derivation has been initiated from the development of Galois theory and the theory of invariants. This theory has been studied very widely by many researchers on various algebraic structures. The author in [1] studied this topic on 𝐻∗-algebra by introducing Jordan ∗-derivation pair. While the authors in [2] considered the topic of BCI-algebras, and the same topic has been investigated on BCC-algebras by the authors in [3]. Moreover, some other works with different algebraic structures can be found in [4-5]. On the other hand, some other studies have studied this topic with some types of rings such as prime and semiprime rings, see [6-8]. [12] presented a new definition of derivation pair instead of Jordan ∗-derivation pair which was provided by [1]. In this paper, we extended the results of [12] by introducing the notion of (𝜃1,𝜃2)-derivation pair and studied some of its properties. 2. Basic Concepts This section contains some of the previous results that are needed in this study which are as follows: Definition 2.1[9] Ibn Al Haitham Journal for Pure and Applied Sciences Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index Doi: 10.30526/35.2.2723 Article history: Received 21 November, 2021, Accepted,25, January, 2022, Published in April 2022. https://creativecommons.org/licenses/by/4.0/ mailto:moha861122@yahoo.com Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 109 A non-empty set Γ is said to be an associative ring, if for all 𝑐1, 𝑐2, 𝑐3 ∈ Γ there exist two binary operations defined on Γ and denoted by + and ⋅ respectively, such that i. 𝑐1 + 𝑐2 = 𝑐2 + 𝑐1 ii. (𝑐1 + 𝑐2) + 𝑐3 = 𝑐1 + (𝑐2 + 𝑐3) iii. ∀𝑐1 ∈ Γ ∃ 0 ∈ Γ such that 𝑐1 + 0 = 0 + 𝑐1 = 𝑐1 iv. ∀𝑐1 ∈ Γ ∃ − 𝑐1 ∈ Γ such that −𝑐1 + 𝑐1 = 𝑐1 + (−𝑐1) = 0 v. 𝑐1 ⋅ 𝑐2 ∈ Γ vi. (𝑐1 ⋅ 𝑐2) ⋅ 𝑐3 = 𝑐1 ⋅ (𝑐2 ⋅ 𝑐3) vii. 𝑐1 ⋅ (𝑐2 + 𝑐3) = 𝑐1 ⋅ 𝑐2 + 𝑐1 ⋅ 𝑐3 viii. (𝑐1 + 𝑐2) ⋅ 𝑐3 = 𝑐1 ⋅ 𝑐3 + 𝑐2 ⋅ 𝑐3. Definition 2.2 [9] A ring Γ is said to be a prime ring if for each 𝑐1, 𝑐2 ∈ Γ, 𝑐1Γ𝑐2 = 0 implies that 𝑐1 = 0 or 𝑐2 = 0. Definition 2.3 [9] A ring Γ is said to be 𝑘-torsion-free if whenever 𝑘𝑐 = 0 implies that 𝑐 = 0, where 𝑐 ∈ Γ and 𝑘 ≠ 0. Definition 2.4 [10] Let Γ be a ring, then [𝑐1, 𝑐2] is said to be Lie product and given as [𝑐1, 𝑐2] = 𝑐1𝑐2 − 𝑐2𝑐1 and 𝑐1 ∘ 𝑐2 is said to be Jordan product and given as 𝑐1 ∘ 𝑐2 = 𝑐1𝑐2 + 𝑐2𝑐1. Definition 2.5 [11] The characteristic of a ring Γ (for short char(Γ)) is the smallest positive integer 𝑧 such that 𝑧𝑟 = 0 with 𝑟 ∈ Γ. Otherwise, char(Γ) = 0. Definition 2.6 [12] Let Γ be a ring and let 𝜇, 𝜎: Γ ⟶ Γ be two additive mappings, then 𝜇, 𝜎 are said to be derivation pair (𝜇, 𝜎) if the following equations are holds: 𝜇(𝑢𝑣𝑢) = 𝜇(𝑢)𝑣𝑢 + 𝑢𝜎(𝑣)𝑢 + 𝑢𝑣𝜇(𝑢), for each 𝑢, 𝑣 ∈ Γ 𝜎(𝑢𝑣𝑢) = 𝜎(𝑢)𝑣𝑢 + 𝑢𝜇(𝑣)𝑢 + 𝑢𝑣𝜎(𝑢), for each 𝑢, 𝑣 ∈ Γ and are called Jordan derivation pair if: 𝜇(𝑢3) = 𝜇(𝑢)𝑢2 + 𝑢𝜎(𝑢)𝑢 + 𝑢2𝜇(𝑢), for each 𝑢 ∈ Γ 𝜎(𝑢3) = 𝜎(𝑢)𝑢2 + 𝑢𝜇(𝑢)𝑢 + 𝑢2𝜎(𝑢), for each 𝑢 ∈ Γ. 3. Main Results In this section, we presented the notion of (𝜃1, 𝜃2)-derivation pair on the ring Γ where 𝜃1 and 𝜃2 are two mappings from the ring Γ into itself. Moreover, some properties of this concept have been proved. Definition 3.1 Let Γ be a ring. Let 𝛿1, 𝛿2: Γ ⟶ Γ be additive mappings, then (𝛿1, 𝛿2) is said to be (𝜃1, 𝜃2)-derivation pair, if the following are holds: 𝛿1(𝑢𝑣𝑢) = 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢), for each 𝑢, 𝑣 ∈ Γ 𝛿2(𝑢𝑣𝑢) = 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿2(𝑢), for each 𝑢, 𝑣 ∈ Γ. and are said to be Jordan (𝜃1, 𝜃2)-derivation pair, if the following are holds: 𝛿1(𝑢 3) = 𝛿1(𝑢)𝜃1(𝑢 2) + 𝜃2(𝑢)𝛿2(𝑢)𝜃1(𝑢) + 𝜃2(𝑢 2)𝛿1(𝑢) for all 𝑢 ∈ Γ Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 110 𝛿2(𝑢 3) = 𝛿2(𝑢)𝜃1(𝑢 2) + 𝜃2(𝑢)𝛿1(𝑢)𝜃1(𝑢) + 𝜃2(𝑢 2)𝛿2(𝑢) for all 𝑢 ∈ Γ. Example 3.1 Let Γ be a non-commutative ring, let 𝑐1, 𝑐2 ∈ Γ such that 𝜃2(𝑢)𝑐1 = 𝜃2(𝑢)𝑐2 = 0 (resp. 𝜃2(𝑣)𝑐1 = 𝜃2(𝑣)𝑐2 = 0) for all 𝑢, 𝑣 ∈ Γ .Define 𝛿1, 𝛿2: Γ ⟶ Γ as follows: 𝛿1(𝑢) = 𝑐1𝜃1(𝑢) and 𝛿2(𝑢) = 𝑐2𝜃1(𝑢), ∀𝑢 ∈ Γ, where 𝜃1, 𝜃2: Γ ⟶ Γ are two endomorphism mappings. Then (𝛿1, 𝛿2) is a (𝜃1, 𝜃2)-derivation pair of Γ. Let 𝑢, 𝑣 ∈ Γ, then 𝛿1(𝑢𝑣𝑢) = 𝑐1𝜃1(𝑢𝑣𝑢) = 𝑐1𝜃1(𝑢(𝑣𝑢)) = 𝑐1𝜃1(𝑢)𝜃1(𝑣𝑢) = 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝑐2𝜃1(𝑣𝑢) = 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝑐2𝜃1(𝑣)𝜃1(𝑢) = 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢)𝜃2(𝑣)𝑐1𝜃1(𝑢) = 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢)𝜃2(𝑣)𝛿1(𝑢) = 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢) Also, 𝛿2(𝑢𝑣𝑢) = 𝑐2𝜃1(𝑢𝑣𝑢) = 𝑐2𝜃1(𝑢(𝑣𝑢)) = 𝑐2𝜃1(𝑢)𝜃1(𝑣𝑢) = 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝑐1𝜃1(𝑣𝑢) = 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝑐1𝜃1(𝑣)𝜃1(𝑢) = 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢)𝜃2(𝑣)𝑐2𝜃1(𝑢) = 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢)𝜃2(𝑣)𝛿2(𝑢) = 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿2(𝑢) Thus, (𝛿1, 𝛿2) is a (𝜃1, 𝜃2)-derivation pair of Γ. Remark 3.1: Every (𝜃1, 𝜃2)-derivation pair is a Jordan (𝜃1, 𝜃2)-derivation pair, but the converse is not true in general. Example 3.2: Let Γ be a 2-torsion free non-commutative ring, let 𝑐 ∈ Γ such that 𝜃2(𝑢)𝑐𝜃1(𝑢) = 0, ∀𝑢 ∈ Γ, but 𝜃2(𝑢)𝑐𝜃1(𝑣) ≠ 0 for some 𝑢 ≠ 𝑣 ∈ Γ. Define 𝛿1, 𝛿2: Γ ⟶ Γ as follows: 𝛿1(𝑢) = 𝜃2(𝑢)𝑐 + 𝑐𝜃1(𝑢) and 𝛿2(𝑢) = 𝜃2(𝑢)𝑐 − 𝑐𝜃1(𝑢) , ∀𝑢 ∈ Γ, where 𝜃1, 𝜃2: Γ ⟶ Γ are two endomorphisms. Then (𝛿1, 𝛿2) is a Jordan (𝜃1, 𝜃2)-derivation but not (𝜃1, 𝜃2)-derivation. Let 𝑢, 𝑣 ∈ Γ, then 𝛿1(𝑢 3) = 𝜃2(𝑢 3)𝑐 + 𝑐𝜃1(𝑢 3) and 𝛿1(𝑢 3) = 𝛿1(𝑢)𝜃1(𝑢 2) + 𝜃2(𝑢)𝛿2(𝑢)𝜃1(𝑢) + 𝜃2(𝑢 2)𝛿1(𝑢). Thus, 𝛿1(𝑢)𝜃1(𝑢 2) + 𝜃2(𝑢)𝛿2(𝑢)𝜃1(𝑢) + 𝜃2(𝑢 2)𝛿1(𝑢) = (𝜃2(𝑢)𝑐 + 𝑐𝜃1(𝑢))𝜃1(𝑢 2) + 𝜃2(𝑢)(𝜃2(𝑢)𝑐 − 𝑐𝜃1(𝑢))𝜃1(𝑢) + 𝜃2(𝑢 2)(𝜃2(𝑢)𝑐 + 𝑐𝜃1(𝑢)) = (𝜃2(𝑢)𝑐𝜃1(𝑢)𝜃1(𝑢) + 𝑐𝜃1(𝑢)𝜃1(𝑢)𝜃1(𝑢)) + (𝜃2(𝑢)𝜃2(𝑢)𝑐𝜃1(𝑢) − 𝜃2(𝑢)𝑐𝜃1(𝑢)𝜃1(𝑢)) + (𝜃2(𝑢)𝜃2(𝑢)𝜃2(𝑢)𝑐 + 𝜃2(𝑢)𝜃2(𝑢)𝑐𝜃1(𝑢)) = 𝑐𝜃1(𝑢)𝜃1(𝑢)𝜃1(𝑢) + 𝜃2(𝑢)𝜃2(𝑢)𝜃2(𝑢)𝑐 = 𝜃2(𝑢 3)𝑐 + 𝑐𝜃1(𝑢 3). Also, 𝛿2(𝑢) = 𝜃2(𝑢)𝑐 − 𝑐𝜃1(𝑢) and 𝛿2(𝑢 3) = 𝛿2(𝑢)𝜃1(𝑢 2) + 𝜃2(𝑢)𝛿1(𝑢)𝜃1(𝑢) + 𝜃2(𝑢 2)𝛿2(𝑢). Thus, 𝛿2(𝑢)𝜃1(𝑢 2) + 𝜃2(𝑢)𝛿1(𝑢)𝜃1(𝑢) + 𝜃2(𝑢 2)𝛿2(𝑢) = (𝜃2(𝑢)𝑐 − 𝑐𝜃1(𝑢))𝜃1(𝑢 2) + 𝜃2(𝑢)(𝜃2(𝑢)𝑐 + 𝑐𝜃1(𝑢))𝜃1(𝑢) + 𝜃2(𝑢 2)(𝜃2(𝑢)𝑐 − 𝑐𝜃1(𝑢)) = (𝜃2(𝑢)𝑐𝜃1(𝑢)𝜃1(𝑢) − 𝑐𝜃1(𝑢)𝜃1(𝑢)𝜃1(𝑢)) + (𝜃2(𝑢)𝜃2(𝑢)𝑐𝜃1(𝑢) + 𝜃2(𝑢)𝑐𝜃1(𝑢)𝜃1(𝑢)) +(𝜃2(𝑢)𝜃2(𝑢)𝜃2(𝑢)𝑐 − 𝜃2(𝑢)𝜃2(𝑢)𝑐𝜃1(𝑢)) = −𝑐𝜃1(𝑢)𝜃1(𝑢)𝜃1(𝑢) + 𝜃2(𝑢)𝜃2(𝑢)𝜃2(𝑢)𝑐 = 𝜃2(𝑢 3)𝑐 − 𝑐𝜃1(𝑢 3). Therefore, (𝛿1, 𝛿2) is a Jordan(𝜃1, 𝜃2)-derivation pair. Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 111 Now, 𝛿1(𝑢𝑣𝑢) = 𝜃2(𝑢𝑣𝑢)𝑐 + 𝑐𝜃1(𝑢𝑣𝑢) and 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢) = (𝜃2(𝑢)𝑐 + 𝑐𝜃1(𝑢))𝜃1(𝑣𝑢) + 𝜃2(𝑢)(𝜃2(𝑣)𝑐 − 𝑐𝜃1(𝑣))𝜃1(𝑢) + 𝜃2(𝑢𝑣)(𝜃2(𝑢)𝑐 + 𝑐𝜃1(𝑢)) = 𝑐𝜃1(𝑢𝑣𝑢) + 𝜃2(𝑢𝑣𝑢)𝑐 + 2𝜃2(𝑢𝑣)𝑐𝜃1(𝑢) = 𝜃2(𝑢𝑣𝑢)𝑐 + 𝑐𝜃1(𝑢𝑣𝑢). also, 𝛿2(𝑢𝑣𝑢) = 𝜃2(𝑢𝑣𝑢)𝑐 − 𝑐𝜃1(𝑢𝑣𝑢) and 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿2(𝑢) = (𝜃2(𝑢)𝑐 − 𝑐𝜃1(𝑢))𝜃1(𝑣𝑢) + 𝜃2(𝑢)(𝜃2(𝑣)𝑐 + 𝑐𝜃1(𝑣))𝜃1(𝑢) + 𝜃2(𝑢𝑣)(𝜃2(𝑢)𝑐 − 𝑐𝜃1(𝑢)) = −𝑐𝜃1(𝑢𝑣𝑢) + 𝜃2(𝑢𝑣𝑢)𝑐 + 2𝜃2(𝑢)𝑐𝜃1(𝑣𝑢). Since 𝜃2(𝑢)𝑐𝜃1(𝑣) ≠ 0 for some 𝑢 ≠ 𝑣 ∈ Γ, this means that (𝛿1, 𝛿2) is not (𝜃1, 𝜃2)-derivation pair. Theorem 3.1 Let Γ be a prime ring. Let 𝜃1 and 𝜃2 be two automorphisms of Γ. If Γ is a (𝛿1, 𝛿2)- Derivation pair such that 𝛿1(𝑢) = ∓ 𝜃1(𝑢) for each 𝑢 ∈ Γ, then 𝛿2(𝑢) = 0. Proof: Let 𝑢 ∈ Γ. If 𝛿1(𝑢) = 𝜃1(𝑢) for each 𝑢 ∈ Γ (1) Replacing 𝑢 by 𝑢𝑣𝑢 in (1), we get: 𝛿1(𝑢𝑣𝑢) = 𝜃1(𝑢𝑣𝑢) for each 𝑢, 𝑣 ∈ Γ (2) That is: 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢) = 𝜃1(𝑢𝑣𝑢) for each 𝑢, 𝑣 ∈ Γ (3) By using (1) we have: 𝜃1(𝑢𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝜃1(𝑢) − 𝜃1(𝑢𝑣𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (4) That is: 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝜃1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (5) That is: 𝜃2(𝑢)(𝛿2(𝑣) + 𝜃2(𝑣))𝜃1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (6) Replacing 𝛿2(𝑣) + 𝜃2(𝑣) by 𝜃2(𝑣) in (6), and using (1) we get: 𝜃2(𝑢𝑣)𝛿1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (7) Left multiplying of (7) by 𝛿2(𝑢) we have: 𝛿2(𝑢) 𝜃2(𝑢𝑣)𝛿1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (8) Since Γ is a prime ring, (8) gives: 𝛿2(𝑢) = 0 for each 𝑢 ∈ Γ. Now, If 𝛿1(𝑢) = − 𝜃1(𝑢) for each 𝑢 ∈ Γ (9) Replacing 𝑢 by 𝑢𝑣𝑢 in (9), we get: 𝛿1(𝑢𝑣𝑢) = − 𝜃1(𝑢𝑣𝑢) for each 𝑢, 𝑣 ∈ Γ (10) That is: 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢) = − 𝜃1(𝑢𝑣𝑢) for each 𝑢, 𝑣 ∈ Γ (11) By using (9) we have: −𝜃1(𝑢𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) − 𝜃2(𝑢𝑣)𝜃1(𝑢) + 𝜃1(𝑢𝑣𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (12) That is: 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) − 𝜃2(𝑢𝑣)𝜃1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (13) That is: 𝜃2(𝑢)(𝜃2(𝑣) − 𝛿2(𝑣))(−𝜃1(𝑢)) = 0 for each 𝑢, 𝑣 ∈ Γ (14) By using (9) we have: 𝜃2(𝑢)(𝜃2(𝑣) − 𝛿2(𝑣))𝛿1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (15) Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 112 Replacing 𝜃2(𝑣) − 𝛿2(𝑣) by 𝜃2(𝑣) in (15), we get: 𝜃2(𝑢𝑣)𝛿1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (16) Left multiplying of (16) by 𝛿2(𝑢) we have: 𝛿2(𝑢) 𝜃2(𝑢𝑣)𝛿1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (17) Since Γ is a prime ring, (17) gives: 𝛿2(𝑢) = 0 for each 𝑢 ∈ Γ. ∎ Theorem 3.2 Let Γ be a prime ring. Let 𝜃1 and 𝜃2 be two automorphisms of Γ. If Γ is a (𝛿1, 𝛿2)- Derivation pair such that 𝛿2(𝑢) = ∓ 𝜃1(𝑢) for each 𝑢 ∈ Γ, then 𝛿1(𝑢) = 0. Proof: Let 𝑢 ∈ Γ. If 𝛿2(𝑢) = 𝜃1(𝑢) for each 𝑢 ∈ Γ (18) Replacing 𝑢 by 𝑢𝑣𝑢 in (18), we get: 𝛿2(𝑢𝑣𝑢) = 𝜃1(𝑢𝑣𝑢) for each 𝑢, 𝑣 ∈ Γ (19) That is: 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿2(𝑢) = 𝜃1(𝑢𝑣𝑢) for each 𝑢, 𝑣 ∈ Γ (20) By using (18) we have: 𝜃1(𝑢𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝜃1(𝑢) − 𝜃1(𝑢𝑣𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (21) That is: 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝜃1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (22) That is: 𝜃2(𝑢)(𝛿1(𝑣) + 𝜃2(𝑣))𝜃1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (23) Replacing 𝛿1(𝑣) + 𝜃2(𝑣) by 𝜃2(𝑣) in (23), and using (18) we get: 𝜃2(𝑢𝑣)𝛿2(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (24) Left multiplying of (24) by 𝛿1(𝑢) we have: 𝛿1(𝑢) 𝜃2(𝑢𝑣)𝛿2(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (25) Since Γ is a prime ring, (25) gives: 𝛿1(𝑢) = 0 for each 𝑢 ∈ Γ. Now, If 𝛿2(𝑢) = − 𝜃1(𝑢) for each 𝑢 ∈ Γ (26) Replacing 𝑢 by 𝑢𝑣𝑢 in (26), we get: 𝛿2(𝑢𝑣𝑢) = − 𝜃1(𝑢𝑣𝑢) for each 𝑢, 𝑣 ∈ Γ (27) That is: 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿2(𝑢) = − 𝜃1(𝑢𝑣𝑢) for each 𝑢, 𝑣 ∈ Γ (28) By using (26) we have: −𝜃1(𝑢𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) − 𝜃2(𝑢𝑣)𝜃1(𝑢) + 𝜃1(𝑢𝑣𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (29) That is: 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) − 𝜃2(𝑢𝑣)𝜃1(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (30) That is: 𝜃2(𝑢)(𝜃2(𝑣) − 𝛿1(𝑣))(−𝜃1(𝑢)) = 0 for each 𝑢, 𝑣 ∈ Γ (31) By using (26) we have: 𝜃2(𝑢)(𝜃2(𝑣) − 𝛿1(𝑣))𝛿2(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (32) Replacing 𝜃2(𝑣) − 𝛿1(𝑣) by 𝜃2(𝑣) in (32), we get: 𝜃2(𝑢𝑣)𝛿2(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (33) Left multiplying of (33) by δ1(𝑢) we have: Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 113 𝛿1(𝑢) 𝜃2(𝑢𝑣)𝛿2(𝑢) = 0 for each 𝑢, 𝑣 ∈ Γ (34) Since Γ is a prime ring, (34) gives: 𝛿1(𝑢) = 0 for each 𝑢 ∈ Γ. ∎ Theorem 3.3 Let Γ be a prime ring. Let 𝜃1 and 𝜃2 be two automorphisms of Γ. If Γ is a (𝛿1, 𝛿2)- Derivation pair such that 𝛿2(𝑢)𝛿1(𝑣) = 0 (resp. 𝛿1(𝑢)𝛿2(𝑣) = 0) for each 𝑢, 𝑣 ∈ Γ, then 𝛿2(𝑢) = 0 (resp. 𝛿1(𝑢) = 0). Proof: Let 𝑢, 𝑣 ∈ Γ. If 𝛿2(𝑢)𝛿1(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (35) Replacing 𝑢 by 𝑢𝑣𝑢 in (35), we get: 𝛿2(𝑢𝑣𝑢)𝛿1(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (36) That is: (𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿2(𝑢))𝛿1(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (37) That is: 𝛿2(𝑢)𝜃1(𝑣𝑢)𝛿1(𝑣) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢)𝛿1(𝑣) + 𝜃2(𝑢𝑣)𝛿2(𝑢)𝛿1(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (38) By using (35), we have: 𝛿2(𝑢)𝜃1(𝑣𝑢)𝛿1(𝑣) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢)𝛿1(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (39) That is: (𝛿2(𝑢)𝜃1(𝑣) + 𝜃2(𝑢)𝛿1(𝑣))𝜃1(𝑢)𝛿1(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (40) Replacing 𝛿2(𝑢)𝜃1(𝑣) + 𝜃2(𝑢)𝛿1(𝑣) by 𝜃1(𝑣) in (40), we get: 𝜃1(𝑣𝑢)𝛿1(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (41) Left multiplying (41) by 𝛿2(𝑢) we have: 𝛿2(𝑢) 𝜃1(𝑣𝑢)𝛿1(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (42) Since Γ is a prime ring, (42) gives: 𝛿2(𝑢) = 0 for each 𝑢 ∈ Γ. Now, If 𝛿1(𝑢)𝛿2(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (43) Replacing 𝑢 by 𝑢𝑣𝑢 in (43), we get: 𝛿1(𝑢𝑣𝑢)𝛿2(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (44) That is: (𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢))𝛿2(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (45) That is: 𝛿1(𝑢)𝜃1(𝑣𝑢)𝛿2(𝑣) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢)𝛿2(𝑣) + 𝜃2(𝑢𝑣)𝛿1(𝑢)𝛿2(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (46) By using (43), we have: 𝛿1(𝑢)𝜃1(𝑣𝑢)𝛿2(𝑣) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢)𝛿2(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (47) That is: (𝛿1(𝑢)𝜃1(𝑣) + 𝜃2(𝑢)𝛿2(𝑣))𝜃1(𝑢)𝛿2(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (48) Replacing 𝛿1(𝑢)𝜃1(𝑣) + 𝜃2(𝑢)𝛿2(𝑣) by 𝜃1(𝑣) in (48), we get 𝜃1(𝑣𝑢)𝛿2(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (49) Left multiplying of (49) by 𝛿1(𝑢), we have: 𝛿1(𝑢) 𝜃1(𝑣𝑢)𝛿2(𝑣) = 0 for each 𝑢, 𝑣 ∈ Γ (50) Since Γ is a prime ring, (50) gives: 𝛿1(𝑢) = 0 for each 𝑢 ∈ Γ. ∎ Theorem 3.4 Let Γ be a prime ring. Let 𝜃1 and 𝜃2 be two automorphisms of Γ. If Γ is a (𝛿1, 𝛿2)- Derivation pair such that 𝑐𝛿1(𝑢) = 0 or 𝛿1(𝑢)𝑐 = 0 (resp. c𝛿2(𝑢) = 0 or 𝛿2(𝑢)𝑐 = 0) for each 𝑢, 𝑐 ∈ Γ, then either 𝑐 = 0 or 𝛿1(𝑢) = 0 (resp.𝑐 = 0 or 𝛿2(𝑢) = 0). Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 114 Proof: Let 𝑢, 𝑐 ≠ 0 ∈ Γ. If 𝑐𝛿1(𝑢) = 0 for each 𝑐, 𝑢 ∈ Γ (51) Replacing 𝑢 by 𝑢𝑣𝑢 in (51), we get: 𝑐𝛿1(𝑢𝑣𝑢) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (52) That is 𝑐(𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢)) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (53) That is 𝑐𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝑐𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝑐𝜃2(𝑢𝑣)𝛿1(𝑢) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (54) By using (51), we have: 𝑐𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝑐𝜃2(𝑢𝑣)𝛿1(𝑢) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (55) That is 𝑐𝜃2(𝑢)(𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑣)𝛿1(𝑢)) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (56) Replacing 𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑣)𝛿1(𝑢) by 𝛿1(𝑢) in (56), we get: 𝑐𝜃2(𝑢)𝛿1(𝑢) = 0 for each 𝑢, 𝑐 ∈ Γ (57) Since 𝑐 ≠ 0 and Γ is a prime rings, then (57) gives 𝛿1(𝑢) = 0. Now, let 𝑢, 𝑐 ≠ 0 ∈ Γ. If 𝑐𝛿2(𝑢) = 0 for each 𝑐, 𝑢 ∈ Γ (58) Replacing 𝑢 by 𝑢𝑣𝑢 in (58), we get: 𝑐𝛿2(𝑢𝑣𝑢) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (59) That is 𝑐(𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿2(𝑢)) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (60) That is 𝑐𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝑐𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝑐𝜃2(𝑢𝑣)𝛿2(𝑢) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (61) By using (58), we have: 𝑐𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝑐𝜃2(𝑢𝑣)𝛿2(𝑢) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (62) That is 𝑐𝜃2(𝑢)(𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑣)𝛿2(𝑢)) = 0 for each 𝑢, 𝑣, 𝑐 ∈ Γ (63) Replacing 𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑣)𝛿2(𝑢) by 𝛿2(𝑢) in (63), we get: 𝑐𝜃2(𝑢)𝛿2(𝑢) = 0 for each 𝑢, 𝑐 ∈ Γ (64) Since 𝑐 ≠ 0 and Γ is a prime rings, then (64) gives 𝛿2(𝑢) = 0. ∎ Theorem 3.5 Let Γ be a prime ring with char(Γ) ≠ 2. Let 𝜃1and 𝜃2 be two endomorphisms of Γ. If Γ is a (𝛿1, 𝛿2)-Derivation pair such that 𝑐1𝑞𝑐2𝛿1(𝑢) + 𝛿1(𝑢)𝑐2𝑞𝑐1 = 0 (resp. 𝑐1𝑞𝑐2𝛿2(𝑢) + 𝛿2(𝑢)𝑐2𝑞𝑐1 = 0) for each 𝑢, 𝑐1, 𝑐2, 𝑞 ∈ Γ, then 𝑐1 = 0 or 𝑐2 = 0. Proof: From the assumption we have: 𝑐1𝑞𝑐2𝛿1(𝑢) + 𝛿1(𝑢)𝑐2𝑞𝑐1 = 0 for each 𝑢, 𝑐1, 𝑐2, 𝑞 ∈ Γ (65) Replacing 𝑢 by 𝑢𝑣𝑢 in (65), we have: 𝑐1𝑞𝑐2𝛿1(𝑢𝑣𝑢) + 𝛿1(𝑢𝑣𝑢)𝑐2𝑞𝑐1 = 0 for each 𝑢, 𝑣, 𝑐1, 𝑐2, 𝑞 ∈ Γ (66) That is 𝑐1𝑞𝑐2(𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢)) + (𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢))𝑐2𝑞𝑐1 = 0 for each 𝑢, 𝑣, 𝑐1, 𝑐2, 𝑞 ∈ Γ (67) That is (𝑐1𝑞𝑐2𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝑐1𝑞𝑐2𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝑐1𝑞𝑐2𝜃2(𝑢𝑣)𝛿1(𝑢)) + (𝛿1(𝑢)𝜃1(𝑣𝑢)𝑐2𝑞𝑐1 + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢)𝑐2𝑞𝑐1 + 𝜃2(𝑢𝑣)𝛿1(𝑢)𝑐2𝑞𝑐1 ) = 0 for each 𝑢, 𝑣, 𝑐1, 𝑐2, 𝑞 ∈ Γ (68) By setting 𝜃1(𝑣𝑢) = 𝜃2(𝑢𝑣) = 1 in (68), we have: (𝑐1𝑞𝑐2𝛿1(𝑢) + 𝑐1𝑞𝑐2𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝑐1𝑞𝑐2𝛿1(𝑢)) + (𝛿1(𝑢)𝑐2𝑞𝑐1 + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢)𝑐2𝑞𝑐1 + 𝛿1(𝑢)𝑐2𝑞𝑐1 ) = 0 for each 𝑢, 𝑣, 𝑐1, 𝑐2, 𝑞 ∈ Γ (69) By using (65), we get: 𝑐1𝑞𝑐2𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢)𝑐2𝑞𝑐1 = 0 for each 𝑢, 𝑣, 𝑐1, 𝑐2, 𝑞 ∈ Γ (70) Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 115 By setting 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) = 1 in (70), we have: 𝑐1𝑞𝑐2 + 𝑐2𝑞𝑐1 = 0 for each 𝑢, 𝑣, 𝑐1, 𝑐2, 𝑞 ∈ Γ (71) Replacing 𝑞 by 𝑥𝑐1𝑦 in (71), we get: 𝑐1𝑥𝑐1𝑦𝑐2 + 𝑐2𝑥𝑐1𝑦𝑐1 = 0 for each 𝑥, 𝑦, 𝑐1, 𝑐2 ∈ Γ (72) That is 𝑐1𝑦𝑐2 = −𝑐2𝑦𝑐1 and 𝑐2𝑥𝑐1 = −𝑐1𝑥𝑐2 (73) Substituting (73) in (72) we have: −𝑐1𝑥𝑐2𝑦𝑐1 − 𝑐1𝑥𝑐2𝑦𝑐1 = 0 (74) That is 2𝑐1Γ𝑐2Γ𝑐1 = (0) (75) Since char(Γ) ≠ 2 and Γ is a prime, then (75) gives 𝑐1 = 0 or 𝑐2 = 0. ∎ Theorem 3.6 Let Γ be a 2-torsion free ring with an identity element. Furthermore, let (𝛿1, 𝛿2) be a Jordan (𝜃1, 𝜃2)-derivation pair such that 𝛿1(1) = 𝛿2(1). Then 𝛿1(𝑢) = 𝛿2(𝑢), ∀𝑢 ∈ Γ where 𝜃1 and 𝜃2 are two mappings of Γ. Proof: Let 𝜑: Γ → Γ be a mapping given by 𝜑(𝑢) = 𝛿1(𝑢) − 𝛿2(𝑢), ∀𝑢 ∈ Γ. By Definition 3.1, we have: 𝛿1(𝑢 3) = 𝛿1(𝑢)𝜃1(𝑢 2) + 𝜃2(𝑢)𝛿2(𝑢)𝜃1(𝑢) + 𝜃2(𝑢 2)𝛿1(𝑢) for all 𝑢 ∈ Γ (76) 𝛿2(𝑢 3) = 𝛿2(𝑢)𝜃1(𝑢 2) + 𝜃2(𝑢)𝛿1(𝑢)𝜃1(𝑢) + 𝜃2(𝑢 2)𝛿2(𝑢) for all 𝑢 ∈ Γ (77) Subtracting (77) from (76), we get: 𝜑(𝑢3) = 𝜑(𝑢)𝜃1(𝑢 2) − 𝜃2(𝑢)𝜑(𝑢)𝜃1(𝑢) + 𝜃2(𝑢 2)𝜑(𝑢) for all 𝑢 ∈ Γ (78) Linearizing (78), we have: 𝜑(𝑢2𝑣 + 𝑣𝑢2 + 𝑢𝑣2 + 𝑣2𝑢 + 𝑢𝑣𝑢 + 𝑣𝑢𝑣) = 𝜑(𝑢)𝜃1(𝑢𝑣) + 𝜑(𝑢)𝜃1(𝑣𝑢) + 𝜑(𝑢)𝜃1(𝑣 2) + 𝜑(𝑣)𝜃1(𝑢𝑣) + 𝜑(𝑣)𝜃1(𝑣𝑢) + 𝜑(𝑣)𝜃1(𝑢 2) − 𝜃2(𝑢)𝜑(𝑢)𝜃1(𝑣) − 𝜃2(𝑢)𝜑(𝑣)𝜃1(𝑢) − 𝜃2(𝑣)𝜑(𝑢)𝜃1(𝑢) − 𝜃2(𝑣)𝜑(𝑣)𝜃1(𝑢) − 𝜃2(𝑢)𝜑(𝑣)𝜃1(𝑣) − 𝜃2(𝑣)𝜑(𝑢)𝜃1(𝑣) + 𝜃2(𝑢𝑣)𝜑(𝑢) +𝜃2(𝑣𝑢)𝜑(𝑢) + 𝜃2(𝑣 2)𝜑(𝑢) + 𝜃2(𝑢𝑣)𝜑(𝑣) + 𝜃2(𝑣𝑢)𝜑(𝑣)+𝜃2(𝑢 2)𝜑(𝑣) for all 𝑢, 𝑣 ∈ Γ (79) Replacing 𝑢 by −𝑢 in (79), we get: 𝜑(𝑢2𝑣 + 𝑣𝑢2 − 𝑢𝑣2 − 𝑣2𝑢 + 𝑢𝑣𝑢 − 𝑣𝑢𝑣) = 𝜑(𝑢)𝜃1(𝑢𝑣) + 𝜑(𝑢)𝜃1(𝑣𝑢) − 𝜑(𝑢)𝜃1(𝑣 2) − 𝜑(𝑣)𝜃1(𝑢𝑣) − 𝜑(𝑣)𝜃1(𝑣𝑢) + 𝜑(𝑣)𝜃1(𝑢 2) + 𝜃2(𝑢)𝜑(𝑢)𝜃1(𝑣) + 𝜃2(𝑢)𝜑(𝑣)𝜃1(𝑢) + 𝜃2(𝑣)𝜑(𝑢)𝜃1(𝑢) − 𝜃2(𝑣)𝜑(𝑣)𝜃1(𝑢) − 𝜃2(𝑢)𝜑(𝑣)𝜃1(𝑣) − 𝜃2(𝑣)𝜑(𝑢)𝜃1(𝑣) + 𝜃2(𝑢𝑣)𝜑(𝑢) +𝜃2(𝑣𝑢)𝜑(𝑢) − 𝜃2(𝑣 2)𝜑(𝑢) − 𝜃2(𝑢𝑣)𝜑(𝑣) − 𝜃2(𝑣𝑢)𝜑(𝑣) + 𝜃2(𝑢 2)𝜑(𝑣) for all 𝑢, 𝑣 ∈ Γ (80) According to (79) and (80), we have: 𝜑(𝑢2𝑣 + 𝑣𝑢2 + 𝑢𝑣𝑢) = 𝜑(𝑢)𝜃1(𝑢𝑣) + 𝜑(𝑢)𝜃1(𝑣𝑢) + 𝜑(𝑣)𝜃1(𝑢 2) − 𝜃2(𝑣)𝜑(𝑣)𝜃1(𝑢) − 𝜃2(𝑢)𝜑(𝑣)𝜃1(𝑣) − 𝜃2(𝑣)𝜑(𝑢)𝜃1(𝑣) + 𝜃2(𝑢𝑣)𝜑(𝑢) + 𝜃2(𝑣𝑢)𝜑(𝑢) + 𝜃2(𝑢 2)𝜑(𝑣) for all 𝑢, 𝑣 ∈ Γ (81) Replacing 𝑢 by 1 in (81), we get: 2𝜑(𝑣) = 𝜑(𝑣)𝜃1(1) − 𝜃2(𝑣)𝜑(𝑣)𝜃1(1) − 𝜃2(1)𝜑(𝑣)𝜃1(𝑣) + 𝜃2(1)𝜑(𝑣)for all 𝑣 ∈ Γ (82) By setting 𝜃1(1) = 𝜃2(1) = 0 in (82), then we have: 2𝜑(𝑣) = 0 for all 𝑣 ∈ Γ (83) Since Γ is a 2-torsion free ring, then 𝜑(𝑣) = 0 for all 𝑣 ∈ Γ (84) Therefore, (84) gives 𝛿1(𝑢) = 𝛿2(𝑢) for all 𝑢 ∈ Γ. ∎ Proposition 3.1 Let Γ be a ring, and 𝜃1, 𝜃2 be two mappings of Γ. Then 1- If (𝛿1, 𝛿2) is a (𝜃1, 𝜃2)-derivation pair on Γ, then 𝛿1 + 𝛿2 is a (𝜃1, 𝜃2)-derivation. Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 116 2- If (𝛿1, 𝛿2) is a Jordan (𝜃1, 𝜃2)-derivation pair on Γ, then 𝛿1 + 𝛿2 is a Jordan (𝜃1, 𝜃2)- derivation. Proof: (1) Since 𝛿1 and 𝛿2 is a (𝜃1, 𝜃2)-derivation pair, then by Definition 3.1, we have: 𝛿1(𝑢𝑣𝑢) = 𝛿1(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿2(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿1(𝑢) for all 𝑢, 𝑣 ∈ Γ (85) 𝛿2(𝑢𝑣𝑢) = 𝛿2(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)𝛿1(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)𝛿2(𝑢) for all 𝑢, 𝑣 ∈ Γ (86) By adding (85) and (86), we have: (𝛿1 + 𝛿2)(𝑢𝑣𝑢) = (𝛿1 + 𝛿2)(𝑢)𝜃1(𝑣𝑢) + 𝜃2(𝑢)(𝛿1 + 𝛿2)(𝑣)𝜃1(𝑢) + 𝜃2(𝑢𝑣)(𝛿1 + 𝛿2)(𝑢) Thus, 𝛿1 + 𝛿2 is a (𝜃1, 𝜃2)-derivation. By a similar way to prove (2). ∎ 4. Conclusion As a conclusion, this article presented the notion of (𝜃1, 𝜃2)-derivation pair with some of its properties. This study displayed that the sum of two (𝜃1, 𝜃2)-derivation pair is a (𝜃1, 𝜃2)- derivation and the sum of two Jordan (𝜃1, 𝜃2)-derivation pair is a Jordan (𝜃1, 𝜃2)-derivation. 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