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Orthogonal Generalized Symmetric Reverse Bi-(𝜎, 𝝉)-Derivations of 

Semi Prime Ring 

 
V.S.V. Krishna Murty1, K.Chennakesavulu2, C. Jaya Subba Reddy3 

krishnamurty.vadrevu@gmail.com1, intell.chenna@gmail.com2, cjsreddysvu@gmail.com3 
1Research Scholar, Department of Mathematics, S.V.University, Tirupati, Andhra Pradesh, India. 

2Professor, Department of Mathematics, PVKKIT, Anantapur, Andhra Pradesh, India. 
3Professor, Department of Mathematics, S. V. University, Tirupati, Andhra Pradesh, India. 

 

 

Article History: 

Received: 05-04-2024 

Revised: 25-05-2024 

Accepted: 10-06-2024 

Abstract:  

Let R be a semi prime ring. Suppose that 𝜎, 𝜏 are automorphisms on R.  A symmetric bi-

additive mapping 𝛿1: 𝑅𝑋 𝑅 → 𝑅 is said to be a generalized symmetric reverse bi-(𝜎,𝜏)-

derivation on R if there exists a symmetric reverse bi-(𝜎,𝜏)-derivation D on R such that 

𝛿1(𝑢𝑣, 𝑤) = 𝛿1(𝑣, 𝑤)𝜎(𝑢) + 𝜏(v)D(𝑢, 𝑤) holds ∀ 𝑢, 𝑣, 𝑤 ∈ 𝑅. Let  [𝛿1, 𝐷1] and [𝛿2, 𝐷2] 

be two generalized symmetric reverse bi-(𝜎,𝜏)-derivations of R  with associated  reverse  

bi-(𝜎, 𝜏)-derivations 𝐷1 , 𝐷2. In this paper, we establish some equivalent conditions for 

the orthogonality between two symmetric generalized reverse bi-(𝜎,𝜏)-derivations of 

semiprime ring R. 

Keywords: Semiprime ring, Generalized reverse  biderivation, Generalized reverse bi-

(𝜎,𝜏)-derivation, Orthogonal biderivation. 

 

1. INTRODUCTION:  

The concept of orthogonal derivation was introduced by M. Bresar and J. Vukman [14]  and 

proved some  results on the orthogonal derivations of semiprime rings which were related to  

Posner’s First Theorem [9]. Some results on (𝜎, 𝜏)-derivations in prime rings were studied by M. 

Ashraf  [13] and K. Kaya et al. [12] . J.C. Chang [11] introduced the notion of a generalized (𝛼, 𝛽)-

derivation of a ring R and investigated some properties of such derivations. Argac et al. [17] 

introduced the notion of orthogonality for a pair (D, d), (G, g) of generalized derivations on 

semiprime rings and gave several necessary and sufficient conditions for (D, d) and (G, g) to be 

orthogonal. O.Golbasi and N. Aydin [18] extended the results of Argac to orthogonal generalized 

(𝜎, 𝜏)-derivations. Orthogonality of generalized (𝜎, 𝜏)-derivations on ideals of semiprime rings was 

studied in [10]. Several studies were established on the orthogonality of derivations, biderivations by 

M.N. Daif et al. [15] and C. Jaya Subba Reddy et al. [2,4, 5]. A.Ali et al. [1] and M.N.Daif et al. [16] 

established some results on biderivations of prime and semiprime rings and the study of 

orthogonality of symmetric bi-(𝜎, 𝜏)-derivations in semi  prime rings  was carried out  in  [3,6]. 

Recently, C. Jaya Subba Reddy et al. [7, 8] have studied orthogonal symmetric reverse bi-(𝜎,𝜏)-

derivations in semi prime rings and orthogonal generalized  reverse (σ, τ)- derivations in semiprime 

Γ-rings. In the present paper, we extended the results of orthogonality on generalized symmetric bi-

(𝜎, 𝜏)-derivations established in [6] to generalized symmetric reverse bi-(𝜎, 𝜏)-derivations. 

 

 



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2. PRELIMINARIES: 

Throughout this paper, R will denote an associative ring with center Z(R). A ring R is known 

to be semiprime if 𝑢𝑅𝑢 = {0} implies 𝑢 = 0, ∀ 𝑢 ∈ 𝑅. We say that R is 2-torsion-free if 2𝑢 =

0 implies 𝑢 = 0, ∀ 𝑢 ∈ 𝑅. An additive mapping 𝑑: 𝑅 → 𝑅 is said to be a derivation (respectively, 

reverse derivation) on R if 𝑑(𝑢𝑣) = 𝑑(𝑢)𝑣 + 𝑢𝑑(𝑣) (respectively, 𝑑(𝑢𝑣) = 𝑑(𝑣)𝑢 + 𝑣𝑑(𝑢) holds 

for all  𝑢, 𝑣 ∈ 𝑅. Suppose that 𝜎 and 𝜏 are automorphisms of R. An additive mapping 𝑑: 𝑅 → 𝑅 is 

said to be a (𝜎,𝜏)-derivation (respectively, reverse (𝜎,𝜏)-derivation) on R if  𝑑(𝑢𝑣) = 𝑑(𝑢)𝜎(𝑣) +

𝜏(𝑢)𝑑(𝑣) (respectively,  𝑑(𝑢𝑣) = 𝑑(𝑣)𝜎(𝑢) + 𝜏(𝑣)𝑑(𝑢)) holds, ∀ 𝑢, 𝑣 ∈ 𝑅.  

An additive mapping 𝐷1: 𝑅 → 𝑅  is called a generalized derivation (respectively, generalized 

reverse derivation) if there exists a derivation (respectively, reverse derivation) ‘d’ such that 

𝐷1(𝑢𝑣) = 𝐷1(𝑢)𝑣 + 𝑢𝑑(𝑣)  (respectively,𝐷1(𝑢𝑣) = 𝐷1(𝑣)𝑢 + 𝑣𝑑(𝑢) holds ∀ 𝑢, 𝑣 ∈ 𝑅. An additive 

mapping 𝐷1: 𝑅 → 𝑅  is called a generalized (𝜎,𝜏)-derivation (respectively, generalized reverse (𝜎, 𝜏)-

derivation) if there exists a (𝜎,𝜏)-derivation (respectively, reverse (𝜎,𝜏)-derivation) ‘d’ such that 

𝐷1(𝑢𝑣) = 𝐷1(𝑢)𝜎(𝑣) + 𝜏(𝑢)𝑑(𝑣) (respectively, 𝐷1(𝑢𝑣) = 𝐷1(𝑣)𝜎(𝑢) + 𝜏(𝑣)𝑑(𝑢) holds ∀ 𝑢, 𝑣 ∈

𝑅. Thus, the concept of generalized (𝜎, 𝜏)-derivation covers the concept of (𝜎,𝜏)-derivation.  

A bi additive mapping  D1: RX R → R is said to be symmetric if  D1(u, v) =  D1(v, u).  A 

symmetric bi additive mapping  D1: RX R → R is said to be a symmetric biderivation on R if 

D1(uv, w) = uD1(v, w) + D1(u, w)v holds ∀ u, v , w ∈ R. A symmetric biadditive mapping 

 D1: RX R → R is said to be a symmetric bi-(𝜎,τ)-derivation (respectively, symmetric reverse bi-(𝜎,τ)-

derivation) on R if D1 (uv, w)=  D1(u, w)σ(v) + τ(u) D1(v, w) (respectively, D1(uv, w) 

= D1 (v, w)σ(u) +  τ(v) D1(u, w) holds ∀ u, v , w ∈ R. A symmetric biadditive mapping δ 1: RX R → R 

is said to be a generalized symmetric biderivation (respectively, generalized symmetric reverse 

biderivation) on R if there exists a symmetric biderivation (respectively, symmetric reverse 

biderivation) D1on R such that δ1(uv, w) = δ1(u, w)v + uD1(v, w)(respectively,δ1(uv, w)= δ1(v, w)u 

+ vD1(u, w), ∀ u, v , w ∈ R. A symmetric biadditive mapping δ1: RX R → R is said to be a 

generalized symmetric bi-(𝜎,τ)-derivation (respectively, generalized symmetric reverse bi-(𝜎,τ)-

derivation) on R if there exists a symmetric bi-(𝜎,τ)-derivation (respectively, symmetric reverse bi-

(𝜎,τ)-derivation) D1 on R such that δ1(uv, w) =   δ1(u, w) σ(v) + τ(u) D1(v, w) (respectively, 

δ1(uv, w) = δ1(v, w)σ(u) + τ(v)D1(u, w) holds ∀ u, v , w ∈ R. Two symmetric reverse bi-(σ,τ)-

derivations D1, D2 are said to be orthogonal if D1(u, v)RD2(v, w) = {0} = D2(v, w)RD1(u, v), for all 

u, v, w ∈ R. Two generalized symmetric reverse bi-(σ,τ)-derivations δ1, δ2 are said to be orthogonal if 

δ1(u, v)Rδ2(v, w) = {0} =  δ2(v, w)Rδ1(u, v),  for all  u, v, w ∈ R. 

 We assume throughout the paper that R is a 2-torsion-free semiprime ring, while σ and τ are 

automorphisms of R. Also D1, D2 are  reverse bi-(σ,τ)-derivations of R such that D1τ = τD1, D2τ = 

τD2, σD1 = D1σ, σD2 = D2σ. We denote two generalized reverse bi-(𝜎, 𝜏)-derivations δ1: RX R → R 

and  δ2: RX R → R determined by reverse  bi-(σ,τ)-derivations D1, D2 of R be such that  δ1τ = τδ1, 

δ2τ = τδ2, σδ1 = δ1σ, σδ2 = δ2σ. 

Lemma 1: [Lemma 1,[14]] 

If R is a 2-torsion free semi prime ring and 𝑢, 𝑣 ∈ 𝑅, then the following conditions are equivalent: 



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1.𝑢𝑟𝑣 = 0, for all  𝑟 ∈ 𝑅. 

2.𝑣𝑟𝑢 = 0, for all  𝑟 ∈ 𝑅. 

3. 𝑢𝑟𝑣 +  𝑣𝑟𝑢 = 0, for all   𝑟 ∈ 𝑅. 

If anyone of the above conditions is fulfilled, then  𝑢𝑣 = 𝑣𝑢 =  0. 

LEMMA 2: [ LEMMA 2, [5]] 

Let R be a semiprime ring. Suppose that two bi-additive mappings D1: R X R → R and D2: R X R →

R satisfies D1(u, v)RD2(v, u) = {0}, ∀ u, v ∈ R, then  D1(u, v)RD2(v, w) = {0},  ∀ u, v , w ∈ R. 

LEMMA 3: [THEOREM 1, [7]] 

Let R be a 2 torsion free semi prime ring. Then the following conditions are equivalent: 

1.Two symmetric reverse bi-(σ, τ)-derivations D1 and D2  are orthogonal. 

2.D1(u, v) D2(v, w) +  D2(u, v)D1(v, w) = 0,  ∀ u , v , w ∈ R. 

LEMMA 4: Let R be a 2 torsion free semi prime ring. Then two symmetric reverse bi-(𝜎, 𝜏)-

derivations  D1 and D2 are orthogonal  if and only if  D1D2 = 0. 

Proof:  Suppose that  𝐷1  and 𝐷2 are orthogonal.    

Since 𝐷1, 𝐷2 are orthogonal, we can have  𝐷1(𝑢, 𝑣)𝑟𝐷2(𝑣, 𝑤) = 0, ∀ 𝑢 , 𝑣 , 𝑤, 𝑟 ∈ 𝑅 

𝐷1(𝐷1(𝑢, 𝑣) 𝑟𝐷2(𝑣, 𝑤), 𝑚) = 0, ∀ 𝑢 , 𝑣 , 𝑤, 𝑟, 𝑚 ∈ 𝑅 

𝐷1(𝐷2(𝑣, 𝑤), 𝑚)𝜎(𝑟)𝜎(𝐷1(𝑢, 𝑣) + 𝜏(𝐷2(𝑣, 𝑤))𝐷1(𝑟, 𝑚)𝜎(𝐷1(𝑢, 𝑣)) +  𝜏(𝑟𝐷2(𝑣, 𝑤) 

𝐷1(𝐷1(𝑢, 𝑣), 𝑚)=0. 

Using  𝐷1𝜎 = 𝜎𝐷1, 𝜏𝐷2 = 𝐷2𝜏  and  𝜎 and 𝜏 are  automorphisms of R, we have  

𝐷1(𝐷2(𝑣, 𝑤), 𝑚)𝑟𝐷1(𝑢, 𝑣) + 𝐷2(𝑣, 𝑤) 𝐷1(𝑟, 𝑚) 𝐷1(𝑢, 𝑣) + 𝑟𝐷2(𝑣, 𝑤) 𝐷1(𝐷1(𝑢, 𝑣), 𝑚)=0. 

Using the condition of orthogonality of  𝐷1 ,𝐷2, we get  

𝐷1𝐷2(𝑣, 𝑤)𝑟𝐷1(𝑢, 𝑣)=0.    

In Particular if we put  𝑢 = 𝐷2(𝑣, 𝑤) in the above equation, we get  

𝐷1𝐷2(𝑣, 𝑤)𝑟𝐷1(𝐷2(𝑣, 𝑤), 𝑣) = 0    

𝐷1𝐷2(𝑣, 𝑤)𝑟𝐷1𝐷2(𝑣, 𝑤) =  0  

𝐷1𝐷2(𝑣, 𝑤) = 0   ( By Semiprimeness  of R.)  

𝐷1𝐷2 = 0. 

Conversely, 

Let 𝐷1 and 𝐷2 be two reverse  bi-(𝜎, 𝜏)-derivations such that 𝐷1𝐷2 = 0. 

𝐷1𝐷2(𝑢𝑣, 𝑤) = 𝐷1(𝐷2(𝑢𝑣, 𝑤), 𝑚) 

=𝐷1(𝐷2( 𝑣, 𝑤)𝜎(𝑢) + 𝜏(𝑣) 𝐷2(𝑢, 𝑤), 𝑚) 



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=𝐷1(𝐷2( 𝑣, 𝑤)𝜎(𝑢), 𝑚 ) + 𝐷1(𝜏(𝑣)𝐷2(𝑢, 𝑤), 𝑚)    

=𝐷1(𝜎(𝑢), 𝑚)𝜎(𝐷2(𝑣, 𝑤)) + 𝜏(𝜎(𝑢))𝐷1(𝐷2(𝑣, 𝑤), 𝑚)+𝐷1(𝐷2(𝑢, 𝑤), 𝑚)𝜎(𝜏(𝑣)) + 

                                     𝜏(𝐷2(𝑢, 𝑤))𝐷1(𝜏(𝑣), 𝑚). 

Using 𝜎𝐷2 = 𝐷2𝜎, 𝜏𝐷2 = 𝐷2𝜏  ; 𝜎 and 𝜏 are  automorphisms of R and  𝐷1𝐷2 = 0, we obtain 

0 = 𝐷1(𝑢, 𝑚) 𝐷2( 𝑣, 𝑤) + 𝐷2(𝑢, 𝑤)𝐷1(𝑣, 𝑚).      

In particular,  𝐷1(𝑢, 𝑤) 𝐷2( 𝑣, 𝑤) + 𝐷2(𝑢, 𝑤)𝐷1(𝑣, 𝑤)  = 0 . Therefore   

𝐷1(𝑢, 𝑤) 𝐷2(𝑤, 𝑣) + 𝐷2(𝑢, 𝑤)𝐷.1 (𝑤, 𝑣)  = 0, ∀ 𝑢 , 𝑣 , 𝑤 ∈ 𝑅. (Since  𝐷1,𝐷2  are symmetric) 

By Lemma 3, we can conclude that 𝐷1 and 𝐷2 are orthogonal. 

3. MAIN RESULTS: 

THEOREM 1: 

If (δ1, D1) and (δ2, D2) are  two  orthogonal generalized symmetric reverse  bi-(σ, τ)-derivations  of 

R, then (δ1, D1) and (δ2, D2) are orthogonal  if and only if the following conditions are satisfied: 

(i)  δ1(u, v)δ2(v, w) +  δ2 (u, v) δ1(v, w) = 0, ∀ u , v , w, r ∈ R. 

(ii) D1(u, v)δ2(v, w) + D2 (u, v) δ1(v, w) = 0, ∀ u , v , w, r ∈ R. 

Proof: Suppose that (δ1, D1) and (δ2, D2) are orthogonal generalized symmetric reverse bi-(σ, τ)-

derivations of R. 

By the definition of orthogonality δ1 and  δ2, we have  

δ1(u, v)rδ2(v, w) = 0 = δ2(u, v)rδ1(v, w)                                       (3.1) 

Hence, δ1(u, v)δ2(v, w) = 0 =  δ2 (v, w)δ1(u, v)  ( By Lemma 1)                         (3.2) 

and  so  δ1(u, v)δ2(v, w) + δ2(v, w)δ1(u, v) = 0 

δ1(u, v)δ2(v, w) + δ2(w, v)δ1(v, u) = 0  (Since δ1, δ2 are symmetric) 

δ1(u, v)δ2(v, w) + δ2(u, v)δ1(v, w) = 0. 

Hence  condition (i)  is proved . 

Now, Suppose that   δ1(u, v)δ2(v, w) = 0. (From 3.2) 

Again replacing u by ur, r ∈ R in the above equation  and using (3.2), we get  

δ1(r, v)σ(u) δ2(v, w) + τ(r)D1(u, v)δ2(v, w) = 0.  

Since  σ ,  τ are automorphisms, we get  

δ1(r, v)uδ2(v, w) + r D1(u, v)δ2(v, w) = 0, ∀ u, v, r, w ∈ R. 

Using the equation (3.1), we get 

rD1(u, v) δ2(v, w) = 0, ∀ u, v, r, w ∈ R. 

Left  multiplying  the above equation by  D1(u, v)δ2(v, w)  and using the semiprimeness of R,  



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we get D1(u, v)δ2(v, w) = 0.                              (3.3) 

Replacing u by ru, r ∈ R in (3.3) , we obtain 

D1(u, v)σ(r) δ2(v, w) + τ(u)D1(r, v)δ2(v, w) = 0. 

Using (3.3) and  σ is an automorphism  of R, we get  

D1(u, v)r δ2(v, w) = 0, ∀ u, v, r, w ∈ R . 

In view of Lemma 1, we can have  

D1(u, v)rδ2(v, w) = 0 = δ2(v, w)rD1(u, v)                      (3.4) 

and  D1(u, v)δ2(v, w) = 0 =  δ2(v, w)D1(u, v).                (3.5) 

Again by the definition of orthogonality  of  δ1 and δ2, we have   

δ1(u, v)rδ2(v, w) = 0 = δ2(u, v)rδ1(v, w), for all  u, v, w ∈ R.                         (3.6) 

By Lemma  (1), we can have  δ1(u, v)δ2(v, w) = δ2(v, w)δ1(u, v) = 0   

and also  δ2(u, v)δ1(v, w) = 0. 

Consider  δ2(u, v)δ1(v, w)=0, ∀ u, v, w ∈ R .                                               (3.7) 

Replacing u by ur, r ∈ R in the equation ( 3.7)  and using (3.6), we obtain 

τ(r)D2(u, v)δ1(v, w) = 0, ∀ u, v, w, r ∈ R. 

Left Multiplying the above equation by D2(u, v)δ1(v, w)  and using the semiprimeness of R,  

we get  D2(u, v)δ1(v, w) = 0.                     (3.8) 

Replacing u by ru,𝑟 ∈ 𝑅  in (3.8) and using (3.8), we get  

D2(u, v)σ(r)δ1(v, w) = 0, for all  u, v, w, r ∈ R.  

D2(u, v)σ(r)δ1(v, w) = 0. 

Since  σ  is an   automorphism, we obtain 

D2(u, v) rδ1(v, w) = 0 = δ1(v, w) rD2(u, v), ∀ u, v, w, r ∈ R.                                    (3.9) 

By lemma 1,  D2(u, v)δ1(v, w) = 0 = δ1(v, w)D2(u, v), for all  u, v, w ∈ R.                                (3.10) 

From (3.5)  and  (3.10), we can have  D1(u, v)δ2(v, w) + D2(u, v)δ1(v, w)  = 0. 

 Hence condition (ii) is  Proved. 

Conversely, 

Suppose  the conditions  (i)  δ1(u, v)δ2 (v, w) + δ2 (u, v) δ1(v, w) = 0  and            (3.11) 

                                       (ii) D1(u, v)δ2 (v, w) + D2 (u, v) δ1(v, w) = 0  holds good.          (3.12) 

We prove that (δ1, D1) and (δ2, D2) are orthogonal generalized symmetric reverse bi-(σ, τ)-

derivations  of R. 

Replacing u by ru, r ∈ R in (3.11) and using (3.12), we get 



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δ1(u, v)σ(r)δ2(v, w) + δ2 (u, v)σ(r)δ1(v, w)= 0,  for all  u, v, w, r ∈ R.             

Since σ is an automorphism, we get  

δ1(u, v)rδ2 (v, w) + δ2 (u, v)rδ1(v, w) = 0, for all  u, v, w, r ∈ R.             

By Lemma 1, we can conclude that  δ1and  δ2 are orthogonal. 

THEOREM 2: 

If (δ1, D1) and (δ2, D2) are two orthogonal generalized symmetric reverse bi-(σ, τ)-derivations  of R, 

then  (δ1, D1) and (δ2, D2) are orthogonal iff  δ1(u, v)δ2 (v, w) = 0 = D1 (u, v) δ2(v, w).  

Proof: Suppose that (δ1, D1) and  (δ2, D2) are orthogonal generalized symmetric reverse bi-(σ, τ)-

derivations of  R. 

By the definition of orthogonality, we have  δ1(u, v)Rδ2(v, w) = {0}. 

δ1(u, v)rδ2(v, w) = 0 and so δ1(u, v)δ2(v, w) = 0, for all u, v, w ∈ R. ( By Lemma1)              (3.13) 

From the equation (3.5) of   Theorem 1, we have D1(u, v)δ2(v, w) = 0.    

Hence, we conclude that  δ1(u, v)δ2(v, w) = 0 =  D1(u, v)δ2(v, w), for all u, v, w ∈ R.  

Conversely, 

Suppose that δ1(u, v)δ2 (v, w) = D1 (u, v) δ2(v, w) = 0,  for all u, v, w ∈ R.          (3.14) 

We have to prove that δ1and  δ2 are orthogonal 

Consider δ1(u, v)δ2 (v, w) = 0. ( From ( 3.14))             (3.15) 

Replacing u by ru, 𝑟 ∈ 𝑅  in (3.15) and using (3.14), we get  

δ1(u, v)σ(r) δ2(v, w) = 0.  

Since σ is an automorphism, we get   

δ1(u, v) rδ2(v, w) = 0, for all u, v, w, r ∈ R. 

Therefore, δ1and  δ2 are orthogonal. 

THEOREM 3: 

 If (δ1, D1) and (δ2, D2) are two orthogonal generalized  symmetric reverse bi-(σ, τ)-derivations   of 

R, then (δ1, D1) and (δ2, D2) are orthogonal if and only if δ1(u, v)δ2 (v, w) = 0 and  D1δ2 = 0 =

D1D2. 

Proof: Suppose that (δ1, D1) and (δ2, D2) are orthogonal generalized  symmetric reverse  bi-(σ, τ)-

derivations of R. 

By the definition of orthogonality, it is  evident that  

δ1(u, v)rδ2(v, w) = 0 and so  δ1(u, v)δ2(v, w) = 0 . ( By using Lemma 1) 

To Prove that   D1δ2 = 0: 

Consider  δ2(v, w)rD1(u, v)=0   (By  the equation (3.4) of Theorem1) 



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 δ1(δ2(v, w)rD1(u, v) , m ) = 0 

δ1(rD1(u, v) , m) σ(δ2(v, w)) +  τ(rD1(u, v))D1(δ2(v, w), m)= 0 

(δ1(D1(u, v), m)σ(r) + τ(D1(u, v))D1(r, m)) σ(δ2(v, w) + τ(rD1(u, v))D1(δ2(v, w), m)= 0. 

Using D1τ =  τD1 , σδ2 = δ2σ  and σ  and τ  are automorphisms, we  can have  

δ1(D1(u, v), m )rδ2(v, w) + D1(u, v)D1(r, m)δ2(v, w) +  rD1(u, v)D1(δ2(v, w), m) = 0.          (3.16) 

Using the condition of orthogonality of  δ1  and δ2 , we  can have δ1(D1(u, v ), m )rδ2(v, w) = 0 

and by  Theorem 2, we can have D1(r, m) δ2(v, w) = 0,  for all v, w, r, m ∈ R. 

By applying the above conditions in (3.16), we get  

 rD1(u, v)D1(δ2(v, w), m) = 0, for all r, u, v, w ∈ R. 

Left Multiplying the above equation by D1(u, v)D1(δ2(v, w), m) and using the semiprimeness of R, 

we get  D1(u, v)D1(δ2(v, w), m) =  0 

D1(u, v)D1δ2(v, w) = 0.                            (3.17) 

Replacing u by  uδ2(v, w)  in (3.17), we get  

(D1(δ2(v, w), v)σ(u) +  τ(δ2(v, w)) D1(u, v))D1δ2(v, w) = 0. 

Using (3.17) and σ is an automorphism of R, we obtain 

D1(δ2(v, w), v)uD1δ2(v, w) = 0, for all u, v, w ∈ R 

D1δ2(v, w) u D1δ2(v, w) = 0     

D1δ2(v, w)RD1δ2(v, w) = 0     

D1 δ2 = 0.     (By the semiprime ness of R) 

To Prove that  D1D2 = 0: 

Let (δ1, D1) and (δ1, D2) are  two orthogonal generalized symmetric reverse bi-(σ, τ)-derivations  of 

a semi prime ring  R. 

First  we  prove  D1 and D2 are orthogonal. 

By the definition of orthogonality  of  δ1 and δ2 , we have   

δ1(u, v)rδ2(v, w) = 0, for all  u, v, w ∈ R.  

By  Lemma 1, we have  δ1(u, v)δ2(v, w) = 0.               (3.18) 

Replacing u by ru, r ∈ R in the  equation (3.18), we get  

δ1(u, v)σ(r)δ2(v, w) +  τ(u)D1(r, v)δ2(v, w) = 0, for all  u, v, w, r ∈ R. 

Since  σ  and τ  are automorphisms of R, we can have  

δ1(u, v)r δ2(v, w) +  uD1(r, v)δ2(v, w) = 0,  for all u, v, w, r ∈ R.                                  (3.19) 

Replacing w by rw, r ∈ R in equation (3.19) and using the fact that σ  and τ  are automorphisms ,  



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we get  

δ1(u, v)r δ2(v, w)r + δ1(u, v)rwD2(v, r) + uD1(r, v)δ2(v, w)r + uD1(r, v) wD2(v, r) = 0. 

Using the condition of orthogonality of  δ1  and δ2 , equation (3.9) of Theorem 2 and also by  

Theorem 2, the first three terms of the above equation are  zero. 

The above equation reduces to  uD1(r, v) wD2(v, r)  = 0, for all  u, v, r ∈ R. 

Left Multiplying the above equation  by  D1(r, v) wD2(v, r)  and using the semipriness of R, we 

obtain  D1(r, v)wD2(v, r) = 0 .  

In particular, D1(u, v) wD2(v, u) = 0 

D1(u, v) wD2(u, v) = 0 

D1(u, v) wD2(v, w) = 0. ( By lemma 2) 

Therefore,  D1 and D2  are orthogonal. 

By Lemma 4, we can have   D1D2 = 0. 

Hence, the two conditions are proved.  

Conversely 

Suppose that  δ1(u, v)δ2 (v, w) = 0  and  D1δ2 = 0 = D1D2 .            (3.20) 

Let D1δ2 = 0. 

D1δ2(uv, w) = D1(δ2(uv, w), m), for all u, v, w, m ∈ R  

= D1(δ2(v, w)σ(u) +  τ(v)D2(u, w), m). 

Since  σ  and τ  are automorphisms, we get  

=D1(δ2(v, w)u + vD2(u, w), m) 

=D1(u, m)σ(δ2(v, w)) + τ (u)D1(δ2(v, w), m) + D1(D2(u, w), m)σ(v) + 

τ(D2(u, w))D1(v, m), for all  u, v, w, m ∈ R . 

Again using the fact that  σ  and τ  are automorphisms, δ2σ = σ δ2, τD2 = D2τ we get  

=D1(u, m)δ2(v, w) + uD1(δ2(v, w), m) + D1(D2(u, w), m)v +  D2(u, w)D1(v, m) 

=D1(u, m)δ2(v, w) + uD1δ2(v, w) + D1D2(u, w)v + D2(u, w)D1(v, m), ∀ u, v, w, m ∈ R.         (3.21) 

By hypothesis, we have D1δ2 = 0 =  D1D2. 

By Lemma 4, we have  D1D2 = 0 implies  D1, D2 are orthogonal and hence 

 D2(u, w)D1(v, m) = 0, ∀ u, v, w, m ∈ R. 

Therefore, equation (3.21) becomes   

D1δ2(uv, w) = D1(u, m)δ2(v, w), for all u, v, w, m ∈ R 

But D1δ2 = 0  and hence we have  D1(u, m)δ2(v, w)=0.                        (3.22) 



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Replacing u by ru, r ∈ R in the equation (3.22) and using the fact that  σ and τ are automorphisms of 

R, we  obtain,  

D1(u, m) rδ2(v, w) +  u1D1(r, m)δ2(v, w) = 0   

D1(u, m) r δ2(v, w) = 0.   ( By  the equation (3.22), ∀ u, m, r ∈ R 

By Lemma (1), we have  D1(u, m) δ2(v1, w) = 0  =  δ2(v, w)D1(u, m).  

In Particular,  D1(u, v) δ2(v, w) = 0 =  δ2(v, w) D1(u, v).                                    (3.23) 

From (3.20) and (3.23), we have  δ1(u, v)δ2(v, w) = 0 = D1(u, v) δ2(v, w).  

By  Theorem 2, we have δ1 and δ2 are orthogonal. 

THEOREM 4: If (δ
1

, D1) and (δ
2

, D2) are two orthogonal generalized  symmetric reverse  bi-(σ, τ)-

derivations  of R, then (i)  D1 and D2 are orthogonal reverse  bi-(σ, τ)-derivations (ii) δ2D1 = 0      

(iii) δ1D2 = 0    (iv) D2δ
1

= 0 (v) δ1δ
2

= 0 (vi) δ2δ
1

= 0. 

Proof:  (i)  To Prove δ2D1 = 0 : 

Suppose that δ1, δ2  are two orthogonal generalized  symmetric reverse bi-(σ, τ)-derivations of R. 

Then, by the  definition of orthogonality, we have  δ1(u, v)rδ2 (v, w)  =  0,  ∀ u, v, w, r ∈ R. 

By Lemma 1, we can write δ1(u, v)δ2(v, w) =  0. 

Replacing u by ru, r ∈ R in the above equation, we get  

δ1(ru, v)δ2 (v, w) =  0 

δ1(u, v)σ(r)δ2 (v, w) + τ(u)D1(r, v)δ2 (v, w) =  0.                                                                      (3.24) 

Replacing w by rw, r ∈ R in equation (3.24),  

δ1(u, v)σ(r)δ2 (v, rw) + τ(u)D1(r, v)δ2(v, rw) =  0 

δ1(u, v)σ(r)δ2(v, w)σ(r) +  δ1(u, v)σ(r)τ(w)D2(v, r) +  τ(u)D1(r, v)δ2 (v, w)σ(r) 

+ τ(u)D1(r, v)τ(w)D2(v, r) = 0, ∀ u, v, w, r ∈ R.                                                                         (3.25) 

Since  σ ,  τ  are automorphisms of R,  

Using equations (3.1), (3.9) and (3.5), the first three terms are zero, then equation (3.25) reduces to    

uD1(r, v)wD2(v, r) = 0, ∀ u, v, w, r ∈ R 

Then, D1(r, v)wD2(v, r)uD1(r, v)wD2(v, r) = 0, ∀ u, v, w, r ∈ R. 

By the semiprimeness of R, we get  

D1(r, v)wD2(v, r) = 0  which is  same as  D1(r, v)wD2(r, v) = 0. 

Using Lemma 2, we can write D1(r, v)wD2(v, u) = 0, ∀ u, v, w, r ∈ R. 

By Lemma1,  D1(r, v)D2(v, u) = 0 = D2(v, u)D1(r, v) = 0, ∀ u, v, r ∈ R.                                   (3.26) 

which shows that D1 , D2  are orthogonal. 



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(ii)To Prove δ2D1 = 0: 

Since, δ1, δ2  are two orthogonal generalized  symmetric reverse bi-(σ, τ)-derivation of R. 

By equation (3.4) of Theorem 1, we have D1(u, v)r δ2(v, w)  = 0 =  δ2(v, w)rD1(u, v). 

Consider,  δ2(v, w)rD1(u, v) = 0,  ∀ u, v, w, r ∈ R 

Then, δ2(δ
2

(v, w)rD1 (u, v), m) = 0, ∀ u, v, w, r, m ∈ R 

(δ
2

(rD1(u, v), m)σ(δ2(v, w)) +  τ(rD1(u, v)D2(δ2(v, w), m) = 0 

δ2(D1(u, v), m)σ(r)σ(δ2(v, w)) + τ(D1(u, v))D2(r, m)σ(δ2(v, w)) +

 τ(rD1 (u, v)D2( δ2(v, w), m) = 0. 

Using σδ2 = δ2σ ; τD1 =  D1τ;   σ , τ  are automorphisms of R and using the  fact that  D1,D2  are 

orthogonal, we have  

δ2(D1(u, v), m)r δ2(v, w) = 0, ∀ u, v, w, r ∈ R 

δ2D1(u, v)r δ2(v, w) = 0. 

Replacing w by  wδ2(v, w) in the above equation, we obtain 

δ2D1(u, v)r δ2(D1(u, v), v)w + δ2D1(u, v)r D1(u, v)D2(v, w) = 0. 

Since D1 , D2  are orthogonal  are orthogonal, we get 

δ2D1(u, v)r δ2D1(u, v)w = 0, ∀ u, v, w, r ∈ R 

δ2D1(u, v)r δ2D1(u, v)wδ2D1(u, v)r δ2D1(u, v) = 0. 

Using the semiprimeness of R, we get  

δ2D1 (u, v)r δ2D1 (u, v) = 0  (Since R is semiprime) 

δ2D1 (u, v) = 0 

δ2D1 = 0. 

(iii)To Prove  δ1D2 = 0: 

By (3.9) of Theorem 1, we have  δ1(v, w) rD2(u, v) = 0,  ∀ u, v, w, r ∈ R 

δ1(δ1(v, w)rD2(u, v), m) = 0,  ∀ u, v, w, m, r ∈ R 

δ1(rD2(u, v), m)σ(δ1(v, w)) + τ(rD2(u, v))D1(δ1(v, w), m) = 0 

δ1(D2(u, v), m)σ(r)σ(δ1(v, w)) + τ(D2(u, v))D1(r, m)σ(δ1(v, w)) + τ(rD2(u, v))D1(δ1(v, w), m) =

0,  ∀ u, v, w, r ∈ R. 

Using  σδ1 = δ1σ; τD2 =  D2τ; and  σ  and  τ are automorphisms of R, we get 

δ1(D2(u, v), m)rδ1(v, w) + D2(u, v)D1(r, m)δ1(v, w) + rD2(u, v)D1(δ1(v, w), m) = 0,  

                                                                                                                                ∀ u, v, w, r ∈ R. 

Using (3.26), the above equation reduces to δ1(D2(u, v), m)rδ1(v, w) = 0 



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δ1D2(u, v)rδ1(v, w) = 0, ∀ u, v, w, r ∈ R. 

Replacing w by wD2(u, v) in the above equation and using the fact that σ and τ are automorphisms of 

R, we obtain 

δ1D2(u, v)rδ1(v, D2(u, v))w + δ1D2(u, v)rD2(u, v)D
1
(v, w) = 0,  ∀ u, v, w, r ∈ R. 

Using (3.26), the above equation reduces to  

δ1D2(u, v)rδ1(v, D2(u, v))w = 0 

δ1D2(u, v)rδ1(v, D2(u, v))wδ1D2(u, v)rδ1(v, D2(u, v)) = 0. 

Using the Semiprimeness of R, we get  

δ1D2(u, v)rδ1(v, D2(u, v)) = 0 

δ1D2(u, v)rδ1(D2(u, v), v) = 0 

δ1D2(u, v)rδ1D2(u, v) = 0 

Again using the semiprimeness of R, we get δ1D2 = 0. 

(iv)To Prove D2δ
1

= 0: 

Since,  δ1, δ2  are two orthogonal generalized  symmetric reverse(σ, τ) biderivations  of R. 

By (3.9) of Theorem 1, we  can have δ1(v, w) rD2(u, v) = 0   

δ2(δ1(v, w) rD2(u, v), m) = 0.   

By expanding the above equation  and using the  fact that σ  and τ  are automorphisms of R, we 

obtain    δ2(D2(u, v), m)rδ1(v, w) +  D2(u, v)D2(r, m)δ1(v, w) + rD2(u, v)D2δ1(v, w) = 0. 

Using (3.1) and (3.10), the first two terms are zero and hence we get  

rD2(u, v)D2δ1(v, w) = 0 

D2(u, v)D2δ1(v, w)rD2(u, v)D2δ1(v, w) = 0,  ∀ u, v, w, r ∈ R. 

By the semiprimeness of R, we get  

D2(u, v)D2δ1(v, w) = 0,∀ u, v, w ∈ R.                                                                                            (3.27) 

Replacing v = vδ1(v, w) in the above equation ,  we get  

D2(u, vδ1(v, w))D2δ1(v, w) = 0 

D2(δ1(v, w), u)vD2δ1(v, w) + δ1(v, w)D2(u, v)D2δ1(v, w) = 0. 

Using (3.27), we get  

D2(δ1(v, w), u)vD2δ1(v, w) = 0 

D2δ1(v, w)vD2δ1(v, w) = 0,   ∀ v, w ∈ R. Hence, D2δ
1

= 0. 

(v)To Prove δ1δ
2
=0 : 



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Since δ1, δ2 are orthogonal, we can  have δ1(u, v) rδ2(v, w) = 0, ∀ u , v , w, r ∈ R 

δ1(δ1(u, v) rδ2(v, w) , m) = 0, ∀ u , v , w, r, m ∈ R 

δ1(δ2(v, w) , m)σ(r)σ(δ1(u, v) + τ(δ2(v, w))D1(r, m)σ(δ
1
(u, v)) +  τ(rδ2(v, w) D1(δ1(u, v), m) = 0. 

Using  σδ1 = δ1σ ; τδ2 = δ2τ ; and   σ and τ are  automorphisms of R. 

δ1(δ2(v, w), m)rδ1(u, v)+ δ2 (v, w)D1(r, m) δ1(u, v) +  rδ2(v, w) D1(δ1(u, v), m)=0. 

Using  (3.5) of Theorem 1, the last two terms of the above equation becomes zero, then we get  

δ1δ2(v, w)r δ1(u, v) = 0.    

In Particular if we put  u = δ2(v, w) in the above equation, we get  

δ1δ2(v, w)rδ1(δ2(v, w), v)=0    

δ1δ2(v, w)rδ1δ2(v, w) = 0  

δ1δ2(v, w) =  0   ( By Semiprimeness  of R.)  and so δ1δ2 = 0. 

(vi)To Prove δ2δ
1

= 0: 

Since δ1, δ2 are orthogonal , we can  have  δ2(u, v) rδ1(v, w) = 0 , ∀ u , v , w, r ∈ R 

δ2(δ2(u, v) rδ1(v, w) , m) = 0, ∀ u , v , w, r, m ∈ R. 

By following the similar procedure as we adopted in the previous case, we  can easily obtain the 

result.  

4. DISCUSSION:  

The study of derivations and their generalizations plays a significant role in ring theory. In 

this manuscript, we focus on the concept of generalized symmetric reverse bi-(𝜎, 𝜏)-derivations and 

explore the conditions for their orthogonality within the framework of semi prime rings. Let R be a 

semi prime ring. A symmetric bi-additive mapping δ1:R×R→R is termed a generalized symmetric 

reverse bi-(σ,τ)-derivation if there exists a symmetric reverse bi-(𝜎, 𝜏)-derivation 𝐷1 on R such that 

for all 𝑢, 𝑣, 𝑤 ∈ 𝑅, 𝛿1(uv,w)= 𝛿1(v,w)σ(u) + τ(v) 𝐷1 (u,w). This definition extends the classical 

notion of  generalized  symmetric reverse bi-derivations by incorporating the actions of two 

automorphisms, σ and τ, thus providing a richer and more flexible structure for analysis. 

We consider two generalized symmetric reverse bi-(𝜎, 𝜏)-derivations [δ1,D1] and [δ2,D2] of R 

with associated reverse bi-(𝜎, 𝜏)-derivations 𝐷1 an 𝐷2 The primary goal of this paper is to establish 

equivalent conditions for the orthogonality between these two mappings. Orthogonality in this 

context refers to the condition where the products of the mappings and their associated derivations 

satisfy specific nullity conditions. 

Through our analysis, we derive several equivalent conditions that characterize the 

orthogonality of generalized symmetric reverse bi-(𝜎, 𝜏)-derivations in semi prime rings. These 

conditions provide insights into the underlying algebraic structures and their inter relationships. 

Specifically, we show that orthogonality can be characterized in terms of commutativity and specific 

interaction properties between the mappings and their associated derivations. 



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The exploration of generalized symmetric reverse bi-(σ,τ)-derivations opens several avenues 

for future research. One potential direction is the investigation of these derivations in the context of 

non-associative algebras or rings with additional structures or other ring-theoretic properties, such as 

ideals and radicals etc. While this study focuses on semiprime rings, the concepts can be extended to 

other classes of rings offering a broader applicability of the results.  

5. CONCLUSION:  

In conclusion, this manuscript contributes to the field of ring theory by providing a detailed 

analysis of the orthogonality conditions for generalized symmetric reverse bi-(𝜎, 𝜏)-derivations in 

semi prime rings. The equivalence conditions established here deepen our understanding of these 

mappings and pave the way for further explorations into the rich landscape of derivations and their 

generalizations in other algebraic structures. 

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