\section{Introduction}
Recall that a ring $R$ is said to be a prime ring if $aRb=0$ for some $a, b\in R$ implies either $a=0$ or $b=0,$ and a semi-prime ring if $aRa=0$ for some $a\in R$ implies that $a=0$. Let $n$ be a positive integer. A ring $R$ is called $n$-torsion free if $na=0$ for some $a\in R$ implies that $a=0$. Let $R=(R,+,\cdot)$ be a ring. Then the opposite ring of $R$, denoted by $R^{\circ}=(R,+,\star)$, is same as $R$ except the multiplication $\star$ is defined by $a\star b=b \cdot a$, for all $a, b\in R$.
Throughout this paper, let $R$ be a ring with the identity element $1$ and a non-trivial idempotent element $e$. Let $e_1=e$ and $e_2=1-e_1$. Then $R=R_{11} \oplus R_{12} \oplus R_{21} \oplus R_{22}$ where $R_{ij}=e_i R e_j,$ and $1\leq i, j \leq 2$. A map $f: R \rightarrow R$ is said to be additive if \begin{align*}
f(a+b)=f(a)+f(b),
\end{align*}
for all $a,b \in R$. An additive bijective map $\varphi: R \rightarrow R$ is said to be an automorphism if
\begin{align*}
\varphi(ab)=\varphi(a)\varphi( b),
 \end{align*}
 for all $a,b \in R$; and an anti-automorphism if
\begin{align*}
\varphi(ab)=\varphi(b)\varphi( a),
 \end{align*}
 for all $a,b \in R$. Note that without assuming the additivity condition of $\varphi$, the map is known as a multiplicative automorphism and multiplicative anti-automorphism, respectively. An additive map $d: R \rightarrow R$ is said to be a derivation if
 \begin{align*}
 d(a b)=d(a) b+a d(b),
  \end{align*}
 for all $a,b\in R$. Let $\varphi$ be an automorphism of $R$.
 In 1957 \cite{herstein1957jordan}, Herstein proved that any Jordan derivation (a generalization of ordinary derivation) over a prime ring become an ordinary derivation with some torsion restriction of the ring.
 An additive map $D: R \rightarrow R$ is said to be a skew derivation if
  \begin{align*}
D(ab)=D(a) b+\varphi(a) D(b),
  \end{align*}
for all $a,b \in R$. It is also known as $\varphi$-derivation. For example, we consider identity homomorphism $1_R$ on $R$. Then the map $\varphi-1_R$ is an example of skew derivation.
Similarly, we can now define generalized skew derivation. An additive map $g: R \rightarrow R$ is said to be a generalized skew derivation if
 \begin{align*}
g(ab)=g(a)b+\varphi(a) D(b),
  \end{align*}
for all $a, b \in R$ where $\varphi$ is an automorphism of $R$ and $D$ is a skew derivation associated with $g$. Thus we can think of the above two derivations as a generalization of both derivation and automorphism. Researchers studied skew derivations in ring theory and the theory of operator algebras. If we remove the condition of additivity, then it is called multiplicative derivation. Similarly, $D$ and $g$ are called multiplicative skew derivation and multiplicative generalized skew derivation, respectively, without assuming the additivity of the maps. So it is natural to ask, `when are some multiplicative maps additive?'.
In 1948, Rickart \cite{rickart1948one} first raised this question. He showed that a bijective and multiplicative onto map $h: B \rightarrow R^{\prime}$ is additive, where $B$ is a Boolean ring and $R^{\prime}$ is any ring. Also he proved that any bijective multiplicative mapping from a ring $R$ onto a ring $S$ is additive, where $R$ contains a family of minimal ideals satisfying some conditions.
In 1958 \cite{johnson1958rings}, Johnson extended Rickart's result to a larger class of rings.
In 1969, Martindale \cite{martindale1969multiplicative} proved that every multiplicative isomorphism from a ring $R$ onto a ring $R^{\prime}$ is additive. Using Martindale conditions, in 1991, Daif \cite{daif1991multiplicative} proved that any multiplicative derivation on a ring $R$ is additive.
In 2009 \cite{wang2009additivity}, Wang proved that any multiplicative isomorphism from a ring $R$ onto a ring $S$ is additive, where $R$ contains a family of idempotents satisfying some conditions.
In 2012 \cite{jing2012additivity}, Jing and Lu proved that every multiplicative Jordan derivation and Jordan triple derivation over a ring $R$ with a non-trivial idempotent satisfying some conditions, is additive.
In 2014 \cite{ferreira2014multiplicative}, Ferreira proved that every $m$-multiplicative isomorphism from a triangular $n$-matrix ring onto another ring is additive and every multiplicative $m$-derivation over any triangular $n$-matrix ring is additive, where triangular $n$-matrix ring satisfies some conditions in both the cases.
In 2015 \cite{ferreira2015jordan}, he also proved that every multiplicative Jordan derivation on a triangular ring with some conditions is additive.
In 2017, Yadav and Sharma \cite{yadav2017additivity} proved that any multiplicative generalized Jordan derivation on a ring $R$ with a non-trivial idempotent is additive. This result raises a question for us: When are multiplicative automorphisms, multiplicative skew derivations, and multiplicative generalized skew derivations additive?'. In this paper, we find an affirmative answer to this question. Also, in the case of multiplicative skew derivation $D$, we consider $\varphi$ as a multiplicative automorphism. Similarly, for multiplicative generalized skew derivation $g$, we think of $D$ as multiplicative skew derivation.
 \\

Let $R$ satisfy the following condition.

\begin{equation}
  \tag{A}
  \text{If} ~a_{i j} x_{j k}=0~\text{ for all} ~x_{j k} \in R_{j k}~ \text{and for some }~a_{ij}\in R_{ij}, ~\text{then}~ a_{i j}=0.
  \label{eqA}
\end{equation}

	
	
	
	
	
	\section{Multiplicative Automorphism}

\begin{theorem}
\label{thm2.1}
If $\varphi$ is a multiplicative automorphism on $R$, then $\varphi$ is additive. Moreover, $\varphi$ is an automorphism on $R$.
\end{theorem}

Before proving Theorem \ref{thm2.1}, we have several lemmas.

\begin{lemma}
\label{lem2.2}
$\varphi(0)=0$.
\end{lemma}

\begin{proof}
Since $\varphi$ is onto, there exists $ x \in R$ such that $\varphi(x)=0$. Now, $\varphi(0)=\varphi(0\cdot x)=\varphi(0)  \varphi(x)=0$.
\end{proof}

\begin{lemma}
\label{lem2.3}
$\varphi(1)=1$.
\end{lemma}

\begin{proof}
Let $\varphi(1)=p$ and $q \in R$ be an arbitrary element. Since $\varphi$ is onto, there exists $ x \in R$ such that $\varphi(x)=q.$ Now $p q=\varphi(1) \varphi(x)=\varphi(1 \cdot x)=\varphi(x)=q$. Also, $q p=\varphi(x)  \varphi(1)=\varphi(x \cdot 1)=\varphi(x)=q$. Therefore
\begin{align*}
p q=q p=q \implies p=1 \implies \varphi(1)=1.
\end{align*}
\end{proof}
	
	
\begin{lemma}
\label{lem2.4}
$ \varphi (e_1)+\varphi (e_2) =1.$
\end{lemma}


\begin{proof}
Since $\varphi$ is onto, there exists $ z \in R$ such that $\varphi(z)=\varphi (e_1)+\varphi (e_2)$.
Since $\varphi$ is  one-one, we have
	$$
	\begin{aligned}	
		 \varphi (z e_1)&=\varphi(z) \varphi (e_1) = [\varphi (e_1)+\varphi (e_2)] \varphi (e_1)=\varphi(e_1) \varphi (e_1)+\varphi (e_2) \varphi(e_1)\\
		 &=\varphi (e_1e_1)+\varphi (e_2 e_1)=\varphi(e_1)+\varphi(0)=\varphi(e_1)~\text{(By Lemma \ref{lem2.2})}\\	
	\implies & z e_1= e_1
\end{aligned}
$$
	$$
	\begin{aligned}	
		\text{Also,}~ \varphi (z e_2)	&=\varphi(z) \varphi (e_2)= [\varphi (e_1)+\varphi (e_2)] \varphi (e_2)=\varphi (e_1) \varphi (e_2)+\varphi (e_2) \varphi (e_2)\\
	&=\varphi (e_1 e_2)+\varphi (e_2  e_2)=\varphi(0)+\varphi (e_2)=\varphi (e_2)~\text{(By Lemma \ref{lem2.2})}\\
\implies 	& z e_2= e_2
		\end{aligned}
$$
	Adding both, we get
		$$
	\begin{aligned}
	 &z e_1+ ze_2=e_1+e_2 \implies z (e_1+e_2)=e_1+e_2  \implies z=1 \\
	 & \implies \varphi(z)=\varphi(1) \implies \varphi (e_1)+\varphi (e_2) = 1~\text{(By Lemma \ref{lem2.3})}.
	\end{aligned}
	$$
	Hence, the result follows.
\end{proof}


	\begin{lemma}
	\label{lem2.5}
	Let $a_{ij}, b_{ij} \in R_{ij}$. Then
			$$
		\begin{aligned}
	(i)& ~\varphi (a_{11}+b_{12})=\varphi (a_{11})+\varphi (b_{12}),\\
	(ii)&~ \varphi (a_{22}+b_{21})=\varphi (a_{22})+\varphi (b_{21}),\\
	(iii)&~ \varphi (a_{11}+b_{21})=\varphi (a_{11})+\varphi (b_{21}),\\
	(iv)&~ \varphi (a_{22}+b_{12})=\varphi (a_{22})+\varphi (b_{12}).
		\end{aligned}
$$
	\end{lemma}
	
	
\begin{proof}
	 Let $x_{22} \in R_{22}$. Since $\varphi$ is onto, there exists $z \in R$ such that $\varphi(z)=\varphi (a_{11})+\varphi (b_{12})$. Since $\varphi$ is  one-one,
	\begin{equation}
	\label{eq2.1}
	\begin{aligned}
		\varphi (z x_{22}) &=\varphi(z) \varphi (x_{22}) = [\varphi (a_{11})+\varphi (b_{12})] \varphi (x_{22}) =\varphi (a_{11}) \varphi (x_{22})+\varphi (b_{12}) \varphi (x_{22}) \\
		&=\varphi (a_{11} x_{22})+\varphi (b_{12} x_{22}) =\varphi(0)+\varphi (b_{12} x_{22}) =\varphi (a_{11} x_{22}+b_{12} x_{22}) \\
\implies		& z x_{22} = (a_{11}+b_{12}) x_{22}
 \implies   [z- (a_{11}+b_{12})] x_{22}=0\\
\implies		&[z- (a_{11}+b_{12})]_{22}=0 ~\&~  [z- (a_{11}+b_{12})]_{12}=0 \\
& ~\text{(By the assumption \eqref{eqA} on R)}.
	\end{aligned}
	\end{equation}
Let $x_{12}\in R_{12}$. Then 	
	\begin{equation}
	\label{eq2.2}
\begin{aligned}
	 \varphi (z x_{12}) &= [\varphi (a_{11})+\varphi (b_{12})] \varphi (x_{12})=\varphi (a_{11}) \varphi (x_{12})+\varphi (b_{12}) \varphi (x_{12})\\
	&=\varphi (a_{11} x_{12})+\varphi (b_{12} x_{12})=\varphi (a_{11} x_{12})+\varphi(0)=\phi (a_{11} x_{12}+b_{12} x_{12})\\
	\implies & z x_{12}= (a_{11}+b_{12}) x_{12} \implies [z- (a_{11}+b_{12})] x_{12}=0\\
\implies	& [z- (a_{11}+b_{12})]_{11}=0 ~\text{and}~  [z- (a_{11}+b_{12})]_{21}=0.
\end{aligned}
\end{equation}
By \eqref{eq2.1} and \eqref{eq2.2},
\begin{align*}
&z=a_{11}+b_{12} \implies \varphi (a_{11}+b_{12})=\varphi(z)=\varphi (a_{11})+\varphi (b_{12}).
\end{align*}	
	
Hence, we get $(i)$. Similarly, we can prove $(ii)$. Now,
$$
		 \begin{aligned}
		 	\varphi (e_1) \varphi (a_{11}+b_{21})&=\varphi (e_1 (a_{11}+b_{21}))=\varphi (e_1 a_{11}+0)=\phi (e_1 a_{11})+\phi (e_1 b_{21})\\
	&=\varphi (e_1) \varphi (a_{11})+\varphi (e_1) \varphi (b_{21}).\\
\end{aligned}
$$

Therefore,
\begin{align*}
\varphi (e_1) [\varphi (a_{11}+b_{21})-\varphi (a_{11})-\varphi (b_{21})]=0.
\end{align*}

Similarly, we have
\begin{align*}
\varphi (e_2) [\varphi (a_{11}+b_{21})-\varphi (a_{11})-\varphi (b_{21})]=0.
\end{align*}

	Adding both the above,
\begin{align*}
	&[\varphi (e_1)+\varphi (e_2)] [\varphi (a_{11}+b_{21})-\varphi (a_{11})-\varphi (b_{21})]=0\\
	& \implies \varphi (a_{11}+b_{21})=\varphi (a_{11})+\varphi (b_{21})~\text{(By Lemma \ref{lem2.4})}.
\end{align*}
	Hence, we have $(iii)$. Similarly, we can prove $(iv)$.
	\end{proof}
	
	
	
		\begin{lemma}
		\label{lem2.6}
		Let $a_{ij}, b_{ij}\in R_{ij}$. Then
			 $$
			\begin{aligned}
			(i) &~\varphi (a_{12}+b_{12})=\varphi (a_{12})+\varphi (b_{12})\\
	\text{and} (ii) ~ \varphi (a_{21}+b_{21})=\varphi (a_{21})+\varphi (b_{21})\\
\end{aligned}
.$$
		\end{lemma}
	\begin{proof}
		Let $x_{22} \in R_{22}$. Since $\varphi$ is onto, there exists $z\in R$ such that $\varphi(z)=\varphi (a_{12})+\varphi (b_{12})$. Now,
	$$
	\begin{aligned}
	\varphi (z x_{22})&=\varphi(z) \varphi (x_{22})= [\varphi (a_{12})+\varphi (b_{12})] \varphi (x_{22})\\
	&=\varphi (a_{12}) \varphi (x_{22})+\varphi (b_{12}) \varphi (x_{22})=\varphi (a_{12} x_{22})+\varphi (b_{12} x_{22})\\
	&=\varphi (e_1 x_{22})+\varphi (e_1 a_{12} x_{22})+\varphi (b_{12} x_{22})+\varphi (b_{12} a_{12} x_{22})\\
	&= [\varphi (e_1)+\varphi (b_{12})] [\varphi (x_{22})+\varphi (a_{12} x_{22})]\\
	&=\varphi (e_1+b_{12}) \varphi (x_{22}+a_{12} x_{22})~\text{(By Lemma \ref{lem2.5})} \\
	&=\varphi ( (e_1+b_{12}) (x_{22}+a_{12} x_{22}))\\
	&=\varphi (a_{12} x_{22}+b_{12} x_{22}).
\end{aligned}
$$

Since $\varphi$ is one-one,
\begin{align*}
z x_{22}=a_{12} x_{22}+b_{12} x_{22} \implies [z- (a_{12}+b_{12})] x_{22}=0.
\end{align*}



By the assumption \eqref{eqA} on $R$,
	\begin{equation}
	\label{eq2.3}
	\begin{aligned}
& [z- (a_{12}+b_{12})]_{12}=0~ \text{and} ~[z- (a_{12}+b_{12})]_{22}=0.
\end{aligned}
	\end{equation}

Let $x_{11}\in R_{11}$. Similarly, we have
	\begin{equation}
	\label{eq2.4}
	\begin{aligned}
		 \varphi (z x_{11})&=\varphi(z) \varphi (x_{11})= [\varphi (a_{12})+\varphi (b_{12})] \varphi (x_{11})=\varphi (a_{12}) \varphi (x_{11})+\varphi (b_{12}) \varphi (x_{11})\\
	&=\varphi (a_{12} x_{11})+\varphi (b_{12} x_{11})=\varphi(0)+\varphi(0)=\varphi (a_{12} x_{11}+b_{12} x_{11})\\
	\implies & z x_{11}=a_{12} x_{11}+b_{12} x_{11}\\
	 \implies & [z- (a_{12}+b_{12})] x_{11}=0\\
	\implies & [z- (a_{12}+b_{12})]_{11}=0
	~ \text{and} [z- (a_{12}+b_{12})]_{21}=0.
\end{aligned}
	\end{equation}
	
Therefore, by \eqref{eq2.3} and \eqref{eq2.4},
\begin{align*}
 z=a_{12}+b_{12} \implies \varphi (a_{12}+b_{12})=\varphi(z)= \varphi (a_{12})+\varphi (b_{12}).
\end{align*}
	
	
Hence, we have $(i)$. Similarly, we can prove $(ii).$
\end{proof}


	\begin{lemma}
	\label{lem2.7}
	Let $a_{ij}, b_{ij} \in R_{ij}$. Then
			$$
		\begin{aligned}
		 (i)&~ \varphi (a_{11}+b_{11})=\varphi (a_{11})+\varphi (b_{11})\\
	(ii) &~\varphi (a_{22}+b_{22})=\varphi (a_{22})+\varphi (b_{22}).
\end{aligned}
$$
\end{lemma}


	\begin{proof}
 Since $\varphi$ is onto, there exists $z \in R$ such that $\varphi(z)=\varphi (a_{11})+\varphi(b_{11})$. Let $ x_{12} \in R_{12}$. Then, we have
\begin{equation}
\label{eq2.5}
\begin{aligned}
			\varphi (z x_{12}) &=\varphi(z) \varphi (x_{12})= [\varphi (a_{11})+\varphi (b_{11})] \varphi (x_{12}) =\varphi (a_{11}) \varphi (x_{12})+\varphi (b_{11}) \varphi (x_{12}) \\
			&=\varphi (a_{11} x_{12})+\varphi (b_{11} x_{12}) =\varphi (a_{11} x_{12}+b_{11} x_{12})~\text{(By Lemma \ref{lem2.6})}\\
		\implies & z x_{12}= (a_{11}+b_{11}) x_{12}  ~\text{(Since}~ \varphi ~\text{is one-one)}\\
		\implies & [z- (a_{11}+b_{11})] x_{12}=0 \\
		\implies & [z- (a_{11}+b_{11})]_{11}=0 ~\text{and} [z- (a_{11}+b_{11})]_{21}=0.
	\end{aligned}
\end{equation}

Similarly, by taking $ x_{21} \in R_{21}$ we obtain
\begin{equation}
\label{eq2.6}
 [z- (a_{11}+b_{11})]_{12}=0
		~\text{and}~  [z- (a_{11}+b_{11})]_{22}=0.
\end{equation}

Therefore, by \eqref{eq2.5} and \eqref{eq2.6}, we have
\begin{align*}
& \varphi (a_{11}+b_{11})=\varphi (a_{11})+\varphi (b_{11}).
\end{align*}	
	
Similarly, we can prove that
\begin{align*}
& \varphi (a_{22}+b_{22})=\varphi (a_{22})+\varphi (b_{22}).
\end{align*}	
\end{proof}



	\begin{lemma}
	\label{lem2.8}
	Let $a_{ij}, b_{ij}, c_{ij}, d_{ij} \in R_{ij}$. Then
	\begin{align*}
	\varphi (a_{11}+b_{12}+c_{21}+d_{22})= \varphi (a_{11})+\varphi (b_{12}) +\varphi (c_{21})+\varphi (d_{22}).
	\end{align*}
		\end{lemma}
		
		
		\begin{proof}
		 Since $\varphi$ is onto, there exists $z \in R$ such that
\begin{align*}
\varphi(z) =  \varphi (a_{11})+\varphi (b_{12})+\varphi (c_{21})+\varphi (d_{22}).
\end{align*}		
Let $x_{11}\in R_{11}$. Then		
			 	$$
			 \begin{aligned}
			  \varphi(zx_{11}) &=\varphi(z)\varphi(x_{11})=  	 (\varphi (a_{11})+\varphi (b_{12})+\varphi (c_{21})+\varphi (d_{22}))\varphi(x_{11})\\
			  &= \varphi (a_{11})\varphi(x_{11}) +\varphi (b_{12})\varphi(x_{11}) +\varphi (c_{21})\varphi(x_{11}) +\varphi (d_{22})\varphi(x_{11})\\
			  &= \varphi (a_{11}x_{11}) + \varphi (b_{12}x_{11}) + \varphi (c_{21}x_{11}) + \varphi (d_{22}x_{11})\\
			  &=\varphi (a_{11}x_{11})+\varphi(0) + \varphi (c_{21}x_{11})+\varphi(0)\\
			  &=\varphi (a_{11}x_{11} + c_{21}x_{11} )~\text{(By Lemma \ref{lem2.2} and \ref{lem2.5})}\\
			  &=\varphi (a_{11}x_{11} + b_{12}x_{11}+ c_{21}x_{11} + d_{22}x_{11} )\\
		\implies	  & z x_{11}= (a_{11}+b_{12}+c_{21}+d_{22}) x_{11} ~\text{(Since}~\varphi~\text{is one-one)}.
			 \end{aligned}
		 $$
		 	By the condition \eqref{eqA} on $R$, we obtain
	\begin{equation}
	\label{eq2.7}
	\begin{aligned}
		& [z- (a_{11}+b_{12}+c_{21}+d_{22})] x_{11}=0 \\
	\implies 	& [z- (a_{11}+b_{12}+c_{21}+d_{22})]_{11}=0 \text{and} ~[z- (a_{11}+b_{12}+c_{21}+d_{22})]_{21}=0.
	\end{aligned}
	\end{equation}
	
Similarly, by taking $x_{22} \in R_{22}$, we get
	\begin{equation}
	\label{eq2.8}
	{ [z- (a_{11}+b_{12}+c_{21}+d_{22})]_{12}=0 } ~
		\text{and} ~ { [z- (a_{11}+b_{12}+c_{21}+d_{22})]_{22}=0.}
	\end{equation}
	
	Therefore, by \eqref{eq2.7} and \eqref{eq2.8}
	\begin{align*}
	\varphi (a_{11}+b_{12}+c_{21}+d_{22})=\varphi (a_{11})+\varphi (b_{21})+\varphi (c_{21}) +\varphi (d_{22}).
	\end{align*}
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.1}]
Let $a, b \in R$. Then
\begin{align*}
& a=a_{11}+a_{12}+a_{21}+a_{2 2} ~\text{and}~ b=b_{11}+b_{12}+b_{21}+b_{22}~\text{for some}~a_{ij}, b_{ij} \in R_{ij}.
\end{align*}
Then			
	$$
 \begin{aligned}
		\varphi(a+b)&=\varphi (a_{11}+a_{12}+a_{21}+a_{22}+b_{11}+b_{12}+b_{21}+b_{22}) \\
		&=\varphi ( (a_{11}+b_{11})+ (a_{12}+b_{12})+ (a_{21}+b_{21})+ (a_{22}+b_{22})) \\
		&=\varphi (a_{11}+b_{11})+\varphi (a_{12}+b_{12})+\varphi (a_{21}+b_{21}) + \varphi (a_{22}+b_{22})\\
		&~(\text{By Lemma \ref{lem2.8}})\\
		&=\varphi (a_{11}) + \varphi (b_{11}) + \varphi (a_{12}) + \varphi (b_{12}) + \varphi (a_{21}) + \varphi (b_{21})
		+ \varphi (a_{22}) + \varphi (b_{22})\\
		&~(\text{By Lemma \ref{lem2.6} and \ref{lem2.7}})\\
		&= \varphi (a_{11} + a_{12}  + a_{21} + a_{22} ) + \varphi (b_{11} + b_{12} + b_{21} + b_{22} )\\
		&~(\text{By Lemma \ref{lem2.8}})\\
		&= \varphi (a) + \varphi (b).
		\end{aligned}
	$$
	Hence, $\varphi$ is additive.
\end{proof}

With motivation of the Corollaries in \cite{martindale1969multiplicative}, we have the following Corollaries.
\begin{corollary}
\label{cor2.9}
If $\varphi$ is a multiplicative automorphism on a prime ring $R$ with a non-trivial idempotent $e$, then $\varphi$ is additive. Moreover, $\varphi$ is an automorphism on $R$.
\end{corollary}

\begin{proof}
Let $e_1=e$ and $e_2=1-e_1$. Then $R_{ij}=e_i R e_j$. Then
\begin{align*}
& a_{ij}x_{jk}=0,~\text{for some}~a_{ij}\in R_{ij}~\text{and for all}~x_{jk}\in R_{jk}\\
\implies & e_i a e_j e_j x e_k=0, ~\text{for some}~a\in R~\text{and for all}~x\in R\\
\implies & (e_i a e_j) x (e_k)=0, ~\text{and for all}~x\in R\\
\implies & e_i a e_j=0 ~\text{(Since R is a prime ring)}\\
\implies & a_{ij}=0.
\end{align*}
Thus, $R$ satisfies condition \eqref{eqA}. Hence, we get the desired result by Theorem \ref{thm2.1}.
\end{proof}


	\begin{corollary}
	\label{cor2.10}
		If $\varphi$ is a multiplicative anti-automorphism on a prime ring $R$ with a non-trivial idempotent $e$, then $\varphi$ is additive. Moreover, $\varphi$ is an anti-automorphism on $R$.
	\end{corollary}
	
	\begin{proof}
	Let $\tau: R \rightarrow R^{\circ}$ be the map defined by
	\begin{align*}
	\tau(a)=a,
	\end{align*}
	for all $a\in R$. Then $\tau$ is an anti-isomorphism. Let $\sigma=\tau \circ \varphi$. Then $\sigma: R \rightarrow R^{\circ}$ is a multiplicative isomorphism. Then $\sigma$ is additive by a result in \cite{martindale1969multiplicative}. Therefore, $\varphi$ is additive (Since $\tau$ is additive and one-one).
	\end{proof}

	
	\section{Multiplicative Skew Derivation}
\begin{theorem}
\label{thm3.1}
If $D$ is a multiplicative skew derivation on $R$, then $D$ is additive. Moreover, $D$ is a skew derivation on $R$.
\end{theorem} 	
	
Let $\varphi$ be the associated multiplicative automorphism on $R$. Hence, by Theorem \ref{thm2.1}, $\varphi$ is additive. Before proving Theorem \ref{thm3.1}, we have several lemmas.

\begin{lemma}
\label{lem3.2}
\begin{align*}
D(0)=0.
\end{align*}
\end{lemma}

\begin{proof}
\begin{align*}
D(0)=D(0\cdot 0)=D(0)0+\varphi(0)D(0)=0 ~(\text{By Lemma \ref{lem2.2}}).
\end{align*}
\end{proof}
	
	\begin{lemma}
	\label{lem3.3}
		Let $a_{ij}, b_{ij} \in R_{ij}$. Then
		$$
		\begin{aligned}
			(i) &~ D (a_{11}+b_{12} )=D (a_{11} )+D (b_{12} )\\
			(ii) &~  D (a_{22}+b_{21} )=D (a_{22} )+D (b_{21} )\\
			(iii) &~  D (a_{11}+b_{21} )=D (a_{11} )+D (b_{21} ) \\
			(iv) &~  D (a_{22}+b_{12} )=D (a_{22} )+D (b_{12} ) .
		\end{aligned}
		$$
	\end{lemma}
	
	\begin{proof}
		Let $x_{22} \in R_{22}$. Then
		
		\begin{equation}
		\label{eq3.1}
		 D ( (a_{11}+b_{12} ) x_{22} ) = D (a_{11}+b_{12} ) x_{22}+\varphi (a_{11}+b_{12} ) D (x_{22} ).
		\end{equation}
	On the other hand,	
		\begin{equation}
		\label{eq3.2}
		\begin{aligned}
			& D ( (a_{11}+b_{12} ) x_{22} )\\
			=&~D (b_{12} x_{22} ) \\
			=&~D (a_{11} x_{22} )+D (b_{12} x_{22} ) \\
			=&~D (a_{11} ) x_{22}+\varphi (a_{11} ) D (x_{22} )+D (b_{12} ) x_{22}+\varphi (b_{12} ) D (x_{22} ).
		\end{aligned}
		\end{equation}
Comparing \eqref{eq3.1} and \eqref{eq3.2}, we get
		$$
		\begin{aligned}
			&{ [D (a_{11}+b_{12} )-D (a_{11} )-D (b_{12} ) ] x_{22}=0} ~\text{(Since} ~\varphi~\text{is additive)}\\
			\implies~& [D (a_{11}+b_{12} )-D (a_{11} )-D (b_{12} ) ]_{12} x_{22}=0\\
			\text{and}~&[D (a_{11}+b_{12} )-D (a_{11} )-D (b_{12} ) ]_{22} x_{22}=0.
		\end{aligned}
		$$
		By the assumption \eqref{eqA} on $R$,
		$$
		\begin{gathered}
			{ [D (a_{11}+b_{12} )-D (a_{11} )-D (b_{12} ) ]_{12}=0 } \\
			\text{and} ~[D (a_{11}+b_{12} )-D (a_{11} )-D (b_{12} ) ]_{22}=0.
		\end{gathered}
		$$
		Let $x_{12} \in R_{12}$. Then
		\begin{equation}
		\label{eq3.3}
		D ( (a_{11}+b_{12} ) x_{12} ) = D (a_{11}+b_{12} ) x_{12}+\varphi (a_{11}+b_{12} ) D (x_{12}).
		\end{equation}
		Also,
		\begin{equation}
		\label{eq3.4}
		\begin{aligned}
			D ( (a_{11}+b_{12} ) x_{12} )
			&=D (a_{11} x_{12} ) \\
			&=D (a_{11} x_{12} )+D (b_{12} x_{12} ) \\
			&=D (a_{11} ) x_{12}+\varphi (a_{11} ) D (x_{12} )+D (b_{12} ) x_{12}+\varphi (b_{12} ) D (x_{12} ).
		\end{aligned}
		\end{equation}
		Comparing \eqref{eq3.3} and \eqref{eq3.4},
		$$
		[D (a_{11}+b_{12} )-D (a_{11} )-D (b_{12} ) ] x_{12}=0.
		$$
		Again, by the condition \eqref{eqA} on $R$, we have
		$$
		\begin{aligned}
			&{ [D (a_{11}+b_{12} )-D (a_{11} )-D (b_{12} ) ]_{11}=0} \\
			\text{and}~ & [D (a_{11}+b_{12} )-D (a_{11} )-D (b_{12} ) ]_{21}=0.
		\end{aligned}
		$$
		Hence,
		\begin{align*}
		D (a_{11}+b_{12} )=D (a_{11} )+D (b_{12} ).
		\end{align*}
		Similarly, we can prove
		\begin{align*}
		D (a_{22}+b_{21} )=D (a_{22} )+D (b_{21} ).
		\end{align*}
				Now,
		\begin{equation}
		\label{eq3.5}
		D(e_1(a_{11}+b_{21}))=D(e_1)(a_{11}+b_{21})+\varphi(e_1)D(a_{11}+b_{21}).
		\end{equation}
		Also,
\begin{equation}
\label{eq3.6}
\begin{aligned}
D(e_1(a_{11}+b_{21}))&=D(e_1 a_{11})\\
&=D(e_1 a_{11})+D(e_1 b_{21})\\
&=D(e_1)a_{11}+\varphi(e_1)D(a_{11})+D(e_1)b_{21}+\varphi(e_1)D(b_{21}).
\end{aligned}
\end{equation}
Comparing \eqref{eq3.5} and \eqref{eq3.6},		
		\begin{equation}
		\label{eq3.7}
		\varphi(e_1)(D(a_{11}+b_{21})-D(a_{11})-D(b_{21}))=0.
		\end{equation}
		Similarly,
		\begin{equation}
		\label{eq3.8}
		\varphi(e_2)(D(a_{11}+b_{21})-D(a_{11})-D(b_{21}))=0.
		\end{equation}
		Adding \eqref{eq3.7} and \eqref{eq3.8}, we have
		\begin{align*}
		&(\varphi(e_1)+\varphi(e_2))(D(a_{11}+b_{21})-D(a_{11})-D(b_{21}))=0\\
		&\implies D(a_{11}+b_{21})=D(a_{11})+D(b_{21}) ~(\text{By Lemma \ref{lem2.3}}).
		\end{align*}
		Similarly, we can prove
		\begin{align*}
		D (a_{22}+b_{12} )=D (a_{22} )+D (b_{12} ).
		\end{align*}
	\end{proof}

	
	\begin{lemma}
		\label{lem3.4}
		Let $a_{ij}$, $b_{ij}, c_{ij} \in R_{ij}$. Then
		$$
		\begin{aligned}
			(i) &~ D (a_{12}+b_{12} c_{22} )=D (a_{12} )+D (b_{12} c_{22} ) \\
			(ii) &~ D (a_{21}+b_{22} c_{21} )=D (a_{21} )+D (b_{22} c_{21} ).
		\end{aligned}
		$$
	\end{lemma}
	
	\begin{proof}
		Note that
		\begin{align*}
		a_{12}+b_{12} c_{22}= (e_1+b_{12} )(a_{12}+c_{22} ).
\end{align*}		
	Therefore,	
		$$
		\begin{aligned}
			 D (a_{12}+b_{12} c_{22} )
			=~& D ( (e_1+b_{12} ) (a_{12}+c_{22} ) ) \\
			=~& D (e_1+b_{12} ) (a_{12}+c_{22} )+\varphi (e_1+b_{12} ) D (a_{12}+c_{22} ) \\
			=~& { [D (e_1 )+D (b_{12} ) ] (a_{12}+c_{22} )+ [\varphi (e_1 )+\varphi (b_{12} ] } [D (a_{12} )+D (c_{22} ) ]\\
			&~\text{(By Lemma \ref{lem3.3})} \\
			=~& D (e_1 a_{12} )+D (e_1 c_{22} )+D (b_{12} c_{22} ) +D (b_{12} a_{12} ) \\
			=~& D (a_{12} )+D (b_{12} c_{22} )~\text{(By Lemma \ref{lem3.2})} .
		\end{aligned}
		$$
		Similarly, we can prove that
		\begin{align*}
		D (a_{21}+b_{22} c_{21} )=D (a_{21} )+D (b_{22} c_{21} ).
		\end{align*}
	\end{proof}
	
	\begin{lemma}
		\label{lem3.5}
		Let $a_{12}, b_{12} \in R_{12}$ and $a_{21}, b_{21} \in R_{21}$. Then
		$$
		\begin{aligned}
			(i) &~ D (a_{12}+b_{12} )=D (a_{12} )+D (b_{12} ) \text {. } \\
			(ii) &~ D (a_{21}+b_{21} )=D (a_{21} )+D (b_{21} ) \text {. }
		\end{aligned}
		$$
	\end{lemma}
	
	\begin{proof}
		Let $x_{22} \in R_{22}$. Then
		$$
		\begin{aligned}
			&~D(a_{12}+b_{12} ) x_{22}+\varphi (a_{12}+b_{12} ) D (x_{22} )\\
			=&~D [ (a_{12}+b_{12} ) x_{22} ]\\
			=&~D (a_{12} x_{22} + b_{12} x_{22} ) \\
			=&~D (a_{12} x_{22} )+D (b_{12} x_{22} ) ~\text{(By Lemma \ref{lem3.4})}\\
			=&~D (a_{12} ) x_{12}+\varphi (a_{12} ) D (x_{22} )+D (b_{12} ) x_{22}+\varphi (b_{12} ) D (x_{22} ).
		\end{aligned}
		$$
		Consequently, we obtain
		\begin{equation}
		\label{eq3.9}
		\begin{aligned}
			&[D (a_{12}+b_{12} )-D (a_{12} )-D (b_{12} ) ] x_{22}=0\\
			\implies ~&[D (a_{12}+b_{12} )-D (a_{12} )-D (b_{12} ) ]_{12}=0\\
			\text{and} & [D (a_{12}+b_{12} )-D (a_{12} )-D (b_{12} ) ]_{22}=0.
		\end{aligned}
		\end{equation}
		Let $x_{12} \in R_{12}$. Then
		$$
		\begin{aligned}
			 &D(a_{12}+b_{12} ) x_{12}+\varphi (a_{12}+b_{12} ) D (x_{12} )\\
			 &=D [ (a_{12}+b_{12} ) x_{12} ] \\
			&=D(0)=0 \\
			&=D (a_{12} x_{12} )+D (b_{12} x_{12} ) \\
			&=D (a_{12} ) x_{12}+\varphi (a_{12} ) D (x_{12} )+D (b_{12} ) x_{12}+\varphi (b_{12} ) D (x_{12} ),
		\end{aligned}
		$$
		which gives
		\begin{equation}
		\label{eq3.10}
		\begin{aligned}
			&[D (a_{12}+b_{12} )-D (a_{12} )-D (b_{12} ) ] x_{12}=0\\
			\implies ~&[D (a_{12}+b_{12} )-D (a_{12} )-D (b_{12} ) ]_{11}=0\\
			\text{and}~ & [D (a_{12}+b_{12} )-D (a_{12} )-D (b_{12} ) ]_{21}=0.
		\end{aligned}
		\end{equation}
		Hence, by \eqref{eq3.9} and \eqref{eq3.10},
		\begin{align*}
		D (a_{12}+b_{12} )=D (a_{12} )+D (b_{12} ).
        \end{align*}
		Similarly, we can get
		\begin{align*}
		D (a_{21}+b_{21} )=D (a_{21} )+D (b_{21} ).
\end{align*}		
	\end{proof}
	
	\begin{lemma}
	\label{lem3.6}
		Let $a_{ii}, b_{ii} \in R_{ii}$, where $i \in \{1,2\}$. Then
		$$
		\begin{aligned}
			& (i) ~D (a_{11}+b_{11} )=D (a_{11} )+D (b_{11} )\\
			& (ii) ~D (a_{22}+b_{22} )=D (a_{22} )+D (b_{22} ).
		\end{aligned}
		$$
	\end{lemma}
	
	\begin{proof} Let $x_{22} \in R_{22}$. Then
		$$
		\begin{aligned}
			&D (a_{11}+b_{11} ) x_{22}+\varphi (a_{11}+b_{11} ) D (x_{22} )\\
			=& D [ (a_{11}+b_{11} ) x_{22} ] \\
			=& D(0)=0 \\
			=& D (a_{11} x_{22} )+D (b_{11} x_{22} ) \\
			=& D (a_{11} ) x_{22}+\varphi (a_{11} ) D (x_{22} )+D (b_{11} ) x_{22}+\varphi (b_{11} ) D (x_{22} )
		\end{aligned}
		$$
		Therefore,
		\begin{align*}
		[D (a_{11}+b_{11} )-D (a_{11} )-D (b_{11} ) ] x_{22}=0,
		\end{align*}
		 which implies
		 \begin{equation}
		 \label{eq3.11}
		 \begin{gathered}
			{ [D (a_{11}+b_{11} )-D (a_{11} )-D (b_{11} ) ]_{12}=0} \\
			\text{and} ~[D (a_{11}+b_{11} )-D (a_{11} )-D (b_{11} ) ]_{22}=0
		\end{gathered}
		 \end{equation}
		Let $x_{12} \in R_{12}$. Then
		$$
		\begin{aligned}
			&D (a_{11}+b_{11}) x_{12}+\varphi (a_{11}+b_{11} ) D (x_{12} )\\
			& =D [( a_{11}+b_{11} ) x_{12} ]\\
			&= D (a_{11} x_{12}+b_{11} x_{12} ) \\
			&=D (a_{11} x_{12} )+D (b_{11} x_{12} )~\text{(By Lemma \ref{lem3.5})} \\
			&=D (a_{11} ) x_{12}+\varphi (a_{11} ) D (x_{12} )+D (b_{11} ) x_{12}+\varphi (b_{11} ) D (x_{12} )
		\end{aligned}
		$$
		Therefore,
		\begin{align*}
		[D (a_{11}+b_{11} )-D (a_{11} )-D (b_{11} ) ] x_{12}=0,
		\end{align*}
		which implies
		\begin{equation}
		\label{eq3.12}
		\begin{gathered}
			[D (a_{11}+b_{11} )-D (a_{11} )-D (b_{11} ) ]_{11}=0\\
			\text{and}~ [D (a_{11}+b_{11} )-D (a_{11} )-D (b_{11} ) ]_{21}=0.
		\end{gathered}
		\end{equation}
		Hence, by \eqref{eq3.11} and \eqref{eq3.12},
		\begin{align*}
		D (a_{11}+b_{11} )=D (a_{11} )+D (b_{11} ).
\end{align*}		
		Similarly, we can prove that
		\begin{align*}
		D (a_{22}+b_{22} )=D (a_{22} )+D (b_{22} ).
\end{align*}		
	\end{proof}
	
	\begin{lemma}
	\label{lem3.7}
		Let $a_{11} \in R_{11},~ b_{12} \in R_{12},~ c_{21} \in R_{21},~ d_{22} \in R_{22}$. Then
		\begin{align*}
		D (a_{11}+b_{12}+c_{21}+d_{22} )= D (a_{11} )+D (b_{12} ) +D (c_{21} )+D (d_{22} ).
\end{align*}		 		
	\end{lemma}
	
	\begin{proof}
		Let $x_{11} \in R_{11}$. Then
		$$
		\begin{aligned}
			& D (a_{11}+b_{12}+c_{21}+d _{22} ) x_{11} +\varphi (a_{11}+b_{12}+c_{21}+d_{22} ) D (x_{11} ) \\
			=&D [ (a_{11}+b_{12}+c_{21}+ d_{22} ) x_{11} ]\\
			=& D (a_{11} x_{11}+c_{21} x_{11} ) \\
			=& D (a_{11} x_{11} )+D (b_{12} x_{11} )+D (c_{21} x_{11} )+D (d_{22} x_{11} )\\
			&~\text{(By Lemma \ref{lem3.2} and \ref{lem3.3})} \\
			=& D (a_{11} ) x_{11}+\varphi (a_{11} ) D (x_{11} )+D (b_{12} ) x_{11}+\varphi (b_{12} ) D (x_{11} ) \\
			&+D (c_{21} ) x_{11}+\varphi (c_{21} ) D (x_{11} )+D (d _{22} ) x_{11}+\varphi (d_{22} ) D (x_{11} ).
		\end{aligned}
		$$
		Comparing both sides and using the additivity of $\varphi$, we have
		\begin{equation}
		\label{eq3.13}
		\begin{aligned}
			&  [D (a_{11}+b_{12}+c_{21}+d_{22} )-D (a_{11} )-D (b_{12} )-D (c_{21} )-D (d_{22} ) ] x_{11}=0\\
			\implies &   [D (a_{11}+b_{12}+c_{21}+d_{22} )-D (a_{11} )-D (b_{12} )-D (c_{21} )-D (d_{22} ) ]_{11}=0\\
			&  \text{and} ~[D (a_{11}+b_{12}+c_{21}+d_{22} )-D (a_{11} )-D (b_{12} )-D (c_{21} )-D (d_{22} ) ] _{21}=0.
		\end{aligned}
		\end{equation}
		Similarly, by taking $x_{22}$ from $R_{22}$, we can get
		\begin{equation}
		\label{eq3.14}
		\begin{aligned}
			& [D (a_{11}+b_{12}+c_{21}+d_{22} )-D (a_{11} )-D (b_{12} )-D (c_{21} )-D (d_{22} ) ]_{12}=0\\
			& \& ~  [D (a_{11}+b_{12}+c_{21}+d_{22} )-D (a_{11} )-D (b_{12} )-D (c_{21} )-D (d_{22} ) ] _{22}=0.
		\end{aligned}
		\end{equation}
		Hence, by \eqref{eq3.13} and \eqref{eq3.14},
		\begin{align*}
		D (a_{11}+b_{12}+c_{21}+d_{22} )= D (a_{11} )+D (b_{12} ) +D (c_{21} )+D (d_{22} ).
\end{align*}	
	\end{proof}
	
\begin{proof}[Proof of Theorem \ref{thm3.1}]
		Let $a, b \in R$. Then
		\begin{align*}
		& a=a_{11}+a_{12}+a_{21}+a_{2 2},\\
		& b=b_{11}+b_{12}+b_{21}+b_{22},
\end{align*}		
for some $a_{ij}, b_{ij} \in R_{ij}$.
		Now,
		$$
		\begin{aligned}
			D(a+b) =&D (a_{11}+a_{12}+a_{21}+a_{22}+b_{11}+b_{12}+b_{21}+b_{22} ) \\
			=& D [ (a_{11}+b_{11} )+ (a_{12}+b_{12} )+ (a_{21}+b_{21} )+ (a_{22}+b_{22} ) ] \\
			=& D (a_{11}+b_{11} )+D (a_{12}+b_{12} )+D (a_{21}+b_{21} )+D (a_{22}+b_{22} ) \\
			&~\text{(By Lemma \ref{lem3.7})}\\
			=& D (a_{11} )+D (b_{11} )+D (a_{12} )+D (b_{12} )\\
			&+D (a_{21} ) +D (b_{21} )+D (a_{22} )+D (b_{22} ) ~\text{(By Lemma \ref{lem3.5} and \ref{lem3.6})}\\
			=& D (a_{11}+a_{12}+a_{21}+a_{22} ) +D(b_{11} +b_{12}+b_{21}+b_{22} ) \\
			&~\text{(By Lemma \ref{lem3.7})}\\
			=& D(a) + D(b).
		\end{aligned}
		$$
		Hence, $D$ is additive.
	\end{proof}

\begin{corollary}
\label{cor3.8}
Let $R$ be a $2$-torsion free semi-prime ring with a non-trivial idempotent $e$ and satisfies the condition
	\begin{equation}
	 \tag{B}
		a_{i i} x_{i j}=0,~ \text{for some }~a_{ii}\in R_{ii}~\text{and for all} ~x_{i j} \in R_{i j}~(i\neq j), ~\text{then}~ a_{i i}=0.
		\label{eqB}
	\end{equation}	
Then every multiplicative skew derivation $D$ over $R$ is additive. Moreover, $D$ is a skew derivation on $R$.
\end{corollary}

\begin{proof}
Since $R$ satisfies \eqref{eqB}, by Lemma $1.5$ in \cite{jing2012additivity}, $R$ also satisfies the condition \eqref{eqA}. Hence, we have the desired result by Theorem \ref{thm3.1}.
\end{proof}


\section{Multiplicative Generalized Skew Derivation}
\begin{theorem}
\label{thm4.1}
If $g$ is a multiplicative generalized skew derivation on $R$, then $g$ is additive. Moreover, $g$ is a generalized skew derivation on $R$.
\end{theorem} 	
	
Let $\varphi$ and $D$ be the associated multiplicative automorphism and associated multiplicative skew derivation on $R$, respectively. Hence, by Theorem \ref{thm2.1} and \ref{thm3.1}, $\varphi$ and $D$ are additive. Before proving Theorem \ref{thm4.1}, we have several lemmas.
	
\begin{lemma}
\label{lem4.2}
\begin{align*}
g(0)=0.
\end{align*}
\end{lemma}
	
\begin{proof}
\begin{align*}
g(0)=g(0\cdot 0)=g(0)0+\varphi(0)D(0)=0~(\text{By Lemma \ref{lem2.2} and \ref{lem3.2}}).
\end{align*}
\end{proof}	

\begin{lemma}
\label{lem4.3}
\begin{align*}
D(1)=0.
\end{align*}
\end{lemma}

\begin{proof}
\begin{align*}
&g(1)=g(1\cdot 1)=g(1)1+\varphi(1)D(1)=g(1)+D(1)\\
& \implies D(1)=0.
\end{align*}
\end{proof}

\begin{lemma}
\label{lem4.4}
\begin{align*}
g(e_1+e_2)=g(e_1)+g(e_2).
\end{align*}
\end{lemma}

\begin{proof}
\begin{align*}
& g(e_1)=g(1\cdot e_1)=g(1)e_1+\varphi(1)D(e_1)\\
 ~\text{and}~&g(e_2)=g(1\cdot e_2)=g(1)e_2+\varphi(1)D(e_2).
\end{align*}
Adding these,
\begin{align*}
g(e_1)+g(e_2)&=g(1)(e_1+e_2)+\varphi(1)(D(e_1)+D(e_2))\\
&=g(1)1+\varphi(1)D(e_1+e_2)\\
&=g(1)+D(1)=g(1)~(\text{By Lemma \ref{lem4.3}})\\
&=g(e_1+e_2).
\end{align*}
\end{proof}
	
	\begin{lemma}
	\label{lem4.5}
	Let $a_{ij} \in R_{ij}$ and $b_{ij} \in R_{ij}$. Then
		$$
		\begin{aligned}
			(i) & ~ g(a_{11}+b_{12} )=g (a_{11} )+g (b_{12} )\\
			(ii)&~  g (a_{22}+b_{21} )=g (a_{22} )+g (b_{21} )\\
			 (iii) & ~  g (a_{11}+b_{21} )=g (a_{11} )+g (b_{21} ) \\
			 (iv)&~  g (a_{22}+b_{12} )=g (a_{22} )+g (b_{12} ) .
		\end{aligned}
		$$
	\end{lemma}
	
	\begin{proof}
		Let $x_{22} \in R_{22}$. Then
		$$
		\begin{aligned}
			&g (a_{11}+b_{12} ) x_{22}+\varphi (a_{11}+b_{12} ) D (x_{22} ) \\
			&=g [ (a_{11}+b_{12} ) x_{22} ]\\
			&=g (b_{12} x_{22} ) \\
			&=g (a_{11} x_{22} )+g (b_{12} x_{22} ) ~(\text{By Lemma \ref{lem4.2}})\\
			&=g (a_{11} ) x_{22}+\varphi (a_{11} ) D (x_{22} )+g (b_{12} ) x_{22}+\varphi (b_{12} ) D (x_{22} ).
		\end{aligned}
		$$
		Comparing both sides,
		$$
		[g (a_{11}+b_{12} )-g (a_{11} )-g (b_{12} ) ] x_{22}=0~(\text{By Theorem \ref{thm2.1}}).
		$$
		Using the assumption \eqref{eqA} on $R$, we get
		\begin{equation}
		\label{eq4.1}
		\begin{aligned}
			& { [g (a_{11}+b_{12} )-g (a_{11} )-g (b_{12} ) ]_{12}=0 } \\
			\text{and} & ~{ [g (a_{11}+b_{12} )-g (a_{11} )-g (b_{12} ) ]_{22}=0 . }
		\end{aligned}
		\end{equation}
		Similarly, for $x_{12} \in R_{12}$, we have
		$$
		\begin{aligned}
			&g (a_{11}+b_{12} ) x_{12}+\varphi (a_{11}+b_{12} ) D (x_{12} )\\
			&=g [ (a_{11}+b_{12} ) x_{12} ]\\
			& =g (a_{11} x_{12} )\\
			&=g (a_{11} x_{12} )+g (b_{12} x_{12} )\\
			&=g (a_{11} ) x_{12}+\varphi (a_{11} ) D (x_{12} )+g (b_{12} ) x_{12}+\varphi (b_{12} ) D (x_{12} ).
		\end{aligned}
		$$
		This yields that,
		\begin{equation}
		\label{eq4.2}
		\begin{aligned}
			& [g (a_{11}+b_{12} )-g (a_{11} )-g (b_{12} ) ] x_{12}=0 \\
			& \implies [g (a_{11}+b_{12} )-g (a_{11} )-g (b_{12} ) ]_{11}=0 \\
			&\text{and}~ [g (a_{11}+b_{12} )-g (a_{11} )-g (b_{12} ) ]_{21}=0.
		\end{aligned}
		\end{equation}
		Hence, by \eqref{eq4.1} and \eqref{eq4.2},
		\begin{align*}
		g (a_{11}+b_{12} )=g (a_{11} )+g (b_{12} ).
		\end{align*}
		Similarly, we can prove that
		\begin{align*}
		g (a_{22}+b_{21} )=g (a_{22} )+g (b_{21} ).
		\end{align*}
		Now,
		\begin{align*}
		&g(a_{11})=g(e_1(a_{11}+b_{21}))=g(e_1)(a_{11}+b_{21})+\varphi(e_1)D(a_{11}+b_{21})\\
		\text{and}~& g(b_{21})=g(e_2(a_{11}+b_{21}))=g(e_2)(a_{11}+b_{21})+\varphi(e_2)D(a_{11}+b_{21}).
		\end{align*}
	Adding these,
	\begin{align*}
	g(a_{11})+g(b_{21})&=(g(e_1)+g(e_2))(a_{11}+b_{21})+(\varphi(e_1)+\varphi(e_2))D(a_{11}+b_{21})\\
	&=g(1)(a_{11}+b_{21})+\varphi(1)D(a_{11}+b_{21})~\text{(By Lemma \ref{lem4.4})}\\
	&=g(1(a_{11}+b_{21}))=g(a_{11}+b_{21}).
\end{align*}		
Similarly, we can prove that
\begin{align*}
g (a_{22}+b_{12} )=g (a_{22} )+g (b_{12} ) .
\end{align*}
	\end{proof}
	
	\begin{lemma}
		\label{lem4.6}
		Let $a_{ij}, b_{ij}, c_{ij} \in R_{ij}$. Then
		$$
		\begin{aligned}
			(i) ~ g (a_{12}+b_{12} c_{22} )=g (a_{12} )+g (b_{12} c_{22} )\\
			(ii) ~ g (a_{21}+b_{22} c_{21} )=g (a_{21} )+g (b_{22} c_{21} ).
		\end{aligned}
		$$
	\end{lemma}
	
	\begin{proof}
		Note that,
		\begin{align*}
	a_{12}+b_{12} c_{22}= (e_1+b_{12} ) (a_{12}+c_{22} ).	
\end{align*}		
		Therefore,
		$$
		\begin{aligned}
			 g(a_{12}+b_{12} c_{22})
			 =&g( (e_1+b_{12} ) (a_{12}+c_{22} ) ) \\
			=& g (e_1+b_{12} ) (a_{12}+c_{22} )+\varphi (e_1+b_{12} ) D (a_{12}+c_{22} ) \\
			=& (g (e_1) + g(b_{12} )) (a_{12}+c_{22} )+ (\varphi (e_1 )+\varphi (b_{12} )) (D (a_{12} )+D (c_{22} ) ) \\
			&\text{(By Lemma \ref{lem4.5})}\\
			=& g (e_1 ) a_{12}+\varphi (e_1 ) D (a_{12} )+g (e_1 ) c_{22}+\varphi (e_1 ) D (c_{22} ) \\
			& +g (b_{12} ) a_{12}+\varphi (b_{12} ) D (a_{12} )+g (b_{12} ) c_{22}+\varphi (b_{12} ) D (c_{22} ) \\
			=&  g(e_1 a_{12}) +g(e_1 c_{22})+ g (b_{12} a_{12} )+g (b_{12} c_{22} ) \\
			=& g (a_{12}) +g (b_{12} c_{22} )~\text{(By Lemma \ref{lem4.2})}.
		\end{aligned}
		$$
		Similarly, we can prove that
		\begin{align*}
		g (a_{21}+b_{22} c_{21} )=g (a_{21} )+g (b_{22} c_{21} ).
		\end{align*}
	\end{proof}
	
	\begin{lemma}
		\label{lem4.7}
		Let $a_{ij}, b_{ij} \in R_{ij}$. Then
		$$
		\begin{aligned}
			 (i)& ~ g (a_{12}+b_{12} )=g (a_{12} )+g (b_{12} )\\
			 (ii)&~ g (a_{21}+b_{21} )=g (a_{21} )+g (b_{21} ).
		\end{aligned}
		$$
	\end{lemma}
	
	\begin{proof}
		Let  $x_{22} \in R_{22}$. Then
		$$
		\begin{aligned}
			& g (a_{12}+b_{12} ) x_{22}+\varphi (a_{12}+b_{12} ) D (x_{22} ) \\
			&=g [ (a_{12}+b_{12} ) x_{22} ]\\
			&= g (a_{12} x_{22}+b_{12} x_{22} ) \\
			&= g (a_{12} x_{22} )+g (b_{12} x_{22} ) ~(\text{By Lemma \ref{lem4.6}})\\
			&=g (a_{12} ) x_{22}+\varphi (a_{12} ) D (x_{22} )+g (b_{12} ) x_{22}+\varphi (b_{12} ) D (x_{22} ).
		\end{aligned}
		$$
		Comparing both sides, we get
		\begin{equation}
		\label{eq4.3}
		\begin{aligned}
			& { [g (a_{12}+b_{12} )-g (a_{12} )-g (b_{12} ) ] x_{22}=0 } \\
			\implies &  { [g (a_{12}+b_{12} )-g (a_{12} )-g (b_{12} ) ]_{12}=0 } \\
			& \text{and} ~ [g (a_{12}+b_{12} )-g (a_{12} )-g (b_{12} ) ]_{22}=0 .
		\end{aligned}
		\end{equation}
		Let $x_{12} \in R_{12}$. Then
		$$
		\begin{aligned}
			& g (a_{12}+b_{12} ) x_{12}+\varphi (a_{12}+b_{12} ) D (x_{12} ) \\
			&=g [ (a_{12}+b_{12} ) x_{12} ]\\
			&= g(0)=0 \\
			&=g (a_{12} x_{12} )+ g(b_{12}x_{12} ) \\
			&= g (a_{12} ) x_{12}+\varphi (a_{12} ) D (x_{12} )+g (b_{12} ) x_{12}+\varphi (b_{12} ) D (x_{12} ),
		\end{aligned}
		$$
		which yields,
		\begin{equation}
		\label{eq4.4}
		\begin{aligned}
			& { [g (a_{12}+b_{12} )-g (a_{12} )-g (b_{12} ) ] x_{12}=0 } \\
			\implies & { [g (a_{12}+b_{12} )-g (a_{12} )-g (b_{12} ) ]_{11}=0 } \\
			&\text{and} ~ [g (a_{12}+b_{12} )-g (a_{12} )-g (b_{12} ) ]_{21}=0 .
		\end{aligned}
		\end{equation}
	Hence, by \eqref{eq4.3} and \eqref{eq4.4},
\begin{align*}
g (a_{12}+b_{12} ) = g (a_{12} ) + g (b_{12} ).
\end{align*}	
	Similarly, we can prove that
	\begin{align*}
g (a_{21}+b_{21} ) = g (a_{21} ) + g (b_{21} ).
\end{align*}	
		\end{proof}

	\begin{lemma}
	\label{lem4.8}
		Let $a_{ii}, b_{ii} \in R_{ii}$. Then
		$$
		\begin{aligned}
			& (i)~ g (a_{11}+b_{11} )=g (a_{11} )+g (b_{11} )\\
			& (ii)~ g (a_{22}+b_{22} )=g (a_{22} )+g (b_{22} ).
		\end{aligned}
		$$
	\end{lemma}
	
	\begin{proof}
	Let $x_{22} \in R_{22}$. Then
	\begin{align*}
	 & g (a_{11}+b_{11} ) x_{22}+\varphi (a_{11}+b_{11} ) D (x_{22} )\\
	 &=g [ (a_{11}+b_{11} ) x_{22} ]\\
	 &=g [ (a_{11}+b_{11} ) x_{22} ]\\
	 &=g(0)=0\\
	 &=g (a_{11} x_{22} )+g (b_{11} x_{22} )\\
	 &=g (a_{11} ) x_{22}+\varphi (a_{11} ) D (x_{22} )+g (b_{11} ) x_{22}+\varphi (b_{11} ) D (x_{22} ).
	\end{align*}
		Comparing both sides,
		\begin{equation}
		\label{eq4.5}
		\begin{aligned}
			& { [g (a_{11}+b_{11} )-g (a_{11} )-g (b_{11}) ] x_{22}=0 } \\
			\implies & { [g (a_{11}+b_{11} )-g (a_{11} )-g (b_{11} ) ]_{12}=0 } \\
			& \text{and} ~ [g (a_{11}+b_{11} )-g (a_{11} )-g (b_{11} ) ]_{22}=0
		\end{aligned}
		\end{equation}
		Similarly, by taking $x_{12} \in R_{12}$,
		\begin{equation}
		\label{eq4.6}
		\begin{aligned}
			& { [g (a_{11}+b_{11} )-g (a_{11} )-g (b_{11} ) ]_{11}=0 } \\
			& \text{and} ~ [g (a_{11}+b_{11} )-g (a_{11} )-g (b_{11} ) ]_{21}=0.
		\end{aligned}
		\end{equation}
		Hence, by \eqref{eq4.5} and \eqref{eq4.6},
		\begin{align*}
		g (a_{11}+b_{11} )=g (a_{11} )+g (b_{11} ).
\end{align*}		
		Similarly, we can prove that
		\begin{align*}
		g (a_{22}+b_{22} )=g (a_{22} )+g (b_{22} ).
\end{align*}		
	\end{proof}
	
	
	
	\begin{lemma}
	\label{lem4.9}
		Let $a_{11} \in R_{11}$, $b_{12} \in R_{12}$, $c_{21} \in R_{21}$ and $d_{22} \in R_{22}$. Then
		$$
		\begin{aligned}
			g (a_{11}+b_{12}+c_{21}+d_{22} )= g (a_{11} )+g (b_{12} ) +g (c_{21} )+g (d_{22} ).
		\end{aligned}
		$$
	\end{lemma}
	
	\begin{proof} Let $x_{11} \in R_{11}$. Then
		$$
		\begin{aligned}
			& g (a_{11}+b_{12}+c_{21}+d_{22} ) x_{11} +\varphi (a_{11}+b_{12}+c_{21}+d_{22} ) D (x_{11} ) \\
			&=g [ (a_{11}+b_{12}+c_{21}+d_{22} ) x_{11} ]\\
			&=g (a_{11} x_{11}+c_{21} x_{11} ) \\
			&=g (a_{11} x_{11} )+g (b_{12} x_{11} )+g (c_{21} x_{11} )+g (d_{22} x_{11} )~(\text{By Lemma \ref{lem4.2} and \ref{lem4.5}}) \\
			&=g (a_{11} ) x_{11}+\varphi (a_{11} ) D (x_{11} )+g (b_{12} ) x_{11}+\varphi (b_{12} ) D (x_{11} )\\
			& +g (c_{21} ) x_{11}+\varphi (c_{21} ) D (x_{11} )+g (d_{22} ) x_{11}+\varphi (d_{22} ) D (x_{11} ).
		\end{aligned}
		$$
		Comparing both sides, we get
		\begin{equation}
		\label{eq4.7}
		\begin{aligned}
			& { [g (a_{11}+b_{12}+c_{21}+d_{22} )-g (a_{11} )-g (b_{12} )-g (c_{21} )-g (d_{22}) ] x_{11}=0 .} \\
			\implies & { [g (a_{11}+b_{12}+c_{21}+d_{22} )-g (a_{11} )-g (b_{12} )-g (c_{21} )-g (d_{22} ) ]_{11}=0 } \\
			\text{and}~ & { [g (a_{11}+b_{12}+c_{21}+d_{22} )-g (a_{11} )-g (b_{12} )-g (c_{21} )-g (d_{22} ) ]_{21}=0 }. \\
		\end{aligned}
		\end{equation}
		Similarly, by taking $x_{22}\in R_{22}$,
		\begin{equation}
		\label{eq4.8}
		\begin{aligned}
			& { [g (a_{11}+b_{12}+c_{21}+d_{22} )-g (a_{11} )-g (b_{12} )-g (c_{21} )-g (d_{22} ) ]_{12}=0 } \\
			\text{and} ~& { [g (a_{11}+b_{12}+c_{21}+d_{22} )-g (a_{11} )-g (b_{12} )-g (c_{21} )-g (d_{22} ) ]_{22}=0 }. \\
		\end{aligned}
		\end{equation}
		Hence, by \eqref{eq4.7} and \eqref{eq4.8},
		\begin{align*}
		g (a_{11}+b_{12}+c_{21}+d_{22} )= g (a_{11} )+g (b_{12} ) +g (c_{21} )+g (d_{22} ).
		\end{align*}
	\end{proof}
	
\begin{proof}[Proof of Theorem \ref{thm4.1}]
		Let $a, b \in R$. Then
		\begin{align*}
		& a=a_{11}+a_{12}+a_{21}+a_{22},\\
		& b=b_{11}+b_{12}+b_{21}+b_{22},~\text{for some}~a_{ij},~b_{ij}\in R_{ij}.
		\end{align*}
		Now,
		$$
		\begin{aligned}
			&g(a+b)=g (a_{11}+a_{12}+a_{21}+a_{22}+b_{11}+b_{12}+b_{21}+b_{22} )\\
			&=g [ (a_{11}+b_{11} )+ (a_{12}+b_{12} )+ (a_{21}+b_{21} )+ (a_{22}+b_{22} ) ]\\
			&=g (a_{11}+b_{11} )+g (a_{12}+b_{12} )+g (a_{21}+b_{21} )+g (a_{22}+b_{22} )~\text{(By Lemma \ref{lem4.9})}\\
			&=g (a_{11} )+g(b_ {11})+g (a_{12} )+g (b_{12} )+g (a_{21} )+g (b_{21} ) +g (a_{22} )+g (b_{22} )\\
			& ~\text{(By Lemma \ref{lem4.7} and \ref{lem4.8})}\\
			&=g (a_{11}+a_{12}+a_{21}+a_{22} )+g (b_{11}+b_{12}+b_{21}+b_{22} )~\text{(By Lemma \ref{lem4.9})}\\
			 &=g(a)+g(b).
		\end{aligned}
		$$
		Hence, $g$ is additive.
	\end{proof}
	
\begin{corollary}
\label{cor4.10}
	Let $R$ be a $2$-torsion free semi-prime ring with a non-trivial idempotent $e$ and satisfies the condition \eqref{eqB} given in Corollary \ref{cor3.8}. Then every multiplicative generalized skew derivation $g$ over $R$ is additive. Moreover, $g$ is a generalized skew derivation on $R$.
\end{corollary}

\begin{proof}
We have the desired result by Lemma $1.5$ in \cite{jing2012additivity} and Theorem \ref{thm4.1}.
\end{proof}

	
	\section*{Acknowledgement}
	The first author is thankful to the Indian Institute of Technology Patna for providing the research facilities.
	
	
	
	