We define Jordan derivation of an inverse semiring *S* as an additive map *d* : *S* → *S* such that
$$\begin{array}{}d({x}^{2})+d(x){x}^{\prime}+{x}^{\prime}d(x)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in S\end{array}$$(11)
holds. For example, if *R* is a commutative ring and *I*(*R*) is semiring of all two sided ideals of *R* with respect to ordinary addition and product of ideals, and *T* is subsemiring of *I*(*R*) then *S* = {(*r*, *I*) : *r* ∈ *R*, *I* ∈ *T*} is an inverse semiring with respect to ⊕ and ⊙ defined as (*r*_{1}, *I*) ⊕(*r*_{2}, *J*) = (*r*_{1} + *r*_{2}, *I* + *J*) and (*r*_{1}, *I*)⊙(*r*_{2}, *J*) = (*r*_{1}r_{2}, *IJ*) (see [4]).

If we fix *a* = (*r*, {0}) ∈ *S* then *d*(*x*) = [*a*, *x*], ∀*x* ∈ *S* is a Jordan derivation.

#### Lemma 4.1

*If d is a Jordan derivation of S then for all x*, *y* ∈ *S the following statements hold*:

*d*(*xyx*) = *d*(*x*)*yx* + *xd*(*y*)*x* + *xyd*(*x*)

*d*(*xyz* + *zyx*) = *d*(*x*)*yz* + *xd*(*y*)*z* + *xyd*(*z*) + *d*(*z*)*yx* + *zd*(*y*)*x* + *zyd*(*x*).

#### Proof

Linearization of (11) gives
$$\begin{array}{}d(xy+yx)+d(y){x}^{\prime}+d(x){y}^{\prime}+{x}^{\prime}d(y)+{y}^{\prime}d(x)=0\end{array}$$(12)

Replacing *y* by *xy* + *yx* in (12) we get
$$\begin{array}{}d(x(xy+yx)+(xy+yx)x)+d(xy+yx){x}^{\prime}+d(x)(x{y}^{\prime}+y{x}^{\prime})+\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{x}^{\prime}d(xy+yx)+(x{y}^{\prime}+y{x}^{\prime})d(x)=0\end{array}$$(13)

By (12),(13) and Lemma 1.1 we have
$$\begin{array}{}d({x}^{2}y+y{x}^{2})+2d(\mathit{x}\mathit{y}\mathit{x})+2d(x){y}^{\prime}x+2{x}^{\prime}d(y)x+2{x}^{\prime}\mathit{y}\mathit{d}(x)+d(y{)}^{\prime}{x}^{2}+\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}yd(x){x}^{\prime}+d(x)x{y}^{\prime}+{x}^{\prime}d(x)y+{x}^{2}d(y{)}^{\prime}+y{x}^{\prime}d(x)=0\end{array}$$(14)

Replacing *x* by *x*^{2} in (12) we have
$$\begin{array}{}d({x}^{2}y+y{x}^{2})+d(y{)}^{\prime}{x}^{2}+d({x}^{2}){y}^{\prime}+{x}^{2}d(y{)}^{\prime}+{y}^{\prime}d({x}^{2})=0\end{array}$$

In the view of (11) and Lemma 1.1, we can replace *d*(*x*^{2}) by *d*(*x*)*x* + *xd*(*x*) . Thus
$$\begin{array}{}d({x}^{2}y+y{x}^{2})+d(y{)}^{\prime}{x}^{2}+d(x)x{y}^{\prime}+xd(x){y}^{\prime}+{x}^{2}d(y{)}^{\prime}+{y}^{\prime}d(x)x+{y}^{\prime}xd(x)=0\end{array}$$(15)

From (14) and (15) we have
$$\begin{array}{}d(xyx)+d(x){y}^{\prime}x+xd(y){x}^{\prime}+{x}^{\prime}yd(x)=0\end{array}$$(16)
which gives (i). To obtain (ii), linearize (16) we have
$$\begin{array}{}d(xyz+zyx)+d(x)yz+xd(y{)}^{\prime}+x{y}^{\prime}d(z)+d(z)y{x}^{\prime}+zd(y){x}^{\prime}+z{y}^{\prime}d(x)=0\end{array}$$(17)
and hence (ii) follows by Lemma 1.1. This completes the proof. □

If we write *x*^{y} for *d*(*xy*) + *d*(*x*)*y*′ + *x*′*d*(*y*) then by (12) and Lemma 1.1, we have
$$\begin{array}{}{x}^{y}={y}^{{x}^{\prime}}\end{array}$$(18)

Moreover,
$$\begin{array}{}{x}^{y+z}={x}^{y}+{x}^{z}\end{array}$$(19)
holds for all *x*, *y* ∈ *S*.

#### Lemma 4.2

*Let S be a* 2-*torsion free inverse semiring and d* : *S* → *S be a Jordan derivation*. *Then*
$$\begin{array}{}{x}^{y}s[x,y]+[x,y]s{x}^{y}=0.\end{array}$$(20)
∀*x*, *y*, *s* ∈ *S*.

#### Proof

Replacing *y* by *ysy* in (16) we get
$$\begin{array}{}d(xysyx)+d(x)ys{y}^{\prime}x+xd(ysy){x}^{\prime}+xy{s}^{\prime}yd(x)=0\end{array}$$(21)

Replacing *x* by *y* and *y* by *xsx* in (16) gives
$$\begin{array}{}d(yxsxy)+d(y)xs{x}^{\prime}y+yd(xsx){y}^{\prime}+yx{s}^{\prime}xd(y)=0\end{array}$$(22)

Adding (21) and (22), we have
$$\begin{array}{}d(xysyx+yxsxy)+d(x)ys{y}^{\prime}x+xd(ysy){x}^{\prime}+xy{s}^{\prime}yd(x)+d(y)xs{x}^{\prime}y+yd(xsx){y}^{\prime}+yx{s}^{\prime}xd(y)=0\end{array}$$
By (i) of Lemma 4.1, we have
$$\begin{array}{}d(xysyx+yxsxy)+d(x)ys{y}^{\prime}x+xd(y)s{y}^{\prime}x+xyd(s){y}^{\prime}x+xysd(y){x}^{\prime}+xy{s}^{\prime}yd(x)+\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}d(y)xs{x}^{\prime}y+yd(x)s{x}^{\prime}y+yxd(s){x}^{\prime}y+yxsd(x){y}^{\prime}+yx{s}^{\prime}xd(y)=0\end{array}$$(23)

Replacing *x* by *xy*, *y* by *s* and *z* by *yx* in (17) we get
$$\begin{array}{}d(xysyx+yxsxy)+d(xy)s{y}^{\prime}x+xyd(s){y}^{\prime}x+x{y}^{\prime}sd(yx)+d(yx)s{x}^{\prime}y+yxd(s){x}^{\prime}y+y{x}^{\prime}sd(xy)=0\end{array}$$

From this, we have
$$\begin{array}{}d(xysyx+yxsxy)+xyd(s){y}^{\prime}x+yxd(s){x}^{\prime}y=d(xy)syx+xysd(yx)+d(yx)sxy+yxsd(xy)\end{array}$$(24)
(24) (23) and (24) give
$$\begin{array}{}d(xy)syx+xysd(yx)+d(yx)sxy+yxsd(xy)+d(x)ys{y}^{\prime}x+xd(y)s{y}^{\prime}x+xysd(y){x}^{\prime}+\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}xy{s}^{\prime}yd(x)+d(y)xs{x}^{\prime}y+yd(x)s{x}^{\prime}y+yxd(d){x}^{\prime}y+yx{s}^{\prime}xd(y)=0\end{array}$$
which can be written as 0 = *x*^{y}syx + *xysy*^{x} + *yxsx*^{y} + *y*^{x}sxy = *x*^{y}syx + *xysx*^{y′} + *yxsx*^{y} + *x*^{y′}*sxy*.

Hence, *x*^{y}s[*x*, *y*] + [*x*, *y*]*sx*^{y} = 0. □

#### Theorem 4.3

*A Jordan derivation d on* 2-*torsion free inverse semiring S is a derivation*.

#### Proof

To prove the theorem, we have to show that *x*^{y} = 0 ∀*x*, *y* ∈ *S*.

By Lemmas 1.2 and 4.2, we get
$$\begin{array}{}{x}^{y}s[x,y]=0=[x,y]s{x}^{y}\end{array}$$(25)

Replacing *y* by *y* + *z* in *x*^{y}s[*x*, *y*] = 0 and using it again, we have
$$\begin{array}{}{x}^{y}s[x,z]+{x}^{z}s[x,y]=0\end{array}$$(26)
which gives *x*^{y}s[*x*, *z*] = *x*^{z}ś[*x*, *y*]. Thus (*x*^{y}s[*x*, *z*])*t*(*x*^{y}s[*x*, *z*]) = *x*^{z}ś[*x*, *y*]*tx*^{y}s[*x*, *z*] = 0, ∀ *t* ∈ *S*.

Semiprimeness of *S* gives
$$\begin{array}{}{x}^{y}s[x,z]=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}x,y,z,s\in S\end{array}$$(27)

Similarly, [*x*, *z*]*sx*^{y} = 0.

By using the above technique on (27), we get
$$\begin{array}{}{x}^{y}s[w,z]=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}x,y,z,s,w\in S\end{array}$$(28)

Pre multiplying (28) by [*w*, *z*] and post multiplying it by *x*^{y}, we get
$$\begin{array}{}[w,z]{x}^{y}=0,\phantom{\rule{1em}{0ex}}x,y,w,z\in S\end{array}$$(29)

Therefore,
$$\begin{array}{}[{x}^{y},t]r[{x}^{y},t]={x}^{y}tr[{x}^{y},t]+{t}^{\prime}{x}^{y}r[{x}^{y},t]=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}r\in S\end{array}$$
which implies that [*x*^{y}, *t*] = 0 or *x*^{y}t + *t*′*x*^{y} = 0. Adding *tx*^{y} on both sides of the last expression we have *x*^{y}t + (*t* + *t*′)*x*^{y} = *x*^{y}t or *x*^{y}t + *x*^{y}(*t* + *t*′) = *tx*^{y}, ∀ *t* ∈ *S*. Thus *x*^{y} is in the center of *S*.

From (18), we have
$$\begin{array}{}2({x}^{y}{)}^{2}={x}^{y}({x}^{y}+{y}^{{x}^{\prime}})={x}^{y}\{d(xy)+d(yx{)}^{\prime}+[y,d(x)]+[d(y),x]\}\end{array}$$(30)
or
$$\begin{array}{}2({x}^{y}{)}^{2}={x}^{y}\{d[x,y]+[y,d(x)]+[d(y),x]\}\end{array}$$

Using (29) and the fact that *x*^{y} is in center, we get
$$\begin{array}{}2({x}^{y}{)}^{2}={x}^{y}d[x,y]\end{array}$$(31)

Also, *x*^{y}[*x*, *y*] + [*x*, *y*]*x*^{y} = 0 which gives
$$\begin{array}{}d({x}^{y}[x,y]+[x,y]{x}^{y})=0\end{array}$$(32)

Put *x* = *x*^{y} and *y* = [*x*, *y*] in (12) and using (32), we obtain
$$\begin{array}{}d({x}^{y})[x,y]+{x}^{y}d[x,y]+d[x,y]{x}^{y}+[x,y]d({x}^{y})=0\end{array}$$

By (31), we have *d*(*x*^{y})[*x*, *y*] + [*x*, *y*]*d*(*x*^{y}) + 4 (*x*^{y})^{2} = 0. Post multiplying last equation by *x*^{y}, we have 4 (*x*^{y})^{3} = 0. Thus (*x*^{y})^{3} = 0.

Since *x*^{y} is in the center of *S* so for *t* ∈ *S*, (*x*^{y})^{2}*r*(*x*^{y})^{2} = (*x*^{y})^{3}(*x*^{y})*r* = 0, *r* ∈ *S*. Thus (*x*^{y})^{2} = O. This implies that *x*^{y} = 0, ∀*x*, *y* ∈ *S*, as required. □

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