An element *a* of an AG-groupoid *S* is called a left (*right*) regular element of *S* if there exists some *x*∈*S* such that *a*=*a*^{2}*x* (*a*=*x**a*^{2}) and *S* is called left (*r**i**g**h**t*) regular if every element of *S* is left (*r**i**g**h**t*) regular.

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**Remark 4**

Let *S* be *an AG-groupoid with left identity*. Then, the concepts of left and right regularity coincide in *S*.

*Indeed*,*for every* *a*∈*S**there exist some* *x,y*∈*S**such that* *a*=*x**a*^{2}=*a*^{2}*y*. *As* *a*=*x**a*^{2}=*e**x*·*a**a*=*a**a*·*x**e*=*a*^{2}*y,and* *a*=*a*^{2}*y*=*x**a*^{2}*also holds in a similar way.*

Let us give an example of an AG-groupoid which will be used for the converse parts of various problems in this section.

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**Example 2**

Let us consider an AG-groupoid *S*={1,2,3,4,5} with left identity 4 defined in the following multiplication table.

*It is easy to check that*
*S*
*is non-commutative and non-associative.*

An AG-groupoid *S* is called left (right) duo if every left (right) ideal of *S* is a two-sided ideal of *S* and is called duo if it is both left and right duo. Similarly an AG-groupoid *S* is called an SI-left (SI-right) duo if every SI-left-ideal (SI-right-ideal) of *S* over *U* is an SI-two-sided-ideal of *S* over *U*, and *S* is called an SI-duo if it is both an SI-left and an SI-right duo.

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**Lemma 8**

If every SI-left-ideal of an AG-groupoid *S* with left identity over *U* is anSI-interior-ideal of *S* over *U*, then *S* is left duo.

###
*Proof*

Let *I* be any left ideal of *S* with left identity. Now by Lemma 1, the identity soft mapping *X*_{I} is an SI-left-ideal of *S* over *U*. Thus, by hypothesis, *X*_{I} is anSI-interior-ideal of *S* over *U*, and by using Lemma 1 again, *I* is an interior ideal of *S*. Thus *I**S*=*I*·*S**S*=*S*·*I**S*=*S**S*·*I**S*=*S**I*·*S**S*=*S**I*·*S*⊆*I*. This shows that *S* is left duo. □

The converse part of Lemma 8 is not true in general. Let us consider an AG-groupoid *S* (from Example 2). It is easy to see that *S* is left duo because the only left ideal of *S* is {1,5} which is also a right ideal of *S*. Let *A*=*S* and define a soft set *f*_{A} of *S* over *U*={*p*_{1},*p*_{2},*p*_{3},*p*_{4,}*p*_{5},*p*_{6}} as follows:

\(f_{A}(x)=\left \{ \begin {array}{c} U\ \text {if}\ x=1 \\ \{p_{1},p_{2},p_{3},p_{4}\}\ \text {if}x\ =2 \\ \{p_{2},p_{3},p_{4,},p_{5}\}\ \text {if}\ x=3 \\ \{p_{3},p_{4,},p_{5}\}\ \text {if}\ x=4 \\ \{p_{1},p_{2},p_{3},p_{4,}p_{5}\}\ \text {if}\ x=5 \end {array} \right \}.\)

Then, it is easy to see that *f*_{A} is anSI -left-ideal of *S* over *U* but it is not anSI -interior-ideal of *S* over *U* because \(f_{A}(42\ast 4)\varsupsetneq f_{A}(2).\)

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**Corollary 2**

Every interior ideal of an AG-groupoid *S* with left identity is a right ideal of *S*.

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**Theorem 1**

Every SI-right-ideal of an AG-groupoid *S* with left identity is anSI-interior-ideal of *S* over *U* if and only if *S* is right duo.

###
*Proof*

It is simple. □

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**Theorem 2**

Let *S* be a right regular AG-groupoid with left identity. Then, *S* is left duo if and only if every SI-left-ideal of *S* over *U* is anSI-interior-ideal of *S* over *U*.

###
*Proof*

Necessity. Let a right regular *S* with left identity be a left duo, and assume that *f*_{A} is any SI-left-ideal of *S* over *U*. Let *a,b*,*c*∈*S*, then *b*≤*x**b*^{2} for some *x*∈*S*. Since *Sa* is a left ideal of *S*, therefore by hypothesis, *Sa* is a two-sided ideal of *S*. Thus, *ab*·*c*=*a*(*x*·*b**b*)·*c*=*a*(*b*·*x**b*)·*c*=*b*(*a*·*x**b*)·*c*=*c*(*a*·*x**b*)·*b*. It follows that *ab*·*c*∈*S*(*a*·*S**S*)·*b*⊆(*S*·*a**S*)*b*=(*S**S*·*a**S*)*b*=(*S**a*·*S**S*)*b*⊆(*S**a*·*S*)*b*⊆*S**a*·*b*. Thus, *ab*·*c*=*t**a*·*b* for some *t*∈*S*, and therefore *f*_{A}(*ab*·*c*)=*f*_{A}(*t**a*·*b*)⊇*f*_{A}(*b*), implies that *f*_{A} is anSI-interior-ideal of *S* over *U*.

Sufficiency. It can be followed from Lemma 8. □

By left-right dual of above Theorem, we have the following Theorem :

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**Theorem 3**

Let *S* be a right regular AG-groupoid with left identity. Then,*S* is right duo if and only if every SI-right-ideal of *S* over *U* is anSI-interior-ideal of *S* over *U*.

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**Lemma 9**

A non-empty subset *A* of a right regular AG-groupoid *S* with left identity is a two-sided ideal of *S* if and only if it is an interior ideal of *S*.

###
*Proof*

It is simple. □

###
**Lemma 10**

Every left ideal of an AG-groupoid *S* with left identity is an interior ideal of *S* if *S* is an SI-left duo.

###
*Proof*

It can be followed from Lemmas 1 and 9. □

The converse of Lemma 10 is not true in general. The only left ideal of *S* (from Example 2) is {1,2} which is also an interior ideal of *S*. Let *A*={2,3,4,5} and define a soft set *f*_{A} of *S* over \( U= \mathbb {Z} \) as follows:

\(f_{A}(x)=\left \{ \begin {array}{c} 4 \mathbb {Z} \ \text {if}\ x=2 \\ 8 \mathbb {Z} \ \text {if}\ x=3 \\ 16 \mathbb {Z} \ \text {if}\ x=4 \\ 2 \mathbb {Z} \ \text {if}\ x=5 \end {array} \right \} \).

Then, it is easy to see that *f*_{A} is anSI -left-ideal of *S* over *U* but it is not anSI -right-ideal of *S* over *U* because \(f_{A}(2\ast 4)\varsupsetneq f_{A}(2).\)

It is easy to see that every SI-right-ideal of *S* with left identity over *U* is anSI-left-ideal of *S* over *U*.

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**Remark 5**

Every SI-right-ideal of an AG-groupoid *S* with left identity is anSI-left-ideal of *S* over *U*, but the converse is not true in general.

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**Theorem 4**

Every right ideal of an AG-groupoid *S* with left identity is an interior ideal of *S* if and only if *S* is an SI-right duo.

###
*Proof*

It is straightforward. □

###
**Theorem 5**

Let *S* be a right regular AG-groupoid with left identity. Then,*S* is an SI-left duo if and only if every left ideal of *S* is an interior ideal of *S*.

###
*Proof*

The direct part can be followed from Lemma 10. The converse is simple. □

By left-right dual of above Theorem, we have the following Theorem.

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**Theorem 6**

Let *S* be a right regular AG-groupoid with left identity. Then,*S* is an SI-right duo if and only if every right ideal of *S* is an interior ideal of *S*.

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**Theorem 7**

Let *S* be an AG-groupoid with left identity and *E*={*x*∈*S*:*x*=*x*^{2}}⊆*S*. Then the following assertions hold :

(*i*)*E**forms a semilattice* ;

(*i**i*)*E**is a singleton set*,*if* *a*=*a**x*·*a*,∀*a,x*∈*S*.

###
*Proof*

It is simple. □

###
**Theorem 8**

For an AG-groupoid *S* with left identity, the following conditions are equivalent :

(*i*)*S**is right regular* ;

(*i**i*)*For any interior ideal**I**of* *S*;

(*a*) *I*⊆*I*^{2},

(*b*)*I**is semiprime*.

(*i**i**i*)*For any SI-interior-ideal* *f*_{A}*of**S**over* *U*;

(*a*)\(f_{A}\overset {\sim }{\subseteq }f_{A}\circ f_{A},\)

(*b*) *f*_{A}*is soft semiprime over* *U*.

(*i**v*)*S**is right regular and* |*E*|=1, (*a*=*a**x*·*a*, ∀ *a,x*∈*E*);

(*v*)*S**is right regular and* *∅*≠*E*⊆*S**is semilattice*.

###
*Proof*

(*i*)⇒(*v*)⇒(*i**v*) can be followed from Theorem 7.

(*i**v*)⇒(*i**i**i*):(*a*). Let *f*_{A} be any SI -interior-ideal of a right regular *S* with left identity. Thus, for each *a*∈*S*, there exists some *x*∈*S* such that *a*=*x*·*a**a*=*a*·*x**a*=*a*·*x*(*x*·*a**a*)=*a*·(*e**x*)(*a*·*x**a*)=*a*·(*x**a*·*a*)(*x**e*). Therefore,

$$\begin{array}{@{}rcl@{}} (f_{A}\circ f_{A})(a) &=&\bigcup \limits_{a=a\cdot (xa\cdot a)(xe)}\left \{ f_{A}(a)\cap f_{A}((xa\cdot a)(xe))\right \} \\ &\supseteq &f_{A}(a)\cap f_{A}((xa\cdot a)(xe))\supseteq f_{A}(a)\cap f_{A}(a)=f_{A}(a)\text{.} \end{array} $$

This shows that \(f_{A}\overset {\sim }{\subseteq }f_{A}\circ f_{A}.\)

(*b*). Also,

$$\begin{array}{@{}rcl@{}} a &=&x\cdot aa\leq ex\cdot aa=aa\cdot xe=(a\cdot xa^{2})(xe)=(x\cdot aa^{2})(xe)=x(ea\cdot aa)\cdot (xe) \\ &=&x(aa\cdot ae)\cdot (xe)=(aa)(x\cdot ae)\cdot (xe)=(ae\cdot x)(aa)\cdot (xe)=(ae\cdot x)a^{2}\cdot (xe). \end{array} $$

This implies that *f*_{A}(*a*)=*f*_{A}((*a**e*·*x*)*a*^{2}·(*x**e*))⊇*f*_{A}(*a*^{2}). Hence, *f*_{A} is *soft*semiprime.

(*i**i**i*)⇒(*i**i*):(*a*). Assume that *I* is any interior ideal of *S*, then by using Lemma 1, *X*_{I} is an SI-interior-ideal of *S* over *U*. Let *i*∈*I*, then by using Lemma 2, we have *X*_{I}(*i*)⊆(*X*_{I}∘*X*_{I})(*i*)=(*X*_{I})(*i*)=*U*. Hence, *I*⊆*I*^{2}.

(*b*). Let *i*^{2}∈*I*. Then, by given assumption, we have *X*_{I}(*i*)⊇*X*_{I}(*i*^{2})=*U*. This implies that *i*∈*I*, and therefore *I* is semiprime.

(*i**i*)⇒(*i*): Let *a*∈*S* with left identity. Since *S**a*^{2} is an interior ideals of *S*, and clearly *a*^{2}∈*S**a*^{2}. Thus, by using given assumption, *a*∈*S**a*^{2}. Hence, *S* is right regular. □

###
**Corollary 3**

Every SI-interior-ideal of a right regular AG-groupoid *S* with left identity is *soft*semiprime over *U*.

###
*Proof*

Let *f*_{I} be any SI-interior-ideal of a right regular *S* with left identity. Then, for each *a*∈*S*, there exists some *x*∈*S* such that *f*_{I}(*a*)=*f*_{I}(*x*·*a**a*)=*f*_{I}(*a*·*x**a*)=*f*_{I}(*x**a*^{2}·*x**a*)⊇*f*_{I}(*a*^{2}). □

###
**Corollary 4**

Let *I* be an interior ideal of an AG-groupoid *S*. Then, *I* is semiprime if and only if *X*_{I} is *soft*semiprime over *U*.

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**Theorem 9**

Let *S* be an AG-groupoid with left identity. Then, *S* is right regular if and only if every SI-interior-ideal *f*_{A} of *S* over *U* is *soft*idempotent and *soft*semiprime.

###
*Proof*

Necessity : Let *f*_{A} be any SI-interior-ideal of a right regular *S* with left identity over *U*. Then, clearly \(f_{A}\circ f_{A} \overset {\sim }{\subseteq }f_{A}\). Now for each *a*∈*S*, there exists some *x*∈*S* such that *a*=*x*·*a**a*=*a*·*x**a*=*e**a*·*x**a*=*a**x*·*a**e*=(*a**e*·*x*)*a*. Thus,

$$\begin{array}{@{}rcl@{}} (f_{A}\circ f_{A})(a) &=&\bigcup \limits_{a=(ae\cdot x)a}\left \{ f_{A}(ae\cdot x)\cap f_{A}(a)\right \} \supseteq f_{A}(ae\cdot x)\cap f_{A}(a) \\ &\supseteq &f_{A}(a)\cap f_{A}(a)=(f_{A}\cap f_{A})(a)=f_{A}(a)\text{.} \end{array} $$

This shows that *f*_{A} is *soft*idempotent over *U*. Again *a*=*e**x*·*a**a*=*a**a*·*x**e*=*a*^{2}·*x**e*. Therefore, *f*_{A}(*a*)=*f*_{A}(*a*^{2}·*x**e*)⊇*f*_{A}(*a*^{2}). Hence, *f*_{A} is *soft*semiprime over *U*.

Sufficiency : Since *S**a*^{2} is an interior ideal of *S*, therefore by Lemma 1, its soft characteristic function \(\phantom {\dot {i}\!}X_{Sa^{2}}\) is an SI-interior-ideal of *S* over *U* such that \(\phantom {\dot {i}\!}X_{Sa^{2}}\) is *soft*idempotent over *U*. Since by given assumption, \(\phantom {\dot {i}\!}X_{Sa^{2}}\) is *soft*semiprime over *U* so by Corollary 4,*S**a*^{2} is semiprime. Since *a*^{2}∈*S**a*^{2}, therefore, *a*∈*a*^{2}*S*. Thus, by using Lemma 2, we have \(\phantom {\dot {i}\!}X_{Sa^{2}}\circ X_{Sa^{2}}=X_{Sa^{2}},\) and \(\phantom {\dot {i}\!}X_{Sa^{2}}\circ X_{Sa^{2}}=X_{(Sa^{2}\cdot Sa^{2})}.\) Thus, we get \(\phantom {\dot {i}\!}X_{(Sa^{2}\cdot Sa^{2})}=X_{Sa^{2}}.\) This implies that \(\phantom {\dot {i}\!}X_{(Sa^{2}\cdot Sa^{2})}(a)=X_{Sa^{2}}(a)=U.\) Therefore, *a*∈*S**a*^{2}·*S**a*^{2}=*a*^{2}*S*·*S**a*^{2}=(*S**a*^{2}·*S*)*a*^{2}⊆*S**a*^{2}. Hence,*S* is right regular. □

###
**Lemma 11**

Every SI-interior-ideal of a right regular AG-groupoid *S* with left identity over *U* is *soft*idempotent.

###
*Proof*

Let *f*_{A} be any SI-interior-ideal of a right regular *S* with left identity over *U*. Then, by using Lemma 4, \(f_{A}\circ f_{A}\overset {\sim }{\subseteq }f_{A}\). Since *S* right regular, therefore for every *a*∈*S* there exists some *x*∈*S* such that *a*=*x*·*a**a*=*a*·*x**a*=*x**a*^{2}·*x**a*=*a**x*·*a*^{2}*x*=(*a*^{2}*x*·*x*)*a*=(*x**x*·*a**a*)*a*=(*a**a*·*x*^{2})*a*. Therefore,

$$\begin{array}{@{}rcl@{}} (f_{A}\circ f_{A})(a) &=&\bigcup \limits_{a=(aa\cdot x^{2})a}\left \{ f_{A}(aa\cdot x^{2})\cap f_{A}(a)\right \} \supseteq f_{A}(aa\cdot x^{2})\cap f_{A}(a) \\ &\supseteq &f_{A}(a)\cap f_{A}(a)=(f_{A}\cap f_{A})(a)\text{.} \end{array} $$

Thus, *f*_{A}∘*f*_{A}=*f*_{A}. □

###
**Theorem 10**

Let *S* be an AG-groupoid with left identity and *f*_{A} be any SI-interior-ideal of\(\ S\mathcal {\ }\)over *U*. Then,*S* is right regular if and only if *f*_{A}=(*X*_{S}∘*f*_{A})^{2} and *f*_{A} is *soft*semiprime.

###
*Proof*

Necessity : Let *f*_{A} be any SI-interior-ideal of a right regular *S* with left identity over *U*. Then, by using Lemmas 4 and 2, we have

$$(X_{S}\circ (X_{S}\circ f_{A}))\circ X_{S}=(X_{S}\circ f_{A})\circ X_{S}=(f_{A}\circ X_{S})\circ X_{S}=(X_{S}\circ X_{S})\circ f_{A}=X_{S}\circ f_{A}. $$

This shows that *X*_{S}∘*f*_{A} is anSI -interior-ideal of *S* over *U*. Now by using Lemmas 11 and 4, we have (*X*_{S}∘*f*_{A})^{2}=*X*_{S}∘*f*_{A}=*f*_{A}. It is easy to see that *f*_{A} is *soft*semiprime.

Sufficiency : Let *f*_{A}=(*X*_{S}∘*f*_{A})^{2} holds for any SI-interior-ideal *f*_{A} of *S* over *U*. Then, by given assumption and Lemma 14, we get, \(f_{A}=(X_{S}\circ f_{A})^{2}=f_{A}^{2}.\) Thus, by using Theorem 9, *S* is right regular. □

###
**Corollary 5**

Let *S* be an AG-groupoid with left identity and *f*_{A} be any SI-interior-ideal of\(\ S\mathcal {\ }\)over *U*. Then, *S* is right regular if and only if \(f_{A}=X_{S}\circ f_{A}^{2}\) and *f*_{A} is *soft* semiprime.

###
*Proof*

From above theorem, \(f_{A}=(X_{S}\circ f_{A})^{2}=(X_{S}\circ f_{A})(X_{S}\circ f_{A})=(X_{S}\circ f_{A})\circ f_{A}=(f_{A}\circ f_{A})\circ X_{S}=(f_{A}\circ f_{A})\circ (X_{S}\circ X_{S})=(X_{S}\circ X_{S})\circ (f_{A}\circ f_{A})=X_{S}\circ f_{A}^{2}.\) □

###
**Lemma 12**

Let *S* be an AG-groupoid with left identity and *f*_{A} be any SI-left-ideal *(*SI -right-ideal,SI-two-sided-ideal)of\( \ S\mathcal {\ }\)over *U*. Then, *S* is right regular if and only if *f*_{A} is *soft*idempotent.

###
*Proof*

Necessity : Let *f*_{A} be anSI-left-ideal of a right regular *S* with left identity over *U*. Then, it is easy to see that \( f_{A}\circ f_{A}\overset {\sim }{\subseteq }f_{A}.\) Let *a*∈*S*, then there exists *x*∈*S* such that *a*=*a**a*·*x*=*x**a*·*a*. Thus

$$(f_{A}\circ f_{A})(a)=\bigcup \limits_{a=xa\cdot a}\{f_{A}(xa)\cap f_{A}(a)\} \supseteq f_{A}(a)\cap f_{A}(a)=f_{A}(a), $$

which implies that *f*_{A} is *soft*idempotent.

Sufficiency : Assume that *f*_{A}∘*f*_{A}=*f*_{A} holds for all SI-left-ideal of *S* with a left identity over *U*. Since *S**a* is a left ideal of *S*, therefore by Lemma 1, it follows that *X*_{Sa} is anSI-left-ideal of *S* over *U*. Since *a*∈*S**a*, it follows that (*X*_{Sa})(*a*)=*U*. By hypothesis and Lemma 2, we obtain (*X*_{Sa})∘(*X*_{Sa})=*X*_{Sa} and (*X*_{Sa})∘(*X*_{Sa})=*X*_{Sa·Sa}. Thus, we have (*X*_{Sa·Sa})(*a*)=*X*_{Sa}(*a*)=*U*, which implies that *a*∈*S**a*·*S**a*. Therefore, *a*∈*S**a*·*S**a*=*S*^{2}*a*^{2}=*S**a*^{2}. This shows that *S* is right regular. □

###
**Theorem 11**

Let *S* be an AG-groupoid with left identity and *f*_{A} be any SI-left-ideal *(*SI -right-ideal,SI-two-sided-ideal) of\(\ S \mathcal {\ }\)over *U*. Then, *S* is right regular if and only if *f*_{A}=(*X*_{S}∘*f*_{A})∘(*X*_{S}∘*f*_{A}).

###
*Proof*

Necessity : Let *S* be a right regular *S* with left identity and let *f*_{A} be any SI-left-ideal of *S* over *U*. It is easy to see that *X*_{S}∘*f*_{A} is also anSI -left-ideal of *S* over *U*. By Lemma 12, we obtain \((X_{S}\circ f_{A})\circ (X_{S}\circ f_{A})=(X_{S}\circ f_{A})\overset {\sim }{\subseteq } f_{A}.\) Let *a*∈*S*, then there exists *x*∈*S* such that *a*=*a**a*·*x*=*x**a*·*a*=(*x**a*)(*a**a*·*x*)=(*x**a*)(*x**a*·*a*). Therefore,

$$\begin{array}{@{}rcl@{}} \left((X_{S}\circ f_{A})\circ (X_{S}\circ f_{A})\right) (a) &\supseteq &(X_{S}\circ f_{A})(xa)\cap (X_{S}\circ f_{A})(xa\cdot a) \\ &\supseteq &X_{S}(x)\cap f_{A}(a)\cap X_{S}(xa)\cap f_{A}(a)=f_{A}(a), \end{array} $$

which is what we set out to prove.

Sufficiency : Suppose that *f*_{A}=(*X*_{S}∘*f*_{A})∘(*X*_{S}∘*f*_{A}) holds for all SI-left-ideal *f*_{A} of *S* over *U*. Then \(f_{A}=(X_{S}\circ f_{A})\circ (X_{S}\circ f_{A})\overset {\sim }{ \subseteq }f_{A}\circ f_{A}\overset {\sim }{\subseteq }X_{S}\circ f_{A} \overset {\sim }{\subseteq }f_{A}.\) Thus, by Lemma 12, it follows that *S* is right regular. □

###
**Lemma 13**

Let *f*_{A} be any SI-interior-ideal of a right regular AG-groupoid *S* with left identity\(\mathcal {\ }\)over *U*. Then, *f*_{A}(*a*)=*f*_{A}(*a*^{2}), for all *a*∈*S*.

###
*Proof*

Let *f*_{A} be any SI-interior-ideal of a right regular *S* with left identity over *U*. For *a*∈*S*, there exists some *x* in *S* such that *a*=*e**x*·*a**a*=*a**a*·*x**e*=(*x**e*·*a*)*a*=(*x**e*·*a*)(*e**x*·*a**a*)=(*x**e*·*a*)(*a**a*·*x**e*)=*a**a*·(*x**e*·*a*)(*x**e*)=*e**a*^{2}·(*x**e*·*a*)(*x**e*). Therefore *f*_{A}(*a*)=*f*_{A}(*e**a*^{2}·(*x**e*·*a*)(*x**e*))⊇*f*_{A}(*a*^{2})=*f*_{A}(*a**a*)=*f*_{A}(*a*(*e**x*·*a**a*))=*f*_{A}(*a*(*a**a*·*x**e*))=*f*_{A}((*a**a*)(*a*·*x**e*))=*f*_{A}((*x**e*·*a*)(*a**a*))⊇*f*_{A}(*a*). That is, *f*_{A}(*a*)=*f*_{A}(*a*^{2}),∀*a*∈*S* □

The converse of Lemma 13 is not true in general. Let us consider an AG-groupoid *S* (from Example 2). Let *A*={1,2,4,5} and define a soft set *f*_{A} of *S* over \(U=\left \{ \left [ \begin {array}{cc} 0 & 0 \\ x & x \end {array} \right ] /x\in \mathbb {Z} _{3}\right \} (\)the set of all 2×2 matrices with entries from \( \mathbb {Z} _{3})\) as follows:

\(f_{A}(x)=\left \{ \begin {array}{c} \left \{ \left [ \begin {array}{cc} 0 & 0 \\ 0 & 0 \end {array} \right ],\left [ \begin {array}{cc} 0 & 0 \\ 1 & 1 \end {array} \right ],\left [ \begin {array}{cc} 0 & 0 \\ 2 & 2 \end {array} \right ] \right \}\ \text {if}\ x=1 \\ \left \{ \left [ \begin {array}{cc} 0 & 0 \\ 1 & 1 \end {array} \right ],\left [ \begin {array}{cc} 0 & 0 \\ 2 & 2 \end {array} \right ] \right \}\ \text {if}\ x=2 \\ \left \{ \left [ \begin {array}{cc} 0 & 0 \\ 2 & 2 \end {array} \right ] \right \}\ \text {if}\ x=4 \\ \left \{ \left [ \begin {array}{cc} 0 & 0 \\ 1 & 1 \end {array} \right ],\left [ \begin {array}{cc} 0 & 0 \\ 2 & 2 \end {array} \right ] \right \}\ \text {if}\ x=\mathit {5} \end {array} \right \}.\)

It is easy to see that *f*_{A} is anSI -interior-ideal of *S* such that *f*_{A}(*x*)⊇*f*_{A}(*x*^{2}),∀*x*∈*S* but *S* is not right regular.

On the other hand, it is easy to see that every SI -two-sided-ideal of *S* over *U* is anSI -interior-ideal of *S* over *U*.

###
**Remark 6**

Every SI-two-sided-ideal of a right regular AG-groupoid *S* with left identity\(\mathcal {\ }\)over *U* is an SI-interior-ideal of *S* over *U* but the converse is not true in general.

###
**Theorem 12**

For an AG-groupoid *S* with left identity, the following conditions are equivalent :

(*i*)*S**is right regular* ;

(*i**i*)*Every interior ideal of**S**is semiprime* ;

(*i**i**i*)*Every SI-interior-ideal of**S**over**U**is soft semiprime* ;

(*i**v*)*For every SI-interior-ideal* *f*_{A}*of**S**over* *U*, *f*_{A}(*a*)=*f*_{A}(*a*^{2}), ∀ *a*∈*S*.

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*Proof*

(*i*)⇒(*i**v*) can be followed from Lemma 13.

(*i**v*)⇒(*i**i**i*) and (*i**i**i*)⇒(*i**i*) are obvious.

(*i**i*)⇒(*i*): Since *S**a*^{2} is an interior ideal of *S* with left identity such that *a*^{2}∈*S**a*^{2}, therefore by given assumption, we have *a*∈*S**a*^{2}. Thus, *S* is right regular. □