# Some Study of Semigroups ofh-Bi-Ideals of Semirings

Authors: Rukhshanda Anjum; Fairouz Tchier; Zeeshan Saleem Mufti; Qin Xin; Syed Irfan Ali Shah; Yae Ulrich Gaba
Journal: Computational and Mathematical Methods in Medicine
 (2021)

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## Abstract

Semigroups are generalizations of groups and rings. In the semigroup theory, there are certain kinds of band decompositions which are useful in the study of the structure of semigroups. This research will open up new horizons in the field of mathematics by aiming to use semigroup ofh-bi-ideal of semiring with semilattice additive reduct. With the course of this research, it will prove that subsemigroup, the set of all right h-bi-ideals, and set of all left h-bi-ideals are bands for h-regular semiring. Moreover, it will be demonstrated that if semigroup of all h-bi-ideals BH,∗ is semilattice, then H is h-Clifford. This research will also explore the classification of minimal h-bi-ideal.


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## Body

## 1. Introduction

Primary idea of semigroup and monoid is given by Wallis [1]. Wallis expressed that a set which satisfies the associative law under some binary operations is called semigroup, and a set which is semigroup with identity is called monoid. Several researchers work on semigroup theory like [2, 3]. [4] discusses the actual concept of group and commutative group, as well as other major points. Vandiver [5] discussed semiring and elucidates several main concepts. He believed that an algebraic structure is made up of a nonempty set that can be manipulated using the binary operations (+) and (·). Semirings play a significant role in geometry, but they also play a role in pure mathematics. Several ideas and results relating semiring have been presented by different researchers like [6, 7]. Semirings with additive inverse are studied by Karvellas Goodearl, Petrich, and Reutenauer [8–11]. In Applied Mathematics and Information Sciences, semirings have been established for solving the different problems. Semirings with commutative addition and zero element are also very necessary in theoretical computer science.Ideals of semiring are used in structure theory and have a wide variety of applications like in [12, 13]. In [14], Gan and Jiang investigated the notion of an “ordered semiring containing 0,” as well as several other definitions such as the maximal ideal, ordered ideals, and the minimum ideal of an ordered semiring. In [15], Han and others explored ordered semiring. Ma and Zhan uses the principle of ideal in [16], but he used the h-ideals, a new class of ideals proposed by Iizuka [17]. Zhan and others use this particular class of ideals in their study [18–22] for a variety of purposes specific to their analysis.The concept of regular semirings was introduced by Von Neumann [23] and Bourne [24]. Von Neumann showed that the ring H,· would also be regular if the semigroup is regular. Bourne showed that if ∀t∈H, there exists h1,h2∈H such that t+th1t=th2t, then the semiring is regular.It was in 1952 that Good and Hughes [25] first described the definition of a bi-ideal of semigroup. Lajos suggested the m,n-ideals in [26]. As a generalization of k-clifford semirings, Bhuniya introduced the left k-clifford semiring [27]. New research in semigroups and semirings is advancing day by day in different fields of asymptotics, control theory, biology, and medicine [28–31].We compare a semigroupBH of h-bi-ideal whose additive reduct is a semilattice in this article. As an extension of the inscription of bi-ideals in semigroup, we introduce h-bi-ideal of a semiring. Further, we have shown that the h-bi-ideal semigroup BH elegantly illustrates different sectors of a semiring.This research representsh-regular semiring along their h-ideals. We will define the different subclass of the h-regular semiring by their semigroup of h-bi-ideals. Also, we will show that BH=RHLH for h-regular semiring H. We will characterize a new class of semigroup of h-bi-ideals in h-Clifford semiring. Lastly, we will work on minimal h-ideal along their h-bi-ideals and will prove that BmH is a rectangular band. This article is arranged in such a way that after passing through introduction in Section 1, we will discuss basic definitions in Section 2. We will work on construction of semigroups of h-bi-ideals in h-regular semiring in Section 3, and Section 4 will consist of semigroup of h-bi-ideals in h-Clifford semiring, a discussion on semigroup of minimal h-bi-ideal. Then, we will conclude our research in the last.


## 2. Preliminaries

SemiringH,+,· is an algebra with two binary operation + and · such that both the additive reduct H,+ and the multiplicative reduct H,· are semigroups and are subject to the following distributive laws:
(1)a1a2+a3=a1a2+a1a3,a1+a2a3=a1a3+a2a3,∀ai∈H.LetM is a nonempty subset of semiring H, then M is said to be a band if M is a semigroup, and every element of M is an idempotent one.A band which is commutative is called semilattice. Throughout this, except as otherwise specified,H is a semiring such that the additive reduct is a semilattice, and this class of all such semiring is denoted by HL+.LetL be nonempty subset of semiring H, then L is said to be left ideal of H if l1+l2∈L for all l1,l2∈L, and let a∈H, then ∃l∈L such that al∈L. The right ideals can be define dually.In the case of a nonempty subsetN of semiring H, the h-closure of N is defined by
(2)N¯=j∈H∣j+n1+i=n2+i,forsomen1,n2∈N,i∈H.SinceH is commutative, so n+n=n for all n∈H, N⊆N.¯ Let j∈N¯, then j+n1+i=n2+i, for some n1,n2∈N, i∈H. Also, j+j+n1+i=j+n2+i⇒j+n1+i=j+n2+i⇒n2+i=j+n2+i. Since H,+ is supposed to be idempotent, N¯=j∈H∣j+n+i=n+i;forsome,n∈N,i∈H and N¯¯=N. N is called a h-set if N¯=N.A left idealN of a semiring H is called a left h-ideal if for any i∈H, there exists n1,n2∈N such that i+n1+j=n2+j, for j∈H.A left (resp.right) idealN of semiring H is a left (resp.right) h-ideal if it is h-set. We can say that, if N is the left h-ideal of H, then N¯ is the smallest h-ideal of H containing N. We can also write N¯¯=N¯ for every N⊆H. Further, N⊆M⊆H⇒N¯⊆M¯.The intersection of left (resp. right)h-ideal is again h-ideal (if it is nonempty). There is the smallest left (resp. right) h-ideal that contains n∈H. This is called the left (resp. right) h-ideal generated by n and is denoted by Lhnresp.Rhn.Definition 1.
LetH be a semiring and N be the subsemiring of H, then N is h-bi-ideal of H if NHN⊆N and N=N¯.A nonempty subsetN of H is called a generalized h-bi-ideal of H if NHN⊆N and N=N¯.Letu∈H. We define Bu =x1+x2+x3⋯+xn∈u∪u2∪uHu. Then, Bu be a subsemiring of semiring H. Further, ∀h∈H and x,y∈u∪u2∪uHu. We get xhy∈Hu which refers that BuHBu⊆Bu and thus Bu be a bi-ideal of H.Lemma 2.
LetH be a semiring and let take u from H. Then, the principal h-bi-ideal of H generated by u is given by:
(3)Bhu=v∈H∣v+u+u2+utu+w=u+u2+utu+w,forsomet∈H;w∈H.Proof.
Letx,y∈Bhu. Then, ∃m,n∈H and u∈H such that x+u+u2+umu+w=u+u2+umu+w and y+u+u2+unu+w′=u+u2+unu+w′ for w∈H. Then, we have:
(4)x+u+u2+umu+w=u+u2+umu+w,xy+u+u2+umu+wy=u+u2+umu+wy,xy+u+u2+umu+wy+u+u2+unu+w′=u+u2+umu+wy+u+u2+unu+w′,xy+u+u2+umu+wu+u2+unu+w′=u+u2+umu+wu+u2+unu+w′,xy+u+u2+uzu+w′′=u+u2+uzu+w′′,since H,+ is semillatice, where
(5)z=u+u2+un+mu+u2n+mu2+mu2n+mw′+wnandww′=w′′,and so, xy∈Bhu. Also, x+y∈Bhu. Thus, Bhu is a subsemiring of H. Similarly, xmy∈Bhu for all x,y∈Bhu and m∈H. Thus, Bhu is a bi-ideal of H. Indeed Bhu=Bu¯, the h−closure of Bu. Hence, Bhu is a h-bi-ideal of H. Let u∈B and let B be a h-bi-ideal of semiring H. Let x∈Bhu, then there exists m∈H such that:
(6)x+u+u2+umu+w=u+u2+umu+wforw∈H.
Now,u∈B implies that u+u2+umu+w∈B and so, x∈B. Hence, Bhu⊆B. Thus, Bhu is the least h-bi-ideal of H, which contains u.Theorem 3.
The set of allh-bi-ideals, the set of all left h-ideals, and the set of all right h-ideals shall be each semigroup with respect to the product of subsets of H specified in the usual way: ∗:PH×PH⟶PH by M∗N=MN.¯Proof.
Instead of defining binary operation on these three sets individually, we consider them to be semigroups of all subsets ofH. Acknowledge receipt by PH of all subsets of H. We represent set of all h-bi-ideal by BH, set of all left h-ideal by LH, and set of right h-ideal by RH. We define a binary operation ∗:PH×PH⟶PH by M∗N=MN¯.
AsMN=mn∣m∈M,n∈N; so,
(7)MN¯=i∈H∣i+m1n1+j=m2n2+j,mi∈M,ni∈N;i=1,2,..andj∈H.
We will check thatPH is a semigroup under the usual operation ∗. Also, since L1 and L2 are left h-ideal, then L1L2¯ will become left h-ideal. So, ∗:PH×PH⟶PH induced a binary operation on LH; in the interests of comfort, we signify with the same symbol “∗.” Thus, LH is a semigroup under the defined operation “∗.” Analogously, RH is a semigroup under the same binary operation “∗.” Now, we can show that the same holds for BH.
LetM and N be the h-bi-ideals of semiring H. Let u,v∈M∗N=MN¯, then there exists m∈M and n∈N such that
(8)u+mn+j=mn+j;v+mn+j′=mn+j′;j,j′∈H,then
(9)u+v+mn+mn+j+j′=mn+mn+j+j′,u+v+mn+j′′=mn+j′′&u+mn+j=mn+j,where j+j′=j′′and mn+mn=mn. Since H,+ is semillatice, so
(10)⇒u+mn+jv=mn+jv,⇒uv+mn+jv=mn+jv,⇒uv+mn+jv+mn+j′=mn+jv+mn+j′,⇒uv+mn+jmn+j′=mn+jmn+j′,⇒uv+mnmn+mnj′+jmn+jj′=mnmn+mnj′+jmn+jj′,⇒uv+mnmn+j1=mnmn+mnj+j1,where j1=mnj′+jmn+jj′. ⇒u+v & uv both belong to MN¯. Also, for h∈H, we have uhv+mnhmn+j1h=mnhmn+j1h implies that uhv∈MNHMN¯⊆MN¯⇒MN¯HMN¯⊆MN¯. As MN¯ is h-closure of MN & MN¯=MN, i.e., h-subset of H. Thus, MN¯ be h-bi-ideal of H. This shows that BH is a semigroup.Example 4.
Now, consider the semiringS=0,d,e,f defined by the following tables.
(11)+0def00defddefdeefdeffdef⋅0def00000d0defe0eeff0fef
The onlyh-ideal is S itself. Obviously, this h-ideal is h-bi-ideal which forms a semigroup.


## 3. Semigroup ofh-Bi-Ideals in h-Regular Semirings

SupposeH be a semigroup, then H is said to be a regular semigroup if for each t∈H, there exists u′∈H such that t=tu′t. Bourne [24] has described a semiring S to be regular if u′,v′∈H exists for each a∈H that t+tu′t=tv′t. Adhikari [32] defined k-regularity; let H be a semiring, if H,+ is a semillatice, then t+tu′t=tv′t. Adding tu′t+tv′t to both sides, that t+tu′+v′t=tu′+v′t. We can say that a semiring H∈HL+ is called k-regular if and only if for any t∈H, there is u′∈H such that t+tu′t=tu′t.Definition 5.
A semiringH is called h-regular if ∀t′∈H, then ∃u′,v′∈H such that t′+t′u′t′+w=t′v′t′+w, for w∈H.
LetH be a semiring, if H,+ is a semillatice, then t+tu′t+w=tv′t+w. Adding tu′t+tv′t to both sides, then t+tu′+v′t+w=tu′+v′t+w. We can say that a semiring H∈HL+ is called h-regular if and only if for any t∈H, there is u′∈H such that t+tu′t+w=tu′t+w.Theorem 6.
The following conditions are identical for the semiringH.
(i)
H is h-regular(ii)
For everyh-bi-ideal B of H, B=BHB¯(iii)
For every generalizedh-bi-ideal M of H, M=MSM¯Proof.
iii⇒ii We know that every h-bi-ideal is a generalized h-bi-ideal from definition, so it is obvious.
i⇒iii Assume that H is h-regular and let M be the generalized h-bi-ideal of H. Then, MHM⊆M refers to MHM¯⊆M¯HM¯⊆M¯⊆M. Let u∈M, since H is h-regular, then there exists m′∈H such that u+um′u+w=um′u+w for w∈H. Now, um′u∈MHM implies that u∈MHM¯ and so M⊆MHM¯. Thus, M=MHM¯
ii⇒i Let u′∈H, consider the h-bi-ideal Bhu′. Then, u′∈Bhu′=Bhu′HBhu′¯ implies that there exists n1,n2,n3,n4∈Bhu′, and h1,h2∈H such that
(12)u′+n1h1n2+w=n3h2n4+w,forw∈H,⇒u′+n1+n2+n3+n4h1+h2n1+n2+n3+n4+w=n1+n2+n3+n4h1+h2n1+n2+n3+n4+w,⇒u′+nhn+w=nhn+w,where n=n1+n2+n3+n4∈Bhu′ and h=h1+h2∈H. Hence, ∃h∈H such that
(13)n+u′+u′2+u′hu′+w=u′+u′2+u′hu′+w,then we have
(14)u′+nhn+w=nhn+w⇒u′+n+u′+u′2+u′hu′+w′hn+u′+u′2+u′hu′+w′+w=n+u′+u′2+u′hu′+w′hn+u′+u′2+u′hu′+w′+w⇒u′+u′+u′2+u′hu′+w′hu′+u′2+u′hu′+w′+wu′+u′2+u′hu′+w′hu′+u′2+u′hu′+w′+w⇒u′+u′tu′+w=u′tu′+w,forsomet∈H,where t=u′+u′2+u′h2+u′2h2+h3u′2+h2w′+w′h2, and w′w′=w. Thus, H is h-regular.Lemma 7.
SubsemigroupRH and LH are bands for h-regular semiring H.Proof.
LetRH and LH are two subsemigroups. Let R′∈RH and u∈R′, then ∃m∈H such that u+umu+w=umu+w for w∈H. Now,
um∈R′H⊆R′ implies that u∈R′2=R′R′¯, and so, R′⊆R′2. Also, it is obvious that R′2⊆R′. Thus, it can be written as R′=R′2. Hence, RH is a band.
Similarly,LH is a band that can prove dually.In the following theorem, we will prove that semigroup of allh-bi-ideals BH is a product of RH and LH semigroups. Bear in mind that RHLH=RL∣R′∈RH,L′∈LH.Theorem 8.
Show thatBH=RHLH for h-regular semiring H.Proof.
To prove thatBH=RHLH, we have to show that BH⊆RHLH and RHLH⊆BH. Suppose R′∈RH and L′∈LH and we represent B=R′∗L′=R′L′¯. So, B is h-subsemiring of H. Now, BHB=R′L′¯HR′L′¯⊆R′L′HR′L′¯⊆R′L′¯=B. This shows B∈BH. Hence, RHLH⊆BH.
Now, supposeB∈BH. We represent L′=HBUB¯ and R′=BHUB¯. Then, HL=HHBUB¯⊆HHBUB¯=HB¯⊆HBUB¯=L′ refers that L′ is left h-ideal of H. Now, R′H=BHUB¯H⊆BHUBH¯=BH¯⊆BHUB¯=R′ implies that R′ is right h-ideal of H. Now, R′L′¯=BHUB¯HBUB¯¯⊆BHUBHBUB¯⊆BHBUB2¯⊆B¯=B. Also, B⊆HBUB⊆HBUB¯=L′ and B⊆BHUB⊆BHUB¯=R′. Since H is h-regular, and we know that H is h-regular iffB=BHB¯ for every h-bi-ideal B of H. So, B=BHB¯ implies that B⊆R′HL′¯⊆R′L′¯. Hence, B=R′L′¯ and so BH⊆RHLH. Thus, BH=RHLH. Hence, proved.


## 4. Semigroup ofh-Bi-Ideals in h-Clifford and Left h-Clifford Semiring

In this segment, we are characterizing the semigroup of allh-bi-ideals of the h-Clifford semiring and the left h-Clifford semiring.Definition 9.
LetH be a semiring, then H is called a h-Clifford semiring if
(i)
H be a h-regular semiring, i.e., for every t∈H, then ∃u,v∈H such that t+tut+w=tvt+w; for h∈H(ii)
Ht¯=tH¯ for every t∈HDefinition 10.
An elemente of semiring H is h-idempotent of H if e+e2+w=e2+w for w∈H. EhH is representation of set of all h-idempotents.Theorem 11.
Inh-regular semiring H,+,·, the given following statements are identical
(i)
H be h-Clifford semiring(15)L¯=R¯,(ii)
uH¯=Hu¯ for all u∈H(iii)
All lefth-ideals as well as all right h-ideals are two sided and T1∩T2=T2T1¯ for any two of them(iv)
L∩R=LR¯ for all left h-ideal and for all right h-ideals R(v)
L1∩L2=L1L2¯ for any two left h-ideals L1,L2 of H and R1∩R2=R1R2¯ for any two right h-ideals R1,R2 of HProof.
i⇒ii Suppose H is h-Clifford and we want to show that L¯=R¯. For this, we have to prove that L¯ and R¯ are subsets of each other. For this, let u,v∈L¯ for u,v∈H, then ∃t∈H such that
(16)u+tv+w=tv+w,forw∈H,(17)v+tu+w′=tu+w′,forw′∈H,since H is h-regular, then there is h1,h2∈H such that u+uh1u+w=uh1u+w and v+vh2v+w′=vh2v+w′. Now, u+uh1u+uh1tv+w+w′=uh1u+uh1tv+w+w′; this implies that u+uh1u+uh1tv+w=uh1u+uh1tv+w, where w+w′=w. Hence, by (16), u+uh1tv+w=uh1tv+w. Similarly, v+vh2tu+w′=vh2tu+w′. Since H,+ is a semillatice, these implies that
(18)u+uxu+w=uxu+w,(19)v+vxv+w′=vxv+w′,(20)u+uxv+w=uxv+w,(21)v+vxu+w′=vxu+w′,where x=h1+h2+h1t+h2t. Thus, ux and vx are h-idempotents. Since H is a h-Clifford semiring, there are s1,s2 in H such that u+vs1v+w=vs1v+w and vxu+us2u+w′′=us2u+w′′for w′′∈H. Hence, (20) and (21) refers that u+vs1v+w=vs1v+w and v+us2u+w′=us2u+w′. Therefore, uR¯v. Hence, L¯⊆R¯. Similarly, R¯⊆L¯. Thus, L¯=R¯.
ii⇒iii Suppose that L¯=R¯. We will prove that uH¯=Hu¯ for all u∈H. For this, let v∈uH¯, then there is t1∈H such that v+ut1+w′=ut1+w′ for w′∈H. Since H is h-regular, there is t2∈H for which u+ut2u+w=ut2u+w. Thus, there is t=t1+t2∈H such that v+ut+w′=ut+w′ and u+utu+w=utu+w. This implies that uR¯ut, so utL¯u. Therefore, there is x∈H such that ut+xu+w=xu+w. Now, v+ut+xu+w′=ut+xu+w′ implies that v+xu+w′=xu+w′, that is, v∈Hu¯. Thus, uH¯⊆Hu¯. Similarly, Hu¯⊆uH¯. Therefore, uH¯=Hu¯.
iii⇒iv Lets take u from left h-ideal L, and t∈H, then ut∈uH¯=Hu¯⊆L implies that L is two-sided h-ideal. Similarly, this will be hold for right h-ideal R. Let M and N are two h-ideals of H. Then, MN contain in M∩N. Let u∈M∩N, since H is h-regular, there is h∈H such that u+utu+w=utu+w for w∈H. So, u will be in MN¯, thus M∩N⊆MN¯. Therefore, M∩N⊆MN¯.
iv⇒v This is trivial.
v⇒i Let u∈H and e be h-idempotent in H. Then, from v, we can write Hu¯.eH¯¯=Hu¯∩eH¯. Since H is h-regular, there is t∈H such that ue+uetue+w1=uetue+w1 for w1∈H and so,
(22)ue+uetuetue+w1=uetuetue+w1.
Now,ue+uetuetue+w1=uetuetue+w1 implies that ue∈Hu¯.eH¯¯ and so, ue∈Hu¯. Hence, there is p∈H such that ue+pu+w2=pu+w2. Now, (22) implies that
(23)ue+uetuetue+pu+w2+w1=uetuetue+pu+w2+w1,that is,
(24)ue+uetuetpu+uetuetw2+w1=uetuetpu+uetuetw2+w1.
Thus,ue+us1u+w3=us1u+w3, where s1=etuetp and uetuetw2+w1=w3. Similarly, there is s2∈H such that eu+us2u+w4=us2u+w4. Therefore, ue+usu+w=usu=w and eu+usu+w=usu+w, where s=s1+s2 and w3+w4=w. Hence, H is h-Clifford.
iv⇒vi This is trivial.
vi⇒i Let u∈H and e∈EhH. Since H is h-regular semiring, there is t∈H such that ue+uetue+w5=uetue+w5 for w5∈H, and so,
(25)ue+uetuetue+w5=uetuetue+w5.
Now,ue+uetuetue+w=uetuetue+w implies that ue∈Hu¯.He¯¯=Hu¯∩He¯. Then, ue∈Hu¯ implies that there is p∈H such that ue+pu+w6=pu+w6. Hence, (25) implies that
(26)ue+uetuetue+pu+w6+w5=uetuetue+pu+w6+w5,
that is,(27)ue+uetuetpu+uetuetw6+w5=uetuetpu+uetuetw6+w5.
Thus,ue+us3u+w7=us3u+w7, where s3=etuetp and uetuetw6+w5=w7. Similarly, there is s2∈H such that eu+us4+w8=us4+w8. Therefore, ue+usu+w=usu+w and eu+usu+w=usu+w, where s3+s4=s and w7+w8=w. Hence, H is h-Clifford.Theorem 12.
LetH be h-regular semiring. Then, H is h-Clifford if and only if for all u,v∈H, there is t∈H such that
(28)uv+utu+w1=utu+w1,forw1∈H,vu+utu+w2=utu+w2,forw2∈H.Proof.
LetH be a semiring, and for all u,v∈H, there is t∈H such that
(29)uv+utu+w1=utu+w1,(30)vu+utu+w2=utu+w2.
Letu∈H and m∈uH¯. Then, there is x∈H such that m+ux+w3=ux+w3 for w3∈H. Now, (29) implies that ux+utu+w1=utu+w1 for some t∈H. Therefore, m+utu+w1=utu+w1. Hence, m∈Hu¯. Thus, uH¯⊆Hu¯. Similarly, Hu¯⊆uH¯. Therefore, uH¯=Hu¯. Hence, H is h-Clifford.
Conversely, letH be h-Clifford and u,v∈H. Since H is h-regular, there is t∈H such that
(31)uv+uvtuv+w1=uvtuv+w1,forw∈H.
Now,uv∈uH¯=Hu¯ implies that for some x∈H,uv+xu+w1=xu+w1. Now, (31) implies that
(32)uv+uvtuv+xu+w1+w4=uvtuv+xu+w1+w4,forw4∈H,and this implies that
(33)uv+uvtxu+uvtw1+w4=uvtxu+uvtw1+w4.
Thus,s1=vtx∈H and uvtw1+w4=w5∈H such that uv+us1u+w5=us1u+w5. Similarly, there is s2∈H such that v+us2u+w6=us2+w6, and then, uv+usu+w=usu+w and vu+usu+w=usu+w, where s=s1+s2 and w5+w6=w∈H.Theorem 13.
LetH be h-regular semiring. Then, H is h-Clifford if and only if for all u,v∈H, there is n∈H such that
(34)uv+vnu+w=vnu+w,forw∈H.Proof.
Suppose thatH be h-Clifford and let ∀u,v∈H. Since H is a h-regular, there exist t∈H such that
(35)uv+uvtuv+w1=uvtuv+w1,forw1∈H.
Now,uv∈uH¯=Hu¯, then there is x1∈H, such that uv+x1u+w2=x1u+w2, for w2∈H. Similarly, uv∈Hv¯=vH¯; there is x2∈H, such that uv+vx2+w3=vx2+w3, for w3∈H. Then, uv+vnu+w=vnu+w, for w∈H implies that
(36)uv+uv+vx2+w3tuv+x1u+w2+w=uv+vx2+w3tuv+x1u+w2+w,or
(37)uv+vx2tx1u+w3tw3+w=vx2tx1u+w3tw3+w.
Hence, there isn=x2tx1∈H and w=w3tw3+w∈H such that uv+vnu+w=vnu+w.
Conversely, letu∈H and t∈uH¯. Then, there exists v∈H such that t+uv+w=uv+w; w∈H. By our hypothesis, there is n∈H for which uv+vnu+w=vnu+w. Therefore, t+vnu+w=vnu+w. So, t∈Hu¯. Thus, Hu¯⊆uH¯. Similarly, uH¯⊆Hu¯. Therefore, uH¯=Hu¯. Hence, H is h-Clifford.Theorem 14.
Inh-Clifford semiring H every h-bi-ideal is an h-ideal of H.Proof.
Suppose thatH be h-Clifford semiring and let B∈H and we are to prove that B is h-bi-ideal of H. For this, let m∈B and u∈H. Then, ∃x,y∈H such that
(38)um+mxm+w=mxm+w,forw∈H,mu+mym+w=mym+w.
Thus,um,mu∈BHB¯⊆B¯=B which shows that B is h-ideal of H.. Hence, proved.Theorem 15.
LetH be semiring if semigroup of all h-bi-ideals BH,∗ is semilattice, then prove that H is h-Clifford.Proof.
Assume thatBH is semillatice, then H is h-regular semiring. Let u,v∈H, then uv∈BhuBhv=BhvBhu implies that uv+rs+w=rs+w for some r∈Bhv and s∈Bhu and for w∈H; since H is h-regular, there are x,y∈H such that r+vxv+w′=vxv+w′ and s+uyu+w′′=uyu+w′′ for w′,w′′∈H. Then, we have uv+rs+w=rs+w(39)⇒uv+r+vxv+w′s+uyu+w′′+w=r+vxv+w′s+uyu+w′′+w,⇒uv+vxvuyu+vxvw′+w′uyu+w′w′′=vxvuyu+vxvw′+w′uyu+w′w′′,⇒uv+vzu+w′′′=vzu+w′′′,where z=xvuy and w′′′=vxvw′+w′uyu+w′w′′∈H.
Hence,H is a h-Clifford semiring.Theorem 16.
The set of allh-bi-ideals BH is a left normal band if and only if semiring is a left h-Clifford.Proof.
LetH be h-Clifford semiring. Let L,M,N∈BH and t∈L∗M∗N=LMN¯. Then, for some l∈L,m∈M, and n∈N, we have t+lmn+w=lmn+w for w∈H. Since H is h-regular; there is s∈H such that lmn+lmnslmn+w1=lmnslmn+w1 for w1∈H, which implies that t+lmnslmn+w=lmnslmn+w, from which we get t+lxlmnxnsls1n+w=lxlmnxnsls1n+w as l+lxl+w′=lxl+w′,n+nxn+w′′=nxn+w′′, and mn+s1m+w′′′=s1m+w′′′ for some x,s1∈H and w′,w′′,w′′′∈H. Again, since lm+s2l+w1=s2l+w1 and nsls1+s3n+w2=s3n+w2; for some s2,s3∈H and w1,w2∈H, after rearranging the above expression, we get t+lxs2lnxs3nm+w=lxs2lnxs3nm+w. Thus, t∈LMN¯=L∗M∗N. Therefore, L∗M∗N⊆L∗N∗M; similarly, it can be proved that L∗N∗M is subset of L∗M∗N. Hence, L∗M∗N=L∗N∗M, and so, BH is a left normal band.
Conversely, suppose that set of allh-bi-ideals BH be a left normal band and we are to prove that H is a left h-Clifford. Thus, H is h-regular, we know that semigroup BH is regular under ∗ iff H is h-regular. Using this result, let u,v∈H, then there is t∈H such that uv+uvtuv+w=uvtuv+w which implies that uv∈BhuvtBhuBhv¯=BhuvtBhvBhu¯; since BH is a left normal band, then uv+opq+w′=opq+w′, where o∈Bhuvt,p∈Bhv,q∈Bhu, and w′∈H. Since H is h-regular, there is h∈H such that q+uhu+w′′=uhu+w′′. Now, uv+opq+w′=opq+w′ implies that uv+opuhu+w′′′=opuhu+w′′′, for w′′′∈H. Thus, H is left h-Clifford.


## 5. Semigroup of Minimalh-Bi-Ideals

We will define semiring in terms of minimalh-ideals in this section and will work on semiring of minimal left h-ideals and minimal right h-ideals. Also, we will look at those semirings H∈HL+ that have a minimal leftrighth-ideal, and a minimal h-bi-ideal. BmH stands for the class of all minimal h-bi-ideals. Likewise, we will denote minimal left and minimal right h-ideals with LmH and RmH.Definition 17.
LetH be a semiring and N be the h-bi-ideal of H, then N be a minimal h-bi-ideal if for every nontrivial h-bi-ideal U of H, U⊆N follows that U=N.Theorem 18.
Let∅≠N⊆H is a minimal h-bi-ideal of H if and only if N⊆R1L1¯ and R1L1¯⊆N for minimal right and left h-ideal R1, L1 of H, respectively.Proof.
Let∅≠N⊆H, suppose that N is a minimal h-bi-ideal of H, then R1=NH¯ and L1=HN¯ follows directly right h-ideal and left h-ideal, respectively. Let us take R2⊆R1 is a right h-ideal of H, then R2L¯ will be h-bi-ideal of H, this follows that R2L1¯⊆R1L1¯⊆NH¯HN¯¯⊆NHN¯⊆N¯HN¯⊆N¯=N, by definition of minimality of N; N=R2L1¯. This implies that R1=NH¯=R2L1¯H¯⊆R2H¯⊆R2, and so, R2=R1. Thus, R1=NH¯ will be a minimal right h-ideal of H. Likewise, L1=HN¯ is a minimal left h-ideal of H, since R1L1¯⊆N be a h-bi-ideal and N is a minimal h-b-ideal of H.. Finally, N=R1L1¯. Hence, proved.Theorem 19.
IfBmH is set of all minimal h-bi-ideals, then BmH is a subsemigroup of BH.Proof.
Suppose two minimalh-bi-ideal M1and M2 in BmH. Let G∈BH be such that G⊆M1M2¯ then GM1¯,M2G¯∈BH. Now, GM1¯⊆M1M2M1⊆M1GM2⊆M1. Then, by minimality of M1,we have GM1¯=M1. Similarly, M2G¯=M2.Then, M1M2¯=GM1¯M2G¯¯⊆GM1M2G¯⊆GHG¯⊆G¯=G. Therefore, G=M1M2¯. Hence, the set BmH would be subsemigroup of BH.As the same pattern, the setLmH and RmH are subsemigroup of BH which can justify easily.Theorem 20.
(a) If a nonempty class of all minimal lefth-ideals is LmH and L1∈LH, then L1∗Lm′=Lm′, where Lm′ is in LmH. Hence, LmH be a left ideal of semigroup, i.e., LH,∗, in particular a subsemigroup. Further, LmH,∗ be a right zero band.
(b) If a nonempty class of all minimal righth-ideals is RmH and R1∈RH, then R1∗Rm′=Rm′, where Rm′ is in RmH. Hence, RmH be a right ideal of semigroup, i.e., RH,∗, in particular, a subsemigroup. Further, RmH,∗ be a left zero band.Proof.
(a) For someL1∈LH and Lm′∈LmH, take the left h-ideal L1∗Lm′. Then, L1∗Lm′ implies L1∗Lm′=L1Lm′¯, and this follows L1Lm′¯⊆Lm′¯=Lm′ and by minimality of Lm′ shows L1∗Lm′=Lm′.
(b) This proceeds dually.Theorem 21.
If set of all minimalh-bi-ideal BmH for a semiring H is nonempty, then ∀Mm,Gm∈BmH, and B∈BH; there are Mm∗B∗Gm∈BmH, Mm∗Mm=Mm, and Mm∗B∗Mm=Mm. As a consequence, semigroup’s bi-ideal is the set of all minimal h-bi-ideals BmH,∗, in particular, a subsemigroup. Further, BmH,∗ is rectangular band.Proof.
LetBmH is set of all minimal h-bi-ideal and BH denotes the set of h-bi-ideal, then suppose Mm,Gm∈BmH and B∈BH. Denote the bi-ideal Mm∗B∗Gm by G. Let F∈BH such that F⊆G, then F∗Mm⊆MmBGmMm¯⊆Mm, and minimality of Mm shows that Mm=F∗Mm. Similarly, Gm=Gm∗F; therefore, G=Mm∗B∗Gm=F∗Mm∗B∗Gm∗F=FMmBGmF¯⊆F, and so, G=F. Thus, the h-b-ideal Mm∗B∗Gm∈BmH. Furthermore, BmH be a subsemigroup of BH. As a result, BmH is bi-ideal of BH. For some Mm∈BmH, let us assume the h-bi-ideal Mm∗Mm. Then, Mm∗Mm=MmMm¯⊆Mm¯=Mm and minimality of Mm shows that Mm∗Mm=Mm. Since, also, Mm∗Mm∗Mm is h-bi-ideal, from Mm∗Mm∗Mm=MmBMm¯⊆Mm and minimality of Mm it means that Mm∗B∗Mm=Mm.


## 6. Conclusion

This research representsh-regular semiring along their h-ideals. The different subclasses of the h-regular semiring are described by their h-bi-ideal semigroup. We have shown that BH=RHLH for h-regular semiring H. We have characterized a new class of semigroup of h-bi-ideals in h-Clifford semiring. Lastly, we worked on minimal h-ideal along their h-bi-ideals and proved that BmH is a rectangular band. This research further can be extended with h-quasi ideals and k-ideals. The other idea of its extension is in the field of biology by using in growth-fragmentation-coagulation equation.



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