Robust kernel regression function with uncertain scale parameter for high dimensional ergodic data using $ k $-nearest neighbor estimation

Fatimah Alshahrani; Wahiba Bouabsa; Ibrahim M. Almanjahie; Mohammed Kadi Attouch; Fatimah Alshahrani; Wahiba Bouabsa; Ibrahim M. Almanjahie; Mohammed Kadi Attouch

doi:10.3934/math.2023655

AIMS Mathematics

2023, Volume 8, Issue 6: 13000-13023. doi: 10.3934/math.2023655

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Research article

Robust kernel regression function with uncertain scale parameter for high dimensional ergodic data using $k$ -nearest neighbor estimation

1.
Department of Mathematical Sciences, College of Science, Princess Nourah bint Abdulrahman University, Riyadh 11671, Saudi Arabia
2.
Laboratory of Statistics and Stochastic Processes, University of Djillali Liabes BP 89, Sidi Bel Abbes 22000, Algeria
3.
Department of Mathematics, College of Science, King Khalid University, Abha 62223, Saudi Arabia

Received: 09 January 2023 Revised: 17 February 2023 Accepted: 22 February 2023 Published: 03 April 2023
MSC : 62H12, 62G07, 62G35, 62G20

In this paper, we consider a new method dealing with the problem of estimating the scoring function $\gamma_a$ , with a constant $a$ , in functional space and an unknown scale parameter under a nonparametric robust regression model. Based on the $k$ Nearest Neighbors ( $k$ NN) method, the primary objective is to prove the asymptotic normality aspect in the case of a stationary ergodic process of this estimator. We begin by establishing the almost certain convergence of a conditional distribution estimator. Then, we derive the almost certain convergence (with rate) of the conditional median (scale parameter estimator) and the asymptotic normality of the robust regression function, even when the scale parameter is unknown. Finally, the simulation and real-world data results reveal the consistency and superiority of our theoretical analysis in which the performance of the $k$ NN estimator is comparable to that of the well-known kernel estimator, and it outperforms a nonparametric series (spline) estimator when there are irrelevant regressors.

Keywords:

Citation: Fatimah Alshahrani, Wahiba Bouabsa, Ibrahim M. Almanjahie, Mohammed Kadi Attouch. Robust kernel regression function with uncertain scale parameter for high dimensional ergodic data using $k$ -nearest neighbor estimation[J]. AIMS Mathematics, 2023, 8(6): 13000-13023. doi: 10.3934/math.2023655

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Abstract

1. Introduction

Von Neumann gave the idea of regularity in rings in $1935$ ^[1] and showed that if the semigroup, $(S, \cdot)$ is regular, then the ring $(S, +, \cdot)$ is also regular ^[1]. In 1951, Bourne showed if $\forall x\in S$ there exist $a, b\in S$ such that $x+xax = xbx$ , then semiring $(S, +, \cdot)$ is also regular ^[2]. In structure theory of semirings, ideals play a vital role^[3]. In ^[4], Xueling Ma and Jianming Zhan used the concept of $h$ -ideals. They used the basic and main concept of $h$ -ideals to prove many properties and results. Similarly, Jianming Zhan et al., in ^[5] also used $h$ -ideals in their researches. This class of $h$ -ideals has been used in many researches by different researchers. Ideals of semirings used in the structure theory play an important role in many aspects. Some properties of ideals are discussed in ^[6,7,8]. Gan and Jiang ^[9] studied the ordered semirings containing 0. Han and others in ^[10] discussed also the ordered semirings. Iizuka ^[11] introduced a new type of ideals namely $h$ -ideals. In ^[12,13,14] they used $h$ -ideals for many purposes related to their researches.

Main and basic concepts related to ordered semirings are given by Gan and Jiang ^[9]. The authors also derived some ideas related to minimal ideal, maximal ideal, ordered ideal of an ordered semiring and simple ordered semirings. Han, Kim and Neggers ^[10] also worked on semirings by partial ordered set. Munir and Shafiq ^[19] characterized the regular semirings through $m$ -ideals. Satyt Patchakhieo and Bundit Pibalijommee ^[15] gave the basic definition of ordered semirings and left and right ordered ideal of the ordered semirings. They used two definitions in their properties and applications to prove their result.

Keeping in view the different characterizations of the regular semirings by the properties of the $h$ -ideals, we were motivated to characterize the ordered $h$ -regualr semirings by the properties of their ordered $h$ -ideals. For this purpose, this paper represents ordered $h$ -regular semirings along with their ordered $h$ -ideals. In Section 2, we give some basic definitions which will be used in our further course of work. In Section 3, we characterize the ordered $h$ -ideals semirings by their ordered $h$ -ideals. In Section 4, we characterize the ordered $h$ -regular semirings, and in Section 5, the characterization of the ordered $h$ -weakly regular semirings is given. The conclusion of the paper is presented in the final Section 6.

2. Preliminaries

Definition 1. A non-empty set $S$ together with two binary operations $+$ and $\cdot$ satisfying the following properties:

(C $_{\bf{1}}$ ) $(S, +)$ is a semigroup,

(C $_{\bf{2}}$ ) $(S, \cdot)$ is a semigroup,

(C $_{\bf{3}}$ ) Distributive laws hold in S, that is

$\begin{equation*} t_{1}\cdot (t_{2}+t_{3}) = t_{1}\cdot t_{2}+t_{1}\cdot t_{3}, \end{equation*}$

and

$\begin{equation*} (t_{1}+t_{2})\cdot t_{3} = t_{1}\cdot t_{3}+t_{2}\cdot t_{3}\; for\; all \;t_{1},t_{2},t_{3}\in S, \end{equation*}$

is called a semiring, which is denoted by $(S, +, \cdot).$

Definition 2. $(S, +, \cdot)$ is additively commutative iff for all $x_{1}, y_{1}\in S, x_{1}+y_{1} = y_{1}+x_{1}$ . $S$ is multiplicatively commutative iff for all $x_{1}, y_{1}\in S, x_{1}\cdot y_{1} = y_{1}\cdot x_{1}. \; (S, +, \cdot)$ is called a commutative semiring iff it is both additively commutative and multiplicatively commutative. Suppose $(S, +, \cdot)$ is a semiring, if $\forall a\in S; a+0 = a = 0+a$ and $a\cdot 0 = 0 = 0\cdot a$ , then $0\in S$ is called absorbing zero in S.

Definition 3. ^[15] Let $E\neq \emptyset$ and $(S, +, \cdot)$ is a semiring, $E\subseteq S$ , is a left ideal or right ideal if these properties are satisfied:

(I $_{\bf{1}}$ ) $t_{1}+t_{2}\in E$ for all $t_{1}, t_{2}\in E.$

(I $_{\bf{2}}$ ) $SE\subseteq E$ or $ES\subseteq E.$

If $E$ is left ideal and right ideal of S, then E is an ideal of $S$ .

Definition 4. ^[15] Suppose $(S, \leq)$ is a partially ordered set satisfying the following properties:

(T $_{1}$ ) $(S, +, \cdot)$ is semiring,

(T $_{2}$ ) if $x_{1}\leq x_{2}$ , then $x_{1}+e\leq x_{2}+e,$

(T $_{3}$ ) if $x_{1}\leq x_{2}$ , then $x_{1}e\leq x_{2}e$ and $ex_{1}\leq ex_{2}$ ,

for all $x_{1}, x_{2}, e\in S$ , then, $(S, +, \cdot, \leq)$ is an ordered semiring.

Definition 5. ^[15] Suppose $(S, +, \cdot, \leq)$ is an ordered semiring. Let $E\neq \emptyset$ , $F\neq \emptyset$ be subsets of $S$ , then we denote $(E] = \{g\in S|g\leq r$ for some $r\in E\}$ and $EF = \{gh|g\in E, h\in F\}$ .

We can write $(S, +, \cdot, \leq)$ as $S$ .

Definition 6. ^[15] Suppose $S$ is an ordered semiring, $E\neq \emptyset$ and $E\subseteq S$ satisfies the following properties:

(T $_{\bf{1}}$ ) E is left ideal or right ideal of S;

(T $_{\bf{2}}$ ) if $g\leq w$ for some $w\in E$ , then $g\in E.$

Then E is a left ordered ideal or right ordered ideal.

If E is both left ordered ideal and right ordered ideal of S, then E is ordered ideal of S.

Definition 7. Suppose S is an ordered semiring, if $x_{1}\in S$ , there exist $t\in S$ such that $x_{1}\leq x_{1}tx_{1}$ , then $S$ is called a regular ordered semiring.

3. Ordered $h$ -ideals semirings

In this section, we characterize the ordered $h$ -ideals semirings by their ordered $h$ -ideals.

Definition 8. Suppose $E$ is a nonempty subset of an ordered semiring $S$ , then $E$ is a left ordered $h$ -ideal of $S$ if the following properties are satisfied:

(1) $E$ is a left ordered ideal of $S$ ,

(2) if $e+x_{1}+t = x_{2}+t$ for some $x_{1}, x_{2} \in E, t\in E$ , then $e\in E$ .

Similarly, we define the right ordered $h$ -ideal.

If $E$ is both a left ordered $h$ -ideal and a right ordered $h$ -ideal of $S$ , then $E$ is said to be an ordered $h$ -ideal of S.

Definition 9. Suppose $E\neq \emptyset$ , $E\subseteq S$ and $S$ is an ordered semiring, then the h-closure of $E$ , denoted by $\overline{E}$ , is defined by

$\begin{equation*} \overline{E} = \{g\in S, \;there\; exist \;x_{1},x_{2}\in E, g+x_{1}+h\leq x_{2}+h, h\in E\}. \end{equation*}$

Definition 10. Suppose S is an ordered semiring. If for every $x_{1}\in S$ , there exist $e, h, c\in S$ such that $x_{1}+x_{1}ex_{1}+c\leq x_{1}hx_{1}+c, .$ Then S is called $h$ -regular ordered semiring.

Definition 11. Suppose S is an ordered semiring then $x_{1}\in S$ is said to an ordered $h$ -regular if $x_{1}\in \overline{(x_{1}Sx_{1}]}$ . If each element of S is ordered $h$ -regular, then S is said to be an ordered $h$ -regular semiring.

It is easy to see that each $h$ -regular ordered semiring is an ordered $h$ -regular semiring but converse does not hold. We see this by the following example.

Example 1. Suppose $S = \left\{ t_{1}, t_{2}, t_{3}\right\}$ . Define binary operations $\cdot$ and $+$ on $S$ as:

We define order relation $\leq$ on $S$ as follows :

$\begin{equation*} \leq = \left\{ \left( t_{1,}t_{1}\right) ,\left( t_{2,}t_{2}\right) ,\left( t_{3,}t_{3}\right) ,\left( t_{1,}t_{2}\right) ,\left( t_{1,}t_{3}\right) ,\left( t_{2,}t_{3}\right) \right\}. \end{equation*}$

Then $\left(S, +, \cdot, \leq \right)$ is an ordered semiring. Furthermore, forall $a\in S$ (1) $a+t_{1}+c\leq t_{2}+c$ , $c\in S$ (2) $t_{1}, t_{2}\in \left(aSa\right]$ i.e. $t_{1}\leq asa$ , $t_{2}\leq asa$ , for some $asa\in aSa$ . Hence $S$ is an ordered $h$ -regular semiring. On the other hand $t_{3}+t_{3}at_{3}+t_{2}\leq t_{3}ct_{3}+t_{2}$ has no solution, so $S$ is not an $h$ -regular ordered semiring.

Lemma 1. Suppose $S$ is an ordered semiring and $E\subseteq S$ and $F\subseteq S$ , where E and F are nonempty, then

(1) $(\overline{E}]\subseteq \overline{\left(E\right] }.$

(2) If $E\subseteq F$ , then $\overline{E}\subseteq \overline{F}.$

(3) $\overline{\left(E\right] }F\subseteq \overline{\left(EF \right] }$ and $E\overline{\left(F\right] }\subseteq \overline{\left(EF \right] }.$

Proof. (1) Let $g\in (\overline{E}]$ . Then there exists $h\in \; \overline{E}$ such that $g\leq h.$ Since $h\in \; \overline{E}$ , then there exist $r_{1}, r_{2}\in E$ such that $h+r_{1}+k\leq r_{2}+k$ , $k\in E.$ It follows that $g+r_{1}+k\leq h+r_{1}+k\leq r_{2}+k.$ Since $E\subseteq \; \left(E \right], \; r_{1}, r_{2}\in \left(E\right]$ , $k\in \left(E\right]$ , $g\in \overline{\left(E\right] },$ i.e. $(\overline{E}]\subseteq \overline{ \left(E\right] }.$

(2) Consider $E\subseteq F$ . Let $g\in \overline{E}.$ Then, there exist $r_{1}, r_{2}\in E$ such that $g+r_{1}+k\leq r_{2}+k$ , $k\in E.$ By the assumption, we get $r_{1}, r_{2}, k\in \; F.$ This implies $g\in \overline{F}$ , so $\overline{E}\subseteq \overline{F}.$

(3) Let $g\in \overline{\left(E\right] }$ and $w\in F$ . So, there exist $p, q\in \left(E\right]$ such that $g+p+s\leq q+s, \; s\in \left(E \right]$ . So, $gw+pw+sw\leq qw+sw$ . Since $p, q, s\in \left(E\right]$ , $p\leq r_{1}$ and $q\leq r_{1}^{/}$ and $s\leq r_{1}^{//},$ for some $r_{1}, r_{1}^{/}, r_{1}^{//} \; \in E$ , so $pw\leq r_{1}w\in EF$ and $qw\leq r_{1}^{/}w\in EF$ and $sw\leq r_{1}^{//}w\in EF.$ This implies that $gw\in \overline{\left(EF\right] }$ . So $\overline{\left(E\right] }F\subseteq \overline{\left(EF\right]}$ . Similarly we get $E\overline{\left(F\right] } \subseteq \overline{\left(EF\right] }.$

Lemma 2. ^[15] Suppose $E\subseteq S$ , where $E$ is nonempty and $S$ is an ordered semiring. If $E$ is closed under addition, then so are $(E]$ , $\overline{(E]}$ .

Now we will use further throughout the section $N$ (set of all positive integers). Let S be ordered semiring, $E\neq \emptyset$ and $E\subseteq S$ , suppose $\sum_{finite}E$ be set of all finite sum of elements of $E$ , and for $x\in S$ , let $Nx = \{nx|n\in N\}.$

Lemma 3. Suppose $E$ and $F$ are nonempty subsets of an ordered semiring $S$ , with $E+E\subseteq E$ and $F+F\subseteq F$ . Then

(1) $E\subseteq (E]\subseteq \overline{E}\subseteq \overline{(E]},$

(2) $\overline{(E]}$ = $\overline{\overline{(E]}}$ , if $E$ is left ordered $h$ -ideal (or right ordered $h$ -ideal) of $S$ ,

(3) $E+F\subseteq \overline{E}+\overline{F}\subseteq \overline{E+F},$

(4) $\overline{(E]}+\overline{(F]}\subseteq \overline{\overline{(E]} +\overline{(F]}}\subseteq \overline{(E+F]},$

(5) $\overline{E} \; \overline{F}\subseteq \overline{(E]} \; \overline{(F]}$ ,

(6) If $E$ and $F$ are two left ordered $h$ -ideal and right ordered $h$ -ideal of $S$ , respectively then $\overline{(E]} \; \overline{(F]} \subseteq \overline{\left(\sum_{finite}EF\right]}.$

Proof. (1) We see that $E\subseteq \left(E\right].$

Let $g\in \left(E\right]$ , so by definition of “ $(\; ]$ ”, there exists $r\in E$ such that

$\begin{eqnarray*} g &\leq &r \\ g+r+r &\leq &r+r+r. \end{eqnarray*}$

This implies that $g\in \overline{E}\Longrightarrow (E]\subseteq \overline{E}.$

Since $E\subseteq (E]$ ,

$\begin{equation*} \Longrightarrow \overline{E}\subseteq \overline{(E]}. \end{equation*}$

(2) Let $E$ is left ordered $h$ -ideal (or right ordered $h$ -ideal) of $S$ .

By(ⅰ), $(E]\subseteq \overline{(E]}; \Longrightarrow \overline{(E]}\subseteq \overline{\overline{(E]}}.$

Let $g\in \overline{\overline{\left(E\right]}}$ , then by definition of $h$ -closure, there exist $h, k\in \overline{\left(E\right]}$ such that

$\begin{equation*} g+h+s\leq k+s, s\in \overline{\left( E\right]}. \end{equation*}$

Since $h, k, s\in \overline{\left(E\right]},$ then by definition of $h$ -closure, there exist $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}, r_{6}\in (E],$ such that

$\begin{equation*} h+r_{1}+s_{1}\leq r_{2}+s_{1}, s_{1}\in \left( E\right] \end{equation*}$

$\begin{equation*} k+r_{3}+s_{2}\leq r_{4}+s_{2}, s_{2}\in \left( E\right] \end{equation*}$

$\begin{equation*} s+r_{5}+s_{3}\leq r_{6}+s_{3}, s_{3}\in \left( E\right] \end{equation*}$

$\Longrightarrow$

$\begin{eqnarray*} g+h+s+r_{1}+s_{1}+r_{3}+s_{2}+r_{5}+s_{3} &\leq &k+s+r_{1}+s_{1}+r_{3}+s_{2}+r_{5}+s_{3} \\ &\leq &r_{4}+s_{2}+s+r_{1}+s_{1}+r_{5}+s_{3} \\ &\leq &r_{4}+s_{2}+r_{6}+s_{3}+r_{1}+s_{1} \\ & = &r_{1}+r_{4}+r_{6}+s_{1}+s_{2}+s_{3} \end{eqnarray*}$

$\Longrightarrow$

$\begin{equation*} g+\left( h+s+r_{1}+r_{3}+r_{5}\right) +\left( s_{1}+s_{2}+s_{3}\right) \leq \left( r_{1}+r_{4}+r_{6}\right) +\left( s_{1}+s_{2}+s_{3}\right). \end{equation*}$

Since $s_{1}, s_{2}, s_{3}\in \left(E\right]$ , then by definition of “ $(\; ]$ ”, there exist $s_{1}^{/}, s_{2}^{/}, s_{3}^{/}\in E$ such that $s_{1}\leq s_{1}^{/}$ and $s_{2}\leq s_{2}^{/}$ and $s_{3}\leq s_{3}^{/}.$

$\begin{equation*} \Longrightarrow s_{1}+s_{2}+s_{3}\leq s_{1}^{/}+s_{2}^{/}+s_{3}^{/} \end{equation*}$

As $E$ is a left ordered $h$ -ideal (or right ordered $h$ -ideal) of $S$ , so $E$ is a left ordered ideal (or right ordered ideal) of $S$ .

Then by definition of left ordered ideal or right ordered ideal, we get

$\begin{equation*} s_{1}+s_{2}+s_{3}\in E. \end{equation*}$

Now, since $r_{1}, r_{4}, r_{6}\in \left(E\right],$ so by using definition of “ $(\; ]$ ”, there exist $r_{1}^{/}, r_{4}^{/}, r_{6}^{/}\in E,$ such that $r_{1}\leq r_{1}^{/}$ and $r_{4}\leq r_{4}^{/}$ and $r_{6}\leq r_{6}^{/}$

$\begin{equation*} \Longrightarrow r_{1}+r_{4}+r_{6}\leq r_{1}^{/}+r_{4}^{/}+r_{6}^{/}. \end{equation*}$

Then by definition of left ordered ideal or right ordered ideal, we get

$\begin{equation*} r_{1}+r_{4}+r_{6}\in E. \end{equation*}$

Then by definition of left ordered ideal or right ordered ideal, we get $r_{1}, r_{2}, r_{5}, r_{6}, s_{1}, s_{3}\in E$

$\begin{eqnarray*} h+s+(r_{1}+r_{5}+s_{1}+s_{3}) &\leq &(r_{2}+r_{6}+s_{1}+s_{3}) \\ g+\left( h+s+r_{1}+r_{3}+r_{5}\right) +\left( s_{1}+s_{2}+s_{3}\right) &\leq &\left( r_{1}+r_{4}+r_{6}\right) +\left( s_{1}+s_{2}+s_{3}\right) \end{eqnarray*}$

where $s_{1}+s_{2}+s_{3}\in E,$ then by definition of $h$ -closure, we get $g\in \overline{E},$ then by (1), we get $g\in \overline{E}\subseteq \overline{(E]}\Longrightarrow g\in \overline{(E]}$ , $\Longrightarrow \overline{\overline{\left(E\right] }}\subseteq \overline{(E] }$

$\begin{equation*} \Longrightarrow \overline{(E]} = \overline{\overline{(E]}} \end{equation*}$

(3) From (1), we have $E\subseteq \overline{E}$ and $F\subseteq \overline{F}$

$\begin{equation*} \Longrightarrow E+F\subseteq \overline{E}+\overline{F}. \end{equation*}$

Now we show $\overline{E}+\overline{F}\subseteq \overline{E+F}.$ Suppose $g\in \overline{E}+\overline{F},$ so there exists $h\in \overline{E}$ and $k\in \overline{F}$ such that $g = h+k$ .

Since $h\in \overline{E}$ , $k\in \overline{F},$ so by using definition of $h$ -closure, there exist $r, r^{/}\in E$ and $w, w^{/}\in F$ such that

$\begin{equation*} h+r+s_{1}\leq r^{/}+s_{1},s_{1}\in E, \end{equation*}$

and

$\begin{equation*} k+w+s_{2}\leq w^{/}+s_{2},s_{2}\in F. \end{equation*}$

$\Longrightarrow$

$\begin{eqnarray*} g+r+s_{1}+w+s_{2} & = &h+k+r+s_{1}+w+s_{2} \\ g+(r+w)+(s_{1}+s_{2}) &\leq &r^{/}+s_{1}+w^{/}+s_{2} \\ & = &r^{/}+w^{/}+s_{1}+s_{2} \end{eqnarray*}$

$\begin{equation*} g+(r+w)+(s_{1}+s_{2})\leq \left( r^{/}+w^{/}\right) +\left( s_{1}+s_{2}\right) \end{equation*}$

As $\left(s_{1}+s_{2}\right) \in E+F,$ then by definition of $h$ -closure, we get $g\in \overline{E+F}$ ,

$\begin{equation*} \Longrightarrow \overline{E}+\overline{F}\subseteq \overline{E+F} \end{equation*}$

(4) Let $g\in \overline{(E]}+\overline{(F]},$ then there exists $p\in \overline{(E]}, \; q\in \overline{(F]},$ such that $g = p+q$ .

Now,

$\begin{equation*} g+\left( p+q\right) +h = \left( p+q\right) +\left( p+q\right) +h\Longrightarrow g+\left( p+q\right) +h = \left( p+p\right) +\left( q+q\right)+h. \end{equation*}$

Since $p+p\in \overline{(E]}$ and $q+q\in \overline{(F]},$ then by definition of $h$ -closure, we get $g\in \overline{\overline{(E]}+\overline{(F]}}$ ,

$\begin{equation*} \Longrightarrow \overline{(E]}+\overline{(F]}\subseteq \overline{\overline{(E]}+\overline{(F]}}. \end{equation*}$

Suppose $g\in \overline{(E]}+\overline{(F]}$ , so there exists $p\in \overline{(E]}, \; q\in \overline{(F]},$ such that $g = p+q$ .

Since $p\in \overline{(E]}$ and $q\in \overline{(F]}$ , so by using definition of $h$ -closure, there exist $r, r^{/}\in \left(E\right]$ and $w, w^{/}\in \left(F\right]$ such that

$\begin{equation*} p+r+s_{1}\leq r^{/}+s_{1},s_{1}\in \left( E\right], \end{equation*}$

and

$\begin{equation*} q+w+s_{2}\leq w^{/}+s_{2},s_{2}\in \left( F\right]. \end{equation*}$

$\begin{equation*} g+r+w+s_{1}+s_{2} = p+q+r+w+s_{1}+s_{2} \end{equation*}$

$\begin{equation*} g+r+w+s_{1}+s_{2}\leq r^{/}+w^{/}+s_{1}+s_{2}. \end{equation*}$

Since $\left(r+w\right), \left(r^{/}+w^{/}\right) \in \left(E+F\right],$

$\begin{equation*} g+\left( r+w\right) +\left( s_{1}+s_{2}\right) \leq \left( r^{/}+w^{/}\right) +\left( s_{1}+s_{2}\right). \end{equation*}$

This implies $g\in \overline{\left(E+F\right] }$ , $\Longrightarrow$

$\begin{eqnarray*} \overline{(E]}+\overline{(F]} &\subseteq &\overline{\left( E+F\right]} \\ \overline{\overline{(E]}+\overline{(F]}} &\subseteq &\overline{\overline{\left( E+F\right] }} \\ \overline{\overline{(E]}+\overline{(F]}} &\subseteq &\overline{\left( E+F\right]}. \end{eqnarray*}$

(5) By Lemma 1, we get

$\begin{equation*} \overline{E}\text{ }\overline{F}\subseteq \overline{(E]}\text{ }\overline{(F]}. \end{equation*}$

(6) Let $E$ , $F$ are two left ordered $h$ -ideal or right ordered $h$ -ideal of S. We will prove that $\overline{\left(E\right] } \; \overline{ \left(F\right] }\subseteq \overline{\left(\sum_{fij}EF\right] }.$

For this, let $g\in \overline{\left(E\right] } \; \overline{\left(F\right] }$ then $g = hk,$ as $h\in \overline{\left(E\right] }, k\in \overline{\left(F\right] },$ then by definition of $h$ -closure, there exist $p, p^{/}\in \left(E\right]$ and $q, q^{/}\in \left(F\right]$ such that

$\begin{equation*} h+p+s_{1}\leq p^{/}+s_{1},s_{1}\in \left( E\right], \end{equation*}$

and

$\begin{equation*} k+q+s_{2}\leq q^{/}+s_{2},s_{2}\in \left( F\right]. \end{equation*}$

$\begin{equation*} hk+pk+s_{1}k\leq p^{/}k+s_{1}k. \end{equation*}$

Also

$\begin{equation*} pk+pq+ps_{2}\leq pq^{/}+ps_{2}\;\text{ and }\;p^{/}k+p^{/}q+p^{/}s_{2}\leq p^{/}q^{/}+p^{/}s_{2}. \end{equation*}$

$\begin{eqnarray*} g & = &hk \\ g+pk+pq+p^{/}q+\left( s_{1}k+p^{/}s_{2}+ps_{2}\right) & = &hk+pk+pq+p^{/}q+s_{1}k+p^{/}s_{2}+ps_{2} \\ &\leq &p^{/}k+s_{1}k+pq+p^{/}q+p^{/}s_{2}+ps_{2}\\ &\leq &p^{/}q^{/}+p^{/}s_{2}+s_{1}k+pq+ps_{2} \\ & = &pq+p^{/}q^{/}+\left( s_{1}k+p^{/}s_{2}+ps_{2}\right). \end{eqnarray*}$

Since $E$ and $F$ are left ordered ideal and right ordered ideal of $S$ respectively, therefore,

$\begin{equation*} pk+pq+p^{/}q\in \sum\limits_{finite}EF,pq+p^{/}q^{/}\in \sum\limits_{finite}EF, \end{equation*}$

and

$\begin{equation*} s_{1}k+p^{/}s_{2}+ps_{2}\in \sum\limits_{finite}EF. \end{equation*}$

This implies that

$\begin{equation*} g+\left( pk+pq+p^{/}q\right) +\left( s_{1}k+p^{/}s_{2}+ps_{2}\right) \leq \left( pq+p^{/}q^{/}\right) +\left( s_{1}k+p^{/}s_{2}+ps_{2}\right). \end{equation*}$

$\begin{equation*} g\in \overline{\sum\limits_{finite}EF}\subseteq \overline{\left( \sum\limits_{finite}EF \right] }\Longrightarrow g\in \overline{\left( \sum\limits_{finite}EF \right] }. \end{equation*}$

$\begin{equation*} \Longrightarrow \overline{\left( E\right] }\text{ }\overline{\left( F\right] }\subseteq \overline{\left( \sum\limits_{finite}EF\right] }. \end{equation*}$

Example 2. (i) Every regular ordered semiring is an ordered $h$ -regular semiring.

(ii) Consider the semiring $(N, +, \cdot, \leq)$ , where $N$ is the set of natural numbers. We define the relation $\preceq$ on $N$ by $g\preceq h\Leftrightarrow g\geq h$ for all $g, h\in N$ . Then, $(N, \preceq)$ is a partially ordered set, furthermore $(N, +, \cdot, \preceq)$ is an ordered semiring. Since $g+ghg+s\preceq ghg+s$ for all $g, h\in N, s\in N$ , $(N, +, \cdot, \preceq)$ is an ordered $h$ -regular semiring. Moreover, since $2\in N, 2\npreceq 2h2 = 4h$ for all $h\in N$ , $(N, +, \cdot, \preceq)$ is not a regular ordered semiring. In addition, we get $(2N]$ is an ordered ideal of $(N, +, \cdot, \preceq)$ which is not an ${h}$ -ideal, for the reason that $2+4+1 = 3+4 \; as \; 1\notin (2N]$ .

Theorem 1. Suppose $S$ is an ordered semiring and $E$ be left ideal or right ideal or ideal, then conditions given below are equivalent:

(1) $E$ is left ordered $h$ -ideal or right ordered $h$ -ideal or ordered $h$ -ideal of $S$ ;

(2) Let $g\in S, g+r_{1}+h\leq r_{2}+h$ for some $r_{1}, r_{2}\in E$ , $h\in E$ then $g\in E$ ;

(3) $\overline{E} = E.$

Proof. (1) $\Longrightarrow$ (2) Suppose $E$ is a left ordered $h$ -ideal. Suppose $g\in S$ such that $g+r_{1}+h\leq r_{2}+h$ for some $r_{1}, r_{2}\in E$ , $h\in E$ then by using definition of left ordered $h$ -ideal, we get $g\in E.$

(2) $\Longrightarrow$ (3) Suppose (2) is true. Consider $g\in \; \overline{E}$ , then there exist $r_{1}, r_{2}\in E$ such that $g+r_{1}+h\leq r_{2}+h$ , $h\in E.$ By condition (2), we get $g\in E$ . So, $\overline{E}\subseteq E$ . Since $E\subseteq \overline{E}$ , therefore $\overline{E} = E.$

(3) $\Longrightarrow$ (1) Assume that $\overline{E} = E$ . Let $g\in S$ be such that $g+r_{1}+h\leq r_{2}+h$ for some $r_{1}, r_{2}\in E$ , $h\in E$ . Then $g\in \overline{E}.$ Since $\overline{E} = E$ , so $g\in \overline{E} = E.$ Thus $g\in E.$ Since $g+r_{1}+h\leq r_{2}+h$ for some $r_{1}, r_{2}\in E$ , $h\in E$ , then $g\in E$ , so by the definition of left ordered $h$ -ideal or right ordered $h$ -ideal or ordered $h$ -ideal, we get $E$ is left ordered $h$ -ideal or right ordered $h$ -ideal or ordered $h$ -ideal of $S.$

Theorem 2. Suppose $S$ is an ordered semiring, then:

(1) Intersection of any family of left ordered $h$ -ideals of $S$ is a left ordered $h$ -ideal.

(2) Intersection of any family of right ordered $h$ -ideals of $S$ is a right ordered $h$ -ideal.

(3) Intersection of any family of ordered $h$ -ideals of $S$ is an ordered $h$ -ideal.

Proof. (1) Suppose $E_{n}$ is a left ordered $h$ -ideal of $S$ for all $n\in J$ , as $\cap _{n\in J}E_{n}\neq \emptyset$ . Since $E_{n}$ is a left ordered $h$ -ideal, we get $E_{n}$ is a left ordered ideal for all $n\in J$ . Then $\cap _{n\in J}E_{n}\$ is left ordered ideal. Consider there exist $g\in S$ and $r_{1}, r_{2}\in \cap _{n\in J}E_{n}, \; h\in \cap _{n\in J}E_{n}$ is such that $g+r_{1}+h = r_{2}+h.$ Since $\cap _{n\in J}E_{n}\subseteq E_{n}$ for all $n\in J$ , we get, $r_{1}, r_{2}, h\in E_{n}$ . Since $E_{n}$ is a left ordered $h$ -ideal and $r_{1}, r_{2}\in E_{n}, \; g+r_{1}+h = r_{2}+h, \; h\in E_{n}$ for all $n\in J,$ so by using definition of left ordered $h$ -ideal, we get $g\in E_{n}$ for all $n\in J$ . So $g\in \cap _{n\in J}E_{n}$ . Therefore, $r_{1}, r_{2}\in \cap _{n\in J}E_{n}, \; g+r_{1}+h = r_{2}+h, \; h\in \cap _{n\in J}E_{n}$ . Then $g\in \cap _{n\in J}E_{n}$ . By definition of left ordered $h$ -ideal, we get $\cap _{n\in J}E_{n}$ is a left ordered $h$ -idealof S.

(2) Suppose that $E_{n}$ is a right ordered $h$ -ideal of $S$ for all $n\in J$ , as $\cap _{n\in J}E_{n}\neq \emptyset$ . Since $E_{n}$ is right ordered $h$ -ideal, we get, $E_{n}$ is right ordered ideal for all $n\in J$ . Then $\cap _{n\in J}E_{n}\$ is right ordered ideal. Consider that there exist $g\in S$ and $r_{1}, r_{2}\in \cap _{n\in J}E_{n}, \; h\in \cap _{n\in J}E_{n}$ such that $g+r_{1}+h = r_{2}+h$ . Since $\cap _{n\in J}E_{n}\subseteq E_{n}$ for all $n\in J$ , we have $r_{1}, r_{2}, h\in E_{n}$ . Since $E_{n}$ is a right ordered $h$ -ideal and $r_{1}, r_{2}\in E_{n}, \; g+r_{1}+h = r_{2}+h, \; h\in E_{n}$ for all $n\in J$ . So by using the definition of right ordered $h$ -ideal, we get $g\in E_{n}$ for all $n\in J$ . So $g\in \cap _{n\in J}E_{n}$ . Therefore, $r_{1}, r_{2}\in \cap _{n\in J}E_{n}, \; g+r_{1}+h = r_{2}+h, \; h\in \cap _{n\in J}E_{n}$ . Then $g\in \cap _{n\in J}E_{n}$ . By definition of right ordered $h$ -ideal, we have $\cap _{n\in J}E_{n}$ is a right ordered $h$ -idealof S.

(3) From (1) and (2), we get $\cap _{n\in J}E_{n}$ is a left and right ordered $h$ -ideal of S. Therefore, $\cap _{n\in J}E_{n}$ is an ordered $h$ -ideal of S. Hence proved.

Remark 1. $(1)$ The sum of two left ordered $h$ -ideals or right ordered $h$ -ideals is a left ordered $h$ -ideal or right ordered $h$ -ideal.

$(2)$ The sum of two left ordered ideals or right ordered ideals is a left ordered ideal or right ordered ideal.

$(3)$ The sum of two left ideals or right ideals is a left ideal or right ideal.

Theorem 3. Suppose $S$ is an ordered semiring and $E\neq \emptyset, F\neq \emptyset , E\subseteq S, F\subseteq S.$

(1) Consider $E, F$ be two left ordered $h$ -ideals, then $\overline{ (E+F]}$ is smallest left ordered $h$ -ideal containing $E\cup F$ .

(2) Consider $E, F$ be two right ordered $h$ -ideals, then $\overline{ (E+F]}$ is smallest right ordered $h$ -ideal containing $E\cup F$ .

(3) Consider $E, F$ be two ordered $h$ -ideals, then $\overline{(E+F]}$ is smallest ordered $h$ -ideal containing $E\cup F$ .

Proof. (1) Suppose $E$ , $F$ are two left ordered $h$ -ideal of $S$ . Suppose $g, h\in \overline{\left(E+F\right] }$ , $s\in S$ .

By the definition of $h$ -closure, there exist $r, r^{/}, w, w^{/}\in \left(E+F \right]$ such that

$\begin{equation*} g+r+f_{1}\leq r^{/}+f_{1},\text{ }f_{1}\in \left( E+F\right], \end{equation*}$

and

$\begin{equation*} h+w+f_{2}\leq w^{/}+f_{2},\text{ }f_{2}\in \left( E+F\right]. \end{equation*}$

Hence

$\begin{equation*} g+h+r+w+f_{1}+f_{2}\leq r^{/}+w^{/}+f_{1}+f_{2} \end{equation*}$

and

$\begin{equation*} sg+sr+sf_{1}\leq sr^{/}+sf_{1} \end{equation*}$

As $\left(sf_{1}\right) \in \left(E+F\right]$ , so by using the definition of $h$ -closure, we get $\left(g+h\right) \in \overline{\left(E+F\right] }$ and $\left(sg\right) \in \overline{\left(E+F\right]}$ .

This implies

$\begin{equation*} \overline{\left( E+F\right] }\subseteq \overline{\overline{\left( E+F\right] }} \end{equation*}$

Let $g\in \overline{\overline{\left(E+F\right] }},$ then by definition of $h$ -closure, there exist $i, v\in \overline{\left(E+F\right] }$ such that

$\begin{equation*} g+i+f\leq v+f,\;f\in \overline{\left( E+F\right]}. \end{equation*}$

Since $i, v, f\in \overline{\left(E+F\right]}$ , then by definition of $h$ -closure, there exist $r, r^{/}, w, w^{/}, d, d^{/}\in (E+F],$ such that

$\begin{equation*} i+r+f_{1}\leq r^{/}+f_{1},\text{ }f_{1}\in \left( E+F\right] \end{equation*}$

$\begin{equation*} v+w+f_{2}\leq w^{/}+f_{2},\text{ }f_{2}\in \left( E+F\right] \end{equation*}$

and

$\begin{equation*} f+d+f_{3}\leq d^{/}+f_{3},\text{ }f_{3}\in \left( E\right]. \end{equation*}$

Now,

$\begin{eqnarray*} g+i+f+r+w+d+f_{1}+f_{2}+f_{3} &\leq &v+f+r+w+d+f_{1}+f_{2}+f_{3} \\ &\leq &w^{/}+f_{2}+f+r+d+f_{1}+f_{3} \\ & = &f+d+f_{3}+w^{/}+f_{2}+r+f_{1} \\ &\leq &d^{/}+f_{3}+w^{/}+f_{2}+r+f_{1} \\ & = &r+w^{/}+d^{/}+f_{1}+f_{2}+f_{3}. \end{eqnarray*}$

Since $\left(r+w^{/}+d^{/}\right), \left(i+f+r+w+d\right) \in E+F$ and $\left(f_{1}+f_{2}+f_{3}\right) \in E+F$ , then by definition of $h$ -closure, we get $g\in \overline{E+F}\subseteq \overline{\left(E+F\right] }\Longrightarrow \overline{\overline{\left(E+F \right] }}\subseteq \overline{\left(E+F\right]}$ .

So, we get $\overline{\overline{\left(E+F\right] }} = \overline{\left(E+F\right]}$ .

This shows that $\overline{\left(E+F\right] }$ is a left ordered $h$ -ideal.

Suppose $g\in E\cup F,$ then $g\in E$ or $g\in F$

As $g\in E,$ then $g+\left(g+w\right) = \left(g+g\right) +w\in E+F$ , for all $w\in F$ . Thus $g\in \overline{\left(E+F\right]}$

As $g\in F,$ then $\left(r+g\right) +g = r+\left(g+g\right) \in E+F$ , for all $r\in E$ . Thus $g\in \overline{\left(E+F\right] }$

Hence,

$\begin{equation*} E\cup F\subseteq \overline{\left( E+F\right] } \end{equation*}$

Suppose $L$ is a left ordered $h$ -ideal containing $E\cup F.$

Then $E+F\subseteq L$ and hence $\left(E+F\right] \subseteq \left(L\right] = L$ implies that $\overline{\left(E+F\right] }\subseteq \overline{L} = L$

Therefore, $\overline{\left(E+F\right] }$ is the smallest left ordered $h$ -ideal containing $E\cup F.$

(2) This is similar to (1).

(3) From (1) and (2), we prove that $\overline{ \left(E+F\right] }$ is smallest left and right ordered $h$ -ideal containing $E\cup F$ . Therefore, $\overline{\left(E+F\right] }$ is smallest ordered $h$ -ideal containing $E\cup F$ .

Theorem 4. Suppose S is an ordered semiring and $E\neq \emptyset$ , $E\subseteq S$ . Then these properties hold.

(1) Consider $E$ a left ideal, then $\overline{(E]}$ is the smallest left ordered $h$ -ideal containing $E$ .

(2) Consider $E$ a right ideal, then $\overline{(E]}$ is the smallest right ordered $h$ -ideal containing $E$ .

(3) Consdier $E$ an ideal, then $\overline{(E]}$ is the smallest ordered $h$ -ideal containing $E$ .

Proof. Suppose $E$ is a left ideal. We know that $\overline{(E]}$ is closed with respect to the operation of addition.

Suppose $g\in \overline{(E]},$ and $k\in E,$ then by using definition of $h$ -closure, there exist $r, w\in \left(E\right]$ such that

$\begin{equation*} g+r+h\leq w+h,h\in \left( E\right]. \end{equation*}$

Hence

$\begin{equation*} kg+kr+kh\leq kw+kh. \end{equation*}$

So by using definition of “ $(\; ]$ ”, we have $kh\in \left(E\right].$ Since $\left(kr\right), \left(kw\right) \in \left(E\right]$ , $kg+\left(kr\right) +\left(kh\right) \leq \left(kw\right) +\left(kh\right), \; \left(kh\right) \in \left(E\right]$ .

Then by definition of $h$ -clousre, we get $kg\in \overline{\left(E\right]}.$ Therefore, $\overline{\left(E\right] }$ is a left ordered $h$ -ideal.

We know that $\overline{\left(E\right] }$ is a left ordered $h$ -ideal containing $E$ .

Suppose $Q$ is a left ordered $h$ -ideal containing $E$ . So $(E]\subseteq (Q] = Q$ . Then, $\overline{(E]}\subseteq \overline{Q} = Q.$ Therefore, $\overline{(E]}$ is the smallest left ordered $h$ -ideal containing $E$ .

(2) This is similar to (1).

(3) From (1) and (2), we prove that $\overline{ \left(E\right] }$ is the smallest left and right ordered $h$ -ideal containing $E$ . Therefore, $\overline{\left(E\right] }$ is the smallest ordered $h$ -ideal containing $E$ .

Corollary 1. Suppose S is an ordered semiring, let $\emptyset \neq E\subseteq S$ . We denote the smallest left ordered $h$ -ideal containing $E$ by $L_{h}(E)$ , the smallest right ordered $h$ -ideal containing $E$ by $R_{h}(E)$ , and the smallest ordered $h$ -ideal of $S$ containing $E$ by $M_{h}(E)$ . Then, the following results follows:

(1) $L_{h}(E) = \overline{(\sum_{finite}E+\sum_{finite}SE]},$

(2) $R_{h}(E) = \overline{(\sum_{finite}E+\sum_{finite}ES]},$

(3) $M_{h}(E) = \overline{(\sum_{finite}E+\sum_{finite}SE+ \sum_{finite}ES+\sum_{finite}SES]}.$

Proof. We want to prove $\sum_{finite}E+\sum_{finite}SE$ is a left ideal.

For this,

$\left(I_{1}\right)$ Let $a, b\in \sum_{finite}E+\sum_{finite}SE.$ Then

$\begin{equation*} a+b\in \sum\limits_{finite}E+\sum\limits_{finite}SE \end{equation*}$

$\left(I_{2}\right)$ Let $a\in \sum_{finite}E+\sum_{finite}SE, r\in S$

$\Longrightarrow$

$\begin{eqnarray*} ra &\in &r\left( \sum\limits_{finite}E+\sum\limits_{finite}SE\right) \\ & = &\sum\limits_{finite}rE+\sum\limits_{finite}rSE \\ &\subseteq &\sum\limits_{finite}E+\sum\limits_{finite}SE \end{eqnarray*}$

$\begin{equation*} \Longrightarrow ra\in \sum\limits_{finite}E+\sum\limits_{finite}SE \end{equation*}$

Therefore, $\sum_{finite}E+\sum_{finite}SE$ is a left ideal. By Theorem 4, we get

$\begin{equation*} L_{h}(E) = \overline{(\sum\limits_{finite}E+\sum\limits_{finite}SE]} \end{equation*}$

We see that the proofs of (2) and (3) are similar to that of (1).

Corollary 2. Suppose $S$ is an ordered semiring, let $r\in S$ . Then

(1) $L_{h}(r) = \overline{(Nr+Sr]};$

(2) $R_{h}(r) = \overline{(Nr+rS]};$

(3) $M_{h}(r) = (Nr+Sr+rS+\sum_{finite}SrS],$ where $N$ = Set of natural numbers.

4. Ordered h-regular semirings

We now give the characterization of ordered $h$ -regular semirings by their ordered $h$ -ideals.

Theorem 5. Consider $S$ is an ordered semiring. Then $S$ is an ordered $h$ -regular iff $E\cap F = \overline{(EF]}$ , for all right ordered $h$ -ideals $E$ , left ordered $h$ -ideals $F$ of $S$ .

Proof. Suppose $S$ is an ordered $h$ -regular semiring and $E$ is right ordered $h$ -ideal, F is left ordered $h$ -ideal of S. Then, we have $EF\subseteq E$ and $EF\subseteq F$ . Thus, $\left(EF\right] \subseteq \left(E\right] = E$ and $\left(EF\right] \subseteq \left(F\right] = F$ .

This implies $\overline{\left(EF\right] }\subseteq \overline{E} = E$ and $\overline{\left(EF\right] }\subseteq \overline{F} = \overline{F}.$ Thus $\overline{\left(EF\right] }\subseteq E\cap F$

Let $p\in E\cap F.$ As $S$ is an ordered $h$ -regular, there exist $h, k\in \left(pSp\right],$ such that

$\begin{equation*} p+h+o\leq k+o,o\in \left( pS\;p\right]. \end{equation*}$

Since $h, k, o\in \left(pSp\right]$ , then by definition of “ $(\; ]$ ”, there exist $s, j, j_{1}\in S$ such that

$\begin{equation*} h\leq psp,\text{ }k\leq pjp,\text{ }o\leq pj_{1}p. \end{equation*}$

Since $E$ is a right ordered $h$ -ideal, $F$ is a left ordered $h$ -ideal, we have $psp, pjp, pj_{1}p\in EF.$ Since $h\leq psp\in EF, \; k\leq pjp\in EF, \; o\leq pj_{1}p\in EF$ , so by using definition of “ $(\; ]$ ”, we have $h, k, o\in \left(EF\right]$ so $p\in \overline{ \left(EF\right]}\Longrightarrow E\cap F\subseteq \; \overline{\left(EF\right]}$

$\begin{equation*} \Longrightarrow E\cap F = \overline{\left( EF\right] } \end{equation*}$

Conversely, Consider $E\cap F = \overline{\left(EF\right] }$ for all right ordered $h$ -ideals $E$ of $S$ , left ordered $h$ -ideals $F$ of $S$ . Suppose $d\in S,$ then by above Corollary 2 we get

$\begin{equation*} L_{h}(d) = \overline{(Nd+Sd]}\text{ and }R_{h}(d) = \overline{(Nd+dS]}. \end{equation*}$

By assumption, $(R_{h}(d)\cap L_{h}(d) = \overline{\left(R_{h}(d)L_{h}(d) \right] }).$ Now we show that $R_{h}(d)L_{h}(d)\subseteq \overline{(Sd]}\cap \overline{(dS]}$

Let $p\in R_{h}(d)$ and $h\in L_{h}(d).$

Since $p\in R_{h}(d)$ , then by the definition of $h$ -closure,

$\begin{equation*} \text{There exist} i,i^{/}\in (Nd+dS],\text{ such that }\;p+i+o_{1}\leq i^{/}+o_{1},o_{1}\in (Nd+dS]. \end{equation*}$

Since $h\in L_{h}(d)$ , then by the definition of $h$ -closure,

$\begin{equation*} \text{There exist} \; v,v^{/}\in (Nd+Sd],\text{ such that }\;h+v+o_{2}\leq v^{/}+o_{2},o_{2}\in (Nd+Sd], \end{equation*}$

$\begin{equation*} ph+ih+o_{1}h\leq i^{/}h+o_{1}h. \end{equation*}$

Since $h, k, o\in \left(pSp\right]$ , then by definition of “ $(\; ]$ ”, there exist $s, j, j_{1}\in S$ such that

$\begin{equation*} i\leq ed+ds,\;i^{/}\leq fd+dj,\;o_{1}\leq gd+dj_{1} \end{equation*}$

It follows that

$\begin{equation*} ih\leq edh+dsh = d\left( eh+sh\right) \in dS \end{equation*}$

$\begin{equation*} i^{/}h\leq fdh+djh = d\left( fh+jh\right) \in dS \end{equation*}$

$\begin{equation*} o_{1}h\leq gdh+dj_{1}h = d\left( gh+j_{1}h\right) \in dS. \end{equation*}$

Then by definition of “ $(\; ]$ ”, we get $ih, i^{/}h, o_{1}h\in \left(dS\right]$ . Since $ih, i^{/}h\in \left(dS\right]$ ,

$\begin{equation*} ph+ih+o_{1}h\leq i^{/}h+o_{1}h,\text{ }o_{1}h\in \left( dS\right]. \end{equation*}$

Then by definition of h-closure, we get $ph\in \overline{\left(dS\right]}.$ So,

$\begin{equation*} R_{h}(d)L_{h}(d)\subseteq \overline{(dS]} \end{equation*}$

Similarly, we can show that

$\begin{equation*} R_{h}(d)L_{h}(d)\subseteq \overline{(Sd]} \end{equation*}$

Therefore,

$\begin{equation*} R_{h}(d)L_{h}(d)\subseteq \overline{(Sd]}\cap \overline{(dS]}. \end{equation*}$

Since $\overline{(dS]}$ is a right ordered $h$ -ideal, $\overline{(Sd]}$ is a left ordered $h$ -ideal and by assumption, we have,

$\begin{equation*} \overline{(dS]}\cap \overline{(Sd]}\subseteq \overline{\left( \overline{(dS]} \cap \overline{(Sd]}\right]}. \end{equation*}$

Now we will show that $\overline{(dS]}\overline{(Sd]}\subseteq \overline{ (dSd]}.$ For this let $m\in \overline{(dS]}$ and $k\in \overline{(Sd]},$ then by definition of $h$ -closure, there exist $l, l_{1}\in (dS]$ and $q, q_{1}\in (Sd],$ such that

$\begin{equation*} m+l+o_{3}\leq l_{1}+o_{3},o_{3}\in (dS] \end{equation*}$

and

$\begin{equation*} k+q+o_{4}\leq q_{1}+o_{4},o_{4}\in (Sd] \end{equation*}$

From above equations we get,

$\begin{eqnarray*} mk+lk+o_{3}k &\leq &l_{1}k+o_{3}k \\ lk+lq+lo_{4} &\leq &lq_{1}+lo_{4} \\ l_{1}k+l_{1}q+l_{1}o_{4} &\leq &l_{1}q_{1}+l_{1}o_{4} \\ o_{3}k+o_{3}q+o_{3}o_{4} &\leq &o_{3}q_{1}+o_{3}o_{4}. \end{eqnarray*}$

Since S is a multiplicatively commutative ordered semiring,

$\begin{eqnarray*} mk+\left( lk+lq+l_{1}q\right) +\left( o_{3}k+lo_{4}+l_{1}o_{4}\right) &\leq &l_{1}k+o_{3}k+lq+l_{1}q+lo_{4}+l_{1}o_{4} \\ &\leq &l_{1}q_{1}+l_{1}o_{4}+o_{3}k+lq+lo_{4} \\ & = &\left( l_{1}q_{1}+lq\right) +\left( o_{3}k+lo_{4}+l_{1}o_{4}\right). \end{eqnarray*}$

Since $l, l_{1}, o_{3}\in (dS], q, q_{1}, o_{4}\in (Sd],$ then by definition of “ $(\; ]$ ”, there exist $s_{1}, s_{2}, s_{3}, l^{/}, l^{//}, l^{///}\in S,$ such that

$\begin{equation*} l\leq ds_{1},\text{ }l_{1}\leq ds_{2},\text{ }o_{3}\leq ds_{3}\text{ and } q\leq l^{/}d,\text{ }q_{1}\leq l^{//}d,\text{ }o_{4}\leq l^{///}d. \end{equation*}$

Hence, we obtained,

$\begin{eqnarray*} lk+lq+l_{1}q &\leq &l_{1}q+lo_{4}+lq_{1}+lo_{4} \\ &\leq &ds_{2}l^{/}d+ds_{1}l^{///}d+ds_{1}l^{//}d+ds_{1}l^{///}d\in dSd \end{eqnarray*}$

$\Longrightarrow$

$\begin{equation*} lk+lq+l_{1}q\in \left( dSd\right] \end{equation*}$

and

$\begin{eqnarray*} o_{3}k+lo_{4}+l_{1}o_{4} &\leq &o_{3}q+o_{3}o_{4}+lo_{4}+l_{1}o_{4}+o_{3}q_{1}+o_{3}o_{4} \\ &\leq &ds_{3}l^{/}d+ds_{3}l^{///}d+ds_{1}l^{///}d+ds_{2}l^{///}d+ds_{3}l^{//}d+ds_{3}l^{///}d\in dSd. \end{eqnarray*}$

Then by definition of “ $(\; ]$ ”,

$\begin{equation*} o_{3}k+lo_{4}+l_{1}o_{4}\in \left( dSd\right] \text{ and }l_{1}q_{1}+lq\leq ds_{2}l^{//}d+ds_{1}l^{/}d\in dSd \end{equation*}$

$\Longrightarrow$

$\begin{equation*} l_{1}q_{1}+lq\in \left( dSd\right]. \end{equation*}$

So, $mk\in \overline{\left(dSd\right] }$ . Hence $\overline{(dS]}\overline{(Sd]}\subseteq \overline{(dSd]}.$

Hence

$\begin{equation*} \overline{(dS]}\cap \overline{(Sd]}\subseteq \overline{\left( \overline{(dS]} \cap \overline{(Sd]}\right] }\subseteq \overline{\overline{(dSd]}} = \overline{(dSd]}. \end{equation*}$

Thus,

$\begin{equation*} R_{h}(d)\cap L_{h}(d) = \overline{\left( R_{h}(d)L_{h}(d)\right] }\subseteq \overline{\left( \overline{(dS]}\cap \overline{(Sd]}\right] }\subseteq \overline{\overline{(dSd]}} = \overline{(dSd]} \end{equation*}$

It turns out, $d\in \overline{\left(dSd\right] }.$ Hence, $S$ is an ordered $h$ -regular.

Corollary 3. Suppose $S$ is a commutative ordered semiring. Then $S$ is an ordered $h$ -regular iff for each ordered $h$ -ideal $E$ of $S$ , $E = \overline{(E^{2}]}.$

Proof. Suppose $S$ is an ordered $h$ -regular. Consider $E$ is an ordered $h$ -ideal of S. Obviously we have $E = E\cap E = \overline{(E^{2}]}.$

Conversely, Suppose that for each ordered $h$ -ideal E of S, $E = \overline{ (E^{2}]}.$ Consider $r\in S$ . As $S$ is a commutatively multiplication ordered semiring, we get

$\begin{equation*} M_{h}(r) = L_{h}(r) = R_{h}(r). \end{equation*}$

So,

$\begin{eqnarray*} r &\in &M_{h}(r) = \overline{(M_{h}(r)M_{h}(r)]} \\ & = &\overline{(R_{h}(r)L_{h}(r)]} \\ & = &\overline{\left( \overline{(Nr+rS]}\overline{(Nr+Sr]}\right] } \\ &\subseteq &\overline{\left( \sum\limits_{finite}\overline{\left( Nr+rS\right] } \overline{(Nr+Sr]}\right] } \\ &\subseteq &\overline{\left( \overline{(rS]}\right] } = \overline{\left( rS \right]}. \end{eqnarray*}$

Since $S$ is commutative i.e., $\overline{(rS]} = \overline{(Sr]}$ , and $\overline{(rS]}$ is an ordered $h$ -ideal, therefore

$\begin{eqnarray*} r &\in &\overline{(rS]} = \overline{\left( (\overline{(rS]}^{2}\right] } = \overline{\left( \overline{(rS]}\overline{(rS]}\right] } \\ & = &\overline{\left( \overline{(rS]}\overline{(Sr]}\right] } \\ &\subseteq &\overline{\left( \overline{(\sum\limits_{finite}rSSr]}\right] }\subseteq \overline{\left( \overline{(rSr]}\right] } = \overline{(rSr]}. \end{eqnarray*}$

Thus, $S$ is an ordered $h$ -regular.

Definition 12. Suppose $S$ is an ordered semiring, let $r\in S$ . Suppose $r\in \overline{ (Sr^{2}]},$ then r is said to be left ordered $h$ -regular. Suppose $r\in \overline{(r^{2}S]},$ then $r$ is called right ordered $h$ -regular. Suppose each element of $S$ is left or right ordered $h$ -regular. Then, ordered semiring $S$ is said to be a left or right ordered $h$ -regular.

Theorem 6. Suppose $S$ is left ordered $h$ -regular semiring. Then

(1) for each left ordered $h$ -ideal $E$ of S, $\overline{(E^{2}]} = E$ ;

(2) $Q\cap E = \overline{(QE]}$ , for each left ordered $h$ -ideal $E$ and each ordered $h$ -ideal $Q$ of $S$ .

Proof. (1) Suppose $E$ is left ordered $h$ -ideal of $S$ . Then, we get $\overline{(E^{2}]}\subseteq \overline{(E]} = E.$

Suppose $r\in E$ . As $S$ is a left ordered $h$ -regular, so $r\in \overline{(Sr^{2}]}.$

Since $\overline{(Sr^{2}]}\subseteq \overline{(SE^{2}]}\subseteq \overline{(E^{2}]}, \Longrightarrow r\in \overline{(E^{2}]}$ . Hence, $E\subseteq \overline{(E^{2}]}\Longrightarrow \overline{(E^{2}]} = E$ .

(2) Let $E$ is left ordered $h$ -ideal and $Q$ is ordered $h$ -ideal of S. Then, we get, $\overline{(QE]}\subseteq \overline{(Q]} = Q$ and $\overline{(QE]}\subseteq \overline{(E]} = E.$ Hence, $\overline{(QE]}\subseteq Q\cap E$ . Let $a\in Q\cap E$ . As $S$ is left ordered $h$ -regular,

$\begin{equation*} a\in \overline{(Sa^{2}]}\subseteq \overline{(SQE]}\subseteq \overline{(QE]}. \end{equation*}$

$\Longrightarrow a\in \overline{(QE]}$ . Hence, $Q\cap E\subseteq \overline{(QE]}.$ Thus,

$\begin{equation*} \Longrightarrow Q\cap E = \overline{(QE]}. \end{equation*}$

Theorem 7. Suppose $S$ is a right ordered $h$ -regular semiring. Then

(1) for each right ordered $h$ -ideal E of S, $\overline{(E^{2}]} = E$ ;

(2) $E\cap Q = \overline{(EQ]}$ , for each right ordered $h$ -ideal $E$ and each ordered $h$ -ideal $Q$ of $S$ .

Theorem 8. Suppose $S$ is an ordered semiring, then the conditions given below are equivalent:

(1) for each left ordered $h$ -ideal $E$ , $F$ of $S$ , $E\cap F = \overline{(EF]}.$

(2) for each left ordered $h$ -ideal $E$ and each ordered $h$ -ideal $Q$ of $S$ , $E\cap Q = \overline{(EQ]}.$

(3) S is left ordered $h$ -regular and $R_{h}(E)\subseteq L_{h}(E)$ for all $\emptyset \neq E\subseteq S$

Proof. (1) $\Rightarrow$ (2) Let $E$ is left ordered $h$ -ideal of $S$ and $F$ is ordered $h$ -ideal of $S$ . Then $E\cap F = \overline{(EF]}, \; F$ being left ordered $h$ -ideal of $S$ , we get $E\cap F = \overline{(EF]}.$

(2) $\Rightarrow$ (3) Let $\emptyset \neq E\subseteq S$ . By assumption, we get $L_{h}(E) = L_{h}(E)\cap S = \overline{(L_{h}(E)S]}$ . We have

$\begin{eqnarray*} R_{h}(E) & = &\overline{(\sum\limits_{finite}E+\sum\limits_{finite}ES]} \\ &\subseteq &\overline{(\sum\limits_{finite}L_{h}(E)+\sum\limits_{finite}L_{h}(E)S]} \\ & = &\overline{\left( \sum\limits_{finite}\overline{(L_{h}(E)S]}+\sum\limits_{finite}\overline{ (L_{h}(E)S]}S\right] } \\ &\subseteq &\overline{\left( \sum\limits_{finite}\overline{(L_{h}(E)S]}+\sum\limits_{finite} \overline{(L_{h}(E)SS]}\right] } \\ & = &\overline{\left( \sum\limits_{finite}\overline{(L_{h}(E)S]}\right] } \\ & = &\overline{\sum\limits_{finite}L_{h}(E)} \\ & = &L_{h}(E). \end{eqnarray*}$

Moreover, we show that $L_{h}(E) = M_{h}(L_{h}(E))$ . Since $R_{h}(E)\subseteq L_{h}(E)$ , we get, $L_{h}(E)\subseteq R_{h}(L_{h}(E))\subseteq L_{h}(L_{h}(E)) = L_{h}(E).$

Hence, $L_{h}(E) = R_{h}(L_{h}(E))$ . It follows that $L_{h}(E) = M_{h}(L_{h}(E))$

Let $p\in S.$ From assumption, we get

$\begin{eqnarray*} p &\in &L_{h}(p)\cap M_{h}(p) = \overline{(L_{h}(p)M_{h}(p)]} \\ & = &\overline{(L_{h}(p)M_{h}(L_{h}(p))]} \\ & = &\overline{(L_{h}(p)L_{h}(p)]} \\ &\subseteq &\overline{(Np^{2}+\sum\limits_{finite}Sp^{2}+\sum\limits_{finite}pSp+ \sum\limits_{finite}SpSp]} \\ &\subseteq &\overline{(Np^{2}+\sum\limits_{finite}Sp^{2}+\sum\limits_{finite}R_{h}(p)p+ \sum\limits_{finite}SR_{h}(p)p]} \\ &\subseteq &\overline{(Np^{2}+\sum\limits_{finite}Sp^{2}+\sum\limits_{finite}L_{h}(p)p+ \sum\limits_{finite}SL_{h}(p)p]} \\ &\subseteq &\overline{(Np^{2}+\sum\limits_{finite}Sp^{2}+\sum\limits_{finite}L_{h}(p)p+ \sum\limits_{finite}L_{h}(p)p]} \\ & = &\overline{(Np^{2}+\sum\limits_{finite}Sp^{2}+\sum\limits_{finite}L_{h}(p)p]} \\ & = &\overline{(Np^{2}+Sp^{2}+L_{h}(p)p]} \\ & = &\overline{\left( Np^{2}+Sp^{2}+\overline{(Np+Sp]}p\right] } \\ &\subseteq &\overline{\left( Np^{2}+Sp^{2}+\overline{(Np^{2}+Sp^{2}]}\right] } \\ & = &\overline{\left( \overline{(Np^{2}+Sp^{2}]}\right] } \\ & = &\overline{(Np^{2}+Sp^{2}]} \end{eqnarray*}$

$\Longrightarrow$

$\begin{equation*} p\in \overline{(Np^{2}+Sp^{2}]} \end{equation*}$

Since $p\in \overline{(Np^{2}+Sp^{2}]}$ , then by definition of $h$ -closure, there exist $v, v^{/}\in (Np^{2}+Sp^{2}],$ such that

$\begin{equation*} p+v+t_{1}\leq v^{/}+t_{1},\text{ where }t_{1}\in (Np^{2}+Sp^{2}]. \end{equation*}$

Since $v, v^{/}, t_{1}\in (Np^{2}+Sp^{2}],$ then by definition of “ $(\; ]$ ”, there exist $e, f, g\in N$ and $\ s, r, r_{1}\in S$ such that

$\begin{equation*} v\leq ep^{2}+sp^{2},\text{ }v^{/}\leq fp^{2}+rp^{2},\text{ }t_{1}\leq gp^{2}+r_{1}p^{2} \end{equation*}$

In a similar way, we obtain

$\begin{equation*} p^{2}\in \overline{(Np^{4}+Sp^{4}]}. \end{equation*}$

Then by definition of $h$ -closure, there exist $u, u^{/}\in (Np^{4}+Sp^{4}],$ such that

$\begin{equation*} p^{2}+u+t_{2}\leq u^{/}+t_{2} \quad ,t_{2}\in (Np^{4}+Sp^{4}]. \end{equation*}$

Since $u, u^{/}, t_{2}\in (Np^{4}+Sp^{4}],$ then by definition of “ $(\; ]$ ”, there exist $e^{/}, f^{/}, g^{/}\in N$ and $\ s^{/}, r^{/}, r_{1}^{/}\in S$ such that

$\begin{equation*} u\leq e^{/}p^{4}+s^{/}p^{4},\text{ }u^{/}\leq f^{/}p^{4}+r^{/}p^{4},\text{ } t_{2}\leq g^{/}p^{4}+r_{1}^{/}p^{4}. \end{equation*}$

From $p^{2}+u+t_{2}\leq u^{/}+t_{2}$ we get

$\begin{equation*} ep^{2}+eu+et_{2}\leq eu^{/}+et_{2}. \end{equation*}$

Now we have

$\begin{eqnarray*} v+eu+et_{2} &\leq &ep^{2}+sp^{2}+eu+et_{2} \\ &\leq &eu^{/}+et_{2}+sp^{2} \\ &\leq &ef^{/}p^{4}+er^{/}p^{4}+et_{2}+sp^{2} \\ v+eu+et_{2}+fu+ft_{2} &\leq &ef^{/}p^{4}+er^{/}p^{4}+et_{2}+sp^{2}+fu+ft_{2} \\ &\leq &ef^{/}p^{4}+er^{/}p^{4}+sp^{2}+f(e^{/}p^{4}+s^{/}p^{4})+et_{2}+ft_{2} \\ &\leq &ef^{/}p^{4}+er^{/}p^{4}+sp^{2}+fe^{/}p^{4}+fs^{/}p^{4}+e(gp^{2}+r_{1}p^{2})+f(gp^{2}+r_{1}p^{2}) \\ &\leq &ef^{/}p^{4}+er^{/}p^{4}+sp^{2}+fe^{/}p^{4}+fs^{/}p^{4}+egp^{2}+er_{1}p^{2}+fgp^{2}+fr_{1}p^{2}\in Sp^{2}. \end{eqnarray*}$

Then by definition of “ $(\; ]$ ”, $v+eu+et_{2}+fu+ft_{2}\in \left(Sp^{2}\right].$

Now

$\begin{eqnarray*} v^{/}+fu+ft_{2} &\leq &fp^{2}+rp^{2}+fu+ft_{2} \\ v^{/}+eu+et_{2}+fu+ft_{2} &\leq &fp^{2}+rp^{2}+eu+et_{2}+sp^{2}+fu+ft_{2} \\ v^{/}+eu+et_{2}+fu+ft_{2} &\leq &fp^{2}+rp^{2}+e(e^{/}p^{4}+s^{/}p^{4})+e(g^{/}p^{4}+r_{1}^{/}p^{4})+sp^{2}+f(e^{/}p^{4}+s^{/}p^{4}) \\ &&+f(g^{/}p^{4}+r_{1}^{/}p^{4}) \\ v^{/}+eu+et_{2}+fu+ft_{2} &\leq &fp^{2}+rp^{2}+ee^{/}p^{4}+es^{/}p^{4}+eg^{/}p^{4}+er_{1}^{/}p^{4}+sp^{2}+fe^{/}p^{4}+fs^{/}p^{4} \\ +fg^{/}p^{4}+fr_{1}^{/}p^{4} &\in &Sp^{2}. \end{eqnarray*}$

Then by definition of “ $(\; ]$ ”,

$\begin{equation*} \Longrightarrow v^{/}+eu+et_{2}+fu+ft_{2}\in \left( Sp^{2}\right]. \end{equation*}$

Now,

$\begin{eqnarray*} et_{2}+ft_{2} &\leq &e(g^{/}p^{4}+r_{1}^{/}p^{4})+f(g^{/}p^{4}+r_{1}^{/}p^{4}) \\ & = &eg^{/}p^{4}+er_{1}^{/}p^{4}+fg^{/}p^{4}+fr_{1}^{/}p^{4}\in Sp^{2}. \end{eqnarray*}$

Then by definition of “ $(\; ]$ ”,

$\begin{equation*} et_{2}+ft_{2}\in \left( Sp^{2}\right]. \end{equation*}$

Now

$\begin{equation*} p+(v+eu+et_{2}+fu+ft_{2})+\left( et_{2}+ft_{2}\right) \leq \left( v^{/}+eu+et_{2}+fu+ft_{2}\right) +\left( et_{2}+ft_{2}\right) \end{equation*}$

$\begin{equation*} \Longrightarrow p\in \overline{\left( Sp^{2}\right]}. \end{equation*}$

Hence, $S$ is a left ordered $h$ -regular.

(3) $\Rightarrow$ (1)

Let $E$ , $F$ are left ordered $h$ -ideals of S, then we get $\overline{(EF]} \subseteq \overline{(F]} = F$ We see that $E\subseteq R_{h}(E)\subseteq L_{h}(E) = E$ . Hence, $E$ is an ordered $h$ -ideal. Thus $\overline{(EF]}\subseteq \overline{(E]} = E$ . So, $\overline{(EF]}\subseteq E\cap F$

Suppose $p\in E\cap F$ . By assumption, we get $p\in \overline{(Sp^{2}]}.$ Since $\overline{(Sp^{2}]}\subseteq \overline{(SEF]}\subseteq \overline{(EF]}, \; p\in (EF]$ . It turns out that $E\cap F\subseteq \overline{(EF]}.$

Therefore,

$\begin{equation*} E\cap F = \overline{(EF]} \end{equation*}$

Theorem 9. Suppose S is an ordered semiring. Then the conditions given below are equivalent:

(1) for each right ordered $h$ -ideal $E$ , $F$ of $S$ , $E\cap F = \overline{(EF]}.$

(2) for each right ordered $h$ -ideal $E$ of $S$ , each ordered $h$ -ideal $Q$ of $S$ , $Q\cap E = \overline{(QE]}.$

(3) $S$ is right ordered $h$ -regular, $L_{h}(E)\subseteq R_{h}(E)$ for all $\emptyset \neq E\subseteq S.$

5. Ordered h-weakly regular semirings

Definition 13. Suppose $S$ is an ordered semiring, let $r\in S$ . Suppose $r\in \overline{(\sum_{finite}(Sr)^{2}]}$ , then $r$ is said to be a left ordered $h$ -weakly regular. Suppose $r\in \overline{(\sum_{finite}(rS)^{2}]}$ , then $r$ is said to be a right ordered $h$ -weakly regular. Suppose each element in $S$ is left or right ordered $h$ -weakly regular, then ordered semiring $S$ is said to be left or right ordered $h$ -weakly regular.

Theorem 10. Suppose $S$ is an ordered semiring, then the conditions given below are equivalent:

(1) $S$ is a left ordered $h$ -weakly regular.

(2) for each left ordered $h$ -ideal $E$ of $S$ , $\overline{ (\sum_{finite}E^{2}]} = E.$

(3) for each left ordered $h$ -ideal $E$ of $S$ and each ordered $h$ -ideal $Q$ of $S$ , $Q\cap E = \overline{(\sum_{finite}QE]}.$

Proof. (1) $\Rightarrow$ (2) Suppose $E$ is a left ordered $h$ -ideal of S. Then, we get, $\overline{(\sum_{finite}E^{2}]}\subseteq \overline{(E]} = E$

Let $a\in E$ . By assumption, we have

$\begin{equation*} a\in \overline{(\sum\limits_{finite}SaSa]}\subseteq \overline{(\sum\limits_{finite}SESE]} \subseteq \overline{(\sum\limits_{finite}E^{2}]}. \end{equation*}$

Hence,

$\begin{equation*} E\subseteq \overline{(\sum\limits_{finite}E^{2}]}. \end{equation*}$

Thus,

$\begin{equation*} \overline{(\sum\limits_{finite}E^{2}]} = E \end{equation*}$

(2) $\Rightarrow$ (1) Let $r\in S$ . From assumption, Lemma 3, and Corollary 2. We get,

$\begin{eqnarray*} r &\in &L_{h}(r) = \overline{(\sum\limits_{finite}L_{h}(r)^{2}]} \\ & = &\overline{\left( \sum\limits_{finite}\overline{(Nr+Sr]}\overline{(Nr+Sr]}\right] } \\ &\subseteq &\overline{\left( \sum\limits_{finite}\overline{ (\sum\limits_{finite}(Nr+Sr)(Nr+Sr)]}\right] } \\ &\subseteq &\overline{\left( \sum\limits_{finite}\overline{(Sr]}\right] } \\ & = &\overline{\left( \overline{(Sr]}\right] } \\ & = &\overline{(Sr]}. \end{eqnarray*}$

Since $\overline{(Sr]}$ is left ordered $h$ -ideal, we get $r\in \overline{ (Sr]} = \overline{\left(\sum_{finite}\overline{(Sr]}^{2}\right] }.$

By Lemma 2 and Theorem 4, we have

$\begin{eqnarray*} \overline{\left( \sum\limits_{finite}\overline{(Sr]}^{2}\right] } &\subseteq & \overline{\left( \sum\limits_{finite}\overline{(\sum\limits_{finite}SrSr]}\right] } \\ & = &\overline{\left( \overline{(\sum\limits_{finite}SrSr]}\right] } \\ & = &\overline{(\sum\limits_{finite}(Sr)^{2}]}. \end{eqnarray*}$

Hence,

$\begin{equation*} r\in \overline{(\sum\limits_{finite}(Sr)^{2}]}. \end{equation*}$

Therefore, $S$ is a left ordered $h$ -weakly regular.

(2) $\bf{\Rightarrow }$ (3) Suppose $E$ is a left ordered $h$ -ideal of $S$ and $Q$ is an ordered $h$ -ideal of $S$ . Then,

$\begin{equation*} \overline{(\sum\limits_{finite}QE]}\subseteq \overline{(\sum\limits_{finite}Q]} = Q, \end{equation*}$

and

$\begin{equation*} \overline{(\sum\limits_{finite}QE]}\subseteq \overline{(\sum\limits_{finite}E]} = E. \end{equation*}$

Hence, $\overline{(\sum_{finite}QE]}\subseteq Q\cap E.$

Let $a\in Q\cap E.$ By assumption, we get,

$\begin{eqnarray*} a &\in &L_{h}(a) = \overline{(\sum\limits_{finite}L_{h}(a)^{2}]} \\ &\subseteq &\overline{\left(\sum\limits_{finite}M_{h}(a)L_{h}(a)\right] } \\ &\subseteq &\overline{(\sum\limits_{finite}QE]}. \end{eqnarray*}$

Hence,

$\begin{equation*} Q\cap E\subseteq \overline{(\sum\limits_{finite}QE]}. \end{equation*}$

Thus

$\begin{equation*} Q\cap E = \overline{(\sum\limits_{finite}QE]} \end{equation*}$

(3) $\Rightarrow$ (2) Suppose $E$ is a left ordered $h$ -ideal of S, then, we get $\overline{(\sum_{finite}E^{2}]}\subseteq \overline{(E]} = E.$

By Lemma 1, 2, Theorem 4 and Corollary 1, we get

$\begin{eqnarray*} E & = &M_{h}(E)\cap E \\ & = &\overline{(\sum\limits_{finite}M_{h}(E)E]} \\ & = &\overline{\left( \sum\limits_{finite}\overline{(\sum\limits_{finite}E+\sum\limits_{finite}SE+ \sum\limits_{finite}ES+\sum\limits_{finite}SES]}E\right] } \\ &\subseteq &\overline{\left( \sum\limits_{finite}\overline{(\sum\limits_{finite}EE+ \sum\limits_{finite}SEE+\sum\limits_{finite}ESE+\sum\limits_{finite}SESE]}\right] } \\ &\subseteq &\overline{\left( \sum\limits_{finite}\overline{(\sum\limits_{finite}E^{2}]} \right] } \\ & = &\overline{\left( \overline{(\sum\limits_{finite}E^{2}]}\right] } = \overline{(\sum\limits_{finite}E^{2}]}. \end{eqnarray*}$

Thus,

$\begin{equation*} \overline{(\sum\limits_{finite}E^{2}]} = E. \end{equation*}$

Theorem 11. Suppose $S$ is an ordered semiring, then the conditions given below are equivalent:

(1) $S$ is a right ordered $h$ -weakly regular.

(2) for each right ordered $h$ -ideal $E$ of $S$ , $\overline{ (\sum_{finite}E^{2}]} = E.$

(3) for each right ordered $h$ -ideal $E$ of $S$ and each ordered $h$ -ideal $Q$ of $S$ , $E\cap Q = \overline{(\sum_{finite}EQ]}.$

Proof. Straightforward.

6. Conclusions

Concepts of the ordered $h$ - ideals in semirings, alongside their essential properties, were presented. The classes of the semirings like ordered $h$ -regular and ordered $h$ -weakly regular semirings were characterized by the properties of the ordered $h$ -ideals.

The ideas of the ordered $h$ -ideals can be extended to the non associative structures like the ones in (^{[16,17,18,20,21,22]}). Moreover, ordered $h$ -ideals can be extended for fuzzification in semiring theory.

Acknowledgments

The research was supported by the National Natural Science Foundation of China (Grant Nos. 11971142, 11871202, 61673169, 11701176, 11626101, 11601485).

Conflict of interest

The authors declare no conflict of interest.

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