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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# On some varieties of ai-semirings satisfying xp+1 ≈ x

Aifa Wang
• School of Mathematics, Northwest University, Xi’an 710127, Shaanxi, China
• School of Science, Chongqing University of Technology, Chongqing, 400054, China
• Email
• Other articles by this author:
/ Yong Shao
Published Online: 2018-08-16 | DOI: https://doi.org/10.1515/math-2018-0079

## Abstract

The aim of this paper is to study the lattice of subvarieties of the ai-semiring variety defined by the additional identities

$xp+1≈xandzxyz≈(zxzyz)pzyxz(zxzyz)p,$

where p is a prime. It is shown that this lattice is a distributive lattice of order 179. Also, each member of this lattice is finitely based and finitely generated.

Keywords: Burnside ai-semiring; Identity; Lattice; Variety

MSC 2010: 08B15; 08B05; 16Y60; 20M07

## 1 Introduction

Semirings (see [9]) abound in the mathematical world around us. The set of natural numbers, the first mathematical structure we encounter, is a semiring. The intensive study of semiring theory was initiated during the late 1960’s when their real and significant applications were found. Nowdays, semiring theory is an enormously broad topic and has advanced on a very broad front. Semirings (S, +, ⋅) occurring in the literature satisfy at least the following axioms: (S, +) and (S, ⋅) are semigroups, and the multiplication distributes over addition from both sides. It is often assumed that (S, +) is idempotent and/or commutative. A semiring S is an additively idempotent semiring, or shortly ai-semiring, if (S, +) is a semilattice (it is also called a semilattice-ordered semigroup in [8, 10, 11]). It is well-known that the endomorphism semiring of a semilattice is an ai-semiring. Also every ai-semiring can be embeded into the endomorphism semiring of some semilattice (see [8, 12]). Important role in mathematics as well as broad applications (in theoretical computer science, optimization theory, quantum physics and many other areas of science [9, 13, 14, 15]) make ai-semirings and, especially, their varieties to be among the favourite subjects for the researchers in the algebraic theory of semirings.

The variety of all ai-semirings is denoted by AI. Let X be a fixed countably infinite set of variables and X+ the free semigroup on X. Then Pf(X+) is free in AI on X (see [8]). An AI-identity means an identity uv, where u = u1+⋯+uk, v = v1+⋯+ v, ui, vjX+, ik, j, k = {1, 2, …, k}. Recall that an ai-semiring is called a Burnside ai-semiring if its multiplicative reduct is a Burnside semigroup, i.e., it satisfies the identity xnxm with m < n (see [16, 17, 18]). The variety of all Burnside ai-semirings (resp., Burnside semigroups) satisfying the identities xnxm will be denoted by Sr(n, m) (resp., Sg(n, m)). There are many papers in the literature considering Burnside semigroups and Burnside ai-semirings (see [1, 2, 4, 5, 6, 7, 8, 10, 16, 17, 18, 19, 20]). In particular, in 1979, McKenzie and Romanowska [19] studied the lattice of subvarieties of the subvariety Bi of Sr(2, 1) defined by the additional identity xyyx. They showed that this lattice contains precisely 5 elements: the trivial variety T, the variety D of distributive lattices, the ai-semiring variety M defined by the additional identity x + yxy, the varieties DM and Bi (see Figure 1).

Fig. 1

The lattice of subvarieties of Bi

In 2002, Zhao [7] studied the variety Sr(2, 1), which need not satisfy xyyx. They provided a model of the free object in this variety by introducing the notion of closed subsemigroup of a semigroup. In 2005, Ghosh et al. [1], Pastijn [2] and Pastijn and Zhao [3] studied the lattice of subvarieties of Sr(2, 1). They showed that this lattice is a distributive lattice of order 78 and that every member of this lattice is finitely based and is generated by a finite number of finite ordered bands.

Along this research route, some authors studied the subvarieties of Sr(n, 1). In 2005, Kuřil and Polák generalized the notion of closed subsemigroups introduced by Zhao [7] to that of n-closed subset of a semigroup. They provided a construction of the free object in Sr(n, 1) by using n-closed subset of the free object in Sg(n, 1). Moreover, Gajdoš and Kuřil [10] showed that Sr(n, 1) is locally finite if and only if Sg(n, 1) is locally finite. In 2015, Ren and Zhao [21] introduced the notion of (n, m)-closed subsets of a semigroup and gave a model of the free object in Sr(n, m) by using (n, m)-closed subset of the free object in Sg(n, m). In 2016, Ren, Zhao and Shao [20] proved that the multiplicative semigroup of each member of Sr(n, 1) is a regular orthocryptogroup. As an application, a model of the free object in such a variety is given. In the same year, Ren and Zhao [4] studied the lattice of subvarieties of the subvariety of Sr(3, 1) defined by the additional identity xyyx. They showed that it is a 9-element distributive lattice. As a continuation of [4], Ren et al. [5] studied the lattice of subvarieties of the subvariety of Sr(n, 1) defined by the additional identity xyyx. They showed that if n − 1 is square-free, then this lattice is a 2+2r+1+ 3r-element distributive lattice, where r denotes the number of prime divisors of n − 1. They also proved that this lattice is finitely based and finitely generated. In 2017, Ren et al. [6] studied that the lattice of the subvarieties of Sr(3, 1). They showed that this lattice is a 179-element distributive lattice. They also showed that every member of this lattice is finitely based and finitely generated. This paper is another contribution to this line of investigation. We shall characterize the lattice of subvarieties of the subvariety of Sr(p + 1, 1) defined by the additional identities zxyz ≈ (zxzyz)pzyxz(zxzyz)p.

This paper is organized as follows. After this introductory section, in Sect. 2 we shall give some auxiliary results and notations that are needed in the sequel. In Sect. 3 we shall study the subvariety of Sr(p + 1, 1) defined by the additional identity zxyz ≈ (zxzyz)pzyxz(zxzyz)p. We shall show that its lattice of subvarieties is a distributive lattice of order 179. Also, all members of this lattice are finitely based and finitely generated. In particular, Sr(3, 1) is just the case of above variety when p = 2. Thus our main results generalize and extend the main results in [1, 2, 3, 4, 6, 19].

For notation and terminology not given in this paper, the reader is referred to [22, 23, 24, 25].

## 2 Preliminaries and some notations

Let Ap+1 denote the group variety defined by the identities

$xp+1≈x,$(1)

$xy≈yx.$(2)

For any SSr(3, 1), every subgroup of (S, ⋅) satisfies the identity x3x. This shows that every subgroup of (S, ⋅) is an abelian group in A3. That is to say, (S, ⋅) is a union of abelian groups which belong to A3.

For any semigroup SSg(p + 1, 1), by [22, Proposition II.7.1 and Exercises V.5.8 (V)]), we have

#### Lemma 2.1

If S satisfies the identity zxyz ≈ (zxzyz)pzyxz(zxzyz)p, then S is a union of abelian groups which belong to Ap+1.

We denote by ROBAp+1 the subvariety of Sg(p + 1, 1) defined by the additional identity zxyz ≈ (zxzyz)pzyxz(zxzyz)p. As usual, for a semigroup variety V, we denote by V the semiring variety consisting of all ai-semirings whose multiplicative reduct belongs to V. It is easy to see that $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ is just the variety of all ai-semirings whose multiplicative reduct is a union of abelian groups which belong to Ap+1, i.e., the ai-semiring variety defined by the additional identities

$xp+1≈xandzxyz≈(zxzyz)pzyxz(zxzyz)p.$

In particular, Sr(3, 1) is equal to $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{3}^{\circ }.\end{array}$

Let SAp+1 denote the semigroup variety defined by the identities (1) and (2), and S denote the variety of semilattices. By [22, Lemma IV.2.3, Theorem IV. 2.4 and Exercise IV.2.16 (xii(γ))], we have

#### Lemma 2.2

SAp+1 = SAp+1.

Let ω be an element of X+. The following notions and notation are needed for solving the word problem for SAp+1 and ROBAp+1:

• i(ω) denotes the initial part of ω, i.e., the word obtained from ω by retaining only the first occurrence of each variable.

• f(ω) denotes the final part of ω, i.e., the word obtained from ω by retaining only the last occurrence of each variable.

• c(ω) denotes the content of ω, i.e., the set of all variables occurring in ω.

• m(x, ω) denotes the multiplicity of x in ω, i.e., the number of occurrence of x in ω.

• ri(ω) denotes the set {xc(ω) | im(x, ω) (mod p)}, where i ∈ {0, 1, …, p − 1}.

• rp+1(ω) denotes the set {xk | xc(ω), xp − 1, km(x, ω) (mod p)}.

• ω denotes the word obtaining from ω by deleting all occurrences of variables which belong to r0(ω).

It is easy to check that a semigroup identity uv is satisfied by Ap+1 if and only if ri(u) = ri(v) for all ip − 1. Thus, for any semigroup identity uv,

$SAp+1⊨u≈v⇔c(u)=c(v),ri(u)=ri(v)(∀i∈p−1_),$(3)

$ROBAp+1⊨u≈v⇔i(u)=i(v),f(u)=f(v),ri(u)=ri(v)(∀i∈p−1_),$(4)

where ROBAp+1 = ReBAp+1 (see [22, Theorem V.5.3]), and ReB the variety of regular bands. Now we have

#### Lemma 2.3

$\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ satisfies the following identities

$(xy)p≈xpyp,$(5)

$xpyx≈xyxp,$(6)

$xyzx≈xyxpzx,$(7)

$xy+z≈xy+z+xzpy,$(8)

$x1+x2+⋯+xp+1≈x1+x2+⋯+xp+1+x1x2⋯xp+1.$(9)

#### Proof

By (4), it follows immediately that (5), (6) and (7) hold in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Also,

$x1+x2+⋯+xp+1≈(x1+x2+⋯+xp+1)p+1(sinceROBAp+1∘⊨(1))≈∑i1,i2,…,ip+1∈p+1_xi1xi2⋯xip+1≈∑i1,i2,…,ip+1∈p+1_xi1xi2⋯xip+1+x1x2⋯xp+1(sinceROBAp+1∘⊨x+x≈x)≈(x1+x2+⋯+xp+1)p+1+x1x2⋯xp+1≈x1+x2+⋯+xp+1+x1x2⋯xp+1.$

Thus, $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. In the remainder we need only to prove that (8) holds in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. The following is derivable from the identities determining $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ and identities (4), (5), (9):

$xy+z≈xy+z+xyzp+zp−1xyz(by(9))≈xy+z+xp+1yzp+zp−1xyz(xy)pzp(by(5))≈xy+z+(xp+zp−1xyz(xy)p−1)xyzp≈xy+z+(xp+zpxpyp+xpzpxpyp)xyzp(by(4),(9))≈xy+z+xp+1yzp+zpxyzp+xpzpxyzp.$

Thus, $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ satisfies the identity

$xy+z≈xy+z+xpzpxyzp.$

In a left-right dual way we can show that $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ satisfies the identity

$xy+z≈xy+z+zpxyzpyp.$

Furthermore, we have

$xy+z≈xy+z+xpzpxyzp+zpxyzpyp+xyzp(by(9))≈xy+z+xpzpxyzp+zpxyzpyp+xyzp+xpzpxyzp⋅(xyzp)p−1⋅zpxyzpyp(by(9))≈xy+z+xpzpxyzp+zpxyzpyp+xyzp+xpzpxyzpyp(by(4))≈xy+z+xpzpxyzp+zpxyzpyp+xyzp+xpzpxzpyzpyp(by(7))≈xy+z+xpzpxyzp+zpxyzpyp+xyzp+(xz)pxzpy(zy)p(by(5))≈xy+z+xpzpxyzp+zpxyzpyp+xyzp+xzpy.$

Thus, $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ satisfies identity (8). □

#### Lemma 2.4

Let S be an ai-semiring in Sr(p + 1, 1). Then the following is true,

$(∀a∈S)(∀i,j∈p_,i≠j)ai+aj=a+a2+⋯+ap.$(10)

#### Proof

Suppose that S is an ai-semiring in Sr(p + 1, 1). Then for any aS, ap+1 = a. Without loss of generality, suppose that i, jp and i < j. We have

$ai+aj=ai(ap+aj−i)=ai(ap+aj−i)p+1(sinceS⊨xp+1≈x)=ai(aj−i+a2(j−i)+⋯+ap(j−i))=ai(a+a2+⋯+ap)(sincepis a prime)=a+a2+⋯+ap.$

This completes the proof. □

Let uv be an AI-identity and Σ a set of identities which include the identities determining AI. Under the presence of the identities determining AI, it is easy to verify that uv gives rise to the identities uu+vj, vv+ui, ik, j. Conversely, the latter k + identities give rise to uu + vv. Thus, to show that uv is derivable from Σ, we only need to show that the simpler identities uu+vj, vv+ui, ik, j are derivable from Σ.

## 3 The lattice 𝓛$\begin{array}{}\left({\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\right)\end{array}$

In the current section, we shall characterize the lattice 𝓛$\begin{array}{}\left({\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\right)\end{array}$ of subvarieties of $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$, and prove that each member of this lattice is finitely based and finitely generated.

Let (Zp, ⋅) be the cyclic group of order p and $\begin{array}{}{Z}_{p}^{0}\end{array}$ the 0-group obtained from Zp by adjoining an extra element 0, where a ⋅ 0 = 0 ⋅ a = 0 for every aZp ∪ {0}. Define an additive operation on $\begin{array}{}{Z}_{p}^{0}\end{array}$ as follows:

$a+b=a,if a=b;0,otherwise.$

Then ($\begin{array}{}{Z}_{p}^{0}\end{array}$,+, ⋅ ) forms an ai-semiring. It is easy to check that $\begin{array}{}{Z}_{p}^{0}\end{array}$ is in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$, but not in Sr(2, 1). In fact, we have

#### Lemma 3.1

Let S be an ai-semiring in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Then S is a member of Sr(2, 1) if and only if it does not contain a copy of $\begin{array}{}{Z}_{p}^{0}\end{array}$.

#### Proof

The direct part is obvious. Conversely, suppose that S is an ai-semiring in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$, but not in Sr(2, 1). Then there exists aS such that a is not equal to ap. Thus by Lemma 2.4, we can show that {a, a2, …, ap, a + a2+ ⋯ +ap} is a subsemiring of S and is a copy of $\begin{array}{}{Z}_{p}^{0}\end{array}$. This completes the proof. □

Let Mp denote $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\cap \left[{x}^{p}+{y}^{p}\approx {x}^{p}{y}^{p}\right].\end{array}$ From [5, Lemma 4.12], HSP($\begin{array}{}{Z}_{p}^{0}\end{array}$) = Mp. Also, by [5, Proposition 4.13], the interval [M, Mp] consists of the two varieties M and Mp. Thus we have

#### Lemma 3.2

The interval [T, Mp] consists of the three varieties T, M and Mp. □

#### Proof

Let V ∈ [T, Mp] and VT. Then MV since M is minimal nontrivial subvariety of Mp. It follows that V ∈ [M, Mp] and so V = M or V = Mp. □

Let u = u1+⋯+uk, where uiX+, ik. We put C(u) = ⋃ik c(ui). To solve the word problem of the ai-semiring $\begin{array}{}{Z}_{p}^{0}\end{array}$ we need the following Lemmas, which are the special cases of [5: Corollary 4.15 and Lemma 5.1] when n = p+1.

#### Lemma 3.3

Let uu+q be an AI-identity, where u = u1+⋯+uk, ui, qX+, ik. Then $\begin{array}{}{Z}_{p}^{0}\end{array}$ satisfies uu+q if and only if c(q) ⊆ C(u) and rp+1(q) = rp+1(ui1ui(p+1)) for some positive integer ℓ and some ui1, …, ui(p+1) ∈ {ui | ik}.

#### Lemma 3.4

Let uu+q be an AI-identity, where u = u1+⋯+uk, ui, qX+, ik. If $\begin{array}{}{Z}_{p}^{0}\end{array}$ satisfies uu+q, then there exists q1 in X+ with rp+1(q1) = rp+1(q) and c(q) ⊆ c(q1) ⊆ C(u) such that uu+q1 is satisfied in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$.

For a semiring variety V, we denote by 𝓛(V) the lattice of all subvarieties of V. In the following we shall characterize the lattice 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$). Suppose that S be a member of $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. It is shown in [20, Lemma 2.1] that E(S) forms a member of Sr(2, 1). Define a mapping φ as follows $φ:L(ROBAp+1∘)→L(Sr(2,1)), V↦V∩Sr(2,1).$

Then it is easy to see that φ is subjective. For any V ∈ 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$), if Sφ(V), then SVSr(2, 1) and so S ∈ {E(S)|SV}. It follows that φ(V) ⊆ {E(S)|SV}. On the other hand, if E(S) ∈ {E(S)|SV}, then SV. Since E(S) is a subsemiring of S, we have that E(S) ∈ V and so E(S) ∈ VSr(2, 1). It follows that {E(S)|SV} ⊆ φ(V). Therefore, φ(V) = {E(S)|SV}. We use t(x1, …, xn) to denote the AI-term t which contains no other variables than x1, …, xn (but not necessarily all of them). If W is the subvariety of Sr(2, 1) determined by the additional identities $u(x1,…,xn)≈v(x1,…,xn),$

then we denote by $\begin{array}{}\stackrel{^}{{\mathbf{W}}_{p}}\end{array}$ the subvariety of $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ determined by the additional identities $u(x1p,…,xnp)≈v(x1p,…,xnp).$

Then for any V ∈ [W, Ŵp], $W=φ(W)⊆φ(V)⊆φ(W^p)=W^p∩(Sr(2,1))=W,$

it follows that Vφ−1(W) and so [W, Ŵp] ⊆ φ−1(W). Conversely, if Vφ−1(W), then V ∈ [W, Ŵp]. Otherwise, φ(V) ≠ W, a contradiction. Thus, for any W ∈ 𝓛 (Sr(2, 1)), φ−1(W) is the interval [W, $\begin{array}{}\stackrel{^}{{\mathbf{W}}_{p}}\end{array}$] of 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$). Moreover, if W1, W2 ∈ 𝓛(Sr(2, 1)) such that W1W2, then it is easy to check that Ŵ1Ŵ2. Suppose that (Vi)iI is a family of varieties in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Then φ(Vi) ⊆ Vi$\begin{array}{}\stackrel{^}{\phi \left({\mathbf{V}}_{i}\right)}\end{array}$ for all iI. Furthermore, $⋁i∈Iφ(Vi)⊆⋁i∈IVi⊆⋁i∈Iφ(Vi)^⊆⋁i∈Iφ(Vi)^.$

This implies that φ(⋁iIVi) = ⋁iIφ(Vi) and so φ is a complete ∨-epimorphism. Moreover, it is clear that φ is a complete ∧-epimorphism. It follows that φ is a complete epimorphism. Thus, 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$) = ⋃W ∈ 𝓛 (Sr(2, 1))[W, $\begin{array}{}\stackrel{^}{{\mathbf{W}}_{p}}\end{array}$].

Notice that M is the subvariety of Sr (2, 1) determined by the additional identities xyyx and x+yxy. Thus $\begin{array}{}\stackrel{^}{{\mathbf{M}}_{p}}\end{array}$ is the subvariety of $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ determined by the additional identities xpypypxp and xp+ypxpyp. By [22, Lemma IV.2.3], $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ ∩ [xpypypxp] = $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$ and so $\begin{array}{}\stackrel{^}{{\mathbf{M}}_{p}}\end{array}$ = Mp.

Recall (see [1]) that the lattice 𝓛(Sr (2, 1)) can be divided into five intervals: [T, NPSr (2, 1)], [D, NSr (2, 1)], [M, PSr (2, 1)], [DM, KSr(2, 1)] and [Bi, Sr(2, 1)], where NSr(2, 1) is the subvariety of Sr(2, 1) determined by the identity $x≈x+xyx,$(11)

PSr(2, 1) is the subvariety of Sr(2, 1) determined by the identity $xyx≈x+xyx,$(12)

and KSr(2, 1) is the subvariety of Sr(2, 1) determined by the identity $x+xyx+xyzyx≈x+xyzyx.$(13)

Thus we have

#### Theorem 3.5

Let V ∈ [T, NPSr(2, 1)] ∪ [D, NSr(2, 1)]. Then $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$ = V.

#### Proof

From NSr(2, 1) ⊨ (11), we have that $\begin{array}{}\stackrel{^}{{\mathbf{N}\cap \mathbf{S}\mathbf{r}\left(2,1\right)}_{p}}\end{array}$ satisfies the identity $xp≈xp+xpypxp.$(14)

Notice that $\begin{array}{}{Z}_{p}^{0}\end{array}$(14). Then $\begin{array}{}{Z}_{p}^{0}\notin \stackrel{^}{{\mathbf{N}\cap \mathbf{S}\mathbf{r}\left(2,1\right)}_{p}}\end{array}$ and so by Lemma 3.1, $\begin{array}{}\stackrel{^}{{\mathbf{N}\cap \mathbf{S}\mathbf{r}\left(2,1\right)}_{p}}\end{array}$Sr(2, 1). It follows from $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\subseteq \stackrel{^}{{\mathbf{N}\cap \mathbf{S}\mathbf{r}\left(2,1\right)}_{p}}\text{\hspace{0.17em}that\hspace{0.17em}}\stackrel{^}{{\mathbf{V}}_{p}}\subseteq \mathbf{S}\mathbf{r}\left(2,1\right).\end{array}$

Assume that V ∈ 𝓛(Sr(2, 1)) and Vuu+q, where u = u1+⋯+uk, ui, qX+, ik. From $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$(5) it follows that $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$ satisfies the identity $u1p+⋯+ukp≈u1p+⋯+ukp+qp.$(15)

#### Lemma 3.6

Let V ∈ [M, PSr(2, 1)]. Then VHSP( $\begin{array}{}{Z}_{p}^{0}\end{array}$) = $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$.

#### Proof

Since PSr(2, 1) ⊨ (12), it follows that $\begin{array}{}\stackrel{^}{{\mathbf{P}\cap \mathbf{S}\mathbf{r}\left(2,1\right)}_{p}}\end{array}$ satisfies the identity $xpypxp≈xp+xpypxp.$(16)

We immediately have that this is the case for $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$, too. Notice that HSP( $\begin{array}{}{Z}_{p}^{0}\end{array}$) = $\begin{array}{}\stackrel{^}{{\mathbf{M}}_{p}}\end{array}$. Thus VHSP( $\begin{array}{}{Z}_{p}^{0}\end{array}$) ⊆ $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$. It remains to show that every identity which is satisfied in VHSP( $\begin{array}{}{Z}_{p}^{0}\end{array}$) can be deduced from the identities which hold in $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$.

Let uu+q be an AI-identity which is satisfied in VHSP( $\begin{array}{}{Z}_{p}^{0}\end{array}$), where u = u1+⋯+uk, ui, qX+, ik. Since $\begin{array}{}{Z}_{p}^{0}\end{array}$ satisfies this identity, by Lemma 3.4 there exists q1 in X+ with rp+1(q1) = rp+1(q) (and so ri(q1 = ri(q) for all ip−1) and c(q) ⊆ c(q1) such that uu+q1 is satisfied in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Thus the following are derivable from the identities which hold in $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$: $u≈u+q1≈u+q1+u1pq1+⋯+ukpq1 (by(9))≈u+q1+(u1p+⋯+ukp)q1≈u+q1+(u1p+⋯+ukp+qp)q1 (by(15))≈u+q1+(u1p+⋯+ukp)q1+qpq1.$

This derives the identity $u≈u+qpq1.$

In a left-right dual way we have $u≈u+q1qp.$

Therefore the following are satisfied in $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$: $u≈u+qpq1+q1qp≈u+qpq1+q1qp+qpq1(q1qp)p (by(9))≈u+qpq1+q1qp+qpq1qp (by(1),(5))≈u+qpq1qp.$

Furthermore, we have the following are derivable from the identities which are satisfied in $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$: $u≈u+qpq1qp≈u+qq1pqp (Vp^⊨qpq1qp≈qq1pqp)≈u+q(qpq1pqp+qp) (by (16))≈u+qqpq1pqp+qqp≈u+qqpq1pqp+q.$

This shows that uu+q is satisfied in $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$ and so VHSP( $\begin{array}{}{Z}_{p}^{0}\end{array}$) = $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$. □

By Lemma 3.1 and 3.6, we can establish the following result.

#### Theorem 3.7

Let V ∈ [M, PSr(2, 1)]. Then the interval [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$] of 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$) consists of two varieties V and VHSP( $\begin{array}{}{Z}_{p}^{0}\end{array}$).

For an ai-semiring S we denote by S0 the ai-semiring obtained from S by adding an extra element 0, where a + 0 = a, a0 = 0a = 0 for every aS. Written Bp as $\begin{array}{}{Z}_{p}^{0}\end{array}$, then we have

#### Lemma 3.8

Let S$\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Then S satisfies the identity $x+(x+xp)yp(x+xp)≈xp+(x+xp)yp(x+xp)$(17)

if and only if it does not contain a copy of $\begin{array}{}{B}_{p}^{0}\end{array}$.

#### Proof

Necessary. It is obvious.

Sufficiency. Suppose that S$\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$, but that S does not satisfy identity (17). Then there exist a, bS such that $a+(a+ap)bp(a+ap)≠ap+(a+ap)bp(a+ap).$

It is easily verified that c, a+c, a2+c, …, ap+c and a+ap+c are not equal to each other, where c = (a+ap)bp(a+ap). Put S1 = {c, a+c, a2+c, …, ap+c, a+ap+c}. By identities (1) and (10), it is routine to verify that S1 forms a subsemiring of S, which is a copy of $\begin{array}{}{B}_{p}^{0}\end{array}$. □

Let u = u1+⋯+um, where uiX+, im, and Z ⊆ ⋃imc(ui). If c(ui) ∩ Z ≠ ∅ for every i, then we write DZ(u) = ∅. Otherwise, DZ(u) is the sum of terms ui for which c(ui) ∩ Z = ∅. The proof of the following result is easy and omitted.

#### Lemma 3.9

Let S$\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ and uv be an AI-identity, where u = u1+⋯+um, v = v1+⋯+vn, ui, vjX+, im, jn. Then S0 satisfies uv if and only if for every ZC(u) ⋃ C(v), either DZ(u) = DZ(v) = ∅ or DZ(u) ≠ ∅ ≠ DZ(v) and DZ(u) ≈ DZ(v) is satisfied in S.

#### Lemma 3.10

HSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) = $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$.

#### Proof

It is obvious that HSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) ⊆ $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$. To prove that $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$HSP( $\begin{array}{}{B}_{p}^{0}\end{array}$), we need only to show that every identity satisfied in HSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) can be derived from the identities determining $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$.

Let uu+q be any identity which is satisfied in HSP( $\begin{array}{}{B}_{p}^{0}\end{array}$), where u = u1+⋯+un, ui, qX+, in. Choose Z = C(u) ∖ c(q). Since DZ(u+q) ≠ ∅, by Lemma 3.9, DZ(u) ≠ ∅. Assume that DZ(u) = u1+⋯+uk, kn. By using Lemma 3.4 again, BpDZ(u) ≈ DZ(u)+q. It follows from Lemma 3.3 that C(DZ(u)) = c(q). By Lemma 3.4, there exists q1 in X+ with rp+1(q1) = rp+1(q) (and so ri(q1) = ri(q) for all ip − 1) and c(q1) ⊆ C(DZ(u)) such that DZ(u) ≈ DZ(u)+q1 is satisfied in $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Thus, it is easy to check that $DZ(u)≈DZ(u)+q1≈DZ(u)+q1+u1p⋯ukpq1 (by(9))$

also is satisfied in $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Notice that $\begin{array}{}{u}_{1}^{p}\cdots {u}_{k}^{p}{q}_{1}\approx q\end{array}$ is satisfied in SAp+1. Thus DZ(u) ≈ DZ(u)+q is satisfied in $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$ and so uu+q is also satisfied in $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$. This shows that HSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) = $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$. □

Suppose that S$\begin{array}{}\stackrel{^}{{\mathbf{B}\mathbf{i}}_{p}}\end{array}$. Then S$\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ and E(S) ∈ Bi. This means that (S, ⋅) is a semilattice of abelian groups which belong to Ap+1 and so by [22, Exercise IV.2.16 (v), p177], S$\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$. It follows that $\begin{array}{}\stackrel{^}{{\mathbf{B}\mathbf{i}}_{p}}\end{array}$$\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Thus, $\begin{array}{}\stackrel{^}{{\mathbf{B}\mathbf{i}}_{p}}\end{array}$ = $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$ since $\begin{array}{}{\mathbf{S}\mathbf{A}}_{p+1}^{\circ }\end{array}$$\begin{array}{}\stackrel{^}{{\mathbf{B}\mathbf{i}}_{p}}\end{array}$ is clear.

#### Lemma 3.11

Let V ∈ [Bi, Sr(2, 1)]. Then VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) = $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$.

#### Proof

Let V ∈ [Bi, Sr(2, 1)]. Since it is easily seen that VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) ⊆ $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$, we only need to show that $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$). Let uu+q be an AI-identity which is satisfied in VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$), where u = u1+⋯ +un, ui, qX+, in. In the following we show that this identity is derivable from the identities determining $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$.

Let Z = C(u) ∖ c(q). Since DZ(u+q) ≠ ∅, by Lemma 3.9, DZ(u) ≠ ∅. Assume that DZ(u) = u1+⋯ +uk, kn. By Lemma 3.9, BpDZ(u) ≈ DZ(u)+q and so ⋃ikc(ui) = c(q). By Lemma 3.4, there exists q1 in X+ with rp+1(q1) = rp+1(q) (and so ri(q1) = ri(q) for all ip − 1) and c(q) ⊆ c(q1) ⊆ ⋃ikc(ui) such that DZ(u) ≈ DZ(u)+q1 is satisfied in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Therefore, uu+q1 is also satisfied in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Proceeding as in the proof of Lemma 3.6, we have that uu+qpq1qp is derivable from the identities determining $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$. Notice that $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$qpq1qpq. Thus uu+q is derivable from the identities determining $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$. Thus, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) and so $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$ = VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$). □

#### Lemma 3.12

The subvariety of $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ determined by (17) satisfies the identity $xypzp+xpz≈xpypz+xzp.$(18)

#### Proof

We only need to show that identity (18) is derivable from (17) and the identities determining $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. On one hand, we have $xpypzp+xp−1z≈xpypzp+xp−1z+xp−1(xpypzp)pz+(xp−1z)pxpypzp(xp−1z)p(by(8),(9))≈xpypzp+xp−1z+xp−1ypz+(xp−1z)pyp(xp−1z)p(by(1),(5),(7))≈xpypzp+xp−1z+xp−1ypz+(xp−1z)pyp(xp−1z)p +(xp−1z)((xp−1z)pyp(xp−1z)p)p+((xp−1z)pyp(xp−1z)p)p(xp−1z) +(xp−1z)(xpypzp)p−1(xp−1z)(by(9))≈xpypzp+xp−1z+xp−1ypz+(xp−1z)pyp(xp−1z)p +(xp−1z)yp(xp−1z)p+(xp−1z)pyp(xp−1z)+(xp−1z)yp(xp−1z)(by(1),(5),(7))≈xpypzp+xp−1ypz+xp−1z +(xp−1z+(xp−1z)p)yp(xp−1z+(xp−1z)p)≈xpypzp+xp−1ypz+(xp−1z)p +(xp−1z+(xp−1z)p)yp(xp−1z+(xp−1z)p).(by(17))$

This derives the identity $xypzp+xpz≈x(xpypzp+xp−1z)≈x(xpypzp+xp−1z+xp−1ypz+(xp−1z)p).$(19)

On the other hand, we also have $xp−1ypz+xpzp≈xp−1ypz+xpzp+xp(xp−1ypz)pzp(by(8))≈xp−1ypz+xpzp+xpypzp(by(1),(5))≈xp−1ypz+(xp−1z)p+xpypzp+(xp−1z)pxpypzp(xp−1z)p(by(1),(5),(9))≈xp−1ypz+(xp−1z)p+xpypzp+(xp−1z)pyp(xp−1z)p(by(1),(5),(7))≈xp−1ypz+(xp−1z)p+xp−1((xp−1z)p)pyp((xp−1z)p)pz +xpypzp+(xp−1z)pyp(xp−1z)p(by(8))≈xp−1ypz+(xp−1z)p+xp−1zpypxpz+xpypzp +(xp−1z)pyp(xp−1z)p(by(1),(5)≈xp−1ypz+(xp−1z)p+xp−1zpypxpz+xp−1zypxpzp +xpypzp+(xp−1z)pyp(xp−1z)p(by(6))≈xp−1ypz+(xp−1z)p+xp−1zpypxpz+(xp−1z)yp(xp−1z)p +xpypzp+(xp−1z)pyp(xp−1z)p(by(1),(5))≈xp−1ypz+(xp−1z)p+xp−1zpypxpz+(xp−1z)yp(xp−1z)p +(xp−1z)pyp(xp−1z)+xpypzp+(xp−1z)pyp(xp−1z)p(by(6))≈xp−1ypz+(xp−1z)p+xp−1zpypxpz+(xp−1z)yp(xp−1z)p +(xp−1z)pyp(xp−1z)+xpypzp+(xp−1z)pyp(xp−1z)p +(xp−1z)yp(xp−1z)p(xpypzp)p−1(xp−1z)pyp(xp−1z)(by(9))≈xp−1ypz+(xp−1z)p+xp−1zpypxpz+(xp−1z)yp(xp−1z)p +(xp−1z)pyp(xp−1z)+(xp−1z)yp(xp−1z) +xpypzp+(xp−1z)pyp(xp−1z)p(by(1),(5),(7))≈xp−1ypz+(xp−1z)p+xp−1zpypxpz+xpypzp +(xp−1z+(xp−1z)p)yp(xp−1z+(xp−1z)p)≈xp−1ypz+xp−1z+xp−1zpypxpz+xpypzp +(xp−1z+(xp−1z)p)yp(xp−1z+(xp−1z)p).(by(17))$

This deduces the identity $xpypz+xzp≈x(xp−1ypz+xpypzp)≈x(xp−1ypz+xpypzp+xp−1z+(xp−1z)p).$(20)

This shows that the subvariety of $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$ determined by (17) satisfies identity (18). □

#### Lemma 3.13

Let V ∈ 𝓛(Sr(2, 1)) such that DMV. Then VHSP(Bp) is the subvariety of [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$] determined by the identity (18).

#### Proof

It is easily seen that VHSP(Bp) ⊆ $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$. Since both V and Bp satisfy (18), VHSP(Bp) also satisfies this identity. It remains to show that every AI-identity satisfied in VHSP(Bp) is derivable from (18) and the identities determining $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$.

Let uu+q be an AI -identity which is satisfied in VHSP(Bp), where u = u1+⋯ +un, ui, qX+, in. Since D2uu+q, there exists ui such that c(ui) ⊆ c(q), where in. Since Bpuu+q, by Lemma 3.4 there exists q1 in X+ with rp+1(q1) = rp+1(q) (and so ri(q1) = ri(q) for all ip − 1) and c(q) ⊆ c(q1) such that uu+q1 is satisfied in $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$. Proceeding as in the proof of Lemma 3.6, $Vp^⊨u≈u+qpq1qp.$

Similarly, $Vp^⊨u≈u+qpuiqp.$

Thus the following are derivable from identity (18) and the identities which hold in $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$: $u≈u+qpq1qp+qpuiqp≈u+qq1pqp+qpuiqp≈u+qq1p(uiqp)p+qpuiqp (since Vp^⊨qpq1qp≈qq1p(uiqp)p)≈u+q(uiqp)p+qpq1puiqp (by (18))≈u+q+qpq1puiqp.$

Hence, uu+q is derivable from (18) and the identities determining $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$. □

#### Theorem 3.14

Let V ∈ [Bi, Sr(2, 1)]. Then the interval [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$] of 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$) consists of the three varieties V, VHSP(Bp) and VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$).

#### Proof

By Lemma 3.11, VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) = $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$ and so VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) ∈ [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$]. Since $\begin{array}{}{B}_{p}^{0}\end{array}$VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) and $\begin{array}{}{B}_{p}^{0}\end{array}$VHSP(Bp), VHSP(Bp) ≠ VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$). Further, it is easy to see that V, VHSP(Bp) and VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$) are different members of [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$]. Suppose that W ∈ [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$] and WV. Then it follows from Lemma 3.1 that VHSP(Bp) ⊆ W. If W is a proper subvariety of VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$), then, by Lemma 3.8 and 3.12, W satisfies identity (18). It follows from Lemma 3.13 that VHSP(Bp) = W. □

#### Theorem 3.15

Let V ∈ [DM, KSr(2, 1)]. Then the interval [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$] consists of the two varieties V and VHSP(Bp).

#### Proof

Since KSr(2, 1) ⊨ (13), it follows immediately that $\begin{array}{}\stackrel{^}{\mathbf{K}\cap \mathbf{S}\mathbf{r}\left(2,1{\right)}_{p}}\end{array}$ satisfies the identity $xp+xpypxp+xpypzpypxp≈xp+xpypzpypxp.$(21)

Notice that $\begin{array}{}{B}_{p}^{0}\end{array}$(21). Then, by Lemma 3.8, $\begin{array}{}\stackrel{^}{\mathbf{K}\cap \mathbf{S}\mathbf{r}\left(2,1{\right)}_{p}}\end{array}$(17). From Lemma 3.12 we have that $\begin{array}{}\stackrel{^}{\mathbf{K}\cap \mathbf{S}\mathbf{r}\left(2,1{\right)}_{p}}\end{array}$(18). Suppose now that V ∈ [DM, KSr(2, 1)]. Then $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$(18). Thus, by Lemma 3.13, VHSP(Bp) = $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$. Thus, by Lemma 3.1, the result holds. □

#### Theorem 3.16

Each member of 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$) is finitely based and finitely generated.

#### Proof

Since each member of 𝓛(Sr(2, 1)) is finitely based and finitely generated, it follows from Lemmas 3.6, 3.11, 3.13, Theorems 3.5, 3.7, 3.14 and 3.15 that this is the case for each member of 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$), too. □

#### Theorem 3.17

𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$) is a 179-element distributive lattice.

#### Proof

Notice that both [T, NPSr(2, 1)] and [M, PSr(2, 1)] have 4 elements, [D, NSr(2, 1)] has 9 elements, [DM, KSr(2, 1)] has 25 elements, and [Bi, Sr(2, 1)] has 36 elements (see [2, Section 4]). By Theorems 3.5, 3.7, 3.14 and 3.15, we have that 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$) has 179 elements.

Assume that V1, V2, V3 ∈ 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$) such that V1V2 = V1V3 and V1V2 = V1V3. Then we have that φ(V1) ∨ φ(V2 = φ(V1) ∨ φ(V3) and φ(V1) ∧ φ(V2) = φ(V1) ∧ φ(V3) and so φ(V2) = φ(V3) since 𝓛(Sr(2, 1)) is distributive. Let V denote the variety φ(V2). Then V2 and V3 are members of [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$]. If V ∈ [T, NPSr(2, 1)] ∪ [D, NSr(2, 1)]. Then, by Theorem 3.5, V2 = V3. If V ∈ [M, PSr(2, 1)] ∪ [DM, KSr(2, 1)]. Then, by Theorem 3.7 and 3.15, we have that [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$] = {V, VHSP(Bp)}. Suppose that V2V3. Then V1V2 = V1V3 and V1V2 = V1V3 can not hold at the same time, by Lemma 3.1. This implies that V2 = V3. If V ∈ [Bi, Sr(2, 1)]. Then, by Theorem 3.14, we have that [V, $\begin{array}{}\stackrel{^}{{\mathbf{V}}_{p}}\end{array}$] = {V, VHSP(Bp), VHSP( $\begin{array}{}{B}_{p}^{0}\end{array}$)}. Suppose that V2V3. Then V1V2 = V1V3 and V1V2 = V1V3 can not hold at the same time, by Lemma 3.1 and 3.8. Thus, V2 = V3. This shows that 𝓛( $\begin{array}{}{\mathbf{R}\mathbf{O}\mathbf{B}\mathbf{A}}_{p+1}^{\circ }\end{array}$) is a distributive lattice. □

## Acknowledgement

This paper is supported by Natural Science Foundation of China (11571278, 11701449) and Scientific and Technological Research Program of Chongqing Municipal Education Commission (KJ1600930).

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Accepted: 2018-06-07

Published Online: 2018-08-16

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 913–923, ISSN (Online) 2391-5455,

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