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     {\huge СИБИРСКИЕ \ ЭЛЕКТРОННЫЕ} \\[2mm]  %Headings in Russian
     {\huge МАТЕМАТИЧЕСКИЕ ИЗВЕСТИЯ} \\[2mm]
     {\large Siberian Electronic Mathematical Reports} \\[1mm]
     {\LARGE\tt{http://semr.math.nsc.ru}}\\[0.5mm]
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     {\flushleft\it Том 19, стр. 144--144 (2022) \hspace{\fill}{\rm\small УДК \udcs}} %Volume, pages in Russian,
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         {\rm\small DOI~10.17377/semi.2015.12.xxx}\hphantom{aaaaaaaaaaaaaaaaaaaaaaaa}{\rm\small MSC\ \ \mscs }
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\title[The quasivariety $\mathbf{S}\mathbf{P}(L_6)$. I]{The quasivariety $\mathbf{S}\mathbf{P}(L_6)$. I. An equational basis}
\author{A.~O. Basheyeva}
\address{Aynur Orynbasarovna Basheyeva
\newline\hphantom{iii} %${}^{\rm (a)}$ 
L.~N. Gumilev Eurasian National University,
\newline\hphantom{iii} Kazhymukan str., 13,
\newline\hphantom{iii} 
010008, Nur-Sultan, Kazakhstan}%
\email{basheeva@mail.ru}%
\author{M.~V. Schwidefsky}
\address{Marina Vladimirovna Schwidefsky
\newline\hphantom{iii} %${}^{\rm (a)}$ 
Sobolev Institute of Mathematics SB RAS,
\newline\hphantom{iii} Acad. Koptyug ave., 4,
\newline\hphantom{iii} 
630090, Novosibirsk, Russia}%
\email{semenova@math.nsc.ru}%
\author{K.~D. Sultankulov}
\address{Kuanysh Dauletbekovich Sultankulov
\newline\hphantom{iii} %${}^{\rm (a)}$ 
L.~N. Gumilev Eurasian National University,
\newline\hphantom{iii} Kazhymukan str., 13,
\newline\hphantom{iii} 
010008, Nur-Sultan, Kazakhstan}%
\email{kuanysh.sultankulov@edu.kz}%

\thanks{\sc Basheyeva, A.~O., Schwidefsky, M.~V., Sultankulov, K.~D., The quasivariety $\mathbf{S}\mathbf{P}(L_6)$. I. An equational basis}
\thanks{\copyright \ 2022 Basheyeva, A.~O., Schwidefsky, M.~V., Sultankulov, K.~D}
\thanks{The research was carried out in the framework of the state contract of Sobolev Institute of Mathematics SB RAS,
project no. FWNF-2022-0012. The results of Section 4 were obtained under the support of the Russian Science Foundation, project number 22-21-00104.
The first author was supported by the Science Committee of the Ministry of Science and Higher Education of the Republic of Kazakhstan (project no. AP13268735).}
\thanks{\it Received  March 20, 2022, published  ????.}%


\semrtop \vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } We prove that the quasivariety $\mathbf{S}\mathbf{P}(L_6)$ is a variety and find an equational basis for this variety.
\medskip

\noindent{\bf Keywords:} lattice, quasivariety, variety, poset.
 \end{quote}
}

\section{Introduction}

\noindent
In the present paper, we consider the finite lattice $L_6$, see Figure \ref{F:0}, which is isomorphic to the suborder lattice of a three-element chain.
Suborder lattices were in the focus in a number of articles as they provide a convenient tool for proving certain embeddability results.

By a theorem by D. Bredikhin and B. Schein \cite{BS}, suborder lattices are \emph{lattice universal};
that is, each lattice is embeddable into a suitable suborder lattice.
By a theorem of B. Siv\'ak \cite{Si}, a lattice $L$ is embeddable into the suborder lattice of a finite partial order if and only if $L$ is finite and lower bounded
in the sense of R. McKenzie \cite{McK}.
Suborder lattices were used for embedding lattices into the subsemigroup lattices in V.\,B. Repnitski\v{\i} \cite{R1,R2} as well as in \cite{S3}.

Suborder lattices were also studied in papers \cite{S99,S05}.
A general construction to embed an arbitrary lattice into a suitable suborder lattice was suggested in \cite{S99}.
Based on this construction, it was shown in \cite{S05} that for arbitrary $n<\omega$, the class $\mathbf{SO}_n$ of lattices embeddable into suborder lattices of
posets of length at most $n$ forms a finitely based variety.
An equational basis for this variety was found in \cite{S05}.
There are still a number of unsolved problems which concern suborder lattices.
In particular, Question 2 in \cite{S05} asks if the quasivariety generated by a finite suborder lattice is a variety.

In this paper, we give a positive answer to this question  in a particular case.
More specifically, we prove that the quasivariety $\mathbf{Q}(L_6)$ generated by the lattice $L_6$ is a finitely based variety and find a finite basis
for this variety. The method we use was developed in \cite{N5}. In a subsequent article, the main result of this paper, Theorem \ref{5-7}, will be applied
for proving a duality result for the quasivariety $\mathbf{Q}(L_6)$.

\section{Definitions and auxiliary results}

\noindent
We assume all classes to be \emph{abstract}; that is, closed under taking isomorphic copies of structures.

\subsection{(Quasi)varieties}
A \emph{quasi-identity} is a universal Horn sentence of the form
\[
\forall\bar{x}\ A_0(\overline{x})\ \&\ \ldots\ \&\ A_n(\overline{x})\ \longrightarrow\ A(\overline{x}),
\]
where $n<\omega$ and $A_0(\overline{x})$, \ldots, $A_n(\overline{x})$, $A(\overline{x})$ are atomic formulas of a fixed type.
An \emph{identity} is a sentence of the form
\[
\forall\bar{x}\ A(\overline{x}),%\ \&\ \ldots\ \&\ A_n(\overline{x}),
\]
%where $n<\omega$ and $A_0(\overline{x})$, \ldots, $A_n(\overline{x})$ are atomic formulas of a fixed type.
where $A(\overline{x})$ is an atomic formula of a fixed type.
A \emph{quasivariety} is the class $\operatorname{Mod}(\Sigma)$ of models for a set $\Sigma$ of quasi-identities.
A \emph{variety} is the class $\operatorname{Mod}(\Sigma)$ of models for a set $\Sigma$ of identities. In this case,
$\Sigma$ is called a \emph{quasi-equational basis} [an \emph{equational basis}, respectively] of $\mathbf{K}$.
It is clear that each variety is a quasi-variety.

For a type $\sigma$, let $\mathbf{K}(\sigma)$ denote the class of all structures of type $\sigma$.
For an arbitrary class $\mathbf{K}\subseteq\mathbf{K}(\sigma)$ of structures, let $\mathbf{S}(\mathbf{K})$ denote the class of structures
from $\mathbf{K}(\sigma)$ embeddable into structures from the class $\mathbf{K}$,
and let $\mathbf{P}(\mathbf{K})$ denote the class of structures from $\mathbf{K}(\sigma)$ isomorphic to Cartesian products of structures from $\mathbf{K}$.
Whenever $\mathbf{K}$ contains only one structure $\mathcal{A}$ (up to isomorphism),
we write $\mathbf{O}(\mathcal{A})$ instead of $\mathbf{O}(\{\mathcal{A}\})$ for a class operator $\mathbf{O}$.
Let $\mathbf{Q}(\mathbf{K})$ denote the smallest quasivariety containing $\mathbf{K}$.
It is well known \cite{M} that for a finite structure $\mathcal{A}$, the class $\mathbf{S}\mathbf{P}(\mathcal{A})$ is a quasivariety.
Thus, $\mathbf{Q}(\mathcal{A})=\mathbf{S}\mathbf{P}(\mathcal{A})$ for each finite structure $\mathcal{A}$.

For all the notions concerning (quasi)varieties of structures which are not defined here, we refer to A.\,I. Maltsev \cite{M}, V.\,A. Gorbunov \cite{G}, and
J. Hyndman and J.\,B. Nation \cite{HN}.

\subsection{General lattices}
Most of the following definitions correspond to R. Freese, J. Je\v{z}ek, and J.\,B. Nation \cite{FJN}.

Let $L$ be a lattice. For arbitrary two sets $A,B\subseteq L$, we say that $A$ \emph{refines} $B$ and write $A\ll B$ if for each $a\in A$, there is $b\in B$ such that
$a\leq b$. If $x\in L$, then $A$ is a \emph{join cover} of $x$ if $\bigvee A$ exists and $x\leq\bigvee A$;
we also call the inequality $x\leq\bigvee A$ a \emph{join cover} in this case.
A join cover $x\leq\bigvee A$ is \emph{nontrivial} if
$x\nleq a$ for all $a\in A$; $x\leq\bigvee A$ is \emph{finite} if
the set $A$ is finite.
A join cover $x\leq\bigvee A$ is \emph{irredundant} if
$x\nleq\bigvee B$ for all proper subsets $B\subset A$.
A join cover $x\leq\bigvee A$ is \emph{minimal} if
$A\subseteq B$ for each join cover $x\leq\bigvee B$ such that $B\ll A$.
The lattice $L$ has the \emph{complete minimal join cover refinement property} $(\mathrm{CR})_X$ for a set $X\subseteq L$ if for each nontrivial join cover
$A$ of an element $x\in X$, there is a minimal nontrivial join cover $B$ of $x$ such that $B\ll A$.

A non-zero element $a$ of a lattice $L$ is said to be \emph{join-irreducible} if $a=b\vee c$ implies that $a\in\{b,c\}$ for all $b,c\in L$;
$a$ is said to be \emph{completely join-irreducible} if $a=\bigvee B$ implies that $a\in B$ for all nonempty sets $B\subseteq L$.
Let $\mathrm{J}(L)$ denote the set of all join-irreducible elements in $L$ and let
$\mathrm{CJ}(L)$ denote the set of all completely join-irreducible elements in $L$.

\begin{definition}\cite{N5}
For a set $J\subseteq\mathrm{J}(L)$, we say that $L$ is a $J$-\emph{lattice} if $L$ possesses the following properties:
\begin{enumerate}
\item
for each element $a\in L$, there is a subset $J_a\subseteq J$
with $a=\bigvee J_a$;
\item
for each element $a\in J$ and each nontrivial join cover $a\leq a_0\vee\ldots\vee a_n$ with $n<\omega$ and $a_0,\ldots,a_n\in L$, there is a finite set $F\subseteq J$
such that $a\leq\bigvee F$ is a minimal join cover and $F\ll\{a_0,\ldots,a_n\}$.
\end{enumerate}
We say that $L$ is a $CJ$-\emph{lattice} if $L$ possesses the following properties:
\begin{enumerate}
\item
for each element $a\in L$, there is a subset $J_a\subseteq\mathrm{CJ}(L)$
with $a=\bigvee J_a$;
\item
$L$ has the property $(\mathrm{CR})_{\mathrm{CJ}(L)}$.
\end{enumerate}
\end{definition}

\noindent
%It follows from the definition above that each upper continuous $CJ$-lattice is a $J$-lattice for $J=\mathrm{CJ}(L)$.
In what follows, we consider the following identity of $n$-\emph{distributivity}, where $1<n<\omega$, which we denote by $(\mathrm{D}_n)$:
\[
x\wedge(y_0\vee y_1\vee\ldots\vee y_n)=\bigvee_{i\leqslant n}\bigl[x\wedge\bigvee_{j\ne i}y_j\bigr].
\]
This identity was introduced and considered by A.\,P. Huhn \cite{H}. It is clear that $(\mathrm{D}_1)$ is just the identity of distributivity.

The following lemma is folklore and straightforward to prove, see for example J.\,B. Nation \cite{N}.

\begin{lem}\label{D2}
Let $n>0$, let $L$ be a lattice, let a set $J\subseteq\mathrm{J}(L)$ be such that for each element $a\in L$, there is a subset $J_a\subseteq J$
with $a=\bigvee J_a$.
The following conditions are equivalent.
\begin{enumerate}
\item
$(\mathrm{D}_n)$ holds in $L$.
\item
If $a\leq b_0\vee b_1\vee\ldots\vee b_n$ for some $a\in J$ and some $b_0,b_1,\ldots,b_n\in L$, then there is $i\leqslant n$ such that $a\leq\bigvee_{j\ne i}b_j$.
\end{enumerate}
\end{lem}

\begin{cor}\label{cD2}
Let $n>1$ and let $L$ be a $J$-lattice for some set $J\subseteq\mathrm{J}(L)$.
The following conditions are equivalent.
\begin{enumerate}
\item
$(\mathrm{D}_n)$ holds in $L$.
\item
If $a\leq b_0\vee\ldots\vee b_m$ is a minimal nontrivial join cover for some $a,b_0,\ldots,b_m\in J$ then $0<m<n$.
\end{enumerate}
\end{cor}

\begin{prop}\label{ba}\cite{N5}
Let $L$ be a complete dually algebraic lattice. Then the following statements hold.
\begin{enumerate}
\item
If $L$ is $n$-distributive then $L$ is a $\mathrm{J}(L)$-lattice.
\item
If $L$ is in addition algebraic then $L$ is a $CJ$-lattice.
\end{enumerate}
\end{prop}

\noindent
Following \cite{N5}, we denote the next identity by $(\mathrm{C})$:
\begin{align*}
x\wedge(y_0\vee y_1)\wedge(z_0\vee z_1)=&\bigvee_{i<2}\bigl[x\wedge y_i\wedge(z_0\vee z_1)\bigr]\ \vee\ \bigvee_{i<2}
\bigl[x\wedge z_i\wedge(y_0\vee y_1)\bigr]\ \vee\\
&\vee\ \bigvee_{i<2}\bigl[x\wedge\bigl((y_0\wedge z_i)\vee(y_1\wedge z_{1-i})\bigr)\bigr].
\end{align*}

\noindent
The next four statements were established in \cite{N5}. Since \cite{N5} does not contain complete proofs, we present here sketches of proofs
of Lemma \ref{C} and Lemma \ref{N51} for the sake of completeness.
We emphasize that these proofs are due to the authors of \cite{N5}.

\begin{lem}\label{C}\cite{N5}
Let $L$ be a lattice, let a set $J\subseteq\mathrm{J}(L)$ be such that for each element $a\in L$, there is a subset $J_a\subseteq J$
with $a=\bigvee J_a$.
The following conditions are equivalent.
\begin{enumerate}
\item
$(\mathrm{C})$ holds in $L$.
\item
If $a\leq a_0\vee a_1$ and $a\leq b_0\vee b_1$ are nontrivial join covers for some $a\in J$ and some $a_0,a_1,b_0,b_1\in L$,
then there are $c_0,c_1\in L$ such that $a\leq c_0\vee c_1$, $\{c_0,c_1\}\ll\{a_0,a_1\}$, and $\{c_0,c_1\}\ll\{b_0,b_1\}$.
\end{enumerate}
\end{lem}

\begin{proof}
To prove that (1) implies (2), we
assume that the assumptions of (2) hold. Since $(\mathrm{C})$ holds in $L$, we have
\begin{align*}
a=a\wedge(a_0\vee a_1)\wedge(b_0\vee b_1)=&\bigvee_{i<2}\bigl[a\wedge a_i\wedge(b_0\vee b_1)\bigr]\ \vee\ \bigvee_{i<2}
\bigl[a\wedge b_i\wedge(a_0\vee a_1)\bigr]\ \vee\\
&\vee\ \bigvee_{i<2}\bigl[a\wedge\bigl((a_0\wedge b_i)\vee(a_1\wedge b_{1-i})\bigr)\bigr].
\end{align*}
As $a$ is a join-irreducible element, it equals one of the joinands on the right-hand side of the equality above.
This implies that the conclusion of (2) also holds.

To prove that (2) implies (1), we note that the inequality
\begin{align*}
\bigvee_{i<2}\bigl[x\wedge y_i\wedge(z_0\vee z_1)\bigr]\ &\vee\ \bigvee_{i<2}\bigl[x\wedge z_i\wedge(y_0\vee y_1)\bigr]\ \vee\\
&\vee\ \bigvee_{i<2}\bigl[x\wedge\bigl((y_0\wedge z_i)\vee(y_1\wedge z_{1-i})\bigr)\bigr]\leq\\
&\leq x\wedge(y_0\vee y_1)\wedge(z_0\vee z_1)
\end{align*}
holds in each lattice. Therefore, in order to prove that $(\mathrm{C})$ holds in $L$, we have to establish that the reverse inequality holds in $L$.
To this end, choose arbitrary elements $u,a_0,a_1,b_0,b_1\in L$. We put
\[
w=\bigvee_{i<2}\bigl[u\wedge a_i\wedge(b_0\vee b_1)\bigr]\ \vee\ \bigvee_{i<2}\bigl[u\wedge b_i\wedge(a_0\vee a_1)\bigr]\ \vee
\ \bigvee_{i<2}\bigl[u\wedge\bigl((a_0\wedge b_i)\vee(a_1\wedge b_{1-i})\bigr)\bigr].
\]
We have to show that
\[
u\wedge(a_0\vee a_1)\wedge(b_0\vee b_1)\leq w.
\]
According to our assumption about $L$, it suffices to show that each element $a\in J$ which is below $u\wedge(a_0\vee a_1)\wedge(b_0\vee b_1)$
is also below $w$.
But this follows from (2).
\end{proof}

\begin{cor}\label{cC}\cite{N5}
Let $L$ be a $2$-distributive $J$-lattice for some set $J\subseteq\mathrm{J}(L)$.
The following conditions are equivalent.
\begin{enumerate}
\item
$(\mathrm{C})$ holds in $L$.
\item
If $a\leq a_0\vee a_1$ and $a\leq b_0\vee b_1$ are minimal join covers for some elements $a,a_0,a_1,b_0,b_1\in J$, then $\{a_0,a_1\}=\{b_0,b_1\}$.
\end{enumerate}
\end{cor}

\noindent
As in \cite{N5}, we denote the following identity by $(\mathrm{N}_5^1)$:
\[
x\wedge\bigl[\bigl(y_0\wedge(z_0\vee z_1)\bigr)\vee y_1\bigr]=\bigl[x\wedge y_0\wedge(z_0\vee z_1)\bigr]\vee\bigl[x\wedge y_1\bigr]\vee
\bigvee_{i<2}\bigl[x\wedge\bigl((y_0\wedge z_i)\vee y_1\bigr)\bigr].
\]

\begin{lem}\label{N51}\cite{N5}
Let $L$ be a lattice, let a set $J\subseteq\mathrm{J}(L)$ be such that for each element $a\in L$, there is a subset $J_a\subseteq J$
with $a=\bigvee J_a$.
The following conditions are equivalent.
\begin{enumerate}
\item
$(\mathrm{N}_5^1)$ holds in $L$.
\item
If $a\leq a_0\vee a_1$ is a nontrivial join cover and $a_0\leq b_0\vee b_1$ for some $a\in J$ and some $a_0,a_1,b_0,b_1\in L$,
then $a\leq (a_0\wedge b_i)\vee a_1$ for some $i<2$.
\end{enumerate}
\end{lem}

\begin{proof}
To prove that (1) implies (2), we assume that
$a\leq a_0\vee a_1$ for some $a\in J$ and some $a_0,a_1\in L$ and that $a_0\leq b_0\vee b_1$ for some $b_0,b_1\in L$.
Since $(\mathrm{N}_5^1)$ holds in $L$, we have
\[
a=a\wedge\bigl[\bigl(a_0\wedge(b_0\vee b_1)\bigr)\vee a_1\bigr]=
\bigl[a\wedge a_0\wedge(b_0\vee b_1)\bigr]\vee\bigl[a\wedge a_1\bigr]\vee
\bigvee_{i<2}\bigl[a\wedge\bigl((a_0\wedge b_i)\vee a_1\bigr)\bigr].
\]
As $a$ is a join-irreducible element, it equals one of the joinands on the right-hand side of the equality above.
This implies that the conclusion of (2) also holds.

To prove that (2) implies (1), we again notice that
\[
\bigl[x\wedge y_0\wedge(z_0\vee z_1)\bigr]\vee\bigl[x\wedge y_1\bigr]\vee
\bigvee_{i<2}\bigl[x\wedge\bigl((y_0\wedge z_i)\vee y_1\bigr)\bigr]\leq
x\wedge\bigl[\bigl(y_0\wedge(z_0\vee z_1)\bigr)\vee y_1\bigr]
\]
holds in each lattice.
Therefore, in order to prove that $(\mathrm{N}_5^1)$ holds in $L$, we have to establish that the reverse inequality holds in $L$.
In order to do this, we choose arbitrary elements $u,a_0,a_1,b_0,b_1\in L$ and put
\[
w=\bigl[u\wedge a_0\wedge(b_0\vee b_1)\bigr]\vee\bigl[u\wedge a_1\bigr]\vee
\bigvee_{i<2}\bigl[u\wedge\bigl((a_0\wedge b_i)\vee a_1\bigr)\bigr].
\]
We have to show that
\[
u\wedge\bigl[\bigl(a_0\wedge(b_0\vee b_1)\bigr)\vee a_1\bigr]\leq w.
\]
According to our assumption about $L$, it suffices to show that each element $a\in J$
which is below $u\wedge\bigl[\bigl(a_0\wedge(b_0\vee b_1)\bigr)\vee a_1\bigr]$
is also below $w$.
But this conclusion follows from (2).
\end{proof}

\begin{cor}\label{cN51}\cite{N5}
Let $L$ be a $2$-distributive $J$-lattice for some set $J\subseteq\mathrm{J}(L)$.
The following conditions are equivalent.
\begin{enumerate}
\item
$(\mathrm{N}_5^1)$ holds in $L$.
\item
If $a\leq a_0\vee a_1$ is a minimal join cover for some $a,a_0,a_1\in J$, then $a_0$ and $a_1$ are join-prime elements.
\end{enumerate}
\end{cor}

\noindent
We use the following notation, cf. R. Freese, J, Je\v{z}ek, and J.\,B. Nation \cite[Lemma 2.33]{FJN}.

\begin{definition}\label{D:cl}
Consider a $J$-lattice $L$, where $J\subseteq\mathrm{J}(L)$.
For an element $x\in J$, we put
\[
\mathfrak{M}(x)=\Bigl\{A\subseteq J\mid 1<|A|<\omega,\ x\leq\bigvee A\ \text{is a minimal nontrivial join cover of}\ x\Bigr\}.
\]
For a set $S\subseteq J$, we put
\begin{align*}
&S^{[0]}=S;\\
&S^{[{n+1}]}=\bigcup\bigl\{A\in\mathfrak{M}(x)\mid x\in S^{[n]}\bigr\},\quad n<\omega;\\
&\langle S\rangle_\mathfrak{M}=\bigcup_{i<\omega}S^{[i]}.
\end{align*}
For an element $x\in J$, we write $\langle x\rangle_\mathfrak{M}$ instead of $\langle\{x\}\rangle_\mathfrak{M}$.
It is straightforward that $\langle S\rangle_\mathfrak{M}=\bigcup_{x\in S}\langle x\rangle_\mathfrak{M}\subseteq J$,
whence $\langle\ \rangle_\mathfrak{M}$ is an algebraic closure operator on $J$.
A set $S\subseteq J$ is $\mathfrak{M}$-\emph{closed}, if $S=\langle S\rangle_\mathfrak{M}$.
\end{definition}

\noindent
It is clear that $\mathfrak{M}(x)=\varnothing$ whence $\langle x\rangle_\mathfrak{M}=\{x\}$ for each element $x\in J$ which is join-prime.

For a set $S\subseteq J$, we define a binary relation $\Gamma_S$ on $L$ as follows.
If $a,b\in L$ then we put
\[
(a,b)\in\Gamma_S\quad\text{if and only if}\ S\cap \dnw a=S\cap\dnw b.
\]

\begin{lem}\label{J-0}\cite{N5}
Let $L$ be a $J$-lattice and let $A\subseteq J$.
The following statements hold.
\begin{enumerate}
\item
If $A\subseteq B$ for some $B\subseteq J$ then $\Gamma_B\subseteq\Gamma_A$.
\item
If $A=\bigcup_{i\in I}A_i$ for some $A_i\subseteq J$, $i\in I$, then $\Gamma_A=\bigcap_{i\in I}\Gamma_{A_i}$.
\item
If $A$ is an $\mathfrak{M}$-closed set then $\Gamma_A$ is a congruence on $L$.
\end{enumerate}
\end{lem}


\subsection{Suborder lattices}
Let $X$ be a set and let $R\subseteq X^2$ be a partial order on $X$; that is a reflexive, antisymmetric, and transitive binary relation.
In this case, we say that $(X;R)$ is a \emph{partially ordered set} or a \emph{poset} for short.
A subset $R^\prime\subseteq R$ is a \emph{suborder} of $R$ if the structure $(X;R^\prime)$ is also a poset.
The set $\operatorname{O}(X,R)$  of all suborders of a partial order $R$ on $X$ is a partially ordered set with respect to the relation $\subseteq$
of set-theoretic inclusion.
Obviously, $\Delta=\bigl\{(a,a)\mid a\in X\bigr\}$ is the least suborder of $R$.
Thus, $\Delta$ is the smallest element in $\operatorname{O}(X,R)$.
It is also obvious that $R$ is the largest element in $\operatorname{O}(X,R)$.
It is straightforward to check that for an arbitrary family $\{R_i\mid i\in I\}\subseteq\operatorname{O}(X,R)$,
the relation $\bigcap_{i\in I}R_i$ is also a suborder of $R$; that is,
\[
\bigwedge_{i\in I}R_i=\bigcap_{i\in I}R_i\in\operatorname{O}(X,R).
\]
Thus, $\operatorname{O}(X,R)$ is a complete lattice, where
\[
\bigvee_{i\in I}R_i=\bigl(\bigcup_{i\in I}R_i\bigr)^t
\]
and $Y^t$ denotes the transitive closure of a binary relation $Y\subseteq X^2$.

For more information on suborder lattices, we refer to D. Bredikhin and B. Schein \cite{BS}, B. Siv\'{a}k \cite{Si} as well as to \cite{S99,S05}.



\section{The lattice $L_6$}

\begin{figure}
	\begin{center}
		\setlength{\unitlength}{1mm}
		\begin{picture}(50,45)
			\put(0,7){$0$}
			\put(5,7){\circle{2}}
			\put(5,8){\line(0,1){8}}
			\put(0,17){$a$}
			\put(5,17){\circle{2}}
			\put(5,18){\line(0,1){8}}
			\put(0,27){$1$}
			\put(5,27){\circle{2}}
			\put(0,0){$M_1$}
			%%%%%%%%%%%%%%%%%%%%%
			
\put(34,1){$O$}
\put(35,7){\circle{2}}
\put(34,40){$I$}
\put(35,37){\circle{2}}
\put(20,27){$X$}
\put(25,27){\circle{2}}
\put(25.8,27.8){\line(1,1){8.4}}
\put(20,17){$A$}
\put(25,17){\circle{2}}
\put(25,18){\line(0,1){8}}
\put(25.8,16.2){\line(1,-1){8.4}}

\put(48,27){$Y$}
\put(45,27){\circle{2}}
\put(44.2,27.8){\line(-1,1){8.4}}
\put(48,17){$C$}
\put(45,17){\circle{2}}
\put(45,18){\line(0,1){8}}
\put(44.2,16.2){\line(-1,-1){8.4}}

\put(34,20){$B$}
\put(35,17){\circle{2}}
\put(35,8){\line(0,1){8}}
\put(34.2,17.8){\line(-1,1){8.4}}
\put(35.8,17.8){\line(1,1){8.4}}

			
			%%%%%%%%%%%%%%%%%%%%
			
			%\put(31,5){$0$}
			%\put(35,7){\circle{2}}
			%\put(31,28){$1$}
			%\put(35,27){\circle{2}}
			%\put(20,17){$a_1$}
			%\put(25,17){\circle{2}}
			%\put(25.8,17.8){\line(1,1){8.4}}
			%\put(25.8,16.2){\line(1,-1){8.4}}
			%\put(30,17){$a_2$}
			%\put(35,17){\circle{2}}
			%\put(35,18){\line(0,1){8}}
			%\put(35,8){\line(0,1){8}}
			
			%\put(40,17){\circle*{1}}
			%\put(42,17){\circle*{1}}
			%\put(44,17){\circle*{1}}
			
			%\put(47,17){$a_n$}
			%\put(55,17){\circle{2}}
			%\put(54,17.3){\line(-2,1){18}}
			%\put(54,16.7){\line(-2,-1){18}}
			%\put(35,0){$M_n$}
			
			%%%%%%%%%%%%%%%%%%%%
			
			%\put(81,5){$0$}
			%\put(85,7){\circle{2}}
			%\put(81,28){$1$}
			%\put(85,27){\circle{2}}
			%\put(69,17){$a_1$}
			%\put(75,17){\circle{2}}
			%\put(75.8,17.8){\line(1,1){8.4}}
			%\put(75.8,16.2){\line(1,-1){8.4}}
			%\put(80,17){$a_2$}
			%\put(85,17){\circle{2}}
			%\put(85,18){\line(0,1){8}}
			%\put(85,8){\line(0,1){8}}
			
			%\put(90,17){\circle*{1}}
			%\put(92,17){\circle*{1}}
			%\put(94,17){\circle*{1}}
			
			
			%\put(97,17){$a_n$}
			%\put(105,17){\circle{2}}
			%\put(104,17.3){\line(-2,1){18}}
			%\put(104,16.7){\line(-2,-1){18}}
			%\put(88,0){$M_\omega$}
			
			%\put(112,17){\circle*{1}}
			%\put(114,17){\circle*{1}}
			%\put(116,17){\circle*{1}}
			
		\end{picture}
	\end{center}
	\caption{ Partially ordered set $M_1$ and lattice $L_6\cong\mathrm{O}(M_1)$}\label{F:0}
\end{figure}

\begin{lem}\label{O1}
The suborder lattice $\mathrm{O}(M_1)$ is isomorphic to $L_6$.
\end{lem}

\begin{proof}
The order relation on $M_1$ is $R=\bigl\{(0,0),(0,a),(a,a),(a,1),(0,1),(1,1)\bigr\}$.
Then the suborders of $R$ are exactly the sets
	\begin{align*}
	&O=\bigl\{(0,0),(a,a),(1,1)\bigr\};\\
	&A=\bigl\{(0,a)\bigr\}\cup O;\\
	&B=\bigl\{(0,1)\bigr\}\cup O;\\
	&C=\bigl\{(a,1)\bigr\}\cup O;\\
	&X=\bigl\{(0,a),(0,1)\bigr\}\cup O;\\
	&Y=\bigl\{(a,1),(0,1)\bigr\}\cup O;\\
	&I=R=\bigl\{(0,a),(a,1),(0,1)\bigr\}\cup O. 
	\end{align*}
Then it follows that $\mathrm{O}(M_1)\cong L_6$, cf. Figure \ref{F:0}.
\end{proof}

\section{An equational basis for $\mathbf{S}\mathbf{P}(L_6)$}

\noindent
We put $\Sigma=\{(\mathrm{C}),(\mathrm{D}_2),(\mathrm{N}_5^1)\}$.

\begin{prop}\label{5-6}
Let $L$ be a dually algebraic lattice such that $L\models\Sigma$.
Then for each element $b\in\mathrm{J}(L)$ which is not join-prime, we have
\[
\langle b\rangle_\mathfrak{M}=\{a,b,c\},\ \text{where}\ b\leq a\vee c\ \text{is a minimal join cover}.
\]
Moreover, $L\in\mathbf{S}\mathbf{P}(L_6)$.
\end{prop}

\begin{proof}
According to Proposition \ref{ba}(1), $L$ is a $J$-lattice, where $J=\mathrm{J}(L)$ is the set of all join-irreducible elements of $L$.
If $b\in J$ is not join-prime, then according to Corollary \ref{cC},
$\mathfrak{M}(b)=\bigl\{\{a,c\}\bigr\}$ for some $a,c\in J$ such that $b\leq a\vee c$ is a minimal nontrivial join cover.
According to Corollary \ref{cN51}, elements $a$ and $c$ are join-prime, whence $\langle b\rangle_\mathfrak{M}=\{a,b,c\}$.

According to Lemma \ref{J-0}(3), $\Gamma_{\langle b\rangle_\mathfrak{M}}$ is a congruence on $L$ for each $b\in J$.
Since $L$ is a $J$-lattice and $J$ is an $\mathfrak{M}$-closed set, $\Gamma_J=\Delta_L$ is the least congruence on $L$, whence $L{\slash}\Gamma_J\cong L$.
As $J=\bigcup_{x\in J}\langle x\rangle_\mathfrak{M}$, we conclude by Lemma \ref{J-0}(2) that
$L\leq_s\prod_{x\in J}L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}$.
We fix an element $x\in J$.
In what follows, let $[z]$ denote the $\Gamma_{\langle x\rangle_\mathfrak{M}}$-equivalence class of an element $z\in L$.

If $x\in J$ is join-prime, then $\langle x\rangle_\mathfrak{M}=\{x\}$.
Therefore, there are only two $\Gamma_{\langle x\rangle_\mathfrak{M}}$-equivalence classes:
$[0_L]$ and $[x]$. Hence $L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}\cong 2$.
If $x\in J$ is not join-prime, then
$\langle x\rangle_\mathfrak{M}=\{x,u,v\}$, where $u,v\in J$ are join-prime and $x\leq u\vee v$ is a minimal nontrivial join cover.
This implies in particular that $u$ and $v$ are incomparable and that $x\notin\upw u\cap\upw v$.
The following three cases are therefore possible.

\Case{1}: $u\nleq x$ and $v\nleq x$. In this case, elements $x$, $u$, and $v$ are pairwise incomparable.
This implies that $x\nleq u\wedge v$, $u\nleq x\wedge v$, and $v\nleq x\wedge u$.
Therefore, $[u\wedge v]=[x\wedge u]=[x\wedge v]=[0_L]$.
If $v\leq x\vee u$, then $v\leq x$ or $v\leq u$ as $v$ is join-prime. Both cases are impossible as $\{x,u,v\}$ is an anti-chain.
This implies that $[x\vee u]$ and $[x\vee v]$ are incomparable elements in the lattice $L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}$.
Moreover, $[u\vee v]=[1_L]$ as $x\leq u\vee v$.
Therefore, for an arbitrary element $z\in L$, we have in \emph{Case 1} that one of the following cases occurs:
\begin{figure}
	\begin{center}
		\setlength{\unitlength}{1mm}
		\begin{picture}(40,45)
						
\put(21,1){$[0]$}
\put(22,7){\circle{2}}
\put(21,40){$[1]$}
\put(22,37){\circle{2}}
\put(0,27){$[x\vee u]$}
\put(12,27){\circle{2}}
\put(12.8,27.8){\line(1,1){8.4}}
\put(5,17){$[u]$}
\put(12,17){\circle{2}}
\put(12,18){\line(0,1){8}}
\put(12.8,16.2){\line(1,-1){8.4}}

\put(35,27){$[x\vee v]$}
\put(32,27){\circle{2}}
\put(31.2,27.8){\line(-1,1){8.4}}
\put(35,17){$[v]$}
\put(32,17){\circle{2}}
\put(32,18){\line(0,1){8}}
\put(31.2,16.2){\line(-1,-1){8.4}}

\put(20,21){$[x]$}
\put(22,17){\circle{2}}
\put(22,8){\line(0,1){8}}
\put(21.2,17.8){\line(-1,1){8.4}}
\put(22.8,17.8){\line(1,1){8.4}}

			
		\end{picture}
	\end{center}
	\caption{ Lattice $L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}\cong L_6$}\label{F:1}
\end{figure}

\begin{align*}
&\langle x\rangle_\mathfrak{M}\cap\dnw z=\varnothing;\quad
\langle x\rangle_\mathfrak{M}\cap\dnw z=\{x,u,v\};\\
&\langle x\rangle_\mathfrak{M}\cap\dnw z=\{x\};\quad\langle x\rangle_\mathfrak{M}\cap\dnw z=\{u\};\quad\langle x\rangle_\mathfrak{M}\cap\dnw z=\{v\};\\
&\langle x\rangle_\mathfrak{M}\cap\dnw z=\{x,u\};\quad\langle x\rangle_\mathfrak{M}\cap\dnw z=\{x,v\},
\end{align*}
see Figure \ref{F:1}.
This implies that $L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}$ has the following elements:
$[0_L]$, $[x]$, $[u]$, $[v]$, $[x\vee u]$, $[x\vee v]$, $[1_L]$. Hence $L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}\cong L_6$.

\Case{2}: $u<x$. In this case, we have $v\nleq x$, whence $x$ and $v$ are incomparable.
Moreover, $x\nleq u\wedge v$ and $u\nleq x\wedge v$, whence $[u\wedge v]=[x\wedge v]=[0_L]$.
It is also clear that $[x\vee v]=[u\vee v]=[1_L]$.
Therefore, for an arbitrary element $z\in L$, we have in \emph{Case 2} that one of the following cases occurs:
\begin{align*}
&\langle x\rangle_\mathfrak{M}\cap\dnw z=\varnothing;\quad
\langle x\rangle_\mathfrak{M}\cap\dnw z=\{x,u,v\};\\
&\langle x\rangle_\mathfrak{M}\cap\dnw z=\{x\};\quad\langle x\rangle_\mathfrak{M}\cap\dnw z=\{u\};\quad\langle x\rangle_\mathfrak{M}\cap\dnw z=\{v\},
\end{align*}
see Figure \ref{F:2}.
This implies that $L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}$ has the following elements:
$[0_L]$, $[x]$, $[u]$, $[v]$, $[1_L]$. Hence $L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}\cong N_5\leq L_6$.

\Case{3}: $v<x$. This case is symmetric to \emph{Case 2} and therefore, $L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}\cong N_5\leq L_6$.

\vskip5mm
\noindent
The above implies that $L$ is a subdirect product of lattices isomorphic either to $2$ or to $N_5$, or to $L_6$.
Since both lattices, $2$ and $N_5$, embed into $L_6$, we obtain that $L\in\mathbf{S}\mathbf{P}(L_6)$.
\begin{figure}
	\begin{center}
		\setlength{\unitlength}{1mm}
		\begin{picture}(40,45)
						
\put(21,1){$[0]$}
\put(22,7){\circle{2}}
\put(21,40){$[1]$}
\put(22,37){\circle{2}}
\put(5,27){$[x]$}
\put(12,27){\circle{2}}
\put(12.8,27.8){\line(1,1){8.4}}
\put(5,17){$[u]$}
\put(12,17){\circle{2}}
\put(12,18){\line(0,1){8}}
\put(12.8,16.2){\line(1,-1){8.4}}

\put(35,17){$[v]$}
\put(32,17){\circle{2}}
\put(32,18){\line(-1,2){9.2}}
\put(31.2,16.2){\line(-1,-1){8.4}}

		\end{picture}
	\end{center}
	\caption{ Lattice $L{\slash}\Gamma_{\langle x\rangle_\mathfrak{M}}\cong N_5$}\label{F:2}
\end{figure}
\end{proof}

\begin{cor}\label{5-6-0}
Let $L$ be a bi-algebraic lattice such that $L\models\Sigma$.
Then for each element $x\in\mathrm{CJ}(L)$ which is not join-prime, we have
\[
[x]_\mathfrak{M}=\{x,a,b\},\ \text{where}\ a,b\in\mathrm{CJ}_P(L),\ x\leq a\vee b.
\]
In particular, $L\in\mathbf{S}\mathbf{P}(L_6)$.
\end{cor}

\begin{proof}
The argument is similar to the one in the proof of Proposition \ref{5-6} and uses Proposition \ref{ba}(2).
\end{proof}

\begin{thm}\label{5-7}
The quasivariety $\mathbf{S}\mathbf{P}(L_6)$ is a variety and
$\Sigma$ forms an equational basis for $\mathbf{S}\mathbf{P}(L_6)$.
\end{thm}

\begin{proof}
Let $L\models\Sigma$ and let $F$ be the dual filter lattice of $L$. Then $F$ is dually algebraic and $F\models\Sigma$.
By Proposition \ref{5-6}, $F\in\mathbf{S}\mathbf{P}(L_6)$ whence $L\in\mathbf{S}\mathbf{P}(L_6)$.
This proves that $\operatorname{Mod}(\Sigma)\subseteq\mathbf{S}\mathbf{P}(L_6)$.
On the other hand, the lattice $L_6$ has the only nontrivial join cover $b\leq a\vee c$ of a join-irreducible element.
Thus, $L_6$ is $2$-distributive by Corollary \ref{cD2}.
Moreover, $L_6$ satisfies the condition (2) of Corollaries \ref{cC} and \ref{cN51}.
This implies that $L_6\models\Sigma$ and that $\mathbf{S}\mathbf{P}(L_6)\subseteq\operatorname{Mod}(\Sigma)$
which proves the desired statement.
\end{proof}


\noindent
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\end{document}