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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]
%     {\small Институт математики им. С.Л.Соболева СО РАН}\\[-1mm]
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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.2022.12.xxx}\hphantom{aaaaaaaaaaaaaaaaaaaaaaaa}{\rm\small MSC\ \ \mscs }
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\title[Quasivarieties generated by small suborder lattices. I]{\sc Quasivarieties generated by small suborder lattices. I. Equational bases}
\author{O.~A. Kadyrova}
\address{Olga Aleksandrovna Kadyrova
\newline\hphantom{iii} %${}^{\rm (a)}$ 
Novosibirsk State University,
\newline\hphantom{iii} Pirogova str., 1,
\newline\hphantom{iii} 
630090  Novosibirsk, Russia}%
\email{o.kadyrova@g.nsu.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{m.schwidefsky@g.nsu.ru}%

\thanks{\sc Kadyrova, O.~A., Schwidefsky, M.~V., Quasivarieties generated by small suborder lattices. I. Equational bases}
\thanks{\copyright \ 2022 Kadyrova, O.~A., Schwidefsky, M.~V.}
\thanks{The research was carried out under the support of the Russian Science Foundation, project no. 22-21-00104.}
\thanks{\it Received  August 17, 2022, published  ????.}%


\semrtop \vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } For each cardinal $\kappa>0$, the quasivariety generated by the suborder lattice of $M_\kappa$
is a finitely based variety. An equational basis for this variety is found.
\medskip

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

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Introduction}

\noindent
Suborder lattices were studied by several authors; we refer to D. Bredikhin and B. Schein \cite{BS} and to B. Siv\'ak \cite{Si} as well as to \cite{S99,S05,BSS1}.
Suborder lattices were used as a convenient tool in establishing some deep results for subsemigroup lattices
which are presented in the papers of V.\,B. Repnitski\v{\i} \cite{R1,R2};
see also \cite{S3}.

For a positive integer $n$, let $\mathbf{SO}_n$ denote the class of lattices embeddable into suborder lattices of partial orders of length at most $n$.
It was established in \cite{S05} that $\mathbf{SO}_n$ is a finitely based variety and a particular finite equational basis was found for this variety in \cite{S05}.

There are still some unsolved problems which concern suborder lattices.
For example, Question 2 in \cite{S05} asks if the quasivariety generated by a finite suborder lattice is a variety.
A positive answer to this question was given in \cite{BSS1} for the suborder lattice $\operatorname{O}(M_1)$. Moreover, it was established in \cite{BSS1} that
the quasivariety generated by $\operatorname{O}(M_1)$ is a variety and a particular finite equational basis was found for this variety.

In this paper, we extend the results from \cite{BSS1} to a more general case and consider lattices $\operatorname{O}(M_\kappa)$ for an arbitrary cardinal $\kappa>0$,
see Figure \ref{F:1}.
Specifically, we prove that the quasivariety $\mathbf{Q}\bigl(\operatorname{O}(M_\kappa)\bigr)$ generated by the
suborder lattice $\operatorname{O}(M_\kappa)$ is a finitely based variety and find a finite basis
for this variety, see Theorem \ref{5-7}, Theorem \ref{9}, and Corollaries \ref{10} and \ref{11}. In a subsequent article, the results of this paper will be used
for establishing categorical dualities for the quasivarieties $\mathbf{Q}\bigl(\operatorname{O}(M_\kappa)\bigr)$ where $1<\kappa\leqslant\omega$.


\section{Basic concepts}

\noindent
For all the notions which are not defined in this section, we refer to A.\,I. Maltsev \cite{M} and V.\,A. Gorbunov \cite{G}.

\subsection{Lattices}
Most of the following definitions concerning join covers are in accordance with 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 $x\leq\bigvee A$ 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 any proper subset $B\subset A$.
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 each nontrivial join cover
of each element from $X$ can be refined to a minimal one.

By $\mathrm{J}(L)$, we denote the set of all join-irreducible elements of $L$ and by $\mathrm{CJ}(L)$~--- the set of all completely join-irreducible elements of $L$.
Similarly, by $\mathrm{P}(L)$, we denote the set of all join-prime elements of $L$ and by $\mathrm{CP}(L)$~--- the set of all completely join-prime elements of $L$.

\begin{definition}
Let $L$ be a lattice and let $J\subseteq\mathrm{J}(L)$. 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 $CJ$-lattice is a $J$-lattice for $J=\mathrm{CJ}(L)$.
$J$-lattices were considered in \cite{ADKS,DS}, see also \cite{S05}.

For a $J$-lattice $L$ and an element $x\in\mathrm{J}(L)$, let $\mathfrak{M}(x)$ denote the set of all finite minimal join covers of $x$.
%For a $CJ$-lattice $L$ and an element $x\in\mathrm{CJ}(L)$, let $\mathfrak{M}_c(x)$ denote the set of all minimal join covers of $x$.

\begin{remark}
We note that in an upper continuous lattice $L$, each minimal join cover of an element $x\in\mathrm{CJ}(L)$ belongs to $\mathfrak{M}(x)$.
\end{remark}

\begin{prop}\label{ba}\cite{DS}
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 $J$-lattice.
\item
If $L$ is in addition algebraic then $L$ is a $CJ$-lattice.
\end{enumerate}
\end{prop}

\subsection{Suborder lattices}
Let $X$ be a set and let $R\subseteq X^2$ be a \emph{strict partial order} on $X$; that is an irreflexive, antisymmetric, and transitive binary relation.
In this case, we also say that $\langle X;R\rangle$ is a \emph{partially ordered set} or a \emph{poset} for short.
A subset $R^\prime\subseteq R$ is a (\emph{strict}) \emph{suborder} of $R$ if the structure $\langle X;R^\prime\rangle$ is also a poset.
The set $\operatorname{O}(X,R)$  of all (strict) suborders of a partial order $R$ on $X$ is a partially ordered set with respect to the relation $\subseteq$
of set-theoretic inclusion.
Obviously, $\varnothing$ is a least suborder of $R$.
Thus, $\varnothing$ is a least element in $\operatorname{O}(X,R)$.
It is also obvious that $R$ is a greatest 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, $\operatorname{O}(X,R)$ forms a complete lattice, where
\[
\bigvee_{i\in I}R_i=\bigl(\bigcup_{i\in I}R_i\bigr)^t;
\]
here $Y^t$ denotes the transitive closure of a binary relation $Y\subseteq X^2$.
It is clear that
\[
\mathrm{J}\bigl(\operatorname{O}(X,R)\bigr)=\mathrm{CJ}\bigl(\operatorname{O}(X,R)\bigr)=\Bigl\{\bigl\{(a,b)\bigr\}\mid(a,b)\in R\Bigr\}.
\]
We consider here strict partial orders instead of ordinary partial orders for the sake of simplicity only; the least element of a suborder lattice is in this case $\varnothing$ and not the set $\{(x,x)\mid x\in X\}$.

In this article, we consider suborder lattices of posets $M_n$, $0<n\leqslant\omega$, which all have length $2$, see Figure \ref{F:1}.

\begin{figure}
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(115,30)
\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(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{Posets $M_1$, $M_n$, and $M_\omega$}\label{F:1}
\end{figure}


\section{An equational basis for $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)$}

\subsection{Identity $(\mathrm{D}_n)$}
We consider the identity of $n$-distributivity, where $0<n<\omega$, which we denote by $(\mathrm{D}_n)$:
\[
x\wedge(y_0\vee y_1\vee\ldots\vee y_n)=\bigvee_{i\leqslant n}\bigl[x\wedge\bigvee_{j\ne i}y_j\bigr].
\]

\noindent
This identity was introduced by A. Huhn in \cite{H} as a generalization of distributivity---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 \cite{N}.

\begin{lem}\label{D2}
Let $L$ be a lattice, let $J\subseteq\mathrm{J}(L)$ be a set 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 $L$ 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 elements $a$ and $b_0,\ldots,b_m\in J$ then $0<m<n$.
\end{enumerate}
\end{cor}

\subsection{Identity $(\mathrm{P})$}
We denote the following identity by $(\mathrm{P})$:
\[
x\wedge\bigl[\bigl(y_0\wedge(z_0\vee z_1)\bigr)\vee y_1\bigr]=\bigl[x\wedge y_0\wedge(z_0\vee z_1)\bigr]\vee\bigl[x\wedge y_1\bigr]\vee
\bigvee_{i<2}\bigl[x\wedge\bigl((y_0\wedge z_i)\vee y_1\bigr)\bigr].
\]
This identity was introduced in \cite{DS} under the name $(\mathrm{N}_5^1)$.
It was used in \cite{DS} as one of four identities which constitute an equational basis for the [quasi]variety $\mathbf{S}\mathbf{P}(N_5)$.
It was also used in \cite{BSS1} as one of three identities which form an equational basis of the [quasi]variety $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_1)\bigr)$,
also under the name $(\mathrm{N}_5^1)$.
For the next two statements, we refer to \cite{DS}, see also \cite[Lemma 6, Corollary 7]{BSS1}.

\begin{lem}\label{N51}\cite{DS}
Let $L$ be a lattice, let $J\subseteq\mathrm{J}(L)$ be a set 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
$(\mathrm{P})$ 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{cor}\label{cN51}\cite{DS}
Let $L$ be a $2$-distributive $J$-lattice for some set $J\subseteq\mathrm{J}(L)$.
The following conditions are equivalent.
\begin{enumerate}
\item
$(\mathrm{P})$ 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}

\subsection{Identity $(\mathrm{C}_n)$}
We denote the following identity by $(\mathrm{C}_n)$:
\begin{align*}
&x\wedge\bigwedge_{i\leqslant n}(y_i\vee z_i)=\bigvee_{i\leqslant n}\Bigl[x\wedge y_i\wedge\bigwedge_{j\ne i}(y_j\vee z_j)\Bigr]\ \vee\ \bigvee_{i\leqslant n}
\Bigl[x\wedge z_i\wedge\bigwedge_{j\ne i}(y_j\vee z_j)\Bigr]\ \vee\\
&\vee\ \bigvee_{i<j\leqslant n}\Bigl[x\wedge\bigl((y_i\wedge y_j)\vee(z_i\wedge z_j)\bigr)\wedge\bigwedge_{k\notin\{i,j\}}(y_k\vee z_k)\Bigr]\ \vee\\
&\vee\ \bigvee_{i<j\leqslant n}\Bigl[x\wedge\bigl((y_i\wedge z_j)\vee(y_j\wedge z_i)\bigr)\wedge\bigwedge_{k\notin\{i,j\}}(y_k\vee z_k)\Bigr].
\end{align*}
The identity $(\mathrm{C}_1)$ was introduced in \cite{DS} and used there, under the name $(\mathrm{C})$, as a member of an equational basis
of the [quasi]variety $\mathbf{S}\mathbf{P}(N_5)$.
It was also used in \cite{BSS1} as one of three identities which form an equational basis of the [quasi]variety $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_1)\bigr)$,
also under the name $(\mathrm{C})$.
For the next two statements, we refer to \cite{DS}, see also \cite[Lemma 4, Corollary 5]{BSS1}.

\begin{lem}\label{C}
Let $L$ be a lattice, let $J\subseteq\mathrm{J}(L)$ be a set 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
$(\mathrm{C}_n)$ holds in $L$.
\item
If $a\leq a_0\vee b_0$, \ldots, $a\leq a_n\vee b_n$ are nontrivial join covers for some $a\in J$ and some $a_0,\ldots,a_n$, $b_0,\ldots,b_n\in L$,
then there are $c,d\in L$ such that $a\leq c\vee d$ and $\{c,d\}\ll\{a_i,b_i\}$, $\{c,d\}\ll\{a_j,b_j\}$ for some $i<j\leqslant n$.
\end{enumerate}
\end{lem}

\begin{proof}
We prove first that (i) implies (ii).
Indeed, let the assumptions of (ii) hold. Since $(\mathrm{C}_n)$ holds in $L$, we have
\begin{align*}
a=a&\wedge\bigwedge_{i\leqslant n}(a_i\vee b_i)=\bigvee_{i\leqslant n}\Bigl[a\wedge a_i\wedge\bigwedge_{j\ne i}(a_j\vee b_j)\Bigr]\ \vee\ \bigvee_{i\leqslant n}
\Bigl[a\wedge b_i\wedge\bigwedge_{j\ne i}(a_j\vee b_j)\Bigr]\ \vee\\
&\vee\ \bigvee_{i<j\leqslant n}\Bigl[a\wedge\bigl((a_i\wedge a_j)\vee(b_i\wedge b_j)\bigr)\wedge\bigwedge_{k\notin\{i,j\}}(a_k\vee b_k)\Bigr]\ \vee\\
&\vee\ \bigvee_{i<j\leqslant n}\Bigl[a\wedge\bigl((a_i\wedge b_j)\vee(a_j\wedge b_i)\bigr)\wedge\bigwedge_{k\notin\{i,j\}}(a_k\vee b_k)\Bigr].
\end{align*}
As $a$ is a join-irreducible element, $a$ equals one of the joinands on the right-hand side of the equality above.
Therefore, the following cases are possible.

\Case{1}: $a=a\wedge a_i\wedge\bigwedge_{j\ne i}(a_j\vee b_j)$.
In this case, $a\leq a_i$ which contradicts the assumption that $a\leq a_i\vee b_i$ is a nontrivial join cover.
Therefore, this case is impossible.

\Case{2}: $a=a\wedge b_i\wedge\bigwedge_{j\ne i}(a_j\vee b_j)$.
In this case, $a\leq b_i$ which again contradicts the assumption that $a\leq a_i\vee b_i$ is a nontrivial join cover.
Therefore, this case is also impossible.

\Case{3}: there are $i<j\leqslant n$ such that
$a=a\wedge\bigl((a_i\wedge a_j)\vee(b_i\wedge b_j)\bigr)\wedge\bigwedge_{k\notin\{i,j\}}(a_k\vee b_k)$.
In this case, $a\leq c\vee d$, where $c=a_i\wedge a_j$ and $d=b_i\wedge b_j$.
Moreover, $\{c,d\}\ll\{a_i,b_i\}$ and $\{c,d\}\ll\{a_j,b_j\}$ whence
we get the desired conclusion.

\Case{4}: there are $i<j\leqslant n$ such that
$a=a\wedge\bigl((a_i\wedge b_j)\vee(a_j\wedge b_i)\bigr)\wedge\bigwedge_{k\notin\{i,j\}}(a_k\vee b_k)$.
In this case, we put $c=a_i\wedge b_j$ and $d=b_i\wedge a_j$ and obtain the desired conclusion as above in \emph{Case 3}.

We prove now that (ii) implies (i). Let $u$ denote the value of the left-hand side and $v$
denote the value of the right-hand side of the identity $(\mathrm{C}_n)$ under interpretation $\gamma$, where
\[
\gamma(x)=a,\quad\gamma(y_i)=a_i,\quad\gamma(z_i)=b_i,\quad i\leqslant n.
\]
As inequality $v\leq u$ holds in each lattice, in order to prove that $(\mathrm{C}_n)$ holds in $L$, we have to prove that $u\leq v$.
According to our assumption about $L$, it suffices to show that for each element $a^\prime\in J$, the inequality $a^\prime\leq u$
implies that $a^\prime\leq v$.
Indeed, $a^\prime\leq u$ means that
$a^\prime\leq a$ and $a^\prime\leq a_i\vee b_i$ for all $i\leqslant n$.
If $a^\prime\leq a_i$ for some $i\leqslant n$ then $a^\prime\leq u\wedge a_i\leq v$.
If $a^\prime\leq b_i$ for some $i\leqslant n$ then $a^\prime\leq u\wedge b_i\leq v$.
Assume therefore that $a^\prime\leq a_i\vee b_i$ is a nontrivial join cover for all $i\leqslant n$.
Applying (ii), we obtain that there are elements $c,d\in L$ such that
$a^\prime\leq c\vee d$ and $\{c,d\}\ll\{a_i,b_i\}$, $\{c,d\}\ll\{a_j,b_j\}$ for some $i<j\leqslant n$.
As $a^\prime\leq a_i\vee b_i$ and $a^\prime\leq a_j\vee b_j$ are nontrivial join covers, we conclude that $a^\prime\leq c\vee d$
is also a nontrivial join cover.
Therefore, the following cases are possible.

\Case{1}: $c\leq a_i\wedge a_j$ and $d\leq b_i\wedge b_j$ or $d\leq a_i\wedge a_j$ and $c\leq b_i\wedge b_j$.
In this case, $a^\prime\leq u\wedge\bigl((a_i\wedge a_j)\vee(b_i\wedge b_j)\bigr)\leq v$.

\Case{2}: $c\leq a_i\wedge q$ and $d\leq a_i\wedge p$ for some $p,q\in\{a_j,b_j\}$.
In this case, $a^\prime\leq c\vee d\leq a_i$ which is impossible by our assumption as the join cover $a^\prime\leq a_i\vee b_i$ is nontrivial.

\Case{3}: $c\leq a_j\wedge q$ and $d\leq a_j\wedge p$ for some $p,q\in\{a_i,b_i\}$.
In this case, $a^\prime\leq c\vee d\leq a_j$ which is impossible as the join cover $a^\prime\leq a_j\vee b_j$ is nontrivial.

\Case{4}: $c\leq a_i\wedge b_j$ and $d\leq b_i\wedge a_j$ or $d\leq a_i\wedge b_j$ and $c\leq b_i\wedge a_j$.
In this case, $a^\prime\leq u\wedge\bigl((a_i\wedge b_j)\vee(a_j\wedge b_i)\bigr)\leq v$.

\Case{5}: $c\leq b_i\wedge q$ and $d\leq b_i\wedge p$ for some $p,q\in\{a_j,b_j\}$.
In this case, $a^\prime\leq c\vee d\leq b_i$ which is impossible by our assumption as the join cover $a^\prime\leq a_i\vee b_i$ is nontrivial.

\Case{6}: $c\leq b_j\wedge q$ and $d\leq b_j\wedge p$ for some $p,q\in\{a_i,b_i\}$.
In this case, $a^\prime\leq c\vee d\leq b_j$ which is impossible as the join cover $a^\prime\leq a_j\vee b_j$ is nontrivial.

Therefore, $a^\prime\leq v$ in any case and the desired conclusion follows.
\end{proof}


\begin{cor}\label{cC}
Let $L$ be a $2$-distributive $J$-lattice for some set $J\subseteq\mathrm{J}(L)$.
The following conditions are equivalent.
\begin{enumerate}
\item
$(\mathrm{C}_n)$ holds in $L$.
\item
If $a\leq a_0\vee b_0$, \ldots, $a\leq a_m\vee b_m$ are distinct minimal join covers for some $a$, $a_0,\ldots,a_m$, $b_0,\ldots,b_m\in J$, then $m<n$.
\end{enumerate}
\end{cor}

\begin{proof}
We prove that (i) implies (ii).
Indeed, suppose that $m\geqslant n$. Then, applying Lemma \ref{C}, we obtain that there are $i<j\leqslant n$ and elements $c,d\in L$ such that
$\{c,d\}\ll\{a_i,b_i\}$, $\{c,d\}\ll\{a_j,b_j\}$. As $a\leq a_i\vee b_i$ and $a\leq a_j\vee b_j$ are minimal join covers, we conclude that
$a\leq c\vee d$ is a nontrivial join cover and $\{a_i,b_i\}=\{c,d\}=\{a_j,b_j\}$ which contradicts our assumptions. Therefore, $m<n$.

To prove that (ii) implies (i), we show that statement (ii) of Lemma \ref{C} holds.
So let $a\leq a_0\vee b_0$, \ldots, $a\leq a_n\vee b_n$ be nontrivial join covers for some $a\in J$ and some $a_0,\ldots,a_n$, $b_0,\ldots,b_n\in L$.
As $L$ is a $J$-lattice for some set $J\subseteq\mathrm{J}(L)$, there are finite minimal join covers
$a\leq\bigvee F_0$, \ldots, $a\leq\bigvee F_n$ such that $F_i\ll\{a_i,b_i\}$ for all $i\leqslant n$.
As $L$ is $2$-distributive, we apply Lemma \ref{D2} and obtain that $|F_i|=2$ for all $i\leqslant n$.
Applying our assumption (ii) to finite minimal join covers $a\leq\bigvee F_0$, \ldots, $a\leq\bigvee F_n$, we obtain that
$F_i=F_j=\{c,d\}$ for some $i<j\leqslant n$ and some $c,d\in L$. This means that
$a\leq c\vee d$ and $\{c,d\}\ll\{a_i,b_i\}$, $\{c,d\}\ll\{a_j,b_j\}$ which is our desired conclusion.
\end{proof}

\subsection{An equational basis}
For $0<n<\omega$, we put $\Sigma_n=\{(\mathrm{C}_n),(\mathrm{D}_2),(\mathrm{P})\}$ and $\mathbf{S}_n=\operatorname{Mod}\Sigma_n$.

\begin{prop}\label{5-6}
Let $L$ be a dually algebraic lattice such that $L\models\Sigma_n$, where $0<n<\omega$.
Then for all elements $x\in\mathrm{J}(L)$, each element of the set
$\mathfrak{M}(x)$ is of the form $\{a,b\}$, where $\{a,b\}\subseteq\mathrm{P}(L)$ is an antichain.
Moreover, $|\mathfrak{M}(x)|\leqslant n$.

In particular, $L\in\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)$.
\end{prop}

\begin{proof}
It follows from Proposition \ref{ba}(i) that $L$ is a $J$-lattice. Corollary \ref{cD2} implies that each minimal
nontrivial join cover of an element $x\in\mathrm{J}(L)$ contains exactly two elements.
Corollary \ref{cN51} implies that each minimal
nontrivial join cover of $x$ consists of join-prime elements.
Moreover, $|\mathfrak{M}(x)|\leqslant n$ by Corollary \ref{cC}.
Thus, the first statement follows.

To prove the second statement, we use the method developed in \cite{S99,S05}.
We fix an element $x\in\mathrm{J}(L){\setminus}\mathrm{P}(L)$.
According to the first statement,
\[
\mathfrak{M}(x)=\bigl\{\{b_1(x),c_1(x)\},\ldots,\{b_{n(x)}(x),c_{n(x)}(x)\}\bigr\}
\]
for some natural number $n(x)$ such that $0<n(x)\leqslant n$ and some join-prime elements
$b_1(x),\ldots,b_{n(x)}$, $c_1(x),\ldots,c_{n(x)}$.
We denote by $P_x$ an isomorphic copy of $M_{n(x)}$.
We denote the elements of $P_x$ by $0(x),a_1(x),\ldots,a_{n(x)}(x),1(x)$ respectively, see Figure \ref{F:1}.
As $n(x)\leqslant n$, $P_x$ is a subposet of $M_n$.
We define a mapping $\psi_x\colon\mathrm{J}(L)\to\operatorname{O}(P_x)$ as follows:
\begin{align*}
\psi_x\colon x\mapsto&\bigl\{\bigl(0(x),1(x)\bigr)\bigr\};\\
\psi_x\colon y\mapsto&\bigl\{\bigl(0(x),a_i(x)\bigr)\mid y=b_i(x)\ \text{for some}\ i\in\{1,\ldots,n\}\bigr\}\cup\\
\cup&\bigl\{\bigl(a_i(x),1(x)\bigr)\mid y=c_i(x)\ \text{for some}\ i\in\{1,\ldots,n\}\bigr\},\quad\text{for all}\ y\in\bigcup\mathfrak{M}(x);\\
\psi_x\colon y\mapsto&\varnothing\quad\text{for all}\ y\notin\{x\}\cup\bigcup\mathfrak{M}(x).
\end{align*}
Let $\mathrm{P}^\prime(L)$ denote the set of all join-prime elements of $L$ which do not belong to any minimal nontrivial join cover of any element $x\in\mathrm{J}(L){\setminus}\mathrm{P}(L)$.
For each element $x\in\mathrm{P}^\prime(L)$,
we put $P_x=\bigl\{0(x),1(x)\bigr\}$, where $0(x)<1(x)$ and consider the mapping
\begin{align*}
&\psi_x\colon\mathrm{J}(L)\to\operatorname{O}(P_x);\\
&\psi_x\colon x\mapsto\bigl\{\bigl(0(x),1(x)\bigr)\bigr\};\\
&\psi_x\colon y\mapsto\varnothing\quad\text{for all}\ y\ne x.
\end{align*}
Finally, let $I=\bigl(\mathrm{J}(L){\setminus}\mathrm{P}(L)\bigr)\cup\mathrm{P}^\prime(L)$. We consider the following mapping:
\begin{align*}
&\psi\colon L\to\prod_{x\in I}\operatorname{O}(P_x);\\
&\pi_x\psi(a)=\bigcup\bigl\{\psi_x(y)\mid y\in\mathrm{J}(L),\ y\leq a\bigr\}\quad\text{for all}\ a\in L\ \text{and all}\ x\in I.
\end{align*}

\setcounter{claim}{0}
\begin{claim}
$\psi$ is well-defined.
\end{claim}

\begin{scproof}
We have to prove that $\pi_x\psi(a)$ is a suborder in $P_x$ for all $x\in I$ and all $a\in L$.
As $P_x\cong M_{n(x)}$, it suffices to show that if $\bigl(0(x),a_i(x)\bigr),\bigl(a_i(x),1(x)\bigr)\in\pi_x\psi(a)$ for some $i\in\{1,\ldots,n\}$ then
$\bigl(0(x),1(x)\bigr)\in\pi_x\psi(a)$.
Indeed, suppose that $\bigl(0(x),a_i(x)\bigr),\bigl(a_i(x),1(x)\bigr)\in\pi_x\psi(a)$ for some $i\in\{1,\ldots,n\}$.
In other words, $\psi_x(b_i)\cup\psi_x(c_i)\subseteq\pi_x\psi(a)$ where $x\leq b_i\vee c_i$ is a minimal nontrivial join cover.
This means that $b_i,c_i\leq a$ whence $x\leq b_i\vee c_i\leq a$.
By our definition of $\psi_x$ this implies that
$\bigl\{\bigl(0(x),1(x)\bigr)\bigr\}=\psi_x(x)\subseteq\pi_x\psi(a)$ which is our desired conclusion.
\end{scproof}

\begin{claim}
$\psi$ is a $(0,1)$-lattice homomorphism.
\end{claim}

\begin{scproof}
In order to prove the desired claim, it suffices to show that $\pi_x\psi$ is a $(0,1)$-lattice homomorphism for each $x\in I$.
Indeed, we fix an element $x\in I$ and elements $u,v\in L$.
If $u$ is a least element of $L$ then $y\leq u$ for no element $y\in\mathrm{J}(L)$. Therefore, $\pi_x\psi(u)=\varnothing$.
If $u$ is a greatest element of $L$ then $y\leq u$ for each element $y\in\mathrm{J}(L)$. Therefore, $\pi_x\psi(u)$ is obviously the greatest element of
$\operatorname{O}(P_x)$.
Therefore, $\pi_x\psi$ preserves the bounds.

If $u\leq v$ then $y\leq u$ implies $y\leq v$ for all $y\in\mathrm{J}(L)$.
Therefore, $\pi_x\psi$ is monotone. We prove that $\pi_x\psi$ preserves meets and joins.

Since $\pi_x\psi$ is monotone, $\pi_x\psi(u)\vee\pi_x\psi(v)\subseteq\pi_x\psi(u\vee v)$.
We have to establish that $\pi_x\psi(u\vee v)\subseteq\pi_x\psi(u)\vee\pi_x\psi(v)$.
So suppose that $(z_0,z_1)\in\pi_x\psi(u\vee v)$. This means that $(z_0,z_1)\in\psi_x(y)\ne\varnothing$ for some $y\in\mathrm{J}(L)$ such that $y\leq u\vee v$.
If $y\leq u$ or $y\leq v$ then $(z_0,z_1)\in\pi_x\psi(u)\cup\pi_x\psi(v)$.
Otherwise, $y\leq u\vee v$ is a nontrivial join cover.
As $L$ is a $J$-lattice, we can refine this join cover to a minimal one.
This implies that $y\in\mathrm{J}(L){\setminus}\mathrm{P}(L)$. As $\psi_x(y)\ne\varnothing$, we conclude by the definition of $\psi_x$ that $y=x$.
Moreover, there is $i$ such that $1\leqslant i\leqslant n(x)$ and
$y=x\leq b_i\vee c_i$ is a minimal nontrivial join cover with $\{b_i,c_i\}\ll\{u,v\}$.
Inclusion $(z_0,z_1)\in\psi_x(y)=\psi_x(x)$ implies that $z_0=0(x)$ and $z_1=1(x)$.
Furthermore, $\bigl(0(x),a_i(x)\bigr)\in\psi_x(b_i)\subseteq\pi_x\psi(u)\cup\pi_x\psi(v)$ and
$\bigl(a_i(x),1(x)\bigr)\in\psi_x(c_i)\subseteq\pi_x\psi(u)\cup\pi_x\psi(v)$
as $\{b_i,c_i\}\ll\{u,v\}$.
Hence, $(z_0,z_1)\in\psi_x(b_i)\vee\psi_x(c_i)\subseteq\pi_x\psi(u)\cup\pi_x\psi(v)$.
This proves that $\pi_x\psi$ preserves joins.

Since $\pi_x\psi$ is monotone, $\pi_x\psi(u\wedge v)\subseteq\pi_x\psi(u)\cap\pi_x\psi(v)$.
We have to establish that $\pi_x\psi(u)\cap\pi_x\psi(v)\subseteq\pi_x\psi(u\wedge v)$.
Indeed, let $(z_0,z_1)\in\pi_x\psi(u)\cap\pi_x\psi(v)$. This means that
$(z_0,z_1)\in\psi_x(y)\cap\psi_x(y^\prime)\ne\varnothing$ for some $y,y^\prime\in\mathrm{J}(L)$ such that $y\leq u$ and $y^\prime\leq v$.
If $y\ne y^\prime$ then $\psi_x(y)\cap\psi_x(y^\prime)=\varnothing$ by the definition of $\psi_x$, a contradiction.
Therefore, $y=y^\prime\leq u\wedge v$ and $(z_0,z_1)\in\psi_x(y)\subseteq\pi_x\psi(u\wedge v)$.
This proves that $\pi_x\psi$ preserves meets.
\end{scproof}

\begin{claim}
$\psi$ is an embedding.
\end{claim}

\begin{scproof}
Suppose that $u\nleq v$ in $L$. As $L$ is a $J$-lattice, there is $y\in\mathrm{J}(L)$ such that $y\leq u$ and $y\nleq v$.
By our definition, there is $x\in I$ such that $\psi_x(y)\ne\varnothing$. But then $\varnothing\ne\psi_x(y)\subseteq\pi_x\psi(u)$ and
$\psi_x(y)\cap\pi_x\psi(v)=\varnothing$. This implies that $\pi_x\psi(u)\nsubseteq\pi_x\psi(v)$ whence $\psi(u)\nleq\psi(v)$.
\end{scproof}

\noindent
It follows from the claims above that
\[
L\in\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(P_x)\mid x\in I\bigr)
\subseteq\mathbf{S}\mathbf{P}\mathbf{S}\bigl(\operatorname{O}(M_n)\bigr)
\subseteq\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr).\]
The proof of Proposition \ref{5-6} is complete.
\end{proof}

\begin{prop}\label{5-6-0}
Let $L$ be a bi-algebraic lattice such that $L\models\Sigma_n$, where $0<n<\omega$.
Then for all elements $x\in\mathrm{CJ}(L)$, each element of the set
$\mathfrak{M}(x)$ is of the form $\{a,b\}$, where $\{a,b\}\subseteq\mathrm{P}(L)$ is an antichain.
Moreover, $|\mathfrak{M}(x)|\leqslant n$.

In particular, $L\in\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)$.
\end{prop}

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


\begin{thm}\label{5-7}
$\Sigma_n$ forms an equational basis for $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)$.
In particular, the class $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)=\mathbf{S}_n$ is a lattice variety.
\end{thm}

\begin{proof}
Let $L\models\Sigma_n$ and let $F$ be the dual filter lattice of $L$. It is well-known that $F$ is dually algebraic and it follows that $F\models\Sigma_n$.
By Proposition \ref{ba}, $F$ is a $J$-lattice.
By Proposition \ref{5-6}, $F\in\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)$
whence $L\in\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)$ as $L$ embeds into $F$.
On the other hand, the lattice $\operatorname{O}(M_n)$ has the only minimal join covers:
\begin{align*}
&A\leq A_i\vee B_i,\quad 1\leqslant i\leqslant n,\ \text{where}\\
&A=\bigl\{(0,1)\bigr\},\ A_i=\bigl\{(0,a_i)\bigr\},\ B_i=\bigl\{(a_i,1)\bigr\},\quad 1\leqslant i\leqslant n,
\end{align*}
see Figure \ref{F:1}.
Thus, $\operatorname{O}(M_n)$ is $2$-distributive by Corollary \ref{cD2}.
Moreover, $\operatorname{O}(M_n)$ satisfies the condition (ii) of Corollaries \ref{cC} and \ref{cN51}.
This implies that $\operatorname{O}(M_n)\models\Sigma_n$.
\end{proof}

\noindent
Let $\mathbf{L}_{01}$ denote the variety of $(0,1)$-lattices and let $\mathbf{S}^{01}_n=\mathbf{L}_{01}\cap\operatorname{Mod}\Sigma_n$.

\begin{thm}\label{5-9}
The set $\Sigma_n$ forms an equational basis for $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)$ within the variety $\mathbf{L}_{01}$.
In particular, $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)=\mathbf{S}^{01}_n$ is a variety of $(0,1)$-lattices.
\end{thm}

\begin{proof}
If $L$ is a $(0,1)$-lattice then taking in the proof of Theorem \ref{5-7} the dual lattice of nonempty filters as $F$, we obtain that
$L$ is a $(0,1)$-sublattice of $F$ and $F\in\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)$ by Proposition \ref{5-6}.
Therefore, $L$ belongs in this case to the variety of $(0,1)$-lattices generated by $\operatorname{O}(M_n)$.
\end{proof}

\section{An equational basis for $\mathbf{Q}\bigl(\operatorname{O}(M_\omega)\bigr)$}

\noindent
We put $\Sigma=\{(\mathrm{D}_2),(\mathrm{P})\}$.

\begin{prop}\label{8}
Let $L$ be a dually algebraic lattice such that $L\models\Sigma$.
The following statements hold.
\begin{enumerate}
\item
For all $x\in\mathrm{J}(L)$, each element of the set
$\mathfrak{M}(x)$ is of the form $\{a,b\}$, where $\{a,b\}\subseteq\mathrm{P}(L)$ is an antichain.
\item
If $L$ is bi-algebraic then for all $x\in\mathrm{CJ}(L)$, each element of the set
$\mathfrak{M}(x)$ is of the form $\{a,b\}$, where $\{a,b\}\subseteq\mathrm{P}(L)$ is an antichain.
\end{enumerate}
In particular, $L\in\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_\kappa)\bigr)\subseteq\mathbf{Q}\bigl(\operatorname{O}(M_\omega)\bigr)$ for some cardinal $\kappa$.
\end{prop}

\begin{proof}
Applying the same argument as in the proof of Proposition \ref{5-6}, we obtain that $L\in\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_\kappa)\bigr)$
for some infinite cardinal $\kappa\geqslant|L|$.
As $M_\kappa$ embeds into an ultrapower of $M_\omega$, we conclude that
$\operatorname{O}(M_\kappa)\in\mathbf{S}\mathbf{P}_u\bigl(\operatorname{O}(M_\omega)\bigr)$ and
\[
\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_\kappa)\bigr)\subseteq\mathbf{S}\mathbf{P}\mathbf{P}_u\bigl(\operatorname{O}(M_\omega)\bigr)=
\mathbf{Q}\bigl(\operatorname{O}(M_\omega)\bigr),
\]
which is our desired conclusion.
\end{proof}


\begin{thm}\label{9}
The following statements hold.
\begin{enumerate}
\item
The quasivariety $\mathbf{Q}\bigl(\operatorname{O}(M_\omega)\bigr)$ is a lattice variety and $\Sigma$ forms an equational basis for this variety.
\item
The class $\mathbf{Q}\bigl(\operatorname{O}(M_\omega)\bigr)$ of $(0,1)$-lattices is a variety of $(0,1)$-lattices and $\Sigma$ forms an equational basis for this variety.
\end{enumerate}
\end{thm}

\begin{proof}
(i) If $L\models\Sigma$, then $L\in\mathbf{Q}\bigl(\operatorname{O}(M_\omega)\bigr)$ by Proposition \ref{8}.
Hence, $\operatorname{Mod}\Sigma\subseteq\mathbf{Q}\bigl(\operatorname{O}(M_\omega)\bigr)$. Conversely,
the lattice $\operatorname{O}(M_\omega)$ has the only minimal join covers:
\begin{align*}
&A\leq A_i\vee B_i,\quad 1\leqslant i<\omega,\ \text{where}\\
&A=\bigl\{(0,1)\bigr\},\ A_i=\bigl\{(0,a_i)\bigr\},\ B_i=\bigl\{(a_i,1)\bigr\},\quad 1\leqslant i<\omega,
\end{align*}
see Figure \ref{F:1}.
Thus, $\operatorname{O}(M_n)$ is $2$-distributive by Corollary \ref{cD2}.
Moreover, $\operatorname{O}(M_n)$ satisfies the condition (ii) of Corollary \ref{cN51}.
Therefore, $\operatorname{O}(M_\omega)\models\Sigma$ and
\[
\mathbf{Q}\bigl(\operatorname{O}(M_\omega)\bigr)=\mathbf{S}\mathbf{P}\mathbf{P}_u\bigl(\operatorname{O}(M_\omega)\bigr)\models\Sigma
\]
as identities are stable with respect to the operators $\mathbf{S}$, $\mathbf{P}$, and $\mathbf{P}_u$.
It follows that $\operatorname{Mod}\Sigma=\mathbf{Q}\bigl(\operatorname{O}(M_\omega)\bigr)$.

The proof of (ii) is similar.
\end{proof}

\begin{cor}\label{10}
The following equalities hold for an arbitrary infinite cardinal $\kappa$:
\[
\mathbf{SO}_2=\mathbf{Q}\bigl(\operatorname{O}(M_n)\mid 0<n<\omega\bigr)=\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_\kappa)\bigr).
%=\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_\omega)\bigr).
\]
\end{cor}

\begin{proof}
By \cite[Theorem 4.8]{S05}, $\Sigma$ forms an equational basis for $\mathbf{SO}_2$. Taking into account Theorem \ref{9}, we conclude that
$\mathbf{SO}_2=\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_\omega)\bigr)$.
Furthermore, each algebraic structure embeds into an ultraproduct of its finitely generated substructures, see for example \cite[Theorem 1.2.8]{G}.
Therefore, $M_\kappa\in\mathbf{S}\mathbf{P}_u(M_n\mid 0<n<\omega)$ for each infinite cardinal kappa whence
$\operatorname{O}(M_\kappa)\in\mathbf{S}\mathbf{P}_u\bigl(\operatorname{O}(M_n)\mid 0<n<\omega\bigr)$ and
\begin{align*}
\mathbf{SO}_2=&\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_\omega)\bigr)=\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_\kappa)\bigr)
\subseteq\mathbf{S}\mathbf{P}\mathbf{P}_u\bigl(\operatorname{O}(M_n)\mid 0<n<\omega\bigr)=\\
=&\mathbf{Q}\bigl(\operatorname{O}(M_n)\mid 0<n<\omega\bigr)\subseteq\mathbf{SO}_2.
\end{align*}
The desired conclusion follows.
\end{proof}

\noindent
The following problem was raised in \cite{S05}.

\begin{problem}\label{05}\cite[Question 2]{S05}
If $\langle P;\leq\rangle$ is a finite poset, is it true that the quasivariety $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(P;\leq)\bigr)$ is a variety?
\end{problem}

\noindent
The next statement solves Problem \ref{05} in the positive for finite posets of length at most two.

\begin{cor}\label{11}
If $\langle P;\leq\rangle$ is a finite poset of length at most two then $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(P;\leq)\bigr)$ is a
finitely based variety.
\end{cor}

\begin{proof}
It follows from Corollary \ref{10} and the fact that the poset $\langle P;\leq\rangle$ is finite that
$\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(P;\leq)\bigr)=\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(M_n)\bigr)$
or $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(P;\leq)\bigr)$ is the variety of distributive lattices.
In the first case, $\mathbf{S}\mathbf{P}\bigl(\operatorname{O}(P;\leq)\bigr)$ is a finitely based variety by Theorem \ref{5-7}.
\end{proof}
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