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\title[Algebras of binary formulas]{\large Algebras of binary formulas for circularly ordered
structures: piecewise monotonic case}


\author[B.Sh. KULPESHOV]{{\bf B.Sh. Kulpeshov}\protect\orcidicon{0000-0001-8827-2630}}



\address{Beibut Shaiykovich Kulpeshov  % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Institute of Mathematics and Mathematical Modeling,
\newline\hphantom{iii} Shevchenko str., 28,
\newline\hphantom{iii} 050010, Almaty, Kazakhstan
\newline\hphantom{iii} Kazakh-British Technical University,
\newline\hphantom{iii} Tole bi str., 59,
\newline\hphantom{iii} 050000, Almaty, Kazakhstan}%
\email{\textcolor{blue}{kulpesh@mail.ru, b.kulpeshov@kbtu.kz}}%
%\email{\textcolor{blue}{}}%

\thanks{\sc Kulpeshov B.Sh.,
Algebras of binary formulas for circularly ordered structures: piecewise monotonic case}
\thanks{\copyright \ 2026 Kulpeshov B.Sh.}
\thanks{\rm The work is supported by Science Committee of Ministry of Science and Higher Education
of the Republic of Kazakhstan (grant BR31714735)}
\thanks{\it Received January, 1, 2023, Published December, 31, 2023}%


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\maketitle {\small

\vspace{-12pt}
\centerline{{\it Communicated by} {\sc P.P. Petrov}
%({\it filled by Editorial board})
}

\bigskip
\bigskip

\begin{quote}
\noindent{\bf Abstract:} This article concerns the notion of weak circular mi\-ni\-ma\-li\-ty being a variant of o-minimality
for circularly ordered structures. Algebras of binary isolating formulas are studied for
$\aleph_0$-categorical 1-transitive non-primitive weakly circularly minimal theories of
convexity rank greater than 1 with a trivial definable closure having a piecewise monotonic-to-left
function to the de\-fi\-nable completion of a structure. On the basis of the study, we present
a description of these algebras. It is shown that for this case there exist only non-commutative
algebras. A strict $s$-de\-ter\-mi\-nis\-ti\-ci\-ty of such algebras for some natural number $s$ is also established.\medskip

\noindent{\bf Keywords:} algebra of binary formulas, weak circular minimality, $\aleph_0$-categorical theory,
circularly ordered structure, convexity rank.
 \end{quote}
}

\bigskip

\section{Preliminaries}
\noindent

Algebras of binary formulas are a tool for describing
relationships between elements of the sets of realizations of an one-type at the
binary level with respect to the superposition of binary definable
sets.  A {\it binary isolating formula} is a formula of the form $\varphi(x,y)$
such that for some parameter $a$ the formula $\varphi(a,y)$
isolates a complete type in $S(\{a\})$. The concepts and notations
related to these algebras can be found in the papers
\cite{ShS, Sud2018}. In recent years, algebras of binary formulas
have been studied intensively and have been continued in the works
\cite{AKS_2021}, \cite{E16}--\cite{SMJ_2024}.

Let $L$ be a countable first-order language. Throughout we
consider $L$-structures and assume that $L$ contains a ternary
relational symbol $K$, in\-ter\-pre\-ted as a circular order in these
structures (unless otherwise stated).

The {\it circular order} is described by a ternary
relation $K$ satisfying the following conditions:

(co1) $\forall x\forall y \forall z (K(x,y,z)\to K(y,z,x))$;

(co2) $\forall x\forall y \forall z (K(x,y,z)\land K(y,x,z)
 \Leftrightarrow x=y \lor y=z \lor z=x)$;

 (co3) $\forall x\forall y \forall z(K(x,y,z)\to \forall t[K(x,y,t)
 \lor K(t,y,z)])$;

 (co4) $\forall x\forall y \forall z (K(x,y,z)\lor K(y,x,z))$.



The following observation relates linear  and circular orders.
 \begin{fact}{\rm \cite{bmmn}} \label{compat}
(i) If $\langle M,\le\rangle$ is a linear ordering and $K$ is the ternary relation
derived from $\le$ by the rule
$$K(x,y,z):\Leftrightarrow (x\le y\le z )\lor (z\le x \le y)\lor (y\le z\le x)$$
then $K$ is a circular order relation on $M$.

(ii) If $\langle N, K\rangle$ is a circular ordering and
$a \in N$, then the relation $\le_a$ defined on $M:= N\setminus \{a\}$ by
the rule
$y\le_a z :\Leftrightarrow K(a, y,z)$
is a linear order.
\end{fact}

Thus, any linearly ordered structure is circularly ordered,
since the relation of circular order is $\emptyset$-definable in an arbitrary linearly ordered structure.
However, the opposite is not true. The following example shows that there are circularly ordered structures
not being linearly ordered.

\begin{example}\rm \cite{pcam, dr}
Let $\mathbb{Q}_2^*:=\langle \mathbb{Q}_2, K, L\rangle$ be a circularly ordered structure,
where $L=\{\sigma^2_0, \sigma^2_1\}$, for which the following conditions hold:

(i) its domain $\mathbb{Q}_2$ is a countable dense subset of the unit circle,
 no two points making the central angle $\pi$;

(ii) for distinct $a,b\in \mathbb{Q}_2$
$$(a,b)\in \sigma_0 \Leftrightarrow 0<\arg(a/b)<\pi,\;(a,b)\in \sigma_1 \Leftrightarrow \pi<\arg(a/b)<2\pi,$$
where $\arg(a/b)$ means the value of the central angle between $a$ and $b$ clock\-wise.
Indeed,  one can check that the linear order relation is not $\emptyset$-definable in this structure.
\end{example}


The notion of {\it weak circular minimality} was studied initially
in \cite{KulMac}. Let $A\subseteq M$, where $M$ is a
circularly ordered structure. The set $A$ is called {\it convex}
if for any $a, b\in A$ the following property is satisfied:  for
any $c\in M$ with $K(a,c,b)$, $c\in A$ holds, or for any $c\in M$
with $K(b,c,a)$, $c\in A$ holds. A {\it weakly circularly
minimal structure} is a circularly ordered structure
$M=\langle M, K,\ldots\rangle$ such that any definable
(with parameters) subset of $M$ is a union of finitely many convex
sets in $M$.  The study of weakly circularly
minimal structures was continued in the papers
\cite{AK_2021}, \cite{KulMac}--\cite{Kul09}.

Let $M$ be an $\aleph_0$-categorical weakly circularly minimal
structure, $G:={\rm Aut}(M)$. Following the standard
group theory terminology, the group $G$ is called {\it
$k$-transitive} if for any pairwise distinct $a_1, a_2, \ldots,
a_k\in M$ and pairwise distinct $b_1, b_2, \ldots, b_k\in M$ there
exists $g \in G$ such that $g(a_1)=b_1, g(a_2)=b_2, \ldots,
g(a_k)=b_k$. A {\it congruence} on $M$ is an arbitrary
$G$-invariant equivalence relation on $M$. The group $G$
is called {\it primitive} if $G$ is $1$-transitive and there are
no non-trivial proper congruences on $M$.

\begin{notat}\rm    \label{notation}

(1)   $K_0(x,y,z):=K(x,y,z)\land y\ne x \land y \ne z \land x\neq
z$.

 (2) $K(u_1,\ldots,u_n)$ denotes a formula saying that all subtuples of the tuple
$\langle u_1, \ldots, u_n\rangle$ having the length 3 (in
ascending order) satisfy $K$; similar notations are used for
$K_0$.

(3) Let $A, B, C$ be disjoint convex subsets of a circularly
ordered structure $M$. We write $K(A,B,C)$ if for any
$a,b,c\in M$ with $a\in A$, $b\in B$, $c\in C$ we have $K(a,b,c)$.
We extend naturally that notation using, for instance, the
notation $K_0(A,d,B,C)$ if $d\not\in A\cup B\cup C$ and
$K_0(A,d,B)\wedge K_0(d,B,C)$ holds.
\end{notat}

Further we need  the notion of the definable completion of a circularly
ordered structure, introduced in \cite{KulMac}.
Its linear analog was introduced in \cite{mms}. A {\it cut} $C(x)$ in a circularly
ordered structure $M$ is maximal consistent set of formulas of the form $K(a,x,b)$, where $a,b\in M$.
A cut is said to be {\it algebraic} if there exists $c\in M$ that realizes it. Otherwise, such a cut
is said to be {\it non-algebraic}.
Let $C(x)$ be a non-algebraic cut. If there is some $a\in M$ such that
either for all $b\in M$ the formula $K(a,x,b)\in C(x)$, or for all $b\in M$ the formula $K(b,x,a)\in C( x)$,
then $C(x)$ is said to be {\it rational}.
Otherwise, such a cut is said to be {\it irrational}.
A {\it definable cut} in $M$ is a cut $C(x)$ with the following property:
there exist $a,b\in M$ such that $K(a,x,b)\in C(x)$ and the set $\{c\in M\mid K(a,c,b)$ and $ K(a,x,c)\in C(x)\}$ is definable.
The {\it definable completion} $\overline{M}$ of a structure $M$ consists of $M$ together with all definable
cuts in $M$ that are irrational (essentially $\overline{M}$ consists of endpoints of definable subsets of the structure $M$).



\begin{notat}\rm \cite{KulMac}
Let $F(x,y)$ be an $L$-formula such that $F(M,b)$ is convex
infinite co-infinite for each $b\in M$.
Let $F^\ell(y)$ be the formula saying $y$ is a left endpoint of $F(M,y)$:
$$ \exists z_1 \exists z_2 [K_0(z_1,y,z_2)\land \forall t_1(K(z_1,
t_1,y)\land t_1\ne y \to \neg F(t_1,y))\land$$
$$\forall t_2 (K(y,t_2,z_2)\land t_2\ne y\to F(t_2,y))].$$


We say that $F(x,y)$ is {\em convex-to-right} if
$$M\models \forall y \forall x [F(x,y)\to F^l(y)\land \forall z(K(y,z,x)\to
F(z,y))].$$
If $F_1(x,y), F_2(x,y)$ are arbitrary convex-to-right
formulas we say $F_2$ is {\it bigger than} $F_1$ if there is $a\in
M$ with $F_1(M,a)\subset F_2(M,a)$. If $M$ is 1-transitive and this
holds for some $a$, it holds for all $a$. This gives a total ordering on the
(finite) set of all convex-to-right formulas $F(x,y)$ (viewed up to equivalence modulo
$Th(M)$).

Consider $F(M,a)$ for arbitrary $a\in M$. In general,  $F(M,a)$ has no the right endpoint in $M$.
For example, if $dcl(\{a\})=\{a\}$ holds for some $a\in M$ then for any convex-to-right
formula $F(x,y)$ and any $a\in M$ the formula $F(M,a)$ has no the right endpoint in $M$. We write
$f(y):={\rm rend}\; F(M,y)$, assuming that $f(y)$ is the right endpoint of the set $F(M,y)$ that
lies in general in the definable completion   $\overline{M}$ of  $M$.
Then $f$ is a function mapping $M$ in  $\overline{M}$.
\end{notat}



\begin{notat}\label{n1}\rm Let $E(x,y)$ be an $\emptyset$--definable equivalence relation
partitioning $M$ into infinite convex classes. Suppose that $y$ lies in $\overline{M}$
(non-obligatory in $M$). Then $$E^*(x,y):=\exists y_1\exists y_2[y_1\ne y_2\land
\forall t(K(y_1,t,y_2)\to E(t,x))\land K_0(y_1, y, y_2)].$$
 \end{notat}





\vskip 3mm
Let $M$, $N$ be circularly ordered structures.
The {\it $2$-reduct} of $M$ is a circularly ordered
structure with the same universe of $M$ and consisting
of predicates for each $\emptyset$-definable relation on
$M$ of arity $\leq 2$ as well as of the ternary
predicate $K$ for the circular order, but does not have other
predicates of arities more than two. We say that the structure
$M$ is {\it isomorphic} to $N$ {\it up to
binarity} or {\it binarily isomorphic to} $N$ if the
$2$-reduct of $M$ is isomorphic to the $2$-reduct of
$N$.

\vskip 2mm


Let $f$ be a unary function from $M$ to $\overline{M}$.
We say that $f$ is {\it monotonic-to-right (left) on $M$} if it preserves (reverses) the relation
$K_0$, i.e. for any $a,b,c \in M$ such that $K_0(a,b,c)$, we have
$K_0(f(a),f(b),f(c))$ ($K_0(f(c),f(b),f(a))$).

We also say that $f$
is {\it piecewise monotonic-to-right (left) on $M$} if there exists an
$\emptyset$-definable non-trivial equivalence relation $E(x,y)$ partitioning $M$
into finitely many infinite convex classes so that $f$ is mo\-no\-to\-nic-to-right on each $E$-class
and $f$ is not monotonic-to-left (right) on $M/E$, where by $M/E$ we denote the set of
representatives of $E$-classes in $M$.

\begin{example}\rm \cite{Kul06} Let $M:=\langle M, K, E^2, f^1\rangle$ be a circularly ordered structure,
where $M$ is a disjoint union of $\mathbb{Q}_1$, $\mathbb{Q}_2$, $\ldots$, $\mathbb{Q}_6$, where
$\mathbb{Q}_i$ is a copy of the ordering of rational numbers $\mathbb{Q}$. The symbol $E$ interprets
an equivalence relation on $M$ as follows: $E(a,b)$ iff there is $1\le i\le 6$ with $a,b\in\mathbb{Q}_i$.
The symbol $f$ interprets a function on $M$ as follows: $f(\mathbb{Q}_i)=\mathbb{Q}_{i+3}$ for each $1\le i\le 3$,
$f(\mathbb{Q}_j)=\mathbb{Q}_{j-3}$ for each $4\le j\le 6$, and $f(q)=-q$ for all $q\in \mathbb{Q}$.

It can be proved that $M$ is an $\aleph_0$-categorical 1-transitive weakly circularly minimal structure,
$f$ is a bijection on $M$ so that $f^2(a)=a$ for all $a\in M$, $f$ is monotonic-to-left on each $E$-class and
$f$ is monotonic-to-right on $M/E$, i.e. $f$ is piecewise monotonic-to-left on $M$.
\end{example}

The following definition can be used in a circular ordered structure as well.
\begin{definition}\rm \cite{k1}, \cite{SMJ2021}
Let $T$ be a weakly o-minimal theory, $M$ be a sufficiently sa\-tu\-ra\-ted model of $T$, $A\subseteq M$.
{\it The rank of convexity of the set}
$A$ $(RC(A))$ is defined as follows:

1) $RC(A)=-1$ if $A=\emptyset$.

2) $RC(A)=0$ if $A$ is finite and non-empty.

3) $RC(A) \geq 1$ if $A$ is infinite.

4) $RC(A) \geq  \alpha  +  1$ if there exist a parametrically
definable equivalence relation $E(x,y)$ and an infinite sequence of elements
$b_i\in A, i\in \omega$, such that:
\begin{itemize}
\item For every  $i, j\in \omega$ whenever $i\ne j$ we have  $M \models \neg E(b_i, b_j)$;
\item For every $i\in \omega$, $RC(E(x,  b_i))
\geq \alpha$ and $E(M,b_i)$ is a convex subset of $A$.
\end{itemize}

5) $RC(A) \geq \delta$ if $RC(A)
\geq \alpha$ for all $\alpha < \delta$, where $\delta$ is a limit ordinal.

If $RC(A) = \alpha$ for some $\alpha$, we say that $RC(A)$ is defined. Otherwise (i.e. if $RC(A))$ $\geq$
$\alpha$ for all $\alpha$), we put $RC(A) =\infty$.

{\it The rank of convexity of a formula} $\phi(x,\bar a)$, where $\bar a\in M$, is defined as the rank
of convexity of the set $\phi(M,\bar a)$, i.e. $RC(\phi(x,\bar a)):=RC(\phi(M,\bar a))$.

{\it The rank of convexity of an 1-type} $p$ is defined as the rank of convexity of the set $p(M)$, i.e. $RC(p):=RC(p(M))$.
\end{definition}

In particular, a theory has convexity rank 1 if there is no definable (with parameters)
equivalence relations with infinitely many infinite convex classes.
\vskip 2mm

The following theorem  characterizes up to binarity
$\aleph_0$--catego\-ri\-cal 1-tran\-si\-tive non-pri\-mi\-tive weak\-ly cir\-cu\-lar\-ly minimal structures
$M$ of convexity rank greater  than 1 having both a trivial definable closure and
a convex-to-right  formula $R(x,y)$ such that
$r(y):=\;${\rm rend} $R(M,y)$ is piecewise monotonic-to-left on $M$:

\begin{theorem}\label{th2} {\rm \cite{Kul09}}
Let $M$ be an $\aleph_0$--categorical 1-transitive non-primitive weakly circularly
minimal structure of convexity rank greater than 1,  $dcl(\{a\})$ $=\{a\}$
for some $a\in M$. Sup\-pose that there exists a convex-to-right  formula $R(x,y)$ such that
$r(y):=$ {\rm rend} $\; R(M,y)$ is piecewise monotonic-to-left on $M$.
Then  $M$ is isomorphic up to binarity to  $$M'_{s, m, k}:=\langle M,K^3,E_1^2, E^2_2, \ldots, E^2_s, E^2_{s+1}, R^2\rangle,$$
where $M$ is a circularly ordered structure, $M$ is densely ordered, $s\ge 1$;
$E_{s+1}$ is an equivalence relation par\-ti\-tio\-ning $M$ into $m$ infinite convex classes without endpoints;
$E_i$ for every $1\le i\le s$ is an equivalence relation par\-ti\-tio\-ning every $E_{i+1}$-class
into infinitely many infinite convex $E_i$-subclasses without endpoints so that the induced order on
$E_i$-subclasses is dense without endpoints; $R(M,a)$ has no right endpoint in $M$ and $r^k(a)=a$
for all $a\in M$ and some $k\ge 2$, where   $r^k(y):=r(r^{k-1}(y))$;
for every $1\le i\le s+1$ and any $a\in M$
$$M'_{s,m,k}\models \neg E^*_i(a,r(a))\land \forall y (E_i(y,a)\to \exists u [E^*_i(u,r(a))\land E^*_i(u,r(y))]),$$
$m\ge 4$, $k$ is even and $k$ divides $m$; $r$ is monotonic-to-left on every $E_{s+1}$-class and $r$ is monotonic-to-right on $M/E_{s+1}$.
\end{theorem}


In \cite{KS_2022} algebras of binary isolating formulas
are described for $\aleph_0$-categorical weakly circularly minimal theories of
con\-ve\-xi\-ty rank 1 with a 1-transitive non-primitive automorphism
group and a non-trivial definable closure. In \cite{K_2023}--\cite{KS_2024} algebras of binary isolating formulas
are described for $\aleph_0$-categorical weakly circularly minimal theories of
con\-ve\-xi\-ty rank greater than 1 with a 1-transitive non-primitive automorphism
group and a non-trivial definable closure.
In \cite{SMJ_2024} algebras of binary isolating formulas
are described for $\aleph_0$-categorical weakly circularly minimal theories of
con\-ve\-xi\-ty rank 1 with a 1-transitive non-primitive automorphism
group and a trivial definable closure.
Here we describe algebras of binary isolating formulas
 for $\aleph_0$-categorical weakly circularly minimal theories of
convexity rank greater than 1 with a 1-transitive non-primitive automorphism
group and  a trivial definable closure having a piecewise monotonic-to-left function
to the definable completion of a structure for arbitrary $s$ and $m=k=4$.




\section{Results}
\noindent

Let $M$ be an 1-transitive structure. We denote every binary isolating formula
acting in $M$ by a label $u\in\rho_{M}$, where $\rho_{M}$ denotes the set of all labels for the
algebra $\mathcal{P}_{M}$ of binary isolating formulas of the structure $M$.
\begin{definition}\rm \cite{Sud2018} The algebra $\mathcal{P}_{M}$ is said to be
{\it de\-ter\-mi\-nis\-tic} if $u_1\cdot u_2$ is a singleton for any labels $u_1, u_2\in \rho_{M}$.
\end{definition}

Generalizing the last definition, we say that the algebra $\mathcal{P}_{M}$ is {\it $m$-de\-ter\-mi\-nis\-tic}
if the product $u_1\cdot u_2$ consists of at most $m$ elements for any labels $u_1, u_2\in \rho_{M}$. We also say
that an $m$-deterministic algebra $\mathcal{P}_{M}$ is {\it strictly $m$-deterministic} if it is not $(m-1)$-deterministic.



\begin{example} \rm Consider the structure $M'_{1,4,4}:=\langle M,K^3,E^2_1, E^2_2, R^2\rangle$ from Theorem \ref{th2}
with the condition that the function $r(y):=$ {\rm rend} $R(M,y)$ is piecewise monotonic-to-left on $M$.


We assert that $Th(M'_{1,4,4})$ has seventeen binary isolating formulas:
$$\theta_0(x,y):=x=y,$$ $$\theta_1(x,y):=K_0(x,y,r(x))\land E_1(x,y),$$
$$\theta_2(x,y):=K_0(x,y,r(x))\land E_2(x,y) \land \neg E_1(x,y),$$
$$\theta_3(x,y):=K_0(x,y,r(x))\land E^*_2(y, r(x)) \land \neg E^*_1(y,r(x)),$$
$$\theta_4(x,y):=K_0(x,y,r(x))\land E^*_1(y,r(x)),$$
$$\theta_5(x,y):=K_0(r(x),y,r^2(x))\land E^*_1(y,r(x)),$$
$$\theta_6(x,y):=K_0(r(x),y,r^2(x))\land E^*_2(y,r(x))\land \neg E^*_1(y, r(x)),$$
$$\theta_7(x,y):=K_0(r(x),y,r^2(x))\land E^*_2(y,r^2(x))\land \neg E^*_1(y, r^2(x)),$$
$$\theta_8(x,y):=K_0(r(x),y,r^2(x))\land E^*_1(y,r^2(x)),$$
$$\theta_9(x,y):=K_0(r^2(x),y,r^3(x))\land E^*_1(y,r^2(x)),$$
$$\theta_{10}(x,y):=K_0(r^2(x),y,r^3(x))\land E^*_2(y,r^2(x))\land \neg E^*_1(y, r^2(x)),$$
$$\theta_{11}(x,y):=K_0(r^2(x),y,r^3(x))\land E^*_2(y,r^3(x))\land \neg E^*_1(y, r^3(x)),$$
$$\theta_{12}(x,y):=K_0(r^2(x),y,r^3(x))\land E^*_1(y,r^3(x)),$$
$$\theta_{13}(x,y):=K_0(r^3(x),y,x)\land E^*_1(y,r^3(x)),$$
$$\theta_{14}(x,y):=K_0(r^3(x),y,x)\land E^*_2(y,r^3(x))\land \neg E^*_1(y, r^3(x)),$$
$$\theta_{15}(x,y):=K_0(r^3(x),y,x)\land E_2(x,y)\land \neg E_1(x,y),$$
$$\theta_{16}(x,y):=K_0(r^3(x),y,x)\land E_1(x,y)$$

and the following holds for any $a\in M$:
$$K_0(\theta_0(a,M),\theta_1(a,M),\theta_2(a,M),\ldots, \theta_{14}(a,M),\theta_{15}(a,M), \theta_{16}(a,M)).$$

It can be proved that the algebra $\mathfrak{P}_{M'_{1,4,4}}$ is non-commutative and strictly 5-deterministic.

\end{example}



\begin{theorem}\label{prop_s44}
The algebra $\mathfrak{P}_{M'_{s,4,4}}$ of binary
isolating formulas with piecewise monotonic-to-left func\-tion $r$ has $8s+9$ labels, is non-commutative and strictly
$(2s+3)$-deterministic for every $s\ge 1$.
\end{theorem}

\begin{proof}
We assert that the algebra $\mathfrak{P}_{M'_{s,4,4}}$
has $8s+9$ binary isolating formulas:
$$\theta_0(x,y):=x=y,$$
$$\theta_1(x,y):=K_0(x,y,r(x))\land E_1(x,y),$$
$$\theta_{l_1}(x,y):=K_0(x,y,r(x))\land E_{l_1}(x,y)\land \neg E_{l_1-1}(x,y), \mbox{ where } 2\le l_1\le s+1,$$
$$\theta_{l_2}(x,y):=K_0(x,y,r(x))\land E^*_{2s+3-l_2}(y,r(x))\land \neg E^*_{2s+2-l_2}(y,r(x)),$$ $$ \mbox{ where } s+2\le l_2\le 2s+1,$$
$$\theta_{2s+2}(x,y):=K_0(x,y,r(x))\land E^*_1(y,r(x)),$$
$$\theta_{2s+3}(x,y):=K_0(r(x),y,r^2(x))\land E^*_1(y,r(x)),$$
$$\theta_{l_3}(x,y):=K_0(r(x),y,r^2(x))\land E^*_{l_3-(2s+2)}(y, r(x))\land \neg E^*_{l_3-(2s+3)}(y, r(x)),$$ $$\mbox{where } 2s+4\le l_3\le 3s+3,$$
$$\theta_{l_4}(x,y):=K_0(r(x),y,r^2(x))\land E^*_{4s+5-l_4}(y,r^2(x))\land \neg E^*_{4s+4-l_4}(y,r^2(x)),$$ $$\mbox{where } 3s+4\le l_2\le 4s+3,$$
$$\theta_{4s+4}(x,y):=K_0(r(x),y,r^2(x))\land E^*_1(y,r^2(x)),$$
$$\theta_{4s+5}(x,y):=K_0(r^2(x),y,r^3(x))\land E^*_1(y,r^2(x)),$$
$$\theta_{l_5}(x,y):=K_0(r^2(x),y,r^3(x))\land E^*_{l_5-(4s+4)}(y, r^2(x))\land \neg E^*_{l_5-(4s+5)}(y, r^2(x)),$$ $$\mbox{where } 4s+6\le l_5\le 5s+5,$$
$$\theta_{l_6}(x,y):=K_0(r^2(x),y,r^3(x))\land E^*_{6s+7-l_6}(y,r^3(x))\land \neg E^*_{6s+6-l_6}(y,r^3(x)),$$ $$\mbox{where } 5s+6\le l_6\le 6s+5,$$
$$\theta_{6s+6}(x,y):=K_0(r^2(x),y,r^3(x))\land E^*_1(y,r^3(x)),$$
$$\theta_{6s+7}(x,y):=K_0(r^3(x),y,x)\land E^*_1(y,r^3(x)),$$
$$\theta_{l_7}(x,y):=K_0(r^3(x),y,x)\land E^*_{l_7-(6s+6)}(y, r^3(x))\land \neg E^*_{l_7-(6s+7)}(y, r^3(x)),$$ $$\mbox{where } 6s+8\le l_7\le 7s+7,$$
$$\theta_{l_8}(x,y):=K_0(r^3(x),y,x)\land E_{8s+9-l_8}(y,x)\land \neg E_{8s+8-l_8}(y,x),$$ $$\mbox{where } 7s+8\le l_8\le 8s+7,$$
$$\theta_{8s+8}(x,y):=K_0(r^3(x),y,x)\land E_1(x,y).$$

Thus, we have $2+2s+2+2s+2+2s+2+2s+1=8s+9$ binary isolating formulas.
Moreover, we have defined the formulas so that
for any $a\in M$ the following holds:
$$K_0(\theta_0(a,M),\theta_1(a,M),\theta_2(a,M),\ldots, \theta_{8s+6}(a,M),\theta_{8s+7}(a,M),\theta_{8s+8}(a,M)).$$

Prove now that the algebra $\mathfrak{P}_{M'_{s,4,4}}$ is non-commutative and strictly $(2s+3)$-deterministic
 for every $s\ge 1$.

 Firstly, obviously that $0\cdot l=l\cdot 0=\{l\}$ for any $0\le l\le 8s+8$.

Suppose further that  $l_1> 0$  and $l_2> 0$.
Consider the following formula: $$\exists t [\theta_{l_1}(x,t)\land \theta_{l_2}(t,y)].$$

{\it Case 1:} $l_1=1$.

We have: $K_0(x,t, r(x))$ and $E_1(x,t)$.
Let also $l_2=1$, i.e. $K_0(t,y, r(t))$ and $E_1(t,y)$, whence we obtain:
$K_0(x,y,r(x))\land E_1(x,y)$, i.e. $l_1\cdot l_2=\{l_2\}$.

Let now $2\le l_2\le s+1$, i.e. $K_0(t,y,r(t))\land E_{l_2}(t,y)\land \neg E_{l_2-1}(t,y)$.
We have: $l_1\cdot l_2=\{l_2\}$.
Similarly, we have that $l_2\cdot l_1=\{l_2\}$.

Let now $s+2\le l_2\le 2s+1$, i.e. $K_0(t,y,r(t))\land E^*_{2s+3-l_2}(y,r(t))\land \neg E^*_{2s+2-l_2}(y,r(t))$.
We have: $l_1\cdot l_2=\{l_2\}$.
Similarly, we have that $l_2\cdot l_1=\{l_2\}$.

Let now $l_2=2s+2$, i.e. $K_0(t,y,r(t))\land E^*_1(y,r(t))$.
We have: $l_1\cdot l_2=\{2s+2\}$.
Consider $l_2\cdot l_1$. We have that $l_2\cdot l_1=\{2s+2, 2s+3\}$.

Thus, we conclude that $\mathfrak{P}_{M'_{s,4,4}}$ is not commutative for every $s\ge 1$.

Let now $l_2=2s+3$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r(t))$.
We have: $l_1\cdot l_2=\{2s+2, 2s+3\}$.
Consider $l_2\cdot l_1$. We have that $l_2\cdot l_1=\{2s+3\}$.

Let now $2s+4\le l_2\le 3s+3$, i.e. $$K_0(r(t),y,r^2(t))\land E^*_{l_2-(2s+2)}(y, r(t))\land \neg E^*_{l_2-(2s+3)}(y, r(t)).$$

We have: $l_1\cdot l_2=\{l_2\}$.
Similarly, we have that $l_2\cdot l_1=\{l_2\}$.

Let now $3s+4\le l_2\le 4s+3$, i.e. $$K_0(r(t),y,r^2(t))\land E^*_{4s+5-l_2}(y, r^2(t))\land \neg E^*_{4s+4-l_2}(y, r^2(t)).$$

We have: $l_1\cdot l_2=\{l_2\}$.
Similarly, we have that $l_2\cdot l_1=\{l_2\}$.

Let now $l_2=4s+4$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r^2(t))$.
We have: $l_1\cdot l_2=\{4s+4, 4s+5\}$.
Consider $l_2\cdot l_1$. We also have that $l_2\cdot l_1=\{4s+4, 4s+5\}$.

Let now $l_2=4s+5$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^2(t))$.
We have: $l_1\cdot l_2=\{4s+5\}$.
Consider $l_2\cdot l_1$. We also have that $l_2\cdot l_1=\{4s+5\}$.

Let now $4s+6\le l_2\le 5s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)}(y, r^2(t))\land \neg E^*_{l_2-(4s+5)}(y, r^2(t)).$$

We have: $l_1\cdot l_2=\{l_2\}$.
Similarly, we have that $l_2\cdot l_1=\{l_2\}$.

Let now $5s+6\le l_2\le 6s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y, r^3(t))\land \neg E^*_{6s+6-l_2}(y, r^3(t)).$$

We have: $l_1\cdot l_2=\{l_2\}$.
Similarly, we have that $l_2\cdot l_1=\{l_2\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$.
We have: $l_1\cdot l_2=\{6s+6\}$.
Consider $l_2\cdot l_1$. We have that $l_2\cdot l_1=\{6s+6, 6s+7\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$.
We have: $l_1\cdot l_2=\{6s+6, 6s+7\}$.
Consider $l_2\cdot l_1$. We have that $l_2\cdot l_1=\{6s+7\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e. $$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y, r^3(t))\land \neg E^*_{l_2-(6s+7)}(y, r^3(t)).$$

We have: $l_1\cdot l_2=\{l_2\}$.
Similarly, we have that $l_2\cdot l_1=\{l_2\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e. $K_0(r^3(t),y,t)\land E_{8s+9-l_2}(y, t)\land \neg E_{8s+8-l_2}(y, t)$.
We have: $l_1\cdot l_2=\{l_2\}$.
Similarly, we have that $l_2\cdot l_1=\{l_2\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(t,y)$.
We have: $l_1\cdot l_2=\{0,1, 8s+8\}$.
Consider $l_2\cdot l_1$. We have that $l_2\cdot l_1=\{0,1, 8s+8\}$.

\medskip

{\it Case 2:} $2\le l_1\le s+1$.

We have: $K_0(x,t,r(x))\land E_{l_1}(x,t)\land\neg E_{l_1-1}(x,t)$.

Let also $2\le l_2\le s+1$, i.e. $K_0(t,y,r(t))\land E_{l_2}(t,y)\land\neg E_{l_2-1}(t,y)$.
Then we have the following: if $l_1\ge l_2$ then $l_1\cdot l_2=\{l_1\}$; if
$l_1<l_2$ then $l_1\cdot l_2=\{l_2\}$.

Let now $s+2\le l_2\le 2s+1$, i.e. $$K_0(t,y,r(t))\land E^*_{2s+3-l_2}(y,r(t))\land\neg E_{2s+2-l_2}(y,r(t)).$$

We have the following: if $l_1<2s+3-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2\}$.

If $l_1=2s+3-l_2$ then $l_1\cdot l_2=\{l_2\}$ and $l_2\cdot l_1=\{l_2, \ldots, 4s+5-l_2\}$.

If $l_1>2s+3-l_2$ then $l_1\cdot l_2=\{2s+3-l_1\}$ and $l_2\cdot l_1=\{2s+2+l_1\}$.

Let now $l_2=2s+2$, i.e. $K_0(t,y,r(t))\land E^*_1(y,r(t))$.
Then $l_1\cdot l_2=\{2s+3-l_1\}$ and $l_2\cdot l_1=\{2s+2+l_1\}$.
Let now $l_2=2s+3$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r(t))$.
Also, then $l_1\cdot l_2=\{2s+3-l_1\}$ and $l_2\cdot l_1=\{2s+2+l_1\}$.

Let now $2s+4\le l_2\le 3s+3$, i.e. $$K_0(r(t),y,r^2(t))\land E^*_{l_2-(2s+2)}(y,r(t))\land \neg E^*_{l_2-(2s+3)}(y,r(t)).$$

We have the following: if $l_1<l_2-(2s+2)$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2\}$.

If $l_1=l_2-(2s+2)$ then $l_1\cdot l_2=\{2s+3-l_1, \ldots, l_2\}$ and $l_2\cdot l_1=\{l_2\}$.

If $l_1>l_2-(2s+2)$ then $l_1\cdot l_2=\{2s+3-l_1\}$ and $l_2\cdot l_1=\{2s+2+l_1\}$.

Let now $3s+4\le l_2\le 4s+3$, i.e. $$K_0(r(t),y,r^2(t))\land E^*_{4s+5-l_2}(y,r^2(t))\land \neg E^*_{4s+4-l_2}(y,r^2(t)).$$

We have the following: if $l_1<4s+5-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2\}$.

If $l_1=4s+5-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2, \ldots, 8s+9-l_2\}$.

If $l_1>4s+5-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $l_2=4s+4$, i.e. $K_0(r(t), y,r^2(t))\land E^*_1(y,r^2(t))$.
Then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.
Let now $l_2=4s+5$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^2(t))$.
Also, then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $4s+6\le l_2\le 5s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)}(y,r^2(t))\land \neg E^*_{l_2-(4s+5)}(y,r^2(t)).$$

We have the following: if $l_1\le l_2-(4s+4)$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2\}$.

If $l_1>l_2-(4s+4)$ then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $5s+6\le l_2\le 6s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y,r^3(t))\land \neg E^*_{6s+6-l_2}(y,r^3(t)).$$

We have the following:
if $l_1<6s+7-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2\}$.

If $l_1=6s+7-l_2$ then $l_1\cdot l_2=\{l_2\}$ and $l_2\cdot l_1=\{l_2, \ldots, 6s+6+l_1\}$.

If $l_1>6s+7-l_2$ then $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{6s+6+l_1\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t), y,r^3(t))\land E^*_1(y,r^3(t))$.
Then $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{6s+6+l_1\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t), y,t)\land E^*_1(y,r^3(t))$.
Then also $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{6s+6+l_1\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e. $$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following:
if $l_1<l_2-(6s+6)$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2\}$.

If $l_1=l_2-(6s+6)$ then $l_1\cdot l_2=\{6s+7-l_1, \ldots, l_2\}$ and $l_2\cdot l_1=\{l_2\}$.

If $l_1>l_2-(6s+6)$ then $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{6s+6+l_1\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e. $$K_0(r^3(t),y,t)\land E_{8s+9-l_2}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We have the following:
if $l_1<8s+9-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2\}$.
If $l_1=8s+9-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2, \ldots, 8s+8, 0,1, \ldots, l_1\}$.
If $l_1>8s+9-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.
Let now $l_2=8s+8$, i.e. $K_0(r^3(t), y,t)\land E_1(t,y)$.
Then we have $l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.

\medskip

{\it Case 3.} $s+2\le l_1\le 2s+1$.

We have: $K_0(x,t,r(x))\land E^*_{2s+3-l_1}(t,r(x))\land \neg E^*_{2s+2-l_1}(t,r(x))$.

Let also $s+2\le l_2\le 2s+1$, i.e.
$$K_0(t,y,r(t))\land E^*_{2s+3-l_2}(y,r(t))\land \neg E^*_{2s+2-l_2}(y,r(t)).$$

Then we have the following: if $l_1<l_2$ then $l_1\cdot l_2=\{6s+7-l_1\}$ and
$l_2\cdot l_1=\{2s+2+l_1\}$.
If $l_1=l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{2s+2+l_1, \ldots, 6s+7-l_1\}$.
If $l_1>l_2$ then $l_1\cdot l_2=\{ 2s+2+l_2\}$ and $l_2\cdot l_1=\{6s+7-l_2\}$.

Let now $l_2=2s+2$, i.e. $K_0(t,y,r(t))\land E^*_1(y,r(t))$. Then we have
$l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{2s+2+l_1\}$.

Let now $l_2=2s+3$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r(t))$. Then we also have
$l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{2s+2+l_1\}$.

Let now $2s+4\le l_2\le 3s+3$, i.e.

$$K_0(r(t),y,r^2(t))\land E^*_{l_2-(2s+2)}(y,r(t))\land \neg E^*_{l_2-(2s+3)}(y,r(t)).$$

We have the following: if $2s+3-l_1=l_2-(2s+2)$ then $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{2s+2+l_1, \ldots, 6s+7-l_1\}$.
If $2s+3-l_1< l_2-(2s+2)$ then $l_1\cdot l_2=\{2s+2+l_2\}$
and $l_2\cdot l_1=\{6s+7-l_2\}$.
If $2s+3-l_1>l_2-(2s+2)$ then $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{2s+2+l_1\}$.

Let now $3s+4\le l_2\le 4s+3$, i.e.

$$K_0(r(t),y,r^2(t))\land E^*_{4s+5-l_2}(y,r^2(t))\land \neg E^*_{4s+4-l_2}(y,r^2(t)).$$

We have the following: if $2s+3-l_1= 4s+5-l_2$ then $l_1\cdot l_2=\{2s+2+l_2\}$ and
$l_2\cdot l_1=\{2s+2+l_1, \ldots, 10s+11-l_2\}$.

If $2s+3-l_1< 4s+5-l_2$ then $l_1\cdot l_2=\{2s+2+l_2\}$ and
$l_2\cdot l_1=\{10s+11-l_2\}$.

If $2s+3-l_1>4s+5-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $l_2=4s+4$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r^2(t))$. Then we have
$l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $l_2=4s+5$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^2(t))$. Then we also have
$l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let  now $4s+6\le l_2\le 5s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)}(y,r^2(t))\land \neg E^*_{l_2-(4s+5)}(y,r^2(t)).$$

We have the following: if $2s+3-l_1>l_2-(4s+4)$ then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.
If $2s+3-l_1=l_2-(4s+4)$ then $l_1\cdot l_2=\{4s+4+l_1, \ldots, 2s+2+l_2\}$ and $l_2\cdot l_1=\{4s+4+l_1\}$.
If $2s+3-l_1<l_2-(4s+4)$ then $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let  now $5s+6\le l_2\le 6s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y,r^3(t))\land \neg E^*_{6s+6-l_2}(y,r^3(t)).$$

We have the following: if $2s+3-l_1>6s+7-l_2$ then $l_1\cdot l_2=\{2s+3-l_1\}$ and $l_2\cdot l_1=\{6s+6+l_1\}$.
If $2s+3-l_1=6s+7-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{2s+2+l_2, \ldots, 8s+8, 0, 1, \ldots, 6s+7-l_2\}$.
If $2s+3-l_1<6s+7-l_2$ then $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{6s+7-l_2\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$. Then we have
$l_1\cdot l_2=\{2s+3-l_1\}$ and $l_2\cdot l_1=\{6s+6+l_1\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$. Then we also have
$l_1\cdot l_2=\{2s+3-l_1\}$ and $l_2\cdot l_1=\{6s+6+l_1\}$.

Let  now $6s+8\le l_2\le 7s+7$, i.e. $$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: if $2s+3-l_1\ge l_2-(6s+6)$ then $l_1\cdot l_2=\{2s+3-l_1\}$ and $l_2\cdot l_1=\{6s+6+l_1\}$.
If $2s+3-l_1<l_2-(6s+6)$ then $l_1\cdot l_2=\{l_2-(6s+6)\}$ and $l_2\cdot l_1=\{14s+15-l_2\}$.

Let  now $7s+8\le l_2\le 8s+7$, i.e. $$K_0(r^3(t),y,t)\land E^*_{8s+9-l_2}(y,t)\land \neg E^*_{8s+8-l_2}(y,t).$$

We have the following: if $2s+3-l_1> 8s+9-l_2$ then $l_1\cdot l_2=\{2s+3-l_1\}$ and $l_2\cdot l_1=\{l_1\}$.
If $2s+3-l_1=8s+9-l_2$ then $l_1\cdot l_2=\{l_1\}$ and $l_2\cdot l_1=\{l_1, \ldots, 4s+5-l_1\}$.
If $2s+3-l_1<8s+9-l_2$ then $l_1\cdot l_2=\{l_2-(6s+6)\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have
$l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.

\medskip

{\it Case 4.} $l_1=2s+2$.

We have: $K_0(x,t,r(x))\land E^*_1(t,r(x))$.

Let also $l_2=2s+2$, i.e. $K_0(t,y,r(t))\land E^*_1(y,r(t))$. Then we have $l_1\cdot l_2=\{4s+4, 4s+5\}$.
Let now $l_2=2s+3$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r(t))$. Then we have
$l_1\cdot l_2=\{4s+5\}$ and $l_2\cdot l_1=\{4s+4\}$.

Let now $2s+4\le l_2\le 3s+3$, i.e. $$K_0(r(t),y,r^2(t))\land E^*_{l_2-(2s+2)}(y,r(t))\land \neg E^*_{l_2-(2s+3)}(y,r(t)).$$

We have the following: $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{6s+7-l_2\}$.

Let now $3s+4\le l_2\le 4s+3$, i.e. $$K_0(r(t),y,r^2(t))\land E^*_{4s+5-l_2}(y,r^2(t))\land \neg E^*_{4s+4-l_2}(y,r^2(t)).$$

We have the following: $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let now $l_2=4s+4$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r^2(t))$. Then we have
$l_1\cdot l_2=\{6s+6\}$ and $l_2\cdot l_1=\{6s+6, 6s+7\}$.

Let now $l_2=4s+5$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^2(t))$. Then we have
$l_1\cdot l_2=\{6s+6, 6s+7\}$ and $l_2\cdot l_1=\{6s+6\}$.

Let now $4s+6\le l_2\le 5s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)}(y,r^2(t))\land \neg E^*_{l_2-(4s+5)}(y,r^2(t)).$$

We have the following: $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let now $5s+6\le l_2\le 6s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y,r^3(t))\land \neg E^*_{6s+6-l_2}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{6s+7-l_2\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$. Then we have
$l_1\cdot l_2=l_2\cdot l_1=\{8s+8, 0, 1\}$.
Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$. Then we have
$l_1\cdot l_2=\{1\}$ and $l_2\cdot l_1=\{8s+8\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e. $$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=\{l_2-(6s+6)\}$ and $l_2\cdot l_1=\{14s+15-l_2\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e. $$K_0(r^3(t),y,t)\land E_{8s+9-l_2}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We have the following: $l_1\cdot l_2=\{l_2-(6s+6)\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have
$l_1\cdot l_2=\{2s+2\}$ and $l_2\cdot l_1=\{2s+2,2s+3\}$.

\medskip

{\it Case 5.} $l_1=2s+3$.

We have: $K_0(r(x),t,r^2(x))\land E^*_1(t,r(x))$.

Let also $l_2=2s+3$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r(t))$. Then we have
$l_1\cdot l_2=\{4s+4, 4s+5\}$.
Let now $2s+4\le l_2\le 3s+3$, i.e. $$K_0(r(t),y,r^2(t))\land E^*_{l_2-(2s+2)}(y,r(t))\land \neg E^*_{l_2-(2s+3)}(y,r(t)).$$

We have the following: $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{6s+7-l_2\}$.

Let now $3s+4\le l_2\le 4s+3$, i.e. $$K_0(r(t),y,r^2(t))\land E^*_{4s+5-l_2}(y,r^2(t))\land \neg E^*_{4s+4-l_2}(y,r^2(t)).$$

We have the following: $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let now $l_2=4s+4$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r^2(t))$. Then we have
$l_1\cdot l_2=\{6s+6, 6s+7\}$ and $l_2\cdot l_1=\{6s+7\}$.

Let now $l_2=4s+5$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^2(t))$. Then we have
$l_1\cdot l_2=\{6s+7\}$ and $l_2\cdot l_1=\{6s+6, 6s+7\}$.

Let now $4s+6\le l_2\le 5s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)}(y,r^2(t))\land \neg E^*_{l_2-(4s+5)}(y,r^2(t)).$$

We have the following: $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let now $5s+6\le l_2\le 6s+5$, i.e. $$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y,r^3(t))\land \neg E^*_{6s+6-l_2}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{6s+7-l_2\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$. Then we have
$l_1\cdot l_2=\{8s+8\}$ and $l_2\cdot l_1=\{1\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$. Then we have
$l_1\cdot l_2=l_2\cdot l_1=\{8s+8, 0, 1\}$.
Let now $6s+8\le l_2\le 7s+7$, i.e. $$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=\{l_2-(6s+6)\}$ and $l_2\cdot l_1=\{14s+15-l_2\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e. $$K_0(r^3(t),y,t)\land E_{8s+9-l_2}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We have the following: $l_1\cdot l_2=\{l_2-(6s+6)\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have
$l_1\cdot l_2=\{2s+2, 2s+3\}$ and $l_2\cdot l_1=\{2s+3\}$.

\medskip

{\it Case 6.} $2s+4\le l_1\le 3s+3$.

We have:  $K_0(r(x),t,r^2(x))\land E^*_{l_1-(2s+2)}(t,r(x))\land \neg E^*_{l_1-(2s+3)}(t,r(x))$.

Let also $2s+4\le l_2\le 3s+3$, i.e.
$$K_0(r(t),y,r^2(t))\land E^*_{l_2-(2s+2)}(y,r(t))\land \neg E^*_{l_2-(2s+3)}(y,r(t)).$$

We have the following: if $l_1=l_2$ then $l_1\cdot l_2=\{6s+7-l_2, \ldots, 2s+2+l_2\}$.

If $l_1<l_2$ then $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{6s+7-l_2\}$.

If $l_1>l_2$ then $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{2s+2+l_1\}$.

Let now $3s+4\le l_2\le 4s+3$, i.e.
$$K_0(r(t),y,r^2(t))\land E^*_{4s+5-l_2}(y,r^2(t))\land \neg E^*_{4s+4-l_2}(y,r^2(t)).$$

We have the following: if $l_1-(2s+2)<4s+5-l_2$ then $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.
If $l_1-(2s+2)=4s+5-l_2$ then $l_1\cdot l_2=\{2s+2+l_2, \ldots, 10s+11-l_2\}$ and $l_2\cdot l_1=\{4s+4+l_2\}$.
If $l_1-(2s+2)>4s+5-l_2$ then $l_1\cdot l_2=\{4s+4+l_1\}$ and $l_2\cdot l_1=\{8s+9-l_1\}$.

Let now $l_2=4s+4$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r^2(t))$. Then we have
$l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $l_2=4s+5$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^2(t))$. Then we also have
$l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $4s+6\le l_2\le 5s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)}(y,r^2(t))\land \neg E^*_{l_2-(4s+5)}(y,r^2(t)).$$

We have the following: if $l_1-(2s+2)<l_2-(4s+4)$ then $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.
If $l_1-(2s+2)=l_2-(4s+4)$ then $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{10s+11-l_2, \ldots, 2s+2+l_2\}$.
If $l_1-(2s+2)>l_2-(4s+4)$ then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $5s+6\le l_2\le 6s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y,r^3(t))\land \neg E^*_{6s+6-l_2}(y,r^3(t)).$$

We have the following: if $l_1-(2s+2)\le 6s+7-l_2$ then $l_1\cdot l_2=\{2s+2+l_2\}$ and $l_2\cdot l_1=\{6s+7-l_2\}$.
If $l_1-(2s+2)> 6s+7-l_2$ then $l_1\cdot l_2=\{10s+11-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$. Then we have $l_1\cdot l_2=\{10s+11-l_1\}$
and $l_2\cdot l_1=\{l_1-(2s+2)\}$.
Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$. Then we also have $l_1\cdot l_2=\{10s+11-l_1\}$
and $l_2\cdot l_1=\{l_1-(2s+2)\}$.
Let now $6s+8\le l_2\le 7s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: if $l_1-(2s+2)<l_2-(6s+6)$ then $l_1\cdot l_2=\{l_2-(6s+6)\}$ and $l_2\cdot l_1=\{14s+15-l_2\}$.
If $l_1-(2s+2)=l_2-(6s+6)$ then $l_1\cdot l_2=l_2\cdot l_1=\{10s+11-l_1, \ldots, 0, \ldots, l_1-(2s+2)\}$.
If $l_1-(2s+2)>l_2-(6s+6)$ then $l_1\cdot l_2=\{10s+11-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e.
$$K_0(r^3(t),y,t)\land E_{8s+9-l_2}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We have the following: if $l_1-(2s+2)<8s+9-l_2$ then $l_1\cdot l_2=\{l_2-(6s+6)\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.
If $l_1-(2s+2)=8s+9-l_2$ then $l_1\cdot l_2=\{4s+5-l_1, \ldots, l_1\}$ and $l_2\cdot l_1=\{l_1\}$.
If $l_1-(2s+2)>8s+9-l_2$ then $l_1\cdot l_2=\{10s+11-l_1\}$ and $l_2\cdot l_1=\{l_1\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have
$l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.

\medskip

{\it Case 7.} $3s+4\le l_1\le 4s+3$.

We have: $K_0(r(x),t,r^2(x))\land E^*_{4s+5-l_1}(t,r^2(x))\land \neg E^*_{4s+4-l_1}(t,r^2(x))$.

Let also $3s+4\le l_2\le 4s+3$, i.e.
$$K_0(r(t),y,r^2(t))\land E^*_{4s+5-l_2}(y,r^2(t))\land \neg E^*_{4s+4-l_2}(y,r^2(t)).$$

We have the following: if $l_1\le l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

If $l_1> l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_2\}$.

Let now $l_2=4s+4$, i.e. $K_0(r(t),y,r^2(t))\land E^*_1(y,r^2(t))$. Then we have
$l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $l_2=4s+5$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^2(t))$. Then we also have
$l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.
Let now $4s+6\le l_2\le 5s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)}(y,r^2(t))\land \neg E^*_{l_2-(4s+5)}(y,r^2(t)).$$

We have the following: if $4s+5-l_1<l_2-(4s+4)$ then $l_1\cdot l_2=\{14s+5-l_2\}$ and
$l_2\cdot l_1=\{l_2-(4s+4)\}$.
If $4s+5-l_1=l_2-(4s+4)$ then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1, \ldots, 40, 0, \ldots, 4s+5-l_1\}$.
If $4s+5-l_1>l_2-(4s+4)$ then $l_1\cdot l_2=l_2\cdot l_1=\{4s+4+l_1\}$.

Let now $5s+6\le l_2\le 6s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y,r^3(t))\land \neg E^*_{6s+6-l_2}(y,r^3(t)).$$

We have the following: if $4s+5-l_1<6s+7-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.
If $4s+5-l_1=6s+7-l_2$ then $l_1\cdot l_2=\{l_2-(4s+4), \ldots, 6s+7-l_1\}$ and $l_2\cdot l_1=\{l_2-(4s+4)\}$.
If $4s+5-l_1>6s+7-l_2$ then $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$. Then we have
$l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$. Then we also have
$l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: if $4s+5-l_1<l_2-(6s+6)$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.
If $4s+5-l_1=l_2-(6s+6)$ then $l_1\cdot l_2=\{l_2-(4s+4)\}$ and $l_2\cdot l_1=\{8s+9-l_2, \ldots, l_2-(4s+4)\}$.
If $4s+5-l_1>l_2-(6s+6)$ then $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e.
$$K_0(r^3(t),y,t)\land E_{8s+9-l_2}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We have the following: if $4s+5-l_1<8s+9-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.
If $4s+5-l_1\ge 8s+9-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have
$l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.
\medskip

{\it Case 8.} $l_1=4s+4$.

We have: $K_0(r(x),t,r^2(x))\land E_1^*(t,r^2(x))$.

Let also $l_2=4s+4$, i.e. $K_0(r(t),y,r^2(t))\land E_1^*(y,r^2(t))$. Then we have $l_1\cdot l_2=\{8s+8\}$.
Let now $l_2=4s+5$, i.e. $K_0(r^2(t),y,r^3(t))\land E_1^*(y,r^2(t))$. Then we have $l_1\cdot l_2=\{0,1,8s+8\}$
and $l_2\cdot l_1=\{2s+2\}$.

Let now $4s+6\le l_2\le 5s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)} (y,r^2(t))\land \neg E^*_{l_2-(4s+5)}(y,r^2(t)).$$

We have the following: $l_1\cdot l_2=\{6s+7-l_2\}$ and $l_2\cdot l_1=\{l_2-(4s+4)\}$.

Let now $5s+6\le l_2\le 6s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2} (y,r^3(t))\land \neg E^*_{6s+6-l_2}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=\{l_2-(4s+4)\}$ and $l_2\cdot l_1=\{l_2-(3s+2)\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E_1^*(y,r^3(t))$. Then we have $l_1\cdot l_2=\{2s+2, 2s+3\}$
and $l_2\cdot l_1=\{2s+2\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E_1^*(y,r^3(t))$. Then we have $l_1\cdot l_2=\{2s+3\}$
and $l_2\cdot l_1=\{2s+2, 2s+3\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e.

$$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e.

$$K_0(r^3(t),y,t)\land E_{8s+9-l_2}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We also have the following: $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have $l_1\cdot l_2=l_2\cdot l_1=\{4s+4\}$.
\medskip

{\it Case 9.} $l_1=4s+5$.

We have: $K_0(r^2(x),t,r^3(x))\land E^*_1(t,r^2(x))$.

Let also $l_2=4s+5$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^2(t))$. Then we have $l_1\cdot l_2=\{1\}$.
Let now $4s+6\le l_2\le 5s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)}(y,r^2(t))\land \neg E^*_{l_2-(4s+5)}(y,r^2(t)).$$

We have the following: $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.

Let now $5s+6\le l_2\le 6s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y,r^3(t))\land \neg E^*_{6s+6-l_2}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$. Then we have
$l_1\cdot l_2=\{2s+2\}$ and $l_2\cdot l_1=\{2s+2, 2s+3\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$. Then we have
$l_1\cdot l_2=\{2s+2, 2s+3\}$ and $l_2\cdot l_1=\{2s+3\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e.
$$K_0(r^3(t),y,t)\land E_{8s+9-l_2}(y,t)\land \neg E^*_{8s+8-l_2}(y,t).$$

We have the following: $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$.
Then we have $l_1\cdot l_2=l_2\cdot l_1=\{4s+4, 4s+5\}$.
\medskip

{\it Case 10.} $4s+6\le l_1\le 5s+5$.

We have: $K_0(r^2(x),t,r^3(x))\land E^*_{l_1-(4s+4)}(t,r^2(x))\land \neg E^*_{l_1-(4s+5)}(t,r^2(x))$.

Let also $4s+6\le l_2\le 5s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{l_2-(4s+4)}(y,r^2(t))\land \neg E^*_{l_2-(4s+5)}(y,r^2(t)).$$

We have the following: if $l_1<l_2$ then $l_1\cdot l_2=\{l_2-(4s+4)\}$, if $l_1\ge l_2$ then
$l_1\cdot l_2=\{l_1-(4s+4)\}$.
Let now $5s+6\le l_2\le 6s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y,r^3(t))\land \neg E^*_{6s+6-l_2}(y,r^3(t)).$$

We have the following: if $l_1-(4s+4)<6s+7-l_2$ then $l_1\cdot l_2=\{l_2-(4s+4)\}$,
if $l_1-(4s+4)\ge 6s+7-l_2$ then $l_1\cdot l_2=\{l_1-(3s+4)\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$.

We have: $l_1\cdot l_2=\{l_1-(3s+4)\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$.

We have: $l_1\cdot l_2=\{l_1-(3s+4)\}$ and $l_2\cdot l_1=\{6s+7-l_2\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: if $l_1-(4s+4)=l_2-(6s+6)$ then $l_1\cdot l_2=\{l_1-(4s+3), l_1-(4s+2), \ldots, l_1-(2s+2)\}$
and $l_2\cdot l_1=\{l_2-(4s+4)\}$.

If $l_1-(4s+4)<l_2-(6s+6)$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.

If $l_1-(4s+4)>l_2-(6s+6)$ then $l_1\cdot l_2=\{6s+7-l_1\}$ and $l_2\cdot l_1=\{l_2-(2s+2)\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{8s+9-l_2}(y,t)\land \neg E^*_{8s+8-l_2}(y,t).$$

We have the following: if $l_1-(4s+4)=8s+9-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{8s+9-l_1, 8s+8-l_1, \ldots, l_1\}$.

If $l_1-(4s+4)<8s+9-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_2-(4s+4)\}$.

If $l_1-(4s+4)>8s+9-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$.
Then we have $l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.
\medskip

{\it Case 11.} $5s+6\le l_1\le 6s+5$.

We have: $K_0(r^2(x),t,r^3(x))\land E^*_{6s+7-l_1}(t,r^2(x))\land \neg E^*_{6s+6-l_1}(t,r^2(x))$.

Let also $5s+6\le l_2\le 6s+5$, i.e.
$$K_0(r^2(t),y,r^3(t))\land E^*_{6s+7-l_2}(y,r^2(t))\land \neg E^*_{6s+6-l_2}(y,r^2(t)).$$

We have the following: if $l_1=l_2$ then $l_1\cdot l_2=\{10s+11-l_1\}$.
If $l_1<l_2$ then $l_1\cdot l_2=\{10s+11-l_1\}$.
If $l_1>l_2$ then $l_1\cdot l_2=\{l_1-(2s+2)\}$.

Let now $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$.

We have: $l_1\cdot l_2=\{10s+11-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$.

We have: $l_1\cdot l_2=\{10s+11-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: if $6s+7-l_1=l_2-(6s+6)$ then
$l_1\cdot l_2=\{10s+11-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.
If $6s+7-l_1<l_2-(6s+6)$ then $l_1\cdot l_2=\{l_2-(2s+2)\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.
If $6s+7-l_1>l_2-(6s+6)$ then $l_1\cdot l_2=\{10s+11-l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{8s+9-l_2}(y,t)\land \neg E^*_{8s+8-l_2}(y,t).$$

We have the following: if $6s+7-l_1=8s+9-l_2$ then $l_1\cdot l_2=\{10s+11-l_1\}$ and
$l_2\cdot l_1=\{l_1, l_1+1, \ldots, 14s+15-l_2 (12s+13-l_1)\}$.
If $6s+7-l_1<8s+9-l_2$ then $l_1\cdot l_2=\{l_2-(2s+2)\}$ and $l_2\cdot l_1=\{14s+15-l_2\}$.
If $6s+7-l_1>8s+9-l_2$ then $l_1\cdot l_2=\{l_1\}$ and $l_2\cdot l_1=\{l_1-(2s+2)\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$.
Then we have $l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.
\medskip

{\it Case 12.} $l_1=6s+6$.

We have: $K_0(r^2(x),t,r^3(x))\land E^*_1(t,r^3(x))$.
Let also $l_2=6s+6$, i.e. $K_0(r^2(t),y,r^3(t))\land E^*_1(y,r^3(t))$. Then we have $l_1\cdot l_2=\{4s+4, 4s+5\}$.

Let now $l_2=6s+7$, i.e. $K_0(r^3(t),y,t)\land E^*_1(y,r^3(t))$.

We have the following: $l_1\cdot l_2=\{4s+5\}$ and $l_2\cdot l_1=\{4s+4\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=\{l_2-(2s+2)\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e.
$$K_0(r^3(t),y,t)\land E_{8s+9-l_2)}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We have the following: $l_1\cdot l_2=\{l_2-(2s+2)\}$ and $l_2\cdot l_1=\{14s+15-l_2\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have $l_1\cdot l_2=\{6s+6\}$ and
$l_2\cdot l_1=\{6s+6, 6s+7\}$.
\medskip

{\it Case 13.} $l_1=6s+7$.

We have: $K_0(r^3(x),t,x)\land E^*_1(t, r^3(x))$. Let also $l_2=6s+7$,

i.e.
$K_0(r^3(t),y,t)\land E^*_1(y, r^3(t))$. Then we have $l_1\cdot l_2=\{4s+4, 4s+5\}$.

Let now $6s+8\le l_2\le 7s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: $l_1\cdot l_2=\{l_2-(2s+2)\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e.
$$K_0(r^3(t),y,t)\land E_{8s+9-l_2)}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We have the following: $l_1\cdot l_2=\{l_2-(2s+2)\}$ and $l_2\cdot l_1=\{14s+15-l_2\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have $l_1\cdot l_2=\{6s+6, 6s+7\}$ and
$l_2\cdot l_1=\{6s+7\}$.
\medskip

{\it Case 14.} $6s+8\le l_1\le 7s+7$.

We have: $K_0(r^3(x),t,x)\land E^*_{l_1-(6s+6)}(t,r^3(x))\land \neg E^*_{l_1-(6s+7)}(t,r^3(x))$.

Let also $6s+8\le l_2\le 7s+7$, i.e.
$$K_0(r^3(t),y,t)\land E^*_{l_2-(6s+6)}(y,r^3(t))\land \neg E^*_{l_2-(6s+7)}(y,r^3(t)).$$

We have the following: if $l_1=l_2$ then $l_1\cdot l_2=\{10s+11-l_1, \ldots, l_1-(2s+2)\}$.

If $l_1<l_2$ then $l_1\cdot l_2=\{l_2-(2s+2)\}$ and $l_2\cdot l_1=\{10s+11-l_2\}$.

If $l_1>l_2$ then $l_1\cdot l_2=\{10s+11-l_2\}$ and $l_2\cdot l_1=\{l_2-(2s+2)\}$.

Let now $7s+8\le l_2\le 8s+7$, i.e.
$$K_0(r^3(t),y,t)\land E_{8s+9-l_2)}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We have the following:  if $l_1-(6s+6)=8s+9-l_2$ then $l_1\cdot l_2=\{12s+13-l_1, \ldots, l_1\}$
and $l_2\cdot l_1=\{14s+15-l_2\}$.
If $l_1-(6s+6)<8s+9-l_2$ then $l_1\cdot l_2=\{l_2-(2s+2)\}$ and $l_2\cdot l_1=\{14s+15-l_2\}$.
If $l_1-(6s+6)>8s+9-l_2$ then $l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have $l_1\cdot l_2=\{10s+11-l_1\}$ and
$l_2\cdot l_1=\{l_1\}$.
\medskip

{\it Case 15.} $7s+8\le l_1\le 8s+7$.

We have: $K_0(r^3(x),t,x)\land E_{8s+9-l_1)}(t,x)\land \neg E_{8s+8-l_1}(t,x)$.

Let also $7s+8\le l_2\le 8s+7$, i.e.
$$K_0(r^3(t),y,t)\land E_{8s+9-l_2)}(y,t)\land \neg E_{8s+8-l_2}(y,t).$$

We have the following: if $l_1=l_2$ then $l_1\cdot l_2=\{l_1\}$.
If $l_1<l_2$ then $l_1\cdot l_2=\{l_2\}$. If $l_1>l_2$ then $l_1\cdot l_2=\{l_1\}$.

Let now $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have $l_1\cdot l_2=l_2\cdot l_1=\{l_1\}$.
\medskip

{\it Case 16.} $l_1=8s+8$.

We have: $K_0(r^3(x),t,x)\land E_1(t,x)$. Let also $l_2=8s+8$, i.e. $K_0(r^3(t),y,t)\land E_1(y,t)$. Then we have $l_1\cdot l_2=\{l_1\}$.

Thus, we established that the algebra $\mathfrak{P}_{M'_{s,m,k}}$ is non-commutative and strictly $(2s+3)$-deterministic
 for all valid values $s, m$ and $k$.
\end{proof}

\bigskip

\section{Conclusion}
We investigated algebras of binary isolating formulas for $\aleph_0$-categorical 1-transitive
non-primitive weakly circularly
minimal theories of convexity rank greater than 1 with a trivial definable closure having a
non-trivial piecewise monotonic-to-left function  acting on the universe of a structure with an arbitrary $s$ and $m=k=4$. We also proved
their non-commutativity and established their strict  $l$-deterministicity  for some natural $l$.
It would now be interesting to describe the corresponding algebras for such theories having
arbitrary $s, m$ and $k$.





\bigskip
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