Free Algebras of Full Terms Generated by Order-Preserving Transformations

Abstract

Full terms, which serve as tools for classifying algebras into subclasses, can be studied using an algebraic approach. For a natural number n, this paper introduces the algebra of order-preserving full terms under the $(n+1)$-superposition operation satisfying the superassociativity using mappings on the set $O_n$ of all order-preserving transformations on a finite chain $\{1\le \cdots \le n\}$. We prove the freeness property of such algebra with respect to the variety of superassociative algebras. Additionally, binary operations on the powerset of order-preserving full terms whose elements are called tree languages are discussed. To define order-preserving identities and order-preserving varieties, the left-seminearring of full hypersubstitutions is determined. The required characteristics for any identity to be an order-preserving identity are considered. Furthermore, we also discuss the homomorphism of full hypersubstitutions with other algebraic structures.

1. Introduction

Recall from [1] that a full term, a term in which all of the operations of the given signature are allowed and the structure is unrestricted, is one of the generalizations of the words that are constructed from an alphabet $X_n:=\{x_1,\dots ,x_n\}$ for a natural number $n\in \mathbb{N}:=\{1,2,\dots \}$ and a given signature, also known as operation symbols. Basically, we consider two concrete examples. In a Boolean algebra, if we consider the Boolean algebra signature $\{\vee ,\wedge ,\neg ,0,1\}$, $(x\wedge y)\vee (\neg z)$ is a full term because it uses the binary operations ∨ and ∧, the unary operation ¬, and the variables $x,y,z$. In contrast, $x\wedge y$ is not a full term because it does not use $\vee ,\neg ,0$, or 1. Another example is the expression $(x·y)·z^{-1}$ on a group G with the signature $\{·,e{,}^{-1}\}$, where · is a binary associative operation on G, e is the identity, and ⁻¹ denotes inversion. On the other hand, $x·e$ is not a full term.

By definition, for a type ${\tau }_n={\left(n_i\right)}_{i\in I}$ with $n_i=n$ for all n-ary operation symbols $f_i$ indexed by some non-empty set I, the set $W_{{\tau }_n}^F\left(X_n\right)$ of all n-ary full terms of type ${\tau }_n$ is the smallest set closed under finite application of the following inductive steps: (1) $f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ is an n-ary full term of type ${\tau }_n$ where $\alpha$ is a mapping on a finite set $\{1,\dots ,n\}$ and $f_i$ is an operation symbol of type ${\tau }_n$ and (2) $f_i(t_1,\dots ,t_n)$ is an n-ary full term of type ${\tau }_n$ where $t_1,\dots ,t_n$ are n-ary full terms of type ${\tau }_n$ and $f_i$ is an n-ary operation symbol of type ${\tau }_n$. For more details, we refer the reader to the comprehensive monograph [2]. It is known that an algebra consisting of the set $W_{{\tau }_n}^F\left(X_n\right)$ and the superposition $S^n$ of arity $(n+1)$ satisfies the axiom of superassociativity and is a member of a variety of superassociative algebras [3,4,5,6]. The monograph [2] provides interesting facts in the study of superassociative algebras of multiplace functions. Furthermore, algebralization of terms in various classes has been revealed in many papers; for instance, see [7,8].

Actually, full terms always play a crucial rule in solid varieties, which are closed varieties of algebras under hypersubstitutions. In essence, hypersubstitutions modify terms by systematically replacing the operation symbols with corresponding terms. This provides a way to study the behavior of the terms under structural changes or transformations. When applying a hypersubstitution, the full terms are transformed, and their behavior determines whether certain algebraic identities or properties hold in the corresponding algebraic system. For instance, on a group G with the signature $\{·,e{,}^{-1}\}$, a hypersubstitution $\sigma$ defined by $\sigma (·)=x·e,\sigma \left(e\right)=e$ and $\sigma {(}^{-1})=e$ can transform a full term $(x·y)·z^{-1}$ into another one. In fact, by using the superposition operation, the extension $\widehat{\sigma }$ of $\sigma$ turns $(x·y)·z^{-1}$ into another term. Particularly, we call a hypersubstitution whose images are full terms a full hypersubstitution of type $\tau$. Algebraically, the set of all full hypersubstitutions forms a monoid under a binary operation ${\circ }_h$ and an identity mapping ${\sigma }_{id}$ defined by ${\sigma }_{id}\left(f_i\right)=f_i(x_1,\dots ,x_n)$. Recent trends in this domain can be found in [9,10].

Although a full term is defined by a transformation $\alpha$ in a full transformation semigroup $T_n:=\{\alpha \mid \alpha :\overline{n}\to \overline{n}\}$ under the usual composition where $\overline{n}:=\{1,\dots ,n\}$, there are other possibilities of applying other classes of transformation semigroups for subclasses of full terms. The following are some developments in this line. The set $W_{{\tau }_n}^{SF}\left(X_n\right)$ of all strongly full terms induced by an identity transformation $\left(\begin{array}{cccc}1& 2& \cdots & n\\ 1& 2& \cdots & n\end{array}\right)$ is a subalgebra of the set of full terms. In [11], a subsemigroup $K^{*}(n,r):=\{\alpha \in T_n\mid \left|im\left(\alpha \right)\right|\le r\}\cup \left\{1_{id}\right\}$ for constructing an n-ary $K^{*}(n,r)$-full term of type ${\tau }_n$. $O{D}_n:=\{\alpha \in T_n\mid \alpha \left(k\right)\le k,$$\forall k\in \{1,\dots ,n\}\}$. Thus, the set $W_{{\tau }_n}^{O{D}_n}\left(X_n\right)$ of all order-decreasing full terms of type ${\tau }_n$ is examined. Full terms determined by transformations with a restricted range are discussed in [12]. In 2023, Kumduang and Sriwongsa introduced a full term preserving a partition. The algebra for the quantifier-free formulas generated by such a term is constructed. See the paper [9] for this topic. Recently, for a fixed subset Y of $\{1,\dots ,n\}$, $Fix(\overline{n},Y):=\{\alpha \in T_n\mid \alpha \left(a\right)=a,\forall a\in Y\}$ has been defined as a transformation semigroup with a fixed set. A full term constructed from transformations with a fixed set called a $Fix(I_n,Y)$-full term is presented using such a transformation. See [13] for more information.

Recall that a transformation $\alpha \in T_n$ is called an order-preserving transformation if $x\le y$, and then $\alpha \left(x\right)\le \alpha \left(y\right)$ for all $x,y\in \overline{n}$. The symbol $O_n$ stands for the set of all order-preserving transformations on $\overline{n}$, i.e.,

$$O_n:=\{\alpha \in T_n\mid \mathrm{if}x\le y,\mathrm{then}\alpha \left(x\right)\le \alpha \left(y\right)\}$$

for all $x,y\in \overline{n}$. Refer to [14,15,16,17,18,19] for more details pertaining to order-preserving transformations. This paper aims to apply order-preserving transformations to the set $O_n$ to define a novel class of full terms and study several algebraic properties. The presentation of this paper is organized as follows. In Section 2, the algebra of n-ary order-preserving full terms of type ${\tau }_n$, together with the appropriate operation, is established. We also study the free algebra for full terms constructed from order-preserving transformations. In Section 3, we consider tree languages over the set of all order-preserving full terms of type ${\tau }_n$. Moreover, we define the binary operations on the set of all n-ary order-preserving full terms of type ${\tau }_n$ derived from the superposition operation and demonstrate that these operations possess the associative property. Algebras of mapping whose images are order-preserving full terms, which include an identity element, are constructed and investigated in Section 4. Subclasses of identities arising from order-preserving full terms and varieties of algebras that satisfy these identities are examined. Finally, we discuss the homomorphisms of the algebras of the order-preserving full terms.

2. The Superassociative Algebra of Full Terms Defined by Order-Preserving Transformations

In this section, we aim to construct a novel algebra which satisfies the superassociative law using the concept of order-preserving transformations. Firstly, we inductively define the set of n-ary order-preserving full terms of type ${\tau }_n$ as follows:

Definition 1. 

Let n be a positive integer. An n-ary order-preserving full term of type ${\tau }_n$ is inductively defined by

(1) 

$f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ is an n-ary order-preserving full term of type ${\tau }_n$ if $f_i$ is an n-ary operation symbol and α is an order-preserving transformation on $\overline{n}$.

(2) 

$f_i(t_1,\dots ,t_n)$ is an n-ary order-preserving full term of type ${\tau }_n$ if $f_i$ is an n-ary operation symbol and $t_1,\dots ,t_n$ are n-ary order-preserving full terms of type ${\tau }_n$.

The set of all n-ary order-preserving full terms of type ${\tau }_n$, closed under finite applications of (2), is denoted by $W_{{\tau }_n}^{O_n}\left(X_n\right)$.

Some examples of n-ary order-preserving full terms of type ${\tau }_n$ are shown in the following examples.

Example 1. 

Consider a type ${\tau }_3=\left(3\right)$ with a ternary operation g. Since $O_3=\{\left(\begin{array}{ccc}1& 2& 3\\ 1& 1& 1\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 1& 1& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 1& 1& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 2& 2\end{array}\right)$ $\left(\begin{array}{ccc}1& 2& 3\\ 2& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 3& 3\end{array}\right)\}$, then $g(x_1,x_1,x_1),g(x_1,x_1,x_2),g(x_1,x_1,x_3),$ $g(x_1,x_2,x_2),g(x_1,x_2,x_3),g(x_1,x_3,x_3),g(x_2,x_2,x_2),g(x_2,x_2,x_3),g(x_2,x_3,x_3),g(x_3,x_3,x_3),$ $g(g(x_1,x_2,x_3),g(x_3,x_3,x_3),g(x_1,x_1,x_2))\in W_{{\tau }_3}^{O_3}\left(X_3\right)$. On the other hand, $g(x_1,x_3,x_2),g(x_2,x_1,x_3)\notin W_{{\tau }_3}^{O_3}\left(X_3\right)$ because $\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right)\notin O_3$.

Example 2. 

Let ${\tau }_4=\left(4\right)$ be a type with a quaternary operation symbol f. Then, we have $f(x_1,x_2,x_3,x_4),f(f(x_1,x_2,x_2,x_3),f(x_2,x_2,x_2,x_2),f(x_1,x_3,x_4,x_4),f(x_2,x_3,x_3,x_3))\in W_{{\tau }_4}^{O_4}\left(X_4\right)$ whereas $f(x_1,x_2,x_1,x_3),f(x_4,x_3,x_2,x_1)\notin W_{{\tau }_4}^{O_4}\left(X_4\right).$

It turns out that for a fixed $n\in \overline{n}$, each order-preserving full term of type ${\tau }_n$ can be drawn using a tree diagram, which means that when a type is finitary, i.e., the number of each operation symbol is finite, each order-preserving full term t has a tree representation. From Example 1, a tree representation of

$$t=g(g(x_1,x_2,x_3),g(x_3,x_3,x_3),g(x_1,x_1,x_2))$$

can be shown in the following Figure 1.

We remark that an order-preserving full term t is a completely expanded term given in [8] because the maximum depth of all of the variables in t is equal and equals 2.

Now, we observe that $W_{{\tau }_n}^{SF}\left(X_n\right)\subseteq W_{{\tau }_n}^{O_n}\left(X_n\right)\subseteq W_{{\tau }_n}^F\left(X_n\right)$ where n is a positive integer.

For the set $W_{{\tau }_n}^{O_n}\left(X_n\right)$ of all n-ary order-preserving full terms of type ${\tau }_n$, we define the superposition operation

$$S^n:{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^{n+1}\to W_{{\tau }_n}^{O_n}\left(X_n\right)$$

by

(1)

$S^n(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)}),t_1,\dots ,t_n):=f_i(t_{\alpha \left(1\right)},\dots ,t_{\alpha \left(n\right)})$;

(2)

$S^n(f_i(s_1,\dots ,s_n),t_1,\dots ,t_n):=f_i(S^n(s_1,t_1,\dots ,t_n),\dots ,S^n(s_n,t_1,\dots ,t_n)).$

Consider the following example.

Example 3. 

Let $\alpha =\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 2\end{array}\right)$ be an order-preserving transformation on $\overline{3}$. If we set $t=g(x_{\alpha \left(1\right)},x_{\alpha \left(2\right)},x_{\alpha \left(3\right)}),t_1=f(x_1,x_1,x_1),t_2=f(x_2,x_3,x_3)$, and $t_3=f(x_2,x_2,x_2)$ as ternary order-preserving full terms of type ${\tau }_3=(3,3)$, then we have

$$\begin{array}{ccc}\hfill S^3(t,t_1,t_2,t_3)& =& S^3(g(x_{\alpha \left(1\right)},x_{\alpha \left(2\right)},x_{\alpha \left(3\right)}),f(x_1,x_1,x_1),f(x_2,x_3,x_3),f(x_2,x_2,x_2))\hfill \\ & =& g(f(x_1,x_1,x_1),f(x_2,x_3,x_3),f(x_2,x_3,x_3))\in W_{{\tau }_3}^{O_3}\left(X_3\right).\hfill \end{array}$$

On the other hand,

$$\begin{array}{ccc}\hfill S^3(t,t_3,t_2,t_1)& =& S^3(g(x_{\alpha \left(1\right)},x_{\alpha \left(2\right)},x_{\alpha \left(3\right)}),f(x_2,x_2,x_2),f(x_2,x_3,x_3),f(x_1,x_1,x_1))\hfill \\ & =& g(f(x_2,x_2,x_2),f(x_2,x_3,x_3),f(x_2,x_3,x_3))\in W_{{\tau }_3}^{O_3}\left(X_3\right).\hfill \end{array}$$

We now obtain the following algebra of type $(n+1)$:

$$\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}:=(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n).$$

Recall from [3] that a superassociative algebra of rank n, where $n\in \mathbb{N}$, sometimes called a Menger algebra of rank n, is a pair of the form $(G,\circ )$ where G is a non-empty set and ∘ is an $(n+1)$-ary operation defined on G which satisfies the axiom of superassociative law, i.e.,

$$\circ (\circ (a,b_1,\dots ,b_n),c_1,\dots ,c_n)=\circ (a,\circ (b_1,c_1,\dots ,c_n),\dots ,\circ (b_n,c_1,\dots ,c_n))$$

for all $a,b_1,\dots ,b_n,c_1,\dots ,c_n\in G$.

Then, we prove the following theorem.

Theorem 1. 

$\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$ is a superassociative algebra of rank n.

Proof. 

We prove that $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$ satisfies the superassociative law by induction on the complexity of the order-preserving full term t such that the equation

$$S^n(t,S^n(u_1,v_1,\dots ,v_n),\dots ,S^n(u_n,v_1,\dots ,v_n))=S^n(S^n(t,u_1,\dots ,u_n),v_1,\dots ,v_n)$$

(1)

holds. Firstly, if $t=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ where $\alpha \in O_n$, then

$$\begin{array}{c}{S}^n(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)}),S^n(u_1,v_1,\dots ,v_n),\dots ,S^n(u_n,v_1,\dots ,v_n))=f_i(S^n(u_{\alpha \left(1\right)},v_1,\dots ,v_n)\hfill \\ ,\dots ,S^n(u_{\alpha \left(n\right)},v_1,\dots ,v_n))=S^n(f_i(u_{\alpha \left(1\right)},\dots ,u_{\alpha \left(n\right)}),v_1,\dots ,v_n)=S^n\left(S^n\right(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\hfill \\ ,u_1,\dots ,u_n),v_1,\dots ,v_n).\hfill \end{array}$$

Inductively, assume that $t=f_i(t_1,\dots ,t_n)\in W_{{\tau }_n}^{O_n}\left(X_n\right)$ such that $t_1,\dots ,t_n$ satisfy (1). Then,

$$\begin{array}{ccc}& & S^n(f_i(t_1,\dots ,t_n),S^n(u_1,v_1,\dots ,v_n),\dots ,S^n(u_n,v_1,\dots ,v_n))\hfill \\ & =& f_i(S^n(t_1,S^n(u_1,v_1,\dots ,v_n),\dots ,S^n(u_n,v_1,\dots ,v_n)),\dots ,\hfill \\ & & S^n(t_n,S^n(u_1,v_1,\dots ,v_n),\dots ,S^n(u_n,v_1,\dots ,v_n)))\hfill \\ & =& f_i(S^n(S^n(t_1,u_1,\dots ,u_n),v_1,\dots ,v_n),\dots ,S^n(S^n(t_n,u_1,\dots ,u_n),v_1,\dots ,v_n))\hfill \\ & =& S^n(f_i(S^n(t_1,u_1,\dots ,u_n),\dots ,S^n(t_n,u_1,\dots ,u_n)),v_1,\dots ,v_n)\hfill \\ & =& S^n(S^n(f_i(t_1,\dots ,t_n),u_1,\dots ,u_n),v_1,\dots ,v_n),\hfill \end{array}$$

which completes the proof. □

The generating system of the algebra $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$ is denoted by $F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}$ and constructed by

$$F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}:=\{f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\mid i\in I,\alpha \in O_n\}.$$

Let $V_{SASS}$ be the variety of all algebras of type $(n+1)$ that satisfy the superassociativity. For the free algebra ${\mathcal{F}}_{V_{SASS}}\left(Z\right)$ generated by $Z:=\{z_j\mid j\in J\}$ with respect to $V_{SASS}$, where Z is an alphabet of variables indexed by the set $J:=\{(i,\alpha )\mid i\in I,\alpha \in O_n\}$, the operation on ${\mathcal{F}}_{V_{SASS}}\left(Z\right)$ will be represented by ${\tilde{S}}^n$.

As a consequence, the following theorem is the main result in this section, which provides the freeness property of the algebra $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$.

Theorem 2. 

The algebra $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$ is free with respect to the variety $V_{SASS}$ of superassociative algebra of rank n, freely generated by the set $Z=\{z_{(i,\alpha )}\mid i\in I,\alpha \in O_n\}$.

Proof. 

We claim that $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$ is isomorphic to ${\mathcal{F}}_{V_{SASS}}\left(Z\right)$. To ensure that this claim is true, we inductively define the mapping

$$\phi :W_{{\tau }_n}^{O_n}\left(X_n\right)\to {\mathcal{F}}_{V_{SASS}}\left(Z\right)$$

by $\phi \left(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right):=z_{(i,\alpha )}$ and

$$\phi (S^n(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)}),t_1,\dots ,t_n):={\tilde{S}}^n(z_{(i,\alpha )},\phi \left(t_1\right),\dots ,\phi \left(t_n\right)).$$

Clearly, the mapping $\phi$ is a surjection. In fact, we let $z_{(i,\alpha )}\in Z$. Then, an order-preserving full term $f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ of type ${\tau }_n$ exists such that $\phi \left(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right)=z_{(i,\alpha )}$.

To show that the mapping $\phi$ is injective, let $i,j\in I$ and $\alpha ,\beta$ be order-preserving transformations on a finite chain $\{1\le \cdots \le n\}$. Suppose that $z_{(i,\alpha )}=z_{(j,\beta )}$. Then, $(i,\alpha )=(j,\beta )$, which implies that $i=j$ and $\alpha =\beta$. Thus, $f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})=f_j(x_{\beta \left(1\right)},\dots ,x_{\beta \left(n\right)})$, which proves the injectivity.

Now, we prove by induction on the complexity of an order-preserving full term t that $\phi$ is a homomorphism, i.e.,

$$\phi \left(S^n(t,t_1,\dots ,t_n)\right)={\tilde{S}}^n(\phi \left(t\right),\phi \left(t_1\right),\dots ,\phi \left(t_n\right))$$

for all $t,t_1,\dots ,t_n\in W_{{\tau }_n}^{O_n}\left(X_n\right)$.

If $t=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$, then

$$\begin{array}{c}\phi \left(S^n(t,t_1,\dots ,t_n)\right)=\phi \left(S^n(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)}),t_1,\dots ,t_n)\right)\hfill \\ ={\tilde{S}}^n(z_{(i,\alpha )},\phi \left(t_1\right),\dots ,\phi \left(t_n\right))={\tilde{S}}^n(\phi \left(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right),\phi \left(t_1\right),\dots ,\phi \left(t_n\right))=\hfill \\ {\tilde{S}}^n(\phi \left(t\right),\phi \left(t_1\right),\dots ,\phi \left(t_n\right)).\hfill \end{array}$$

Now, let $t=f_i(s_1,\dots ,s_n)$ and assume that for all $1\le k\le n$,

$$\phi \left(S^n(s_k,t_1,\dots ,t_n)\right)={\tilde{S}}^n(\phi \left(s_k\right),\phi \left(t_1\right),\dots ,\phi \left(t_n\right)).$$

Then, we obtain

$$\begin{array}{cc}\hfill \phi \left(S^n\right(t,t_1,& \dots ,t_n\left)\right)\hfill \\ & =\phi \left(S^n(f_i(s_1,\dots ,s_n),t_1,\dots ,t_n)\right)\hfill \\ & =\phi (f_i(S^n(s_1,t_1,\dots ,t_n),\dots ,S^n(s_n,t_1,\dots ,t_n))\hfill \\ & ={\tilde{S}}^n(z_{(i,1_n)},\phi \left(S^n(s_1,t_1,\dots ,t_n)\right),\dots ,\phi \left(S^n(s_n,t_1,\dots ,t_n)\right))\hfill \\ & ={\tilde{S}}^n(z_{(i,1_n)},{\tilde{S}}^n(\phi \left(s_1\right),\phi \left(t_1\right),\dots ,\phi \left(t_n\right)),\dots ,{\tilde{S}}^n(\phi \left(s_n\right),\phi \left(t_1\right),\dots ,\phi \left(t_n\right))).\hfill \end{array}$$

Similarly, given the fact that

$$\phi \left(f_i(t_1^{\prime },\dots ,t_n^{\prime })\right)={\tilde{S}}^n(z_{(i,1_n)},\phi \left(t_1^{\prime }\right),\dots ,\phi \left(t_n^{\prime }\right))$$

for all $t_1^{\prime },\dots ,t_n^{\prime }\in W_{{\tau }_n}^{O_n}\left(X_n\right)$ where $1_n$ is an identity mapping on $\overline{n}$, we then have

$$\begin{array}{ccc}\hfill {\tilde{S}}^n(\phi \left(t\right),\phi \left(t_1\right),\dots ,\phi \left(t_n\right))& =& {\tilde{S}}^n(\phi \left(f_i(s_1,\dots ,s_n)\right),\phi \left(t_1\right),\dots ,\phi \left(t_n\right))\hfill \\ & =& {\tilde{S}}^n({\tilde{S}}^n(z_{(i,1_n)},\phi \left(s_1\right),\dots ,\phi \left(s_n\right)),\phi \left(t_1\right),\dots ,\phi \left(t_n\right)).\hfill \end{array}$$

Due to the superassociativity of ${\tilde{S}}^n$, we conclude that $\phi \left(S^n(t,t_1,\dots ,t_n)\right)$ and ${\tilde{S}}^n(\phi \left(t\right),\phi \left(t_1\right),\dots ,\phi \left(t_n\right))$ are equal. □

3. Algebras of Languages Induced by the Set ${\mathit{W}}_{{\mathit{\tau }}_{\mathit{n}}}^{{\mathit{O}}_{\mathit{n}}}\mathbf{\left(}{\mathit{X}}_{\mathit{n}}\mathbf{\right)}$

In the first part of this section, as follows from [2], we define the tree languages over the powerset of order-preserving full terms of type ${\tau }_n$. The symbol $P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$ stands for the set of all non-empty subsets of order-preserving full terms of type ${\tau }_n$. Each element in $P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$ is called a language of order-preserving full terms.

Example 4. 

It is straightforward to see from Example 1 that on the set $W_{{\tau }_3}^{O_3}\left(X_3\right)$,

$\left\{g(x_1,x_1,x_1)\right\},\{g(x_2,x_2,x_3),g(x_3,x_3,x_3)\},\{g(x_2,x_2,x_3),g(g(x_1,x_2,x_3),g(x_1,x_1,x_3),g(x_3,x_3,x_3))\}$ are some elements in $P\left(W_{{\tau }_3}^{O_3}\left(X_3\right)\right)$.

Recall from [2] that the non-deterministic superposition operation on the set of all n-ary full terms of type ${\tau }_n$ is a mapping

$$S_{nd}^n:P{\left(W_{{\tau }_n}^F\left(X_n\right)\right)}^{n+1}\to P\left(W_{{\tau }_n}^F\left(X_n\right)\right)$$

given by

(1)

$S_{nd}^n(T,T_1,\dots ,T_n)=\{f_i(t_{\alpha \left(1\right)},\dots ,t_{\alpha \left(n\right)})\mid t_{\alpha \left(1\right)}\in T_{\alpha \left(1\right)},\dots ,t_{\alpha \left(n\right)}\in T_{\alpha \left(n\right)}\}$

if $T=\left\{f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right\}$ where $\alpha \in T_n$;

(2)

$S_{nd}^n(T,T_1,\dots ,T_n)=\{f_i(s_1,\dots ,s_n)\mid s_k\in S_{nd}^n(\left\{t_k\right\},T_1,\dots ,T_n),1\le k\le n\}$

if $T=\left\{f_i(t_1,\dots ,t_n)\right\}$;

(3)

${\displaystyle S_{nd}^n(T,T_1,\dots ,T_n)=\bigcup _{t\in T}{S}_{nd}^n(\left\{t\right\},T_1,\dots ,T_n)}$ if T is an arbitrary non-empty subset of $W_{{\tau }_n}^F\left(X_n\right)$;

(4)

${\displaystyle S_{nd}^n(T,T_1,\dots ,T_n)=\varnothing }$ if $T=\varnothing$ or $T_j=\varnothing$ for some $1\le j\le n$

where $T,T_1,\dots ,T_n$ are in $P\left(W_{{\tau }_n}^F\left(X_n\right)\right)$.

The following example demonstrates the process of computation of languages with respect to the operation $S^n$ for some n.

Example 5. 

On the set $P\left(W_{{\tau }_3}^{O_3}\left(X_3\right)\right)$ described in Example 4, let $T=\{g(x_1,x_2,x_3),g(x_3,x_3,x_3)\},T_1=\left\{g(x_1,x_1,x_1)\right\},T_2=\{g(x_1,x_1,x_2),g(x_2,x_2,x_2)\}$, and $T_3=\left\{g(x_3,x_3,x_3)\right\}$. According to the definition of the operation $S_{nd}^n$, $S_{nd}^3(T,T_1,T_2,T_3)$ is equal to $S_{nd}^3(\left\{g(x_1,x_2,x_3)\right\},T_1,T_2,T_3)\cup S_{nd}^3(\left\{g(x_3,x_3,x_3)\right\},T_1,T_2,T_3)$, and $S_{nd}^3(\left\{g(x_1,x_2,x_3)\right\},T_1,T_2,T_3)$ equals

$$\{g(g(x_1,x_1,x_1),g(x_1,x_1,x_2),g(x_2,x_3,x_3)),g(g(x_1,x_1,x_1),g(x_2,x_2,x_2),g(x_2,x_3,x_3))\}$$

and $S_{nd}^3(\left\{g(x_3,x_3,x_3)\right\},T_1,T_2,T_3)=\left\{g(g(x_3,x_3,x_3),g(x_3,x_3,x_3),g(x_3,x_3,x_3))\right\}$.

Thus, $S_{nd}^3(T,T_1,T_2,T_3)$ is the set

$$\begin{array}{c}\{g(g(x_1,x_1,x_1),g(x_1,x_1,x_2),g(x_2,x_3,x_3)),g(g(x_1,x_1,x_1),g(x_2,x_2,x_2),g(x_2,x_3,x_3)),\\ g(g(x_3,x_3,x_3),g(x_3,x_3,x_3),g(x_3,x_3,x_3))\}\end{array}$$

belonging to $P\left(W_{{\tau }_3}^{O_3}\left(X_3\right)\right)$.

Some properties of the superposition $S^n$ are discussed.

Proposition 1. 

If $T,T_1,\dots ,T_n\in P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$, then

$$S_{nd}^n(T_{\alpha },T_1,\dots ,T_n)=S_{nd}^n(T,T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)}).$$

Proof. 

Let $T,T_1,\dots ,T_n$ be non-empty subsets of the set of all order-preserving full terms of type ${\tau }_n$. There are three cases to consider. Firstly, if $T=\left\{f_i(x_{\beta \left(1\right)},\dots ,x_{\beta \left(n\right)})\right\}$ where β is an order-preserving transformation, then $S_{nd}^n(T_{\alpha },T_1,\dots ,T_n)=S_{nd}^n({\left\{f_i(x_{\beta \left(1\right)},\dots ,x_{\beta \left(n\right)})\right\}}_{\alpha },T_1,\dots ,T_n)$ $=S_{nd}^n(\left\{f_i(x_{\alpha \left(\beta \right(1\left)\right)},\dots ,x_{\alpha \left(\beta \right(n\left)\right)})\right\},T_1,\dots ,T_n)=\{f_i(t_{\alpha \left(\beta \right(1\left)\right)},\dots ,t_{\alpha \left(\beta \right(n\left)\right)})\mid t_{\alpha \left(\beta \right(k\left)\right)}\in T_{\alpha \left(\beta \right(k\left)\right)},1\le k\le n\}=S_{nd}^n(\left\{f_i(x_{\beta \left(1\right)},\dots ,x_{\beta \left(n\right)})\right\},T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)})=S_{nd}^n(T,T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)})$. For the next case, let $T=\left\{f_i(r_1,\dots ,r_n)\right\}$ be such that $S_{nd}^n({\left\{r_k\right\}}_{\alpha },T_1,\dots ,T_n)=S_{nd}^n(\left\{r_k\right\},T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)})$ for all $1\le k\le n$. Consider

$$\begin{array}{ccc}\hfill S_{nd}^n(T_{\alpha },T_1,\dots ,T_n)& =& S_{nd}^n({\left\{f_i(r_1,\dots ,r_n)\right\}}_{\alpha },T_1,\dots ,T_n)\hfill \\ & =& S_{nd}^n(\left\{f_i(r_{\alpha \left(1\right)},\dots ,r_{\alpha \left(n\right)})\right\},T_1,\dots ,T_n)\hfill \\ & =& \{f_i(u_{\alpha \left(1\right)},\dots ,u_{\alpha \left(n\right)})\mid u_{\alpha \left(k\right)}\in S_{nd}^n({\left\{r_k\right\}}_{\alpha },T_1,\dots ,T_n),1\le k\le n\}.\hfill \end{array}$$

On another side, we obtain

$$\begin{array}{ccc}& & S_{nd}^n(T,T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)})\hfill \\ & =& S_{nd}^n(\left\{f_i(r_1,\dots ,r_n)\right\},T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)})\hfill \\ & =& \{f_i(u_{\alpha \left(1\right)},\dots ,u_{\alpha \left(n\right)})\mid u_{\alpha \left(k\right)}\in S_{nd}^n(\left\{r_k\right\},T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)}),1\le k\le n\}.\hfill \end{array}$$

From our assumption, in this case, we conclude that the set $S_{nd}^n(T_{\alpha },T_1,\dots ,T_n)$ equals $S_{nd}^n(T,T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)})$.

Suppose that $\left|T\right|>1$. Because $S_{nd}^n(\left\{t_{\alpha }\right\},T_1,\dots ,T_n)=S_{nd}^n({\left\{t\right\}}_{\alpha },T_1,\dots ,T_n)$, then $S_{nd}^n(T_{\alpha },T_1,\dots ,T_n)={\bigcup }_{t\in T}{S}_{nd}^n(\left\{t\right\},T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)})=S_{nd}^n(T,T_{\alpha \left(1\right)},\dots ,T_{\alpha \left(n\right)})$. □

Proposition 2. 

If $T,T_1,\dots ,T_n\in P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$, then

$${\left(S_{nd}^n(T,T_1,\dots ,T_n)\right)}_{\alpha }=S_{nd}^n(T_{\alpha },T_1,\dots ,T_n).$$

Proof. 

Let $T,T_1,\dots ,T_n\in P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$. Firstly, let $T=\left\{f_i(x_{\beta \left(1\right)},\dots ,x_{\beta \left(n\right)})\right\}$ where $\beta \in O_n$. Consider ${\left(S_{nd}^n(T,T_1,\dots ,T_n)\right)}_{\alpha }={\left(S_{nd}^n(\left\{f_i(x_{\beta \left(1\right)},\dots ,x_{\beta \left(n\right)})\right\},T_1,\dots ,T_n)\right)}_{\alpha }={\left(\{f_i(t_{\beta \left(1\right)},\dots ,t_{\beta \left(n\right)})\mid t_{\beta \left(k\right)}\in T_{\beta \left(k\right)},1\le k\le n\}\right)}_{\alpha }=\left(\{f_i{(t_{\beta \left(1\right)},\dots ,t_{\beta \left(n\right)})}_{\alpha }\mid t_{\beta \left(k\right)}\in T_{\beta \left(k\right)},1\le k\le n\}\right)=\{f_i(t_{\alpha \left(\beta \right(1\left)\right)},\dots ,t_{\alpha \left(\beta \right(n\left)\right)})\mid t_{\alpha \left(\beta \right(k\left)\right)}\in T_{\alpha \left(\beta \right(k\left)\right)},1\le k\le n\}$. According to Proposition 1, we obtain ${\left(S_{nd}^n(T,T_1,\dots ,T_n)\right)}_{\alpha }=S_{nd}^n(T_{\alpha },T_1,\dots ,T_n)$.

Now, let $T=\left\{f_i(r_1,\dots ,r_n)\right\}$ and suppose that

$${\left(S_{nd}^n(\left\{r_k\right\},T_1,\dots ,T_n)\right)}_{\alpha }=S_{nd}^n({\left\{r_k\right\}}_{\alpha },T_1,\dots ,T_n)$$

for all $1\le k\le n.$ Consider

${\left(S_{nd}^n(T,T_1,\dots ,T_n)\right)}_{\alpha }$

$$\begin{array}{ccc}& =& {\left(S_{nd}^n(\left\{f_i(r_1,\dots ,r_n)\right\},T_1,\dots ,T_n)\right)}_{\alpha }\hfill \\ & =& {\left(\{f_i(u_1,\dots ,u_n)\mid u_k\in S_{nd}^n\left(\{u_k,T_1,\dots ,T_n\}\right),1\le k\le n\}\right)}_{\alpha }\hfill \\ & =& {\left(\{f_i{(u_1,\dots ,u_n)}_{\alpha }\mid {\left(u_k\right)}_{\alpha }\in S_{nd}^n(\left\{u_k\right\},T_1,\dots ,T_n),1\le k\le n\}\right)}_{\alpha }\hfill \\ & =& {\left(\{f_i(u_{\alpha \left(1\right)},\dots ,u_{\alpha \left(n\right)})\mid {\left(u_k\right)}_{\alpha }\in S_{nd}^n(\left\{u_k\right\},T_1,\dots ,T_n),1\le k\le n\}\right)}_{\alpha }\hfill \\ & =& \left(\{f_i(u_{\alpha \left(1\right)},\dots ,u_{\alpha \left(n\right)})\mid {\left(u_k\right)}_{\alpha }\in S_{nd}^n({\left\{u_k\right\}}_{\alpha },T_1,\dots ,T_n),1\le k\le n\}\right).\hfill \end{array}$$

Using Proposition 1, we have ${\left(S_{nd}^n(T,T_1,\dots ,T_n)\right)}_{\alpha }=S_{nd}^n(T_{\alpha },T_1,\dots ,T_n).$

Finally, if T is an arbitrary non-empty subset of the set of all order-preserving full terms of type ${\tau }_n$, we then have ${\left(S_{nd}^n(T,T_1,\dots ,T_n)\right)}_{\alpha }={\left({\bigcup }_{t\in T}{S}_{nd}^n(\left\{t\right\},T_1,\dots ,T_n)\right)}_{\alpha }={\bigcup }_{t\in T}{\left(S_{nd}^n(\left\{t\right\},T_1,\dots ,T_n)\right)}_{\alpha }={\bigcup }_{t\in T}{S}_{nd}^n({\left\{t\right\}}_{\alpha },T_1,\dots ,T_n)={\bigcup }_{t\in T}{S}_{nd}^n(\left\{t_{\alpha }\right\},T_1,\dots ,T_n)=S_{nd}^n(T_{\alpha },T_1,\dots ,T_n)$. □

Now, the relationship between the set of all full terms and the set of all order-preserving full terms of the same type is shown by the following theorem.

Theorem 3. 

$(P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right),S_{nd}^n)$ is a subalgebra of $(P\left(W_{{\tau }_n}^F\left(X_n\right)\right),S_{nd}^n)$.

Proof. 

Clearly, $P(W_{{\tau }_n}^{O_n}\left(X_n\right)\subseteq P(W_{{\tau }_n}^F\left(X_n\right)$. For $A,B_1,\dots ,B_n\in P(W_{{\tau }_n}^{O_n}\left(X_n\right)$, we show on a characteristic of a set A that $S_{nd}^n(A,B_1,\dots ,B_n)$ is an element in $P(W_{{\tau }_n}^{O_n}\left(X_n\right)$. There are three cases to consider. If $A=\left\{f_i(x_{{\alpha }_1},\dots ,x_{{\alpha }_n})\right\}$ where $\alpha \in O_n$, then $S_{nd}^n(A,B_1,\dots ,B_n)=S_{nd}^n(\left\{f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right\},B_1,\dots ,B_n)=\{f_i(t_{\alpha \left(1\right)},\dots ,t_{\alpha \left(n\right)})\mid t_{\alpha \left(k\right)}\in T_{\alpha \left(k\right)},1\le k\le n\}$ is an element in $P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$. For the second case, let $A=\left\{f_i(t_1,\dots ,t_n)\right\}$. Assume that $S_{nd}^n(\left\{t_k\right\},B_1,\dots ,B_n)\in P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$ for all $1\le k\le n$. Consider $S_{nd}^n(A,B_1,\dots ,B_n)=S_{nd}^n(\left\{f_i(t_1,\dots ,t_n)\right\},B_1,\dots ,B_n)=\{f_i(t_1,\dots ,t_n)\mid t_k\in S_{nd}^n(\left\{t_k\right\},B_1,\dots ,B_n),1\le k\le n\}$. Through this assumption, we obtain $S_{nd}^n(A,B_1,\dots ,B_n)$, belonging to the set $P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$. For the last case, if A is an arbitrary non-empty subset of $W_{{\tau }_n}^{O_n}\left(X_n\right)$, then $S_{nd}^n(A,B_1,\dots ,B_n)={\bigcup }_{a\in A}{S}_{nd}^n(\left\{a\right\},B_1,\dots ,B_n)$. Since each $S_{nd}^n(\left\{a\right\},B_1,\dots ,B_n)$ is already in $P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$, then $S_{nd}^n(A,B_1,\dots ,B_n)\in P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right).$ The proof is completed. □

The second aim of this section is to define and study the binary operations on $W_{{\tau }_n}^{O_n}\left(X_n\right)$ and $P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$.

Now, let $s,t\in W_{{\tau }_n}^{O_n}\left(X_n\right)$. The binary operation + is defined as follows.

$$+:W_{{\tau }_n}^{O_n}\left(X_n\right)\times W_{{\tau }_n}^{O_n}\left(X_n\right)\to W_{{\tau }_n}^{O_n}\left(X_n\right)$$

by

$$s+t=S^n(s,\underset{n\mathrm{terms}}{\underbrace{t,\dots ,t}}).$$

We define the binary operation on the Cartesian product of n-tuples of order-preserving full terms of type ${\tau }_n$ as follows. For all $(s_1,\dots ,s_n),(t_1,\dots ,t_n)\in W_{{\tau }_n}^{O_n}\left(X_n\right)$, define

$$*:{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n\times {\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n\to {\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n$$

by

$$(s_1,\dots ,s_n)*(t_1,\dots ,t_n)=(S^n(s_1,t_1,\dots ,t_n),\dots ,S^n(s_n,t_1,\dots ,t_n)).$$

Naturally, for $A,B\in P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$, we define the binary operation $\widehat{+}$ on $P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$ by

$$A\widehat{+}B=S_{nd}^n(A,\underset{n\mathrm{terms}}{\underbrace{B,\dots ,B}}).$$

On the Cartesian product $P{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n$, the binary operation

$$\widehat{*}:P{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n\times P{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n\to P{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n$$

is defined by

$$(A_1,\dots ,A_n)\widehat{*}(B_1,\dots ,B_n)=(S_{nd}^n(A_1,B_1,\dots ,B_n),\dots ,S_{nd}^n(A_n,B_1,\dots ,B_n)).$$

for all $(A_1,\dots ,A_n),(B_1,\dots ,B_n)\in P{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n$.

The following theorem presents the connection between $(W_{{\tau }_n}^{O_n}\left(X_n\right),+)$ and $({\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n,*).$

Theorem 4. 

The semigroup $(W_{{\tau }_n}^{O_n}\left(X_n\right),+)$ is embedded into $({\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n,*).$

Proof. 

We first show that $(W_{{\tau }_n}^{O_n}\left(X_n\right),+)$ is a semigroup. To this end, let $a,b,c$ be the elements in $W_{{\tau }_n}^{O_n}\left(X_n\right).$ Then, according to the superassociativity of $S^n$, we obtain $(a+b)+c=S^n(a,b,\dots ,b)+c=S^n(S^n(a,b,\dots ,b),c,\dots ,c)=S^n(a,S^n(b,c,\dots ,c),\dots ,S^n(b,c,\dots ,c))=a+S^n(b,c,\dots ,c)=a+(b+c)$. Therefore, $(W_{{\tau }_n}^{O_n}\left(X_n\right),+)$ is a semigroup. For the associativity of *, let $(a_1,\dots ,a_n),(b_1,\dots ,b_n),(c_1,\dots ,c_n)\in {\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n$. Consider

$$\begin{array}{c}((a_1,\dots ,a_n)*(b_1,\dots ,b_n))*(c_1,\dots ,c_n)\hfill \\ =(S^n(a_1,b_1,\dots ,b_n),\dots ,S^n(a_n,b_1,\dots ,b_n))*(c_1,\dots ,c_n)=\left(S^n\right(S^n(a_1,b_1,\dots ,b_n),\hfill \\ c_1,\dots ,c_n),\dots ,\left(S^n(S^n(a_n,b_1,\dots ,b_n),c_1,\dots ,c_n)\right))=(S^n(a_1,S^n(b_1,c_1,\dots ,c_n),\dots ,\hfill \\ S^n(b_n,c_1,\dots ,c_n)),\dots ,S^n(a_n,S^n(b_1,c_1,\dots ,c_n),\dots ,S^n(b_n,c_1,\dots ,c_n)))=(a_1,\dots ,a_n)*\hfill \\ (S^n(b_1,c_1,\dots ,c_n),\dots ,S^n(b_n,c_1,\dots ,c_n))=(a_1,\dots ,a_n)*((b_1,\dots ,b_n)*(c_1,\dots ,c_n)).\hfill \end{array}$$

Thus, $({\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n,*)$ is a semigroup.

Now, for any $t\in W_{{\tau }_n}^{O_n}\left(X_n\right)$, we define a mapping $\psi :W_{{\tau }_n}^{O_n}\left(X_n\right)\to {\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n$ by $\psi \left(t\right)={\underbrace{(t,\dots ,t)}}_{n\mathrm{terms}}$. Assume that $s,t\in W_{{\tau }_n}^{O_n}\left(X_n\right)$. Consider

$$\begin{array}{ccc}\hfill \psi (s+t)& =& \psi \left(S^n(s,t,\dots ,t)\right)\hfill \\ & =& (\underset{n\mathrm{terms}}{\underbrace{S^n(s,t,\dots ,t),\dots ,S^n(s,t,\dots ,t)}})\hfill \\ & =& (s,\dots ,s)*(t,\dots ,t)\hfill \\ & =& \psi \left(s\right)*\psi \left(t\right).\hfill \end{array}$$

Since ψ is injective, we can conclude that $(W_{{\tau }_n}^{O_n}\left(X_n\right),+)$ is embedded into $({\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n,*).$ □

Moreover, we yield the following relationship.

Theorem 5. 

$(P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right),\widehat{+})$ is embedded into $(P{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n,\widehat{*}).$

Proof. 

Clearly, reasoning dual to the proof of Theorem 4 shows that the mapping

$A\mapsto (A,\dots ,A)$ is a monomorphism from the semigroup $(P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right),\widehat{+})$ to

$(P{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n,\widehat{*})$. □

Other embeddabilities are described as follows:

Theorem 6. 

The following statements hold:

(1) 

$(W_{{\tau }_n}^{O_n}\left(X_n\right),+)$ embeds into $(P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right),\widehat{+}).$

(2) 

$({\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n,\ast )$ embeds into $(P{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n,\widehat{\ast }).$

Proof. 

Let t be an n-ary order-preserving full term. We define a mapping $\gamma :W_{{\tau }_n}^{O_n}\left(X_n\right)\to P\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)$ by $\gamma \left(t\right)=\left\{t\right\}$. It is clear that $\gamma$ is injective. Moreover, $\gamma (s+t)=\gamma \left(S_{nd}^n(s,t,\dots ,t)\right)=\left\{S_{nd}^n(s,t,\dots ,t)\right\}=S_{nd}^n(\left\{s\right\},\left\{t\right\},\dots ,\left\{t\right\})=\left\{s\right\}\widehat{+}\left\{t\right\}=\gamma \left(s\right))\widehat{+}\gamma \left(t\right).$ For (2), we define a mapping $\Gamma :{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n\to P{\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^n$ by $\Gamma \left((t_1,\dots ,t_n)\right)=(\left\{t_1\right\},\dots ,\left\{t_n\right\})$. Obviously, $\Gamma$ is injective. Additionally, $\Gamma$ is also a homomorphism. That means $\Gamma ((s_1,\dots ,s_n)\ast (t_1,\dots ,t_n))=\Gamma (S^n(s_1,t_1,\dots ,t_n),\dots ,S^n(s_n,t_1,\dots ,t_n))=(\left\{S^n(s_1,t_1,\dots ,t_n)\right\},\dots ,\left\{S^n(s_n,t_1,\dots ,t_n)\right\})=(S_{nd}^n(\left\{s_1\right\},\left\{t_1\right\},\dots ,\left\{t_n\right\}),\dots ,S_{nd}^n(\left\{s_n\right\},\left\{t_1\right\},\dots ,\left\{t_n\right\}))=(\left\{s_1\right\},\dots ,\left\{s_n\right\})\widehat{\ast }(\left\{t_1\right\},\dots ,\left\{t_n\right\})=\Gamma (s_1,\dots ,s_n)\widehat{\ast }\Gamma (t_1,\dots ,t_n).$ Thus, the proof is completed. □

4. A Variety of Algebras Satisfying the Identities Induced by Order-Preserving Transformations

In this section, our objective is to construct a novel associative algebra that includes an identity element. Each element of this algebra will be applied to defining a class of identity called an order-preserving hyperidentity, and a variety of algebras satisfies such identity. First, we achieve this by defining a mapping that assigns each operation symbol of type ${\tau }_n$ to the set of all n-ary order-preserving full terms of type ${\tau }_n$ as follows.

An order-preserving full hypersubstitution of type ${\tau }_n$ or an $O_n$-full hypersubstitution of type ${\tau }_n$ is a mapping

$$\sigma :\{f_i\mid i\in I\}\to W_{{\tau }_n}^{O_n}\left(X_n\right)$$

that assigns each n-ary operation symbol of type ${\tau }_n$ to the set of all n-ary order-preserving full terms of the same type. The set of all $O_n$-full hypersubstitutions of type ${\tau }_n$ is denoted by $Hy{p}^{O_n}\left({\tau }_n\right)$.

For $t\in W_{{\tau }_n}^{O_n}\left(X_n\right)$ and $\beta \in O_n$, we define the order-preserving full terms arising from a mapping $\beta$ as follows.

(1)

If $t=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$, then $t_{\beta }:=f_i(x_{\beta \left(\alpha \right(1\left)\right)},\dots ,x_{\beta \left(\alpha \right(n\left)\right)})$;

(2)

If $t=f_i(t_1,\dots ,t_n)$, then $t_{\beta }:=f_i({\left(t_1\right)}_{\beta },\dots ,{\left(t_n\right)}_{\beta })$.

It is observed that if t is an order-preserving full term of type ${\tau }_n$, then $t_{\beta }$ is an order-preserving full term of type ${\tau }_n$ for all $\beta \in O_n$.

Any $O_n$-full hypersubstitution $\sigma :\{f_i\mid i\in I\}\to W_{{\tau }_n}^{O_n}\left(X_n\right)$ of type ${\tau }_n$ can be extended to a mapping

$$\widehat{\sigma }:W_{{\tau }_n}^{O_n}\left(X_n\right)\to W_{{\tau }_n}^{O_n}\left(X_n\right)$$

defined as follows

(1)

$\widehat{\sigma }\left[f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right]:={\left(\sigma \left(f_i\right)\right)}_{\alpha }$;

(2)

$\widehat{\sigma }\left[f_i(t_1,\dots ,t_n)\right]:=S^n(\sigma \left(f_i\right),\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])$.

For ${\sigma }_1,{\sigma }_2\in Hy{p}^{O_n}\left({\tau }_n\right)$, we define the binary operation ${\circ }_h$ by $\left({\sigma }_1{\circ }_h{\sigma }_2\right):={\widehat{\sigma }}_1\circ {\sigma }_2$ where ∘ is the usual composition of functions.

We give some properties of such a hypersubstitution.

Lemma 1. 

Let α be an order-preserving transformation on $\overline{n}$. For any $t,t_1,\dots ,t_n\in W_{{\tau }_n}^{O_n}\left(X_n\right)$, the equation

$$S^n(t_{\alpha },\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])=S^n(t,\widehat{\sigma }\left[t_{\alpha \left(1\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(n\right)}\right])$$

holds.

Proof. 

Let $t\in W_{{\tau }_n}^{O_n}\left(X_n\right)$ and $\alpha \in O_n$. There are two cases to consider: Case 1: If $t=f_i(x_{\beta \left(1\right)},\dots ,x_{\beta \left(n\right)})$ where $\beta \in O_n$, then

$$\begin{array}{ccc}\hfill S^n(t_{\alpha },\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])& =& S^n(f_i(x_{\alpha \left(\beta \right(1\left)\right)},\dots ,x_{\alpha \left(\beta \right(n\left)\right)}),\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])\hfill \\ & =& f_i(\widehat{\sigma }\left[t_{\alpha \left(\beta \right(1\left)\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(\beta \right(n\left)\right)}\right])\hfill \\ & =& S^n(f_i(x_{\beta \left(1\right)},\dots ,x_{\beta \left(n\right)}),\widehat{\sigma }\left[t_{\alpha \left(1\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(n\right)}\right])\hfill \\ & =& S^n(t,\widehat{\sigma }\left[t_{\alpha \left(1\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(n\right)}\right]).\hfill \end{array}$$

Case 2: Let $t=f_i(u_1,\dots ,u_n)$ and assume that

$$S^n({\left(u_k\right)}_{\alpha },\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])=S^n(u_k,\widehat{\sigma }\left[t_{\alpha \left(1\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(n\right)}\right])$$

for all $\alpha \in O_n$ and $1\le k\le n$. According to the definition of a term $t_{\beta }$ arising from $\beta \in O_n$, we obtain

$$\begin{array}{ccc}\hfill S^n(t_{\alpha },\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])& =& S^n(f_i({\left(u_1\right)}_{\alpha },\dots ,{\left(u_n\right)}_{\alpha }),\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])\hfill \\ & =& f_i(S^n({\left(u_1\right)}_{\alpha },\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right]),\dots ,S^n({\left(u_n\right)}_{\alpha },\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right]))\hfill \\ & =& f_i(S^n(u_1,\widehat{\sigma }\left[t_{\alpha \left(1\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(n\right)}\right]),\dots ,S^n(u_n,\widehat{\sigma }\left[t_{\alpha \left(1\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(n\right)}\right]))\hfill \\ & =& S^n(f_i(u_1,\dots ,u_n),\widehat{\sigma }\left[t_{\alpha \left(1\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(n\right)}\right])\hfill \\ & =& S^n(t,\widehat{\sigma }\left[t_{\alpha \left(1\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(n\right)}\right]).\hfill \end{array}$$

□

Through Lemma 1, we have the following theorem.

Theorem 7. 

The extension of each $O_n$-full hypersubstitution of type ${\tau }_n$ is an endomorphism on the superassociative algebra $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$.

Proof. 

We will show by induction on the complexity of $t_0$ that for any $t_0,t_1,\dots ,t_n\in W_{{\tau }_n}^{O_n}\left(X_n\right)$,

$$\widehat{\sigma }\left[S^n(t_0,t_1,\dots ,t_n)\right]=S^n(\widehat{\sigma }\left[t_0\right],\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right]).$$

(2)

Firstly, if $t_0=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ where $\alpha$ is an order-preserving transformation, then according to Lemma 1, we have

$$\begin{array}{ccc}\hfill \widehat{\sigma }\left[S^n(t_0,t_1,\dots ,t_n)\right]& =& \widehat{\sigma }\left[S^n(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)}),t_1,\dots ,t_n)\right]\hfill \\ & =& \widehat{\sigma }\left[f_i(t_{\alpha \left(1\right)},\dots ,t_{\alpha \left(n\right)})\right]\hfill \\ & =& S^n(\sigma \left(f_i\right),\widehat{\sigma }\left[t_{\alpha \left(1\right)}\right],\dots ,\widehat{\sigma }\left[t_{\alpha \left(n\right)}\right])\hfill \\ & =& S^n({\left(\sigma \left(f_i\right)\right)}_{\alpha },\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])\hfill \\ & =& S^n(\widehat{\sigma }\left[t_0\right],\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right]).\hfill \end{array}$$

Assume $t_0=f_i(u_1,\dots ,u_n)$ such that

$$\widehat{\sigma }\left[S^n(u_k,t_1,\dots ,t_n)\right]=S^n(\widehat{\sigma }\left[u_k\right],\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])$$

for all $1\le k\le n$. Then, we consider the left-hand side of (2), i.e.,

$$\begin{array}{ccc}\hfill \widehat{\sigma }\left[S^n(t_0,t_1,\dots ,t_n)\right]& =& \widehat{\sigma }\left[S^n(f_i(u_1,\dots ,u_n),t_1,\dots ,t_n)\right]\hfill \\ & =& \widehat{\sigma }[f_i(S^n(u_1,t_1,\dots ,t_n),\dots ,S^n(u_n,t_1,\dots ,t_n)]\hfill \\ & =& S^n(\sigma \left(f_i\right),\widehat{\sigma }\left[S^n(u_1,t_1,\dots ,t_n)\right],\dots ,\widehat{\sigma }\left[S^n(u_n,t_1,\dots ,t_n)\right])\hfill \\ & =& S^n(\sigma \left(f_i\right),S^n(\widehat{\sigma }\left[u_1\right],\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right]),\dots ,S^n(\widehat{\sigma }\left[u_n\right],\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])).\hfill \end{array}$$

For the right-hand side of 2, we have

$$\begin{array}{ccc}\hfill S^n(\widehat{\sigma }\left[t_0\right],\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])& =& S^n(\widehat{\sigma }\left[f_i(u_1,\dots ,u_n)\right],\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])\hfill \\ & =& S^n(S^n(\sigma \left(f_i\right),\widehat{\sigma }\left[u_1\right],\dots ,\widehat{\sigma }\left[u_n\right]),\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right]).\hfill \end{array}$$

Applying the fact that the operation $S^n$ is superassociative given in Theorem 1, we conclude that $\widehat{\sigma }\left[S^n(f_i(u_1,\dots ,u_n),t_1,\dots ,t_n)\right]=S^n(\widehat{\sigma }\left[f_i(u_1,\dots ,u_n)\right],\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right]).$ □

Proposition 3. 

Let $t\in W_{{\tau }_n}^{O_n}\left(X_n\right)$ and $\beta \in O_n$. Then,

$$\widehat{\sigma }{\left[t\right]}_{\beta }=\widehat{\sigma }\left[t_{\beta }\right].$$

Proof. 

We give a proof by induction on the complexity of the term t. If $t=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ where α is an order-preserving transformation, then $\widehat{\sigma }{\left[t\right]}_{\beta }={\left(\widehat{\sigma }\left[f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right]\right)}_{\beta }={\left(\sigma {\left(f_i\right)}_{\alpha }\right)}_{\beta }={\left(\sigma \left(f_i\right)\right)}_{\beta \circ \alpha }=\widehat{\sigma }\left[f_i(x_{\beta \circ \alpha \left(1\right)},\dots ,x_{\beta \circ \alpha \left(n\right)})\right]=\widehat{\sigma }\left[f_i(x_{\beta \left(\alpha \right(1\left)\right)},\dots ,x_{\beta \left(\alpha \right(n\left)\right)})\right]=\widehat{\sigma }\left[f_i{(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})}_{\beta }\right]=\widehat{\sigma }\left[t_{\beta }\right].$ Let $t=f_i(u_1,\dots ,u_n)$ and assume that $\widehat{\sigma }{\left[u_k\right]}_{\beta }=\widehat{\sigma }\left[{\left(u_k\right)}_{\beta }\right]$ for all $1\le k\le n$. Then, ${\left(\widehat{\sigma }\left[f_i(u_1,\dots ,u_n)\right]\right)}_{\beta }={\left(S^n(\sigma \left(f_i\right),\widehat{\sigma }\left[u_1\right],\dots ,\widehat{\sigma }\left[u_n\right])\right)}_{\beta }=S^n(\sigma \left(f_i\right),\widehat{\sigma }{\left[u_1\right]}_{\beta },\dots ,\widehat{\sigma }{\left[u_n\right]}_{\beta })=S^n(\sigma \left(f_i\right),\widehat{\sigma }\left[{\left(u_1\right)}_{\beta }\right],\dots ,\widehat{\sigma }\left[{\left(u_n\right)}_{\beta }\right])=\widehat{\sigma }\left[f_i({\left(u_1\right)}_{\beta },\dots ,{\left(u_n\right)}_{\beta })\right]=\widehat{\sigma }\left[t_{\beta }\right].$ Here, the proof is completed. □

According to Proposition 3, we obtain the following result, which shows that the extensions of the product of two order-preserving full hypersubstitutions with respect to ${\circ }_h$ and the product of two extensions are the same thing.

Proposition 4. 

Let ${\widehat{\sigma }}_1$ and ${\widehat{\sigma }}_2$ be the extensions of $O_n$-full hypersubstitutions of type ${\tau }_n$. Then,

$$\widehat{\left({\sigma }_1{\circ }_h{\sigma }_2\right)}={\widehat{\sigma }}_1\circ {\widehat{\sigma }}_2.$$

Proof. 

We prove by induction on the complexity of the order-preserving full term t. First, if $t=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ where α is an order-preserving transformation, according to the definition of the full terms arising from another mapping on the set $O_n$, we obtain $\widehat{\left({\sigma }_1{\circ }_h{\sigma }_2\right)}\left[t\right]=\widehat{\left({\sigma }_1{\circ }_h{\sigma }_2\right)}\left[f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right]=\widehat{({\widehat{\sigma }}_1\circ {\sigma }_2)}\left[f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right]={\left(({\widehat{\sigma }}_1\circ {\sigma }_2)\left(f_i\right)\right)}_{\alpha }={\left({\widehat{\sigma }}_1\left[{\sigma }_2\left(f_i\right)\right]\right)}_{\alpha }={\widehat{\sigma }}_1{\left[{\sigma }_2\left(f_i\right)\right]}_{\alpha }={\widehat{\sigma }}_1\left[{\widehat{\sigma }}_2\left[f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right]\right]=({\widehat{\sigma }}_1\circ {\widehat{\sigma }}_2)\left[t\right].$ Let $t=f_i(u_1,\dots ,u_n)$ and assume that $\widehat{\left({\sigma }_1{\circ }_h{\sigma }_2\right)}\left[u_k\right]={\widehat{\sigma }}_1\circ {\widehat{\sigma }}_2\left[u_k\right]$ where $1\le k\le n$. Then,

$$\begin{array}{ccc}\hfill \widehat{\left({\sigma }_1{\circ }_h{\sigma }_2\right)}\left[t\right]& =& \widehat{\left({\sigma }_1{\circ }_h{\sigma }_2\right)}\left[f_i(u_1,\dots ,u_n)\right]\hfill \\ & =& \widehat{({\widehat{\sigma }}_1\circ {\sigma }_2)}\left[f_i(u_1,\dots ,u_n)\right]\hfill \\ & =& S^n(({\widehat{\sigma }}_1\circ {\sigma }_2)\left(f_i\right),\widehat{({\widehat{\sigma }}_1\circ {\sigma }_2)}\left[u_1\right],\dots ,\widehat{({\widehat{\sigma }}_1\circ {\sigma }_2)}\left[u_n\right])\hfill \\ & =& S^n(({\widehat{\sigma }}_1\circ {\sigma }_2)\left(f_i\right),({\widehat{\sigma }}_1\circ \widehat{{\sigma }_2})\left[u_1\right],\dots ,({\widehat{\sigma }}_1\circ \widehat{{\sigma }_2})\left[u_n\right])\hfill \\ & =& S^n({\widehat{\sigma }}_1\left[{\sigma }_2\left(f_i\right)\right],{\widehat{\sigma }}_1\left[\widehat{{\sigma }_2}\left[u_1\right]\right],\dots ,{\widehat{\sigma }}_1\left[\widehat{{\sigma }_2}\left[u_n\right]\right]).\hfill \end{array}$$

On the other hand,

$$({\widehat{\sigma }}_1\circ {\widehat{\sigma }}_2)\left[t\right]={\widehat{\sigma }}_1\left[{\widehat{\sigma }}_2\left[f_i(u_1,\dots ,u_n)\right]\right]={\widehat{\sigma }}_1\left[S^n({\sigma }_2\left(f_i\right),{\widehat{\sigma }}_2\left[u_1\right],\dots ,{\widehat{\sigma }}_2\left[u_n\right])\right].$$

Applying the fact that each order-preserving full hypersubstituion is an endomorphism on the algebra $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$ proved in Theorem 7, thus $\widehat{\left({\sigma }_1{\circ }_h{\sigma }_2\right)}\left[f_i(u_1,\dots ,u_n)\right]$ equals $({\widehat{\sigma }}_1\circ {\widehat{\sigma }}_2)\left[f_i(u_1,\dots ,u_n)\right].$ □

Having Proposition 4 at hand, notice that $\left({\sigma }_1{\circ }_h{\sigma }_2\right){\circ }_h{\sigma }_3={\sigma }_1{\circ }_h\left({\sigma }_2{\circ }_h{\sigma }_3\right)$. Indeed, due to the associative property of the composition ∘ of functions, $\left({\sigma }_1{\circ }_h{\sigma }_2\right){\circ }_h{\sigma }_3=\widehat{\left({\sigma }_1{\circ }_h{\sigma }_2\right)}\circ {\widehat{\sigma }}_3=({\widehat{\sigma }}_1\circ {\widehat{\sigma }}_2)\circ {\sigma }_3={\widehat{\sigma }}_1\circ ({\widehat{\sigma }}_2\circ {\sigma }_3)={\sigma }_1{\circ }_h\left({\sigma }_2{\circ }_h{\sigma }_3\right)$. As a result, $(Hy{p}^{O_n}\left({\tau }_n\right);{\circ }_h,{\sigma }_{id})$ is a monoid where ${\sigma }_{id}$ is the identity hypersubstitution which is defined by ${\sigma }_{id}\left(f_i\right):=f_i(x_1,\dots ,x_n)$.

Our next aim is to define the second binary operation on $Hy{p}^{O_n}\left({\tau }_n\right)$. A binary operation ${+}_h$ on $Hy{p}^{O_n}\left({\tau }_n\right)$ is defined by

$$\left({\sigma }_1{+}_h{\sigma }_2\right)\left(f_i\right):=S^n({\sigma }_1\left(f_i\right),{\sigma }_2\left(f_i\right),\dots ,{\sigma }_2\left(f_i\right)).$$

We notice that ${\sigma }_1{+}_n{\sigma }_2$ is an element in $Hy{p}^{O_n}\left({\tau }_n\right)$. Furthermore, because of the satisfaction of $S^n$ in the superassociative law, ${+}_h$ is associative, and thus $(Hy{p}^{O_n}\left({\tau }_n\right),{+}_h)$ forms a semigroup.

By ${\sigma }_t$, we mean an order-preserving hypersubstitution that maps any operation symbol to a term t.

Proposition 5. 

The semigroup $(W_{{\tau }_n}^{O_n}\left(X_n\right),+)$ embeds into $(Hy{p}^{O_n}\left({\tau }_n\right),{+}_h)$.

Proof. 

Let t be an order-preserving full term of type ${\tau }_n$. It is not difficult to prove that a mapping $t\mapsto {\sigma }_t$ is an injective homomorphism. □

The relationship between two associative binary operations ${\circ }_h$ and + is now discussed.

Theorem 8. 

$(Hy{p}^{O_n}\left({\tau }_n\right),{\circ }_h,+)$ forms a left-seminearring.

Proof. 

It is enough to show the left distributive law, i.e., ${\sigma }_1{\circ }_h\left({\sigma }_2{+}_h{\sigma }_3\right)=\left({\sigma }_1{\circ }_h{\sigma }_2\right){+}_h\left({\sigma }_1{\circ }_h{\sigma }_3\right)$, is satisfied. In fact, according to Theorem 7, we have

$$\begin{array}{ccc}\hfill \left({\sigma }_1{\circ }_h\left({\sigma }_2{+}_h{\sigma }_3\right)\right)\left(f_i\right)& =& {\widehat{\sigma }}_1\left[\left({\sigma }_2{+}_h{\sigma }_3\right)\left(f_i\right)\right]\hfill \\ & =& {\widehat{\sigma }}_1\left[S^n({\sigma }_2\left(f_i\right),{\sigma }_3\left(f_i\right),\dots ,{\sigma }_3\left(f_i\right))\right]\hfill \\ & =& S^n({\widehat{\sigma }}_1\left[{\sigma }_2\left(f_i\right)\right],{\widehat{\sigma }}_1\left[{\sigma }_3\left(f_i\right)\right],\dots ,{\widehat{\sigma }}_1\left[{\sigma }_3\left(f_i\right)\right])\hfill \\ & =& S^n(\left({\sigma }_1{\circ }_h{\sigma }_2\right)\left(f_i\right),\left({\sigma }_1{\circ }_h{\sigma }_3\right)\left(f_i\right),\dots ,\left({\sigma }_1{\circ }_h{\sigma }_3\right)\left(f_i\right))\hfill \\ & =& \left(\left({\sigma }_1{\circ }_h{\sigma }_2\right){+}_h\left({\sigma }_1{\circ }_h{\sigma }_3\right)\right)\left(f_i\right).\hfill \end{array}$$

□

Unfortunately, the right distributivity, i.e., $\left({\sigma }_1{+}_h{\sigma }_2\right){\circ }_h{\sigma }_3=\left({\sigma }_1{\circ }_h{\sigma }_3\right){+}_h\left({\sigma }_2{\circ }_h{\sigma }_3\right)$, is not true, as shown in the following counterexample.

Example 6. 

Let I be a singleton index set and ${\tau }_2=\left(2\right)$ with a binary operation symbol f. Assume that ${\sigma }_1,{\sigma }_2,{\sigma }_3$ are elements in $Hy{p}^{O_2}\left(2\right)$ defined by

$$\begin{array}{c}{\sigma }_1\left(f\right)=f(x_{{\alpha }_1\left(1\right)},x_{{\alpha }_1\left(2\right)}) where {\alpha }_1=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right),\\ {\sigma }_2\left(f\right)=f(x_{{\alpha }_2\left(1\right)},x_{{\alpha }_2\left(2\right)}) where {\alpha }_2=\left(\begin{array}{cc}1& 2\\ 2& 2\end{array}\right),\\ {\sigma }_3\left(f\right)=f(f(x_{{\beta }_1\left(1\right)},x_{{\beta }_1\left(2\right)}),f(x_{{\beta }_2\left(1\right)},x_{{\beta }_2\left(2\right)})) where {\beta }_1=\left(\begin{array}{cc}1& 2\\ 1& 2\end{array}\right) and {\beta }_2=\left(\begin{array}{cc}1& 2\\ 2& 2\end{array}\right)\end{array}$$

Consider

$$\begin{array}{ccc}& & \left(\left({\sigma }_1{+}_h{\sigma }_2\right){\circ }_h{\sigma }_3\right)\left(f\right)\hfill \\ & =& \widehat{\left({\sigma }_1{+}_h{\sigma }_2\right)}\left[f(f(x_{{\beta }_1\left(1\right)},x_{{\beta }_1\left(2\right)}),f(x_{{\beta }_2\left(1\right)},x_{{\beta }_2\left(2\right)}))\right]\hfill \\ & =& S^2(\left({\sigma }_1{+}_h{\sigma }_2\right)\left(f\right),\widehat{\left({\sigma }_1{+}_h{\sigma }_2\right)}\left[f(x_{{\beta }_1\left(1\right)},x_{{\beta }_1\left(2\right)})\right],\widehat{\left({\sigma }_1{+}_h{\sigma }_2\right)}\left[f(x_{{\beta }_2\left(1\right)},x_{{\beta }_2\left(2\right)})\right])\hfill \\ & =& S^2(f(f(x_2,x_2),f(x_2,x_2)),f{(f(x_2,x_2),f(x_2,x_2))}_{{\beta }_1},f{(f(x_2,x_2),f(x_2,x_2))}_{{\beta }_2})\hfill \\ & =& S^2(f(f(x_2,x_2),f(x_2,x_2)),f(f(x_1,x_1),f(x_1,x_1)),f(f(x_2,x_2),f(x_2,x_2)))\hfill \\ & =& f(f(f(f(x_2,x_2),f(x_2,x_2)),f(f(x_2,x_2),f(x_2,x_2))),\hfill \\ & & f(f(f(x_2,x_2),f(x_2,x_2)),f(f(x_2,x_2),f(x_2,x_2))))\hfill \end{array}$$

and $\left(\left({\sigma }_1{\circ }_h{\sigma }_3\right){+}_h\left({\sigma }_2{\circ }_h{\sigma }_3\right)\right)\left(f\right)=S^2(\left({\sigma }_1{\circ }_h{\sigma }_3\right)\left(f\right),\left({\sigma }_2{\circ }_h{\sigma }_3\right)\left(f\right),\left({\sigma }_2{\circ }_h{\sigma }_3\right)\left(f\right)).$

Since we know that $\left({\sigma }_1{\circ }_h{\sigma }_3\right)\left(f\right)=f(f(x_1,x_1),f(x_1,x_1))$ and $\left({\sigma }_2{\circ }_h{\sigma }_3\right)\left(f\right)=f(f(x_2,x_2),f(x_2,x_2))$, a term $(\left({\sigma }_1{\circ }_h{\sigma }_3\right)+\left({\sigma }_2{\circ }_h{\sigma }_3\right))\left(f\right)$ does not equal $\left(\left({\sigma }_1{+}_h{\sigma }_2\right){\circ }_h{\sigma }_3\right)\left(f\right)$. Hence, $\left({\sigma }_1{+}_h{\sigma }_2\right){\circ }_h{\sigma }_3\ne \left({\sigma }_1{\circ }_h{\sigma }_3\right){+}_h\left({\sigma }_2{\circ }_h{\sigma }_3\right)$, which means that the right distributivity does not hold.

Since the algebra $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$ is generated by the set

$$F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}:=\{f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\mid i\in I,\alpha \in O_n\},$$

then any mapping

$$\eta :F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}\to W_{{\tau }_n}^{O_n}\left(X_n\right)$$

is called an $O_n$-full substitution. Let $Subs{t}_{O_n}\left({\tau }_n\right)$ be the set of all $O_n$-full substitutions. Each $O_n$-full substitution $\eta$ can be uniquely extended to an endomorphism

$$\overline{\eta }:W_{{\tau }_n}^{O_n}\left(X_n\right)\to W_{{\tau }_n}^{O_n}\left(X_n\right).$$

Define ${\eta }_1\odot {\eta }_2:={\overline{\eta }}_1\circ {\eta }_2$ for all ${\eta }_1,{\eta }_2\in Subs{t}_{O_n}\left({\tau }_n\right)$ where ∘ is the usual composition. Let $i{d}_{F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}}$ be an identity mapping on $F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}$. We note that the equation $\overline{{\overline{\eta }}_1\circ {\eta }_2}={\overline{\eta }}_1\circ {\eta }_2$ holds due to an application of Theorem 2. Hence, we find that $(Subs{t}_{OP}\left({\tau }_n\right);\odot ,i{d}_{F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}})$ is a monoid.

Theorem 9. 

$(Hy{p}^{O_n}\left({\tau }_n\right);{\circ }_h,{\sigma }_{id})$ can be embedded into $(Subs{t}_{O_n}\left({\tau }_n\right);\odot ,i{d}_{F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}})$.

Proof. 

Let $\sigma \in Hy{p}^{O_n}\left({\tau }_n\right)$. Then, according to Theorem 7, $\widehat{\sigma }:W_{{\tau }_n}^{O_n}\left(X_n\right)\to W_{{\tau }_n}^{O_n}\left(X_n\right)$ is an endomorphism of $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$. Since the algebra $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$ is generated by the generating set $F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}$, we obtain $\widehat{\sigma }/F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}$ as an $O_n$-full substitution with

$$\overline{\widehat{\sigma }/F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}}=\widehat{\sigma }.$$

To show that the mapping $\psi :Hy{p}^{O_n}\left({\tau }_n\right)\to Subs{t}_{O_n}\left({\tau }_n\right)$ defined by

$$\psi \left(\sigma \right)=\widehat{\sigma }/F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}$$

is a monomorphism, let ${\sigma }_1,{\sigma }_2\in Hy{p}^{O_n}\left({\tau }_n\right)$. Obviously, $\psi$ is an injection. Given the fact that the extension of the product of two hypersubstitutions is the composite mapping between each extension of the hypersubstitution, as proven in Proposition 4, then $\psi \left({\sigma }_1{\circ }_h{\sigma }_2\right)={\left({\sigma }_1{\circ }_h{\sigma }_2\right)}^{^}/F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}=({\widehat{\sigma }}_1\circ {\widehat{\sigma }}_2)/F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}=\overline{{\widehat{\sigma }}_1/F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}}\circ {\widehat{\sigma }}_2/F_{W_{{\tau }_n}^{O_n}\left(X_n\right)}=\overline{\psi \left({\sigma }_1\right)}\circ \psi \left({\sigma }_2\right)=\psi \left({\sigma }_1\right)\odot \psi \left({\sigma }_2\right).$ Therefore, $\psi$ is a homomorphism. □

Consider a variety of algebras V of type ${\tau }_n$, and let $IdV$ be the set of all identities in V. Now, define $I{d}^{O_n}V$ as the set of identities $s\approx t$ in V where both s and t are order-preserving full terms of type ${\tau }_n$. In other words,

$$I{d}^{O_n}V:={\left(W_{{\tau }_n}^{O_n}\left(X_n\right)\right)}^2\cap IdV.$$

It is well known that $IdV$ forms a congruence on the absolutely free algebra ${\mathcal{F}}_{{\tau }_n}\left(X_n\right)$. However, this is not generally the case for $I{d}^{O_n}V$. The theorem below demonstrates that $I{d}^{O_n}V$ is a congruence on $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$.

Theorem 10. 

$I{d}^{O_n}V$ is a congruence on $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$.

Proof. 

Assume that $r\approx t,r_1\approx t_1,\dots ,r_n\approx t_n\in I{d}^{O_n}V.$ We will prove that $S^n(r,r_1,\dots ,r_n)\approx S^n(t,t_1,\dots ,t_n)\in I{d}^{O_n}V.$ Firstly, we will prove using induction on the complexity of the term $r\in W_{{\tau }_n}^{O_n}\left(X_n\right)$ that

$$S^n(r,r_1,\dots ,r_n)\approx S^n(r,t_1,\dots ,t_n)\in I{d}^{O_n}V.$$

If $r=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ where $\alpha \in O_n$, it follows given that $IdV$ is compatible with the operations ${\overline{f}}_i$ of the absolutely free algebra ${\mathcal{F}}_{{\tau }_n}\left(X_n\right)$ and given the definition of the order-preserving full terms that

$$f_i(r_{\alpha \left(1\right)},\dots ,r_{\alpha \left(n\right)})\approx f_i(t_{\alpha \left(1\right)},\dots ,t_{\alpha \left(n\right)})\in I{d}^{O_n}V.$$

That is,

$$S^n(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)}),r_1,\dots ,r_n)\approx S^n(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)}),t_1,\dots ,t_n)\in I{d}^{O_n}V.$$

And

$$S^n(r,r_1,\dots ,r_n)\approx S^n(r,t_1,\dots ,t_n)\in I{d}^{O_n}V.$$

Suppose that $r=f_i(u_1,\dots ,u_n)$ such that for all $1\le k\le n$,

$$S^n(u_k,r_1,\dots ,r_n)\approx S^n(u_k,t_1,\dots ,t_n)\in I{d}^{O_n}V.$$

Thus, $f_i(S^n(u_1,r_1,\dots ,r_n),\dots ,S^n(u_n,r_1,\dots ,r_n))\approx f_i(S^n(u_1,t_1,\dots ,t_n),\dots ,S^n(u_n,t_1,\dots ,t_n))\in I{d}^{O_n}V.$

Moreover,

$$S^n(f_i(s_1,\dots ,s_n),r_1,\dots ,r_n)\approx S^n(f_i(s_1,\dots ,s_n),t_1,\dots ,t_n)\in I{d}^{O_n}V.$$

This means $S^n(r,r_1,\dots ,r_n)\approx S^n(r,t_1,\dots ,t_n)\in I{d}^{O_n}V$. So, we have the claim.

Given $S^n(r,t_1,\dots ,t_n)\approx S^n(t,t_1,\dots ,t_n)\in I{d}^{O_n}V$, then

$$S^n(r,r_1,\dots ,r_n)\approx S^n(r,t_1,\dots ,t_n)\approx S^n(t,t_1,\dots ,t_n)\in I{d}^{O_n}V.$$

□

Using the concept of $O_n$-full hypersubstitutions, the notions of an $O_n$-full closed identity and an $O_n$-full closed variety are introduced.

Definition 2. 

Let V be a variety of type ${\tau }_n$.

(1) 

An identity $s\approx t\in I{d}^{O_n}V$ is called an$O_n$-full closed identity (or an order-preserving identity) of V if $\widehat{\sigma }\left[s\right]\approx \widehat{\sigma }\left[t\right]\in I{d}^{O_n}V$ for all $\sigma \in Hy{p}^{O_n}\left({\tau }_n\right)$;

(2) 

A variety V is called $O_n$-full closed (or an order-preserving variety) if every identity in $I{d}^{O_n}V$ is an $O_n$-full closed identity.

A necessary condition for any variety V to be $O_n$-full closed is provided.

Theorem 11. 

Let V be a variety of type ${\tau }_n$. If $I{d}^{O_n}V$ is a fully invariant congruence on $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$, then V is $O_n$-full closed.

Proof. 

Assume that $I{d}^{O_n}V$ is a fully invariant congruence on $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$. If $s\approx t\in I{d}^{O_n}V$ and $\sigma \in Hy{p}^{OP}\left({\tau }_n\right)$, then according to Theorem 7, $\widehat{\sigma }$ acts as an endomorphism on $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$. Consequently, $\widehat{\sigma }\left[s\right]\approx \widehat{\sigma }\left[t\right]\in I{d}^{O_n}V$, meaning that V is $O_n$-full closed.

□

For a variety V of type ${\tau }_n$, $I{d}^{O_n}V$ is a congruence on $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$ according to Theorem 10. We then form the quotient algebra

$$M{A}_{OP}\left(V\right):=(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)/I{d}^{O_n}V.$$

The quotient algebra obtained belongs to $V_{SASS}$. Note that we have a natural homomorphism

$$na{t}_{I{d}^{O_n}V}:(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)\to M{A}_{OP}\left(V\right)$$

such that

$$na{t}_{I{d}^{O_n}V}\left(t\right)={\left[t\right]}_{I{d}^{O_n}V}.$$

Finally, we prove the following theorem.

Theorem 12. 

Every $s\approx t\in I{d}^{O_n}V$ is an $O_n$-full closed identity of a variety V of algebras of type ${\tau }_n$.

Proof. 

Assume that $s\approx t\in I{d}^{O_n}V$ and $\sigma \in Hy{p}^{OP}\left({\tau }_n\right)$. According to Theorem 7, we find that $\widehat{\sigma }:W_{{\tau }_n}^{O_n}\left(X_n\right)\to W_{{\tau }_n}^{O_n}\left(X_n\right)$ is an endomorphism on the algebra $(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)$. Thus,

$$na{t}_{I{d}^{O_n}V}\circ \widehat{\sigma }:(W_{{\tau }_n}^{O_n}\left(X_n\right),S^n)\to M{A}_{OP}\left(V\right)$$

is a homomorphism. By assumption,

$$na{t}_{I{d}^{O_n}V}\circ \widehat{\sigma }\left(s\right)=na{t}_{I{d}^{O_n}V}\circ \widehat{\sigma }\left(t\right).$$

That is,

$$na{t}_{I{d}^{O_n}V}\left(\widehat{\sigma }\left[s\right]\right)=na{t}_{I{d}^{O_n}V}\left(\widehat{\sigma }\left[t\right]\right).$$

Thus,

$${\left[\widehat{\sigma }\left[s\right]\right]}_{I{d}^{O_n}V}={\left[\widehat{\sigma }\left[t\right]\right]}_{I{d}^{O_n}V}.$$

That is,

$$\widehat{\sigma }\left[s\right]\approx \widehat{\sigma }\left[t\right]\in I{d}^{O_n}V.$$

Therefore, $s\approx t$ is an $O_n$-full closed identity of V. □

5. Homomorphisms of Algebras of Full Terms Induced by Order-Preserving Full Terms

In this section, we discuss several kinds of homomorphisms on algebras of full terms induced by order-preserving full terms. The first result shows the homomorphism property of the superassociative algebra $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$ and the algebra of term operations on an arbitrary algebra.

Let $\mathcal{A}:=(A,{\left(f_i^{\mathcal{A}}\right)}_{i\in I})$ be an algebra of type ${\tau }_n$, and let t be an n-ary order-preserving full term of type $\tau$ over an alphabet $X_n$. Then, t induces an n-ary operation $t^{\mathcal{A}}$ on $\mathcal{A}$ through the following steps:

(1)

If $t=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ where $\alpha$ is a mapping in $O_n$,

then $t^{\mathcal{A}}=f_i{(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})}^{\mathcal{A}}=f_i^{\mathcal{A}}(x_{\alpha \left(1\right)}^{\mathcal{A}},\dots ,x_{\alpha \left(n\right)}^{\mathcal{A}})={\left(f_i^{\mathcal{A}}\right)}_{\alpha }$;

(2)

If $t=f_i(t_1,\dots ,t_{n_i})$ is an n-ary order-preserving full term of type $\tau$, and $t_1^{\mathcal{A}},\dots ,t_{n_i}^{\mathcal{A}}$ are the term operations which are induced by $t_1,\dots ,t_{n_i},$ then $t^{\mathcal{A}}=f_i^A(t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})$ where $f_i^A(t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})$ is defined by $f_i^{\mathcal{A}}(t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})(a_1,\dots ,a_n)=f_i^{\mathcal{A}}(t_1^{\mathcal{A}}(a_1,\dots ,a_n),\dots ,t_n^{\mathcal{A}}(a_1,\dots ,a_n))$ for every $a_1,\dots ,a_n\in A$.

Consequently, $t^{\mathcal{A}}$ is called the order-preserving full term operation induced by t on the algebra $\mathcal{A}$. The set of all n-ary order-preserving full term operations on $\mathcal{A}$ will be denoted by $W_{{\tau }_n}^{O_n}{\left(X_n\right)}^{\mathcal{A}}$.

On the set $W_{{\tau }_n}^{O_n}{\left(X_n\right)}^{\mathcal{A}}$, an $(n+1)$-ary operation ${\mathcal{O}}^{n,A}$ can be defined by

(1)

${\mathcal{O}}^{n,A}({\left(f_i^{\mathcal{A}}\right)}_{\alpha },t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}}):=f_i^{\mathcal{A}}(t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})$;

(2)

${\mathcal{O}}^{n,A}(u_1^{\mathcal{A}},\dots ,u_n^{\mathcal{A}}),t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}}):=f_i^{\mathcal{A}}({\mathcal{O}}^{n,A}(u_1^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}}),\dots ,{\mathcal{O}}^{n,A}(u_n^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})).$

Normally, the algebra

$$\mathrm{Clone}\mathcal{A}=(W_{{\tau }_n}^{O_n}{\left(X_n\right)}^{\mathcal{A}},{\mathcal{O}}^{n,A})$$

is constructed.

To show that $\mathrm{Clone}\mathcal{A}$ is a homomorphic image of the algebra $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$, the following lemma is needed.

Lemma 2. 

${\left(S^n(s,t_1,\dots ,t_n)\right)}^{\mathcal{A}}={\mathcal{O}}^{n,A}(s^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})$.

Proof. 

We give a proof on a characteristic of s. Assume that $s=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ where $\alpha \in O_n$. Then, we obtain

$$\begin{array}{ccc}\hfill {\left(S^n(s,t_1,\dots ,t_n)\right)}^{\mathcal{A}}& =& {\left(S^n(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)}),t_1,\dots ,t_n)\right)}^{\mathcal{A}}\hfill \\ & =& {\left(f_i(t_{\alpha \left(1\right)},\dots ,t_{\alpha \left(n\right)})\right)}^{\mathcal{A}}\hfill \\ & =& f_i^{\mathcal{A}}(t_{\alpha \left(1\right)}^{\mathcal{A}},\dots ,t_{\alpha \left(n\right)}^{\mathcal{A}})\hfill \\ & =& {\mathcal{O}}^{n,A}({\left(f_i^{\mathcal{A}}\right)}_{\alpha },t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})\hfill \\ & =& {\mathcal{O}}^{n,A}(f_i{(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})}^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})\hfill \\ & =& {\mathcal{O}}^{n,A}(s^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}}).\hfill \end{array}$$

We now inductively suppose that $s=f_i(s_1,\dots ,s_n)$ and ${\left(S^n(s_j,t_1,\dots ,t_n)\right)}^{\mathcal{A}}={\mathcal{O}}^{n,A}(s_j^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})$ for every $1\le j\le n$. Then, we have

$$\begin{array}{ccc}\hfill {\left(S^n(s,t_1,\dots ,t_n)\right)}^{\mathcal{A}}& =& {\left(S^n(f_i(s_1,\dots ,s_n),t_1,\dots ,t_n)\right)}^{\mathcal{A}}\hfill \\ & =& {\left(f_i(S^n(s_1,t_1,\dots ,t_n),\dots ,S^n(s_n,t_1,\dots ,t_n))\right)}^{\mathcal{A}}\hfill \\ & =& f_i^{\mathcal{A}}(S^n{(s_1,t_1,\dots ,t_n)}^{\mathcal{A}},\dots ,S^n{(s_n,t_1,\dots ,t_n)}^{\mathcal{A}})\hfill \\ & =& f_i^{\mathcal{A}}({\mathcal{O}}^{n,A}(s_1^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}}),\dots ,{\mathcal{O}}^{n,A}(s_n^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}}))\hfill \\ & =& {\mathcal{O}}^{n,A}(f_i^{\mathcal{A}}(s_1^{\mathcal{A}},\dots ,s_n^{\mathcal{A}}),t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})\hfill \\ & =& {\mathcal{O}}^{n,A}(f_i{(s_1,\dots ,s_n)}^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})\hfill \\ & =& {\mathcal{O}}^{n,A}(s^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}}).\hfill \end{array}$$

□

As a consequence, we prove the following:

Theorem 13. 

For every algebra $\mathcal{A}$ of type ${\tau }_n$, the algebra $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$ is homomorphic to $Clone\mathcal{A}$.

Proof. 

For $n\in \mathbb{N}$, we define the mapping $\theta :W_{{\tau }_n}^{O_n}\left(X_n\right)\to W_{{\tau }_n}^{O_n}{\left(X_n\right)}^{\mathcal{A}}$ by

$$\theta \left(t\right)=t^{\mathcal{A}}$$

for every order-preserving full term t. Obviously, $\theta$ is well defined. Indeed, from $s=t$, we obtain $Id\mathcal{A}$, which gives $s^{\mathcal{A}}=t^{\mathcal{A}}$. To prove that $\theta$ is a homomorphism, we let $s,t_1,\dots ,t_n$ be the order-preserving full terms of type ${\tau }_n$. In fact, Lemma 2 implies that $\theta \left(S^n(s,t_1,\dots ,t_n)\right)$ and ${\mathcal{O}}^{n,A}(\theta \left(s\right),\theta \left(t_1\right),\dots ,\theta \left(t_n\right))$ are equal, which means that $\theta \left(S^n(s,t_1,\dots ,t_n)\right)={\left(S^n(s,t_1,\dots ,t_n)\right)}^{\mathcal{A}}={\mathcal{O}}^{n,A}(s^{\mathcal{A}},t_1^{\mathcal{A}},\dots ,t_n^{\mathcal{A}})={\mathcal{O}}^{n,A}(\theta \left(s\right),\theta \left(t_1\right),\dots ,\theta \left(t_n\right)).$ Alternatively, we say that $Clone\mathcal{A}$ is a homomorphic image of $\overline{W_{{\tau }_n}^{O_n}\left(X_n\right)}$. □

Our next purpose is to show another homomorphism of the algebra of order-preserving full terms. For this, we apply the notion of the complexity of terms given in [20] to an order-preserving full term.

Definition 3. 

The maximum depth of an order-preserving full term t, denoted by $maxdepth\left(t\right)$, is the longest distance from the first operation symbol that appears in t (from the left) to the variables. It can be inductively defined by

(1) 

$\mathrm{maxdepth}\left(t\right)=1$ if $t=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$;

(2) 

$\mathrm{maxdepth}\left(t\right)=1+\max \{\mathrm{maxdepth}\left(t_j\right)\mid 1\le j\le n\}$ if $t=f_i(t_1,\dots ,t_n)$.

We produce a formula for counting the maximum depth of order-preserving full terms under the superposition operation.

Theorem 14. 

Let $s,t_1\dots ,t_n\in W_{{\tau }_n}^{O_n}\left(X_n\right)$. Then,

$$\mathrm{maxdepth}\left(S^n(s,t_1,\dots ,t_n)\right)=\mathrm{maxdepth}\left(s\right)+\max \{\mathrm{maxdepth}\left(t_j\right)\mid 1\le j\le n\}.$$

Proof. 

We give a proof by induction on the maximum depth of an order-preserving full term s. If $\mathrm{maxdepth}\left(s\right)=1$, then $s=f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ where $\alpha$ is a mapping on $O_n$ and with each $i\in I$, $f_i$ having arity n. It follows from the superposition that

$$\mathrm{maxdepth}\left(S^n(s,t_1,\dots ,t_n)\right)=\mathrm{maxdepth}\left(f_i(t_{\alpha \left(1\right)},\dots ,t_{\alpha \left(n\right)})\right).$$

(3)

Since $\max \{\mathrm{maxdepth}\left(t_{\alpha \left(j\right)}\right)\mid 1\le j\le n\}=\max \{\mathrm{maxdepth}\left(t_j\right)\mid 1\le j\le n\}$ and from (3), we conclude that

$$\mathrm{maxdepth}\left(S^n(s,t_1,\dots ,t_n)\right)=\mathrm{maxdepth}\left(s\right)+\max \{\mathrm{maxdepth}\left(t_j\right)\mid j=1,\dots ,n\}.$$

Now, let $s=f_i(s_1,\dots ,s_n)$, and the formula is already satisfied for $s_1,\dots ,s_n$. Since we know that

$$S^n(f_i(s_1,\dots ,s_n),t_1,\dots ,t_n)=f_i(S^n(s_1,t_1,\dots ,t_n),\dots ,S^n(s_n,t_1,\dots ,t_n)),$$

then we have

$$\begin{array}{ccc}& & \mathrm{maxdepth}\left(S^n(f_i(s_1,\dots ,s_n),t_1,\dots ,t_n)\right)\hfill \\ & =& 1+\max \{\mathrm{maxdepth}\left(S^n(s_k,t_1,\dots ,t_n)\right)\mid 1\le k\le n\}\hfill \\ & =& 1+\max \{\mathrm{maxdepth}\left(s_k\right)+\max \{\mathrm{maxdepth}\left(t_j\right)\mid 1\le j\le n\}\mid 1\le k\le n\}\hfill \\ & =& 1+\max \{\mathrm{maxdepth}\left(s_k\right)\mid 1\le k\le n\}+\max \{\mathrm{maxdepth}\left(t_j\right)\mid 1\le j\le n\}\hfill \\ & =& \mathrm{maxdepth}\left(s\right)+\max \{\mathrm{maxdepth}\left(t_j\right)\mid 1\le j\le n\}.\hfill \end{array}$$

This completes the proof. □

As a corollary, the formula for the maximum depth of $\widehat{\sigma }\left[t\right]$ in terms of the maximum depth t when t is an arbitrary order-preserving full term and $\sigma$ is an order-preserving hypersubstitution is obtained.

Corollary 1. 

Let ${\tau }_n$ be a type and $t\in W_{{\tau }_n}^{O_n}\left(X_n\right)$. Then,

$$\mathrm{maxdepth}\left(\widehat{\sigma }\left[t\right]\right)=\mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)·\mathrm{maxdepth}\left(t\right).$$

Proof. 

If t is an order-preserving full term of the form $f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})$ where $\alpha \in O_n$, then according to the definition of the extension of $\sigma$, we have $\mathrm{maxdepth}\left(\widehat{\sigma }\left[t\right]\right)=\mathrm{maxdepth}\left(\widehat{\sigma }\left[f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right]\right)=\mathrm{maxdepth}\left({\left(\sigma \left(f_i\right)\right)}_{\alpha }\right)=\mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)·\mathrm{maxdepth}\left(f_i(x_{\alpha \left(1\right)},\dots ,x_{\alpha \left(n\right)})\right)=\mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)·\mathrm{maxdepth}\left(t\right).$ Inductively, assume that $t=f_i(t_1,\dots ,t_n)$ and the formulas are satisfied for $t_1,\dots ,t_n$. Then, we obtain

$$\begin{array}{ccc}& & \mathrm{maxdepth}\left(\widehat{\sigma }\left[t\right]\right)\hfill \\ & =& \mathrm{maxdepth}\left(\widehat{\sigma }\left[f_i(t_1,\dots ,t_n)\right]\right)\hfill \\ & =& \mathrm{maxdepth}\left(S^n(\sigma \left(f_i\right),\widehat{\sigma }\left[t_1\right],\dots ,\widehat{\sigma }\left[t_n\right])\right)\hfill \\ & =& \mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)+\max \{\mathrm{maxdepth}\left(\widehat{\sigma }\left[t_j\right]\right)\mid 1\le j\le n\}\hfill \\ & =& \mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)+\max \{\mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)·\mathrm{maxdepth}\left(t_j\right)\mid 1\le j\le n\}\hfill \\ & =& \mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)+\mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)·\max \{\mathrm{maxdepth}\left(t_j\right)\mid 1\le j\le n\}\hfill \\ & =& \mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)·(1+\max \{\mathrm{maxdepth}\left(t_j\right)\mid 1\le j\le n\})\hfill \\ & =& \mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)·\mathrm{maxdepth}\left(f_i(t_1,\dots ,t_n)\right)\hfill \\ & =& \mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)·\mathrm{maxdepth}\left(t\right).\hfill \end{array}$$

□

Moreover, the maximum depth of order-preserving full hypersubstitutions is now defined.

Definition 4. 

The maximum depth of $\sigma \in Hy{p}^{O_n}\left({\tau }_n\right)$, denoted by $\mathrm{maxdepth}\left(\sigma \right)$, is the greatest value of the maximum depth of each order-preserving full term $\sigma \left(f_i\right)$ for $i\in I$, i.e.,

$$\mathrm{maxdepth}\left(\sigma \right)=\max \{\mathrm{maxdepth}\{\sigma \left(f_i\right)\mid i\in I\}.$$

For example, let ${\tau }_3=(3,3)$ be a type with the ternary operation symbols f and g. If we set $\sigma \in Hy{p}^{O_3}\left((3,3)\right)$ using

$$f\mapsto f(x_1,x_1,x_3)\mathrm{and}g\mapsto g(f(x_1,x_2,x_3),g(x_3,x_3,x_3),f(x_1,x_2,x_3)).$$

It can be seen that $\mathrm{maxdepth}\left(\sigma \right(f\left)\right)=1$ and $\mathrm{maxdepth}\left(\sigma \right(g\left)\right)=2$. As a consequence, $\mathrm{maxdepth}\left(\sigma \right)=\max \{1,2\}=2$.

We close this section with the connection between $(Hy{p}^{O_n}\left({\tau }_n\right),{\circ }_h,{+}_h,{\sigma }_{id})$ and the set of natural numbers with respect to the usual addition and usual multiplication in sense of a homomorphism.

Theorem 15. 

On the left-seminearrings $(Hy{p}^{O_n}\left({\tau }_n\right),{\circ }_h,{+}_h,{\sigma }_{id})$ and $(\mathbb{N},·,+,1)$, the mapping

$$\omega :Hy{p}^{O_n}\left({\tau }_n\right)\to \mathbb{N}$$

defined by $\omega \left(\sigma \right)=\mathrm{maxdepth}\left(\sigma \right)$ is an epimorphism.

Proof. 

It is obvious that the mapping $\omega$ is well defined. For every natural number n, there is an order-preserving full hypersubstitution $\sigma$ such that $\mathrm{maxdepth}\left(\sigma \right)=n$. Thus, $\omega$ is surjective. To show that the mapping $\omega$ preserves the operations, we show that the following two equations are satisfied:

$$\omega \left(\sigma {\circ }_h\alpha \right)=\omega \left(\sigma \right)·\omega \left(\alpha \right)$$

(4)

and

$$\omega \left(\sigma {+}_h\alpha \right)=\omega \left(\sigma \right)+\omega \left(\alpha \right)$$

(5)

for every $\sigma ,\alpha \in Hy{p}^{O_n}\left({\tau }_n\right).$ In fact, applying Corollary 1, we have

$$\begin{array}{ccc}\hfill \mathrm{maxdepth}\left(\sigma {\circ }_h\alpha \right)& =& \max \{\mathrm{maxdepth}\left(\left(\sigma {\circ }_h\alpha \right)\left(f_i\right)\right)\mid i\in I\}\hfill \\ & =& \max \{\mathrm{maxdepth}\left(\widehat{\sigma }\left[\alpha \left(f_i\right)\right]\right)\mid i\in I\}\hfill \\ & =& \max \{\mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)·\mathrm{maxdepth}\left(\alpha \left(f_i\right)\right)\mid i\in I\}\hfill \\ & =& \max \{\mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)\mid i\in I\}·\max \{\mathrm{maxdepth}\left(\alpha \left(f_i\right)\right)\mid i\in I\}\hfill \\ & =& \mathrm{maxdepth}\left(\sigma \right)·\mathrm{maxdepth}\left(\alpha \right),\hfill \end{array}$$

which shows that Equation (4) holds. Moreover, it follows from Theorem 14 that

$$\begin{array}{ccc}\hfill \mathrm{maxdepth}\left(\sigma {+}_h\alpha \right)& =& \max \{\mathrm{maxdepth}\left(\left(\sigma {+}_h\alpha \right)\left(f_i\right)\right)\mid i\in I\}\hfill \\ & =& \max \{\mathrm{maxdepth}\left(S^n(\sigma \left(f_i\right),\alpha \left(f_i\right),\dots ,\alpha \left(f_i\right))\right)\mid i\in I\}\hfill \\ & =& \max \{\mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)+\mathrm{maxdepth}\left(\alpha \left(f_i\right)\right)\}\hfill \\ & =& \max \{\mathrm{maxdepth}\left(\sigma \left(f_i\right)\right)\mid i\in I\}+\max \{\mathrm{maxdepth}\left(\alpha \left(f_i\right)\right)\mid i\in I\}\hfill \\ & =& \mathrm{maxdepth}\left(\sigma \right)+\mathrm{maxdepth}\left(\alpha \right),\hfill \end{array}$$

which proves (5). It is not difficult to show that the mapping $\omega$ sends an identity mapping on $Hy{p}^{O_n}\left({\tau }_n\right)$ to a natural number 1 acting as an identity with respect to a usual multiplication · on $\mathbb{N}$. Indeed, from Corollary 1, we can conclude that $\mathrm{maxdepth}\left({\sigma }_{id}\right)=\max \{\mathrm{maxdepth}\left({\sigma }_{id}\left(f_i\right)\right)\mid i\in I\}=1.$ □