数理解析研究所講究録
第 1846 巻 2013 年 80-86

80

Hyperalgebras and hyperclones-different approaches
Gradimir VOJVODI\’{C}Faculty of Sciences
University of Novi Sad
21000 Novi Sad
Serbia and Montenegro

Jovanka PANTOVI\’{C}*
Faculty of Technical Sciences
University of Novi Sad
21000 Novi Sad
Serbia and Montenegro

Abstract
In the current paper, authors give a resume about results on hyperclone lattice on , obtained byembedding a clone lattice into it and by embedding it into the lattice of clones on. Theseresults are used for characterization of minimal hyperclones on two element set and unary minimal
hyperclones on an arbitrary finite set $A.$
$A$

$P(A)\backslash \{\emptyset\}$1

Introduction

There are several synonims appearing for the class of algebraic systems whose clones have been studied
and presented in this paper. They are hyperalgebras, multialgebras and polyalgebras.
The first paper on this topic was written by Marty in 1934. From that time, there have been publishedhundreds of papers devoted to special hyperalgebras. Some attention has also been given to hyperclones
and partial hyperclones by a few authors.
Hyperoperations and hyperalgebras have both mathematical and nonmathematical applications. For
example, they give possibility to describe nondeterministic processes. Rosenberg has studied hyperclones
on a finite universe via natural (not full) order embedding from the hyperclone lattice on to the clone
lattice on. He posed the question of further study of the lattice of hyperclones, inparticular in the simplest two-element case $A=\{0,1\}$ . It has been shown by Machida that the latticeon the two-element set has continuum cardinality ([8]). Drescher, in [5], presented an overview of partial
hyperoperations and relations. Many questions which arise from the process of extending operations arestudied there, too.
$A$

$A$

$P(A)\backslash \{\emptyset\}.([14], [15])$

2

Definitions and preliminaries

Paper explores some definitions, claims and ideas mentioned in [5], [6],[8] and [11]. The authors site
several of them in the current paper. Proofs of some claims are not presented and could be found in [9]and [10].
Letbe the set of positive integers, $n,$a nonempty finite set, $P(A)$ the power set of $A.$
For a positive integer , an -ary operation on is a mapping : $A^{n}arrow A.$
For a positive integer , an -ary hyperoperation on is a mapping $f.$ :We will denote the set of all operations by
and the set of all hyperoperations on by
For a positive integer , an i-th projection onof arity $n,$ $1\leq i\leq n$ , is an $n-$ ary operation$m\in \mathbb{N},$$\mathbb{N}$

$n$

$n$

$n$

$n$

$A$

$A$

$f$

$A$

$A^{n}arrow P(A)\backslash \{\emptyset\}.$$A$$O_{A}$$n$$\pi_{i}^{n}:A^{n}arrow A,$$H_{A}.$$A$$(x_{1}, \ldots, x_{n})\mapsto x_{i}.$For positive integers and $m$ , we define the composition:, where $f(g_{1}, \ldots,g_{n}):A^{m}arrow A,$ $(x_{1}, \ldots, x_{m})\mapsto f(g_{1}(x_{1}, \ldots, x_{m}), \ldots, g_{n}(x_{1}, \ldots, x_{m}))$ .
$n$$S_{m}^{n}$$0_{A}^{(n)}\cross(O_{A}^{(m)})^{n}arrow O_{A}^{(m)},$$(f, g_{1}, \ldots, g_{n})\mapsto$$f(g_{1}, \ldots, g_{n})$’Supported by Research Institute for Mathematical Sciences, Kyoto University and Hitotsubashi University, Tokyo.
The researh was supported by the Ministry for Science, Technology and Development of Republic Serbia under Grant No.
1708.

81
Let

and $g\in O_{A}^{(m)}$ , thenare defined by$f\in O_{A}^{(n)}$$\nabla f\in O_{A}^{(n+1)}$$\sigma f\in O_{A}^{(n)},$$\tau f\in O_{A}^{(n)},$$\triangle f\in O_{A}^{(n-1)},$$fog\in O_{A}^{(m+n-1)}$and$(\sigma f)(x_{1}, x_{2}, \ldots, x_{n})=f(x_{2}, \ldots, x_{n}, x_{1}),$ $n\geq 2$$(\tau f)(x_{1}, x_{2}, x_{3}, \ldots, x_{n})=f(x_{2}, x_{1}, x_{3}, \ldots, x_{n}),$$(\Delta f)(x_{1}, \ldots, x_{n-1})=f(x_{1}, x_{1}, \ldots , x_{n-1}),$$n\geq 2$$n\geq 2,$$n=1,$$(\sigma f)=(\mathcal{T}f)=(\triangle f)=f,$$(f\circ g)(x_{1}, \ldots, x_{m}, x_{m+1}, \ldots, x_{m+n-1})=f(g(x_{1}, \ldots,x_{m}), x_{m+1}, \ldots, x_{m+n-1})$$(\nabla f)(x_{1}, x_{2}, \ldots, x_{n+1})=f(x_{2}, \ldots, x_{n+1})$,$x_{m+n-1}\in A$ . The full algebra of operations is
where. Each subuniverseofis a clone.
A set of operations is a clone iff it contains all projections and is closed with respect to composition.
For a positive integer , an i-th hyperprojection onof arity $n,$ $1\leq i\leq n$ , is an -ary partial
hyperprojection, defined by
For positive integers and $m$ , we define the composition of hyperoperations:$(f, g_{1}, \ldots, g_{n})\mapsto f(g_{1}, \ldots, g_{n})$$(x_{1}, \ldots , x_{m})\mapsto\cup\{f(y_{1},$, where:$\leq t\leq n\}.$:$\Delta f\in H_{A}^{(n-1)},$ $fog\in H_{A}^{(m+n-1)}$
Let $f\in H_{A}^{(n)}$ and $g\in H_{A}^{(m)}$ , thenandare defined by$x_{1},$$\mathcal{O}_{A}=(O_{A}, 0, \sigma, \tau, \triangle, \pi_{1}^{2})$$\ldots,$$\mathcal{O}_{A}$

$A$

$n$

$n$

$e_{i}^{n}(x_{1}, \ldots, x_{n})=\{x_{i}\},$$e_{i}^{n}:A^{n}arrow P(A)\backslash \{\emptyset\}$$(x_{1}, \ldots , x_{n})\in\rho.$$n$$S_{m}^{n}$$f(g_{1}, \ldots, g_{n})$$H_{A}^{(m)},$$y_{n})$$H_{A}^{(n)}\cross(H_{A}^{(m)})^{n}$$A^{m}arrow P(A)\backslash \{\emptyset\},$$\ldots,$$y_{i}\in g_{i}(x_{1}, \ldots, x_{m}),$ $1,$$\sigma f\in H_{A}^{(n)},$$\tau f\in H_{A}^{(n)},$$\nabla f\in H_{A}^{(n+1)}$$(\sigma f)(x_{1}, x_{2}, \ldots, x_{n})=f(x_{2}, \ldots, x_{n}, x_{1}),$ $n\geq 2$$(\tau f)(x_{1}, x_{2}, x_{3}, \ldots, x_{n})=f(x_{2}, x_{1}, x_{3)}x_{n}),$$(\Delta f)(x_{1}, \ldots, x_{n-1})=f(x_{1}, x_{1}, \ldots, x_{n-1}),$$(\sigma f)=(\tau f)=(\Delta f)=f,$$n\geq 2$$n\geq 2,$$n=1,$$(fog)(x_{1}, \ldots, x_{m}, x_{m+1}, \ldots, x_{m+n-1})=\cup\{f(y, x_{m+1}, \ldots, x_{m+n-1}):y\in g(x_{1}, \ldots, x_{m})\}$$(\nabla f)(x_{1}, x_{2}, \ldots, x_{n+1})=f(x_{2}, \ldots, x_{n+1})$ ,$x_{m+n-1}\in A$ . The full algebra of hyperoperations is
where. Eachsubuniverse ofis a yperclone.
A set of hyperoperations is a hyperclone iff it contains all hyperprojections and is closed with respect
to composition.
For a set we denote the set of all -ary elements from by. If is a set of hyperoperations,
we denote the set $\cup\{\rho_{f} : f\in C^{(n)}\}$ by $D(C)$ .
$x_{1},$$\mathcal{H}_{A}=(H_{A}, 0, \sigma, \tau, \Delta, e_{1}^{2})$$\ldots,$$\mathcal{H}_{A}$

$C$

$C$

$n$

$C^{(n)}$$C$$n\geq 1$

Let us denote the set of all clones on by
and the set of all hyperclones on by. Bothsets form algebraic lattices. The atoms (dual atoms) are called minimal (maximal) elements. The least
elements in both lattices, trivial clones, will be denoted by. For a set $F$ of hyperoperations, the least
hyperclone containing $F$ will be denoted by, and the least clone containing the setof operations
on will be denoted by$n\}arrow\{1, \ldots, m\}$ , we define
If $m,$ are positive integers,
and : 1,by$(x_{1)}\ldots, x_{m})\in A^{m}$ . For a set $F$ of hyperoperations-closure of$F$ isthe set $\delta(F)=\{\delta_{\alpha}(f) : f\in F^{(n)}, \alpha : \{1, \ldots , n\}arrow\{1, \ldots)m\}, m, n\in \mathbb{N}\}$ . For a positive integer$n>1$ , the mapping:, whereis defined by $f^{\#}(X_{1}, \ldots, X_{n})=$$\cup\{f(x_{1}, \ldots , x_{n})|x_{i}\cdot\in X_{i}, 1\leq i\leq n\}$ , is an isomorphism fromonto
and there is a hyperoperation $h\in H_{A}$ such that
. ([14], [15]) We say that an operation
is extended from the hyperoperation . If $F$ is an arbitrary set of hyperoperations, then $F^{\neq}=\{f^{\#}|f\in F\}.$
$A$

$A$

$L_{O_{A}}$$L_{H_{A}}$$J_{A}$$F$$\langle F\rangle$$A$$\langle F\rangle_{A}.$$f\in H_{A}^{(n)}$$n$$\alpha$$\delta_{\alpha}(f)\in H_{A}^{(m)}$$\{$$\ldots,$$\delta_{\alpha}(f)(x_{1}, \ldots, x_{m})=f(x_{\alpha(1)}, \ldots, x_{\alpha(n)}),$$(\cdot)^{\#}$$\delta$$H_{A}^{(n)}arrow O_{P(A)\backslash \{\emptyset\}}^{(n)},$$f\mapsto f^{\#}$$f^{\#}$$(H_{A}^{(n)}, *, \sigma, \tau, e_{1}^{2})$$(O_{P(A)\backslash \{\emptyset\}}^{(n)}, *, \sigma, \tau,p_{1}^{2})$$f^{\#}$$\lambda(\Delta h)\neq\Delta\lambda(h)$$f$3Hyperclone embeddings

3.1

From the hyperclone lattice on

The mapping :i.e. there are $F,$ $G\in L_{H_{A}}$ such that
Let be $\{0,1,2, \ldots\},$ $|A|\geq 3,$ $m\geq 2$ and$\lambda$$L_{H_{A}}arrow L_{P(A)\backslash \{\emptyset\}},$$A$

to the clone lattice on$C\mapsto\langle C\#\rangle_{P(A)\backslash \{\emptyset\}}$, is an order embedding, though not a full one,
. ([5]).
the hyperoperation defined by$[\lambda(F), \lambda(G)]\backslash im\lambda\neq\emptyset$$A$$P(A)\backslash \{\emptyset\}$$g_{m}\in H_{A}^{(m)}$$g_{m}(x_{1}, \ldots, x_{m})=\{\begin{array}{ll}\{2\}, (x_{1}, x_{2}, \ldots, x_{m})\in J_{m}\{0\}, otherwise\end{array}$

82
where

is the set of all-tuples with one coordinate equa12 and all others equal 1.
Let us define the hyperoperation $f_{m+1}\in H_{A}^{(m+1)}$ by$J_{m}$$m$$f_{m+1}(x_{1}, x_{2}, \ldots, x_{m+1})=\{\begin{array}{ll}A x_{1}\neq x_{2}g_{m}(x_{2}, \ldots, x_{m+1}) x_{1}=x_{2}.\end{array}$Thus, extended operation from $f_{m+1}$ is operation$f_{m+1}^{\#}\in(P(A)\backslash \{\emptyset\})^{(m+1)}$

defined by$f_{m+1}^{\#}(X_{1}, X_{2}, \ldots, X_{m+1}) = \cup\{f_{m+1}(x_{1}, x_{2}, \ldots, x_{m+1})|x_{i}\in X_{i}\}$$=$$\{\begin{array}{ll}g_{m}^{\#}(X_{2}, \ldots , X_{m+1}) , X_{1}=X_{2}, |X_{1}|=1A , otherwise\end{array}$

Lemma 3.1 For every clone $F$ of hyperopemtions on$A$and for every$\emptyset\neq Q\subseteq\bigcup_{i\geq 2}\{f_{1}\}$holds$\lambda(F)\neq$$\langle Q^{\#}\rangle_{P(A)\backslash \{\emptyset\}}.$

Proof. Let$Q$

be an arbitrary nonvoid subset of$\bigcup_{t\geq 2}\{f_{i}\}$. There is $m\geq 2$ such that $f_{m+1}\in Q.$

Suppose to the contrary, that there is a clone of hyperoperations $F$ such that its imageisthe clone generated by, i.e.
. Then, there is a hyperoperation $h\in F$ withproperty, it follows $f_{m+1}\in F$ (see [5]). Since $F$ is a clone, the. Fromhyperoperation, defined by $g_{m}(x_{1}, \ldots, x_{m})=f_{m+1}(x_{1}, x_{1}, \ldots,x_{m})$ , also belongs to $F(g_{m}=$. However, we shall prove that, because for every $i\geq 2f_{i}^{\#}(A, A, \ldots, A)=A$ , and,$g_{m}\#(A,A,\ldots,A)=\{0,2\}\neq
A.$becauseSo, $g_{m}\#\in F\#$ and(Note thatis not a clone of hyperoperations on $A.$ )$\lambda(F)$$\lambda$$Q^{\#}$$f_{m+1}^{\#}=\delta_{\alpha}h\#$$\lambda(F)=\langle Q^{\#}\rangle_{P(A)\backslash \{\emptyset\}}$$f_{m+1}^{\#}\in H_{A}^{\#}$$g_{m}\in H_{A}^{(m)}$$f_{m+1}(e_{m}^{1}, e_{m}^{1}, e_{m}^{2}, \ldots, e_{m}^{m}))$$g_{m}\#\not\in\langle Q^{\#}\rangle_{P(A)\backslash \{\emptyset\}}.$$\langle Q^{\#}\rangle\subset Pol_{P(A)\backslash \{\emptyset\}}(A)$$g_{m}\#\not\in Pol_{P(A)\backslash \{\emptyset\}}(A)$$g_{m}\not\in\langle Q^{\#}\rangle_{P(A)\backslash \{\emptyset\}}.$$Q$$\square$Lemma 3.2 For every $i\geq 3$ holds$f_{1}^{\#} \not\in\langle\bigcup_{J\geq 3,j\neq a}\{f_{j}^{\#}\}\rangle_{P(A)\backslash \{\emptyset\}}.$

Proof. Let us define for every $m\geq 2$ relation
by $\rho_{m}=A_{m}\cup B_{m}$ , whereis the set of all-tuples with exactly one coordinate equal {2} and all others equal {1} and $B_{m}=\{\{0\}, \{2\}, \{0,2\}, A\}^{m}\backslash$$(\{2\}, \{2\}, \ldots, \{2\})$ . We are going to show that$i\neq m.$and$(\{1\}, \{1\}, \ldots, \{2\})\in$
For $m$ -tuples $(\{2\}, \{1\}, \ldots, \{1\}),$ $(\{2\}, \{1\}, \ldots, \{1\}),$ $(\{1\}, \{2\}, \ldots, \{1\}),$(the first one and the second one are equal), holds $(f_{m+1}^{\#}(\{2\}, \{2\}, \{1\}, \ldots, \{1\}),$$\{1\},$$\{1\}),$$f_{m+1}^{\#}(\{1\},\{1\},\{1\},\ldots,\{2\}))=(\{2\},\{2\},\ldots,\{2\})\not\in\rho_{m}$
{2},
. So,
Suppose that there is $i\neq m$ such that
. Then, there are tuples $X_{1};=(X_{11},$$X_{m1}),$$X_{1+1}$ $:=(X_{1(t+1)}, \ldots, X_{m(i+1)})\in\rho_{m}$ , such that$:=(f_{i+1}^{\#}(X_{11}, \ldots, X_{1(i+1)})$ ,$f_{1+1}^{\#}(X_{m1},\ldots,X_{m(i+1)})\not\in\rho_{m}$. .. ,. Since $imf_{i+1}^{\#}=\{\{0\}, \{2\}, \{0,2\}, A\}$ , it follows that $(Y_{1}, \ldots, Y_{m})=$$(\{2\}, \{2\}, \ldots, \{2\})$ and it is possible only for$X_{\iota+1}\in A_{m},$ $X_{1}=X_{2}$ and $i=m$ . This is acontradiction.
$A_{m}$$\rho_{m}\in P_{A}^{m}$$m$$f_{m+1}^{\#}\not\in Pol_{P(A)\backslash \{\emptyset\}}\rho_{m}$$f_{1+1}^{\#}\in Pol_{P(A)\backslash \{\emptyset\}}\rho_{m},$$\ldots,$$A_{m}$$f_{m+1}^{\#}(\{1\},$$\ldots,$$f_{m+1}^{\#}\not\in Pol_{P(A)\backslash \{\emptyset\}}\rho_{m}.$$\ldots,$$f_{1+1}^{\#}\not\in Pol_{P(A)\backslash \{\emptyset\}}\rho_{m}$$\ldots,$$(Y_{1}, Y_{2}, \ldots , Y_{m})$$\ldots,$$X_{1},$$X_{2},$$\ldots,$$\square$

Theorem 3.1 There are continuumly many painvise distinct dones of operations on
intervalthat are not in the set of all imagesof the operation$[\lambda(Jh_{A}), \lambda(H_{A})]$

Proof. Let$im\lambda$$\lambda$$\langle Q\rangle\leq H_{A}$On the other hand,the following:$\lambda$

is injective and for $F,$ $G\in L_{H_{A}}F\leq G$ is equivalent to$\langle Q^{\#}\rangle_{P(A)\backslash \{\emptyset\}}\leq\lambda(\langle Q\rangle)\leq\lambda(H_{A}),$$Q\subseteq$\lambda(Jh_{A})\leq\langle Q^{\#}\rangle_{P(A)\backslash \{\emptyset\}}$, because.

(see [14],[15])$\lambda(F)\leq\lambda(G)$.

R.$$\lambda(Jh_{A})=J_{P(A)\backslash \{\emptyset\}}$. This being the result of$(e_{1}^{n})^{\#}(X_{1}, \ldots, X_{n})=\bigcup_{x:\in X}.$ $e_{i}^{n}(x_{1}, \ldots, x_{n})=\bigcup_{x_{i}\in x_{:}}\{x_{i}\}=X_{i}=p_{i}^{n,P(A)\backslash \{\emptyset\}}(X_{1},$$X_{n})$

in the$R= \bigcup_{i\geq 3}\{f_{i}\}.$(a) Since is an order embedding,So, forit follows...,$P(A)\backslash \{\emptyset\}$$\lambda.$

83(b) It follows from Lemma3.1 that for every(c) From Lemma3.2 follows that for all
ifthen$Q_{1}^{\#}\neq Q_{2}^{\#}$$Q\subseteq R$

holds$\langle Q^{\#}\rangle_{P(A)\backslash \{\emptyset\}}\not\in im\lambda.$$Q_{1},$ $Q_{2}\subseteq R$$\langle Q_{1}^{\#}\rangle_{P(A)\backslash \{\emptyset\}}\neq\langle Q_{2}^{\#}\rangle_{P(A)\backslash \{\emptyset\}}.$$\square$

3.2

From the clone lattice on

Let us define a map$A^{n}$$\alpha$:$L_{A}arrow L_{H_{A}}$$f(x_{1}, \ldots, x_{n})=\{f’(x_{1}, \ldots, x_{n})\}\}$

Lemma 3.3 The map$\alpha$$A$

to the hyperclone lattice on$P(A)\backslash \{\emptyset\}$by $\alpha(C)=\bigcup_{n\geq 1}\{f\in H_{A}^{(n)}$ : $\exists f’\in C$. It is easy to show that$\alpha(C)$$\forall(x_{1}, \ldots, x_{n})\in$

is a hyperclone.

as full order embedding.

Proof. Obviously, is 1-1 map, and holds
Let $H$ be an arbitrary hyperclone with the property
immediately follows that there is clone such that $\alpha(C)=H.$
Without loss of generality, we will sometimes identify the hyperclone
to simplify the presentation.

$C_{1}\leq C_{2}\Leftrightarrow\alpha(C_{1})\leq\alpha(C_{2})$$\alpha$.

$\alpha(J_{A})\subseteq H\subseteq\alpha(O_{A})$. From the definition of$C$

Lemma.3.4 Letonminimal.
$A$$A$$\alpha$$\square$$\alpha(C)$and the clone$C$, in order

be a finite set with $|A|\geq 2$ . Every hyperclone generated by constant hyperoperation$w$

Corollary 3.1 Let$A$

be a finite set with $|A|\geq 2$ . There are least $2^{|A|}-1$ minimal clones in the lattice$L_{H_{A}}.$

Theorem 3.2 On any finite set
contains all hyperoperations.

$A$, with $|A|\geq 2$ , there are three minimal hyperclones such that their join

Proof. It is proved in [4] that there are two minimal clones such that their join contains all operations
on any finite set . Romov proved in [13] thatis maximal hyperclone. Hence, it is enough to choose
the third minimal hyperclone from the set of minimal hyperclones that are not minimal in the clone
lattice on . From previous lemma follows that such a set is not empty.
$A$$O_{A}$$A$$\square$

Theorem 3.3 The interval

Proof.

$[\alpha(\langle O_{A}^{(1)}\rangle), \alpha(O_{A})]$rs a chain.

It is the chain obtained by Burle in 1967 [11]. He bas shown that interval$[\langle O_{A}^{(1)}\rangle, O_{A}]$$(|A|+1)$ -element chainis$\langle O_{A}^{(1)}\rangle=U_{1}\subset L\subset U_{2}\subset\ldots\subset U_{k}=O_{A}.$is the set of all operations depending on at most one variable and operations taking at most values andis the set of operations depending of one variable and operations,
where : $\{0,1\}arrow A$ and$1\leq i_{1}<\ldots<i_{t}\leq n$ , are arbitrary maps and: $Aarrow\{0,1\},$is addition modulo 2.
$U_{i}$$j$$L$$f(x_{1}, \ldots, x_{n})=\alpha(\psi_{1}(x_{i_{1}})+\ldots\psi_{t}(x_{i_{t}}))$$\alpha$$\psi_{j}$$j\in\{i_{1}, \ldots, i_{t}\},$$+$$\square$Corollary 3.2 There are finite maximal chains in the hyperclone lattice.
Proof. It is known that there are finite maximal chains in the interval. With the maximal
chain from previous theorem and the clone(sinceis maximal in the hyperclone lattice), we get
the finite maximal chain in the hyperclone lattice.
$[J_{A}, \langle O_{A}^{(1)}\rangle]$$H_{A}$$O_{A}$$\square$84
4

Minimal hyperclones

A hyperclone is minimal if it is not trivial and its only subclone is trivial.
A hyperoperation of minimal arity in a minimal hyperclone, that is not a hyperprojection, is called
minimal hyperoperation.
A temary majority hyperoperation on $A,$ $ma\in H_{A}^{(3)}$ , is a ternary hyperoperation on defined by
ma $(x,x, y)=ma(x, y, x)=ma(y, x,x)=\{x\}$ for all $y\in A.$
A temary minority hyperoperation on $A,$ $mi\in H_{A}^{(3)}$ , is a temary hyperoperation on defined by$mi(x, x, y)=mi(x, y, x)=mi(y, x, x)=\{y\}$ for all $y\in A.$
For $n>2$ and $1\leq i\leq n$ , every -ary hyperoperation with $s(x_{1}, \ldots, x_{n})=\{x_{i}\},$ $|\{x_{1}, \ldots, x_{n}\}|<n$is called semi-hyperprojection.
It is easy to show that the theorem analogous to Rosenberg’s classification theorem ([16]) holds forminimal hyperoperations.
$A$

$x,$

$A$

$x,$

$n$

$s$

Theorem 4.1 Every minimal hyperopemtion us one of the followmg types:(1) a unary hyperoperation,(2) a binary idempotent hyperopemtion,$(S)$a temary majonty hyperoperation,(4) a temary minorety hyperopemtion,(5) an n-ary semi-hyperp rojection, $n>2.$

Lemma 4.1 Let
be a minimal clone on, thenis a minimal hyperclone on $A.$$\langle g\rangle_{P(A)\backslash \{\emptyset\}}$$g=f^{\#}$

Proof. Since. If there is a hyperoperation , such that$f$, and the mapping
is a minimal clone in, studied in [5] is an order embedding we prove the claim.

$\langle\{f^{\#}\}\rangle_{P(A)\backslash \{\emptyset\}}$$\alpha(C)=\langle C\#\rangle_{P(A)\backslash \{\emptyset\}}$

4.1$P(A)\backslash \{\emptyset\}$$\langle f\rangle$$L_{P(A)\backslash \{\emptyset\}}$$\alpha$:$L_{H_{A}}arrow L_{P(A)\backslash \{\emptyset\}\rangle}$

Minimal hyperclones on a two-element set

Let $A=\{0,1\}.$
Lemma 4.2 Let be a hyperoperation on A.
a minimal clone on$f$$\langle\{f\}\rangle\iota s$a minimal hyperdone on

$A$

iff

$\langle\{f^{\#}\}\rangle_{P(A)\backslash \{\emptyset\}}$is$P(A)\backslash \{\emptyset\}.$Lemma 4.3 There are 13 minimal hyperclones on $A.$

Lemma 4.4 There as no pair of minimal hyperclones whose join is the clone of all hyperoperations.
Lemma 4.5 There are 12 three-element sets of minimal hyperoperations whose union genemte the doneof all hyperopemtions on $A.$([4],[17]) and
Proof. Since we know all the pairs of minimal clones whose join is
in the hyperclone lattice, it follows that $\langle\{\neg, \min, f.\})=\langle\{\neg, \max, f_{t}\}\rangle=H_{A},$ $1\leq i\leq 6.$$O_{A}$$O_{A}$

is maximal$\square$Lemma 4.6 There are $S6$ four-element sets of minimal hyperopemtions whose union is the clone of all
hyperoperations on $A.$

4.2

Unary minimal hyperclones

Lemma 4.7 Let $f,$$g\in H_{A}^{(1)}$$(a)(f\circ g)^{\#}=f^{\#}\circ g^{\#}.$$(b)(\Delta f)^{\#}=\Delta f^{\#}.$

Proof.

. Then,85(a) Let $X$ be an arbitrary subset of$A$. Then $(fog)^{\neq}(X)=\cup\{(fog)(x) :x\in X\}=\cup\{\cup\{f(y)$$g(x)\}:x\in X\}=\cup\{f(y):y\in\cup\{g(x)\in X\}\}=\cup\{f(y):y\in g^{\#\#\#}(X)\}=f(g(X))$ .

(b) It follows immediately from the definition ofCorollary 4.1 The mapping
Corollary 4.2 Let$f\in H_{A}^{(1)}$$f\mapsto f^{\#}$. Then,$\Delta$,when $f$ andis isomorp hism from$f^{\#}$

are unary, that$(H_{A}^{(1)};\triangle)$onto: $y\in$$(\triangle f)^{\#}=f^{\#}=\Delta f^{\#}.$$(\lambda(H_{A}^{(1)})_{)}\Delta)$.

$\langle\{f^{\#}\}\rangle_{P(A)\backslash \{\emptyset\}}=\langle\{f\}\rangle^{\#}.$

Corollary 4.3 The restriction of the mapping
is a full order embedding.

$\lambda$:$L_{H_{A}}arrow L_{P(A)\backslash \{\emptyset\}},$$C\mapsto\langle C\#\rangle_{P(A)\backslash \{\emptyset\}}$to the interval$[J_{A}, \langle H_{A}^{(1)}\rangle]$

Corollary 4.4 Let

Lemma 4.8 Let$f\in H_{A}^{(1)}$$f\in H_{A}^{(1)}$. Then,$\langle\{f\}\rangle$

is a minimal hyperclone iff$\langle\{f^{\#}\}\rangle_{P(A)\backslash \{\emptyset\}}lS$

a minimal clone.

. Then, it holds$(a)f^{2}=f\Leftrightarrow(f^{\#})^{2}=f^{\#}.$$(b)f^{p}=e_{1}^{1}\Leftrightarrow(f^{\#})^{p}=\pi_{1}^{1}$, for some pnme

$p.$Proof.
(a)If $f^{2}=f$ then $(f^{\#})^{2}=(f^{2})\#=f^{\#}.$
If $f^{\#}(f^{\#}(X))=f^{\#}(X)$ holds for every $X\subseteq A,$
then it also holds for $|X|=1$ . It means that for every $x\in A,$ $f^{\#}(f^{\#}(\{x\}))=f^{\neq}(\{x\})$ , i.e.
$f(f(x))=f(x)$ .

(b)

Ifthen $(f^{\#})^{p}=(f^{p})\#=(e_{1}^{1})^{\#}=\pi_{1}^{1}.$For every $X\subseteq A(f^{\#})^{p}(X)=X$ implies$(f^{\#})^{p}(\{x\})=\{x\}$
$x\in
A$
every
that forholds, i.e. for every $x\in A$ holds $f^{p}(x)=x.$$(arrow)$$(arrow)$$(arrow)$$f^{p}=e_{1}^{1}$$(arrow)$$\square$

Theorem 4.2 Let$f\in H_{A}^{(1)}$. Then,$\langle\{f\}\rangle$

us minimal iff $f^{2}=f$ or $f^{p}=id_{A}$ , for some prime$p.$

Proof. Let $f^{2}=f$ or $f^{p}=id_{A}$ , for some prime . From Lemma4.8 $(f^{2}=f iff (f^{\#})^{2}=f^{\#})$
and $(f^{p}=e_{1}^{1}, for some$ prime $p iff (f^{\#})^{p}=\pi_{1}^{1}$ , for some prime ). It is known ([16], [3],[18]) that$((f^{\#})^{2}=f^{\#})$ or, for some prime p) iff
is minimal clone in. FromLemma4.4,
is minimal clone iniff
is minimal hyperclone in
$p$

$p$

$(f^{\#})^{p}=\pi_{1}^{1})$$\langle\{f^{\#}\}\rangle_{P(A)\backslash \{\emptyset\}}$$\langle\{f^{\#}\}\rangle_{P(A)\backslash \emptyset}$$L_{O_{P(A)\backslash \{\emptyset\}}}$$\langle\{f\}\rangle$$L_{O_{P(A)\backslash \{\emptyset\}}}$$L_{H_{A}}.$$\square$Example 1 There are 6 unary minimal hyperclones on $A=\{0,1\}.$

Example 2 There are 64 unary minimal hyperclones on the set $A=\{0,1,2\}.$

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