CLAUSAL RELATIONS AND C-CLONES
∗Edith Vargas
Institut f¨ ur Algebra, TU Dresden, Germany
e-mail: Edith Mireya.Vargas Garcia@mailbox.tu-dresden.de
Abstract
We introduce a special set of relations called clausal relations. We study a Galois connection Pol − C Inv between the set of all finitary operations on a finite set D and the set of clausal relations, which is a restricted version of the Galois connection Pol − Inv. We define C - clones as the Galois closed sets of operations with respect to Pol − C Inv and describe the lattice of all C -clones for the Boolean case D = {0, 1}.
Finally we prove certain results about C -clones over a larger set.
Keywords: clone, Galois connection, clausal relation, C -clone.
2000 Mathematics Subject Classification: Primary 08A99;
Secondary 08A02.
Introduction
In this paper we introduce a special set C R
Dof relations on a finite set D, called clausal relations (see Definition 1.4). The definition of clausal relations is based on the notion of a clausal constraint as a disjunction of inequalities of the form x ≥ d and x ≤ d, where d ∈ D = {0, 1, . . . , n−1} and x belongs to a set X of variables. The latter were studied by N. Creignou, M. Hermann, A. Krokhin and G. Salzer (see [1]).
A clone on a set D is a set of finitary operations on D that is closed under composition and contains all projections. It is well known (see [3]) that the Galois closed classes of operations on a finite set D of the Galois connection Pol − Inv are exactly all clones on D. In other words every clone F on D can be described as F = Pol Q for some set Q of relations.
∗
Supported by the DAAD-CONACyT grant no. A/06/13410.
In [1] N. Creignou, M. Hermann and collaborators classified the complexity of clausal constraints. In this paper we will not deal with such complexity problems. We are rather interested in describing clones which are determined by sets of clausal relations, i.e. describing C -clones (see Definition 1.14). The restriction to clausal relations implies a restric- tion of the Galois connection Pol − Inv to a Galois connection Pol − C Inv where C Inv F = Inv F ∩ C R
Dfor F ⊆ O
D(see Definition 1.13). This leads to a much smaller set of clones, a fact motivating us to investigate how many C -clones exist on D and to describe them. In particular, this is a contribution to the structure of the lattice of all clones.
The aim of this paper is to give a complete description of Boolean C -clones, i.e. when D = {0, 1}, and to prove that contrary to the Boolean case, we have infinitely many C -clones for 3 ≤ |D| < ∞.
The paper is organized as follows: In Section 1 we provide definitions related to relations, clausal relations, C -clones and the Galois connection Pol − C Inv. Furthermore, we present some properties of clausal relations.
In Section 2 we describe all Boolean C -clones, obtaining an only 5-element sublattice of the lattice described by E. Post (see [2]). Finally, in Section 3 we investigate how many C -clones exist for an arbitrary finite set D. We show that for |D| ≥ 3 there exist infinitely many C -clones by constructing an infinite descending chain of such clones.
Throughout the paper, N = {0, 1, 2, . . .} denotes the set of natural num- bers, and N
+= {1, 2, . . .} denotes the set of positive natural numbers. Fur- thermore, the domain for our clones is the set D = {0, 1, . . . , n − 1} for a fixed natural number n ≥ 2.
1. Clausal relations
In this section we provide definitions and some properties of clausal relations.
Definition 1.1. Let k, m ∈ N
+. An m-ary relation ̺ on D is a subset of the m-fold Cartesian product D
m. It is often convenient to represent
̺ = {r
1, . . . , r
k} as a matrix (r
ij)
1≤i≤m 1≤j≤k∈ D
m×k, whose columns are the tuples in the relation, i.e. r
j= r
1j, r
2j, . . . , r
mjfor all j ∈ {1, . . . , k}.
We define R
(m)D:= {̺ | ̺ ⊆ D
m} as the set of all m-ary relations defined on D and
R
D:=
[
∞ m=1R
(m)Das the set of all finitary relations on D.
Definition 1.2. Let m ∈ N
+, m := {1, . . . , m}, ε be a partition of m and
∼
εbe the corresponding equivalence relation on m. We define d
ε∈ R
(m)Dto be the relation
d
ε:= {(x
1, . . . x
m) ∈ D
m| i ∼
εj ⇒ x
i= x
j}
and call it a trivial or diagonal relation. The set of all diagonal relations together with the empty relation ∅ is denoted by diag(D).
Definition 1.3. Let p, q ∈ N
+. For given parameters a = (a
1, . . . , a
p) ∈ D
pand b = (b
1, . . . , b
q) ∈ D
q, the clausal relation R
abof type (p, q), is the set of all tuples (x
1, . . . , x
p, y
1, . . . , y
q) ∈ D
p+qsatisfying
(1) (x
1≥ a
1) ∨ . . . ∨ (x
p≥ a
p) ∨ (y
1≤ b
1) ∨ . . . ∨ (y
q≤ b
q).
We observe that if a
i= 0 for some i ∈ {1, . . . , p} or b
j= n − 1 for some j ∈ {1, . . . , q}, then the relation R
abis total, i.e. R
ab= D
p+qbecause (1) is always satisfied.
Definition 1.4. Let p, q ∈ N
+. We use
R
pq:= {R
ab| a ∈ D
p, b ∈ D
q}
to denote the set of all clausal relations of arity
∗p + q and C R
D:= [
(p,q)∈N2+
R
pqfor the set of all finitary clausal relations on D.
∗
If we speak of a clausual relation of arity p + q, we implicitely mean also that the
clausal relation is of type (p, q).
We will write R
abfor R
(a)bin the case p = 1 and likewise R
abfor R
a(b)in the case q = 1. We give two examples of clausal relations.
Examples 1.5.
a) Let D = {0, 1}, then
R
01= {(x
1, y
1) ∈ D
2| x
1≥ 0 ∨ y
1≤ 1} = 0 0 1 1 0 1 0 1
!
= D
2b) Let D = {0, 1, 2}, then
R
(2,2)0=
(x
1, x
2, y
1) ∈ D
3| x
1≥ 2 ∨ x
2≥ 2 ∨ y
1≤ 0
=
0 0 0 0 0 1 1 1 1 1 2 2 2 2 2 2 2 2 2 0 1 2 2 2 0 1 2 2 2 0 0 0 1 1 1 2 2 2 0 0 0 1 2 0 0 0 1 2 0 1 2 0 1 2 0 1 2
Clausal relations are closed with respect to union but not with respect to intersection as the next lemmata show.
Lemma 1.6. Let p, q ∈ N
+. If R
aband R
ab′′are clausal relations of arity p + q, where a = (a
1, . . . , a
p), a
′= (a
′1, . . . , a
′p) ∈ D
p, b = (b
1, . . . , b
q) and b
′= (b
′1, . . . , b
′q) ∈ D
q. Then it holds
R
ab∪ R
ab′′= R
cdwhere c = (min{a
1, a
′1}, . . . , min{a
p, a
′p}) and d = (max{b
1, b
′1}, . . . , max{b
q, b
′q}).
P roof. Let z = (x
1, . . . , x
p, y
1, . . . , y
q) ∈ R
ab∪ R
ab′′,
⇐⇒ x
1≥ a
1∨ . . . ∨ x
p≥ a
p∨ y
1≤ b
1∨ . . . ∨ y
q≤ b
q∨
x
1≥ a
′1∨ . . . ∨ x
p≥ a
′p∨ y
1≤ b
′1∨ . . . ∨ y
q≤ b
′q⇐⇒ _
1≤i≤p
(x
i≥ a
i∨ x
i≥ a
′i) ∨ _
1≤j≤q
(y
j≤ b
j∨ y
j≤ b
′j)
⇐⇒ _
1≤i≤p
(x
i≥ min{a
i, a
′i}) ∨ _
1≤j≤q
(y
j≤ max{b
j, b
′j}).
This is equivalent to z ∈ R
cd.
Similarly, following lemma can be proved.
Lemma 1.7. Let p, q ∈ N
+. If R
aband R
ab′′are clausal relations of arity p + q, where a = (a
1, . . . , a
p), a
′= (a
′1, . . . , a
′p) ∈ D
p, b = (b
1, . . . , b
q) and b
′= (b
′1, . . . , b
′q) ∈ D
q. Then it holds
R
ab∩ R
ab′′⊇ R
cdwhere c = (max{a
1, a
′1}, . . . , max{a
p, a
′p}) and d = (min{b
1, b
′1}, . . . , min{b
q, b
′q}).
In general equality does not hold in 1.7, as the following example shows.
Example 1.8. Let D = {0, 1}, consider a = (0, 1), a
′= (1, 0), b
1= 0 and b
′1= 1. Then, R
ab1= D
3= R
ab′′1
. Let z = (x
1, x
2, y
1) = (0, 0, 1). z ∈ R
ab1∩R
ab′′ 1, but z / ∈ R
(1,1)0.
The set C R
Dcan be separated in trivial and non-trivial clausal relations as follows.
Lemma 1.9. The set C R
Dcan be partitioned as C R
D= {D
(p+q)| p, q ∈ N
+} ˙∪C R
∗D, where
{D
(p+q)| p, q ∈ N
+} = C R
D∩ diag(D)
are the trivial clausal relations and
C R
∗D= {R
ab| a ∈ (D \ {0})
p, b ∈ (D \ {n − 1})
q; p, q ∈ N
+} are the non-trivial clausal relations.
P roof. Let p, q ∈ N
+, a = (a
1, . . . , a
p) ∈ D
p, b = (b
1, . . . , b
q) ∈ D
q. We have observed above that if one of the a
1, . . . , a
pequals 0, or one of the b
1, . . . , b
qequals n − 1, then R
abis a total relation, i.e. R
ab= D
p+q.
We have to show C R
∗D∩ diag(D) = ∅. Let us assume the existence of a relation ̺ ∈ C R
∗D∩ diag(D). Then there exist a ∈ (D \ {0})
pand b ∈ (D \ {n − 1})
qsuch that
̺ = R
aband there exists a partition ε of m = {1, . . . , m} (where m := p + q) such that ̺ = d
ε. Let ∼
εbe the corresponding equivalence relation. We show
∼
ε= {(x, x) | x ∈ m} := ∆
m, thus
̺ = d
ǫ= D
m.
This is a contradiction to R
ab= ̺ = D
m, because (0, . . . , 0, n − 1, . . . , n − 1)
∈ R /
ab.
Let i, j ∈ {1, . . . , p} with i 6= j. Then
(0, . . . , 0, 0, 0 . . . , 0,
in − 1, 0, . . . , 0
j| {z }
p
, 0, . . . , 0
| {z }
q
) ∈ R
ab= d
ε.
Thus i 6∼
εj. Let i, j ∈ {1, . . . , q} with i 6= j. Then
(n − 1, . . . , n − 1
| {z }
p
, 0, . . . , 0,
i+p0 , 0, . . . , 0, n − 1, 0, . . . , 0
j+p| {z }
q
) ∈ R
ab= d
ε.
Thus i + p 6∼
εj + p. Let i ∈ {1, . . . , p} and j ∈ {1, . . . , q}. Then
(n − 1, . . . , n − 1, n − 1, n − 1, . . . , n − 1
i| {z }
p
, 0, . . . , 0,
j+p0 , 0, . . . , 0
| {z }
q
) ∈ R
ab= d
ε.
Thus i 6∼
εj + p.
Hence, if i, j ∈ {1, . . . , p + q} with i ∼
εj then i = j. This shows
∼
ε= ∆
m.
An example of a non-trivial clausal relation can be found in 1.5 b).
Definition 1.10. Let k ∈ N
+. A k-ary operation on D is a function f : D
k−→ D. We denote O
(k)D:= {f | f : D
k−→ D} as the set of all k-ary operations on D and O
D:= S
∞k=1
O
(k)Das the set of all finitary operations on D. For each j ∈ {1, . . . , k} we denote e
kj(d
1, . . . , d
k) := d
jas the j-th projection of arity k and J
D:= {e
kj| k ∈ N
+, 1 ≤ j ≤ k} as the set of all projections on D.
An example of a k-ary operation on D is the k-ary constant operation (briefly, constant) c
ka: D
k−→ D given by c
ka(x
1, . . . , x
k) := a for all x
1, . . . , x
k∈ D, where a is an arbitrary element of D.
Definition 1.11. We say that a k-ary operation f ∈ O
(k)Dpreserves an m-ary relation ̺ ∈ R
(m)D, denoted by f ⊲ ̺, if whenever
r
1= (a
11, . . . , a
m1) ∈ ̺, . . . , r
k= (a
1k, . . . , a
mk) ∈ ̺ it follows that also f applied to these tuples belongs to ̺, i.e.
f [r
1, . . . , r
k] := (f (a
11, . . . , a
1k), . . . , f (a
m1, . . . , a
mk)) ∈ ̺.
Definition 1.12. Let F ⊆ O
Dbe a set of operations on D. Then we define Inv
DF as the set of all relations that are invariant for all f ∈ F :
Inv
DF := {̺ ∈ R
D| ∀f ∈ F : f ⊲ ̺}.
Similarly, for a set Q ⊆ R
Dof relations, Pol
DQ is the set of all operations that preserve every relation ̺ ∈ Q:
Pol
DQ := {f ∈ F | ∀̺ ∈ Q : f ⊲ ̺}.
Furthermore, for k ∈ N
+we abbreviate
Pol
(k)DQ := O
(k)D∩ Pol
DQ.
If D is known from the context we write Pol instead of Pol
D, and Inv instead of Inv
D. The operators Pol and Inv define the Galois connection Pol − Inv, which is induced by the relation ⊲.
It is well known that for F ⊆ O
Dthe Galois closed set Pol Inv F of operations on D is precisely the clone hF i
OD
generated by F , i.e. Pol Inv F = hF i
OD
. Let L
Dbe the set of clones on D, we denote L
D:= (L
D, ⊆) as the lattice of all clones on D, with greatest element O
Dand least element J
D.
Note that by virtue of the Galois connection Pol − Inv the study of the Galois closed sets of operations is equivalent to the study of the Galois closed sets of relations, Inv Pol Q =
Q
RD
for Q ⊆ R
D.
Next we present a restriction of the Galois connection Pol − Inv where the relations are confined to be clausal relations. This restriction gives us a much smaller number of Galois closed sets of operations, so called C -clones.
Definition 1.13. For F ⊆ O
Dwe define C Inv F := Inv F ∩ C R
D. The operators
C Inv : P(O
D) −→ P(C R
D) : F 7→ C Inv F and
Pol : P(C R
D) −→ P(O
D) : Q 7→ Pol Q
define a Galois connection Pol − C Inv between operations and clausal rela- tions.
We will call the Galois closed sets of operations of this Galois connection C -clones, more formally:
Definition 1.14. A set F ⊆ O
Dof operations is called a C -clone if F =
Pol Q for some set Q ⊆ C R
Dof clausal relations, and a set Q ⊆ C R
Dis
called relational C -clone if Q = C Inv F for a set F of operations.
Every Galois connection naturally gives rise to a pair of closure operators.
For one of them we introduce a special notation.
Definition 1.15. For any F ⊆ O
Dwe define hF i
C:= Pol C Inv F .
We finish this section with a lemma clarifying the relationship of this closure operator and the clone generation, i.e. the corresponding closure operator of the Galois connection Pol − Inv.
Lemma 1.16. For any F ⊆ O
Dit holds:
(a) hF i
OD
⊆ hF i
C. (b) hhF i
Ci
OD
= hF i
C, in particular every C -clone is a clone.
P roof. The first statement follows from C Inv F ⊆ Inv F for any F ⊆ O
D, hence hF i
OD
= Pol Inv F ⊆ Pol C Inv F = hF i
C. For the second statement observe that Pol Inv hF i
C= hF i
C.
Let C L
D= {Pol Q | Q ⊆ C R
D} be the set of all C -clones on D, we denote CL
D:= (C L
D, ⊆) as the lattice of all C -clones on D, with greatest element O
Dand least element Pol C R
D. The main goal is to describe the lattice of all C -clones on D, the first step towards this goal is to describe the lattice of all C -clones for D = {0, 1}.
2. Boolean C -clones
In this section we will describe all Boolean C -clones, i.e. when D is the set {0, 1} that we also denote by 2 for short.
From Lemma 1.9, the set C R
2can be written as in the following corol- lary.
Corollary 2.1.
C R
2= {2
p+q| p, q ∈ N
+} ˙∪C R
∗2, where
C R
∗2= {R
10,p,q| p, q ∈ N
+}
and
R
10,p,q:= R
(p times
z }| { 1, . . . , 1
) (0, . . . , 0
| {z }
q times
)
.
Observe that
R
10,p,q= {(x, y) ∈ 2
p× 2
q| ∃i ∈ {1, . . . , p} : x
i= 1 ∨ ∃j ∈ {1, . . . , q} : y
j= 0}
= 2
p+q\ {(0, . . . , 0, 1, . . . , 1)}.
The following lemma shows that every Boolean C -clone can be determined by sets of non-trivial Boolean clausal relations.
Lemma 2.2. For Q ⊆ C R
2, it holds Pol(Q) = Pol(Q ∩ C R
∗2).
We shall describe {Pol Q | Q ⊆ C R
2}. Since Pol − C Inv is a Galois connec- tion, this set is dually isomorphic to {C Inv F | F ⊆ O
2}, and furthermore, we have
C Inv F = \
f ∈F
C Inv f
for F ⊆ O
2. Consequently, it suffices to regard the closed relational sets C Inv f for f ∈ O
2. Since there is a one to one correspondence between C Inv f and hf i
Cvia the operators Pol and C Inv, we will first consider one- generated C -clones. By Lemma 2.2 and Definition 1.15,
hf i
C= Pol C Inv f = Pol(C Inv f ∩ C R
∗2)
i.e. hf i
Cis the set of all the functions that preserve all the non-trivial invariant clausal relations of f .
For the rest of this section we are going to characterize, one-generated C -clones for some special functions f ∈ O
2, (namely f ∈ {¬, h, ∨, ∧, g}).
We also use the notation from Figure 1 without further explanation.
Lemma 2.3. Let ¬ : 2 −→ 2 be the negation operation, i.e. ¬(0) = 1 and
¬(1) = 0. Then it holds
h¬i
C= O
2.
P roof. By Definition 1.15 and Lemma 2.2, we have
h¬i
C= {f ∈ O
2| ∀p, q ∈ N
+: ¬ ⊲ R
10,p,q=⇒ f ⊲ R
10,p,q}.
(2)
Because of (2) it is enough to show ¬ ⋫ R
10,p,qfor all p, q ∈ N
+. In- deed, the tuple r = (1, . . . , 1, 0, . . . , 0) ∈ R
10,p,q= 2
p+q\ {(0, . . . , 0, 1, . . . , 1)}
for all p, q ∈ N
+but
¬[r] = (¬(1), . . . , ¬(1), ¬(0), . . . , ¬(0)) = (0, . . . , 0, 1, . . . , 1) / ∈ R
10,p,q.
Note that for p = q = 1 it holds
R
10,p,q= R
10= {(0, 0), (1, 0), (1, 1)} =≧, and hence
Pol R
10= Pol(≧) = Pol(≦) = M, where M is the clone of all monotone Boolean functions.
Lemma 2.4. Let h ∈ O
(3)2be the ternary majority function on 2 (median), i.e. h(x, y, z) := (x ∧ y) ∨ (x ∧ z) ∨ (y ∧ z) for x, y, z ∈ 2. Then it holds
hhi
C= Pol R
10.
P roof. We show that h ⋫ R
10,p,q, unless p = 1, q = 1. We consider several cases:
• p ≥ 2, q ≥ 1 : The schema
h(1 0 0) = 0 h(0 1 0) = 0 h(0 0 0) = 0
.. .
h(0 0 0) = 0 (row p) h(1 1 0) = 1
h(1 1 1) = 1 .. .
h(1 1 1) = 1 (row p + q)
shows that h ⋫ R
10,p,q, because the tuples (columns of the arguments of h) all belong to R
10,p,q, but after applying h to the tuples, one obtains a tuple (column) that does not belong to R
10,p,q.
• p ≥ 1, q ≥ 2 : Likewise, the schema h(1 0 0) = 0 h(0 0 0) = 0
.. .
h(0 0 0) = 0 (row p) h(1 0 1) = 1
h(1 1 0) = 1 h(1 1 1) = 1
.. .
h(1 1 1) = 1 (row p + q) shows that h ⋫ R
10,p,q.
If p = q = 1, then h ⊲ R
10, because h is a monotone operation and M = Pol(R
10).
Lemma 2.5. It holds
h∧i
C= Pol{R
10,p,q| p = 1, q ∈ N
+} and h∨i
C= Pol{R
10,p,q| q = 1, p ∈ N
+}.
P roof. At first we have a look at ∧. We show that ∧ ⋫ R
10,p,q, unless p = 1 and q ∈ N
+. We consider two cases:
• p ≥ 2 : The schema
1 ∧ 0 = 0 0 ∧ 1 = 0 0 ∧ 0 = 0
.. .
0 ∧ 0 = 0 (row p)
1 ∧ 1 = 1 .. .
1 ∧ 1 = 1 (row p + q)
shows that ∧ ⋫ R
10,p,q.
• p = 1, q ∈ N
+: We show
∧ ⊲ R
10,p,q. We assume the existence of tuples
(x
1, y
1, . . . , y
q), (x
2, z
1, . . . , z
q) ∈ R
10,p,qsuch that
x
1∧ x
2= 0 and for all j ∈ {1, . . . , q}
y
j∧ z
j= 1.
Because of x
1∧ x
2= 0 w.l.o.g. x
1= 0. Then our assumption (x
1, y
1, . . . , y
q) ∈ R
10,p,qimplies that there is one j ∈ {1, . . . , q} such that y
j= 0. Thus, y
j∧ z
j= 0, a contradiction.
Similarly, the result for ∨ can be proved.
In 2.1 we saw that clausal relations are either total or total without one tuple, and none of the diagonals except for total relations are clausal relations.
Furthermore, C Inv O
2= Inv O
2∩ C R
2= diag(2) ∩ C R
2, hence we obtain the following lemma.
Lemma 2.6. It holds
C Inv O
2= {2
(p+q)| p, q ∈ N
+}.
Let c
0, c
1be the unary constant operations on 2. For any p, q ∈ N
+we have c
0⊲ R
10,p,qand c
1⊲ R
10,p,qbecause c
0(y
1) = 0 and c
1(x
1) = 1 for any (x
1, . . . , x
p, y
1, . . . , y
q) ∈ R
10,p,q.
In the rest of the section we will freely use ∨ to denote supremum of two clones in Post’s Lattice (see Figure 1). Nevertheless, we hope not to confuse the reader and be clear.
Lemma 2.7. The least C -clone is
h∅i
C= Pol(C R
2) = {f ∈ O
2| ∀p, q ∈ N
+: f ⊲ R
10,p,q}.
It holds
h∅i
C⊇ hc
0i
O2∨ hc
1i
O2 F ig.1= hc
0, c
1i
O2, and
h∅i
C$ h∧i
C$ M, h∅i
C$ h∨i
C$ M.
Furthermore, h∧i
Cand h∨i
Care incomparable C -clones.
P roof. From the previous observation we obtain c
0, c
1∈ h∅i
C. Because h∅i
Cis a Boolean clone, we have
hc
0i
O2, hc
1i
O2⊆ hh∅i
Ci
O2
1.16
= h∅i
C.
Thus, h∅i
Cis an upper bound for hc
0i
O2and hc
1i
O2. Hence, hc
0i
O2∨ hc
1i
O2⊆ h∅i
C.
Because h∅i
C= Pol(C R
2), we have that neither ∧ ∈ h∅i
Cnor ∨ ∈ h∅i
C, because ∧ ⋫ R
10,p,qfor p ≥ 2 and ∨ ⋫ R
10,p,qfor q ≥ 2. Thus,
∧ ∈ h∧i
C\ h∅i
C,
∨ ∈ h∨i
C\ h∅i
C.
Lemma 2.4 implies M = Pol(R
10). Thence, (cf. 2.5) h∧i
C, h∨i
C⊆ M
2.4= hhi
C. This inclusion is proper since
h ∈ (M \ h∧i
C) ∩ (M \ h∨i
C).
This holds because h is a monotone operation and h ⋫ R
10,p,qfor p = 1, q > 1 and for q = 1, p > 1.
Because of ∧ ⋫ R
10,p,qfor p ≥ 2 and q = 1, we have ∧ ∈ h∧i
C\ h∨i
C, and because of ∨ ⋫ R
10,p,qfor p = 1 and q ≥ 2 we have ∨ ∈ h∨i
C\ h∧i
C. Consequently, the two C -clones are incomparable.
Lemma 2.8. For any subset F ⊆ O
2it holds hc
0, c
1i
O2∨ hF i
O2⊆ hF i
C.
P roof. From 2.7 we infer hc
0, c
1i
O2
⊆ h∅i
C⊆ hF i
C. Because of hF i
C∈ L
2and F ⊆ hF i
Cwe have
hF i
O2⊆ hhF i
Ci
O2
1.16
= hF i
C. Consequently, we obtain hc
0, c
1i
O2
∨ hF i
O2
⊆ hF i
C. Lemma 2.9. It holds
hc
0, c
1, ∧i
O2
= h∧i
Cand hc
0, c
1, ∨i
O2
= h∨i
C.
P roof. From 2.8 and Figure 1 we obtain hc
0, c
1, ∧i
O2 F ig.1= hc
0, c
1i
O2∨ h∧i
O2
⊆ h∧i
C. Let us assume
hc
0, c
1, ∧i
O2$ h∧i
C.
Then, h∧i
Chas to be a clone in L
2being above hc
0, c
1, ∧i
O2
. Because the upper cover of hc
0, c
1, ∧i
O2
in L
2is M (see Post’s Lattice, Figure 1), it follows
M ⊆ h∧i
C,
which is a contradiction to M ⊃ h∧i
C(cf. Lemma 2.7). Similarly, the claim for h∨i
Ccan be proved.
s ¬ s J
2u ⊥ c
1s c
0s
s s g
s L s s
s
s
∨ h∨i
Cu
s s
s
∧ uh∧i
Cs s
m
′s s q
′s
T
1,∞s s
s s
s s
s T
1,3s
s T
1,2s O u
2s T
1s
T
0s u M
s s
h s S s
s p
p p p p
p p
p p
p p p
s
q m
s s s T
0,∞s
s T
0,3s T
0,2s s
s s s
p p p
p p p p p p p p p
Figure 1. Post’s Lattice
Lemma 2.10. For any two subsets F, G ⊆ O
2the following implication holds
F ⊆ G ⊆ hF i
C=⇒ hGi
C= hF i
C. P roof.
F ⊆ G =⇒ hF i
C⊆ hGi
CG ⊆ hF i
C=⇒ hGi
C⊆ hhF i
Ci
C= hF i
C.
Lemma 2.11. Let c
0, c
1be the unary constant operations on 2. Then it holds
h∅i
C= hc
0, c
1i
O2.
P roof. h∅i
C⊆ h∨i
C∩ h∧i
C 2.9= hc
0, c
1, ∧i
O2
∩ hc
0, c
1, ∨i
O2
F ig.1
= hc
0, c
1i
O2
2.7
⊆ h∅i
C.
Lemma 2.12. Let g be the ternary minority operation, i.e.
g(x, x, y) = g(x, y, x) = g(y, x, x) = y.
Then it holds:
hgi
C= hLi
C= O
2.
P roof. Because of ¬ ∈ L, it follows that O
22.3
= h¬i
C⊆ hLi
C⊆ O
2, hence hLi
C= O
2. Applying Lemma 2.8 to {g} leads to
L
F ig.1= hc
0, c
1i
O2
∨ hgi
O2
⊆ hgi
C. Together with 2.10 we infer
hgi
C= hLi
C.
Remark 2.13. Let F ≤ O
2denote a clone in Post’s Lattice. Then we have CL
2= {Pol(C Inv(F )) | F ⊆ O
2} = {Pol(C Inv(F )) | F ≤ O
2}.
P roof. It is obvious that
{Pol(C Inv(F )) | F ≤ O
2} ⊆ {Pol(C Inv(F )) | F ⊆ O
2}.
To show the other inclusion we regard F ⊆ O
2. Let G := Pol Inv(F ) ≤ O
2. Pol C Inv(G) = Pol((Inv(G)) ∩ C R
D) = Pol((Inv(Pol Inv(F ))) ∩ C R
D)
= Pol(Inv(F ) ∩ C R
D) = Pol C Inv(F ).
In the following we prove that there are no more Boolean C -clones than the ones already described in the previous Lemmata 2.4, 2.9, 2.11, 2.12.
Theorem 2.14. The lattice of all Boolean C -clones is CL
2= {⊥, h∧i
C, h∨i
C, hhi
C, O
2},
where
⊥ := hc
0, c
1i
O2h∧i
C= hc
0, c
1, ∧i
O2h∨i
C= hc
0, c
1, ∨i
O2hhi
C= M.
P roof. The next six equalities are consequences of Lemma 2.10 and the previous lemmata.
n hCi
C| C ∈
J
2, hc
0, c
1i
O2
L2
o = hc
0, c
1i
O2
n hCi
C| C ∈
h∧i
O2, h∧, c
0, c
1i
O2L2
o =
h∧, c
0, c
1i
O2n hCi
C| C ∈
h∨i
O2
, h∨, c
0, c
1i
O2
L2
o
=
h∨, c
0, c
1i
O2
n hCi
C| C ∈
h¬i
O2, O
2L2
o = {O
2}
n hCi
C| C ∈
hgi
O2, O
2L2
o
= {O
2} n hCi
C| C ∈ hhi
O2, M
L2
o = {M } .
The next four equalities will be shown below.
n hCi
C| C ∈ h∧i
O2
, M
L2
\ h∧i
O2
, h∧, c
0, c
1i
O2
L2
o
= {M } n hCi
C| C ∈
h∧i
O2, O
2L2
\
h∧i
O2, h∧, c
0, c
1i
O2L2
, C 6⊆ M o
= {O
2} n hCi
C| C ∈ h∨i
O2, M
L2
\ h∨i
O2, h∨, c
0, c
1i
O2L2
o
= {M } n hCi
C| C ∈ h∨i
O2, O
2L2
\ h∨i
O2, h∨, c
0, c
1i
O2L2
, C 6⊆ M o
= {O
2} .
Regarding a clone C ≤ O
2with h∧i
O2⊆ C, but C / ∈ h∧i
O2, h∧, c
0, c
1i
O2L2
yields (using Lemma 2.8)
C
1:= hc
0, c
1i
O2∨ C ⊆ hCi
C.
If C ⊆ M then we have that M = C
1⊆ hCi
C, and because of monotonicity of h·i
Cyields hCi
C⊆ hM i
C= M . Thus,
hCi
C= M.
Otherwise, (i. e. C 6⊆ M ) leads to
O
2= C
1⊆ hCi
C⊆ O
2.
The proof for ∨ instead of ∧ is similar. Knowing all those 10 equalities and applying Remark 2.13, one obtains
CL
2= {⊥, h∧i
C, h∨i
C, M, O
2}, the clones of which are described in the previous lemmata.
The previous theorem does not only describes the set of all the Boolean
C -clones but also the operations that these contain. For example, h∨i
Ccontains the operations c
0, c
1, ∨, all the projections and compositions of
these fuctions.
As we mentioned, when we describe C -clones at the same time we are describing relational C -clones. The lattices of C -clones and of relational C -clones are dually isomorphic as is shown in the next figure.
O
2M
h∧i
CD D D D D D D D
z z z z z z z z
h∨i
CDD DD
DD DD
⊥ z z z z z z z z
C Inv ⊥
C Inv h∧i
CN N N N N N N N N N
p p p p p p p p p p p
C Inv h∨i
CNN NN NN NN NN N
C Inv M
p p p p p p p p p p
C Inv O
2For the remainder of this section, we restrict ourselves to the following:
Given a C -clone Pol(Q) with Q ⊆ C R
∗2, find a minimal subset Q
1⊆ Q, such that Pol(Q) = Pol(Q
1). The motivation for the restriction is to establish that all Boolean C -clones can be described by a finite number of clausal relations.
Lemma 2.15. The following equalities hold O
2= Pol (∅) ,
M = Pol R
10,
h∧i
C= Pol{R
10,p,q| p = 1, q ∈ N
+} = Pol R
1(0,0),
h∨i
C= Pol{R
10,p,q| q = 1, p ∈ N
+} = Pol R
(1,1)0,
hc
0, c
1i
O2= Pol{R
10,p,q| p, q ∈ N
+} = Pol R
(1,1)(0,0).
P roof. The characterization of O
2is trivial. The second statement follows
from Lemma 2.4. The arguments for the rest of the equalities are very
similar, so w.l.o.g. we will only deal with the characterization of h∧i
C.
“⊆”: We have
{R
1(0,0)} ⊆ {R
10,p,q| p = 1, q ∈ N
+} hence
W := Pol{R
1(0,0)} ⊇ Pol{R
10,p,q| p = 1, q ∈ N
+} = h∧i
C.
“⊇”: From the proof of Lemma 2.4 we know M = hhi
C* W , applying the above-established facts that hc
0, c
1, ∧i
O2 2.9= h∧i
C⊆ W and C -clones are clones, one can read off of Post’s Lattice that W = hc
0, c
1, ∧i
O2
= h∧i
C.
3. C -clones
In the previous section we showed that there are five different Boolean C -clones. Next we show that for |D| ≥ 3 there are infinitely many C -clones by exhibiting an infinite descending chain of such clones.
Let D ⊇ {0, 1, 2} and m ∈ N
+. Consider the following clausal relation:
R
(1,...,1)(1,...,1)=
n (x
1, . . . , x
m, y
1, . . . , y
m) ∈ D
2m| x
1≥ 1∨. . .∨x
m≥ 1 ∨ y
1≤ 1∨. . .∨y
m≤ 1 o . We define
̺
m:= R
(1,...,1)(1,...,1). Observe that the tuple (0, . . . , 0, 2, . . . , 2) / ∈ ̺
m.
Proposition 3.1. If ̺
mis the 2m-ary relation defined above, then Pol(̺
m−1) % Pol(̺
m)
holds for any m ∈ N
+.
P roof. Let n ∈ N
+and f ∈ Pol
(n)(̺
m). We have to show f ∈ Pol
(n)(̺
m−1).
Let r
1, . . . , r
n∈ ̺
m−1, where r
k=: (x
1k, . . . , x
m−1k, y
1k, . . . , y
m−1k) for k, that belongs to {1, . . . , n}. Then
x
1k≥ 1 ∨ . . . ∨ x
m−1k≥ 1 ∨ y
1k≤ 1 ∨ . . . ∨ y
m−1k≤ 1
⇔ x
1k≥ 1∨. . .∨x
m−1k≥ 1∨x
m−1k≥ 1∨y
1k≤ 1∨. . .∨y
m−1k≤ 1∨y
m−1k≤ 1.
We define for k ∈ {1, . . . , n} a tuple
r
k′= (x
1k, . . . , x
m−1k, x
mk, y
1k, . . . , y
m−1k, y
mk) := (x
1k, . . . , x
m−1k, x
m−1k, y
1k, . . . , y
m−1k, y
m−1k).
The new tuples r
′1, . . . , r
′nbelong to ̺
mbecause of the above expression.
Because f ⊲ ̺
mwe have
(c
1, . . . , c
m−1, c
m, d
1, . . . , d
m−1, d
m) := f [r
1′, . . . , r
′n] ∈ ̺
m. Then, by construction, we have c
m−1= c
m, d
m−1= d
mand
c
m−1≥ 1 ∨ c
m≥ 1 ⇐⇒ c
m−1≥ 1 d
m−1≤ 1 ∨ d
m≤ 1 ⇐⇒ d
m−1≤ 1.
Therefore, (c
1, . . . , c
m−1, d
1, . . . , d
m−1) = f [r
1, . . . , r
n] ∈ ̺
m−1, i.e. f ⊲̺
m−1. Now we show that the inclusion is proper. Let f be an 2m-ary operation such that
f (x
1, . . . , x
m, . . . , x
2m)=
0 if there is only one 1 among the first m entries and 0 in the other entries.
2 if there is only one 1 between the m+1, . . . , 2m entries and 2 in the other entries.
1 otherwise.
That is
f (1, 0, . . . ,
m0, 0, . . . ,
2m0 ) =, . . . , = f (0, . . . , 0,
m1, 0, . . . , 0) = 0
f (2, 2, . . . ,
m+11 , 2, . . . , 2) =, . . . , = f (2, . . . , 2,
m+12 , . . . , 2, 1) = 2.
We show f / ∈ Pol
(2m)(̺
m).
Consider the tuples r
1, . . . r
2m∈ ̺
m, such that
r
1= (1, 0, . . . ,
m0, 2, . . . ,
2m2 ), . . . , r
2m= (0, . . . , 0, 2, . . . , 2,
2m1 ).
f [r
1, . . . , r
2m] / ∈ ̺
mbecause
1 2 . . . m m+1 m+2 . . . 2m
1 f ( 1 0 . . . 0 0 0 . . . 0 ) =0
2 f ( 0 1 . . . 0 0 0 . . . 0 ) =0
. .. .. . .. .
m f ( 0 0 . . . 1 0 0 . . . 0 ) =0
m+1 f ( 2 2 . . . 2 1 2 . . . 2 ) =2
m+2 f ( 2 2 . . . 2 2 1 . . . 2 ) =2
.. . . .. .. .
2m f ( 2 2 . . . 2 2 2 . . . 1 ) =2
∈ ̺ /
mHowever, f ∈ Pol
(2m)(̺
m−1) as the following argument shows. Let r
1′, . . . , r
2m′∈ ̺
m−1, where
r
k′= (x
1k, . . . , x
m−1k, y
1k, . . . , y
m−1k)
for k ∈ {1, . . . , 2m}. Note that, by definition, every tuple r
k′satisfies the following expression:
x
1k≥ 1 ∨ . . . ∨ x
m−1k≥ 1 ∨ y
1k≤ 1 ∨ . . . ∨ y
m−1k≤ 1.
Let us regard the tuples r
k′∈ ̺
m−1as the columns of a matrix A ∈ D
2(m−1)×(2m). We construct a matrix
(b
ij)
1≤i≤2(m−1)1≤j≤2m