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Fuzzy Sets and Systems 236 (2014) 33–49

www.elsevier.com/locate/fss

Type-2 operations on finite chains

Carol Walker ∗, Elbert Walker

Department of Mathematical Sciences, New Mexico State University, Las Cruces, NM 88003, United States

Received 11 October 2011; received in revised form 14 April 2013; accepted 10 June 2013

Available online 14 June 2013

Abstract

The algebra of truth values for fuzzy sets of type-2 consists of all mappings from the unit interval into itself, with operationscertain convolutions of these mappings with respect to pointwise max and min. This algebra generalizes the truth-value algebrasof both type-1 and of interval-valued fuzzy sets, and has been studied rather extensively both from a theoretical and applied pointof view. This paper addresses the situation when the unit interval is replaced by two finite chains. Most of the basic theory goesthrough, but there are several special circumstances of interest. These algebras are of interest on two counts, both as special casesof bases for fuzzy theories, and as mathematical entities per se.© 2013 Elsevier B.V. All rights reserved.

Keywords: Type-2 fuzzy sets; Lattices; Irreducible elements; Automorphisms

1. Introduction

The algebra of truth values for fuzzy sets of type-2 consists of all mappings from the unit interval into itself, withoperations certain convolutions of these mappings with respect to pointwise max and min [1]. This algebra has beenstudied extensively, for example [2–7]. The basic algebraic theory depends on the fact that [0,1] is a complete chain,so lends itself to various generalizations and consideration of particular cases. This paper develops the theory whereeach copy of the unit interval is replaced by a finite chain. These algebras are of interest both as special cases ofbases for fuzzy theories, and as mathematical entities per se. In most practical applications using type-2 fuzzy sets,computations are carried out on discrete universes of discourse. These are finite chains from the unit interval. Thefinite algebras in this paper can be embedded as subalgebras of the usual infinite algebra of truth values for type-2fuzzy sets. For notational simplicity we denote the elements of these chains by integers rather than decimals between0 and 1.

With operations we describe below, we have a finite algebra that we denote by M(mn) with basically the samealgebraic properties as the algebra of truth values of fuzzy sets of type-2, such as those developed in [7]. The purposeof this paper is to establish properties of M(mn) and to determine its automorphism group; that is, its group ofsymmetries.

* Corresponding author.E-mail addresses: hardy@nmsu.edu (C. Walker), elbert@nmsu.edu (E. Walker).

0165-0114/$ – see front matter © 2013 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.fss.2013.06.006

34 C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49

In Section 3, the basic operations � and � are discussed. Each of these operations determines a partial order, bothof which turn out to be lattice orders. A special case of interest is the algebra consisting of all mappings of a finitechain into a two-element set. The basic type-2 operations in this case are not those corresponding to ordinary unionand intersection of sets. This situation has been studied in detail for mappings of [0,1] into a two-element set in [8].

One special feature of the algebra of mappings of a finite set into a two-element set is its connection with deter-mining equational bases for the truth-value algebra of type-2 fuzzy sets and its reducts, which has been studied in [5].That situation is examined in Section 4 for algebras of mappings of finite chains into finite chains in general.

Some special subalgebras of M(mn) are noted in Section 5. In Section 6, the irreducible elements with respect tothe basic operations of M(mn) are determined. These elements are crucial in determining the automorphism group ofM(mn), the subject of Section 7. The result proved there shows that there are no symmetries of the algebra M(mn),which could have interesting implications in the practical applications of discrete type-2 fuzzy sets. We conclude withsome discussion and explicit problems of interest.

2. The algebra M(mn)

Here we introduce some notation. For a positive integer n, let n be the set {1,2, . . . , n}. This set has its usuallinear order which we denote by �, max and min operations denoted by ∨ and ∧, negation given by ¬k = n − k + 1,and the obvious constants 1 and n. With these operations, n becomes a De Morgan algebra; that is, a bounded,distributive lattice with a negation that satisfies the De Morgan laws. In fact it is a Kleene algebra since it also satisfiesa ∧ ¬a � b ∨ ¬b.

We denote by mn the set {f : n → m} of all mappings from the set n into the set m. The algebra M(mn) consistsof the set mn with operations given in the following definition.

Definition 1. On mn, let

(f � g)(i) =∨

j∨k=i

(f (j) ∧ g(k)

)

(f � g)(i) =∨

j∧k=i

(f (j) ∧ g(k)

)

¬f (i) =∨

j=¬i

f (j) = f (¬i)

1̄(i) ={

m if i = n

1 if i �= nand 0̄(i) =

{m if i = 11 if i �= 1

Thus we have the algebra

M(mn) = (

mn,�,�,¬, 0̄, 1̄)

There are two other operations on the functions in mn, namely pointwise max and min. We also denote these by ∨and ∧, respectively. Just as in the case M([0,1][0,1]), these operations help in determining the properties of the algebraM(mn) via the following auxiliary operations.

Definition 2. For f ∈ mn, let f L and f R be the elements defined by

f L(i) =∨j�i

f (j) and f R(i) =∨j�i

f (j)

We write f LR and f RL for (f L)R and (f R)L, respectively.

The operations � and � in M(mn) can be expressed in terms of the pointwise max and min of functions in twodifferent ways, as follows.

C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49 35

Theorem 3. The following hold for all f,g ∈ M(mn).

f � g = (f ∧ gL

) ∨ (f L ∧ g

) = (f ∨ g) ∧ (f L ∧ gL

)f � g = (

f ∧ gR) ∨ (

f R ∧ g) = (f ∨ g) ∧ (

f R ∧ gR)

As an illustration, we prove the first equation.

(f � g)(i) =∨

j∨k=i

(f (j) ∧ g(k)

) =[∨

j�i

(f (j) ∧ g(i)

)] ∨[∨

j�i

(f (i) ∧ g(j)

)]

=[(∨

j�i

f (j)

)∧ g(i)

]∨

[f (i) ∧

(∨j�i

g(j)

)]

= [f L(i) ∧ g(i)

] ∨ [f (i) ∧ gL(i)

]Introducing the operations L and R and using them in this general situation to express other operations in terms ofpointwise ones, as in the theorem above, appears in many papers, for example in [9]. Using these auxiliary operations,it is fairly routine to verify the following properties of the algebra M(mn). The details of the proofs are almost exactlythe same as for the algebra M([0,1][0,1]), which are given for example in [7].

Corollary 4. Let f,g,h ∈ M(mn). Some basic equations follow.

1. f � f = f ; f � f = f

2. f � g = g � f ; f � g = g � f

3. f � (g � h) = (f � g) � h; f � (g � h) = (f � g) � h

4. f � (f � g) = f � (f � g)

5. 1̄ � f = f ; 0̄ � f = f

6. ¬¬f = f

7. ¬(f � g) = ¬f � ¬g; ¬(f � g) = ¬f � ¬g

Some remarks about these equations are in order.

• Most are obvious from either Definition 1 or Theorem 3.• Each of � and �, being idempotent, commutative and associative, gives rise to a partial order that is a lattice order.

These partial orders are discussed in Section 3 below.• Eq. (4) gives a direct connection between the binary operations � and �. Were the quantities also equal to f , this

equation would assert the absorption law for lattices. However, as we will see, with respect to the operations �and �, the algebra M(mn) is not a lattice except for trivial cases.

• Note that the distributive laws for � over � and vice versa are not included. In fact, they do not hold in general aseasy examples show. (See the proof of Theorem 24.)

• This algebra has an absorbing element, namely the mapping in M(mn) that takes each element in n to 1.• These equations do not form an equational basis for M(mn), a situation we will define and address later.

The algebra (mn,�,�), that is, the algebra with mn as elements and operations � and �, is a bisemilattice [10,11]with special properties that we examine in the ensuing sections.

We denote the elements of mn by n-tuples (a1, a2, . . . , an) of elements of m and when no confusion arises, with-out the commas. For example, we write the element (2,1,3) of 33 as 213. Note that with this notation, in mn theelement 1̄ is (1,1, . . . ,1,m) and 0̄ is (m,1,1, . . . ,1). Further note that the absorbing element is (1,1, . . . ,1). Finally,¬(a1, a2, . . . , an) = (an, an−1, . . . , a1). Namely, apply the negation on n to the subscripts in the n-tuple.

3. The partial orders determined by � and �

As asserted above, the operations � and � determine partial orders by f �� g if f �g = g and f �� g if f �g = f .First, we note the following

36 C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49

Proposition 5. Under the partial order ��, any two elements f and g have a least upper bound. That least upperbound is f � g. Similarly, under the partial order ��, any two elements f and g have a greatest lower bound. Thatgreatest lower bound is f � g.

Proof. For two elements f and g, g �� f � g and f �� f � g since g � (f � g) = f � g and f � (f � g) = f � g.Therefore, f � g is an upper bound for f and g under the partial order ��. Suppose that h is an upper bound for f

and g. Then f � h = h = g � h. Then (f � g) � h = f � (g � h) = f � h = h, and so f � g is the least upper boundof f and g. Similarly the greatest lower bound for f and g for �� is f � g. �Theorem 6. The partial orders �� and �� are lattice orders.

Proof. For M(mn), the element (m,1,1, . . . ,1) is a bottom and the absorbing element (1,1, . . . ,1) is a top for theoperation �. Since M(mn) is finite, it has arbitrary sups, and the inf of any set is the sup of all its lower bounds. Now(1,1, . . . ,1,m) is a top and (1,1, . . . ,1) is a bottom for �, and similarly �� is a lattice order. (See Lemma 14, page31, in [12], or Theorem 2.6 in [13] for other proofs that these partial orders are lattices.) �

At this point, we give some details of a specific example, namely of the algebra (23,�,�), which is in some sensethe smallest non-trivial example. First, we exhibit a table for the operation � for the algebra (23,�,�).

� 211 212 221 222 122 121 111 112211 211 212 221 222 122 121 111 112

212 212 222 222 122 122 111 112

221 221 222 122 121 111 112

222 222 122 122 111 112

122 122 122 111 112

121 121 111 112

111 111 111

112 112

The partial orders determined by � and � are pictured below.

Notice that this partial order determined by � is a distributive lattice. However, an examination of the operation � in(33,�) reveals that its lattice structure is not distributive. Which mn do give distributive lattices for � and �? It shouldbe noted here that the algebras (mn,�) and (mn,�) are isomorphic via the negation ¬, and the lattices determined bythe two operations are anti-isomorphic. Thus if one is distributive, so is the other.

The partial orders given by � and � are not equal, so (23,�,�) is not a lattice with respect to the two binaryoperations � and �.

We will look further at the special cases M(2n) in the next section.

C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49 37

4. Equational bases

A nonempty class K of algebras of a type F is called a variety if it is closed under subalgebras, homomorphicimages, and Cartesian products. A class K of algebras of a type F is an equational class if it is the class of all algebrassatisfying the equations in some class E of equations over type F . We call E an equational basis for the equationalclass. A famous theorem of Birkhoff states that K is an equational class if and only if it is a variety [14]. We will useboth characterizations of a variety in this section.

The variety V(M(mn)) is the class of all algebras of type (2,2,1,0,0) that satisfy the equations satisfied byM(mn). There are two basic problems we would like to solve: find an equational basis for the variety V(M(mn)),and find the “simplest” algebra generating the variety V(M(mn)). And of course we would like to solve the sameproblems for the algebra (mn,�,�) of type (2,2).

In [5], these questions were addressed for the algebra M([0,1][0,1]), and results were obtained pertinent to thepresent situation, for example that V(M([0,1][0,1])) = V(M(25)) and that V([0,1][0,1],�,�) = V(23,�,�). In thissection we show that V(M(mn)) = V(M(2n)).

The equations listed in Corollary 4 do not form an equational basis for the variety V(M(mn)). If all equations sat-isfied by M(mn) were consequences of these equations, then V(M(mn)) would consist of all the algebras satisfyingthese equations. Since M(mn), and hence V(M(mn)), are locally finite (finite subsets generate finite subalgebras), itwould follow that any algebra satisfying Eqs. (1)–(7) would be locally finite. But, for example, ortholattices satisfythese equations, and not all ortholattices are locally finite [15]. A similar argument holds when we restrict to the oper-ations �, � since the variety V((mn,�,�)) is locally finite, and not all the algebras satisfying Eqs. (1)–(4) are locallyfinite—all lattices satisfy (1)–(4), but not all lattices are locally finite. Therefore M(mn) satisfies some equations nota consequence of those in Corollary 4. Eqs. (1) are three such [16].

Complex expressions in � and � can get quite long and difficult to decipher. To make these equations more readablewe write + for � and juxtaposition for �.

f (g + h) + fg + f h = f (g + h)(fg + f h)

f (f + g)(f h + g) = f (f + g)(f h + g + h)

h(f + g)(g + f h) = h(f + g)(g + h + f h) (1)

Definition 7. Let m� 3. For each a with 1 � a < m, define ϕa : M(mn) → M(2n) by

ϕa(f )(x) ={

2 if a < f (x)

1 otherwise

Proposition 8. For 1 � a < m the map ϕa is a homomorphism with respect to all of the operations �, �, ¬, 0̄, 1̄, ∧,∨, L, R.

Proof. We abbreviate ϕa to ϕ. Clearly ϕ(m1 . . .1) = (21 . . .1) and ϕ(1 . . .1m) = (1 . . .12). Next, note that the fol-lowing are equivalent

ϕ(f ∧ g)(x) = 2

f (x) ∧ g(x) > a

f (x) > a and g(x) > a

ϕ(f )(x) = 2 and ϕ(g)(x) = 2(ϕ(f ) ∧ ϕ(g)

)(x) = 2

It follows that ϕ(f ∧g) = ϕ(f )∧ϕ(g). A similar argument considering when ϕ(f ∨g)(x) = 1 shows that ϕ(f ∨g) =ϕ(f ) ∨ ϕ(g).

Now, note that the following are equivalent,

38 C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49

ϕ(¬f )(x) = 2

¬f (x) > a

f (¬x) > a

ϕ(f )(¬x) = 2

¬(ϕ(f )

)(x) = 2

thus ϕ(¬f ) = ¬ϕ(f ).Next, note that the following are equivalent.

ϕ(f L

)(x) = 2

f L(x) > a

sup{f (y)

∣∣ y � x}

> a

f (y) > a for some y � x

ϕ(f )(y) = 2 for some y � x

ϕ(f )L(x) = 2

So ϕ(f L) = ϕ(f )L. A similar argument shows that ϕ(f R) = ϕ(f )R .Finally, since � and � can be expressed in terms of L, R, ∧, ∨, it follows that ϕ preserves these two operations as

well. �Theorem 9. For m � 2, the algebras M(mn) and M(2n) generate the same variety and thus satisfy the same equations.

Proof. M(2n) is clearly isomorphic to the subalgebra of M(mn) consisting of those elements that map n into theset {1,m}. Thus V(M(2n)) ⊆ V(M(mn)). For the other containment, if m = 2, there is nothing to do. If m � 3,consider the product map,

∏a∈[1,m) ϕa : M(mn) → ∏

a∈[1,m) M(2n). By general considerations, this map is a homo-morphism. Suppose that f,g ∈ M(mn) with f �= g. Let x be such that f (x) �= g(x) and suppose that f (x) < g(x).Then ϕf (x)(f )(x) �= ϕf (x)(g)(x). It follows that the product map is an embedding. So M(mn) is isomorphic to asubalgebra of a power of M(2n), showing V(M(mn)) ⊆ V(M(2n)). �

The following theorem is proved in [5], Theorem 34, and further investigation shows that there is a twelve-elementalgebra generating V(M(25)). But we do not even know whether or not these various varieties have finite equationalbases, much less know finite bases for them.

Theorem 10. Let n � 5. Then M(2n) and M(25) generate the same variety and thus satisfy the same equations.

5. Some subalgebras of M(mn)

The truth-value algebra for type-1, or ordinary fuzzy sets, is the unit interval [0,1] with operations ∨ and ∧,negation x → 1 − x, and the obvious constants. In the finite case, the corresponding truth-value algebra is the chain nwith the analogous operations. The resulting algebra is isomorphic to the subalgebra of M(mn) consisting of n-tupleswith the element m in one spot and 1’s everywhere else.

In the interval-valued case the subalgebra of M(mn) consists of those n-tuples with 1’s everywhere except forthe element m in some consecutive spots, with at least one m appearing. The point is that in the finite case thetruth-value algebra for type-2 fuzzy sets contains as subalgebras the truth-value algebras for ordinary fuzzy sets,and for interval-valued fuzzy sets. The case of mappings of [0,1] into [0,1] is discussed in detail in [17] and[18].

A mapping f ∈ mn is normal if f (k) = m for some 1 � k � n, and f is convex if for i � j � k, f (j) � f (i) ∧f (k). Equivalently, f is convex if f = f L ∧f R . The subalgebra of M(mn) consisting of the normal convex functionswas investigated in [19]. This is a De Morgan algebra, which in turn contains a special subalgebra which is Kleene.Again, the details are in [19], and the case of M([0,1][0,1]) is discussed in [7,20,21].

C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49 39

6. Irreducibles in M(mn)

One purpose of this paper is to show that the automorphism group of M(mn) is trivial, that is, consists of onlythe identity element. We will actually show more generally that the automorphism group of the algebra (mn,�,�)

is trivial. Automorphisms map irreducible elements to irreducible elements, and a main tool in determining the au-tomorphism group of (mn,�,�) is the determination of the set of irreducibles of this algebra. We do that in thissection.

Definition 11. An element f of (mn,�,�) is

1. join-irreducible if f = g � h implies that f = g or f = h,2. meet-irreducible if f = g � h implies that f = g or f = h, and3. irreducible if it is both join- and meet-irreducible.

Notation 12. We use the following notation for elements of (mn,�,�).

• x is the constant function with value x. That is, x(i) = x for all i.• xi is the (point) function whose i-th component is x and whose other components are 1. That is, xi(i) = x and

xi(j) = 1 if j �= i.• fi denotes the i-th component of f ; that is fi = f (i).

Lemma 13. Let x ∈ m and i ∈ n.

1. If xi = g � h, then g(j) = h(j) = 1 for all j > i and either g or h is a point function of i.2. If xi = g � h, then g(j) = h(j) = 1 for all j < i and either g or h is a point function of i.

Proof. Since xi is a point function of i, then xi(i) = x, and xi(j) = 1 for j �= i. Suppose xi = g � h. Thus

xi(i) = (g(i) ∨ h(i)

) ∧ gL(i) ∧ hL(i) = x

so g(i) ∨ h(i) � x, and both gL(i) and hL(i) � x. For j �= i,

xi(j) = (g(j) ∨ h(j)

) ∧ gL(j) ∧ hL(j) = 1

If j > i, then gL(j) ∧ hL(j) � gL(i) ∧ hL(i) � x which implies g(j) ∨ h(j) = 1, so g(j) = h(j) = 1. Suppose thatfor some j < i, g(j) > 1. Then gL(j) > 1, so we must have hL(j) = 1 which means that h(k) = hL(k) = 1 for allk � j . Then if for some j < k < i, h(k) �= 1, we would need gL(k) = 1. But gL(k) � gL(j) �= 1. So either g or h is1 for all j < i. Since they are both = 1 for all j > i, either g or h is a point function. The proof of the second half issimilar. �Proposition 14. The elements mi are irreducible in (mn,�,�).

Proof. Suppose mi = g � h. By Lemma 13, either g or h is a point function of i, say it is g. Then gL(i) = m impliesthat g(i) = m. Thus g = mi , and thus mi is join-irreducible. Suppose mi = g �h. Again by Lemma 13, either g or h isa point function of i, say it is g. Then gR(i) = m implies that g(i) = m. Thus g = mi . Thus mi is also meet-irreducibleand hence irreducible. �Proposition 15. The function m1 ∨ mn is irreducible in (mn,�,�).

Proof. Let 1 < j � n. If m1 ∨ mj = (f ∨ g) ∧ f L ∧ gL, then f L(1) = gL(1) = m implies f (1) = g(1) = m, whencef L = gL = m. Thus we must have f (x) ∨ g(x) = 1 for all x /∈ {1, j} so that f (x) = g(x) = 1 for all x /∈ {1, j}.But then f (j) ∨ g(j) = m implies that at least one of f or g equals m1 ∨ mj . Thus m1 ∨ mj is join-irreducible.A symmetric argument proves that a two-element set mi ∨ mn is meet-irreducible for 1 � i < n. It follows thatm1 ∨ mn is irreducible. �

40 C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49

Proposition 16. The function 1 is irreducible in (mn,�,�).

Proof. Suppose 1 = f � g = (f ∨ g) ∧ f L ∧ gL. If f (i) > 1 for some i, then so is f L(i) > 1. Then it must be thatgL(i) = 1, which implies that g(j) = 1 for all j � i. Suppose g(k) > 1 for some k > i. Then f L(j) > 1 for all j � i

implies f L(k) > 1 and we would have (f (k) ∨ g(k)) ∧ f L(k) ∧ gL(k) > 1. Thus we must have g = 1. Thus 1 isjoin-irreducible. A similar proof shows that 1 is meet-irreducible. �

When n = 2 and m � 3, there are additional irreducibles.

Proposition 17. In (m2,�,�), the functions f with f (1) ∨ f (2) = m are irreducible.

Proof. Suppose (m, k) = f �g. Then (m, k) = (f ∨g)∧f L ∧gL implies f L1 = gL

1 = m, whence f1 = g1 = m. Also,f L = gL = m forces f2 ∨ g2 = k, so either f = (m, k) or g = (m, k). Thus (m, k) is join-irreducible. A similar proofshows (k,m) is meet-irreducible.

Suppose (m, k) = f � g. Then (m, k) = (f ∨ g) ∧ f R ∧ gR implies (f2 ∨ g2) ∧ f R2 ∧ gR

2 = k, but f2 = f R2 and

g2 = gR2 , so (f2 ∨ g2) ∧ f R

2 ∧ gR2 = f2 ∧ g2 = k, whence either f2 = k or g2 = k. Now (f1 ∨ g1) ∧ f R

1 ∧ gR1 = m

implies f1 = f R1 = g1 = gR

1 = m, so either f = (m, k) or g = (m, k). Thus (m, k) is meet-irreducible. A similar proofshows (k,m) is join-irreducible. �

In the following series of lemmas we show that there are no other irreducibles in (mn,�,�).

Lemma 18. A point function xi with 1 < i < n and 1 < x < m is neither meet- nor join-irreducible in (mn,�,�).

Proof. Let g(j) = x ∧ mRi ; that is, g(j) =

{x if 1 � j � i

1 if i < j � n. Then

g � mi = (g ∨ mi) ∧ gL ∧ mLi

= ((x ∧ mR

i

) ∨ mi

) ∧ x ∧ mLi = xi

since ((x ∧ mRi ) ∨ mi)(j) = 1 for j > i, mL

i (j) = 1 for j < i, and

((x ∧ mR

i

) ∨ mi

)(i) ∧ x(i) ∧ mL

i (i) = m ∧ x ∧ m = x.

Also g �= xi �= mi . Thus xi is join-reducible. Let h = x ∧mLi . A similar argument shows that xi = h�mi , so xi is also

meet-reducible. �Lemma 19. If n = 2, and 1 < x ∨ y < m, then the 2-tuple (x, y) is join- or meet-reducible.

Proof. Let n = 2 and suppose x < y. Then

(x,m) � (y, y) = (x ∨ y,m ∨ y) ∧ (x,m) ∧ (y, y)

= (y,m) ∧ (x,m) ∧ (y, y)

= (y ∧ x ∧ y,m ∧ m ∧ y) = (x, y)

and

(m,x) � (y, y) = (m ∨ y, x ∨ y) ∧ (m,x) ∧ (y, y)

= (m,y) ∧ (m,x) ∧ (y, y)

= (m ∧ m ∧ y, y ∧ x ∧ y) = (y, x)

so (x, y) is join-reducible and (y, x) is meet-reducible.

C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49 41

Suppose 1 < x = y < m. Then

(x,1) � (m,x) = (x ∨ m,1 ∨ x) ∧ (x, x) ∧ (m,m)

= (m,x) ∧ (x, x) ∧ (m,m)

= (m ∧ x ∧ m,x ∧ x ∧ m) = (x, x)

Thus (x, x) is join-reducible. A similar proof shows (x, x) is also meet-reducible. �Lemma 20. If there are three distinct elements i, j, k ∈ n with f (i) �= 1, f (j) �= 1, and f (k) �= 1, then f is meet- orjoin-reducible.

Proof. Choose i such that f (i) = f LR(i). If j < k < i, define g = h = f except g(j) = 1 and h(k) = 1. ThengR = hR = f R , and g ∨ h = f , so

g � h = (g ∨ h) ∧ gR ∧ hR = f ∧ f R ∧ f R = f

so f is meet-reducible. If i < j < k, the symmetric construction gets g � h = f , so f is join-reducible.If neither of the above holds for any choice of points, then j < i < k and f = 1 at all other points. If f (k) �

f (i) ∨ f (j) or f (j) � f (i) ∨ f (k), this was covered above. If f (i) > f (j) ∨ f (k), define g = f except g(i) = 1,h = f except h(k) = f (i). Then hR = f RL and gR = f R , and

g � h = (g ∨ h) ∧ gR ∧ hR = (g ∨ h) ∧ f RL ∧ f R = f

so f is meet-reducible. By symmetric construction f is also join-reducible. �Lemma 21. If n > 2 and if f > 1 at exactly two points i < j , then f is meet- or join-reducible except in the case{i, j} = {1, n} and f (1) = f (n) = m.

Proof. If f (i) � f (j), 1 < i, and f (i) < m, define g = f except g(i) = m and h = f except h(1) = f (i). ThengL(1) = 1 = f (1) and hL(i) = f (i), and

g � h = (g ∨ h) ∧ gL ∧ hL = f

so f is join-reducible. If f (i) � f (j) and j < n, a symmetric construction gives

g � h = (g ∨ h) ∧ gR ∧ hR = f

so f is meet-reducible.If f (i) � f (j), j < n, define g = f except g(i) = 1, and h = f except h(n) = m. Then g ∨ h = f except at n,

gR = f R and hR = 1. Thus

g � h = (g ∨ h) ∧ gR ∧ hR = (g ∨ h) ∧ gR = f

so f is meet-reducible. If f (i) � f (j), 1 < i, a symmetric construction gives

g � h = (g ∨ h) ∧ gL ∧ hL = (g ∨ h) ∧ gL = f

so f is join-reducible.This leaves i = 1 and j = n, and 1 < f (1)∧f (n) �m. If 1 < f (1) < f (n) < m, define g = f except g(1) = f (n),

and h = f except h(1) = m. Then gL = f LR and hL(1) = f (1), and

g � h = (g ∨ h) ∧ gL ∧ hL = (g ∨ h) ∧ f LR ∧ hL = f

so f is join-reducible. If 1 < f (n) < f (1) < m a symmetric construction shows that f is meet-reducible.If 1 < f (1) = f (n) < m, define g = f except g(1) = 1 and h = f except h(1) = m. Then gR = f RL and hR = 1,

and

g � h = (g ∨ h) ∧ gR ∧ hR = (g ∨ h) ∧ f RL = f

so f is meet-reducible. A symmetric construction shows that f is also join-reducible.

42 C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49

If 1 < f (1) � f (n) = m, and f = (g ∨ h) ∧ gL ∧ hL then either g(1) = f (1) or h(1) = f (1). Say g(1) = f (1).Then we must have h(1) � f (1). For 1 < i < n, gL ∧ hL > 1 so we must have g ∨ h = 1, then m = (g(n) ∨ h(n)) ∧gL(n)∧hL(n) implies that gL(n) = hL(n) = m. If f (1) < m, then g(1) = m so g = f . If f (1) = m, then h(1) = f (1).So g(1) ∨ h(1) = m will then force either g = f or h = f . Thus f is join-irreducible.

If 1 < f (1) � f (n) = m, and f = (g ∨h)∧gR ∧hR , then g(n) = h(n) = m, so gR = hR = 1 and f = g ∨h. Thuseither g(1) = m or h(1) = m, implying g = f or h = f . Thus f is also meet-irreducible. �

The preceding lemmas yield the following.

Theorem 22. Let m,n � 2. The irreducible elements of (mn,�,�) are these:

1. The absorbing element 1.2. The elements mi .3. The element m1 ∨ mn.4. If n = 2, all normal elements and the absorbing element 1.

7. Automorphisms of M(mn)

In [18] it was shown that the automorphism group of M([0,1][0,1]) is the product of the automorphism group ofthe algebra ([0,1],�,0,1) with itself. Since the automorphism groups of m and n are trivial, we conjecture that theautomorphism group of M(mn) is trivial. But the theorem cited about M([0,1][0,1]) was very difficult, and we seeka simpler proof for the finite algebra M(mn). As suggested above, we prove more generally that the automorphismgroup of (mn,�,�) is trivial; that is, contains only the identity automorphism. It follows that the same is true forM(mn). This section is devoted to proving this fact.

We will assume that m� 2 and n� 2. The other cases are trivial.Although m1 and mn are not operations in the algebra (mn,�,�) they are preserved by automorphisms of that

algebra: for all f , f � m1 = f and hence m1 �� f for all f . Therefore m1 is the smallest element in the latticedetermined by �, so is fixed by any automorphism of the lattice (mn,�) and hence of (mn,�,�). Similarly, mn isfixed by any automorphism ϕ of (mn,�,�).

In this section, ϕ will always denote an automorphism of (mn,�,�). Recall that f is normal if and only if f (k) =m for some k.

Theorem 23. If f is normal, then so is ϕ(f ).

Proof. Suppose that f is normal. Then,

f � mn = (f ∨ mn) ∧ f L ∧ mLn

= (f ∨ mn) ∧ f L ∧ mn

= mn

ϕ(f � mn) = ϕ(mn) = mn

= (ϕ(f ) ∨ ϕ(mn)

) ∧ ϕ(f )L ∧ ϕ(mn)L

= (ϕ(f ) ∨ mn

) ∧ ϕ(f )L ∧ mn

= ϕ(f )L ∧ mn

Since ϕ(f )L ∧ mn = mn, we get (ϕ(f )L)n = mn, and so ϕ(f ) is normal. �This of course implies that automorphisms of (mn,�,�) induce automorphisms of the subalgebra of normal func-

tions.As mentioned earlier, a function f is convex if f = f L ∧ f R . In the following proof we take advantage of a

different characterization.

C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49 43

Theorem 24. If f is convex, then so is ϕ(f ).

Proof. Suppose that f is convex. We use Theorem 36 in [7], which says that f is convex if and only if f � (g � h) =(f � g) � (f � h) for all g, h in the algebra (mn,�,�). The proof for (mn,�,�) is exactly the same. Therefore,

ϕ(f ) � (ϕ(g) � ϕ(h)

) = ϕ(f � (g � h)

)= ϕ

((f � g) � (f � h)

)= ϕ(f � g) � ϕ(f � h)

= (ϕ(f ) � ϕ(g)

) � (ϕ(f ) � ϕ(h)

)Thus ϕ(f ) is convex. �

This of course implies that automorphisms of (mn,�,�) induce automorphisms of the subalgebra of convex func-tions.

Corollary 25. ϕ(m1 ∨ mn) = m1 ∨ mn.

Proof. The element m1 ∨ mn is the only normal non-convex irreducible element. �Theorem 26. ϕ(m) = m.

Proof. First, note that

(m1 ∨ mn) � m = (m1 ∨ mn ∨ m) ∧ (m1 ∨ mn)R ∧ mR

= m ∧ m ∧ m = m

Thus

ϕ(m) = ϕ((m1 ∨ mn) � m

)= ϕ(m1 ∨ mn) � ϕ(m)

= (m1 ∨ mn ∨ ϕ(m)

) ∧ (m1 ∨ mn)R ∧ ϕ(m)R

= (m1 ∨ mn ∨ ϕ(m)

) ∧ m ∧ ϕ(m)R

= (m1 ∧ ϕ(m)R

) ∨ (mn ∧ ϕ(m)R

) ∨ (ϕ(m) ∧ ϕ(m)R

)= m1 ∨ (

mn ∧ ϕ(m)R) ∨ ϕ(m)

using the normality of ϕ(m) in the last line. Therefore, ϕ(m) has value m at 1. A similar proof, after noting (m1 ∨mn) � m = m, shows that it has value m at n. Since ϕ(m) is convex, ϕ(m) = ϕ(m)L ∧ ϕ(m)R = m. �

We get the following useful corollary.

Corollary 27. ϕ commutes with L and R. That is, for f ∈ Aut((mn,�,�)), ϕ(f )L = ϕ(f L) and ϕ(f )R = ϕ(f R).

Proof. First note that

f � m = (f ∨ m) ∧ f L ∧ mL

= (f ∨ m) ∧ f L ∧ m

= f L ∧ m

= f L

Similarly, f � m = f R .

44 C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49

ϕ(f L

) = ϕ(f � m) = ϕ(f ) � ϕ(m)

= (ϕ(f ) ∨ ϕ(m)

) ∧ ϕ(f )L ∧ ϕ(m)L

= (ϕ(f ) ∨ m

) ∧ ϕ(f )L ∧ mL

= (ϕ(f ) ∨ m

) ∧ ϕ(f )L ∧ m

= m ∧ ϕ(f )L

= ϕ(f )L

The proof is similar for R. �Theorem 28. ϕ(mi) = mi .

Proof. Since the mi are the only convex normal irreducibles, ϕ(mi) = mj for some j . We show that i = j . Supposethat i � j . Then

mi � mj = (mi ∨ mj) ∧ mLi ∧ mL

j

= (mi ∨ mj) ∧ mLj

= mj

Therefore, mi �� mj , and since ϕ preserves �� order, we get ϕ(mi) �� ϕ(mj ). Since ϕ is one-to-one and onto, andthe mi go to themselves and are finite in number, we get ϕ(mi) = mi . �

Next we show that point functions are fixed by automorphisms.

Lemma 29. ϕ(xi) = yi for some y.

Proof. First, note that

xi � mi = (xLi ∧ mi

) ∨ (xi ∧ mL

i

)= xi

Thus

ϕ(xi) = ϕ(xi � mi)

= (ϕ(xi) ∨ ϕ(mi)

) ∧ ϕ(xi)L ∧ ϕ(mi)

L

= (ϕ(xi) ∨ mi

) ∧ ϕ(xLi

) ∧ mLi

Hence, ϕ(xi) has value 1 to the left of i. A similar calculation with � shows that ϕ(xi) has value 1 to the right of i.It follows that ϕ(xi) is a point function at i. �Lemma 30. x � y if and only if xi �� yi , that is, if and only if xi � yi = xi .

Proof. Suppose that x � y. Then

xi � yi = (xi ∨ yi) ∧ xLi ∧ yL

i

= (xi ∨ yi) ∧ xLi

= xi

Now suppose that xi � yi = xi . Then (xi ∨ yi) ∧ xLi ∧ yL

i = xi . It is clear from this that x � y. �Theorem 31. ϕ(xi) = xi .

C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49 45

Proof. ϕ preserves �� order on the point functions at i, which is the � order. The set of point functions at i is finite,and the lemma follows. �Corollary 32. ϕ(xi)

L = xLi , and ϕ(xi)

R = xRi .

Corollary 33. Let xij be the interval of x’s from component i to component j with the remaining components 1. Thenϕ(xij ) = xij .

Proof. ϕ(xij )L = ϕ(xL

ij ) = ϕ(xLi ) = ϕ(xi)

L = xLi , so ϕ(xij ) has values 1 to the left of i and value x at i. Similarly,

ϕ(xij ) has values 1 to the right of j , and value x at j . Also ϕ(xij ) has no values larger than x. Since ϕ(xij ) is convex,the corollary follows. �Corollary 34. ϕ(x) = x. That is, automorphisms fix constant functions.

Lemma 35. ϕ(m1 ∨ mi) = m1 ∨ mi and ϕ(mi ∨ mn) = mi ∨ mn.

Proof. First, note that

(m1 ∨ mn) � mi = ((m1 ∨ mn) ∨ mi

) ∧ (m1 ∨ mn)R ∧ mR

i

= m1 ∨ mi

so

ϕ(m1 ∨ mi) = ϕ((m1 ∨ mn) � mi

)= ϕ(m1 ∨ mn) � ϕ(mi)

= (ϕ(m1 ∨ mn) ∨ ϕ(mi)

) ∧ ϕ(m1 ∨ mn)R ∧ ϕ(mi)

R

= (m1 ∨ mn ∨ mi) ∧ mRi

= m1 ∨ mi

A similar proof shows that ϕ(mi ∨ mn) = mi ∨ mn. �Lemma 36. ϕ(xi ∨ xj ) = xi ∨ xj .

Proof. Let i � j . Then

(mi ∨ mn) � xj = ((mi ∨ mn) ∨ xj

) ∧ (mi ∨ mn)R ∧ xR

j

= (mi ∨ mn ∨ xj ) ∧ xRj

= xi ∨ xj

Thus

ϕ(xi ∨ xj ) = ϕ((mi ∨ mn) � xj

)= (

ϕ(mi ∨ mn) ∨ ϕ(xj )) ∧ ϕ(mi ∨ mn)

R ∧ ϕ(xj )R

= (mi ∨ mn ∨ xj ) ∧ (mi ∨ mn)R ∧ xR

j

= (mi ∨ mn ∨ xj ) ∧ xRj

= xi ∨ xj �Corollary 37. ϕ(mi ∨ mj) = mi ∨ mj .

Lemma 38. ϕ(m1 ∨ xi) = m1 ∨ xi and ϕ(xi ∨ mn) = xi ∨ mn.

46 C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49

Proof. First, note that

(m1 ∨ mi) � (m1 ∨ xi) = (m1 ∨ mi ∨ m1 ∨ xi) ∧ (m1 ∨ mi)L ∧ (m1 ∨ xi)

L

= (m1 ∨ mi) ∧ (m1 ∨ mi)L ∧ (m1 ∨ xi)

L

= (m1 ∨ mi) ∧ m ∧ m

= m1 ∨ mi

Therefore

ϕ(m1 ∨ mi) = (m1 ∨ mi)

= ϕ((m1 ∨ mi) � (m1 ∨ xi)

)= ϕ(m1 ∨ mi) � ϕ(m1 ∨ xi)

= (ϕ(m1 ∨ mi) ∨ ϕ(m1 ∨ xi)

) ∧ ϕ(m1 ∨ mi)L ∧ ϕ(m1 ∨ xi)

L

= (m1 ∨ mi) ∨ ϕ(m1 ∨ xi) ∧ m ∧ m

= (m1 ∨ mi) ∨ ϕ(m1 ∨ xi)

Thus, we have

m1 ∨ mi = (m1 ∨ mi) ∨ ϕ(m1 ∨ xi)

Therefore, since (ϕ(m1 ∨ xi))L = (ϕ(m1 ∨ xi)

L) = ϕ(m), ϕ(m1 ∨ xi) may be written as m1 ∨ yi for some pointfunction yi . So

ϕ(m1 ∨ xi) = m1 ∨ yi

But

(m1 ∨ xi) � (m1 ∨ yi) = ((m1 ∨ xi) ∨ (m1 ∨ yi)

) ∧ m ∧ m

= (m1 ∨ xi) ∨ (m1 ∨ yi)

and so

(m1 ∨ xi) �� (m1 ∨ yi)

if and only if

xi � yi

It follows that xi = yi , and that ϕ(m1 ∨ xi) = m1 ∨ xi .The proof that ϕ(xi ∨ mn) = xi ∨ mn is similar. �

Lemma 39. If xi � f L, then xi � ϕ(f L).

Proof. Note

xi � f L = (xi ∨ f L

) ∧ xLi ∧ f L

= f L ∧ xLi ∧ f L

= xLi ∧ f L

= xLi

Therefore

C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49 47

ϕ(xLi

) = ϕ(xi � f L

)= (

ϕ(xi) ∨ ϕ(f L

)) ∧ ϕ(xi)L ∧ ϕ

(f L

)L

= (ϕ(xi) ∨ ϕ

(f L

)) ∧ ϕ(xLi

) ∧ ϕ(f L

)= ϕ

(xLi

) ∧ ϕ(f L

)Thus

ϕ(xLi

) = ϕ(xLi

) ∧ ϕ(f L

)and we have

xi � xLi = ϕ

(xLi

)� ϕ

(f L

) �Theorem 40. If xi � f L, then ϕ(xi ∨ f ) = xi ∨ ϕ(f ).

Proof. Note

(m1 ∨ xi) � f = (m1 ∨ xi ∨ f ) ∧ (m1 ∨ xi)L ∧ f L

= (m1 ∨ xi ∨ f ) ∧ f L

= xi ∨ f

Thus

ϕ(xi ∨ f ) = ϕ((m1 ∨ xi) � f

)= ϕ(m1 ∨ xi) � ϕ(f )

= (ϕ(m1 ∨ xi) ∨ ϕ(f )

) ∧ ϕ(m1 ∨ xi)L ∧ ϕ(f )L

= ((m1 ∨ xi) ∨ ϕ(f )

) ∧ ϕ(f )L

= (m1 ∧ ϕ(f )L

) ∨ (xi ∧ ϕ(f )L

) ∨ (ϕ(f )L

)= (

m1 ∧ ϕ(f )L) ∨ xi ∨ ϕ(f )

= xi ∨ ϕ(f ) �Similarly, we can prove the following.

Theorem 41. If xi � f R , then ϕ(xi ∨ f ) = xi ∨ ϕ(f ).

Now we are in a position to prove our main result about automorphisms.

Theorem 42. The automorphism group of (mn,�,�) has only one element.

Proof. Let f ∈ (mn,�,�). We induct on the number of components of f that are not 1. The element f is fixed by anyautomorphism if it has only one component not 1 since automorphisms fix points. Suppose that f has k componentsnot 1 and elements with fewer than k components are fixed by automorphisms for some k � 2. Let fi be the smallestcomponent of f that is not 1. Let g be the element f with its component fi replaced by 1. Then fi � gL unless i isthe first component of f that is not 1. In that case, fi � gR . Suppose that fi � gL. Then

ϕ(f ) = ϕ(fi ∨ g)

= ϕ(fi) ∨ ϕ(g)

= fi ∨ ϕ(g)

and since g has fewer than k components not 1, ϕ(g) = g. Thus ϕ(f ) = fi ∨ g = f . Similarly, if fi � gR , ϕ(f ) =fi ∨ g = f . �

48 C. Walker, E. Walker / Fuzzy Sets and Systems 236 (2014) 33–49

This implies, of course, that the automorphism group of M(mn) has only one element, namely, the identity auto-morphism.

8. Comments

One principal result of this paper is that the partial order given by the operation � is a lattice, and analogouslyfor �. For the operation �, the sup of two elements f and g is f � g, but the inf of the two elements is the sup of theset of all elements below both f and g. The elements f and g are n-tuples of elements of m, and the inf is given bysome formula in the elements in these two n-tuples.

Problem 1. Find a formula for the inf of two elements in the lattice determined by �. And similarly, do the same forthe lattice determined by �.

Problem 2. In the case of the algebra ([0,1][0,1],�), determine whether or not the partial order determined by � is alattice.

In the case of 23, the lattices determined by � and by � are both distributive, but this is not true for all mn.

Problem 3. For which mn are the lattices determined by � and � distributive. We conjecture none for m and n� 3.

The proof that Aut(mn,�,�) consists of only the identity automorphism was effected by a long sequence of lem-mas, etc. Hopefully, there is a much shorted and less computational proof.

Problem 4. Find a proof that Aut(mn,�,�) is trivial that is more conceptual, less computational, and shorter.

In showing that the automorphism group of (mn,�,�) consists of only the identity automorphism, we used inthe proof that an automorphism preserved both � and �. But small examples show that the automorphism group of(mn,�) is just the identity automorphism, and we suspect that this is true in general, but have no proof.

Problem 5. Find the automorphism group of (mn,�). (Since (mn,�) and (mn,�) are isomorphic, their automorphismgroups will be isomorphic.)

Finally, there are many ways to specialize and to generalize the truth-value algebra of type-2 fuzzy sets,([0,1][0,1],�,�,¬,0̄, 1̄). We have just taken a finite chain for each interval [0,1]. For example, one could take any twocomplete lattices instead, or substitute one finite chain for one of the intervals [0,1], and so on. Such investigationsmay be of interest.

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