AUTOMORPHISMS OF THE ALGEBRA OF FUZZY TRUTH VALUES
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AUTOMORPHISMS OF THE ALGEBRA OF FUZZY TRUTH VALUES
CAROL L. WALKER and ELBERT A. WALKER Department of Mathematical Sciences New Mexico State University Las Cruces, New Mexico, USA [email protected] This paper is an investigation of the automorphisms of the algebra of truth values of type-2 fuzzy sets. This algebra contains isomorphic copies of the truth value algebras of type-1 and of interval-valued fuzzy sets. It is shown that these subalgebras are characteristic; that is, are carried onto themselves by automorphisms of the containing algebra of truth values of fuzzy sets. Some other relevant subalgebras are proved characteristic, including the subalgebra of convex normal functions. The principal tool in this study is the determination of various irreducible elements. Keywords : Automorphism, characteristic subalgebra, type-2 fuzzy set, interval-valued fuzzy set, algebra of truth values. Received (received date) Revised (revised date)
1. Introduction Type-2 fuzzy sets–that is, fuzzy sets with fuzzy sets as truth values, were introduced by Zadeh, extending the notion of ordinary fuzzy sets.1 There is a fair amount of literature on the subject. We mention Mendel’s book, which discusses both the theoretical and applied aspects of the topic, and its extensive bibliography.2 Some papers of interest are listed in the references.3 4 5 6 7 8 9 10 . In a recent paper, we endeavored to give a straightforward treatment of the mathematical basics of type-2 fuzzy sets.11 In earlier papers, we investigated mathematical basics of type-1 fuzzy sets and interval-valued fuzzy sets.12 13 14 The present paper is an investigation of the automorphisms of the algebra that underlies the basic theory of type-2 fuzzy sets. For any mathematical object, its group of symmetries is an object of interest. For algebraic structures, these symmetries are called automorphisms. The algebra that underlies type-2 fuzzy sets; that is, the algebra of fuzzy truth values, is rather complicated. It contains as subalgebras the algebras of truth values of ordinary fuzzy sets as well as that of interval-valued fuzzy sets.11 Of special interest is the subalgebra of convex normal functions, which form a bounded distributive lattice. A question of interest is whether or not automorphisms induce automorphisms of these various subalgebras; that is, whether or not these subalgebras are characteristic 1
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subalgebras. Central to this question is the determination of irreducible elements, since irreducibles map to irreducibles under automorphisms. Therefore, considerable effort is made in identifying irreducible elements. We begin with some necessary preliminaries. 2. Background A fuzzy subset A of a set S is a mapping A : S → [0, 1]. Such mappings are called fuzzy sets, or type-1 fuzzy sets. Operations on the set [0, 1]S of all fuzzy subsets of S come from operations on [0, 1]. Common operations on [0, 1] of interest in fuzzy theory are ∧, ∨, and 0 given by x ∧ y = min{x, y}
(1)
x ∨ y = max{x, y} x0 = 1 − x
The constants 0 and 1 are generally considered as part of the algebraic structure, technically being nullary operations. The operations ∧ and ∨ are binary operations, and 0 is a unary operation. So the algebra basic to fuzzy set theory is ([0, 1], ∨, ∧,0 , 0, 1). The corresponding operations on the set [0, 1]S of all fuzzy subsets of S are given pointwise by the formulas (A ∧ B)(s) = A(s) ∧ B(s)
(2)
(A ∨ B)(s) = A(s) ∨ B(s) 0
A0 (s) = (A(s))
and the two nullary operations are given by 1(s) = 1 and 0(s) = 0 for all s ∈ S. The following equations hold in the algebra ([0, 1], ∨, ∧,0 , 0, 1). (1) (2) (3) (4) (5) (6) (7) (8) (9)
x ∧ x = x; x ∨ x = x (idempotent) x ∧ y = y ∧ x; x ∨ y = y ∨ x (commutative) x ∧ (y ∧ z) = (x ∧ y) ∧ z; x ∨ (y ∨ z) = (x ∨ y) ∨ z (associative) x ∧ (x ∨ y) = x; x ∨ (x ∧ y) = x (absorption laws) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z); x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) (distributive laws) 0 ∨ x = x; 1 ∧ x = x (identities) x00 = x (involution) 0 0 (x ∧ y) = x0 ∨ y 0 ; (x ∨ y) = x0 ∧ y 0 (De Morgan’s laws) 0 0 x ∧ x ≤ y ∨ y (Kleene inequality)
This is summarized by saying that ([0, 1], ∨, ∧,0 , 0, 1) is a bounded distributive lattice with an involution 0 that satisfies¡ De Morgan’s laws¢and the Kleene inequality, that is, is a Kleene algebra. Thus [0, 1]S , ∨, ∧,0 , 0, 1 is also a Kleene algebra since it satisfies exactly the same equations as does ([0, 1], ∨, ∧,0 , 0, 1). An algebra satisfying equations 1-8 is a De Morgan algebra.
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An interval-valued fuzzy subset A of a set S is a mapping A : S → n o ³ ´S of all interval[0, 1][2] = (x, y) ∈ [0, 1]2 : x ≤ y . Operations on the set [0, 1][2] valued fuzzy subsets of S come pointwise from coordinatewise operations on [2] [0, 1] : (a, b)³ ∧ (c, d) = (a ∧ c, b ∧ d) ´and (a, b) ∨ (c, d) = (a ∨ c, b ∨ d). The al[2] gebra I[2] = [0, 1] , ∧, ∨, (0, 0), (1, 1) is also a bounded distributive lattice, with largest element (1, 1) and smallest element (0, 0). Thus the algebra of interval-valued fuzzy subsets of a set S is also a bounded distributive lattice. With the involution 0 (x, y) = (y 0 , x0 ), this is a De Morgan algebra (but not a Kleene algebra). The situation for type-2 fuzzy sets is similar except that fuzzy subsets of type-2 are mappings into a more complicated object than [0, 1] or [0, 1][2] , namely into the fuzzy subsets [0, 1][0,1] of [0, 1]. The operations on the type-2 fuzzy subsets ¢S ¡ [0, 1][0,1] of a set S will come pointwise from operations on [0, 1][0,1] . Operations are put on [0, 1][0,1] using operations on both the domain [0, 1] and the range [0, 1] of a mapping in [0, 1][0,1] . This is where the difficulty of type-2 fuzzy sets lies. In The algebra of fuzzy truth values, we studied operations on [0, 1][0,1] that are of interest in type-2 fuzzy set theory, and developed some of their algebraic properties.11 We now recall those operations. Let f and g be in [0, 1][0,1] . The elements f t g and f u g of [0, 1][0,1] are defined by the following equations. They are convolutions of the operations ∨ and ∧ on the domain [0, 1] with respect to the operations ∨ and ∧ on the range [0, 1]. W (f t g) (x) = y∨z=x (f (y) ∧ g(z)) (3) W (f u g) (x) = y∧z=x (f (y) ∧ g(z)) Denote by 1 the element of [0, 1][0,1] defined by ½ 0 if x 6= 1 1(x) = 1 if x = 1
(4)
Denote by 0 the map defined by 0(x) =
½
1 if x = 0 0 if x = 6 0
(5)
These elements 0 and 1 of [0, 1][0,1] are nullary operations, and can be obtained by convolution of the nullary operations 0 and 1. 3. The Algebra ([0, 1][0,1] , u, t, 0, 1)
At this point, we have the algebra M = ([0, 1][0,1] , t, u, 0, 1) with the operations t, u, 0, and 1 obtained by convolution of the operations ∨, ∧, 0, 1. This algebra, with the additional operation ∗ obtained by the convolution of 0 , is the basic algebra for type-2 fuzzy set theory, and is analogous to the algebra ([0, 1], ∧, ∨,0 , 0, 1) basic for type-1 or ordinary fuzzy set theory. Our main purpose in The algebra of fuzzy truth values was to study the algebra (M,∗ ) and some of its subalgebras.11
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Although we are interested in the algebra M, the set [0, 1][0,1] also has the pointwise operations ∨, ∧, 0, 1 on it coming from operations on the range [0, 1]: (f ∧ g) (x) ³ = f (x) ∧ g (x), (f ´ ∨ g) (x) = f (x) ∨ g (x), 0 (x) = 0, and 1 (x) = 1. The [0,1] algebra [0, 1] , ∧, ∨, 0, 1 is a lattice with order given by f ≤ g if f = f ∧ g, or equivalently, if g = f ∨ g. The operations t and u can be expressed in terms of the simpler pointwise ones and this was crucial in developing the properties of M. Definition 1. For f ∈ M, let f L and f R be the elements of M defined by f L (x) = ∨y≤x f (y)
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R
f (x) = ∨y≥x f (y) Note that f L is monotone increasing and that f R is monotone decreasing. The following theorem expresses each of the convolution operations t and u directly in terms of pointwise operations in two alternate forms. Theorem 1. The following hold for all f, g ∈ M. ¢ ¢ ¡ ¡ f t g = f ∧ gL ∨ f L ∧ g ¡ ¢ = (f ∨ g) ∧ f L ∧ g L
¡ ¢ ¢ ¡ f u g = f ∧ gR ∨ f R ∧ g ¡ ¢ = (f ∨ g) ∧ f R ∧ g R
(7)
(8)
There are many relations between the various operations introduced above. We refer to The algebra of fuzzy truth values for details.11 The equations in the following proposition are crucial. Proposition 1. The following hold for f, g ∈ M. L
(f t g) = f L t g L R (f t g) = f R t g R
R
(f u g) = f R u g R L (f u g) = f L u g L
(9)
Theorem 2. The basic properties of the algebra M = ([0, 1][0,1] , t, u, 0, 1) follow. (1) (2) (3) (4)
f t f = f ; f u f = f (idempotent) f t g = g t f ; f u g = g u f (commutative) f t (g t h) = (f t g) t h; f u (g u h) = (f u g) u h (associative) 1 u f = f ; 0 t f = f (identities)
Notice that this list does not include the absorption laws or distributive laws. The algebra M is not a lattice. It does, however, contain two familiar subalgebras that are lattices. For a ∈ [0, 1], let a denote the characteristic function of {a}, that is, a (a) = 1 and a (b) = 0 for b 6= a. Let I = ([0, 1], ∨, ∧, 0, 1), the unit interval with its usual lattice structure. Let I[2] = ([0, 1][2] , ∨, ∧, (0, 0) , (1, 1)) denote the algebra of closed subintervals of [0, 1] with coordinatewise operations. The map I → M : a 7→ a
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is a monomorphism. Observe, for example, that a u b = a ∧ b, and a t b = a ∨ b. Thus the image of this map is a subalgebra of M isomorphic to I. We will denote this image by S, and call its elements singletons. The map R
I[2] → M : (a, b) 7→ aL ∧ b
(11) R
is also a monomorphism, and its image will be denoted by D. Its elements aL ∧ b are determined by the doubleton (a, b). Thus we have S⊂D⊂M
(12)
4. Automorphisms Here we investigate automorphisms of the algebra M = ([0, 1][0,1] , t, u, 0, 1). Automorphisms of an algebra are one-to-one mappings of that algebra onto itself that preserve the operations of that algebra. Thus ϕ is an automorphism of the algebra M if ϕ is a one-to-one mapping of [0, 1][0,1] onto itself such that ϕ (f t g) = ϕ (f ) t ϕ (g)
ϕ (f u g) = ϕ (f ) u ϕ (g) ¡ ¢ ϕ 0 =0 ¡ ¢ ϕ 1 =1
Actually, in this case, it is an easy exercise to show that if the first equation is satisfied, then so is the third; if the second is satisfied, so is the last. So to verify that ϕ is an automorphism, we need only that it is one-to-one and onto and preserves t and u. For an algebra A, Aut(A) will denote the set of all automorphisms of A. This set is a group under composition of maps. In particular, the inverse of an automorphism is an automorphism. Here are some pertinent facts needed about automorphisms α of the algebra ([0, 1], ∨, ∧, 0, 1). They are all fairly obvious. (1) A mapping α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1) if and only it is a one-to-one strictly increasing mapping of [0, 1] onto itself. (2) A mapping α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1) if and only it is an automorphism of ([0, 1], ∨, ∧). (3) A mapping α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1) if and only it is one-to-one and onto and satisfies x ≤ y if and only if α(x) ≤ α(y). (4) If α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1], then for any subset S of [0, 1], α (∨S) = ∨ (α (S)) and α (∧S) = ∧ (α (S)). (5) If α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1], then f 7→ αf and f 7→ f α are one-to-one maps of [0, 1][0,1] onto itself.
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4.1. The Automorphisms αL and αR There are two basic automorphisms of M that arise in a very natural way from each automorphism of I. These are defined in the following two propositions. Proposition 2. For α ∈ Aut ([0,¡ 1], ∨, ∧, 0, ¢1), and f ∈ [0, 1][0,1] , let αL (f ) = αf . Then αL is an automorphism of [0, 1][0,1] , • where • is any operation on [0, 1][0,1] that is the convolution of an operation on [0, 1]. Proof. We prove this for binary operations. The proof for n-ary operations is similar. For a binary operation ◦ on [0, 1], the proposition is proved as follows: Ã ! W (αL (f • g)) (x) = α ((f • g) (x)) = α (f (y) ∧ g(z) y◦z=x
W
=
α (f (y) ∧ g(z))
y◦z=x
W
=
y◦z=x
(α (f (y)) ∧ α (g (z)))
y◦z=x
((αf ) (y) ∧ (αg) (z))
=
W
= (αf • αg) (x) = (αL (f ) • αL (g)) (x) So for any set of operations on [0, 1][0,1] that come from convolutions of operations on [0, 1], and any automorphism α of ([0, 1], ∨, ∧, 0, 1), αL is an automorphism of the algebra consisting of [0, 1][0,1] and those operations on it. This is because α preserves ∨, and ∧, the operations used in the convolution process. We have done all this for binary operations, but this proposition holds for operations of any arity. Proposition 3. Let ◦ be an operation on [0, 1] and α an automorphism of (I, ◦). Then αR (f ) = f α is an automorphism of ([0, 1][0,1] , •), where • is the convolution of ◦. Proof. For binary operations, this follows from the following computations. W (αR (f • g)) (x) = (f • g) (α (x)) = (f (y) ∧ g (z)) W
=
y◦z=α(x)
α(y)◦α(z)=α(x)
=
W
α(y◦z)=α(x)
= =
W
(f (α (y)) ∧ g (α (z)))
(f (α (y)) ∧ g (α (z)))
y◦z=x
(f (α (y)) ∧ g (α (z)))
y◦z=x
((f α) (y) ∧ (gα) (z))
W
= (f α • gα) (x) = (αR (f ) • αR (g)) (x)
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As in the previous proposition, this holds for operations of any arity. Corollary 1. For any automorphism α of ([0, 1], ∨, ∧, 0, 1), both αL and αR are automorphisms of M = ([0, 1][0,1] , t, u, 0, 1). Thus M has many automorphisms, and at this point we know of none that do not come from compositions of these. These automorphisms have many special properties, which we now discuss. First some definitions. Definition 2. For any a ∈ [0, 1], let a be the map [0, 1] → [0, 1] given by a (x) = a, that is, a is the constant map with value a. Definition 3. A point function f is a function that is non-zero at exactly one point a. We say that f is a point function of a. A characteristic function f is a function that is 1 on some subset S of [0, 1] and 0 elsewhere. A characteristic function that is also a point function is called a singleton. For singletons, we write a to denote the characteristic function of the set {a}. Thus point functions can be written as a ∧ x. Also, the constant function with value 1 is both the characteristic function of [0, 1] and the constant function 1, and either notation may be used. In general, for a subset X of [0, 1], we identify X with its characteristic function. In particular, intervals such as [a, b], (a, b), and [a, b) will be identified with their respective characteristic functions. For any f ∈ M, f L and f R were defined earlier in Definition 1. The following are routine. Proposition 4. For f ∈ M , the following hold: (1) f t 1 = f L (2) f u 1 = f R (3) If f t g = f L for all f , then g = 1. Because of these identities, automorphisms ϕ that fix the map 1 commute with the L and R operations. For example, ϕ(f t 1) = ϕ(f L ) = ϕ(f ) t ϕ(1) = ϕ(f ) t 1 = ϕ(f )L . Note that an element f is constant, that is, f = a for some a ∈ [0, 1], if and only if f = f L = f R . Thus if ϕ ∈ Aut (M) satisfies ϕ (1) = 1, then f is constant if and only if ϕ (f ) is constant. One important fact that we will develop is that an arbitrary automorphism of M does indeed fix 1. But we need much more information. We note the following. Proposition 5. The following hold for an automorphism α of ([0, 1], ∨, ∧, 0, 1). (1) (2) (3) (4)
αL (a) = a for all singletons a. αR (a) = a for all constants a. αL (1) = αR (1) = 1. αL (f L ) = αL (f )L ; αL (f R ) = αL (f )R .
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(5) αR (f L ) = αR (f )L ; αR (f R ) = αR (f )R . (6) αR (a) = α−1 (a) for all singletons a. (7) αL (a) = α (a) for all constants a. Proof. (αL (a)) (x) = (αa)) (x) = α(a (x)) = 0 if x 6= a and = α(1) = 1 if x = a. And a(x) = 0 if x 6= a and = 1 if x = a. For item 2, (αR (a)) (x) = (aα)(x) = a(α(x)) = a, and a(x) = a. For item 3, (αL (1)) (x) = α(1 (x)) = α(1) = 1, and 1(x) = 1. That αR (1) = 1 follows from item 2. For item 4, αL (f L ) = αL (1 t f ) = αL (1) t αL (f ) = 1 t αL (f ) = (αL (f ))L . Item 5 follows similarly. For item 6, αR (a) (x) = aα (x) = 1 if α (x) = a, and = 0 otherwise. And α−1 (a) (x) = 1 if x = α−1 (a) and = 0 otherwise. For item 7, αL (a) (x) = α (a (x)) = α (a) = α (a) (x) for all x ∈ [0, 1]. The automorphisms αL and αR thus have quite nice properties. As elements of the group Aut(M), they satisfy the following. (1) (2) (3) (4) (5)
(αβ)L = αL βL (αβ)R = βR αR α βR¢ = βR αL ¡ L−1 −1 α = (αL ) ¡ −1 ¢L α = (αR )−1 R
In particular, the automorphisms of M of the form αL βR form a subgroup of Aut(M), and for all we know, may be all of Aut(M). 5. Normal Mappings and Convex Mappings Definition 4. A mapping f : [0, 1] → [0, 1] is normal if f is convex if for every x ≤ y ≤ z, f (y) ≥ f (x) ∧ f (z).
W
f (x) = 1. The map
x∈[0,1]
Note that this definition of normal does not require that f actually achieves its maximum. This is sometimes called “nearly normal.” We list several facts about normal mappings and convex mappings. Their proofs may be found in The algebra of fuzzy truth values.11 A function f is normal if and only if f L (1) = 1. The normal functions form a subalgebra N of M. A function f in M is convex if and only if f = f L ∧ f R . The convex functions form a subalgebra C of M. The subalgebra (L, t, u, 0, 1) of normal convex functions is a bounded distributive lattice. (6) The subalgebra L contains in a natural way isomorphic copies of both I and I[2] . (7) L is a maximal sublattice of M. (1) (2) (3) (4) (5)
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The lattice L plays a special role in type-2 fuzzy theory. The following theorem implies that all automorphisms of M induce automorphisms of L. Theorem 3. Let ϕ ∈ Aut (M) and f ∈ [0, 1][0,1] . If f is normal, then ϕ (f ) is normal; and if f is convex then ϕ (f ) is convex. ¡ ¢ Proof. Assume that f is normal, so that f L (1) = 1. Note that f t 1 = f ∨ 1 ∧ ¡ ¢ ¡ ¢ ¡ ¢ L f L ∧ 1 = 1. Now ϕ f t 1 = ϕ 1 = 1 = ϕ (f ) t 1 = ϕ (f ) ∨ 1 ∧ ϕ (f )L ∧ 1 , so ϕ (f )L (1) = 1, whence ϕ (f ) is normal. Assume that f is convex. Then ϕ (f ) u (ϕ (g) t ϕ (h)) = ϕ (f u (g t h)) = ϕ((f u g) t (f u h)) = (ϕ (f ) u ϕ (g)) t (ϕ (f ) u ϕ (h)). Earlier, we proved that the only functions that distribute over everything are convex functions.11 Therefore ϕ (f ) is convex. Corollary 2. Automorphisms of M induce automorphisms of L. This is a fundamental property that subalgebras may have: automorphisms of the containing algebra inducing automorphisms on the subalgebra. Definition 5. A subalgebra A of an algebra B is a characteristic subalgebra if every automorphism of B induces an automorphism of A. One should notice that if every automorphism of A maps B into B, then automorphisms of A induce automorphisms of B. So to show that a subalgebra is characteristic, it is enough to show that automorphisms of the containing algebra map the subalgebra into itself. Theorem 4. The algebras N, C, and L are characteristic subalgebras of M. Thus, in particular, L is a characteristic subalgebra of the algebra M. We will show later in this paper that S and D are also characteristic subalgebras of M, but we need much more information first. The algebra M has two partial orders of note. Definition 6. f v g if f u g = f ; f 4 g if f t g = g. Since L is a bounded distributive lattice, these two order relations coincide on L, but in general, they are quite different. Note that, since 1 is both normal and convex, by the previous two propositions, ϕ (1) is normal and convex for any automorphism ϕ of M. The function 1 sits between the monotone decreasing and increasing functions in the following sense: Proposition 6. If f is monotone decreasing and g is monotone increasing, then f u 1 = f and g t 1 = g. Thus f v 1 ¹ g. Proof. f u 1 = f R = f , and g t 1 = g L = g. So f v 1 ¹ g.
The notion of “between” is quite nonintuitive in the algebra M, as can be seen in the following proposition.
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Proposition 7. For all f ∈ M, f u 0 = 0 and f t 0 = 0. Thus 0 v f ¹ 0 for all f ∈ M. Theorem 5. The function ϕ defined by ϕ (f ) = f L ∧ f R is a retract of M onto the subalgebra of convex functions. It is also a retract of the subalgebra of normal functions onto L. Proof. We know the image of ϕ is the subalgebra of convex functions, and also that ϕ takes normal functions to normal functions. So we need only show that ϕ is an endomorphism. Let f, g ∈ M. Then ϕ (f t g) = (f t g)L ∧ (f t g)R ¡ ¢ ¡ ¢ = f L t gL ∧ f R t gR ¡ ¢ ¡¡ ¢ ¢ = f L ∧ g L ∧ f R ∨ g R ∧ f RL ∧ g RL ¡ ¢ ¡ ¢ = f R ∧ f L ∧ g L ∧ f RL ∧ g RL ∨ g R ∧ f L ∧ g L ∧ f RL ∧ g RL ¡ ¢ ¡ ¢ = f R ∧ f L ∧ gL ∨ gR ∧ f L ∧ gL
On the other hand,
¡ ¢ ¡ ¢ ϕ (f ) t ϕ (g) = f L ∧ f R t g L ∧ g R ³¡ ¢L ¡ ¢´ ³¡ ¢ ¡ ¢L ´ = f L ∧ f R ∧ gL ∧ gR ∨ f L ∧ f R ∧ gL ∧ gR ¡ ¡ ¢¢ ¡¡ ¢ ¢ = f L ∧ gL ∧ gR ∨ f L ∧ f R ∧ gL
which shows that
ϕ (f ) t ϕ (g) = ϕ (f t g)
(13)
Also, ϕ (f u g) = (f u g)L ∧ (f u g)R ¡ ¢ ¡ ¢ = f L u gL ∧ f R u gR ¡¡ ¢ ¢ ¡ ¢ = f L ∨ g L ∧ f RL ∧ g RL ∧ f R ∧ g R ¡ ¢ ¡ ¢ = f L ∧ f R ∧ g R ∧ f RL ∧ g RL ∨ g L ∧ f R ∧ g R ∧ f RL ∧ g RL ¡ ¢ ¡ ¢ = f L ∧ f R ∧ gR ∨ gL ∧ f R ∧ gR
and on the other hand, ¡ ¢ ¡ ¢ ϕ (f ) u ϕ (g) = f L ∧ f R u g L ∧ g R ³¡ ¢R ¡ ¢´ ³¡ ¢ ¡ ¢R ´ = f L ∧ f R ∧ gL ∧ gR ∨ f L ∧ f R ∧ gL ∧ gR ¡ ¢¢ ¡¡ ¢ ¢ ¡ = f R ∧ gL ∧ gR ∨ f L ∧ f R ∧ gR = ϕ (f u g)
which shows that ϕ (f ) u ϕ (g) = ϕ (f u g)
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¡ ¢ ¡ ¢ Also ϕ 1 = 1 and ϕ 0 = 0. Thus ϕ is an endomorphism of M. The function f → f L ∧ f R is a natural way to “convexify” a function, and the theorem says that this convexification is an idempotent endomorphism of M. 6. Irreducibles in M Definition 7. An element h ∈ M is meet irreducible if h = f u g implies f = h or g = h, join irreducible if h = f t g implies f = h or g = h, and irreducible if both meet and join irreducible. It is routine to verify that the properties join irreducible, meet irreducible, and irreducible are preserved under automorphisms. Identifying these irreducibles provides information about possible automorphisms. In this section, we investigate the irreducibles of M. The unit interval [0, 1] = 1 is of particular interest as an element of M. See, for example, Proposition 4 which implies that automorphisms ϕ that fix the map 1 commute with the L and R operations. However, 1 is not irreducible. Proposition 8. The characteristic function of the interval [0, 1] is neither join nor meet irreducible in M. ½ 0 if x = a Proof. Let a, b ∈ (0, 1) with a 6= b. Define f and g by f (x) = 1 otherwise ½ 0 if x = b and g (x) = . Then f ∨ g = 1 and f L = g L = f R = g R = 1. Thus 1 otherwise f t g = f u g = 1. Note that f and g used in the above proof are not convex. Lemma 1. Let f be a point function of a. (1) If f = g t h, then g (b) = h (b) = 0 for all b > a and either g or h is a point function of a. (2) If f = g u h, then g (b) = h (b) = 0 for all b < a and either g or h is a point function of a. Proof. If f is a point function of a, then f (a) = x 6= 0, and f (b) = 0 for b 6= a. Suppose f = g t h. Then f (a) = (g(a) ∨ h(a)) ∧ g L (a) ∧ hL (a) = x
So g(a) ∨ h(a) ≥ x, and both g L (a) and hL (a) ≥ x. For all b 6= a, ¢ ¡ ¢ ¡ f (b) = g L (b) ∧ h(b) ∨ g(b) ∧ hL (b) = 0
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(16)
If b > a, g L (b) ∧ hL (b) ≥ x which implies g(b) = h(b) = 0. Suppose that for some b < a, g(b) 6= 0. Then g L (b) 6= 0, so we must have hL (b) = 0 which means that h (c) = hL (c) = 0 for all c ≤ b. Then if for some b < c < a, h(c) 6= 0, we would
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need g L (c) = 0. But g L (c) ≥ g L (b) 6= 0. So either g or h is 0 for all b < a. Since they are both = 0 for all b > a, either g or h is a point function. The proof of the second half is similar. Proposition 9. A point function at a ∈ (0, 1) that has value less than 1 is neither meet nor join irreducible in M. Proof. Suppose f = a ∧ x for some x < 1. Let g = x ∧ [0, a]. Then g t a = (g ∨ a) ∧ g L ∧ aL = ((x ∧ [0, a]) ∨ a) ∧ x ∧ [a, 1] = f , since ((x ∧ [0, a]) ∨ a) (b) = 0 for b 6= a and ((x ∧ [0, a]) ∨ a) (a) ∧ x (a) ∧ [a, 1] (a) = 1 ∧ x ∧ 1 = x. Also g 6= f 6= a. Thus f is join reducible. Let h = x∧[a, 1]. A similar argument shows that f = h ua, so f is also meet reducible. Proposition 10. The singletons are irreducible in M. Proof. Suppose a = g th. By Lemma 1, either g or h is a point function of a, say it is g. Then g L (a) = 1 implies that g (a) = 1. Thus g = a. Thus a is join irreducible. Suppose a = g u h. Again by Lemma 1, either g or h is a point function of a, say it is g. Then g R (a) = 1 implies that g (a) = 1. Thus g = a. Thus a is also meet irreducible and hence irreducible. Proposition 11. The characteristic function of the two-element set {0, 1} is irreducible in M. Proof. If {0, a} = (f ∨ g) ∧ f L ∧ g L , then f L (0) = g L (0) = 1 implies f (0) = g (0) = 1, whence f L = g L = [0, 1]. Thus we must have f (x) ∨ g (x) = 0 for all x ∈ (0, 1) with x 6= a so that f (x) = g (x) = 0 for all x ∈ (0, 1) with x 6= a. But then f (a) ∨ g (a) = 1 implies that at least one of f or g equals {0, a}. Thus {0, a} is join irreducible. A symmetric argument proves that a two-element set {a, 1} is meet irreducible. It follows that {0, 1} is irreducible. Proposition 12. The function 0 is irreducible in M. Proof. Suppose 0 = f tg = (f ∨ g)∧f L ∧g L . If f (a) > 0 for some a, then g L (a) = 0 implies that g (b) = 0 for all b ≤ a. Suppose g (c) > 0 for some c > a. Then f L (b) > 0 for all b ≥ a implies f L (c) > 0 and we would have (f (c) ∨ g (c))∧f L (c)∧g L (c) > 0. Thus we must have g = 0. Thus 0 is join irreducible. A similar proof shows that 0 is meet irreducible. In the following series of lemmas we show that there are no other irreducibles in M. Lemma 2. If f (a) < f LR (a) for all a ∈ [0, 1] then f is meet or join reducible.
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Proof. In this case, there is either an increasing sequence a1 ≤ a2 ≤ . . . with lim f (ai ) = f LR or a decreasing sequence a1 ≥ a2 ≥ ... with lim f (bi ) = f LR . If there is an increasing sequence a1 ≤ a2 ≤ . . . with lim f (ai ) = f LR , pick a < b < c = lim ai , with f (a) 6= 0 and f (b) 6= 0. Define g (a) = 0 and g = f otherwise, and h (b) = 0 and h = f otherwise. Then g R = hR = f R and g ∨ h = f , so g u h = (g ∨ h) ∧ g R ∧ hR = f ∧ f R ∧ f R = f
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Since g 6= f and h 6= f , this shows that f is meet-reducible. If there is a decreasing sequence a1 ≥ a2 ≥ ... with lim f (ai ) = f LR , the symmetric construction gets g t h = f , and we know that f is join-reducible. Lemma 3. If there are three distinct elements a, b, c with f (a) = f LR (a) and with f (b) 6= 0 and f (c) 6= 0 then f is meet or join reducible. Proof. If b < c < a, define g = h = f except g (b) = 0 and h (c) = 0. Then g R = hR = f R , and g ∨ h = f , so g u h = (g ∨ h) ∧ g R ∧ hR = f ∧ f R ∧ f R = f
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Thus f is meet-reducible. If b < c < a, the symmetric construction gets g t h = f , so f is join-reducible. If neither of the above holds for any choice of points, then b < a < c and f = 0 at all other points. If f (c) ≥ f (a)∨f (b) or f (a) ≥ f (b)∨f (c), this was covered above. If f (b) > f (a) ∨ f (c), define g = f except g (a) = 0, h = f except h (c) = f (b). Then hR = f RL and g R = f R , and g u h = (g ∨ h) ∧ g R ∧ hR = (g ∨ h) ∧ f RL ∧ f R = f
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so f is meet-reducible. By a symmetric construction, f is also join-reducible. Proposition 13. If f is nonzero at exactly two points a < b, then f is meet or join reducible except in the case {a, b} = {0, 1} and f (a) = f (b) = 1. Proof. If f (a) ≤ f (b), 0 < a, and f (a) < 1, define g = f except g (a) = 1 and h = f except h (0) = f (a). Then g L (0) = 0 = f (0) and hL (a) = f (a), and g t h = (g ∨ h) ∧ g L ∧ hL = f , so f is join-reducible. If f (a) ≥ f (b) and b < 1, a symmetric construction gives g u h = (g ∨ h) ∧ g R ∧ hR = f , so f is meet-reducible. If f (a) ≤ f (b), b < 1, define g = f except g (a) = 0, and h = f except h (1) = 1. Then g ∨ h = f except at 1, g R = f R and hR = 1. Thus g u h = (g ∨ h) ∧ g R ∧ hR = (g ∨ h) ∧ g R = f , so f is meet-reducible. If f (a) ≥ f (b), 0 < a, a symmetric construction gives g t h = (g ∨ h) ∧ g L ∧ hL = (g ∨ h) ∧ g L = f , so f is join-reducible. This leaves a = 0 and b = 1, and 0 < f (0) ∧ f (1) ≤ 1. If 0 < f (0) < f (1) < 1, define g = f except g (0) = f (1), and h = f except h (1) = 1. Then g L = f LR and hL (0) = f (0), and gth = (g ∨ h)∧g L ∧hL = (g ∨ h)∧f LR ∧hL = f , so f is join-reducible. If 0 < f (1) < f (0) < 1 a symmetric construction shows that f is meet-reducible. If 0 < f (0) = f (1) < 1, define g = f except g (0) = 0 and
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h = f except h (1) = 1. Then g R = f RL and hR = 1, and g uh = (g ∨ h)∧g R ∧hR = (g ∨ h) ∧ f RL = f , so f is meet-reducible. A symmetric construction shows that f is also join-reducible. If 0 < f (0) ≤ f (1) = 1, and f = (g ∨ h) ∧ g L ∧ hL then either g (0) = f (0) or h (0) = f (0). Say g (0) = f (0). Then we must have h (0) ≥ f (0). For 0 < a < 1, g L ∧ hL > 0 so we must have g ∨ h = 0, then 1 = (g (1) ∨ h (1)) ∧ g L (1) ∧ hL (1) implies that g L (1) = hL (1) = 1. If f (0) < 1, then g (1) = 1 so g = f . If f (0) = 1, then h (0) = f (0). So g (1) ∨ h (1) = 1 will then force either g = f or h = f . Thus f is join-irreducible. If 0 < f (0) ≤ f (1) = 1, and f = (g ∨ h) ∧ g R ∧ hR , then g (1) = h (1) = 1, so g R = hR = 1 and f = g ∨ h. Thus either g (0) = 1 or h (0) = 1, implying g = f or h = f . Thus f is also meet-irreducible. Theorem 6. A function f in M is irreducible in M if and only if f is a singleton, f = 0, or f = {0, 1}. Because {0, 1} is the only non-convex irreducible, and 0 is the only convex irreducible in M that is not normal, we also have the following. Corollary 3. Every automorphism of M fixes both {0, 1} and 0. The fact that the singletons are the only convex normal irreducibles in M, together with Theorem 3, yields the following. Corollary 4. Every automorphism of M induces an automorphism of S by its action on singletons. That is, S is a characteristic subalgebra of M. The algebra D is also characteristic in M, but to show that, we need that automorphisms of M commute with the unary operations L and R, which is a consequence of the next section. 7. Irreducibles in L
¡ ¢ We now focus on the subalgebra L = L, u, t, 0, 1 , which is a bounded, distributive lattice. Definition 8. Call h ∈ L meet irreducible in L if h = f ug with f, g ∈ L implies f = h or g = h, join irreducible in L if h = f t g with f, g ∈ L implies f = h or g = h, and irreducible in L if it is both meet and join irreducible in L. Note that elements of L that are meet or join irreducible in M have the same property in L, but the converse need not hold. Proposition 14. In the subalgebra L, intervals [0, a] and [0, a) are join irreducible and not meet irreducible. The intervals (a, 1] and [a, 1] are meet irreducible and not join irreducible.
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Proof. Suppose that [0, a] = f t g = (f ∨ g) ∧ f L ∧ g L . Then f (b) ∨ g (b) = 1 for b ∈ [0, a] and f L = g L = [0, 1]. It follows that f (0) = g (0) = 1 and f (b) = g (b) = 0 for all b > a. Suppose that f (b) < 1 for some b ∈ (0, a]. Then, since f is convex, f (z) < 1 for all z ∈ [b, a]. The same would be true for g. It follows that g = [0, a]. With the obvious modifications, the same argument works for [0, a). Symmetric arguments show that (a, 1] and [a, 1] are meet irreducible. If 0 < b < 1, then [0, b] = [0, 1] u [b, b] so [0, b] is not meet irreducible in L. And [0, b] u (0, b) = ([0, b] ∨ (0, b)) ∧ [0, b] ∧ [0, b) = [0, b), so [0, b) is not meet irreducible in L. If 0 < a < 1 then [a, 1] = [0, 1] t [a, a] and (a, 1] = [a, 1] t (a, 1) so [a, 1] and (a, 1] are not join irreducible in L. Corollary 5. The interval 1 = [0, 1] is irreducible in the subalgebra L. We will eventually show that the intervals in Proposition 14 are the only join or meet irreducibles in L that are not irreducible. Lemma 4. If 0 < a < b < 1 then each of the intervals [a, b], (a, b), [a, b), (a, b] is neither join nor meet irreducible in L. Proof. Since a < b, there is a point c with a < c < b. Then [a, c] t [0, b] = ([a, a] ∨ [0, b]) ∧ [a, 1] ∧ [0, 1] = [0, b] ∧ [a, 1] = [a, b]
[a, c] t [0, b) = ([a, a] ∨ [0, b)) ∧ [a, 1] ∧ [0, 1] = [0, b) ∧ [a, 1] = [a, b)
(a, c) t [0, b] = ((a, c) ∨ [0, b]) ∧ (a, 1] ∧ [0, 1] = [0, b] ∧ (a, 1] = (a, b]
(a, c) t [0, b) = ((a, c) ∨ [0, b)) ∧ (a, 1] ∧ (0, 1] = [0, b) ∧ (a, 1] = (a, b) and [a, 1] u [c, b] = ([a, 1] ∨ [b, b]) ∧ [0, 1] ∧ [0, b] = [a, 1] ∧ [0, b] = [a, b]
(a, 1] u [c, b] = ((a, 1] ∨ [b, b]) ∧ [0, 1] ∧ [0, b] = (a, 1] ∧ [0, b] = (a, b]
[a, 1] u (c, b) = ([a, 1] ∨ (c, b)) ∧ [0, 1] ∧ [0, b) = [a, 1] ∧ [0, b) = [a, b)
(a, 1] u (c, b) = ((a, 1] ∨ (c, b)) ∧ [0, 1] ∧ [0, b) = (a, 1] ∧ [0, b) = (a, b) so none of these intervals is either join or meet irreducible in L. Lemma 5. If h is irreducible in L, then either h is a singleton or h = hR or h = hL . Proof. Suppose that h ∈ L is irreducible and h 6= hR and h 6= hL . If a ∈ [0, 1] with h (a) < hR (a), then (h ∧ (a, 1])R = hR . Then hL and h ∧ (a, 1] are convex and normal. Also, because h is convex, (h ∧ (a, 1])R = hR . So ¡ ¢ hL u(h ∧ (a, 1]) = hL ∨ (h ∧ (a, 1]) ∧hLR ∧(h ∧ (a, 1])R = hL ∧hRL ∧hR = h (20)
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Thus h = h ∧ (a, 1] and we see that h (a) = 0. Similarly, if b ∈ [0, 1], with h (b) < hL (b), then hR and h ∧ [0, b) are convex normal and (h ∧ [0, b))L = hL . So ¡ ¢ L hR t(h ∧ [0, b)) = hR ∨ (h ∧ [0, b)) ∧hRL ∧(h ∧ [0, b)) = hR ∧hRL ∧hL = h (21) Thus h = h ∧ [0, b) and we see that h (b) = 0. This implies that if h (c) 6= 0, we must have h (c) = hR (c) = hL (c) = 1. So h is an interval. But since h 6= hR and h 6= hL , by Lemma 4, h must be a singleton. Lemma 6. If h is irreducible in L, then h assumes at most two values. ½ h (a) if 0 ≤ x ≤ a Proof. Suppose h = hL , h (0) < h (a) < h (1). Define f (x) = h (x) if a ≤ x ≤ 1 ½ h (x) if 0 ≤ x ≤ a and g (x) = . Then f L (x) = f (x) and g L (x) = g (x), and h (1) if a ≤ x ≤ 1 f (x) ∨ g (x) ≥ h (x) for all x. Thus f t g = (f (x) ∨ g (x)) ∧ f (x) ∧ g (x) = f (x) ∧ g (x) = h (x)
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Note that both f and g are normal and convex. Thus if h is increasing and join irreducible, ½ h assumes at most two values. Suppose h = hR , h (0) > h (a) > h (1). Define ½ h (a) if a ≤ x ≤ 1 h (x) if a ≤ x ≤ 1 f (x) = and g (x) = . Then f R (x) = f (x) h (x) if 0 ≤ x ≤ a h (1) if 0 ≤ x ≤ a and g R (x) = g (x), and f (x) ∨ g (x) ≥ h (x) for all x. Thus f u g = (f (x) ∨ g (x)) ∧ f (x) ∧ g (x) = f (x) ∧ g (x) = h (x)
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Again, note that both f and g are normal and convex. Thus if h is decreasing and meet irreducible, h assumes at most two values. What are the irreducibles in L? So far we know that the singletons are irreducible in L, since they are irreducible in M, and the constant 1 is irreducible in L. The only other candidates are monotone step functions having the two values x and 1 for some x ∈ [0, 1]. We can eliminate most of such functions. Lemma 7. If f is irreducible in L and f (a) = f (1) = 1 with a < 1, or f (a) = f (0) = 1 with a > 0, then f = 1. Proof. Since f assumes at most two values, then in the first case, f = x ∨ [a, 1] or f = x ∨ (a, 1] for some x, a with a < 1. Define g (b) = f (b) for 0 ≤ b < 1 and g (1) = 0, and note that g is convex and normal. Then g t 1 = (g ∨ 1) ∧ g L ∧ 1 = g L = f , but f 6= g and f 6= 1, so f is not join-irreducible. A similar argument shows that f = x ∨ [0, a] is not irreducible. Let x < 1 and consider f = x ∨ (0, 1]. Let g = x ∨ (a, 1). Then g is convex and normal and g t 1 = (g ∨ 1) ∧ g L ∧ 1 = g L = f . Again, f 6= g and f 6= 1, so f is not irreducible. A similar argument shows that f = x ∨ [0, 1) is not irreducible. This leaves the functions of the form x ∨ 0 and x ∨ 1.
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Lemma 8. The functions of the form x ∨ 0 and x ∨ 1 are irreducible in L. Proof. Suppose that x ∨ 0 = f t g = (f ∨ g) ∧ f L ∧ g L with f, g ∈ L. Then f L (0) = g L (0) = 1 implies that f (0) = g (0) = 1, whence f L = g L = 1. So f t g = f ∨ g. Because f and g are both convex and 1 at 0, they are both decreasing. For a > 0, f (a) ∨ g (a) = x so in particular, f (a) ≤ x and g (a) ≤ x. At least one equals x. Say f (a) = x and g (a) < x. Then for 0 < b ≤ a, we have x ≥ f (b) ≥ f (a) = x. Moreover, for c > a, g (c) ≤ g (a) < x. We conclude that f = x ∨ 0 and it follows that x ∨ 0 is join irreducible. Suppose that x ∨ 0 = f u g = (f ∨ g) ∧ f R ∧ g R with f, g ∈ L. Since f and g are both normal, f R (0) = g R (0) = 1. Then (f ∨ g) (0) = 1 implies f (0) = 1 or g (0) = 1, say f (0) = 1. Then f is a decreasing function and f = f R . For a > 0, (f (a) ∨ g (a)) ∧ f (a) ∧ g R (a) = f (a) ∧ g R (a) = x, so f (a) ≥ x and g R (a) ≥ x. If f 6= x ∨ 0, there exists 0 < a with f (a) > x. Then g R (a) = x and for all c ≥ a, x = g R (a) ≥ g R (c) ≥ x so g (c) = g R (c) = x for all c ≥ a. Moreover, since f is decreasing, f (b) > x for all b ≤ a whence g (c) = g R (c) = x for all c > 0. Thus, since g is normal, it must be that g = x ∨ 0. Thus x ∨ 0 is meet irreducible, and hence irreducible in L. The proof that x ∨ 1 is irreducible is similar. Theorem 7. The irreducibles in L are the singletons a and the functions of the form x ∨ 0 and x ∨ 1 for 0 ≤ x ≤ 1. The functions of the latter two types form a chain. If x ≤ y, then 0vx∨0vy∨0v1vy∨1vx∨1v1
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The following proposition implies that the downset in L of a singleton contains no irreducibles other than singletons. Proposition 15. Let g ∈ L and c ∈ (0, 1). Then g v c if and only if g (a) = 0 for all a > c. If g is irreducible in L and g v c, then g = a for some a ≤ c. Proof. Suppose that g v c. then g = g u c = (g ∨ c) ∧ g R ∧ cR and c = g t c = (g ∨ c) ∧ g L ∧ cL . From the first of these we see that g (a) = 0 for all a > c, and from the second, we get g L (c) = 1. On the other hand, if if g (a) = 0 for all a > c and g L (c) = 1, then cL (a) = 0 for all a < c implies (g t c) (a) = 0 for all a < c. Also (g t c) (c) = 1. Moreover, g (a) = 0 for all a > c implies that (g ∨ c) (a) = 0 for all a > c and consequently (g t c) (a) = 0 for all a > c. Thus g t c = c so g v c. If g is irreducible in L, then by Theorem 7, g = a for some a ∈ [0, 1] or g = x ∨ 0 or g = x ∨ 1 for some 0 ≤ x ≤ 1. If g v c for some c ∈ (0, 1), then g (a) = 0 for all a > c implies either g = a for some a ≤ c or g = 0 ∨ 0 = 0. In either case, g is a singleton. We are now in a position to prove the following. Several of the steps in the sequence of equations in the proof rely on results in a previous paper.11
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Theorem 8. Every automorphism of M fixes 1. Proof. By Corollary 4, every automorphism ϕ of M induces an automorphism of I by its action on singletons. Since ϕ induces an automorphism of L, and the singletons and the functions of the form x ∨ 0 and x ∨ 1 are the only irreducibles in L, it must be that ϕ (1) = x ∨ 0 or ϕ (1) = x ∨ 1 for some x ∈ [0, 1]. We wish to show that x = 1. Assume that ϕ ¡ (1)¢ = x ∨ 1 and let f : [0, 1] → [0, 1]. Because f u 1 = f R , we have ϕ (1 u f ) = ϕ f R . On the other hand, ¡ ¢ ϕ (1 u f ) = ϕ (1) u ϕ (f ) = x ∨ 1 u ϕ (f ) ¡ ¢ = (x u ϕ (f )) ∨ 1 u ϕ (f ) = (x u ϕ (f )) ∨ ϕ (f ) ³ ´ = (x ∨ ϕ (f )) ∧ xR ∧ ϕ (f )R ∨ ϕ (f ) ´ ¡ ¢ ³ = ((x ∨ ϕ (f )) ∨ ϕ (f )) ∧ xR ∨ ϕ (f ) ∧ ϕ (f )R ∨ ϕ (f ) ³ ´ R = (x ∨ ϕ (f )) ∧ (x ∨ ϕ (f )) ∧ ϕ (f ) ³ ´ R = (x ∨ ϕ (f )) ∧ ϕ (f )
Suppose that x < 1, and choose f such that ϕ (f ) > x and f 6=³f R . (For ´ example, R choose ϕ (f ) to be not convex.) Then ϕ (1 u f ) = (x ∨ ϕ (f )) ∧ ϕ (f ) = ϕ (f ) ∧ ¡ R¢ R ϕ (f ) = ϕ (f ), which contradicts ϕ (1 u f ) = ϕ f . We conclude that x = 1 in this case. The proof for ϕ (1) = x ∨ 0 is similar, starting with f t 1 = f L . Corollary 6. Automorphisms of M commute with R and L. ¡ ¢ Proof. ϕ (1 u f ) = ϕ f R ϕ (1) u ϕ(f ) = 1 u ϕ(f )R , and similarly for L.
Since the constant functions are identified by f = f RL , we have the following.
Corollary 7. Automorphisms of M induce automorphisms of the unit interval by restricting to the constant functions, as well as by restricting to the singletons. Theorem 9. Automorphisms of M induce automorphisms of D. That is, D is a characteristic subalgebra of M. R
R
Proof. For a, b ∈ [0, 1], a ≤ b, and ϕ ∈ Aut(M), we need that ϕ(aL ∧ b ) = cL ∧ d L for some c and d ∈ [0, 1], c ≤ d. We show that ³ willRactually ³ ϕ(aR∧ ´ ´ b) = (ϕ(a)) ∧ ¡ ¢R R RL R ϕ(b) . First, note that a t b = a ∨ b ∧ aL ∧ b = a ∨ b ∧ aL = aL ∧ b
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R
since a ≤ b and so a ∨ b = b . Thus ³ ´ R R ϕ(aL ∧ b ) = ϕ a t b ³ R´ = ϕ (a) t ϕ b ³ R ´´ ³ ³ R ´´L ³ L ∧ (ϕ (a)) ∧ ϕ b = ϕ (a) ∨ ϕ b ³ ³ RL ´ ¡ ¡ ¢¢R ´ L = ϕ (a) ∨ ϕ b ∧ (ϕ (a)) ∧ ϕ b ¡ ¡ ¢¢R ∧ (ϕ (a))L = ϕ b We have the characteristic subalgebras N
C -
L ↑ D ↑ S
% (25)
of M = ([0, 1][0,1] , t, u, 0, 1), with inclusions as indicated. If the algebra M is also endowed with the convolution ∗ of the negation 0 of [0, 1], then S, D, L,N, and C together with the ∗ operation are still subalgebras, and in fact characteristic ones. That they are still subalgebras is easy to verify. They are characteristic since they are characteristic without the negation operations on the algebras, and are subalgebras with the negations operations on them. 8. Homomorphisms of Some Automorphism Groups Let S be the set of singletons in M, and S : I → S ⊂ M : a 7→ a
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Let K be the set of constant functions in M, and K : I → K ⊂ M : x 7→ x
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We have already seen that • The map S is an isomorphism from ([0, 1], ∨, ∧, 0, 1) to (S, t, u, 0, 1). • For any automorphism ϕ of M, ϕ (S) = S and ϕ (K) = K. Also, • The map K preserves minimum: xuy = x ∧ y. (But also note that xty = x ∧ y.)
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Define −1 P : Aut (I) × Aut (I) → Aut (M) : (α, β) 7→ αL βR ¡ −1 ¢ Q : Aut (M) → Aut (I) × Aut (I) : ϕ 7→ K ϕK, S −1 ϕS
(28)
Clearly S −1 ϕS ∈ Aut (I). Let x ≤ y. Then K −1 ϕK (x) = K −1¡ϕ (x) ¢and ¡ ¢ K −1 ϕK ¡(y)¢ = K −1 ϕ ¡y ¢. Now x u y = x ∧ y ¡=¢ x and ϕ (x) = ϕ x u y = −1 ϕ (x)uϕ ¡ ¢y = ϕ (x)∧ϕ y implies that ϕ (x) ≤ ϕ y and consequently K ϕ (x) ≤ −1 K ϕ y . Thus Q is well defined. Proposition 16. Both P and Q are group homomorphisms.
Proof. Let α, β, γ, δ ∈ Aut (I) and f ∈ M. Then P ((α, β) (γ, δ)) (f ) = P ((αγ, βδ)) (f ) −1
= (αγ)L (βδ)R (f ) = (αγ) f (βδ)
−1
P (α, β) P (γ, δ) (f ) = P (α, β) γf δ −1 = αγf δ −1 β −1 = (αγ) f (βδ)−1 so that P ((α, β) (γ, δ)) = P (α, β) P (γ, δ)
(29)
Also ¢ ¡ ¢ ¡ Q (ϕψ) = K −1 ϕψK, S −1 ϕψS = K −1 ϕKK −1 ψK, S −1 ϕSS −1 ψS ¡ ¢¡ ¢ = K −1 ϕK, S −1 ϕS K −1 ψK, S −1 ψS = Q (ϕ) Q (ψ)
(30)
Proposition 17. QP = identity. Proof. ¡ ¢ ¢ ¡ −1 −1 −1 For α, β ∈ Aut (I), QP (α, β) = Q αL βR K, S −1 αL βR S . Let = K −1 αL βR −1 −1 x ∈ [0, 1]. Then K −1 αL βR K (x) = K −1 αL βR x = K −1 αxβ −1 ¡= K −1 αx ¢ = −1 −1 −1 K α (x) = α (x) so K αL βR K = α. Let x, a ∈ [0, 1]. Note that a β −1 (x) = 1 when β −1 (x) = a, or x = β (a), and equals 0 otherwise, so aβ −1 = β (a). Then, −1 S −1 αL βR S (a) = S −1 αaβ −1 = S −1 aβ −1 = S −1 β (a) = β (a), which shows that −1 −1 S αL βR S = β. Thus QP (α, β) = (α, β). The situation just described is an instance of the following. Let G and H be ϕ σ groups and H → G → H be homomorphisms such that σϕ = 1, the identity automorphism of H. Then (ϕσ) (ϕσ) = ϕ (σϕ) σ = ϕσ, so the composition ϕσ = τ is an idempotent endomorphism of the group G. Its kernel N is a normal subgroup of G, and G = N τ (G). Every element ¡ ¡ of G ¢¢is uniquely the product of an element of N and one of τ (G). In fact g = gτ g −1 τ (g). This decomposition of G is not a direct product since τ (G) may not be normal, but a product of a normal subgroup with another subgroup.
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In our situation, this translates to Aut(M) being such a product of the kernel of P Q, which are those automorphisms of M that fix elementwise both K and S, and the subgroup {αL βR : α, β ∈ I}. We have noted that for all we know this latter subgroup may be all of Aut(M). This is so if and only if the kernel of P Q is just the identity automorphism. This is also equivalent to an automorphism being the identity if it fixes elementwise both K and S. Thus to get a counterexample, it suffices to find a non-trivial automorphism of M that fixes elementwise both C and S. There is a simpler situation at hand. The subalgebra S is characteristic, so we have a homomorphism Aut(M) → Aut(S), namely the restriction map. This mapping is onto. An automorphism of S is an α, and the map given by α → αL is a homomorphism Aut(S) → Aut(M). This composition Aut(M) → Aut(S) → Aut(M) is an idempotent endomorphism of Aut(M), so Aut(M) = KAut(S) where K is the kernel of the idempotent endomorphism, and Aut(S) is identified with its image in Aut(M). The same considerations apply to D identified as a characteristic subalgebra of M. For other subalgebras, the situation is more problematic. For example, the restriction map Aut(M) → Aut(L) is a homomorphism, but we do not know if it is onto. Its kernel is a normal subgroup consisting of those automorphisms fixing L. 9. Remarks Fuzzy sets of type-2 are generalizations of those of type-1, or ordinary fuzzy sets, and of interval-valued fuzzy sets. In just what precise mathematical sense is this the case? The basis of any kind of fuzzy sets is the algebra of their truth values. The algebra M is the algebra of truth values of fuzzy sets of type-2. The algebra of truth values of type-1 fuzzy sets is isomorphic to the subalgebra S of singletons, and the algebra of truth values of interval-valued fuzzy sets is isomorphic to the subalgebra D. So in fact M is a generalization. But these subalgebras are characteristic subalgebras of M. This means intuitively that there are no other isomorphic copies of these subalgebras sitting in M in the same way. Any automorphism of M takes these subalgebras onto themselves. So in this precise mathematical sense, type-2 is a generalization of type-1 and of interval-valued fuzzy sets. These remarks hold both with and without negations included as part of the algebraic structures. The characteristic subalgebra L of convex normal functions is of special interest. It is a De Morgan algebra, and is a maximal sublattice of M. This means that there are no sublattices between L and M. We do not know whether or not L is complete as a lattice. It was important above to get the element 1 fixed by automorphisms of M. It would be nice to prove the same thing for automorphisms of L, for then automorphisms of L would commute with R and L. To prove that automorphisms ϕ of L fix 1, it will suffice to show that ϕ(1) is not a singleton, and ϕ(1) is neither x ∨ 0 nor x ∨ 1 unless x = 1.
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1. L. Zadeh, “The concept of a linguistic variable and its application to approximate reasoning,” Inform Sci. 8 (1975) 199-249. 2. J. M. Mendel, Uncertain Rule-Based Fuzzy Logic Systems, Prentice Hall PTR, Upper Saddle River, NJ, 2001. 3. R. John, “Type-2 fuzzy sets: an appraisal of theory and applications,” Int. J. Uncertainty, Fuzziness Knowledge-Based Systems 6(6) (1998) 563-576. 4. N. Karnik and J. Mendel, “Operations on type-2 fuzzy sets,” Fuzzy Sets and Systems 122 (2001) 327-348. 5. M. F. Kawaguchi and M. Miyakoshi, “Extended t-norms as logical connectives of fuzzy truth values,” Multi. Val. Logic 8(1) (2002) 53-69. 6. J. Mendel and R. John, “Type-2 fuzzy sets made simple,” IEEE Transactions on Fuzzy Systems 10(2) (2002) 117-127. 7. M. Mizumoto and K. Tanaka, “Some properties of fuzzy sets of type-2,” Information and Control 31 (1976) 312-340. 8. M. Mizumoto and K. Tanaka, “Fuzzy sets of type-2 under algebraic product and algebraic sum,” Fuzzy Sets and Systems 5 (1981) 277-290. 9. M. Emoto and M. Mukaidono, “Necessary and sufficient conditions for fuzzy truth values to form a De Morgan algebra”, International Journal of Uncertainty, Fuzziness and Knowledge-Based Systems 7(4) (1999) 309-316. 10. J. Nieminen, “On the algebraic structure of fuzzy sets of type 2,” Kybernetika 13 (1977) 261-273. 11. C. Walker and E. Walker, “The algebra of fuzzy truth values”, Fuzzy Sets and Systems, 149(2) (2005) 309-347. 12. M. Gehrke, C. Walker, and E. Walker, “Some comments on interval-valued fuzzy sets”, International Journal of Intelligent Systems 11 (1996) 751-759. 13. M. Gehrke, C. Walker, and E. Walker, “De Morgan systems on the unit interval”, International Journal of Intelligent Systems 11 (1996) 733-750. 14. M. Gehrke, C. Walker, and E. Walker, “A mathematical setting for fuzzy logics”, International Journal of Uncertainty, Fuzziness and Knowledge-Based Systems, 5 (1997) 223-238. 15. H. Nguyen and E. Walker, A First Course in Fuzzy Logic (Chapman & Hall/CRC, Boca Raton, 2000).
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AUTOMORPHISMS OF THE ALGEBRA OF FUZZY TRUTH VALUES
CAROL L. WALKER and ELBERT A. WALKER Department of Mathematical Sciences New Mexico State University Las Cruces, New Mexico, USA [email protected] This paper is an investigation of the automorphisms of the algebra of truth values of type-2 fuzzy sets. This algebra contains isomorphic copies of the truth value algebras of type-1 and of interval-valued fuzzy sets. It is shown that these subalgebras are characteristic; that is, are carried onto themselves by automorphisms of the containing algebra of truth values of fuzzy sets. Some other relevant subalgebras are proved characteristic, including the subalgebra of convex normal functions. The principal tool in this study is the determination of various irreducible elements. Keywords : Automorphism, characteristic subalgebra, type-2 fuzzy set, interval-valued fuzzy set, algebra of truth values. Received (received date) Revised (revised date)
1. Introduction Type-2 fuzzy sets–that is, fuzzy sets with fuzzy sets as truth values, were introduced by Zadeh, extending the notion of ordinary fuzzy sets.1 There is a fair amount of literature on the subject. We mention Mendel’s book, which discusses both the theoretical and applied aspects of the topic, and its extensive bibliography.2 Some papers of interest are listed in the references.3 4 5 6 7 8 9 10 . In a recent paper, we endeavored to give a straightforward treatment of the mathematical basics of type-2 fuzzy sets.11 In earlier papers, we investigated mathematical basics of type-1 fuzzy sets and interval-valued fuzzy sets.12 13 14 The present paper is an investigation of the automorphisms of the algebra that underlies the basic theory of type-2 fuzzy sets. For any mathematical object, its group of symmetries is an object of interest. For algebraic structures, these symmetries are called automorphisms. The algebra that underlies type-2 fuzzy sets; that is, the algebra of fuzzy truth values, is rather complicated. It contains as subalgebras the algebras of truth values of ordinary fuzzy sets as well as that of interval-valued fuzzy sets.11 Of special interest is the subalgebra of convex normal functions, which form a bounded distributive lattice. A question of interest is whether or not automorphisms induce automorphisms of these various subalgebras; that is, whether or not these subalgebras are characteristic 1
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subalgebras. Central to this question is the determination of irreducible elements, since irreducibles map to irreducibles under automorphisms. Therefore, considerable effort is made in identifying irreducible elements. We begin with some necessary preliminaries. 2. Background A fuzzy subset A of a set S is a mapping A : S → [0, 1]. Such mappings are called fuzzy sets, or type-1 fuzzy sets. Operations on the set [0, 1]S of all fuzzy subsets of S come from operations on [0, 1]. Common operations on [0, 1] of interest in fuzzy theory are ∧, ∨, and 0 given by x ∧ y = min{x, y}
(1)
x ∨ y = max{x, y} x0 = 1 − x
The constants 0 and 1 are generally considered as part of the algebraic structure, technically being nullary operations. The operations ∧ and ∨ are binary operations, and 0 is a unary operation. So the algebra basic to fuzzy set theory is ([0, 1], ∨, ∧,0 , 0, 1). The corresponding operations on the set [0, 1]S of all fuzzy subsets of S are given pointwise by the formulas (A ∧ B)(s) = A(s) ∧ B(s)
(2)
(A ∨ B)(s) = A(s) ∨ B(s) 0
A0 (s) = (A(s))
and the two nullary operations are given by 1(s) = 1 and 0(s) = 0 for all s ∈ S. The following equations hold in the algebra ([0, 1], ∨, ∧,0 , 0, 1). (1) (2) (3) (4) (5) (6) (7) (8) (9)
x ∧ x = x; x ∨ x = x (idempotent) x ∧ y = y ∧ x; x ∨ y = y ∨ x (commutative) x ∧ (y ∧ z) = (x ∧ y) ∧ z; x ∨ (y ∨ z) = (x ∨ y) ∨ z (associative) x ∧ (x ∨ y) = x; x ∨ (x ∧ y) = x (absorption laws) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z); x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) (distributive laws) 0 ∨ x = x; 1 ∧ x = x (identities) x00 = x (involution) 0 0 (x ∧ y) = x0 ∨ y 0 ; (x ∨ y) = x0 ∧ y 0 (De Morgan’s laws) 0 0 x ∧ x ≤ y ∨ y (Kleene inequality)
This is summarized by saying that ([0, 1], ∨, ∧,0 , 0, 1) is a bounded distributive lattice with an involution 0 that satisfies¡ De Morgan’s laws¢and the Kleene inequality, that is, is a Kleene algebra. Thus [0, 1]S , ∨, ∧,0 , 0, 1 is also a Kleene algebra since it satisfies exactly the same equations as does ([0, 1], ∨, ∧,0 , 0, 1). An algebra satisfying equations 1-8 is a De Morgan algebra.
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An interval-valued fuzzy subset A of a set S is a mapping A : S → n o ³ ´S of all interval[0, 1][2] = (x, y) ∈ [0, 1]2 : x ≤ y . Operations on the set [0, 1][2] valued fuzzy subsets of S come pointwise from coordinatewise operations on [2] [0, 1] : (a, b)³ ∧ (c, d) = (a ∧ c, b ∧ d) ´and (a, b) ∨ (c, d) = (a ∨ c, b ∨ d). The al[2] gebra I[2] = [0, 1] , ∧, ∨, (0, 0), (1, 1) is also a bounded distributive lattice, with largest element (1, 1) and smallest element (0, 0). Thus the algebra of interval-valued fuzzy subsets of a set S is also a bounded distributive lattice. With the involution 0 (x, y) = (y 0 , x0 ), this is a De Morgan algebra (but not a Kleene algebra). The situation for type-2 fuzzy sets is similar except that fuzzy subsets of type-2 are mappings into a more complicated object than [0, 1] or [0, 1][2] , namely into the fuzzy subsets [0, 1][0,1] of [0, 1]. The operations on the type-2 fuzzy subsets ¢S ¡ [0, 1][0,1] of a set S will come pointwise from operations on [0, 1][0,1] . Operations are put on [0, 1][0,1] using operations on both the domain [0, 1] and the range [0, 1] of a mapping in [0, 1][0,1] . This is where the difficulty of type-2 fuzzy sets lies. In The algebra of fuzzy truth values, we studied operations on [0, 1][0,1] that are of interest in type-2 fuzzy set theory, and developed some of their algebraic properties.11 We now recall those operations. Let f and g be in [0, 1][0,1] . The elements f t g and f u g of [0, 1][0,1] are defined by the following equations. They are convolutions of the operations ∨ and ∧ on the domain [0, 1] with respect to the operations ∨ and ∧ on the range [0, 1]. W (f t g) (x) = y∨z=x (f (y) ∧ g(z)) (3) W (f u g) (x) = y∧z=x (f (y) ∧ g(z)) Denote by 1 the element of [0, 1][0,1] defined by ½ 0 if x 6= 1 1(x) = 1 if x = 1
(4)
Denote by 0 the map defined by 0(x) =
½
1 if x = 0 0 if x = 6 0
(5)
These elements 0 and 1 of [0, 1][0,1] are nullary operations, and can be obtained by convolution of the nullary operations 0 and 1. 3. The Algebra ([0, 1][0,1] , u, t, 0, 1)
At this point, we have the algebra M = ([0, 1][0,1] , t, u, 0, 1) with the operations t, u, 0, and 1 obtained by convolution of the operations ∨, ∧, 0, 1. This algebra, with the additional operation ∗ obtained by the convolution of 0 , is the basic algebra for type-2 fuzzy set theory, and is analogous to the algebra ([0, 1], ∧, ∨,0 , 0, 1) basic for type-1 or ordinary fuzzy set theory. Our main purpose in The algebra of fuzzy truth values was to study the algebra (M,∗ ) and some of its subalgebras.11
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Although we are interested in the algebra M, the set [0, 1][0,1] also has the pointwise operations ∨, ∧, 0, 1 on it coming from operations on the range [0, 1]: (f ∧ g) (x) ³ = f (x) ∧ g (x), (f ´ ∨ g) (x) = f (x) ∨ g (x), 0 (x) = 0, and 1 (x) = 1. The [0,1] algebra [0, 1] , ∧, ∨, 0, 1 is a lattice with order given by f ≤ g if f = f ∧ g, or equivalently, if g = f ∨ g. The operations t and u can be expressed in terms of the simpler pointwise ones and this was crucial in developing the properties of M. Definition 1. For f ∈ M, let f L and f R be the elements of M defined by f L (x) = ∨y≤x f (y)
(6)
R
f (x) = ∨y≥x f (y) Note that f L is monotone increasing and that f R is monotone decreasing. The following theorem expresses each of the convolution operations t and u directly in terms of pointwise operations in two alternate forms. Theorem 1. The following hold for all f, g ∈ M. ¢ ¢ ¡ ¡ f t g = f ∧ gL ∨ f L ∧ g ¡ ¢ = (f ∨ g) ∧ f L ∧ g L
¡ ¢ ¢ ¡ f u g = f ∧ gR ∨ f R ∧ g ¡ ¢ = (f ∨ g) ∧ f R ∧ g R
(7)
(8)
There are many relations between the various operations introduced above. We refer to The algebra of fuzzy truth values for details.11 The equations in the following proposition are crucial. Proposition 1. The following hold for f, g ∈ M. L
(f t g) = f L t g L R (f t g) = f R t g R
R
(f u g) = f R u g R L (f u g) = f L u g L
(9)
Theorem 2. The basic properties of the algebra M = ([0, 1][0,1] , t, u, 0, 1) follow. (1) (2) (3) (4)
f t f = f ; f u f = f (idempotent) f t g = g t f ; f u g = g u f (commutative) f t (g t h) = (f t g) t h; f u (g u h) = (f u g) u h (associative) 1 u f = f ; 0 t f = f (identities)
Notice that this list does not include the absorption laws or distributive laws. The algebra M is not a lattice. It does, however, contain two familiar subalgebras that are lattices. For a ∈ [0, 1], let a denote the characteristic function of {a}, that is, a (a) = 1 and a (b) = 0 for b 6= a. Let I = ([0, 1], ∨, ∧, 0, 1), the unit interval with its usual lattice structure. Let I[2] = ([0, 1][2] , ∨, ∧, (0, 0) , (1, 1)) denote the algebra of closed subintervals of [0, 1] with coordinatewise operations. The map I → M : a 7→ a
(10)
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is a monomorphism. Observe, for example, that a u b = a ∧ b, and a t b = a ∨ b. Thus the image of this map is a subalgebra of M isomorphic to I. We will denote this image by S, and call its elements singletons. The map R
I[2] → M : (a, b) 7→ aL ∧ b
(11) R
is also a monomorphism, and its image will be denoted by D. Its elements aL ∧ b are determined by the doubleton (a, b). Thus we have S⊂D⊂M
(12)
4. Automorphisms Here we investigate automorphisms of the algebra M = ([0, 1][0,1] , t, u, 0, 1). Automorphisms of an algebra are one-to-one mappings of that algebra onto itself that preserve the operations of that algebra. Thus ϕ is an automorphism of the algebra M if ϕ is a one-to-one mapping of [0, 1][0,1] onto itself such that ϕ (f t g) = ϕ (f ) t ϕ (g)
ϕ (f u g) = ϕ (f ) u ϕ (g) ¡ ¢ ϕ 0 =0 ¡ ¢ ϕ 1 =1
Actually, in this case, it is an easy exercise to show that if the first equation is satisfied, then so is the third; if the second is satisfied, so is the last. So to verify that ϕ is an automorphism, we need only that it is one-to-one and onto and preserves t and u. For an algebra A, Aut(A) will denote the set of all automorphisms of A. This set is a group under composition of maps. In particular, the inverse of an automorphism is an automorphism. Here are some pertinent facts needed about automorphisms α of the algebra ([0, 1], ∨, ∧, 0, 1). They are all fairly obvious. (1) A mapping α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1) if and only it is a one-to-one strictly increasing mapping of [0, 1] onto itself. (2) A mapping α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1) if and only it is an automorphism of ([0, 1], ∨, ∧). (3) A mapping α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1) if and only it is one-to-one and onto and satisfies x ≤ y if and only if α(x) ≤ α(y). (4) If α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1], then for any subset S of [0, 1], α (∨S) = ∨ (α (S)) and α (∧S) = ∧ (α (S)). (5) If α : [0, 1] → [0, 1] is an automorphism of ([0, 1], ∨, ∧, 0, 1], then f 7→ αf and f 7→ f α are one-to-one maps of [0, 1][0,1] onto itself.
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4.1. The Automorphisms αL and αR There are two basic automorphisms of M that arise in a very natural way from each automorphism of I. These are defined in the following two propositions. Proposition 2. For α ∈ Aut ([0,¡ 1], ∨, ∧, 0, ¢1), and f ∈ [0, 1][0,1] , let αL (f ) = αf . Then αL is an automorphism of [0, 1][0,1] , • where • is any operation on [0, 1][0,1] that is the convolution of an operation on [0, 1]. Proof. We prove this for binary operations. The proof for n-ary operations is similar. For a binary operation ◦ on [0, 1], the proposition is proved as follows: Ã ! W (αL (f • g)) (x) = α ((f • g) (x)) = α (f (y) ∧ g(z) y◦z=x
W
=
α (f (y) ∧ g(z))
y◦z=x
W
=
y◦z=x
(α (f (y)) ∧ α (g (z)))
y◦z=x
((αf ) (y) ∧ (αg) (z))
=
W
= (αf • αg) (x) = (αL (f ) • αL (g)) (x) So for any set of operations on [0, 1][0,1] that come from convolutions of operations on [0, 1], and any automorphism α of ([0, 1], ∨, ∧, 0, 1), αL is an automorphism of the algebra consisting of [0, 1][0,1] and those operations on it. This is because α preserves ∨, and ∧, the operations used in the convolution process. We have done all this for binary operations, but this proposition holds for operations of any arity. Proposition 3. Let ◦ be an operation on [0, 1] and α an automorphism of (I, ◦). Then αR (f ) = f α is an automorphism of ([0, 1][0,1] , •), where • is the convolution of ◦. Proof. For binary operations, this follows from the following computations. W (αR (f • g)) (x) = (f • g) (α (x)) = (f (y) ∧ g (z)) W
=
y◦z=α(x)
α(y)◦α(z)=α(x)
=
W
α(y◦z)=α(x)
= =
W
(f (α (y)) ∧ g (α (z)))
(f (α (y)) ∧ g (α (z)))
y◦z=x
(f (α (y)) ∧ g (α (z)))
y◦z=x
((f α) (y) ∧ (gα) (z))
W
= (f α • gα) (x) = (αR (f ) • αR (g)) (x)
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As in the previous proposition, this holds for operations of any arity. Corollary 1. For any automorphism α of ([0, 1], ∨, ∧, 0, 1), both αL and αR are automorphisms of M = ([0, 1][0,1] , t, u, 0, 1). Thus M has many automorphisms, and at this point we know of none that do not come from compositions of these. These automorphisms have many special properties, which we now discuss. First some definitions. Definition 2. For any a ∈ [0, 1], let a be the map [0, 1] → [0, 1] given by a (x) = a, that is, a is the constant map with value a. Definition 3. A point function f is a function that is non-zero at exactly one point a. We say that f is a point function of a. A characteristic function f is a function that is 1 on some subset S of [0, 1] and 0 elsewhere. A characteristic function that is also a point function is called a singleton. For singletons, we write a to denote the characteristic function of the set {a}. Thus point functions can be written as a ∧ x. Also, the constant function with value 1 is both the characteristic function of [0, 1] and the constant function 1, and either notation may be used. In general, for a subset X of [0, 1], we identify X with its characteristic function. In particular, intervals such as [a, b], (a, b), and [a, b) will be identified with their respective characteristic functions. For any f ∈ M, f L and f R were defined earlier in Definition 1. The following are routine. Proposition 4. For f ∈ M , the following hold: (1) f t 1 = f L (2) f u 1 = f R (3) If f t g = f L for all f , then g = 1. Because of these identities, automorphisms ϕ that fix the map 1 commute with the L and R operations. For example, ϕ(f t 1) = ϕ(f L ) = ϕ(f ) t ϕ(1) = ϕ(f ) t 1 = ϕ(f )L . Note that an element f is constant, that is, f = a for some a ∈ [0, 1], if and only if f = f L = f R . Thus if ϕ ∈ Aut (M) satisfies ϕ (1) = 1, then f is constant if and only if ϕ (f ) is constant. One important fact that we will develop is that an arbitrary automorphism of M does indeed fix 1. But we need much more information. We note the following. Proposition 5. The following hold for an automorphism α of ([0, 1], ∨, ∧, 0, 1). (1) (2) (3) (4)
αL (a) = a for all singletons a. αR (a) = a for all constants a. αL (1) = αR (1) = 1. αL (f L ) = αL (f )L ; αL (f R ) = αL (f )R .
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(5) αR (f L ) = αR (f )L ; αR (f R ) = αR (f )R . (6) αR (a) = α−1 (a) for all singletons a. (7) αL (a) = α (a) for all constants a. Proof. (αL (a)) (x) = (αa)) (x) = α(a (x)) = 0 if x 6= a and = α(1) = 1 if x = a. And a(x) = 0 if x 6= a and = 1 if x = a. For item 2, (αR (a)) (x) = (aα)(x) = a(α(x)) = a, and a(x) = a. For item 3, (αL (1)) (x) = α(1 (x)) = α(1) = 1, and 1(x) = 1. That αR (1) = 1 follows from item 2. For item 4, αL (f L ) = αL (1 t f ) = αL (1) t αL (f ) = 1 t αL (f ) = (αL (f ))L . Item 5 follows similarly. For item 6, αR (a) (x) = aα (x) = 1 if α (x) = a, and = 0 otherwise. And α−1 (a) (x) = 1 if x = α−1 (a) and = 0 otherwise. For item 7, αL (a) (x) = α (a (x)) = α (a) = α (a) (x) for all x ∈ [0, 1]. The automorphisms αL and αR thus have quite nice properties. As elements of the group Aut(M), they satisfy the following. (1) (2) (3) (4) (5)
(αβ)L = αL βL (αβ)R = βR αR α βR¢ = βR αL ¡ L−1 −1 α = (αL ) ¡ −1 ¢L α = (αR )−1 R
In particular, the automorphisms of M of the form αL βR form a subgroup of Aut(M), and for all we know, may be all of Aut(M). 5. Normal Mappings and Convex Mappings Definition 4. A mapping f : [0, 1] → [0, 1] is normal if f is convex if for every x ≤ y ≤ z, f (y) ≥ f (x) ∧ f (z).
W
f (x) = 1. The map
x∈[0,1]
Note that this definition of normal does not require that f actually achieves its maximum. This is sometimes called “nearly normal.” We list several facts about normal mappings and convex mappings. Their proofs may be found in The algebra of fuzzy truth values.11 A function f is normal if and only if f L (1) = 1. The normal functions form a subalgebra N of M. A function f in M is convex if and only if f = f L ∧ f R . The convex functions form a subalgebra C of M. The subalgebra (L, t, u, 0, 1) of normal convex functions is a bounded distributive lattice. (6) The subalgebra L contains in a natural way isomorphic copies of both I and I[2] . (7) L is a maximal sublattice of M. (1) (2) (3) (4) (5)
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The lattice L plays a special role in type-2 fuzzy theory. The following theorem implies that all automorphisms of M induce automorphisms of L. Theorem 3. Let ϕ ∈ Aut (M) and f ∈ [0, 1][0,1] . If f is normal, then ϕ (f ) is normal; and if f is convex then ϕ (f ) is convex. ¡ ¢ Proof. Assume that f is normal, so that f L (1) = 1. Note that f t 1 = f ∨ 1 ∧ ¡ ¢ ¡ ¢ ¡ ¢ L f L ∧ 1 = 1. Now ϕ f t 1 = ϕ 1 = 1 = ϕ (f ) t 1 = ϕ (f ) ∨ 1 ∧ ϕ (f )L ∧ 1 , so ϕ (f )L (1) = 1, whence ϕ (f ) is normal. Assume that f is convex. Then ϕ (f ) u (ϕ (g) t ϕ (h)) = ϕ (f u (g t h)) = ϕ((f u g) t (f u h)) = (ϕ (f ) u ϕ (g)) t (ϕ (f ) u ϕ (h)). Earlier, we proved that the only functions that distribute over everything are convex functions.11 Therefore ϕ (f ) is convex. Corollary 2. Automorphisms of M induce automorphisms of L. This is a fundamental property that subalgebras may have: automorphisms of the containing algebra inducing automorphisms on the subalgebra. Definition 5. A subalgebra A of an algebra B is a characteristic subalgebra if every automorphism of B induces an automorphism of A. One should notice that if every automorphism of A maps B into B, then automorphisms of A induce automorphisms of B. So to show that a subalgebra is characteristic, it is enough to show that automorphisms of the containing algebra map the subalgebra into itself. Theorem 4. The algebras N, C, and L are characteristic subalgebras of M. Thus, in particular, L is a characteristic subalgebra of the algebra M. We will show later in this paper that S and D are also characteristic subalgebras of M, but we need much more information first. The algebra M has two partial orders of note. Definition 6. f v g if f u g = f ; f 4 g if f t g = g. Since L is a bounded distributive lattice, these two order relations coincide on L, but in general, they are quite different. Note that, since 1 is both normal and convex, by the previous two propositions, ϕ (1) is normal and convex for any automorphism ϕ of M. The function 1 sits between the monotone decreasing and increasing functions in the following sense: Proposition 6. If f is monotone decreasing and g is monotone increasing, then f u 1 = f and g t 1 = g. Thus f v 1 ¹ g. Proof. f u 1 = f R = f , and g t 1 = g L = g. So f v 1 ¹ g.
The notion of “between” is quite nonintuitive in the algebra M, as can be seen in the following proposition.
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Proposition 7. For all f ∈ M, f u 0 = 0 and f t 0 = 0. Thus 0 v f ¹ 0 for all f ∈ M. Theorem 5. The function ϕ defined by ϕ (f ) = f L ∧ f R is a retract of M onto the subalgebra of convex functions. It is also a retract of the subalgebra of normal functions onto L. Proof. We know the image of ϕ is the subalgebra of convex functions, and also that ϕ takes normal functions to normal functions. So we need only show that ϕ is an endomorphism. Let f, g ∈ M. Then ϕ (f t g) = (f t g)L ∧ (f t g)R ¡ ¢ ¡ ¢ = f L t gL ∧ f R t gR ¡ ¢ ¡¡ ¢ ¢ = f L ∧ g L ∧ f R ∨ g R ∧ f RL ∧ g RL ¡ ¢ ¡ ¢ = f R ∧ f L ∧ g L ∧ f RL ∧ g RL ∨ g R ∧ f L ∧ g L ∧ f RL ∧ g RL ¡ ¢ ¡ ¢ = f R ∧ f L ∧ gL ∨ gR ∧ f L ∧ gL
On the other hand,
¡ ¢ ¡ ¢ ϕ (f ) t ϕ (g) = f L ∧ f R t g L ∧ g R ³¡ ¢L ¡ ¢´ ³¡ ¢ ¡ ¢L ´ = f L ∧ f R ∧ gL ∧ gR ∨ f L ∧ f R ∧ gL ∧ gR ¡ ¡ ¢¢ ¡¡ ¢ ¢ = f L ∧ gL ∧ gR ∨ f L ∧ f R ∧ gL
which shows that
ϕ (f ) t ϕ (g) = ϕ (f t g)
(13)
Also, ϕ (f u g) = (f u g)L ∧ (f u g)R ¡ ¢ ¡ ¢ = f L u gL ∧ f R u gR ¡¡ ¢ ¢ ¡ ¢ = f L ∨ g L ∧ f RL ∧ g RL ∧ f R ∧ g R ¡ ¢ ¡ ¢ = f L ∧ f R ∧ g R ∧ f RL ∧ g RL ∨ g L ∧ f R ∧ g R ∧ f RL ∧ g RL ¡ ¢ ¡ ¢ = f L ∧ f R ∧ gR ∨ gL ∧ f R ∧ gR
and on the other hand, ¡ ¢ ¡ ¢ ϕ (f ) u ϕ (g) = f L ∧ f R u g L ∧ g R ³¡ ¢R ¡ ¢´ ³¡ ¢ ¡ ¢R ´ = f L ∧ f R ∧ gL ∧ gR ∨ f L ∧ f R ∧ gL ∧ gR ¡ ¢¢ ¡¡ ¢ ¢ ¡ = f R ∧ gL ∧ gR ∨ f L ∧ f R ∧ gR = ϕ (f u g)
which shows that ϕ (f ) u ϕ (g) = ϕ (f u g)
(14)
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¡ ¢ ¡ ¢ Also ϕ 1 = 1 and ϕ 0 = 0. Thus ϕ is an endomorphism of M. The function f → f L ∧ f R is a natural way to “convexify” a function, and the theorem says that this convexification is an idempotent endomorphism of M. 6. Irreducibles in M Definition 7. An element h ∈ M is meet irreducible if h = f u g implies f = h or g = h, join irreducible if h = f t g implies f = h or g = h, and irreducible if both meet and join irreducible. It is routine to verify that the properties join irreducible, meet irreducible, and irreducible are preserved under automorphisms. Identifying these irreducibles provides information about possible automorphisms. In this section, we investigate the irreducibles of M. The unit interval [0, 1] = 1 is of particular interest as an element of M. See, for example, Proposition 4 which implies that automorphisms ϕ that fix the map 1 commute with the L and R operations. However, 1 is not irreducible. Proposition 8. The characteristic function of the interval [0, 1] is neither join nor meet irreducible in M. ½ 0 if x = a Proof. Let a, b ∈ (0, 1) with a 6= b. Define f and g by f (x) = 1 otherwise ½ 0 if x = b and g (x) = . Then f ∨ g = 1 and f L = g L = f R = g R = 1. Thus 1 otherwise f t g = f u g = 1. Note that f and g used in the above proof are not convex. Lemma 1. Let f be a point function of a. (1) If f = g t h, then g (b) = h (b) = 0 for all b > a and either g or h is a point function of a. (2) If f = g u h, then g (b) = h (b) = 0 for all b < a and either g or h is a point function of a. Proof. If f is a point function of a, then f (a) = x 6= 0, and f (b) = 0 for b 6= a. Suppose f = g t h. Then f (a) = (g(a) ∨ h(a)) ∧ g L (a) ∧ hL (a) = x
So g(a) ∨ h(a) ≥ x, and both g L (a) and hL (a) ≥ x. For all b 6= a, ¢ ¡ ¢ ¡ f (b) = g L (b) ∧ h(b) ∨ g(b) ∧ hL (b) = 0
(15)
(16)
If b > a, g L (b) ∧ hL (b) ≥ x which implies g(b) = h(b) = 0. Suppose that for some b < a, g(b) 6= 0. Then g L (b) 6= 0, so we must have hL (b) = 0 which means that h (c) = hL (c) = 0 for all c ≤ b. Then if for some b < c < a, h(c) 6= 0, we would
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need g L (c) = 0. But g L (c) ≥ g L (b) 6= 0. So either g or h is 0 for all b < a. Since they are both = 0 for all b > a, either g or h is a point function. The proof of the second half is similar. Proposition 9. A point function at a ∈ (0, 1) that has value less than 1 is neither meet nor join irreducible in M. Proof. Suppose f = a ∧ x for some x < 1. Let g = x ∧ [0, a]. Then g t a = (g ∨ a) ∧ g L ∧ aL = ((x ∧ [0, a]) ∨ a) ∧ x ∧ [a, 1] = f , since ((x ∧ [0, a]) ∨ a) (b) = 0 for b 6= a and ((x ∧ [0, a]) ∨ a) (a) ∧ x (a) ∧ [a, 1] (a) = 1 ∧ x ∧ 1 = x. Also g 6= f 6= a. Thus f is join reducible. Let h = x∧[a, 1]. A similar argument shows that f = h ua, so f is also meet reducible. Proposition 10. The singletons are irreducible in M. Proof. Suppose a = g th. By Lemma 1, either g or h is a point function of a, say it is g. Then g L (a) = 1 implies that g (a) = 1. Thus g = a. Thus a is join irreducible. Suppose a = g u h. Again by Lemma 1, either g or h is a point function of a, say it is g. Then g R (a) = 1 implies that g (a) = 1. Thus g = a. Thus a is also meet irreducible and hence irreducible. Proposition 11. The characteristic function of the two-element set {0, 1} is irreducible in M. Proof. If {0, a} = (f ∨ g) ∧ f L ∧ g L , then f L (0) = g L (0) = 1 implies f (0) = g (0) = 1, whence f L = g L = [0, 1]. Thus we must have f (x) ∨ g (x) = 0 for all x ∈ (0, 1) with x 6= a so that f (x) = g (x) = 0 for all x ∈ (0, 1) with x 6= a. But then f (a) ∨ g (a) = 1 implies that at least one of f or g equals {0, a}. Thus {0, a} is join irreducible. A symmetric argument proves that a two-element set {a, 1} is meet irreducible. It follows that {0, 1} is irreducible. Proposition 12. The function 0 is irreducible in M. Proof. Suppose 0 = f tg = (f ∨ g)∧f L ∧g L . If f (a) > 0 for some a, then g L (a) = 0 implies that g (b) = 0 for all b ≤ a. Suppose g (c) > 0 for some c > a. Then f L (b) > 0 for all b ≥ a implies f L (c) > 0 and we would have (f (c) ∨ g (c))∧f L (c)∧g L (c) > 0. Thus we must have g = 0. Thus 0 is join irreducible. A similar proof shows that 0 is meet irreducible. In the following series of lemmas we show that there are no other irreducibles in M. Lemma 2. If f (a) < f LR (a) for all a ∈ [0, 1] then f is meet or join reducible.
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Proof. In this case, there is either an increasing sequence a1 ≤ a2 ≤ . . . with lim f (ai ) = f LR or a decreasing sequence a1 ≥ a2 ≥ ... with lim f (bi ) = f LR . If there is an increasing sequence a1 ≤ a2 ≤ . . . with lim f (ai ) = f LR , pick a < b < c = lim ai , with f (a) 6= 0 and f (b) 6= 0. Define g (a) = 0 and g = f otherwise, and h (b) = 0 and h = f otherwise. Then g R = hR = f R and g ∨ h = f , so g u h = (g ∨ h) ∧ g R ∧ hR = f ∧ f R ∧ f R = f
(17)
Since g 6= f and h 6= f , this shows that f is meet-reducible. If there is a decreasing sequence a1 ≥ a2 ≥ ... with lim f (ai ) = f LR , the symmetric construction gets g t h = f , and we know that f is join-reducible. Lemma 3. If there are three distinct elements a, b, c with f (a) = f LR (a) and with f (b) 6= 0 and f (c) 6= 0 then f is meet or join reducible. Proof. If b < c < a, define g = h = f except g (b) = 0 and h (c) = 0. Then g R = hR = f R , and g ∨ h = f , so g u h = (g ∨ h) ∧ g R ∧ hR = f ∧ f R ∧ f R = f
(18)
Thus f is meet-reducible. If b < c < a, the symmetric construction gets g t h = f , so f is join-reducible. If neither of the above holds for any choice of points, then b < a < c and f = 0 at all other points. If f (c) ≥ f (a)∨f (b) or f (a) ≥ f (b)∨f (c), this was covered above. If f (b) > f (a) ∨ f (c), define g = f except g (a) = 0, h = f except h (c) = f (b). Then hR = f RL and g R = f R , and g u h = (g ∨ h) ∧ g R ∧ hR = (g ∨ h) ∧ f RL ∧ f R = f
(19)
so f is meet-reducible. By a symmetric construction, f is also join-reducible. Proposition 13. If f is nonzero at exactly two points a < b, then f is meet or join reducible except in the case {a, b} = {0, 1} and f (a) = f (b) = 1. Proof. If f (a) ≤ f (b), 0 < a, and f (a) < 1, define g = f except g (a) = 1 and h = f except h (0) = f (a). Then g L (0) = 0 = f (0) and hL (a) = f (a), and g t h = (g ∨ h) ∧ g L ∧ hL = f , so f is join-reducible. If f (a) ≥ f (b) and b < 1, a symmetric construction gives g u h = (g ∨ h) ∧ g R ∧ hR = f , so f is meet-reducible. If f (a) ≤ f (b), b < 1, define g = f except g (a) = 0, and h = f except h (1) = 1. Then g ∨ h = f except at 1, g R = f R and hR = 1. Thus g u h = (g ∨ h) ∧ g R ∧ hR = (g ∨ h) ∧ g R = f , so f is meet-reducible. If f (a) ≥ f (b), 0 < a, a symmetric construction gives g t h = (g ∨ h) ∧ g L ∧ hL = (g ∨ h) ∧ g L = f , so f is join-reducible. This leaves a = 0 and b = 1, and 0 < f (0) ∧ f (1) ≤ 1. If 0 < f (0) < f (1) < 1, define g = f except g (0) = f (1), and h = f except h (1) = 1. Then g L = f LR and hL (0) = f (0), and gth = (g ∨ h)∧g L ∧hL = (g ∨ h)∧f LR ∧hL = f , so f is join-reducible. If 0 < f (1) < f (0) < 1 a symmetric construction shows that f is meet-reducible. If 0 < f (0) = f (1) < 1, define g = f except g (0) = 0 and
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h = f except h (1) = 1. Then g R = f RL and hR = 1, and g uh = (g ∨ h)∧g R ∧hR = (g ∨ h) ∧ f RL = f , so f is meet-reducible. A symmetric construction shows that f is also join-reducible. If 0 < f (0) ≤ f (1) = 1, and f = (g ∨ h) ∧ g L ∧ hL then either g (0) = f (0) or h (0) = f (0). Say g (0) = f (0). Then we must have h (0) ≥ f (0). For 0 < a < 1, g L ∧ hL > 0 so we must have g ∨ h = 0, then 1 = (g (1) ∨ h (1)) ∧ g L (1) ∧ hL (1) implies that g L (1) = hL (1) = 1. If f (0) < 1, then g (1) = 1 so g = f . If f (0) = 1, then h (0) = f (0). So g (1) ∨ h (1) = 1 will then force either g = f or h = f . Thus f is join-irreducible. If 0 < f (0) ≤ f (1) = 1, and f = (g ∨ h) ∧ g R ∧ hR , then g (1) = h (1) = 1, so g R = hR = 1 and f = g ∨ h. Thus either g (0) = 1 or h (0) = 1, implying g = f or h = f . Thus f is also meet-irreducible. Theorem 6. A function f in M is irreducible in M if and only if f is a singleton, f = 0, or f = {0, 1}. Because {0, 1} is the only non-convex irreducible, and 0 is the only convex irreducible in M that is not normal, we also have the following. Corollary 3. Every automorphism of M fixes both {0, 1} and 0. The fact that the singletons are the only convex normal irreducibles in M, together with Theorem 3, yields the following. Corollary 4. Every automorphism of M induces an automorphism of S by its action on singletons. That is, S is a characteristic subalgebra of M. The algebra D is also characteristic in M, but to show that, we need that automorphisms of M commute with the unary operations L and R, which is a consequence of the next section. 7. Irreducibles in L
¡ ¢ We now focus on the subalgebra L = L, u, t, 0, 1 , which is a bounded, distributive lattice. Definition 8. Call h ∈ L meet irreducible in L if h = f ug with f, g ∈ L implies f = h or g = h, join irreducible in L if h = f t g with f, g ∈ L implies f = h or g = h, and irreducible in L if it is both meet and join irreducible in L. Note that elements of L that are meet or join irreducible in M have the same property in L, but the converse need not hold. Proposition 14. In the subalgebra L, intervals [0, a] and [0, a) are join irreducible and not meet irreducible. The intervals (a, 1] and [a, 1] are meet irreducible and not join irreducible.
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Proof. Suppose that [0, a] = f t g = (f ∨ g) ∧ f L ∧ g L . Then f (b) ∨ g (b) = 1 for b ∈ [0, a] and f L = g L = [0, 1]. It follows that f (0) = g (0) = 1 and f (b) = g (b) = 0 for all b > a. Suppose that f (b) < 1 for some b ∈ (0, a]. Then, since f is convex, f (z) < 1 for all z ∈ [b, a]. The same would be true for g. It follows that g = [0, a]. With the obvious modifications, the same argument works for [0, a). Symmetric arguments show that (a, 1] and [a, 1] are meet irreducible. If 0 < b < 1, then [0, b] = [0, 1] u [b, b] so [0, b] is not meet irreducible in L. And [0, b] u (0, b) = ([0, b] ∨ (0, b)) ∧ [0, b] ∧ [0, b) = [0, b), so [0, b) is not meet irreducible in L. If 0 < a < 1 then [a, 1] = [0, 1] t [a, a] and (a, 1] = [a, 1] t (a, 1) so [a, 1] and (a, 1] are not join irreducible in L. Corollary 5. The interval 1 = [0, 1] is irreducible in the subalgebra L. We will eventually show that the intervals in Proposition 14 are the only join or meet irreducibles in L that are not irreducible. Lemma 4. If 0 < a < b < 1 then each of the intervals [a, b], (a, b), [a, b), (a, b] is neither join nor meet irreducible in L. Proof. Since a < b, there is a point c with a < c < b. Then [a, c] t [0, b] = ([a, a] ∨ [0, b]) ∧ [a, 1] ∧ [0, 1] = [0, b] ∧ [a, 1] = [a, b]
[a, c] t [0, b) = ([a, a] ∨ [0, b)) ∧ [a, 1] ∧ [0, 1] = [0, b) ∧ [a, 1] = [a, b)
(a, c) t [0, b] = ((a, c) ∨ [0, b]) ∧ (a, 1] ∧ [0, 1] = [0, b] ∧ (a, 1] = (a, b]
(a, c) t [0, b) = ((a, c) ∨ [0, b)) ∧ (a, 1] ∧ (0, 1] = [0, b) ∧ (a, 1] = (a, b) and [a, 1] u [c, b] = ([a, 1] ∨ [b, b]) ∧ [0, 1] ∧ [0, b] = [a, 1] ∧ [0, b] = [a, b]
(a, 1] u [c, b] = ((a, 1] ∨ [b, b]) ∧ [0, 1] ∧ [0, b] = (a, 1] ∧ [0, b] = (a, b]
[a, 1] u (c, b) = ([a, 1] ∨ (c, b)) ∧ [0, 1] ∧ [0, b) = [a, 1] ∧ [0, b) = [a, b)
(a, 1] u (c, b) = ((a, 1] ∨ (c, b)) ∧ [0, 1] ∧ [0, b) = (a, 1] ∧ [0, b) = (a, b) so none of these intervals is either join or meet irreducible in L. Lemma 5. If h is irreducible in L, then either h is a singleton or h = hR or h = hL . Proof. Suppose that h ∈ L is irreducible and h 6= hR and h 6= hL . If a ∈ [0, 1] with h (a) < hR (a), then (h ∧ (a, 1])R = hR . Then hL and h ∧ (a, 1] are convex and normal. Also, because h is convex, (h ∧ (a, 1])R = hR . So ¡ ¢ hL u(h ∧ (a, 1]) = hL ∨ (h ∧ (a, 1]) ∧hLR ∧(h ∧ (a, 1])R = hL ∧hRL ∧hR = h (20)
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Thus h = h ∧ (a, 1] and we see that h (a) = 0. Similarly, if b ∈ [0, 1], with h (b) < hL (b), then hR and h ∧ [0, b) are convex normal and (h ∧ [0, b))L = hL . So ¡ ¢ L hR t(h ∧ [0, b)) = hR ∨ (h ∧ [0, b)) ∧hRL ∧(h ∧ [0, b)) = hR ∧hRL ∧hL = h (21) Thus h = h ∧ [0, b) and we see that h (b) = 0. This implies that if h (c) 6= 0, we must have h (c) = hR (c) = hL (c) = 1. So h is an interval. But since h 6= hR and h 6= hL , by Lemma 4, h must be a singleton. Lemma 6. If h is irreducible in L, then h assumes at most two values. ½ h (a) if 0 ≤ x ≤ a Proof. Suppose h = hL , h (0) < h (a) < h (1). Define f (x) = h (x) if a ≤ x ≤ 1 ½ h (x) if 0 ≤ x ≤ a and g (x) = . Then f L (x) = f (x) and g L (x) = g (x), and h (1) if a ≤ x ≤ 1 f (x) ∨ g (x) ≥ h (x) for all x. Thus f t g = (f (x) ∨ g (x)) ∧ f (x) ∧ g (x) = f (x) ∧ g (x) = h (x)
(22)
Note that both f and g are normal and convex. Thus if h is increasing and join irreducible, ½ h assumes at most two values. Suppose h = hR , h (0) > h (a) > h (1). Define ½ h (a) if a ≤ x ≤ 1 h (x) if a ≤ x ≤ 1 f (x) = and g (x) = . Then f R (x) = f (x) h (x) if 0 ≤ x ≤ a h (1) if 0 ≤ x ≤ a and g R (x) = g (x), and f (x) ∨ g (x) ≥ h (x) for all x. Thus f u g = (f (x) ∨ g (x)) ∧ f (x) ∧ g (x) = f (x) ∧ g (x) = h (x)
(23)
Again, note that both f and g are normal and convex. Thus if h is decreasing and meet irreducible, h assumes at most two values. What are the irreducibles in L? So far we know that the singletons are irreducible in L, since they are irreducible in M, and the constant 1 is irreducible in L. The only other candidates are monotone step functions having the two values x and 1 for some x ∈ [0, 1]. We can eliminate most of such functions. Lemma 7. If f is irreducible in L and f (a) = f (1) = 1 with a < 1, or f (a) = f (0) = 1 with a > 0, then f = 1. Proof. Since f assumes at most two values, then in the first case, f = x ∨ [a, 1] or f = x ∨ (a, 1] for some x, a with a < 1. Define g (b) = f (b) for 0 ≤ b < 1 and g (1) = 0, and note that g is convex and normal. Then g t 1 = (g ∨ 1) ∧ g L ∧ 1 = g L = f , but f 6= g and f 6= 1, so f is not join-irreducible. A similar argument shows that f = x ∨ [0, a] is not irreducible. Let x < 1 and consider f = x ∨ (0, 1]. Let g = x ∨ (a, 1). Then g is convex and normal and g t 1 = (g ∨ 1) ∧ g L ∧ 1 = g L = f . Again, f 6= g and f 6= 1, so f is not irreducible. A similar argument shows that f = x ∨ [0, 1) is not irreducible. This leaves the functions of the form x ∨ 0 and x ∨ 1.
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Lemma 8. The functions of the form x ∨ 0 and x ∨ 1 are irreducible in L. Proof. Suppose that x ∨ 0 = f t g = (f ∨ g) ∧ f L ∧ g L with f, g ∈ L. Then f L (0) = g L (0) = 1 implies that f (0) = g (0) = 1, whence f L = g L = 1. So f t g = f ∨ g. Because f and g are both convex and 1 at 0, they are both decreasing. For a > 0, f (a) ∨ g (a) = x so in particular, f (a) ≤ x and g (a) ≤ x. At least one equals x. Say f (a) = x and g (a) < x. Then for 0 < b ≤ a, we have x ≥ f (b) ≥ f (a) = x. Moreover, for c > a, g (c) ≤ g (a) < x. We conclude that f = x ∨ 0 and it follows that x ∨ 0 is join irreducible. Suppose that x ∨ 0 = f u g = (f ∨ g) ∧ f R ∧ g R with f, g ∈ L. Since f and g are both normal, f R (0) = g R (0) = 1. Then (f ∨ g) (0) = 1 implies f (0) = 1 or g (0) = 1, say f (0) = 1. Then f is a decreasing function and f = f R . For a > 0, (f (a) ∨ g (a)) ∧ f (a) ∧ g R (a) = f (a) ∧ g R (a) = x, so f (a) ≥ x and g R (a) ≥ x. If f 6= x ∨ 0, there exists 0 < a with f (a) > x. Then g R (a) = x and for all c ≥ a, x = g R (a) ≥ g R (c) ≥ x so g (c) = g R (c) = x for all c ≥ a. Moreover, since f is decreasing, f (b) > x for all b ≤ a whence g (c) = g R (c) = x for all c > 0. Thus, since g is normal, it must be that g = x ∨ 0. Thus x ∨ 0 is meet irreducible, and hence irreducible in L. The proof that x ∨ 1 is irreducible is similar. Theorem 7. The irreducibles in L are the singletons a and the functions of the form x ∨ 0 and x ∨ 1 for 0 ≤ x ≤ 1. The functions of the latter two types form a chain. If x ≤ y, then 0vx∨0vy∨0v1vy∨1vx∨1v1
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The following proposition implies that the downset in L of a singleton contains no irreducibles other than singletons. Proposition 15. Let g ∈ L and c ∈ (0, 1). Then g v c if and only if g (a) = 0 for all a > c. If g is irreducible in L and g v c, then g = a for some a ≤ c. Proof. Suppose that g v c. then g = g u c = (g ∨ c) ∧ g R ∧ cR and c = g t c = (g ∨ c) ∧ g L ∧ cL . From the first of these we see that g (a) = 0 for all a > c, and from the second, we get g L (c) = 1. On the other hand, if if g (a) = 0 for all a > c and g L (c) = 1, then cL (a) = 0 for all a < c implies (g t c) (a) = 0 for all a < c. Also (g t c) (c) = 1. Moreover, g (a) = 0 for all a > c implies that (g ∨ c) (a) = 0 for all a > c and consequently (g t c) (a) = 0 for all a > c. Thus g t c = c so g v c. If g is irreducible in L, then by Theorem 7, g = a for some a ∈ [0, 1] or g = x ∨ 0 or g = x ∨ 1 for some 0 ≤ x ≤ 1. If g v c for some c ∈ (0, 1), then g (a) = 0 for all a > c implies either g = a for some a ≤ c or g = 0 ∨ 0 = 0. In either case, g is a singleton. We are now in a position to prove the following. Several of the steps in the sequence of equations in the proof rely on results in a previous paper.11
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Theorem 8. Every automorphism of M fixes 1. Proof. By Corollary 4, every automorphism ϕ of M induces an automorphism of I by its action on singletons. Since ϕ induces an automorphism of L, and the singletons and the functions of the form x ∨ 0 and x ∨ 1 are the only irreducibles in L, it must be that ϕ (1) = x ∨ 0 or ϕ (1) = x ∨ 1 for some x ∈ [0, 1]. We wish to show that x = 1. Assume that ϕ ¡ (1)¢ = x ∨ 1 and let f : [0, 1] → [0, 1]. Because f u 1 = f R , we have ϕ (1 u f ) = ϕ f R . On the other hand, ¡ ¢ ϕ (1 u f ) = ϕ (1) u ϕ (f ) = x ∨ 1 u ϕ (f ) ¡ ¢ = (x u ϕ (f )) ∨ 1 u ϕ (f ) = (x u ϕ (f )) ∨ ϕ (f ) ³ ´ = (x ∨ ϕ (f )) ∧ xR ∧ ϕ (f )R ∨ ϕ (f ) ´ ¡ ¢ ³ = ((x ∨ ϕ (f )) ∨ ϕ (f )) ∧ xR ∨ ϕ (f ) ∧ ϕ (f )R ∨ ϕ (f ) ³ ´ R = (x ∨ ϕ (f )) ∧ (x ∨ ϕ (f )) ∧ ϕ (f ) ³ ´ R = (x ∨ ϕ (f )) ∧ ϕ (f )
Suppose that x < 1, and choose f such that ϕ (f ) > x and f 6=³f R . (For ´ example, R choose ϕ (f ) to be not convex.) Then ϕ (1 u f ) = (x ∨ ϕ (f )) ∧ ϕ (f ) = ϕ (f ) ∧ ¡ R¢ R ϕ (f ) = ϕ (f ), which contradicts ϕ (1 u f ) = ϕ f . We conclude that x = 1 in this case. The proof for ϕ (1) = x ∨ 0 is similar, starting with f t 1 = f L . Corollary 6. Automorphisms of M commute with R and L. ¡ ¢ Proof. ϕ (1 u f ) = ϕ f R ϕ (1) u ϕ(f ) = 1 u ϕ(f )R , and similarly for L.
Since the constant functions are identified by f = f RL , we have the following.
Corollary 7. Automorphisms of M induce automorphisms of the unit interval by restricting to the constant functions, as well as by restricting to the singletons. Theorem 9. Automorphisms of M induce automorphisms of D. That is, D is a characteristic subalgebra of M. R
R
Proof. For a, b ∈ [0, 1], a ≤ b, and ϕ ∈ Aut(M), we need that ϕ(aL ∧ b ) = cL ∧ d L for some c and d ∈ [0, 1], c ≤ d. We show that ³ willRactually ³ ϕ(aR∧ ´ ´ b) = (ϕ(a)) ∧ ¡ ¢R R RL R ϕ(b) . First, note that a t b = a ∨ b ∧ aL ∧ b = a ∨ b ∧ aL = aL ∧ b
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R
since a ≤ b and so a ∨ b = b . Thus ³ ´ R R ϕ(aL ∧ b ) = ϕ a t b ³ R´ = ϕ (a) t ϕ b ³ R ´´ ³ ³ R ´´L ³ L ∧ (ϕ (a)) ∧ ϕ b = ϕ (a) ∨ ϕ b ³ ³ RL ´ ¡ ¡ ¢¢R ´ L = ϕ (a) ∨ ϕ b ∧ (ϕ (a)) ∧ ϕ b ¡ ¡ ¢¢R ∧ (ϕ (a))L = ϕ b We have the characteristic subalgebras N
C -
L ↑ D ↑ S
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of M = ([0, 1][0,1] , t, u, 0, 1), with inclusions as indicated. If the algebra M is also endowed with the convolution ∗ of the negation 0 of [0, 1], then S, D, L,N, and C together with the ∗ operation are still subalgebras, and in fact characteristic ones. That they are still subalgebras is easy to verify. They are characteristic since they are characteristic without the negation operations on the algebras, and are subalgebras with the negations operations on them. 8. Homomorphisms of Some Automorphism Groups Let S be the set of singletons in M, and S : I → S ⊂ M : a 7→ a
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Let K be the set of constant functions in M, and K : I → K ⊂ M : x 7→ x
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We have already seen that • The map S is an isomorphism from ([0, 1], ∨, ∧, 0, 1) to (S, t, u, 0, 1). • For any automorphism ϕ of M, ϕ (S) = S and ϕ (K) = K. Also, • The map K preserves minimum: xuy = x ∧ y. (But also note that xty = x ∧ y.)
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Define −1 P : Aut (I) × Aut (I) → Aut (M) : (α, β) 7→ αL βR ¡ −1 ¢ Q : Aut (M) → Aut (I) × Aut (I) : ϕ 7→ K ϕK, S −1 ϕS
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Clearly S −1 ϕS ∈ Aut (I). Let x ≤ y. Then K −1 ϕK (x) = K −1¡ϕ (x) ¢and ¡ ¢ K −1 ϕK ¡(y)¢ = K −1 ϕ ¡y ¢. Now x u y = x ∧ y ¡=¢ x and ϕ (x) = ϕ x u y = −1 ϕ (x)uϕ ¡ ¢y = ϕ (x)∧ϕ y implies that ϕ (x) ≤ ϕ y and consequently K ϕ (x) ≤ −1 K ϕ y . Thus Q is well defined. Proposition 16. Both P and Q are group homomorphisms.
Proof. Let α, β, γ, δ ∈ Aut (I) and f ∈ M. Then P ((α, β) (γ, δ)) (f ) = P ((αγ, βδ)) (f ) −1
= (αγ)L (βδ)R (f ) = (αγ) f (βδ)
−1
P (α, β) P (γ, δ) (f ) = P (α, β) γf δ −1 = αγf δ −1 β −1 = (αγ) f (βδ)−1 so that P ((α, β) (γ, δ)) = P (α, β) P (γ, δ)
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Also ¢ ¡ ¢ ¡ Q (ϕψ) = K −1 ϕψK, S −1 ϕψS = K −1 ϕKK −1 ψK, S −1 ϕSS −1 ψS ¡ ¢¡ ¢ = K −1 ϕK, S −1 ϕS K −1 ψK, S −1 ψS = Q (ϕ) Q (ψ)
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Proposition 17. QP = identity. Proof. ¡ ¢ ¢ ¡ −1 −1 −1 For α, β ∈ Aut (I), QP (α, β) = Q αL βR K, S −1 αL βR S . Let = K −1 αL βR −1 −1 x ∈ [0, 1]. Then K −1 αL βR K (x) = K −1 αL βR x = K −1 αxβ −1 ¡= K −1 αx ¢ = −1 −1 −1 K α (x) = α (x) so K αL βR K = α. Let x, a ∈ [0, 1]. Note that a β −1 (x) = 1 when β −1 (x) = a, or x = β (a), and equals 0 otherwise, so aβ −1 = β (a). Then, −1 S −1 αL βR S (a) = S −1 αaβ −1 = S −1 aβ −1 = S −1 β (a) = β (a), which shows that −1 −1 S αL βR S = β. Thus QP (α, β) = (α, β). The situation just described is an instance of the following. Let G and H be ϕ σ groups and H → G → H be homomorphisms such that σϕ = 1, the identity automorphism of H. Then (ϕσ) (ϕσ) = ϕ (σϕ) σ = ϕσ, so the composition ϕσ = τ is an idempotent endomorphism of the group G. Its kernel N is a normal subgroup of G, and G = N τ (G). Every element ¡ ¡ of G ¢¢is uniquely the product of an element of N and one of τ (G). In fact g = gτ g −1 τ (g). This decomposition of G is not a direct product since τ (G) may not be normal, but a product of a normal subgroup with another subgroup.
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In our situation, this translates to Aut(M) being such a product of the kernel of P Q, which are those automorphisms of M that fix elementwise both K and S, and the subgroup {αL βR : α, β ∈ I}. We have noted that for all we know this latter subgroup may be all of Aut(M). This is so if and only if the kernel of P Q is just the identity automorphism. This is also equivalent to an automorphism being the identity if it fixes elementwise both K and S. Thus to get a counterexample, it suffices to find a non-trivial automorphism of M that fixes elementwise both C and S. There is a simpler situation at hand. The subalgebra S is characteristic, so we have a homomorphism Aut(M) → Aut(S), namely the restriction map. This mapping is onto. An automorphism of S is an α, and the map given by α → αL is a homomorphism Aut(S) → Aut(M). This composition Aut(M) → Aut(S) → Aut(M) is an idempotent endomorphism of Aut(M), so Aut(M) = KAut(S) where K is the kernel of the idempotent endomorphism, and Aut(S) is identified with its image in Aut(M). The same considerations apply to D identified as a characteristic subalgebra of M. For other subalgebras, the situation is more problematic. For example, the restriction map Aut(M) → Aut(L) is a homomorphism, but we do not know if it is onto. Its kernel is a normal subgroup consisting of those automorphisms fixing L. 9. Remarks Fuzzy sets of type-2 are generalizations of those of type-1, or ordinary fuzzy sets, and of interval-valued fuzzy sets. In just what precise mathematical sense is this the case? The basis of any kind of fuzzy sets is the algebra of their truth values. The algebra M is the algebra of truth values of fuzzy sets of type-2. The algebra of truth values of type-1 fuzzy sets is isomorphic to the subalgebra S of singletons, and the algebra of truth values of interval-valued fuzzy sets is isomorphic to the subalgebra D. So in fact M is a generalization. But these subalgebras are characteristic subalgebras of M. This means intuitively that there are no other isomorphic copies of these subalgebras sitting in M in the same way. Any automorphism of M takes these subalgebras onto themselves. So in this precise mathematical sense, type-2 is a generalization of type-1 and of interval-valued fuzzy sets. These remarks hold both with and without negations included as part of the algebraic structures. The characteristic subalgebra L of convex normal functions is of special interest. It is a De Morgan algebra, and is a maximal sublattice of M. This means that there are no sublattices between L and M. We do not know whether or not L is complete as a lattice. It was important above to get the element 1 fixed by automorphisms of M. It would be nice to prove the same thing for automorphisms of L, for then automorphisms of L would commute with R and L. To prove that automorphisms ϕ of L fix 1, it will suffice to show that ϕ(1) is not a singleton, and ϕ(1) is neither x ∨ 0 nor x ∨ 1 unless x = 1.
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1. L. Zadeh, “The concept of a linguistic variable and its application to approximate reasoning,” Inform Sci. 8 (1975) 199-249. 2. J. M. Mendel, Uncertain Rule-Based Fuzzy Logic Systems, Prentice Hall PTR, Upper Saddle River, NJ, 2001. 3. R. John, “Type-2 fuzzy sets: an appraisal of theory and applications,” Int. J. Uncertainty, Fuzziness Knowledge-Based Systems 6(6) (1998) 563-576. 4. N. Karnik and J. Mendel, “Operations on type-2 fuzzy sets,” Fuzzy Sets and Systems 122 (2001) 327-348. 5. M. F. Kawaguchi and M. Miyakoshi, “Extended t-norms as logical connectives of fuzzy truth values,” Multi. Val. Logic 8(1) (2002) 53-69. 6. J. Mendel and R. John, “Type-2 fuzzy sets made simple,” IEEE Transactions on Fuzzy Systems 10(2) (2002) 117-127. 7. M. Mizumoto and K. Tanaka, “Some properties of fuzzy sets of type-2,” Information and Control 31 (1976) 312-340. 8. M. Mizumoto and K. Tanaka, “Fuzzy sets of type-2 under algebraic product and algebraic sum,” Fuzzy Sets and Systems 5 (1981) 277-290. 9. M. Emoto and M. Mukaidono, “Necessary and sufficient conditions for fuzzy truth values to form a De Morgan algebra”, International Journal of Uncertainty, Fuzziness and Knowledge-Based Systems 7(4) (1999) 309-316. 10. J. Nieminen, “On the algebraic structure of fuzzy sets of type 2,” Kybernetika 13 (1977) 261-273. 11. C. Walker and E. Walker, “The algebra of fuzzy truth values”, Fuzzy Sets and Systems, 149(2) (2005) 309-347. 12. M. Gehrke, C. Walker, and E. Walker, “Some comments on interval-valued fuzzy sets”, International Journal of Intelligent Systems 11 (1996) 751-759. 13. M. Gehrke, C. Walker, and E. Walker, “De Morgan systems on the unit interval”, International Journal of Intelligent Systems 11 (1996) 733-750. 14. M. Gehrke, C. Walker, and E. Walker, “A mathematical setting for fuzzy logics”, International Journal of Uncertainty, Fuzziness and Knowledge-Based Systems, 5 (1997) 223-238. 15. H. Nguyen and E. Walker, A First Course in Fuzzy Logic (Chapman & Hall/CRC, Boca Raton, 2000).
