INTRODUCING MV-ALGEBRAS. Daniele Mundici

Transcription

1 INTRODUCING MV-ALGEBRAS Daniele Mundici

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3 Contents Chapter 1. Chang subdirect representation 5 1. MV-algebras 5 2. Homomorphisms and ideals 8 3. Chang subdirect representation theorem MV-equations MV-chains 14 Chapter 2. Chang completeness theorem The functor Γ Good sequences and Chang group Chang completeness theorem 22 Chapter 3. Free MV-algebras and McNaughton theorem McNaughton functions Piecewise linearity McNaughton theorem 30 Chapter 4. Simple and semisimple MV-algebras Ideals of free MV-algebras Simple MV-algebras Semisimple MV-algebras The Hay-Wójcicki theorem 40 Chapter 5. Γ is a categorical equivalence Inverting the Γ functor Preservation properties of Γ 47 Index 49 3

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5 CHAPTER 1 Chang subdirect representation 1. MV-algebras Definition 1.1. An MV-algebra is an algebra A,,, 0 with a binary operation, a unary operation and a constant 0 satisfying the following equations: MV1): x (y z) = (x y) z MV2): x y = y x MV3): x 0 = x MV4): x = x MV5): x 0 = 0 MV6): ( x y) y = ( y x) x. Axioms MV1) MV3) state that A,, 0 is an abelian monoid. Following tradition, we denote an MV-algebra A,,, 0 by its universe A. A singleton {0} is a trivial example of an MV-algebra. An MV-algebra is said nontrivial provided its universe has more than one element. As a first example of a nontrivial MV-algebra, consider the real unit interval [0, 1] = {x R 0 x 1}, and for all x, y [0, 1], let x y = def min{1, x + y} and x = def 1 x. It is easy to see that [0, 1] = [0, 1],,, 0 is an MV-algebra. As a second example, if A,,,, 0, 1 is a boolean algebra, then A,,, 0 is an MV-algebra, where, and 0 denote, respectively, the join, the complement and the smallest element in A. A subalgebra of an MV-algebra A is a subset S of A such that S, with the 0 and the operations of A, is an MV-algebra. Equivalently, S A is a subalgebra of an MV-algebra A iff 0 S and whenever x, y S then x y S and x S. The rational numbers in [0, 1], and, for each integer n 2, the n-element set L n = {0, 1/(n 1),..., (n 2)/(n 1), 1}, yield examples of subalgebras of [0, 1]. Given an MV-algebra A and a set X, the set A X of all functions f : X A becomes an MV-algebra if the operations and and the element 0 are defined pointwise. The continuous functions from [0, 1] into [0, 1] form a subalgebra of the MV-algebra [0, 1] [0,1]. On each MV-algebra A we define the constant 1 and the operations and as follows: (1) 1 = def 0, (2) x y = def ( x y), (3) x y = def x y. An MV-algebra is nontrivial if and only if 0 1. The following identities are immediate consequences of MV4): 5

6 6 1. CHANG SUBDIRECT REPRESENTATION MV7): 1 = 0, and MV8): x y = ( x y). Axioms MV5) and MV6) can now be written as MV5 ): x 1 = 1, and MV6 ): (x y) y = (y x) x. Setting y = 0 in MV6) we obtain MV9): x x = 1. In the MV-algebra [0, 1] we have x y = max{0, x+y 1} and x y = max{0, x y}. Notational convention. Following common usage, we consider the operation more binding than any other operation, and the operation more binding than and. Lemma 1.2. Let A be an MV-algebra and x, y A. Then the following conditions are equivalent: (i) x y = 1, (ii) x y = 0, (iii) y = x (y x), (iv) There is an element z A such that x z = y. Proof: (i) (ii) By MV4) and MV7). (ii) (iii) Immediate from MV3) and MV6 ). (iii) (iv) Take z = y x. (iv) (i) By MV9), x x z = 1. Let A be an MV-algebra. For any two elements x and y of A let us agree to write x y iff x and y satisfy the equivalent conditions (i)-(iv) in the above lemma. It follows that is a partial order, which is called the natural order of A. Indeed, reflexivity is equivalent to MV9), antisymmetry follows from conditions (ii) and (iii), and transitivity follows from condition (iv). An MV-algebra whose natural order is total is called an MV-chain. By (iv), the natural order of the MV-chain [0, 1] coincides with the natural order of the real numbers. Lemma 1.3. Let A be an MV-algebra. For each a A, a is the unique solution x of the system of equations: (4) { a x = 1 a x = 0 Proof: By Lemma 1.2, these two equations state that a x a. Lemma 1.4. In every MV-algebra A the natural order has the following properties: (i) x y if and only if y x; (ii) If x y then for each z A, x z y z and x z y z; (iii) If x y z then x y z. Proof: (i) This follows from Lemma 1.2 (i), since x y = y x. (ii) The monotonicity of is an easy consequence of Lemma 1.2(iv); using (i), one immediately proves the monotonicity of. (iii) Note that x y z is equivalent to 1 = (x y) z = x y z.

7 1. MV-ALGEBRAS 7 Proposition 1.5. On each MV-algebra A the natural order determines a lattice structure. Specifically, the join x y and the meet x y of the elements x and y are given by (5) x y = (x y) y = (x y) y, (6) x y = ( x y) = x ( x y). Proof: To prove (5), by MV6 ), MV9) and Lemma 1.4(ii), we have x (x y) y and y (x y) y. Suppose x z and y z. By (i) and (iii) in Lemma 1.2 we have x z = 1 and z = (z y) y. Then by MV6 ) we have: ((x y) y) z = ( (x y) y) y (z y) = (y (x y)) (x y) (z y) = (y (x y)) x y (z y) = (y (x y)) x z = 1. It follows that (x y) y z, which completes the proof of (5). We now immediately obtain (6) as a consequence of (5) together with Lemma 1.4(i). Proposition 1.6. The following equations hold in every MV-algebra: (i) x (y z) = (x y) (x z), (ii) x (y z) = (x y) (x z). Proof: By MV6 ) and Lemma 1.4(ii) we have x y x (y z) and x z x (y z). Suppose x y t and x z t. Then by Lemma 1.4(iii), y x t and z x t, whence y z x t. One more application of Lemma 1.4 (iii) yields (y z) x t, which completes the proof of (i). It is now easy to see that (ii) is a consequence of (i), using Lemma 1.4(i), together with MV4) and MV8). Proposition 1.7. Every MV-algebra satisfies the equation (7) (x y) (y x) = 0. Proof: By making repeated use of MV6) and its variants, together with the basic properties of the operations and we obtain: (x y) (y x) = (x y) ( (x y) (y x)) = x y (y x (y x)) = x ( x (y x)) ( ( x (y x)) y) = (y x) ( (y x) x) ( ( x (y x)) y) = y x ( (y x) x) ((x (y x)) y) = x (x (y x)) y ( y (x ( y x))) = x (x (y x)) (x ( y x)) ( (x ( y x)) y) = 0, since, by MV8) and MV9), x x = 0. Let A be an MV-algebra. For each x A, we let 0x = 0, and for each integer n 0, (n + 1)x = nx x. Lemma 1.8. Let x and y be elements of an MV-algebra A. If x y = 0 then for each integer n 0, nx ny = 0.

8 8 1. CHANG SUBDIRECT REPRESENTATION Proof: If x y = 0 then by monotonicity (Lemma 1.4) and distributivity (Proposition 1.6), x = x (x y) = (x x) (x y) 2x y, whence 0 = x y 2x y. It follows that 0 = 2x 2y = 4x 4y = 8x 8y =.... The desired conclusion now follows from nx ny 2 n x 2 n y = Homomorphisms and ideals Let A and B be MV-algebras. A function h: A B is a homomorphism iff it satisfies the following conditions, for each x, y A: H1): h(0) = 0, H2): h(x y) = h(x) h(y), H3): h( x) = h(x). Following current usage, any injective homomorphism will be called a monomorphism; by an isomorphism we will mean a surjective monomorphism. We write A = B iff there is an isomorphism from A onto B. The kernel of a homomorphism h: A B is the set Ker(h) = h 1 (0) = {x A h(x) = 0}. Our next task is to characterize kernels of homomorphisms. An ideal of an MV-algebra A is a subset I of A satisfying the following conditions: I1): 0 I, I2): If x I, y A and y x then y I, I3): If x I and y I then x y I. The intersection of any family of ideals of A is an ideal of A. For every subset W A, the intersection of all ideals I W is said to be the ideal generated by W. In particular, given an element z of an MV-algebra A, the ideal z generated by the singleton {z} is called the principal ideal generated by z, and we have z = {x A nz x for some integer n 0}. In particular, 0 = {0} and 1 = A. An ideal I of an MV-algebra A is proper iff I A. We say that I iff it is proper and satisfies the following condition: I4): For each x and y in A, either (x y) I or (y x) I. is prime An ideal I of an MV-algebra A is called maximal iff it is proper and no proper ideal of A strictly contains I, i.e., for each ideal J I, if I J then J = A. We denote by I(A), P(A) and M(A) the sets of ideals, prime ideals and maximal ideals, respectively, of an MV-algebra A. In the next lemma we summarize, for further reference, some easy relations between ideals and kernels of homomorphisms: Lemma 1.9. Let A, B be MV-algebras, and h: A B a homomorphism. Then the following properties hold: (i) For each J I(B), h 1 (J) = {x A h(x) J} I(A). Thus in particular, Ker(h) I(A). (ii) h(x) h(y) iff x y Ker(h). (iii) h is a monomorphism iff Ker(h) = {0}. (iv) Ker(h) A iff B is nontrivial.

9 2. HOMOMORPHISMS AND IDEALS 9 (v) Ker(h) P(A) iff B is nontrivial and the image f(a), as a subalgebra of B, is an MV-chain. The distance function d: A A A (8) d(x, y) = (x y) (y x). is defined by In particular, on the MV-algebra [0, 1], d(x, y) = x y. On any boolean algebra the distance function coincides with the symmetric difference operation. Proposition In every MV-algebra A we have: (i) d(x, y) = 0 iff x = y, (ii) d(x, y) = d(y, x), (iii) d(x, z) d(x, y) d(y, z), (iv) d(x, y) = d( x, y), (v) d(x s, y t) d(x, y) d(s, t). Proof: Properties (i), (ii) and (iv) immediately follow by definition, recalling the basic properties of the natural order on A (Lemmas 1.2, 1.4). To prove (iii), note first that (x z) (x y) (y z) = ( x y) (z y) y y = 1. Hence, (x z) (x y) (y z). In a similar way we obtain (z x) (y x) (z y), whence (iii) follows from the monotonicity of (Lemma 1.4(ii)). One similarly proves (v) by observing that ((x s) (y t)) (x y) (s t) = (x s) (x y) (t s) (x s) x s = 1. As an immediate consequence we have Proposition Let I be an ideal of an MV-algebra A. Then the binary relation I on A defined by x I y iff d(x, y) I is a congruence relation. (Stated otherwise, I is an equivalence relation such that x I s and y I t imply x I s and x y I s t.) Moreover, I = {x A x I 0}. Conversely, if is a congruence on A, then {x A x 0} is an ideal, and x y iff d(x, y) 0. Therefore, the correspondence I I is a bijection from the set of ideals of A onto the set of congruences on A. Given x A, the equivalence class of x with respect to I will be denoted by x/i and the quotient set A/ I by A/I. Since I is a congruence, defining on the set A/I the operations (x/i) = def x/i and x/i y/i = def (x y)/i, the system A/I,,, 0/I becomes an MV-algebra, called the quotient algebra of A by the ideal I. Moreover, the correspondence x x/i defines a homomorphism h I from A onto the quotient algebra A/I, which is called the natural homomorphism from A onto A/J. Note that Ker(h I ) = I. The next lemma is an easy consequence of Lemma 1.9(ii). Lemma If A, B and C are MV-algebras, and f : A B and g : A C are onto homomorphisms, then Ker(f) Ker(g) if and only if there is an onto homomorphism h: B C such that h f = g, i.e., h(f(x)) = g(x) for all x A. This homomorphism h is an isomorphism if and only if Ker(f) = Ker(g). Since Ker(h) = Ker(h Ker(h) ) we immediately get

10 10 1. CHANG SUBDIRECT REPRESENTATION Theorem Let A and B be MV-algebras. If h is a homomorphism of A onto B, then there is an isomorphism f : A/Ker(h) B such that f(x/ker(h)) = h(x) for all x A. Proposition If A is an MV-chain, then all proper ideals of A are prime. Proof: Let I be a proper ideal of A. Since h I : A A/I is an onto homomorphism, then A/I is also an MV-chain, and hence, by Lemma 1.9(v), I must be a prime ideal. Proposition Let J be an ideal of an MV-algebra A. Then the map I h J (I) determines an inclusion preserving one-to-one correspondence between the ideals of A containing J and the ideals of the quotient MV-algebra A/J. The inverse map also preserves inclusions, and is obtained by taking the inverse image (K) of each ideal K in A. h 1 J Proof: Let I be an ideal of A such that J I. Since h A maps A onto A/J and Ker(h J ) = J I, by Lemma 1.9 (ii) and (MV6 ), we have h J (I) I(A/J) and, moreover, h 1 J (h(j) I. Since the other inclusion holds for all onto mappings, one has that I = h 1 J (h J(I)). On the other hand, by Lemma 1.9 (i), h 1 J (K) I(A) for each K I(A). To complete the proof, it is sufficient to note that J = h 1 J ({0}) (K) and h J(h 1 (K)) = K. h 1 J J Remark If A is an MV-chain, then the set I(A) is totally ordered by inclusion. Indeed, if I, J were ideals of A such that I J and J I, then there would be elements a, b in A such that a I \ J and b J \ I. This would imply that a b and b a, contradicting the hypothesis that A is totally ordered. Theorem The following properties hold in any MV-algebra A: (i) Every proper ideal of A that contains a prime ideal is prime. (ii) For each prime ideal J of A, the set {I I(A) J I} is totally ordered by inclusion. Proof: Let J be a prime ideal of A. By Lemma 1.9 (v), A/J is an MV-chain, and by Proposition 1.14 and the above remark, all the proper ideals of A/J are prime and are totally ordered by inclusion. This, together with Lemma 1.15, implies (ii). To prove (i), note that if I is a proper ideal of A such that J I, then again by Proposition 1.15, I = h 1 J (h J(I)), and this shows that I is the inverse image of a prime ideal of A/J. Corollary Every prime ideal J of an MV-algebra A is contained in a unique maximal ideal of A. Proof: The set H = {I I(A) I A and J I} is totally ordered by inclusion. Therefore, M = I H I is an ideal of A. Further, M is a proper ideal, because 1 M; we conclude that M is the only maximal ideal that contains J. The next proposition will play an important role in the proof of Chang subdirect representation theorem. Proposition Let A be an MV-algebra. If a A and a 0 then there is a prime ideal P of A such that a P.

11 3. CHANG SUBDIRECT REPRESENTATION THEOREM 11 Proof: By hypothesis, a does not belong to the ideal {0}. Then a routine application of Zorn s Lemma shows that there is an ideal I of A which is maximal with respect to the property that a I. We will show that I is a prime ideal. Let x and y be elements of A, and suppose that both x y I and y x I (absurdum hypothesis). Then the ideal generated by I and x y (i.e., the smallest ideal containing I and x y) must contain the element a; stated otherwise, a s p(x y) for some s I and some integer p 1. Similarly, there is an element t I and an integer q 1 such that a t q(y x). Let u = s t and n = max(p, q). Then u I, a u n(x y) and a u n(y x). Hence by (6) and (7), together with Proposition 1.6(ii) and Lemma 1.8, we have a (u n(x y)) (u n(y x)) = u (n(x y) n(y x)) = u, whence a I, a contradiction. 3. Chang subdirect representation theorem The direct product of a family {A i } i I of MV-algebras, denoted by i I A i, is the MV-algebra obtained by endowing the set-theoretical cartesian product of the family with the MV-operations defined pointwise. In other words, i I A i is the set of all functions f : I i I A i such that f(i) A i for all i I, with the operations and defined by ( f)(i) = def f(i) and (f g)(i) = def f(i) g(i), for each i I, and with the function 0(i) = 0 for each i I as the zero element. For each i I, define π i : i I A i A i by π i (f) = def f(i), for all f i I A i. Each π i is a homomorphism onto A i, called the i th projection function. For each MV-algebra A and each set X, the MV-algebra A X is the direct product of the family {A x } x X, where A x = A for all x X. Definition An MV-algebra A is a subdirect product of a family {A i } i I of MV-algebras iff there exists a monomorphism h: A i I A i such that for each i I, the composite map π i h is a homomorphism onto A i. If A is a subdirect product of the family {A i } i I, then A is isomorphic to the subalgebra h(a) of the direct product of the family, and the restriction of each projection to this subalgebra must be an onto mapping. The following result is a particular case of a theorem in Universal Algebra, due to Birkhoff: Theorem An MV-algebra A is a subdirect product of a family {A i } i I of MV-algebras if and only if there is a family {J i } i I of ideals of A such that (i) A i = A/Ji for each i I, and (ii) i I J i = {0}. Proof: Suppose first that A is a subdirect product of a family {A i } i I of MValgebras. Let h: A i I A i be the corresponding monomorphism, and, for each i I, let J i = Ker(π i h). By Theorem 1.13, A i = A/Ji for each i I. If x i I J i, then π i (h(x)) = 0 for each i I. This implies that h(x) = 0, and since h is a monomorphism, x = 0. Therefore i I J i = {0}, and conditions (i) and (ii) hold true.

12 12 1. CHANG SUBDIRECT REPRESENTATION Conversely, suppose {J i } i I to be a family of ideals of A that satisfies conditions (i) and (ii). Let h: A i I A i be the function defined, for each x A, by h(x) = x/j i. It follows from (ii) that Ker(h) = {0}, whence, by Lemma 1.9(iii), h is a monomorphism. Since for each i I and α A/J i there is a A such that α = a/j i, it follows that π i h maps A onto A/J i. Therefore A is a subdirect product of the family {A/J i } i I. The following result is fundamental: Theorem (Chang Subdirect Representation Theorem) Every nontrivial MV-algebra is a subdirect product of MV-chains. Proof: By Theorem 1.21 and Lemma 1.9(v), an MV-algebra A is a subdirect product of a family of MV-chains if and only if there is a family {P i } i I of prime ideals of A such that i I P i = {0}. Now apply Proposition MV-equations An important consequence of Chang subdirect representation theorem is that in order to prove that an equation holds in all MV-algebras it is sufficient to check that the equation holds in all MV-chains. To state this result precisely we need a formal definition of an equation, which we have used several times in a rather informal way. Accordingly, this section is devoted to sketch the basic equational notions in the theory of MV-algebras. Readers familiar with Universal Algebra may skip it. By a string over a finite set S we understand a finite list of elements of S. The latter are often called the symbols of alphabet S. For each natural number t 1, let S t = {0,,, x 1,..., x t, (, )}. An MVterm in the variables x 1,..., x t is a string over S t that is obtained applying a finite number of times the following rules: (T1): The elements 0 and x i, for i = 1,..., t, considered as one-element lists, are MV-terms. (T2): If the string p is an MV-term, then so is p. (T3): If the strings p and q are MV-terms, then so is (p q). In other words, a string p over S t is an MV-term if and only if there is a parsing certificate for p i.e., a finite list of strings over S t, say p 1,..., p n, such that a) p n = p; b) for each i {1,..., n}, p i satisfies at least one of the following conditions: (i) p i = 0 or p i = x j, for some 1 j t, (ii) there is j < i such that p i = p j, (iii) there are j < i and k < i such that p i = (p j p k ). Those strings p i that belong to every parsing certificate for p are said to be the subformulas of p. The unique readability theorem (which is the same as for the classical propositional calculus, and is left as an exercise 1 ) states that every term p i satisfies precisely one of the above conditions. Moreover, in case (iii) the pair (p j, p k ) is uniquely determined. 1 see Theorem 7.3 in D.Mundici, Logic, a Brief Course, Springer, Milan, 2012

13 4. MV-EQUATIONS 13 The following strings are examples of MV-terms in the variables x 1, x 2, x 3 : ( 0 x 1 ), ((x 2 0) (x 3 x 1 )) and (x 2 x 1 ). As usual, for the sake of readability, in writing terms we will omit the outermost pair of brackets. Strings containing the symbols for the operations,,, and for the constant 1 can be easily transformed into MV-terms using formulas (2), (3), (5), (6) and (1); thus, given MV-terms p and q in the variables x 1,..., x t, we will freely refer to the strings p, p q, p q, p q, p q and p q as MV-terms in the same variables. An MV-equation in the variables x 1,..., x t is a pair (p, q) of MV-terms in the variables x 1,..., x t. One usually writes p = q instead of (p, q). Axioms (MV1) (MV6) are examples of MV-equations in the variables x, y, z. Let p be an MV-term in the variables x 1,..., x t, and let A be an MV-algebra. Substituting an element a i A for all occurrences of the variable x i in p, for i = 1,..., t, using the unique readability theorem, and interpreting the symbols 0, and as the corresponding operations in A, we obtain an element of A, denoted p A (a 1,..., a t ). Definition An MV-algebra A satisfies the MV-equation p = q, in symbols, A = p = q, iff for each t-tuple (a 1,..., a t ) of elements of A, p A (a 1,..., a t ) = q A (a 1,..., a t ). For instance, all MV-algebras satisfy the equations x x = 0 and (x x) = 0. As another example, L n = x x x = x x if and only if n = 2 or n = 3. Remark By Proposition 1.10(i), an MV-equation p = q holds in an MValgebra A if and only if the equation (p q) (q p) = 0 holds in A. Therefore we can assume that all MV-equations are of the form p = 0, where p is an MV-term. Lemma Let A, B, A i (for all i I) be MV-algebras; (i) If A = p = q then S = p = q for each subalgebra S of A. (ii) If h: A B is a homomorphism, then for each MV-term p in the variables x 1,..., x s and each s-tuple (a 1,..., a s ) of elements of A we have p B (h(a 1 ),..., h(a s )) = h(p A (a 1,..., a s ). In particular, when h maps A onto B, from A = p = q it follows that B = p = q. (iii) If A i = p = q for each i I, then i I A i = p = q. Proof: (i) and (ii) are immediate from the definition. (iii) Let f 1,..., f s A = i I A i. By hypothesis, for each i I we can write p A (f 1,..., f s )(i) = p Ai (f 1 (i),..., f s (i)) = q Ai (f 1 (i),..., f s (i)) = q A (f 1,..., f s )(i), whence p A (f 1,..., f s ) = q A (f 1,..., f s ). Theorem Let A be a subdirect product of a family {A i } i I of MValgebras, and let p = q be an MV-equation in the variables x 1,..., x s. Then A = p = q if and only if A i = p = q for each i I. Proof: Let h: A i I A i be the monomorphism determining the subdirect product. Suppose that A = p = q. Since for each i I, π i h maps A onto A i, it follows that A i = p = q.

14 14 1. CHANG SUBDIRECT REPRESENTATION Conversely, suppose that A i = p = q for all i I. By the above lemma, i I = p = q, and since h(a) is a subalgebra of i I A i, h(a) = p = q. Finally, since h 1 maps h(a) onto A, we conclude that A = p = q. Corollary An MV-equation is satisfied by all MV-algebras if and only if it is satisfied by all MV-chains. Proof: Suppose that p = q is satisfied by all MV-chains, and let A be an MValgebra. If A = {0} then, trivially, p A (0,..., 0) = 0 = q A (0,... 0), whence A = p = q. If A is nontrivial the desired conclusion follows from Theorems 1.21 and From this result it follows that the underlying lattice of any MV-algebra is distributive. Corollary 1.27 will be considerably improved in the next chapter, by proving that an equation holds in all MV-algebras if and only if it holds in the algebra [0, 1]. 5. MV-chains In this section we collect several results on totally ordered MV-algebras that we will need in the next chapter. Lemma The following properties hold in every MV-chain A: (i) If x y < 1 then x y = 0, (ii) If x y = x z and x y = x z then y = z, (iii) If x y = x z < 1 then y = z, (iv) If x y = x z > 0 then y = z, (v) x y = x iff x = 1 or y = 0, (vi) x y = x iff x y = y, (vii) If x y = 1 and x z < 1 then (x y) z = (x z) y. Proof: (i) By hypothesis, x y, whence y < x. (ii) By hypothesis, max( x, y) = x (y x) = x (z x) = max( x, z). Similarly, min( x, y) = min( x, z), whence y = z. Condition (iii) is an immediate consequence of (i) and (ii). Condition (iv) follows from (iii) by Lemma 1.4(i). Condition (v) follows from (iii). From (v) one immediately obtains (vi). Finally, to prove (vii), since by assumption y x, we get y (x y) z = ( y x) z = x z < 1 and y (y (x z)) = y (x z) = x z, whence (vii) follows from (iii). Remark From Theorem 1.22 it follows that every MV-algebra satisfies conditions (ii) and (vi). Proposition The following equations hold in every MV-algebra A: (9) x y (x y) = x y, (10) (x y) ((x y) y) = x, (11) (x y) ((x y) z) = (x z) ((x z) y). Proof: By Theorem 1.22 we can safely assume that A is a chain. If x y = 1, then (9) follows by MV5 ). If x y < 1, then (9) follows from Lemma 1.28(i). To prove (10), if x y then x y = 0 and x = x y = (x y) y; if, on the other hand, y < x then (x y) (x y) = (x y) y = x y = x. As a prerequisite for the proof of (11) we will prove the following equation: (12) (x y) ((x y) z) = (x y) ((x y) z).

15 5. MV-CHAINS 15 Indeed, if x y = 1 both members of (12) coincide with (x y) z. If x y < 1 then by Lemma 1.28(i) both members coincide with (x y) z. Thus (12) holds for all MV-algebras. From MV4) and MV8) we now obtain: (13) ((x y) ((x y) z)) = ( x y) (( x y) z). To complete the proof of (11) we argue by cases as follows: Case 1 : x y z < 1. Then since A is a chain, by Lemma 1.28(i), both members of (11) are equal to 0. Case 2 : x y z < 1. Same as Case 1, recalling (13). There remains to consider Case 3 : x y z = 1 and x y z = 1. Subcase 3.1 : x y = 1 and x z < 1, or x y < 1 and x z = 1. By symmetry, it is sufficient to consider the case x y = 1 and x z < 1. Then x z = 0, and (11) becomes (x y) z = (x z) y, which follows from Lemma 1.28(vii). Subcase 3.2 : x y = x z = 1. Then equation (11) becomes (14) (x y) z = (x z) y. Note that equation (14) holds in case x y = 0 or x z = 0. Indeed, suppose x y = 0. Since x y = 1, it follows from Lemma 1.3 that x = y, whence from y = x z we obtain (x y) z = z = y z = ( y z) y = (x z) y. By symmetry, (14) holds under the hypothesis x z = 0. We next observe that if one of the members of (14) is equal to 1 then so is the other. Assume, for instance, (x y) z = 1. Since x y z = 1 is equivalent to x y z = 0, it follows from Lemma 1.3 that z = (x y) = x y. Hence, by Proposition 1.6, (x z) y = (x ( x y)) y = (x y) y = (x y) ( y y) = 1. Thus, to complete the analysis of Subcase 3.2 we may restrict to the case when (x y) z < 1, (x z) y < 1, x y > 0, x z > 0. Under these hypotheses, by Lemma 1.28(vii) we obtain: x (z (x y)) = (x z) (x y) > 0, x (y (x z)) = (x y) (x z), thus establishing (14) by Lemma 1.28(iv). Subcase 3.3 : x y < 1 and x z < 1. Then by Lemma 1.28(i), x y = 0 and x z = 0, i.e., x y = 1 and x z = 1. Recalling (13) and arguing as in Subcase 3.2 (with x, y, z instead of x, y, z) we conclude that (11) also holds in this case.

16

17 CHAPTER 2 Chang completeness theorem 1. The functor Γ A (partially) ordered abelian group is an abelian group G, +,, 0 endowed with a partial order relation satisfying the following compatibility condition, for all x, y, t G: if x y then t + x t + y. When the order relation is total, G is said to be a totally ordered abelian group, or o-group for short. When the order of G defines a lattice structure, G is called a lattice-ordered abelian group, or l-group, for short. In any l-group we have t + (x y) = (t + x) (t + y) and t + (x y) = (t + x) (t + y). Note that o-groups are particular cases of l-groups. Indeed, we have x y = max{x, y} and x y = min{x, y}. For each element x of an l-group G, the following notations are currently used: x + = def 0 x, x = def 0 x, and [ x = def x + + x = x x. An order unit u of G (also called a strong order unit) is an element 0 u G such that for each x G, there is an integer n 0 such that x nu. Let G be an l-group. For each u G, u > 0, let [0, u] = def {x G 0 x u}, and for each x, y [0, u], let x y = def u (x + y), and x = def u x. It is not hard to see that [0, u],,, 0 is an MV-algebra, which will be denoted by Γ(G, u). A routine inspection shows that the natural order of Γ(G, u) coincides with the order of [0, u] inherited from G by restriction. In the particular case when G = R = the additive group of real numbers with the natural order, then Γ(R, 1) coincides with the MV-algebra [0,1]. We also have Q [0, 1] = Γ(Q, 1), and for each n 2 ( L n = Γ Z 1 ) n 1, 1, where Z 1 n 1 = { z n 1 z Z}, considered as a subgroup of Q. In particular, the two-element boolean algebra L 2 coincides with Γ(Z, 1). Let G be an l-group, and 0 < u G. Then S = {x G for some integer n 0, x nu} is a subgroup and a sublattice of G containing u, and Γ(G, u) = Γ(S, u). Therefore, in considering the MV-algebras Γ(G, u) we can assume that u is an order unit of G. 17

18 18 2. CHANG COMPLETENESS THEOREM Let G and H be l-groups. By definition, a function h: G H is an l-group homomorphism, or l-homomorphism iff h is both a group homomorphism and a lattice homomorphism; in other words, for each x, y G, h(x y) = h(x) h(y), h(x y) = h(x) h(y) and h(x y) = h(x) h(y). Suppose that 0 < u G and 0 < v H, and let h: G H be an l-group homomorphism such that h(u) = v. Then h is said to be a unital l-homomorphism. If Γ(h) denote the restriction h [0,u] of h to the unit interval [0, u], then Γ(h) is a homomorphism from Γ(G, u) into Γ(H, v). We let A U be the category whose objects are pairs G, u such that G is an l-group and u is a (strong order) unit of G, and whose morphisms are unital l-homomorphisms. It follows that Γ is a functor from the category A U into the category of MValgebras. Indeed, as we will see in a subsequent chapter, Γ is a natural equivalence (i.e., a full, faithful, dense functor) between these two categories. In the next section we will associate to each MV-algebra A an l-group G A with an order unit u A such that A = Γ(G A, u A ). This result will be used to show that an MV-equation holds in all MV-algebras if and only if it holds in the MV-algebra [0, 1]. 2. Good sequences and Chang group A sequence a = (a 1, a 2,... ) of elements of an arbitrary MV-algebra A is said to be good iff a i a i+1 = a i for each i = 1, 2..., and there is an integer n such that a r = 0 for all r > n. Instead of a = (a 1,..., a n, 0, 0,... ) we will often write, more concisely, a = (a 1,..., a n ). Thus, if 0 m denotes an m-tuple of zeros, we have identical good sequences (15) (a 1,..., a n ) = (a 1,..., a n, 0 m ). By pre-pending to the good sequence (a 1,..., a n ) an m-tuple 1 m of consecutive ones, the resulting good sequence (1 m, a 1,..., a n ) is different from (a 1,..., a n ). For each a A, the good sequence (a, 0..., 0,...) will be denoted by (a). The good sequences of a boolean algebra A are the nonincreasing sequences of elements of A having a finite number of nonzero terms. For totally ordered MV-algebras we have the following characterization of good sequences: Proposition 2.1. Each good sequence of an MV-chain A has the form (16) (1 p, a) for some integer p 0 and a A. Proof: Immediate from Lemma 1.28(v). Lemma 2.2. Let A be a subdirect product of a family {A i } i I of MV- algebras. A sequence a = (a 1,..., a n...) of elements of A is a good sequence if and only if for each i I the sequence a i = (π i (a 1 ),..., π i (a n )...) is a good sequence in A i, and there is an integer n 0 0 such that whenever n > n 0 then for all i I, π i (a n ) = 0. Proof: It is sufficient to note that a n a n+1 = a n means that π i (a n a n+1 ) = π i (a n ) for each i I. In the light of Theorem 1.22, the above Lemma 2.2 and Proposition 2.1 yield a very useful tool for dealing with good sequences. As an example, consider the proof of the following result:

19 2. GOOD SEQUENCES AND CHANG GROUP 19 Lemma 2.3. Let A be an MV-algebra. If a = (a 1,..., a n...) and b = (b 1,..., b n...) are good sequences of A, then the sequence c = (c 1,..., c n...) such that c n = a n b n for each n, is a good sequence of A. Proof: Since a and b are both good sequences, there is an integer n 0 such that c n = 0 for all n > n 0. By Theorem 1.22, A is a subdirect product of a family {C i } i I of MV-chains. For each i I the sequences a i = (π i (a 1 ),..., π i (a n )...) and b i = (π i (b 1 ),..., π i (b n )...) are good sequences of C i. Hence, by Proposition 2.1, a i = (1 p, α i ) and b i = (1 q, β i ), where α i and β i are in C i. Therefore, π i (c n ) = 1 if n max{p, q} and π i (c n ) = 0 if n > max{p, q} + 1. For n = max{p, q} + 1, we have π i (c n ) = α i if p > q, π i (c n ) = β i if p < q and π i (c n ) = max{α i, β i } when p = q. Consequently π i (c n ) is a good sequence for each i I, and c is a good sequence of A. Example. To have a better intuition of the meaning of good sequences, for every real number α 0 let α denote the greatest integer α, and let α = α α be the fractional part of α. Then α can be written as α = α , with α many consecutive 1 s. Considered as elements of the MV-algebra [0, 1], the above summands α 1, α 2,... of α trivially satisfy the identity α i α i+1 = α i for every integer i 1. For 0 β R, let similarly β = β β m 1 + β , where β 1 = = β m 1 = 1 = α 1 = = α n 1, 0 = α n+1 = α n+2 =..., and 0 = β m+1 = β m+2 =.... Let γ = α + β. Then γ = γ 1 + γ , where γ 1 = = γ n+m 2 = 1, γ n+m 1 = α β, γ n+m = α β, and 0 = γ n+m+1 = γ n+m+2 =.... In a more compact notation, for each i = 1, 2,..., the summand γ i is given by (17) γ i = α i (α i 1 β 1 ) (α i 2 β 2 ) (α 2 β i 2 ) (α 1 β i 1 ) β i. In the light of (17) and (15), we can now give the following Definition 2.4. For any good sequences a = (a 1,..., a n ) and b = (b 1,..., b m ) their sum c = a + b is defined by c = (c 1, c 2,... ), where for all i = 1, 2,..., (18) c i = a i (a i 1 b 1 )... (a 1 b i 1 ) b i. Since a p = b q = 0 whenever p > n and q > m, then c j identically vanishes for each j > m + n. The notation c = (c 1,..., c n+m ) = (a 1,..., a n ) + (b 1,..., b m ) is self-explanatory. To sum two good sequences of an MV-chain one can use the following trivial identity: (19) (1 p, a) + (1 q, b) = (1 p+q, a b, a b). Since by (9), (a b, a b) is a good sequence, applying Theorem 1.22 and Lemma 2.2 together with (19), one immediately sees that the sum of two good sequences is a good sequence. Let M A be the set of good sequences of A equipped with addition.

20 20 2. CHANG COMPLETENESS THEOREM Proposition 2.5. For every MV-algebra A, the following additional properties: M A is an abelian monoid with (i) (cancellation): For any good sequences a, b, c, if a + b = a + c then b = c. (ii) (zero-law): If a + b = (0) then a = b = (0). Proof: From (18) one sees that a + (0) = a, addition is commutative, and the zero-law holds. To prove associativity, by Theorem 1.22 we can safely assume A to be totally ordered. By Proposition 2.1 and equation (11) in Proposition 1.30, letting a = (1 p, a), b = (1 q, b), c = (1 r, c), we have (b + a) + c = (1 p+q+r, a b c, (a b) ((a b) c), a b c) = (1 p+q+r, a b c, (a c) ((a c) b), a b c) = b + (a + c). Similarly, to prove cancellation, avoiding trivialities, assume that a, b and c are different from 1. If q = r, then by Lemma 1.28(ii), b = c, and we are done. If q < r 1 then from the identity (1 p+q, a b, a b) = (1 p+r, a c, a c) we get a b = 1, i.e., a = b = 1, which is a contradiction. If q = r 1 then a b = a and a b = 1, which is impossible because these two equalities imply that b = 1. The cases corresponding to r < q are similarly shown to lead to contradiction. Proposition 2.6. Let a = (a 1,..., a n ) and b = (b 1,..., b m ) be good sequences. Recalling (15) assume, without loss of generality, m = n. Then the following are equivalent: (i) There is a good sequence c such that b + c = a; (ii) b i a i for all i = 1,..., n. Proof: (i) (ii) is immediate from (18). (ii) (i). By Remark 1.29, ( b n,..., b 1 ) is a good sequence. Let us denote by c = a b the good sequence obtained by dropping the first n terms in (a 1,..., a n )+ ( b n,..., b 1 ). We will prove that c + b = a. Using Theorem 1.22 we can safely assume A to be totally ordered, so that by (16), a = (1 p, a) and b = (1 q, b). To avoid trivialities assume a 1 and b 1, whence q p. Upon rewriting b = (1 q, b, 0 p q ) we see that ( b n,..., b 1 ) = (1 p q, b, 0 q ), and hence c is obtained by dropping the first p + 1 terms from (1 2p q, a b, a b). Case 1: b a. Then a b = 1, c = (a b) and c + b = (1 q, (a b) b, (a b) b) = (1 p, b a, 0) = (1 p, a) = a. Case 2: b > a. Then p > q, a b = 0, c = (1 p q 1, a b) and c + b = (1 p 1, a b b, (a b) b) = (1 p, a b) = (1 p, a) = a. Construction. For any two good sequences a and b we write: (20) b a iff b and a satisfy the equivalent conditions of Proposition 2.6. Then the proof of the above proposition, together with Proposition 2.5 (i), shows that there is a unique good sequence c such that b + c = a. This is given by (21) c = a b = (a 1,..., a n ) + ( b n,..., b 1 ) omitting first n terms.

21 2. GOOD SEQUENCES AND CHANG GROUP 21 In particular, for each a A we have (22) ( a) = (1) (a). It follows that the order is compatible with the addition operation, in the sense that for all good sequences a, b, d, if b a, then b + d a + d. Given good sequences a = (a 1,..., a n,...) and b = (b 1,..., b n,...), by Lemma 2.3 the sequence a b = (a 1 b 1,..., a n b n,...) is a good sequence, and from Proposition 2.6(ii) it follows that a b is the supremum of the set {a, b} with respect to the order defined by (20). Analogously, the element a b = (a 1 b 1,..., a n b n,...) is a good sequence coinciding with the infimum of {a, b}. From (18) it follows that for each a, b, c A, (23) ((a) + (b)) (1) = (a b). If we enrich the monoid M A with the above lattice-order between good sequences, it is easy to find an l-group G A such that M A is isomorphic, both as a monoid and as a lattice, with the positive cone G A +, i.e., the monoid given by all elements 0 x G A, with the order inherited by restriction. Specifically, let G A be the set of equivalence classes of pairs of good sequences (a, b), where (a, b) is equivalent to (a, b ) if and only if a + b = a + b. The zero element 0 is the equivalence class of ((0), (0)). Addition is defined by (a, b) +(c, d) = (a + c, b + d), and (a, b) = (b, a). The correspondence a (a, (0)) is a monoid isomorphism from M A onto the submonoid of G A formed by the equivalence classes of pairs (a, (0)). By Proposition 2.6(i), (a, b) is equivalent to a pair (c, (0)) if and only if a b. Therefore one can identify M A with the submonoid of G A formed by the equivalence classes of pairs (a, b) such that a b. G A becomes a partially ordered group with the order defined by (a, b) (c, d) iff (c, d) (a, b) M A, the latter being identified with the set of equivalence classes of pairs (a, (0)). As a matter of fact, G A is latticeordered. Indeed, it is easy to very that the join and the meet of the equivalence classes of the pairs (a, b) and (c, d) are, respectively, the equivalence classes of the pairs (a, b) (c, d) = ((a + d) (b + c), b + d) and (a, b) (c, d) = ((a + d) (b + c), b + d). Equipping M A with the natural lattice-order between good sequences, it follows that the correspondence a (a, (0)) is also a lattice-isomorphism from M A onto the positive cone of G A. We call the l-group G A the Chang group of the MV-algebra A. Let n 0 0 be an integer, and a = (a 1, a 2,... ) be a good sequence of A such that a n = 0 for each n n 0. In the Z-module G A we trivially have n 0 (1) = (1)+ +(1) = (1 n0 ). Therefore, for any good sequence b we have (a, b) n 0 ((1), (0)), thus showing that the equivalence class of the pair ((1), (0)) is an order unit of G A. A crucial property of the l-group G A is given by the following result.

22 22 2. CHANG COMPLETENESS THEOREM Theorem 2.7. The correspondence a ϕ A (a) = ((a), (0)) defines an isomorphism from the MV-algebra A onto the MV-algebra Γ(G A, ((1), (0))). Proof: Trivially, ((0), (0)) (a, b) ((1), (0)) iff there is c A such that (a, b) is equivalent to ((c), (0)). It follows that ϕ A maps A onto the unit interval [((0), (0)), ((1), (0))]. It is easy to see that this map is one-one. By (23), ϕ A (a b) = (ϕ(a) + ϕ(b)) ((1), (0)), and by (22), ϕ A ( a) = ((1), (0)) ϕ A (a). Therefore, ϕ A is a homomorphism from A to Γ(G A, ((1), (0))). Remark 2.8. An MV-algebra A is a chain if and only if G A is totally ordered. Indeed, if A is totally ordered, then it follows from Proposition 2.6(i) that M A is totally ordered, and this implies that G A is a totally ordered group. The converse is an immediate consequence of the above theorem. 3. Chang completeness theorem An l-group term in the variables x 1,..., x t is a string of symbols over the alphabet {x 1,..., x n, 0,, +,,, (, )} which is obtained by the same inductive procedure used in Chapter 1.4 to define MV-terms. Let τ be an l-group term in the variables x 1,..., x t and G be an l-group. Substituting an element a i G for all occurrences of the variable x i in τ, for i = 1,..., t, and interpreting the symbols 0,, +, and as the corresponding operations in G, we obtain an element of G, denoted τ G (a 1,..., a t ). To each MV-term in the n variables x 1,..., x n we associate an l-group term ˆτ in the n + 1 variables (x 1,..., x n, y), according to the following stipulations: ˆx i = x i, for each i = 1,..., n, ˆ0 = 0, σ = (y ˆσ), (ρ σ) = (y (ˆρ + ˆσ)). Since unique readability also holds for l-group terms, the mapping τ ˆτ is well defined. We then have a purely syntactic counterpart of the mappings (G, u) Γ(G, u) and A G A, in a sense that is made precise by the following two propositions: Proposition 2.9. If G is a totally ordered abelian group, 0 < u G, 0 g 1,... g n u and A = Γ(G, u), then for every MV-term τ(x 1,..., x n ) we have the identity τ A (g 1,..., g n ) = ˆτ G (g 1,..., g n, u). Proof: By induction on the number of operation symbols in τ. The basis is trivial. For the induction step, by definition of Γ we have: ( σ) A (g 1,..., g n ) = (σ A (g 1,..., g n )) = (ˆσ G (g 1,..., g n, u)) = u ˆσ G (g 1,..., g n, u) = (y G ˆσ G )(g 1,..., g n, u) = σ G (g 1,..., g n, u). The -case is similar. Proposition If A is an MV-chain, a 1,..., a n are elements of A, and τ(x 1,..., x n ) is an MV-term, then the one-term sequence (τ A (a 1,..., a n )) G A coincides with ˆτ G A ((a 1 ),..., (a n ), (1)).

23 3. CHANG COMPLETENESS THEOREM 23 Proof: By induction on the number of operation symbols in τ. The basis is trivial. For the induction step, if σ = φ ψ then using (23), together with the definition of the mapping σ ˆσ, and omitting unnecessary superscripts, we can write: (φ ψ(a 1,..., a n )) = (φ(a 1,..., a n ) ψ(a 1,..., a n )) = min((1), (φ(a 1,..., a n )) + (ψ(a 1,..., a n ))) = min((1), ˆφ((a 1 ),..., (a n ), (1)) + ˆψ((a 1 ),..., (a n ), (1))) = (y ( ˆφ + ˆψ))((a 1 ),..., (a n ), (1)) = φ ψ((a 1 ),..., (a n ), (1)). In the -case, one similarly uses (22). Theorem (Completeness Theorem) An equation holds in [0, 1] if and only if it holds in every MV-algebra. Proof: For the nontrivial direction, suppose an equation fails in an MV-algebra A. By Remark 1.24 we may assume that the equation has the form τ(x 1,..., x n ) = 0. By Corollary 1.27, A may be assumed to be totally ordered. There are elements a 1,..., a n A such that τ A (a 1,..., a n ) > 0. Letting G A denote the Chang group of A, by Proposition 2.10 we have 0 < ˆτ G A ((a 1 ),..., (a n ), (1)) (1). Let S = Z(1) + Z(a 1 ) + + Z(a n ) be the subgroup of G A generated by the elements (1), (a 1 ),..., (a n ), with the induced total order. Recalling that the order in G A is compatible with the addition operation, it follows that G A, is torsion-free. Since S is a finitely generated subgroup of G A, by the fundamental theorem on torsion-free abelian groups, 1 we can identify S with the free abelian group Z r, for some integer r 1. Its elements (1), (a 1 ),..., (a n ) are respectively identified with vectors h 0, h 1,..., h n Z r ; the set of nonnegative elements of S then becomes a submonoid P of Z r such that P P = {0} and P P = Z r. For any two vectors h, k Z r let us write h P k iff k h P. Let us display the subterms σ 0, σ 1,..., σ t of ˆτ as follows: σ 0 = y, σ 1 = x 1,..., σ n = x n, σ n+1, σ n+2,..., σ t 1, σ t = ˆτ. We can safely assume that the list contains the zero term. The map y h 0, x 1 h 1,..., x n h n uniquely extends to an interpretation σ j h j (j = 0,..., t) of subterms of ˆτ into elements of the totally ordered group T = (Z r, P ). In particular, by hypothesis we have (24) (25) 0 P h 1,..., h n P h 0, 0 P h t P h 0, 0 h t = ˆτ T (h 1,..., h n, h 0 ). Let ω be a permutation of {0,..., t} such that h ω(0) P h ω(1) P... P h ω(t). Our aim is to replace P by another total order P over Z r in such a way that the above inequalities still hold with respect to P, and (Z r, P ) is isomorphic to a subgroup of the additive group R with the natural order. For each j = 1,..., t, let the vector d j P be defined by d j = h ω(j) h ω(j 1). 1 Theorem 8.4 in S. Lang, Algebra (Revised 3rd ed.), Springer-Verlag, New York, 2002

24 24 2. CHANG COMPLETENESS THEOREM Embeddding Z r into R r, we define the positive and the negative span of the d j s as follows: t (26) P = { λ j d j 0 λ j R}, N = P. j=1 Trivially, P is a closed and convex subset of R r, and whenever h P and 0 α R, then αh P. It is not hard to see that 0 is an extremal point of P, or, equivalently, that no line is contained in P. (For otherwise, let I be a minimal subset of {1,..., t} such that 0 = j I λ jd j for some 0 < λ j R and d j 0. Then the vector (λ j ) j I is uniquely determined up to multiplication by a constant factor 0 < γ R. Since 0 P d j Z r, there are integers 0 < n j such that 0 = j I n jd j. By definition of P, for each i I we have d i P j I n jd j, whence d i = 0, a contradiction). Essentially the same argument yields (27) P N = {0} and P P = P Z r, whence for arbitrary i, j = 0,..., t, we have h i P h j iff h j h i P. We will now show that P and N can be separated by a hyperplane. As usual, we denote by h k the scalar product of vectors h, k R r. Claim. For some vector g R r, the hyperplane π g = {v R r g v = 0} separates P and N, in the sense that π g P = {0} = π g N. The proof is by induction on r. The basis is trivial. For the induction step, assume r 2 and let δ = {v R r v = 1} be the surface of the unit ball in R r, where, as usual, v denotes the Euclidean norm of v. Then for some vector w δ, the line ρ w = {λw λ R} is such that ρ w P = {0}. (For otherwise, since by (27) P contains no line, each vector u δ belongs to exactly one of P and N ; recalling that P and N are closed, we obtain a partition of δ into disjoint closed sets δ P and δ N, thus contradicting the connectedness of δ). Let P w be the projection of P into the (r 1)-dimensional subspace π w = {v R r w v = 0}. Then P w can be identified with a closed convex subset of R r 1 having 0 as an extremal point, and such that whenever h P w and 0 α R, then αh P w. By induction hypothesis, there is a hyperplane π in R r 1 such that π P w = {0}. It follows that the hyperplane π + ρ w of R r intersects P only in 0, as required. Picking now a vector g = (γ 1,..., γ r ) R r such that π + ρ w = π g, our claim is settled. As an equivalent reformulation of our claim, in the light of the convexity of P, we can safely write g d j > 0 for all nonzero vectors d j, j = 1,..., t. By continuity, g can be assumed to be in general position, in the sense that γ 1,..., γ r are linearly independent over Q. Let π + g = {(ζ 1,..., ζ r ) R r γ i ζ i 0}, P = π + g Z r. Then from (26) it follows that (28) P π + g and N π + g. Let us now consider the totally ordered abelian group T = (Z r, P ). Although T need not coincide with T = (Z r, P ), still by (27)-(28), for all i, j = 0,..., t

25 such that h i h j 3. CHANG COMPLETENESS THEOREM 25 we have (29) h i P h j iff h j h i P iff h j h i / N iff h i P h j. For arbitrary vectors k 0,..., k n Z r, the map y k 0, x 1 k 1,..., x n k n uniquely extends to an interpretation σ j k j, j = 0,..., t of all subterms σ j of ˆτ into elements k j of T. In the particular case when k 0 = h 0,..., k n = h n, arguing by induction on the number of operation symbols occurring in σ j, from (29) we obtain k j = h j for all j = 0,..., t; moreover, all inequalities in (24) are still valid with respect to the new total order relation P over Z r. In symbols, 0 P h 1,..., h n P h 0, and 0 P h t P h 0, 0 h t = ˆτ T (h 1,..., h n, h 0 ). As an effect of the independence of the γ s over Q, T is isomorphic, as a totally ordered group, to the subgroup U = Zγ Zγ r of R generated by γ 1,..., γ r, with the natural order. An isomorphism is given by the map θ : b = (b 1,..., b r ) T b g = b 1 γ b r γ r U. Since the inequalities in (24) are preserved under isomorphism, letting κ 0 = θ(h 0 ), κ 1 = θ(h 1 ),..., κ n = θ(h n ), κ t = θ(h t ) we can write 0 κ 1,..., κ n κ 0 and 0 < κ t κ 0. Assuming without loss of generality, κ 0 = 1, we have κ t = ˆτ U (κ 1,..., κ n, 1) > 0. By Proposition 2.9, in the MV-algebra B = Γ(U, 1) we have τ B (κ 1,..., κ n ) 0, whence, a fortiori, the equation τ = 0 fails in the MV-algebra [0, 1].

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27 CHAPTER 3 Free MV-algebras and McNaughton theorem 1. McNaughton functions For any (finite or infinite) cardinal κ 1, we will denote by [0, 1] κ the cartesian product of κ copies of the unit real interval [0, 1], equipped with the product topology. Any point x [0, 1] κ is a sequence (x o, x 1,..., x α,... ), where x α [0, 1] for each ordinal α < κ. Let Proj κ = {π α α < κ, α an ordinal } be the set of projections, where for each ordinal α < κ, the projection π α : [0, 1] κ [0, 1] is defined by π α (x) = x α, for all x [0, 1] κ. Proposition 3.1. Let F be the smallest MV-algebra of [0,1]-valued functions defined over the cube [0, 1] κ containing all projections π α, for each ordinal α < κ, and closed under the pointwise MV-operations. Then F has the following property: For every MV-algebra A and every function r : Proj κ A, r can be uniquely extended to a homomorphism of F into A. Proof: If A is an MV-algebra and r is any function from Proj κ into A, then r can be uniquely extended to an MV-homomorphism of F into A, using the following familiar procedure: For every MV-term τ, by induction on the number of operation symbols in τ, we define τ A by (i) Xα A = r(π α ), for each variable symbol X α, α < κ; (ii) ( ρ) A = (ρ A ); (iii) (ψ φ) A = (ψ A φ A ). Two terms µ and ν denote the same function g : [0, 1] κ [0, 1] iff the equation µ = ν is satisfied by the MV-algebra [0, 1]. By the completeness theorem, this implies that the equation is satisfied by A, i.e., µ and ν are mapped into the same element µ A = ν A A. This gives the required homomorphism of F into A. An algebra A with a distinguished subset Y of generators of F is said to be free over the generating set Y (and is denoted F ree Y ) iff for every MV-algebra B and every function f : Y B, f can be uniquely extended to a homomorphism of A into B. In particular, each MV-algebra F in the above proposition is free over the generating set Proj κ of projections. For each cardinal κ 1, F will be denoted by F ree κ. Trivially, if Y and Y have the same cardinality, then F ree Y = F reey. Accordingly, F ree κ will be called the free MV-algebra over κ many generators. The proof of the following proposition is an immediate consequence of the definition of free algebra, and of Lemma 1.9: 27