COSTAS DROSSOS and DANIELE MUNDICI
MANY-VALUED POINTS AND EQUALITY
We assume familiarity with the theory of MV-algebras as developed, e.g., in (Chang 1958, 1959; Cignoli and Mundici 1997; Mundici 1986). Let E be a nonempty set, A be an MV-algebra, and G its corresponding abelian lattice-ordered group with strong unit u, as given by the 0 functor. Thus we can identify A with the unit interval [0, u] of G, in such a way that the negation and disjunction operations on A are respectively given by (1)
¬x = u − x,
x ⊕ y = u ∧ (x + y).
By a generalized point of E (with values in A) we mean a function δ:E→A such that δ(x) is nonzero only for finitely many x ∈ E, say x1 , . . . , xk , and x1 + . . . + xk = u. In a more condensed form we can write X (2) δ(x) = u. x∈E
The equality degree [[α = β]] of two generalized points α and β is defined by M [[α = β]] = (3) α(x) ∧ β(x), x∈E
where ∧ is the infimum operation in the natural lattice order in A, coinciding on [0, u] with the lattice order of G. By Equations (1) and (2) we get X [[α = β]] = (4) α(x) ∧ β(x). x∈E
When A is a boolean algebra we recover the well known notion of booleanvalued equality. PROPOSITION 1. Under the above notation, for any two generalized points α and β of E we have Synthese 125: 97–101, 2000. © 2000 Kluwer Academic Publishers. Printed in the Netherlands.
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MANY-VALUED POINTS AND EQUALITY
(i) [[α = α]] = u. (ii) [[α = β]] = [[β = α]]. (iii) If [[α = β]] = u then α = β. All these properties are immediate consequences of the definition. Note that reflexivity – property (i) above - would fail if in Equation (3) the ∧ operation were replaced by, say, Łukasiewicz conjunction (5)
x y = ¬(¬x ⊕ ¬y) = 0 ∨ (x + y − u).
Reflexivity would also fail if the ⊕ operation in Equation (3) were replaced by ∨. Aim of this paper is to establish transitivity: THEOREM 2. For any generalized points α, β, γ of a nonempty set E, with values in an MV-algebra A, we have the inequality (6)
[[α = β]] [[β = γ ]] ≤ [[α = γ ]],
where ≤ is the natural underlying order on A Proof. Let R, S, T be the supports of α, β, γ , respectively. Stated otherwise, R = {x ∈ E | α(x) 6 = 0}, and the like for S and T . We must prove " # " # M M (7) α(x) ∧ β(x) β(x) ∧ γ (x) x∈R∩S
x∈S∩T
M
≤
α(x) ∧ γ (x),
x∈R∩T
where, as usual, a sum with no summands is understood to take value zero. It is no loss of generality to assume (8)
S ⊆ R ∪ T.
We have the trivial inequality (9)
M
α(x) ∧ β(x) ≤
x∈R∩S
" ⊕
M x∈(R∩S)\T
M x∈R∩S∩T
# α(x) ∧ β(x) .
β(x)
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COSTAS DROSSOS AND DANIELE MUNDICI
Similarly we have (10)
M
β(x) ∧ γ (x) ≤
x∈S∩T
M
β(x)
x∈(S∩T )\R
"
#
M
⊕
β(x) ∧ γ (x) .
x∈R∩S∩T
In Equations (9) and (10) we can equivalently replace the ⊕ operation of A by the addition operation of its corresponding lattice-ordered group G. Using Equations (8) and (5), together with the monotonicity properties of the operation, we then obtain " # " # M M α(x) ∧ β(x) β(x) ∧ γ (x) x∈R∩S
x∈S∩T
X
≤
X
β(x) +
α(x) ∧ β(x)
x∈(R∩S)\T
x∈R∩S∩T
X
X
β(x) +
β(x) ∧ γ (x)
x∈R∩S∩T
x∈(S∩T )\R
X
= 0 ∨ [−u +
β(x)
x∈((R∩S)\T )∪((S∩T )\R)
+
X
(α(x) ∧ β(x)) + (β(x) ∧ γ (x))]
x∈R∩S∩T
= 0∨
X
−β(x) +
x∈R∩S∩T
X
(α(x) ∧ β(x))
x∈R∩S∩T
+ (β(x) ∧ γ (x)) = 0∨
X
−β(x) + (α(x) ∧ β(x)) + (β(x) ∧ γ (x)).
x∈R∩S∩T
For all x ∈ R ∩ S ∩ T , we shall now show that each summand in the last expression is ≤ α(x) ∧ γ (x). By Chang’s subdirect representation theorem
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MANY-VALUED POINTS AND EQUALITY
(Chang 1959; Cignoli and Mundici 1959), it is no loss of generality to assume A to be totally ordered. Arguing then by cases we have Case 1: α(x) > β(x). Then −β(x) + (α(x) ∧ β(x)) + (β(x) ∧ γ (x)) = β(x) ∧ γ (x) ≤ α(x) ∧ γ (x). Case 2: α(x) ≤ β(x). Subcase 2.1: γ (x) > β(x). Then −β(x) + (α(x) ∧ β(x)) + (β(x) ∧ γ (x)) = α(x) ∧ β(x) = α(x) ≤ α(x) ∧ γ (x). Subcase 2.2: γ (x) ≤ β(x). Then −β(x) + (α(x) ∧ β(x)) + (β(x) ∧ γ (x)) = −β(x) + α(x) + γ (x) ≤ α(x) ∧ γ (x), as one can immediately see arguing by cases according as α(x) > γ (x) or α(x) ≤ γ (x). In conclusion we have obtained "
#
M
"
α(x) ∧ β(x)
x∈R∩S
M
# β(x) ∧ γ (x)
x∈S∩T
X
≤0∨
α(x) ∧ γ (x)
x∈R∩S∩T
≤
X
α(x) ∧ γ (x)
x∈R∩T
=
M
α(x) ∧ γ (x),
x∈R∩T
as required to prove (7).
Example. With the above notation, let E = {a, b} and A = {0, 1/2, 1} be the three-element Łukasiewicz chain. Let further, α(a) = 1 = 1 − α(b), β(a) = β(b) = 1/2, γ (a) = 0 = 1 − L γ (b). Then R = {a}, S = {a, b}, T = {b}. Direct inspection shows that L L x∈R∩S α(x) ∧ β(x) = 1/2 = x∈S∩T β(x) ∧ γ (x). On the other hand, x∈R∩T α(x) ∧ γ (x) = 0, thus yielding a counterexample to the inequality [[α = β]] ∧ [[β = γ ]] ≤ [[α = γ ]].
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REFERENCES
Chang, C. C.: 1958, ‘Algebraic Analysis of Many-Valued Logics’, Transactions of the American Mathematical Society 88, 467–490. Chang, C. C.: 1959, ‘A New Proof of the Completeness of the Łukasiewicz axioms’, Transactions of the American Mathematical Society 93, 74–90. Cignoli, R. and D. Mundici: 1997, ‘An Elementary Proof of Chang’s Completeness Theorem for the Infinite-Valued Calculus of Łukasiewicz’, Studia Logica 58, 79–97. Mundici, D.: 1986, ‘Interpretation of AF C ∗ -algebras in Łukasiewicz Sentential Calculus’, Journal of Functional Analysis 65, 15–63. Costas Drossos Department of Mathematics University of Patras Patras Greece E-mail:
[email protected] Daniele Mundici Department of Computer Science University of Milan Milan Italy E-mail:
[email protected] 