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D-Posets and Effect Algebras 10th Int. Workshop on IFSs, Banská Bystrica, 29 Sept. 2014 Notes on Intuitionistic Fuzzy Sets ISSN 1310–4926 Vol. 20, 2014, No. 4, 32–40 D-posets and effect algebras Martina Paulínyová Department of Mathematics, Faculty of Natural Sciences Matej Bel University Tajovského 40, 974 01 Banská Bystrica, Slovakia Abstract: In the paper two algebraizations of IF-sets families are considered: D-posets [11] and effect algebras [5]. An elementary proof is presented of the fact that D-posets and effect algebras are isomorphic structures [12, 13]. Moreover a product is defined on effect algebras and it is proved that the corresponding algebraic structure is equivalent with the Kôpka D-poset [15, 16]. Keywords: D-poset, Effect algebra, Multiplicative operation. AMS Classification: 03G12, 03B5D. 1 Introduction MV-algebras presents a natural algebraization of IF - set families. It was proved in [24–26]. From the point of view the cathegory theory it was studied in [6–8]. From the MV-algebra theory results one can obtain some results for IF - sets. Especially important applications gives probability theory on MV-algebras (for a review see [28, 25]). Of course, there are other algebraizations of IF-families: new algebras [3], D-posets [10–12, 17], effect algebras [5], A-posets [31]. In the paper we show that D-posets and effect algebras are equivalent. Both structures have been studied in many papers, so D-posets (e.g. [7, 8, 30]) as effect algebras (e.g. [4, 24]). Very imporant are some applications in quantum structures [14, 15, 20], mainly from the point of view of probability theory. A special case of D-posets are MV-algebras, very near strutures to IF-sets families. They play similar role in quantum structures like Boolean algebras in the classical physics. For the study of conditional probability the MV-algebras with product have been defined and announced. They have been defined independently by Riecanˇ [23], and Montgana [21]. A review of probability theory in MV-algebras is contained in [28] (see also [25]). 32 Therefore Kôpka introduced a product in D-posets [16, 17]. The basic properties of probabil- ity in such D-posets (called Kôpka’s D-posets) were introduced in [27] and then used in [19, 29]. Hence we see that analogous problems can be studied in effect algebras with product. We will define multiplicative operations in D-posets and effect algebras and show, that these structures are equivalent. 2 D-posets and effect algebras Definition 1 By a D-poset D is considered an algebraic structure D = (D; ≤; −; 0; 1), where: 1. ≤ is a partial ordering on D with the least element 0 and the greatest element 1, 2. − : D × D ! D is a partial binary operation, where b − a is defined if a ≤ b and there holds: b − a ≤ b; (1) b − (b − a) = a; (2) a ≤ b ≤ c ) c − b ≤ c − a; (c − a) − (c − b) = b − a: (3) Definition 2 An effect algebra E is an algebraic structure E = (E; +; 0; 1), where + : E × E ! E is a partial binary operation on E with fixed elements 0 and 1 and satisfying the following properties: 9 a + b , 9 b + a; a + b = b + a; (4) 9 a + (b + c) , 9 (a + b) + c; a + (b + c) = (a + b) + c; (5) 8a 9! a0; a + a0 = 1 and 00 = 1; (6) 9 a + 1 ) a = 0; (7) Proposition 1 Algebraic structures listed above have different definitions (different binary oper- ations, partial ordering relation in D-posets) but they describe the same mathematical object. Before we prove this statement we notice the partial ordering axioms. Definition 3 Let A be a set. The relation ≤ is a partial ordering on A if satisfies the following conditions: a ≤ a; 8 a 2 A; a ≤ b; b ≤ a ) a = b; a; b 2 A; (8) a ≤ b; b ≤ c ) a ≤ c; a; b; c 2 A: (9) Proof. Let D be a D-poset. We will show now, that there exists corresponding effect algebra. 33 1. Assume, that there exists an element a + b. It means, that 9 c 2 D; c ≥ a; b = c − a, then we define c = a + b: (10) According to (1) we have c − a ≤ c ) b ≤ c and according to (2) we have c − (c − a) = a, since c − a = b after the substitution we get a = c − b and so c = b + a. c = a + b = b + a: We derived the first axiom of effect algebras. 2. Assume, that 9 a + (b + c). Then element b + c exists and we can write 9 d 2 D; d ≥ b; c = d − b: Then according to (10) d = b + c. There holds 9 e 2 D; e ≥ a; d = e − a: Again according to (10) e = a + d. We substitute d = b + c and get e = a + (b + c): Let 9 f 2 D; f ≥ a; b = f − a, then f = a + b. As we mentioned before e = a + d. We use the axiom (4), which we already proved, and get a + d = d + a so d = e − a. We substitute the element d defined before and apply the axiom (4). e − a = b + c = c + b ) (e − a) − c = b Since b = f − a, thus we have (e − a) − c = f − a. We apply the axiom (3) (e − a) − (e − f) = f − a Then c = e − f ) e = f + c ) e = (a + b) + c. e = a + (b + c) = (a + b) + c Property (5) is now proved. 3. Element 1 is defined as the greatest element of D. Then for all a 2 D there holds a ≤ 1. Denote a0 = 1 − a. According to (10) we get 1 = a + a0. Consider a; a0 2 D; a + a0 = 1, a0 = 1 − a and let 9 a00 2 D; a + a00 = 1. Following the commutative law shown before we have a + a00 = a00 + a = 1. Thus a = 1 − a00. Using the axiom (2) we get: a0 = 1 − (1 − a00) = a00 ) a0 = a00 Let a = 0. As proven a + a0 = 1, then a0 = 1 − a = 1 − 0 = 1. 34 4. Assume, that there exists a + 1 2 D. Denote a + 1 = x. We proved before, that the operation + is commutative, thus we have a + 1 = 1 + a = x. It follows x ≥ 1; a = x − 1. According to the definition of D-posets is 1 the greatest element, therefore 8x 2 D; x ≤ 1. On the basis of partial ordering axiom (8) we have x ≤ 1; x ≥ 1 ) x = 1 After the substitution we get a = x − 1 = 1 − 1 = 0: We showed, that it is possible to derive a corresponding effect algebra to any given D-poset. Now let E = (E; +; 0; 1) be any given effect algebra. We will show, that it is possible to create an equivalent D-poset. We define a partial binary relation ≤ as follows: a ≤ c , 9 b 2 E; a + b = c: We will show now, that the relation ≤ satisfies the partial ordering axioms. 1. According to (6) 8a 9! a0; a + a0 = 1. Hence for the element 1 holds 9 1 + 10 = 1. Applying the axiom (7) we have 10 = 0. Thus: 1 = 1 + 0 = (a + a0) + 0 = (a0 + a) + 0 = a0 + (a + 0): But 1 = a + a0 = a0 + a, so a + 0 = a ) a ≤ a: 2. Let a ≤ b and b ≤ a. Then a ≤ b ) 9 x 2 E; a + x = b; b ≤ a ) 9 y 2 E; b + y = a: Following the axiom (5) there holds (a + x) + y = a + (x + y) = a. Denote x + y = z. Then a + z = a. According to the axiom (6) a + a0 = 1. We substitute and compute: 1 = (a + z) + a0 = (z + a) + a0 = z + (a + a0) = z + 1: We got the equation z + 1 = 1. It follows, that z = 0 and so x + y = 0. Now we consider an element x0 2 E. x0 = x0 + 0 = x0 + (x + y) = (x0 + x) + y = 1 + y: Following (7) 9 1 + y ) y = 0. Thus we have x + 0 = 0 and then x = 0. Finally we get the following equation a + 0 = b ) a = b: 35 3. Let a ≤ b and b ≤ c. Then a ≤ b ) 9 x 2 E; a + x = b; b ≤ c ) 9 y 2 E; b + y = c: Substituting b into the second equation and applying the axiom (5) we get (a + x) + y = a + (x + y) = c ) a ≤ c: We proved, that the binary relation ≤ we defined is a partial ordering on E. Now we define a partial binary operation − : E × E ! E. Let a; c 2 E; a ≤ c. Then 9 b 2 E; a + b = c, we denote b = c − a: (11) Based on this definition we will show, that we can apply axioms of D-posets. 1. Let a ≤ b.Then there exists c 2 E; a + c = b. According to (4) a + c = c + a ) c ≤ b: Following the operation − defined in (11) there holds c = b − a. Thus we get b − a ≤ b 2. From the former part of this proof we have c = b − a; a + c = b. We apply (4): b = a + c = c + a. Thus we have a = b − c and substituting into c = b − a we get a = b − (b − a). 3. Now assume, that a ≤ b; b ≤ c, according to (9) is a ≤ c. Thus there exist elements x; y; z 2 E satisfying: a + x = b ) x = b − a; b + y = c ) y = c − b; a + z = c ) z = c − a: Since c = b + y; b = a + x, substituting and applying the equation (5) we get c = (a + x) + y = a + (x + y): But we defined c = a + z.
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