AXIOMATIZATION OF THE SENTENTIAL LOGIC DUAL TO SOBOCIŃSKI S n-valued LOGIC

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1 Bulletin of the Section of Logic Volume 40:1/2 (2011), pp Anetta Górnicka AXIOMATIZATION OF THE SENTENTIAL LOGIC DUAL TO SOBOCIŃSKI S n-valued LOGIC Abstract In [4] we introduced the sentential calculus dual to Sobociński s n valued logic. Here we give an axiomatization of the calculus and prove its completeness. The paper is a continuation of our research on dual logics. In [2], [3], [1] we presented, respectively, axiomatic systems for logics dual to classical logic, Lukasiewicz three-valued logic and nonsense logic W. 1. Introduction The definition of the dual consequence applied here was given by R. Wójcicki [8]. Let Cn be an arbitrary consequence operation in S, where S is the set of all formulas in some language J. The consequence dcn dual to the consequence Cn is defined as follows: α dcn(x) Y [Y X card(y ) < ℵ 0 Cn({β}) Cn({α})] for all α, β S and every X S. The logic (S, dcn) is called dual to the logic (S, Cn). Let J = (S, {, }) be the language of Sobociński s n valued logic described in [7]. The logical matrix of the calculus is of the form: For clarity we use the same symbols both for language functors and corresponding matrix functions. β Y

2 48 Anetta Górnicka M Sob = ({0, 1, 2,..., n 1}, {1, 2,..., n 1}, {, }), (n 3), where functions, are given below: { y if x y, x y = n 1 if x = y, { x + 1 if x < n 1, x = 0 if x = n 1, for any x, y {0, 1,..., n 1}. The calculus dual to Sobociński s n valued logic can be characterized by the following matrix dual to M Sob : M d Sob = ({0, 1, 2,..., n 1}, {0}, {, }), (n 3), where functions and are defined in the same way as in M Sob. To simplify the notation, we introduce a new functor d defined by α d β df = (α β) ( ((α β) (α β))), which is called the dual implication. The functor d has the following values in the matrix M d Sob : { x d n 1 if x 0 and y = 0, y = 0 otherwise. Let us define recursively a generalized negation by 1 α = α, n α = n 1 ( α); n Axiomatization We are going to prove that a dual system to Sobociński s system can be based on the following set of axioms: a1. (α d β) d α, a2. ((α d β) d (α d γ)) d (γ d β), a3. α d (α d (β d α)), a4 i. (β d α) d n i α, 1 i n 1,

3 Axiomatization of the Sentential Logic Dual to Sobociński s n-valued Logic 49 a5. α d n α, a6. n α d α, a7. (γ d (α β)) d α, a8 i. (γ d (α β)) d n i β, 1 i n 1, a9 i. ((α β) d β) d n i α, 1 i n 1, a10 i. (γ d n i (α β)) d n i α, 1 i n 2, a11 i. (γ d n i (α β)) d β, 1 i n 2, a12 i,j. (γ d n i (α β)) d n j β, 1 i n 2, 1 j n 1, i j, a13 i. ( n i (α β) d n i β) d α, 1 i n 1, a14 i,j. ( n i (α β) d n i β) d n j α, 1 i, j n 1, i j. a15 i. ( (α β) d n i β) d n i α, 0 i n 1. Let is denote Ax = {a1,..., a15 i }. As the only rule of inference r d mp we admit: Let R d = {r d mp}. rmp d : α d β, β. α The consequence C R d(x) is the smallest set Y of formulas of the language of Sobociński s n valued logic closed with respect to rmp d and such that Ax X Y. It was proved in [4] that the consequence C R d is dual to Sobociński s consequence based on Modus Ponens. Let T d be the set of all theorems of the logic dual to Sobociński s n valued logic, i.e. T d = C R d( ). By means of a1, a2 and a3 it is easy to prove the following deduction theorem:

4 50 Anetta Górnicka Theorem 1. For any X S and any formulas α, β S, if α C R d(x {β}), then α d β C R d(x). According to a1 and a2 we also get: Theorem 2. For any formulas α, β S and any X S, if α d β C R d(x), then α C R d(x {β}). 3. Completeness Now, we are going to prove the completeness of our axiomatic system, i.e. we prove that the set T d of theorems of the logic dual to Sobociński s n valued logic is identical with the content E(M d Sob ) of the matrix Md Sob. Theorem 3. T d = E(M d Sob). Proof: It is clear that T d E(M d Sob ). In order to prove E(M d Sob ) Td we apply o method [5]. Suppose γ / T d. We are going to show that γ / E(M d Sob ). According to Lindenbaum s theorem there exists a set Y 0 S with the following properties: (1) γ / Y 0, (2) C R d(y 0 ) = Y 0, (3) α/ Y0 γ C R d(y 0 {α}). Let us define Y i We can prove that 1. Y i Y j = for i j, 2. n 1 Y i = S Y 0. i=1 for 1 i n 1 by: Y i = {α : n i α Y 0 }. To prove 2. suppose that α Y k for some 1 k n 1 and α Y 0. Then n k α Y 0 and α Y 0. Applying a4 k and Theorem 2, we get γ C R d(y 0 ) = Y 0, which contradicts (1).

5 Axiomatization of the Sentential Logic Dual to Sobociński s n-valued Logic 51 By means of the same tools we can prove 1. Now, we are going to prove that for any α, β S: (a o ) α Y 0 iff α Y n 1, (a i ) α Y i iff α Y i 1 for every 1 i n 1, (b i ) α β Y i iff (α / Y i and β Y i ) for every 0 i n 2, (b n 1 ) α β Y n 1 iff β Y n 1 or there exists 0 i n 2 such that (α Y i and β Y i ). Equivalences (a o ) and (a i ) for 2 i n 1 follow directly from the definition of Y i for i 1. (a 1 ),,. Let α Y 1. By the definition of Y 1, we have n 1 ( α) Y 0, then n α Y 0 = C R d(y 0 ). Thus, by a5, we get α C R d(y 0 ) = Y 0.,,. Let α Y 0. By a6, we get n α C R d(y 0 ) = Y 0. Since n α = n 1 ( α) Y 0, then α Y 1. (b 0 ),,. Let α β Y 0. If α Y 0, then by a7, we get γ C R d(y 0 ) = Y 0, which contradicts (1). If β / Y 0, then β S Y 0 and by 2. we get β Y i for some 1 i n 1. From the definition of Y i we obtain n i β Y 0. Hence, according to a8 i, we have again γ Y 0.,,. Let us suppose that α / Y 0 and β Y 0. Thus, there exists 1 i n 1 such that α Y i, so n i α Y 0. According to a9 i, we obtain α β Y 0. (b i ) for 1 i n 2,,. Let α β Y i, where 1 i n 2. If α Y i, then n i (α β) Y 0 and n i α Y 0. Therefore, by a10 i, we get γ C R d(y 0 ) = Y 0, which contradicts (1). If β / Y i, then we have two possibilities I β Y 0. In this case we get a contradiction applying a11 i, since n i (α β) Y 0. II β Y j for some 1 j n 2, i j. Then n i (α β) Y 0, n j β Y 0 and a contradiction results from a12 ij.

6 52 Anetta Górnicka,,. Let α / Y i and β Y i, where 1 i n 2. Since n i β Y 0, then in case α Y 0 by a13 i, we get n i (α β) Y 0, so α β Y i. If α Y j, where 1 j n 1, j i, we have n j α Y 0 and n i β Y 0. Applying a14 i,j, we obtain n i (α β) Y 0 and, hence α β Y i. (b n 1 ),,. Let α β Y n 1 and let us suppose that β / Y n 1 and there is no such 0 i n 2 that both α and β belong to Y i. Since β Y j for some 0 j n 2, then α Y k for some 0 k n 1, where j k. Thus (α β) Y 0, n j β Y 0 and n k α Y 0. Using a14 j,k, we get n j (α β) Y 0, so α β Y j, which contradicts our assumption.,,. I Let β Y n 1 and suppose that α β / Y n 1. Then β Y 0 and n i (α β) Y 0 for some 0 i n 2. By a12 i,n 1, we obtain γ Y 0,which contradicts (1). II Let α, β Y i for some 0 i n 2. Then n i α, n i β Y 0 and according to a15 i, we have (α β) Y 0, which means that α β Y n 1. Due to properties 1. and 2. of sets Y i we can define a valuation v of propositional variables p i by (3.1) v(p i ) = k iff p i Y k for 0 k n 1; i 1. Let v be the homomorphical extension of the valuation v on S. We are going to prove that for any δ S (3.2) v (δ) = k iff δ Y k for 0 k n 1. The proof is inductive on the complexity of formulas. In the case δ = p i thesis follows directly from the definition of v. Let us assume that (3.2) is true for all subformulas of δ. If δ = α, then v (δ) = v ( α) = (v (α)). If δ Y 0, then α Y 0, hence α Y n 1 (by (a )), so (v (α)) = (n 1) = 0. If δ Y i, then α Y i, hence α Y i 1 for 1 i n 1 (by (a i )), so:

7 Axiomatization of the Sentential Logic Dual to Sobociński s n-valued Logic 53 (v (α)) = (i 1) = i. Now, let us consider the case δ = (α β). Then v (δ) = v (α β) = (v (α) v (β)). If δ Y 0, then α β Y 0, so, according to (b 0 ), we have α Y i for some 1 i n 1 and β Y 0. Therefore (v (α) v (β)) = (i 0) = 0. If δ Y i, then α β Y i, so α / Y i and β Y i for some 0 i n 2 (by (b i )), hence α Y j and β Y i for some i j. We get (v (α) v (β)) = (j i) = i. If δ Y n 1, then (α β) Y n 1, hence β Y n 1 or (α Y i and β Y i ) for some 0 i n 2. a) If β Y n 1, then (v (α) v (β)) = (v (α) (n 1)) = n 1. b) If α Y i and β Y i for some 0 i n 2 we have which completes the proof of (3.2). (v (α) v (β)) = (i i) = n 1, Therefore, v : S {0,..., n 1} is a homomorphism from (S, {, }) into ({0, 1,..., n 1}, {, }). Since γ / Y 0, then γ n 1 Y i. Thus, by (3.2), we get v (γ) 0. Then v falsifies the formula γ in the matrix M d Sob, which means that γ / E(M d Sob ). i=1 References [1] G. Bryll, A. Górnicka, Dwie formalizacje rachunku dualnego wzglȩdem systemu W (in Polish), Prace naukowe WSP w Czȩstochowie, Matematyka VII, (1999), pp [2] A. Górnicka, Aksjomatyzacja rachunku zdaniowego dualnego wzglȩdem klasycznego implikacyjno-negacyjnego rachunku zdań (in Polish), Prace naukowe WSP w Czȩstochowie, Matematyka VII, (1999), pp

8 54 Anetta Górnicka [3] A Górnicka, Axiomatzation of the Sentential Logic Dual with Respect to Lukasiewicz s Three - valued Logic, Bulletin of the Section of Logic, vol. 35 no. 2/3 (2006), pp [4] A. Górnicka, The logic dual to Sobociński s n valued logic, Scientific Issues, Mathematics XV, Czȩstochowa, (2010)(in print). [5] J. Loś, An algebraic proof of completeness for the two - valued propositional calculus, Colloquium Mathematicum 2 (1951), pp [6] W.A. Pogorzelski, The classical propositional calculus (in Polish), PWN Warszawa, (1975), pp [7] B. Sobociński, Axiomatization of certain many-valued systems of the theory of deduction Roczniki prac naukowych zrzeszenia asystentów Uniwersytetu Józefa Pi lsudskiego w Warszawie, 1 (1936), pp [8] R. Wójcicki, Dual counterparts of consequence operations, Bulletin of the Section of Logic, vol. 2, 1 (1973), pp Institute of Mathematics and Computer Science Jan D lugosz University of Czȩstochowa al. Armii Krajowej 13/ Czȩstochowa Poland a.gornicka@ajd.czest.pl