Proceedings of the World Congress on Engineering 2014 Vol II, WCE 2014, July 2 - 4, 2014, London, U.K. On Algebraization of Classical First Order Logic Nabila M. Bennour, Kahtan H. Alzubaidy Abstract - Algebraization of first order logic and its 3) SSS()()() deduction are introduced according to Halmos approach. Application to functional polyadic algebra is done. 4) S() id id 5) SJSJ()()()() if IJIJ Index Term: Polyadic algebra, Polyadic ideal, Polyadic filter, 6) (JSSJ ) ( ) ( ) ( 1 ( )) if is one to one on Functional polyadic algebra. 1 J .The cardinal number I is called the degree of the I. INTRODUCTION algebra. If I = n we get the so called n -adic algebra. If There are mainly three approaches to the algebraization of I I first order logic. One approach is to develop cylindric I then 2 and I id . Therefore we get algebras[1]. The second approach is by polyadic algebra[2]. only the identity quantifier () id . Thus 0 -adic the third one is by category theory [3]. In 1956 P. R. Halmos introduced polyadic algebra to express algebra is just the Boolean algebra B . If I 1 then first order logic algebraically. I Polyadic algebra is an extension of Boolean algebra with I id. Therefore there are only two quantifiers () operators corresponding to the usual existential and universal and ()I . Thus 1-adic algebra is the monadic algebra. quantifiers over several variables together with endomorphi- sms to represent first order logic algebraically. Certain polyadic filters and ultra filters have been used to Polyadic Ideals And Filters express deduction in polyadic algebra. These ideas are A subset U of a Boolean algebra B is called a Boolean applied to the specific case functional polyadic algebra. ideal if II. POLYADIC ALGEBRA i) 0U General definition [3] ii) a b U for any a, b U Suppose that B is a complete Boolean algebra. An iii) If a U and b a then b U existential quantifier on B is a mapping : BB such that A subset F of a Boolean algebra B is called a Boolean filter if i) (0) 0 i) 1 F ii) a () a for any a B . ii) a b F for any a, b F iii) (a ( b )) ( a ) ( b ) for any a, b B . iii) If a F and b a then b F . (,)B is called monadic algebra. A subset U of a polyadic algebra B is called a polyadic ideal of B if Let II : I I is a transformation. Denote the i) U is a Boolean ideal set of all endomorphisms on B by End() B and the set of ii) If J I and a U then J a U all quantifiers on B by Qant() B . iii) If a U and II then S a U A polyadic algebra is (,,,)BIS where A subset F of a polyadic algebra B is called a polyadic S: II End() B and : 2I Qant ( B ) such that for filter of B if I I any JK, 2 and , I we have i) F is a Boolean filter 1) id ii) If J I and a F then J a F 2) JKJK iii) If a F and II then S a F . Manuscript received February 06 2014; revised March 24 2014 Nabila M. Bennour is a lecturer at Mathematics department, faculty of Proposition 1) [3] science, Benghazi university; e-mail: [email protected] A subset U of a polyadic algebra B is an ideal of it if and Kahtan H. Alzubaidy is a lecturer at Mathematics department, faculty of science, Benghazi university; e-mail: [email protected] only if U is an ideal of the Boolean algebra B and ISBN: 978-988-19253-5-0 WCE 2014 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online) Proceedings of the World Congress on Engineering 2014 Vol II, WCE 2014, July 2 - 4, 2014, London, U.K. I a U for every a U . F is a filter of B if and Therefore b1 b 2 J . If only if is a filter of the Boolean algebra and F B a b x1 x 2 .... xn , then a J . Then J is I a F whenever a F . a Boolean ideal containing . Therefore UJ . Proof If b J , then b x x ... x where x Let U be an ideal of a polyadic algebra , by the definition 1 1 2 n i i.e. x U . Then b U . Thus UJ as a U is an ideal of a Boolean algebra and I a U . i Let U be an ideal of a Boolean algebra B and Boolean ideal. Let a J then a x1 x 2 ... xn I a U for every a U . IaIxx 1 2 ... xn Ix 1 Ix2 ... Ix n If JI , then J a I a, so that Therefore I a U. J a U . Now we verify that S a U , if I a U and I . We have a I a and ii) A similar argument leads to (ii). S a S I a , so it means to show that every A filter F of a polyadic algebra is called ultrafilter if F is S acts as identity transformation from I I since there is maximal with respect to the property that 0 F . nothing out of I .Then Ultrafilters satisfy the following important properties [4]. S I a id I a I a , therefore S a U . Proposition 5) A similar argument proves the second part. Let F be a filter of a polyadic algebra B . Then i) is an ultrafilter of iff for any exactly one Proposition 2) [3] F B a F There is a one to one correspondence between ideals and of a,' a belongs to F . filters : if U is an ideal , then the set UF of all a ii) F is ultrafilter of B iff 0 F and a b F iff with a U is a filter . Analogously, if F is a filter , then a F or b F for any a, b F . U F a : a F is an ideal. iii) If a B F , then there is an ultrafilter L such that F L and a L . Proposition 3) The set of all polyadic ideals and the set of all polyadic Let B . The ultrafilter containing F() is denoted by filters are closed under the arbitrary intersections. UF () Let B be a polyadic algebra and B . LetU () denote A mapping : BB between two Boolean algebras is the least polyadic ideal containing and F() denote the 1 2 called a Boolean homomorphism if least polyadic filter containing .We say that U () and i) a b a b F() are generated by . ii) a a Obviously, a b a b , 0 0 , Proposition 4) Let B be a polyadic algebra and B . Then 1 1. i) A mapping : BB between two polyadic algebras is U b B: b x1 x 2 ... xn for some x1, x 2 ,..., xn 0 1 2 ii) called polyadic homomorphism if F b B: b x1 x 2 ... xn for some x1, x 2 ,..., xn 1 i) is a Boolean homomorphism Proof i) Let ii) I J b B: b x1 x 2 ... xn for some x1, x 2 ,..., xn 0 iii) for any I 0 J . Let b1, b 2 J . Then b1 x 1 x 2 ... x n Obviously, . and b2 y1 y 2 ... y m for some xi, y i . b1 b 2 x 1 x 2 .... xn y1 y 2 ... ym . ISBN: 978-988-19253-5-0 WCE 2014 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online) Proceedings of the World Congress on Engineering 2014 Vol II, WCE 2014, July 2 - 4, 2014, London, U.K. III. DEDUCTION (p q )( a ) p ( a ) q ( a ) , Notation: for B and b B , ⊢ b reads b is (p q )( a ) p ( a ) q ( a ) , p()() a p a , deduced from in B . Now , we define ⊢ b in a 0(a ) 0 , 1(a ) 1 for any a A. We have polyadic algebra B iff b F . By definitions of filter and deduction ⊢ we have : Proposition 8) [2] A Proposition 6) B , , , ,0,1 is a functional Boolean algebra. i) ⊢1 and ⊬ 0 ii) If ⊢ x , ⊢ y then ⊢ x y . Proposition 9) [5] A A A B , is a monadic algebra, where : BB is General properties of deduction are given by the following: given by (p )( a ) sup p ( a ) : a A Theorem 7) i) If b then ⊢ b Proposition 10) [2] ii) If ⊢ b and then ⊢ b A BIS,,, is a polyadic algebra, where iii) If ⊢ a for any a and ⊢ b , then ⊢ b S: II End() B A and : 2I Qant ( B A ) . iv) If ⊢ b , then ⊢ b for any substitution v) If ⊢ b , then ⊢ b for some finite k 0 0 . Now, let BA p p: Ak B is a function; Proof. k 0,1,2,... Define i) b b F ⊢ b I S( ) p ( a1 ,..., ak ) p ( a (1) ,..., a (k ) ) for any I ii) ⊢ b b F b x1 x 2 ... xn for k and (a1 ,..., ak ) A . we have some xi b F ⊢ b Proposition 11) iii) Let ⊢ b b F Ak BIS,,, is a polyadic algebra b x1 x 2 ... xn for some xi Proof ⊢ x1 , ⊢ x2 ,…, ⊢ xn i) ( )p ( a1 ,..., ak ) sup p ( a1 ,..., ak ) p ( a1 ,..., ak ) ⊢ x x ... x by proposition (6) 1 2 n () id x1 x 2 ... xn F b F ii) (J M ) sup p ( a1 ,... ak ) ⊢ b JM iv) ⊢ b b F b x x ..