Study of Pseudo BL Algebras in View of Left Boolean Lifting Property

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1 Available at Appl. Appl. Math. ISSN: Vol. 13, Issue 1 (Jue 2018), pp Applicatios ad Applied Mathematics: A Iteratioal Joural (AAM) Study of Pseudo BL Algebras i View of Left Boolea Liftig Property 1 B. Barai ia ad 2 A. Borumad Saeid 1 Departmet of Mathematics Kerma Brach Islamic Azad Uiversity Kerma, Ira Barai.Bagher@yahoo.com 2 Departmet of Pure Mathematics Faculty of Mathematics ad Computer Shahid Bahoar Uiversity of Kerma Kerma, Ira arsham@uk.ac.ir Received: November 22, 2017; Accepted: March 6, 2018 Abstract I this paper, we defie left Boolea liftig property (right Boolea liftig property) LBLP (RBLP) for pseudo BL algebra which is the property that all Boolea elemets ca be lifted modulo every left filter (right filter) ad ext, we study pseudo BL-algebra with LBLP (RBLP). We show that Quasi local, local ad hyper Archimedea pseudo BL algebra that have LBLP (RBLP) has a iterestig behavior i direct products. LBLP (RBLP) provides a importat represetatio theorem for semi local ad maximal pseudo BL algebra. Keywords: Boolea ceter; (maximal, local, hyper Archimedea, quasi-local, semi local) pseudo BL algebra; liftig property; (prime, maximal) filter MSC 2010 No.: 06F35, 03G25 1. Itroductio I 1998, Hajek preseted BL-algebra; a algebraic sematics of basic fuzzy logic (Wag ad Xi (2011)). They are geerated by cotiuous t-orms o the iterval [0, 1] ad their residuals (Georgescu ad Muresa (2014)). The, Georgescu itroduced pseudo BL-algebra as a o- 354

2 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 355 commutative extesio of BL-algebra (Georgescu ad Leustea (2002)). The idea of pseudo BL algebra origiates ot oly i logic ad algebra, but also algebraic properties that come from the sytax of certai o-classical propositioal logics, ituitioistic logic. A liftig property for Boolea elemets appears i the study of maximal MV-algebras ad maximal BL algebra. The left liftig property for Boolea elemets modulo, the radical, plays a essetial part i the structure theorem for maximal pseudo BL-algebra. I order to exted the previous works, Georgescu ad Muresa studied Boolea liftig property for arbitrary residuated lattice (Georgescu ad Mureşa (2014)). The results of this study were similar to idempotet elemets i the rigs. The aim of this article is to study pseudo BL algebra which satisfy the left (right) liftig property of Boolea elemets modulo every left filter. We have called this property LBLP (RBLP). Also the aim of the curret study is to show that each Boolea algebra ifused pseudo BL-algebra with LBLP (RBLP), ad to show that hyper Archimedea have LBLP (RBLP). It turs out that the algebras, at pseudo BL-algebra with LBLP (RBLP) are exactly the quasi-local pseudo BL-algebras. We will show that arbitrary pseudo BL-algebra has this property iff for each arbitrary elemet x, there exists a Boolea elemet i this pseudo BL-algebra such that it belog to left filter x. Certai results i this paper which refer to properties of rigs with LIP are formulated aalogously to pseudo BLalgebra with LBLP (RBLP) (Georgescu ad Mureşa (2017)). Pseudo BL-algebra with LBLP (RBLP) also coicide with those pseudo BL-algebra whose lattice of filters is dually B- ormal.the study of pseudo BL-algebra also lead to ew properties, with o correspodet for rigs with LIP. These are the mai sources that ispire the research o pseudo BL algebra. Sectio 2 shows theorems that satisfy the coditio of semi local ad cosists of previously kow cocepts about pseudo BL-algebra which are ecessary i the ext sectios. Sectio 3 is related to some results ad examples about pseudo BL algebra. I sectio 4, we defie the LBLP (RBLP) for pseudo BL-algebra ad provide several results related to this property. Sectio 5 is related to characterizatio of the LBLP (RBLP) which obtais several results ad examples cocerig the LBLP (RBLP). 2. Prelimiaries I this sectio, we state a series of kow cocept ad results related to pseudo BL-algebra, all of them will be used i the paper. We make the usual covetio throughout this paper, every algebraic structure will be desigated by its support set. Wheever it is clear which algebraic structure o that set we are referrig to we shall deote by N the set of the atural umbers ad by N the set of ozero atural umbers. Defiitio 2.1. (Georgescu ad Mureşa (2014)) A pseudo BL-algebra is a algebra (A,,,,,,0,1) of type (2,2,2,2,2,0,0) satisfyig the followig (PSBL 1 ) (A,,, 0,1) is a bouded lattice, (PSBL 2 ) (A,, 1) is a mooid, (PSBL 3 ) a b c iff a b c iff b a c, for all a,b,c A, (PSBL 4 ) a b = (a b) a = a (a b), (PSBL 5 ) (a b) (b a) = (a b) (b a) = 1, for all a,b A. Example 2.2.

3 356 B. Barai ia ad A. Borumad Saeid Let a,b,c,d R, where R is the set of all real umbers.we put defiitio (a, b) (c, d) a < c or (a = c ad b d), for ay a,b R R, we defie operatios ad as follows: Let a b = max{a, b} ad a b = mi{a, b}. A = {( 1, b) 2 R2 : b 0} {(a, b) R 2 : 1 < a < 1, b R 2 {(1, b) R 2 : b 0}, for (a, b), (c, d) A, we put: (a, b) (c, d) = ( 1, 0) (ac, bc+d), 2 (a, b) (c, d) = ( 1, 0) 2 [(c, d b ) (1, 0)], a (a, b) (c, d) = ( 1 2, 0) [(c a, ad bc a a ) (1, 0)], the, ( A,,,,,, ( 1, 0), (1, 0)) is a pseudo BL-algebra. 2 Defiitio 2.3. (Georgescu et al. (2002)) A pseudo BL algebra A is commutative iff a b=a b, for all a,b A, ay commutative pseudo BL algebra A is a BL algebra. The, we shall say that a pseudo BL algebra is proper if it is ot commutative. (if it is ot a BL-algebra ). Propositio 2.4. (Georgescu et al. (2002)) If A is a pseudo BL algebra ad a,b, c A, the, ( psbl c 1 ) a (a b) b a (a b ) ad a (a b) a b (b a ), ( psbl c 2 ) (a b) a a b (a b) ad (a b) a b a (b ), ( psbl c 3 ) If a b the, a c b c ad c a c b, (psbl c 4 ) If a b the, c a c b ad c a c b, ( psbl c 5 ) c (a b) = (c a) (c b) ad (a b) c = (a c) (b c ), ( psbl c 6 ) a b iff a b = 1 iff a b = 1, ( psbl c 7 ) a a = a a = 1, ( psbl c 8 ) 1 a = 1 a = a, ( psbl c 9 ) b a b ad b a b, ( psbl c 10 ) a b a b ad a b a, b, ( psbl c 11 ) a b implies b a ad b a, ( psbl c 12 ) (a b ) = a b, (a b ) = b a, ( psbl c 13 ) ( a b ) = a b, ( a b ) = a b, ( psbl c 14 ) ( a b ) = a b, ( a b ) = a b, ( psbl c 15 ) ( a b ) = a b, ( a b ) = a b,

4 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 357 ( psbl c 16 ) ( a b ) = = a = b =, ( a b ) = = a = b =, ( psbl c 17 ) a b = ( (a b ) b ) ( ( b a ) a ), ( psbl c 18 ) a b = (( a b) b ) ((b a) a), ( psbl c 19 ) 1 = 1 = 0, 0 = 0 = 1, ( psbl c 20 ) a a = a a = 0, ( psbl c 21 ) a (b c) (a b) (a c), ( psbl c 22 ) a b b a, a b b a, ( psbl c 23 ) b a iff a b = 0, ( psbl c 24 ) b a iff b a = 0, ( psbl c 25 ) a a b, a a b, (psbl c 26 ) a (a b) b, a (a b) b, hece a (a ), a (a ), ( psbl c 27 ) a a = a a, ( psbl c 28 ) ((a ) ) = a, ((a ) ) = a. Defiitio 2.5. (Lele ad Ngaou (2014)) A o-empty subset F A is called a filter of A, if the followig coditio are satisfied ( F 1 ) If a, b F, the, a b F, ( F 2 ) If a F, b A, a b, the, b F. Defiitio 2.6. (Muresa (2010)) A filter H of A which is called a ormal filter, if (N) For every a, b A, a b H iff a b H. We deote by Ƒ ( A ) the set of all ormal filters of A. Clearly {1} ad A are ormal filter. Defiitio 2.7. (Muresa (2010)) A proper filter P of A is called prime if, for ay a, b A, the coditio a b P implies a P or b P. Propositio 2.8. (Muresa (2010)) If P is a proper filter, the, the followig coditios are equivalet Note: (i) P is prime filter, (ii) For all a, b A, a b P or b a P, (iii) For all a, b A, a b P or b a P. For a A, a 1, there is a prime filter P a such that a P a.

5 358 B. Barai ia ad A. Borumad Saeid Note: Every proper filter F is the itersectio of those filter which a cotai F. I particular, Spec (A) = {1}. Note: A filter of A is maximal if it is proper ad it is ot cotaied i ay other proper filter. Propositio 2.9. (Mohtashamia ad Borumad Saeid (2012)) If F is a proper filter of A, the, the followig coditios are equivalet (i) F is a maximal filter, (ii) For ay a F there exists f F,, m 1 such that ( f a ) m = 0. We shall deote by Max(A) filters of A, it is obvious that Max(A) Spec(A). Ideed, let M Max(A), M is a proper filter of A the, there is a prime filter P of A such that M P. Sice P is proper, it followed that M = P hece M is prime. Defiitio (Wag ad Xi (2011)) The itersectio of the maximal filter of A is called the radical of A ad will be deoted by Rad (A). It is obvious that Rad(A) is filter of A clearly Rad(A) = A iff A is trivial, ad Rad (A) is proper filter of A iff A is o trivial. A elemet a A is said to be dese iff a = a = 0. The set of the dese elemets of A is deoted by D(A), that is D(A) = { a A a = a = 0 }, clearly D(A) sice 1 = 1 = 0. The elemets a A such that a 2 = a a = a are called idempotet elemets clearly, if a elemet a A is idempotet the, a =a for every a N ad thus [a) = { b A b a }. Obviously, the oly elemet of A which is both ilpotet ad idempotet is 0. Notice that, if A has = the, all elemets of A are idempotet. Actually, these two coditios are equivalet: A has = iff all elemet of A are idempotet. The set of the complemeted elemets of the bouded lattice reduct of A is called Boolea ceter the of A ad deoted by B(A). Clearly 0,1 B(A). The elemets of B(A) are called Boolea elemets of A. It is kow that B(A) is a Boolea algebra, with the operatio iduced by those of A together with the complemetatio operatio give by the egatio i A, also it is straightforward that B(A) is a subalgebra of the pseudo BL algebra. Here are some more properties of the Boolea ceter of a pseudo BL algebra. Remark (Ciugu et al. (2017)) (i) e B(A) has uique complemeted, equal to e = e, ad (e ) = (e ) ~ = e. (ii) e, e = 0 iff e = 1.

6 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 359 Propositio (Ciugu et al. (2017)) If A is a pseudo BL algebra, the, for e A, the followig coditios are equivalet (i) e B(A), (ii) e e = e ad (e ) = (e ) ~ = e, (iii) e e = e ad e e = e, (iv) e e = e ad e e = e, (v) e e = 1, (vi) e e = 1. Lemma (Ciugu et al. (2017)) If e B (A), the, for all a A we have (i) e a = e a = a e, (ii) e e = 0 = e e, (iii) e a = e a. Thus, e 2 = e, hece e = e, for every N (all Boolea elemet are idempotet). Therefore, [e) ={a A e a }. Propositio (Georgescu ad Mureşa (2014)) If a A, ad e B(A), the, (i) a e = a e, (ii) a e = e a. Propositio (Cheptea et al. (2015)) For a A ad 1, the followig assertio are equivalet (i) a B(A), (ii) a (a ) = 1, (iii) a (a ) = 1. Defiitio (Cheptea et al. (2015)) A elemet a A is said to be regular iff (a ) = (a ) = a. A is said to be ivolutive iff all its elemets are regular. The elemets a A such that a = 0 for some N are called ilpotet elemets. Clearly, elemet 0 is ilpotet, we shall deote by N(A) the set of ilpotet elemets of A. By the above, for ay a A ad ay filter F of A, (i) [a) = A = [ 0 ) iff a N(A). (ii) If F N (A), the, F = A.

7 360 B. Barai ia ad A. Borumad Saeid Defiitio (Kuhr et al. (2003)) A is said to be local iff it has exactly oe maximal filter. Defiitio above is a equivalet to the fact that Rad(A) is a maximal filter of A, that is A is local iff Rad(A) Max(A) [Max(A)={ Rad(A)}]. Propositio (Kuhr et al. (2003)) The followig coditios are equivalet (i) A is local, (ii) A\N(A) is local, (iii) A\N(A) is proper filter of A, (iv) A\N(A) is a maximal filter of A, (v) A\N(A) is the oly maximal filter of A, (vi) Rad(A) = A\N(A), (vii) A = N(A) Rad(A), (viii) for all a,b A, if a b N(A), the, a N(A) or b N(A). Lemma (Kuhr et al. (2003)) If A is local, the, B(A) ={0,1} ad A =N(A) {a A a, a N(A)}. Remark I Example 2.2, A is a local, hece A = N(A) {a A a, a N(A)}. A is said to be semi local iff Max(A) is fiite. Semi local pseudo BL-algebra iclude the trivial pseudo BL-algebra, local pseudo BL-algebra, fiite BL-algebra, fiite direct product of local or other semi local pseudo BL-algebra. The pseudo BL-algebra A is said to be simple iff it has exactly two filters. That is iff A is o-trivial ad (A) = {1, A}, A is simple iff {1} is a maximal filter of A iff {1} is the uique maximal filter of A iff A is local ad Rad (A) = {1}. A elemet a A is said to be Archimedea iff a B(A) for some N. Equivalet by Propositio 2.15 with a (a ) =1 or a (a ) =1, a pseudo BL-algebra is called hyper Archimedea if all elemets are Archimedea. Clearly, if B(A) = A that is if uderlyig bouded lattice of A is a Boolea algebra, the, A is a hyper Archimedea pseudo BL-algebra. Let us cosider a filter F of A. Georgescu ad Mureşa (2014) defie two biary relatios o A by: L (F) : a L (F) b iff (a b b a ) F. R (F) : a R (F) b iff (a b ) ( b a ) F. For a give filter F, the relatios L (F) ad R (F) are equivalece relatios o A, moreover we have F={ a A, a L (F) 1 }={ a A, a R (F) 1 }. We shall deote by A ( A R, (F)

8 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 361 respectively) the quotiet set associated with L (F) ( R (F), respectively). a L ( a (F) R, (F) respectively) will deote the equivalece class of a A with respect to L (F) ( R (F), respectively). We shall deote the quotiet set A/ (mod ), simply by A, ad its elemets by a with a A, so A = { a a A}, where, for every a A, a ={b A a b (mod )}. Also we shall deote by P :A A the caoical surjectio, P L(F ) ( a) = a for all a A, ad for every X A we shall deote P L(F ) ( X ) = X ={ a a A }. I particular b, 1 L([a)) iff (b 1) (1 b) [a) iff b 1 [a) iff b [a). Especially if F is a ormal filter of A. For all, b A, we deote a b (mod ) ad say that a ad b are cogruet modulo iff (a b) (b a) iff (a b) (b a). It is kow ad easy to check that (mod ) is a cogruece relatio o A. Remark Cosider the pseudo BL-algebra ( A,,,,,, ( 1, 0), (1, 0)) i Example 2.2. The, 2 D(A) = {(1, 0)}, B(A) = {(1, 0), ( 1, 0) }, Max(A)= [(1, 0)), Rad(A) = [(1, 0)) Some results i pseudo BL algebra Lemma 3.1. For all a A ad N, (a ) (a ), (a ) (a ). Accordig to psbl c 23, psbl c 24, psbl c 20 we have 0 = = (a ) (a) hece (a ) (a ) ad similarly (a ) (a ). 0 = (a a) Lemma 3.2. Let H Ƒ ( A ). The, (i) (a ) H iff (a ) H, (ii) a H the, a, a H. (i) (a ) H iff a 0 H iff a 0 H iff (a ) H. (ii) a H, a (a ) (by psbl c 26 ) the, (a ) H hece (a ) ~ H (by(i)). Lemma 3.3. Let M Max (A). a M, if there exists m 1, such that (a m ) = (a m ) = 1.

9 362 B. Barai ia ad A. Borumad Saeid Let a M. Cosider F =[a) ={ b A b a }, F is ot proper sice F is a proper filter there exists M Max(A) such that F M, therefore a M (sice a F), thus F is ot proper therefore 0 F implies there exists N such that a 0, the, 0 0 a 0 (psbl c 4 ), thus (a ) =1, ad similarly (a ) =1. Coversely, if there exists m 1, such that (a m ) = (a m ) =1 thus a m = 0, hece a M. Lemma 3.4. Rad(A)={ a A ( N ), ( k N ) s. t ( ( a ) ) k = ( ( a ) ) k = 0}. a Rad (A) iff a M iff a M, for all M Max (A) iff a M iff (a ), (a ) M iff ( ( a ) ) k = ( ( a ) ) k = 0. Corollary 3.5. If a Rad (A), the, a, a N(A). Propositio 3.6. (i) B(A) Rad (A) = {1}, (ii) D(A) is a filter of A ad D(A) Rad (A), (iii) B(A) D(A) ={1}. (i) Clearly,1 B(A) Rad(A). Coversely, if there exists 1 a B(A) Rad(A), thus a Rad(A) ad a B(A) therefore for all 1, there exists m 1 such that (( a ) ) m = (( a ) ) m =0 ad sice a B(A) so a = a, (a) = a, (a ) =a, therefore, 1 = [(( a ) ) m ] = [( ( a) ) m ] = (a ) = a, which is a cotradictio. Let a, b D(A). Therefore, (a b) = a b = a 0 = a = 0, ad (a b) = a b = a 0= a = 0, so (a b), (a b) D(A), ow a D(A), b A, a b, the, b a ad b a that is b 0 ad b 0 so b = a = 0 cosequetly b D(A), ow assume that a D(A), the, sice D(A) is a filter thus a D(A) cosequetly (a ) =(a ) = 0 that is a Rad(A). (iii) Obviously{1} D(A) B(A), ow let a D(A) B(A). Therefore, a = a = 0 implies (a ) = (a ) = 1, but sice a B(A), (a ) = (a ) = a hece a = 1. Example 3.7. Cosider pseudo BL algebra A = { 0, a 1, a 2, b, a 3,...,1}, with A bouded lattice structure give by the Hasse diagram below, the implicatio ad give by the followig tables

10 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 363 Table 1. The biary operatios of A 0 a 1 a 2 b a a 1 a 3 1 a 3 1 a 3 1 a 2 b b b a 2 b a 3 1 a 3 1 a 3 a 1 a 1 b b a 1 a 2 b a a 1 a 2 b a a 1 a 3 1 b 1 a 3 1 a 2 b a b a 2 b a 3 1 a 3 1 a 3 a 1 a 1 b b a 1 a 2 b a a 1 0 a 1 0 a 1 0 a 1 a a 2 a 2 b 0 a 1 0 a 1 a 2 b a a 2 a 2 a 3 a a 1 a 2 b a a 1 a 2 b a b a1 a3 a5 a6 a4 a2 0 Figure 1. The example of pseudo BL-algebra Accordig to example above, D(A) ={0}, that is elemet 0 is dese. The example above will be used for presetig both couter example ad defiitios throughout the article. Propositio 3.8. (i) A, A {1} are isomorphism,

11 364 B. Barai ia ad A. Borumad Saeid (ii) a F = 1 F iff a F, (iii) A A is trivial, (iv) a F = 1 F iff a F, (v) a F = 0 F iff a, a F, (vii) a F b F iff a b F iff a b F. We prove oly (i) ad others are trivial. (i) Defie: A A {1} ; ad G G / F set a bijectio betwee {G Ƒ (A) F G} ad Ƒ (A/F). Furthermore, the mappig M M/F sets a bijectio betwee {M Max (A) F M} ad Max(A/F). From this we immediately get that, whe F Rad(A), Max(A/F) is a bijectio to{m Max(A) F M } = Max(A). Ad that Rad(A/F) = Rad(A)/F. Cosequetly, Max(A/Rad(A)) = Max(A) ad Rad(A/ Rad(A)) = Rad(A)/Rad (A) ={1/ Rad(A)}(Propositio 2.10). Here is the secod isomorphism theorem pseudo BL-algebra, for all ormal filters F, G of A such that F G, the pseudo BL-algebra A/a ad (A F ) ( G F ) are isomorphic (the pseudo BL-algebra isomorphism maps b a (b F ) ( G for every b A). F ) 4. Left Boolea liftig property Throughout this sectio uless metioed otherwise A will be a arbitrary pseudo BL-algebra ad F will be a arbitrary filter of A. The caoical morphism P A A iduces a Boolea morphism B(P ): B(A) B ( A ). The rage of this Boolea morphism is B(P )( B(A)) = P (B(A)) = B(A). Lemma 4.1. If F will be a arbitrary filter of A the, (i) B(A) = { a A a a = 1 }, (ii) B(A) = { a a A, a a = 1 }, (iii) B ( A ) = { a a A, a a L( F) }, (iv) B(A) B ( A ). (i) Follows from Propositio (ii) By (i) we have B(A) ={ a a B (A) } = { a a A, a a = 1 }.

12 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 365 (iii) Accordig to (i) we have B( A ) = { a A a ( a L(F ) ) = 1 } a a = 1 } = { a A = { a a A, a a / = 1 } = { a a A, a a = 1 }. (iv) Let a B(A). The, a A, a a = 1 thus a A, a a hece a B ( A ) therefore B(A) B ( A ). Defiitio 4.2. We say that a Boolea elemet f B ( A ) ca be left lifted iff there exists a Boolea elemet e B(A) such that e = f. I other words, f B ( A ) ca be left lifted iff f B(A). We say that the equivalece relatio has the left Boolea liftig property (LBLP) iff all Boolea elemets of A ca be left lifted. I other words, has LBLP iff Boolea morphism B( P ): B (A) B ( A ) is surjective. I other words: B( P )(B (A)) = B ( A ) iff B(A) = B ( A ). Remark 4.3. has LBLP iff B (P ) is surjective. We say that pseudo BL-algebra A has the left Boolea liftig property (LBLP) iff all of its left equivalece relatio have LBLP. Remark 4.4. For ay liearly ordered pseudo BL algebra obviously B(A)={0,1}, because let a be a complemet of a that meas a a or a a. The, a a =1, a a = 0, thus a =1 or a = 0. For ay equivalece relatio of A, the pseudo BL-algebra A is also liearly ordered, hece B( A ) = { 0, 1 } hece has LBLP. Example 4.5. Cosider the pseudo BL algebra A={ 0, a 1, a 2, b, a 3,..., 1 } i Example 3.7. The, B(A)={0, 1}, let us cosider the filter L([b))={b,1}.The elemet a 3 B(A), a 3= a 3 0= a 1. Thus

13 366 B. Barai ia ad A. Borumad Saeid a 1 a 3 =1 L([b)) (by Propositio 2.4 ) we have a 3 L( [b)) B( A L([b)) ). Ad a 3 L( [b))= {b,1}. Thus a 3 L( [b)) b L([b)) ad 0 L([b)) 1 L([b)), a 3 0 = (a 3 0) ( 0 a 3 ) = a 1 1= a 1 L( [b))), a 3 b= (a 3 b) (b a 3 )= b a 3 = a 3 L([b), a 3 a 1 = (a 3 a 1 ) (a 1 a 3 ) =a 1 a 3 = 0 L([b). Hece a 3 L([b)) a 1 a, 3 L([b)) L([b)) b, a 3 L([b)) a i L([b)) the above we have, 3 i N, therefore a 3 L([b)) L([b)) = {a 3 }. Summarizig a 3 L([b)) B( A a L([b)) ), while a 3 B(A) ad 3 L([b)) = {a 3 } hece a 3 L([b)) B(A) L([b)) therefore B( A L([b)) ) B(A) L([b)) which mea that R([b)) does ot have LBLP. Hece A does ot have LBLP, otice that the maximal filters of A are L([a i )), i N, hece Rad(A) = L([a i )) = L([b)), i N, thus Rad(A) does ot have LBLP. Propositio 4.6. (i) The left trivial filter ad the left improper filter have LBLP. I the case of the trivial, the image of the caoical morphism through the fuctio B is bijective. (ii) If B ( A ) = { 0, 1 }, the, the equivalece relatio has LBLP. (i) P L({1}) A A L({1)} is a pseudo BL algebra isomorphism. The, B(P(L({1})): B(A) B( A L({1}) ) is a Boolea isomorphism, thus a bijectio. Hece a surjectio, so L({1}) has LBLP. (ii) We have B(A) B ( A ) assume that B ( A )= { 0, 1 } sice {0,1} B(A) it follows that { 0, 1 } B(A) thus B(A) = B ( A ). This meas that has LBLP. Note. From the previous lemma we get that if Ƒ(L(A)) = {L({1}), L(A)}, the, L(A) has LBLP, that is every simple pseudo BL algebra has LBLP. Note. For every a A we have a L(A) = 1 L(A), hece A L(A) = { 1 L(A) } = {P L(A) L({1})} = B ( A L(A) ), therefore L(A) has LBLP. This statemet could also have bee deduced from (ii). Ay Boolea algebra iduces a pseudo BL algebra with LBLP, because, if A is a Boolea algebra, we obtai a pseudo BL-algebra i the usual way, the, B(A)=A, hece, for every

14 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 367 equivalece relatio of A, B(A) = A B ( A ) B(A). Therefore B(A) = B ( A L(F )) so L( F) has LBLP also, B ( A L(F )) = A. Sice every a A has a complemet a = a A (a A = B(A) implies a = a A) it follows that every a A has a complemet ( a ) = ( a ) = a = a A, thus B ( A ) = A. Example 4.7. Let A={0,a,1} be the three elemets chai (0 < a <1). The, L([a))={a,1} is a equivalece relatio of this pseudo BL algebra which is both o trivial ad proper. 0 L([a)) = {0} ad a L([a)) = 1, thus A L( [a)) L([a)) = { 0 L([a)), 1 L([a)) } hece B( A L([a)) ) = { 0 L([a)), 1 }, therefore L([a)) has LBLP. Actually, sice B(A)={0,1}, with 0 1, L([a)) ad 0 L([a)) 1 L([a)), it follows that B( P L([a)) ) bijective. Ƒ (A) ={ [0), [a), [1) }={A, [a),[1)} ad A ad [1) have LBLP (By Propositio 4.6 (i) ). Therefore A has LBLP of course, but A is ot a Boolea algebra. Propositio 4.8. (i) Every prime filter of a pseudo BL algebra has LBLP. (ii) Every maximal filter of a pseudo BL algebra has LBLP. (i) Let P be a prime filter of pseudo BL algebra A. Assume by absurdum that P does ot have LBLP, that is B(A) P B ( A P ) ( By Lemma 4.1 (iv)). This meas that there exists a elemet a A such that a P B (A P ), but a P B(A) P (By Lemma 4.1 (iii) ) a a P (a a P) but a P the, a P (a P) ( sice if a P the, a P = 1 P B(A) P ). Sice P is prime filter, it follows that a P (a P) that is a P = 1 P ( a P = 1 P ), that is a P = (a P ) = 1 P ((a P ) = 1 P ) thus a P =0 P B(A) P which is a cotradictio, ( Sice a P B(A) P ) hece P has LBLP. (ii) Sice Max(A) Spec(A) ad by (i) we get the result. If e, f B(A), the, e f B(A), e f B(A). Cosider (e f ) (e f ) = (e f ) ( e f )=(e f) ( e f ) ( ( e f ) e ) ( ( e f ) f ) =1) the, (e f ) (e f) =1 thus e f B(A) ad e f =( f e) = f e B(A), e B(A), therefore e B(A), (Accordig to psbl c 13 ad Propositio 2.14). Cosequetly (e f ) (f e) B(A), ( e f ) ( f e ) B(A). Propositio 4.9.

15 368 B. Barai ia ad A. Borumad Saeid For every filter F of A, the followig coditios are equivalet (i) B(P ) is ijective, (ii) B(A) = {1}. We have P : A A therefore B(P ): B(A) B ( A ). (ii) ( i). Assume that B(A) A ={1}, a,b B(A) such that B(P ) (a)= B(P )(b), that is P (a) = P (b) which mea that a = b iff a b. (i) (ii). 1 B(A), 1 the, {1} B(A). Assue that a B(A), thus a hece a = 1, therefore P (a)=p (1), so B (PL(F ))(a) = B P (1) thus a=1, hece B(A) L( F) {1}. Corollary If B(A)={0,1}, the, for every proper filter F of A, B(P ) is ijective. Sice 1, therefore B(A) ={1} the, B(P ) is ijective. Corollary If (F i ) i I is a o empty family of filter of A such that B(P L(Fi) ), is ijective for every i I, the, B( P L(F i )) i I is ijective. B(P L(Fi )) is ijective by Propositio 4.9 we have B(A) L(F i )={1} ( i) the, B(A) ( L(F i )) ={1} (for all i I) hece B( P L(F i )) i I is ijective. Corollary (i) Ay filter F of A such that F Rad (A), the, B(P ) is ijective. (ii) B(P D(A) ) is ijective. (i) We have Rad(A) thus B(A) B(A) Rad (A)={1} the, B(A)={1} hece B(P ) is ijective. (ii) By Propositio 3.6 we have D(A) Rad(A) thus D(A) B(A) Rad(A) B(A) ={1} hece B (A) D(A) = {1}, therefore B(P D(A) ) is ijective.

16 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 369 Remark I Example 2.2, B(P R(F) ) is ijective, for every proper filter F of A, sice B(A) = {(1, 0)}. I order to prove the mai result of this part i Propositio 5.7 ad Propositio 5.32 we will prove several lemmas ad propositios. 5. Characterizatio of left (right) Boolea liftig property Throughout this sectio, uless motioal otherwise, A will be a arbitrary pseudo BL-algebra. Lemma 5.1. For all a A, we have (i) a L([a a )) B ( A L([a a )) ), (ii) a L([a a )) B ( A L([a a )) ). (i) By Lemma 4.1 (iii) we have (a a ) L([a a )), the, a L([a a )) B( A L([a a )) ). (ii) Is similar to (i). Lemma 5.2. For every a,b A, we have (i) for all N, b a b ad (b ) b a, (ii) for all N, b a b ad (b ) b a, (iii) there exists k N, such that b k b a iff there exists N, such that b a, (iv) there exists k N, such that b k b a iff there exists N, such that b a, (v) there exists k N, such that (b ) k a b iff there exists N, such that (b ) a, (vi) there exists k N, such that (b ) k a b iff there exists N, such that (b ) a, (vii) a b L([ b b )) iff a L([b)) ad a L( [b )), (viii) a b L( [ b b )) iff a L([b)) ad a L([b )). We prove (i), (iii), (v), (vii).

17 370 B. Barai ia ad A. Borumad Saeid (i) Let b, a A ad N. By psbl - C 10, psbl C 9, psbl C 4, we have b b a b, the, b a b ad (b ) b = b 0 b a hece (b ) b a. (iii) If there exists k N, such that b k b a, the, by psbl C 1, psbl C 2, b k b (b a ) b a the, b k+1 a, take k+1 = hece N such that b a. Coversely, if there exists N, such that b a, the, by psbl c 4, psbl - C 3, we have b b a b b a the, b +1 b a thus take +1 = k N. (v) If there exists k N, such that (b ) k a b, the, by psbl - C 22 ad Lemma 5.2 (iii) we have (b ) k a b b a thus (b ) k b a, the, there exists N, such that (b ) a. (vii) Accordig to (i),(iii), (v), the followig equivalet hold a b L( [ b b )) iff a b L([b) ) L([b )) iff a b L([b)) ad a b L([b )) iff exist k, j N, such that b k a b ad (b ) j a b iff there exist k, j N such that b k a b, b k b a, (b ) j a b ad (b ) j b a iff there exist k, j N such that b k a b ad (b ) j a b ( b 0 ) (a 0 ) = b a ( by psbl - C 28 ) iff there exist m, N such that b m a ad (b ) a iff a L([b)) ad a L([b )). Lemma 5.3. For every a A, the followig coditios are equivalet (i) there exists e B(A) such that e a L([a a )), (ii) there exists e B(A) such that e L([a)) ad e L([a )). By Lemma 5.2 (vii) is clear. Remark 5.4. For every a A, the followig coditios are equivalet (i) there exists e B(A) such that e a L( [ a a )), (ii) there exists e B(A) such that e L([a)) ad e L([a )). Notatio 5.5. We shall deote S(A) = {a A ( e B(A)) e a L([a a )), e a L([ a a )) }. Remark 5.6. Accordig to Lemma 5.3 ad Notatio 5.5 we have S(A)={a A ( e B(A)) such that e L([a)) ad e L([ a )) or e L([a )) }.

18 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 371 Propositio 5.7. The followig statemets are equivalet (i) A has LBLP, (ii) For all a A, there exists e B(A), such that e a L([ a a )), (iii) For all a A, there exists e B(A), such that e a L([ a a )), (iv) For all a A, there exists e B(A), such that e L([a)) ad e L([ a )), (v) For all a A, there exists e B(A), such that e L([a)) ad e L([a )), (vi) S(A) = A. We oly prove that (i) (ii). (i) (ii). Let a A. By hypothesis, the equivalece relatio L([ a a )) of A has LBLP, which meas that B ( A ) = B(A), by Lemma 4.1 we have L([a a )) L([ a a )) a B(A). That is there exists e B(A), such that L([a a)) L([a a )) a L([ a a )) = e L([a a )). Therefore e a L( [a a )). (ii) (i). Let F be a arbitrary filter of A ad a A, such that a B ( A ). The, by Lemma 4.1 (iii), a a L([a a )). Sice a A by hypothesis there exists e B(A), such that e a L([a a )) that is e a thus e = a so a B(A), therefore B ( A ) B(A), the, B ( A ) = B(A) that is has LBLP. Hece, A has LBLP. Corollary 5.8. (i) If all the elemet of A are idempotet, the, S(A) = A. (ii) If A is liearly ordered, the, S(A) = A. (i) I Remark 5.6, take e = a. Remark 5.9. Clearly i Example 2.2, A is liearly ordered, hece S(A)=A, thus A has LBLP. Remark If B(A)={0,1}, the, accordig to Remark 5.6 ad Propositio 2.4 ( psbl c 19 ), S(A) formed of the elemet a A which satisfy oe of the followig coditio. 0 L([a)) ad 0 =1 L([a )), that is a = 0 for some N. 1 L([a)) ad 1 = 0 L([a )), that is (a ) = 0 for some N. Thus, whe B (A) = {0, 1}, it follows that S(A) cotais exactly the elemet a A such that a, a ilpotet that is S(A) = N(A) { a A a, a N(A) }.

19 372 B. Barai ia ad A. Borumad Saeid Remark If B(A)={0,1} ad all the elemet of A are idempotet, hece N(A)={ 0 }, thus by Remark 4.10, S(A)={ 0 } { a A a = a = 0 } = {0} D(A). Example Let A be the pseudo BL-algebra lattice i Example 4.5. The, B(A)={0,1} ad all the elemet of A are idempotet, hece by Remark 5.11 we have S(A) = {0} D(A) = { 0 } { a A a = a = 0 } = { 0 } { a A a 0 = 0 } { a A a 0 = 0 } = { 0 } {a 1, b, a 3,,1 } {a 1, a 2, b, a 3,,1 } = {0, a 1, a 2, b, a 3,,1 }. Remark If Rad (A) = A \ {0}, the, A is local pseudo BL-algebra. Sice Rad (A) =A {0}, the, 0 Rad (A), Rad(A) { 0 } = A ad Rad (A) is proper filter of A, thus A is local. Remark If A is o-trivial, the, by Propositio 2.4 ( psbl c 19 ), we have 0 = 0 = 1 0 hece 0 D(A). Propositio We have (i) B(A) S(A), (ii) If a A such that a S(A), the, a S(A), (iii) If a A such that a S(A) for some N, the, a S(A), (iv) D(A) S(A), (v) Rad (A) S(A), (vi) For ay filter F of A, S(A) S ( A ). S( A )={ a A e B ( A ) s.t e [ a ), ( e ) [ a ), ( e ) ~ [ a ) ~ }.

20 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 373 (i) For all e B(A), by psbl - C 7 we have e e = 1 L([ e e )). Therefore B(A) S(A). (ii) Let a A be such that a S(A). It meas that there exists e B(A) such e L([a )) ad e L([(a ) )), by psbl- C 26 a (a ), thus L([ (a ) )) L([a)) therefore e L([ (a ) )) L([a)), thus e L([a)) but we have e = (e ) L([a )), therefore a S(A). (iii) Let a A ad N be such that a S(A). By Lemma 3.1 we have (a ) (a ) hece L([(a ) )) L([(a ) )) ad sice a S(A) by Remark 5.6 there exists e B(A) such that e L([ (a ))) =L([a)) ad e L( [(a ) )) L([a )) thus e L([a )) ad e L([a ) ) that is a S(A). (iv) Let a D(A). That is a A with a = a = 0 B(A) S(A) (by (i)) hece a S(A), ( by(ii )). (v) Let a Rad(A) ad take =1 i Lemma 3.4.The, there exists k N such that (a ) k = 0. But 0 B(A) S(A) (by i) hece (a ) k S(A) the, by (iii) a S(A) hece by (ii) a S(A). (vi) Let F be a filter of A ad cosider a arbitrary elemet of S(A). That is a S(A), a S(A) by Remark 5.6, e L([a)) ad e L([a )), for some e B(A). So there exist, m, p N such that a e ad L[(a ) ) e, the, e B(A) B ( A ). ( a ) = a e ad ( a ) m = (a )m e, that is e [ a ) ad ( e ) [ ( a ) ), therefore a S ( A ). Corollary A has LBLP iff A has LBLP, for every filter F of A. Let A have LBLP. The, S(A)=A, hece A = S(A) S ( A ) A, thus S ( A ) = A, therefore A has LBLP. Coversely, let A have LBLP. Take ={1}, the, A has LBLP. Propositio For every filter F of A, the followig coditios are equivalet (i) A has LBLP, (ii) for every filter G of A such that F G, A L(G) has LBLP. (ii) (i). Take F = G i (ii). (i) (ii). Let G be a filter of A such that F G. By hypothesis A has LBLP. We kow that G is a filter of A, thus by Corollary 5.16 it

21 374 B. Barai ia ad A. Borumad Saeid ( A ) follows, that L(G) has LBLP. But the pseudo BL-algebra ( A ) L(G) is isomorphic to A L(G), hece A L(G) has LBLP. Corollary Ay hyper Archimedea pseudo BL-algebra has LBLP, but the coverse is ot true. Let A be a hyper Archimedea pseudo BL-algebra ad a A. The, there exists a N, such that a B(A). So a S(A) (By Propositio 5.15 (i)) hece by Propositio 5.15 (iii) a S(A) hece A S(A). By defiitio clearly S(A) A therefore by Propositio 5.7 (( i), (vi )) we have A is a LBLP. Now, let A be a chai with at least there elemets, orgaized as a pseudo BL-algebras i Example 4.7. The, A has LBLP by Remark 4.4 B(A) ={0,1} A ad all elemets of A are idempotet, hece oe of the elemets of A B(A)=A {0,1} is a Archimedea, therefore A is ot hyper Archimedea. Propositio If all the elemet of A are idempotet, the, S(A) = { a A a, a B(A)}. Assume that all the elemet of A are idempotet, so for every a A, [a) = {b A a b} the, by Remark 5.6, S(A) = {a A e B(A), e a, a e }, but a e implies e a, e a (By psbl - C 11 ), which meas that a e ad a e, a e imply a = e, a = e thus a, a B(A) therefore if a A the, a, a B( A) hece a, a B(A) ad we have a, a [a) ad a [a ), a [a ), therefore S(A)= {a A a, a B(A)}. Corollary If all the elemet of A are idempotet, the, C = { a, a a S(A) }= B(A). Let a, a C, for all a S(A). By Propositio 5.19 a, a B(A) hece C B(A). Now, let b B(A). By Propositio 2.14 b = b = b, therefore b = (b ) = (b ) B (A) implies b, b S(A) (by Propositio 5.19) the, (b ), (b ) C, hece b C thus B(A) C. Corollary 5.21.

22 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 375 If A is ivolutive ad all the elemet of A are idempotet, the, S(A) = B(A). I other words, if all elemets of A are both regular ad idempotet, the, S(A) = B(A). By Propositio 5.15 (i) B(A) S(A). Now, assume that A is ivolutive ad all the elemet of A are idempotet, ad let a S(A). The, by Propositio a, a B(A), thus a = a = a B(A) (Propositio 2.10). Hece S(A) B(A), thus S(A) = B(A). Corollary If A is ivolutive ad all the elemet of A are idempotet, the, A has LBLP iff A is Boolea algebra iff A is hyper Archimedea. By Propositio 5.7, A has LBLP iff A = S(A) ad by Corollary 5.21, S(A) = B(A) thus A = B(A) that is A has LBLP iff A is a Boolea. We prove the secod equivalece sice all the elemet of A are idempotet, implies for some N, a = a B(A) hece B(A) = A iff A is hyper Archimedea. Remark D(A) iff A is trivial. By Remark 5.14 ad Propositio 5.15(i), we have D(A)=B(A) iff D(A)= S(A) iff A is trivial. Propositio If D(A) B(A), the, D(A)={1} cosequetly if D(A) {0}=B(A), the, D(A)={1}ad B(A)={0,1}. By Propositio 3.6 (iii), B(A) D(A)={1} hece if D(A) B(A) implies D(A)={1} thus if B(A) = D(A) {0} implies B(A) ={1} {0}={0,1}. Propositio If B(A) = S(A), the, D(A) = Rad(A) ={1}.

23 376 B. Barai ia ad A. Borumad Saeid Accordig to Propositio 3.6 (i) ad Propositio 5.15(v), we show that, if B(A)= S(A) the, {1}=B(A) Rad(A) = S(A) Rad(A) = Rad (A) implies Rad(A) = {1}. Now by Propositio 3.6 (ii) we have D(A) Rad(A)={1}, D(A) is a filter hece D(A) = {1}. Corollary If A is a Boolea algebra, the, D(A) = Rad(A) = {1}. If A is a Boolea algebra, the, B(A) = A ad by accordig to Corollary 5.8 (i) S(A)=A hece B(A) =A= S(A) therefore by Propositio 5.25, D(A) = Rad(A) ={1}. Corollary If A is ivolutive ad all the elemet of A are idempotet, the, D(A) = Rad(A) = {1}. By Corollary 5.21 we have S(A) = B(A), hece by Propositio 5.25 D(A) =Rad(A)={1}. Defiitio A is said to be quasi local iff for all a A there exists e B(A) ad N, such that a e = 0 ad e (a ) = 0, (a ) e = 0. Example Cosider example 3.7 with a i =b=0 ad e=1, the, A is quasi local. Remark Ay local pseudo BL algebra is quasi local pseudo BL algebra. By Lemma 2.19 we have B(A)={0,1} ad A= N(A) { a A a, a N(A)}. Let a A. The, (i) If a N(A) implies there exists N such that a = 0 take e =1 hece a e = 0 ad e (a ) = 0, (a ) e = 0. (ii) If a { a A a, a N(A)}, the, there exists N such that (a ) = 0, (a ) = 0 take e = 0 that a e = 0 ad e (a ) = 0. Note. Cosider Example 4.5, such that a 1, a 2,... are ilpotet, the, A is quasi local, but A is ot local, ( sice [a 1 ), [a 2 ),... are maximal filter ). Remark 5.31.

24 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 377 If (A i ) i I is o-empty family of pseudo BL algebra ad A = i I A i, the, (i) If A is quasi local pseudo BL algebra, the, for all i I, A i is quasi local pseudo BL algebra. (ii) If either I is fiite or all elemets are idempotet i these pseudo BL algebra, the, A is quasi local pseudo BL algebra iff for all i I, A i is quasi local pseudo BL algebra. (i) Let a = ( a 1, a 2, ) A. Hece there exist N, e B(A)= B(A i ) such that ad ad 0 = (0,0, ) = a e = ( a 1, a 2, ) (e 1, e 2, ) = ( a 1 e 1, a 2 e 2, ), 0 = ( 0,0, ) = (a ) e = ( (a ) 1, (a ) 2, ) ( e 1, e 2, ) = ((a ) 1 e 1, (a ) 2 e 2, ), 0 = ( 0,0, ) = e (a ) = ( e 1, e 2, ) (a ) 1, (a ) 2, ) = ( e 1 (a ) 1, e 2 (a ) 2, ), therefore for all i I, a i e i = 0, (a ) i e i= 0, e i (a ) i = 0, implies for all i I, A i is quasi local pseudo BL algebra. (ii) Let for all i I, A i is quasi local. The, for a i A i, there exist N ad e i B(A i ) such that a i e i = 0, (a ) i e i = 0, e i (a ) i = 0, that is ad ad ( 0, 0, 0 ) = ( a 1 e 1, a 2 e 2, a m e m ) = ( a 1, a 2, a m ) (e 1 e 2 e m ), ( 0, 0,... 0 ) = ((a ) 1 e 1, (a ) 2 e 2, (a ) m e m) = (a ) 1, (a ) 2,, (a ) m ) ( e 1, e 2,, e m ), (0, 0,..., 0 ) = (e 1 (a ) 1, e 2 (a ) 2, e 3 (a ) 3,, e m (a ) m ) = ( e 1, e 2, ) (a ) 1, (a ) 2, ), thus a e = 0, e (a ) = 0, (a ) e = 0. Therefore A is quasi-local pseudo BL algebra. Defiitio 5.32.

25 378 B. Barai ia ad A. Borumad Saeid A bouded distributive lattice L is called a B ormal lattice iff for all a,b L, if a b = 1, the, exist e, f B(L) such that e f = 0 ad a e = b f = 1. Propositio The followig coditios are equivalet (i) A is quasi local pseudo BL algebra, (ii) A has LBLP, (iii) For all a, b A, if [a) [b) = A, the, there exist e, f B(A) such that e f = 1 ad [a) [e) = [ b) [f ) = A, (iv) The bouded distributive lattice PF(A) is dually B ormal, (v) The bouded distributive lattice ( A, V,, 0,1 ) is B ormal. (i) (ii). A is quasi local pseudo BL algebra, for all a A, there exists e B(A) ad N such that a e = 0 ad e (a ) = 0, (a ) e = 0 by accordig to psbl - C 23 ad psbl - C 24, we have a e ad (a ) e = (e ) = e, (a ) e = e. By Propositio 5.7, A has LBLP iff for all a A there exists f B(A) such that f L([a)) ad f L([a )), f L([ a )). That is a m f ad (a ) k f, (a ) s f for some m, k, s N. Now for the direct implicatio take m = k = s = ad f = e = e ( sice e B(A) thus e = e ). For the coverse implicatio, take e = f = f ad = max{ m, k, s } the, a a m f = e ad (a ) (a ) k f = e, (a ) (a ) s f = e. (iii) (i). Let a A. By psbl - C 20, we have a a = 0, a a = 0 hece L([a)) L([a )) = L([a a )) = A, L([a )) L([a)) =L([ a a))= A. Now the hypothesis of this implicatio show that there exist e, f B(A) such that e f =1 ad L([a)) L([e)) = L([a )) L([f )) = L([a )) L([f)) = A. From e f = (e ) f = f (e ) = 1 by Propositio 2.14 we have e f = e f = 1 hece e f, e f ( by psbl -C 6 ). [0) = A = L([a)) L([e)) = L([a e )) meas that a e = a e = (a e ) = 0 for some N (Lemma 2.13) therefore by psbl - C 23 ad psbl - C 24, we have a e ad a e hece a f ad e, e L([a)) so f L([a)), e, e L([a)), aalogously A = L([a )) L([ f )) = L([ a )) L([ f )) implies f L([a )), f L([a )). So f B(A), f L([a)) ad f L([a )), f L([a )), therefore accordig to Propositio 5.7 A has LBLP. (i) (iii). Let a,b A be such that L([a b)) = L([a)) L([b))= L([b)) L([a))= L([b a ))=A=[0). Which meas that a b = ( a b ) = 0, b a = (b a ) = 0, for some N. The hypothesis states that A has LBLP, the, accordig to Propositio 5.7 there exists e, f B(A) such that e L([ a )), e L([(a ) )), e L([(a ) )) ad f L([b )), f L([(b ) )), f L([(b ) )). Thus (a ) p e, ((a ) ) q e, ((a ) ) s e ad (b ) m f, ((b ) ) t f, ((b ) ) j f, for some p, q, s, m, t, j N. Let k=max{ p, q, s, m, t, j } N. The, by psbl - C 10, it follows that a k e, ((a ) ) k e, ((a ) ) k e ad b k f, ((b ) ) k f, ((b ) ) k f. By Lemma 3.1 we have a k e, (a ) k e, (a ) k e ad b k f, (b ) k f, (b ) k f. From a b = b a = 0 we obtai a (b ), a (b ) ad b (a ), b (a ) ( by psbl- C 23, psbl - C 24

26 AAM: Iter. J., Vol. 13, Issue 1 (Jue 2018) 379 ) hece a k ((b ) ) k f, a k ((b ) ) k f ad b k ((a ) ) k e, b k ((a ) ) k e. So a k e, f, thus a k f e = f e ad aalogously a k f e while b k f ad b k e so b k e f = e f similarly b k e f. We deote c = f e = f e B(A) ad d = e f = e f B(A), hece c, c, d, d B(A) ( by psbl- C 36 ). Therefore a k c = (c ) = (c ) ad b k d = (d ) = (d ) are equivalet to a k c = c a k = 0 ad b k d = d b k = 0 (by psbl - C 23 ad psbl C 24 ), from which we get that A = [0) = L([a k c )) = L([c a k )) =L([a k ) ) L([c )) = L([a)) L([c )) = L([c )) L([a k ))= L([c )) L([a)) ad A = [0) = L([b k d ))= L([d b k )) =L([b k )) L([d )) = L([b)) L([d ))=L([d )) L([b k )) = L([d )) L([b)). Now accordig to psbl - C 20, Lemma 2.13, ad Remark 2.11, we have c d = f e e f = f e e f = 0 0 = 0 thus 1= 0 = 0 = (c d) = (c d) = (c d) = (c d) = c d = c d. (iv) (v). By the act that the bouded lattice P Ƒ(A) is isomorphic to the dual of (A,V,, 0, 1). (iii) (iv). states exactly the fact that the bouded distributive lattice P Ƒ(A) is dually B ormal, sice e f =1 is equivalet to {1 } = [1) = L([e f )) = L([e)) L([f )) ad we have λ: A P Ƒ(A), defied by λ (a) =L([a)) for all a A is a bouded lattice isomorphism betwee (A,,, 0,1) ad the dual of P Ƒ(A). The otio of Boolea ceter is clearly dual to itself ad the Boolea ceter B(A) of the pseudo BL algebra A coicides with the Boolea ceter of the bouded pseudo BL algebra ( A,,, 0,1 ) hece B(P Ƒ(A)) = λ (B(A)) = {λ (e) e B(A)}. Corollary Ay local pseudo BL algebra has LBLP. Moreover, if A is a local pseudo BL algebra ad F is proper filter of A, the, B(P ) is a bijectio. Let A be a local pseudo BL-algebra. The, by Remark 5.30 A is quasi local hece A is LBLP. (by Propositio 5.33). Sice A has LBLP hece by Remark 5.3, B(P ) is surjective, but for every F Ƒ(A), there exists M Max(A) such that F M, sice A is local so M = Rad(A) that is F Rad(A) hece by Corollary 5.12 B(P ) is ijective. 6. Coclusio As we metioed i the itroductio, BL-algebras are importat tools for certai ivestigatios i algebraic logic sice they ca be cosidered as fragmets of ay propositioal logic cotaiig a logical coective implicatio ad the costat 1 which is cosidered as the logical value true. At the same time, BL-algebras as well as pseudo BL-algebras could be itesively studied from a algebraic poit of view. I the preset paper, we defied ad studied the Boolea liftig property o pseudo BL-algebras.We cosider that our results could cotribute to the Boolea liftig theory o pseudo BL-algebras. The mai fidig of this article is that pseudo BL-algebras with LBLP (RBLP) are exactly the quasi-local pseudo BL-algebras. I our

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