It has been found that the BCL-algebra is more extensive class than BCK/BCI/BCH algebra. In this paper we study some properties of BCL-algebra of type&nbsp;(2,0) We also find deformation of such algebra and illustrate the connection between divisible algebra and deformation function

Al-Kadi, Deena Sobhi

Hosny, Rodyna Ahmed

English

KHALSA PUBLICATIONS

ISSN 2347-1921 184 | P a g e   ON BCL-ALGEBRA Deena Al-Kadi1, Rodyna Hosny1,2 1Department of Mathematics and statistic, Taif University, Taif, SAUDI ARABIA dak12le@hotmail.co.uk 2Department of Mathematics, Zagazig University,Zagazig, EGYPT rodyna.hosny2001@gmail.com  ABSTRACT: It has been found that the BCL-algebra is more extensive class than BCK/BCI/BCH-algebra. In this paper we study some properties of  BCL-algebra of type  We also find deformation of such algebra and illustrate the connection between divisible algebra and deformation function. KEYWORDS: BCL-algebra; d-algebra; BCH-algebra; BCI-algebra; BCK-algebra; deformation SUBJECT CLASSIFICATION: 03G25, 06F35, 14D15                   1. INTRODUCTION      Council for Innovative Research Peer Review Research Publishing System Journal: Journal of Advances in Mathematics Vol 3, No 2 editor@cirworld.com www.cirworld.com, member.cirworld.com  ISSN 2347-1921 185 | P a g e   1. INTRODUCTION A new class of algebra of type    called  BCL-algebra is presented in [1].  Liushowed in [1, Theorem 2.4]  that  a proper  BCL-algebra does exist, if such  BCL-algebra is not BCK/BCI/BCH-algebra.  It also has been shown in [1, Theorem 2.1]  that  any BCK/BCI/BCH-algebra is a BCL-algebra. The aim of this paper is to find  when  the converse of  Theorem 2.1 in [1]  is  true.  That is, to show  when a BCL-algebra could be a BCK/BCI/BCH-algebra. The case  where a BCL-algebra can be a BCH-algebra is studied and given in  [1, Theorem 2.2].  Later  in the paper  we  study deformation of  BCL-algebra. The work in this part is motivated by the results in  [3] on deformations of d/BCK-algebra.  We start in Section 2 by introducing  the notions of BCL/d/BCH/BCI/ BCK-algebra respectively. Then, in Section 3, we investigate the relation between  BCL-algebra and d/BCH/BCI and BCK-algebra. We give examples throughout  the paper.          The main results in this  section  are given in Theorem 3.1 which  shows that a d-algebra  satisfying    for any    is a BCL-algebra, Theorem 3.2 and Theorem 3.5 which  gives the sufficient conditions which make a BCL-algebra become a BCK/BCI-algebra. In the final section of this paper , we define deformation function, deformation point and divisible algebra.We are concerned on the deformationof  BCL-algebra. The main results in this section is Proposition 4.1 which gives a deformation of BCL-algebra and  Theorem 4.1 that  illustrate the connection between divisible BCL-algebra and a given map defined using associators of a non-zero element in . 2. PRELIMINARIES We give here the definitions of BCL/d/BCH/BCI/BCK-algebra from [1,2,3]. We refer the reader to [4] and  [5] for further information on BCI/BCK -algebra. Definition 2.1: [1, Definition 2.1] An algebra  of type   is a BCL-algebra if it satisfies the following conditions for any :  1) BCL-1:   2) BCL-2:   and imply  3) BCL-3:   Definition 2.2: [2, p2] An algebra  of type  is a d-algebra if it satisfies the following conditions for any  : 1) d-1:  2) d-2:  3) d-3:  and imply   Definition 2.3: [1, Definition 1.3] An algebra  of type  is a BCH-algebra if it satisfies the following conditions for any :  1) BCH-1 ; 2) BCH-2: and  imply ; 3) BCH-3: (   Definition 2.4: [1, Definition 1.1] An algebra  of type  is a BCI-algebra if it satisfies the following conditions for any   : 1) BCI-1  2) BCI-2:  imply ; 3) BCI-3:  and imply  ISSN 2347-1921 186 | P a g e   4) BCI-4:  5) BCI-5:  Definition 2.5: [3, p316] An algebra  of  type   is a BCK-algebra  if  it satisfies  the  following  conditions  for any   : 1) BCK-1  2) BCK-2:  3) BCK-3:  and imply  4) BCK-4:  5) BCK-5:  3. RESULTS ON BCL-ALGEBRAS In this section, we give some properties related to BCL-algebra. We give necessary conditions for a BCL-algebra to become a d/BCK/BCI/BCH-algebra. We start  with the following example of a d-algebra which is not a BCL-algebra. Example 3.1: Let    be a set in which  is defined by the following Cayley table:   0 1 2 3 0 0 0 0 0 1 1 0 3 2 2 2 2 0 1 3 3 1 2 0  We can easily see that    is a d-algebra and that BCL-1 and BCL-2 does hold. For  BCL-3, we can see that if and , then   Thus   is not a BCL-algebra. Lemma 3.1: Not every d-algebra is a BCL-algebra. This leads  us to find a sufficient axiom (as shown in  the next  theorem)  if satisfied  then  the d-algebra will become a BCL-algebra.  We  will  label the extra  axiom   by d-  for brevity.  Theorem 3.1: A d-algebra  satisfying d- is a BCL-algebra. Proof:  Let    be a d-algebra. It is clear that BCL-1, BCL-2 are satisfied. We only need  to show  that  BCL-3  is valid. We have  Therefore,  is a BCL-algebra.  In the next  part  we find a sufficient condition  that  makes a BCL-algebra be a d-algebra.  Theorem 3.2: A BCL-algebra   satisfying     for any   is a d-algebra. Proof:  The proof follows immediately from Definition 2.1.  We will apply Theorem 3.2  to the  next  example. Example 3.2: Consider the BCL-algebra   given in  [1, Theorem 2.4]  with the following  table: ISSN 2347-1921 187 | P a g e     0 1 2 3 0 0 0 0 0 1 1 0 3 1 2 2 3 0 2 3 3 0 0 0  It is obvious that d-1, d-2 and d-3 are applied in this example. Then    is a BCL-algebra, which is a d-algebra. Theorem 3.3: (See [1, Theorem 2.6]) If   a BCL-algebra then the following relations are satisfied for any , 1)  2)   imply   .  Theorem 3.4: (See [1, Theorem 2.1])        1)Any BCK-algebra is a BCL-algebra;                          2) Any BCI-algebra is a BCL-algebra;                                                                      3) Any BCH-algebra is a BCL-algebra.  Motivated by Theorem 2.1 in [1]  (stated above in Theorem 3.4)  we  give our  theorem  which  will  show  the  sufficient conditions  that  we  apply on BCL-algebra  to become  BCK/BCI/BCH  respectively. Note  that  the last case were  studied in [1] and the related  theorem  is  given  below. Theorem 3.5: Let   be a BCL-algebra. If   and   for any , then  1) the BCL-algebra is a BCK-algebra; 2) the BCL-algebra is a BCI-algebra. Proof: It is clear that  the  axioms BCK-1, BCK-2, BCK-3  are satisfied. With the assumptions given  above, we have  This proves that  the  axiom BCK-4  is valid. Finally, we know from Theorem 3.3  that a BCL-algebra  satisfies the relation  BCK-5. Thus  the BCL-algebra is a BCK-algebra. Similarly, we can observe that  the axioms BCI-1 and BCI-3  follows directly from Definition 2.1.  Also, BCI-2  and  BCI-5 are valid from Theorem 3.3.  We show  that BCI-4  is valid  using  the assumptions above as we done in  the  first  part. This proves that  the given  BCL-algebra  is  a BCI-algebra.  We remind the reader that d-  is  the axiom   Theorem 3.6: (See [1, Theorem 2.2] )  If  is a BCL-algebra satisfying  d- then  the BCL-algebra is a BCH-algebra. Corollary 3.1:  Any d-algebra  satisfying d-  is a BCH-algebra. Proof: It  is clear  that  BCH-1, BCH-2  are  satisfied  in  any d-algebra and  BCH-3 is d- . Hence any d-algebra satisfying  d- is  a BCH-algebra.  4. DEFORMATION  OF  BCL-ALGEBRA In  this section we study  deformation of  BCL-algebra. We start  with  basic definitions  taken  from  [3]. Definition 4.1: Let  be an algebra. A map  is said to be a deformation function of    if (i)   implies   (ii) there exist  such that   The element   is called a deformation point of   and   is said to be a deformation algebra. Next  we  will  apply  the  notions  in  Definition 4.1  to a given  BCL-algebra. ISSN 2347-1921 188 | P a g e   Example 4.1:  Consider  the algebra given  in Example 3.2. Define a map  by Then  we  have  Similarly  we can see that and .  Furthermore,  there  exists    such that   Therefore,  the  map   is a deformation  function, the element  2  is a deformation  point of  and    is  a deformation  algebra. Proposition 4.1:  Let    be a BCL- algebra with  and let    be a deformation  function of  .  Define a binary operation on    by:  for any , then   is a d-algebra  which  is  no t a  BCL-algebra. Proof: Given    is a  BCL- algebra  and  that  , by using  the  axioms in  Definition  2.1  we have    Also Assume  that    Then   As  is a deformation  function  we  get  . Hence,  Therefore,   is a d-algebra.  We  show  that    is  not  a  BCL-algebra  by  providing  the next  example.  Example 4.2:  Consider  the  BCL-algebra given  in  Example 3.2  and  consider  the  deformation  function    given  in Example  4.1.  If  we  define    then    is  a  deformed  BCL-algebra  (defined below)  which  is not  BCL-algebra  since     0 1 2 3 0 0 0 0 0 1 1 0 3 1 2 3 3 0 3 3 3 0 0 0  Lemma 4.1: If   then  for any  Corollary 4.1: Let     be a BCH- algebra  with   and  let    be  a deformation  function  of  . Define a binary operation on  by:  for any , then  is  a d-algebra  which  is  not  a  BCH-algebra. Proof: The  proof  is  the  same  as  the  proof  of  Proposition 4.1 above  using  Definition  2.3. To  verify  that    is  not  a  BCH-algebra, consider  the  algebra  defined  as  follows:   0 1 2 3 0 0 0 1 0 1 1 0 3 2 2 2 3 0 1 3 3 2 1 0  ISSN 2347-1921 189 | P a g e   It is not difficult  to check  that    is a BCH-algebra. Define the deformation  function  as given in Example 4.1. Then  it  is clear  from  the  following  table  tha t the  algebra    is a d-algebra and  it  is easy  to check  that  the  axiom  BCH-3  fails  as   Hence,    is not  a  BCH-algebra.    0 1 2 3 0 0 0 0 0 1 1 0 3 3 2 3 3 0 1 3 3 3 1 0                                                            Definition 4.2: An algebra  is  said  to be  rigid  if  it  has no  non-trivial  deformation. Example 4.3: Consider  the  BCL-algebra  given  in  Example  3.2  and define a deformation  function  as follows   With direct  calculations  we  can show  that  the deformed algebra   is a BCL-algebra  where  is defined  below. Thus  the BCL-algebra  in  this example  is not  rigid.  Note that  the  algebra    is a d-algebra.  0 1 2 3 0 0 0 0 0 1 3 0 3 3 2 3 3 0 3 3 3 0 0 0  Definition 4.3:  An algebra   is  said  to be  divisible  if  for  any  non-zero  element   there exists an element    such that    The element    is called  an  associator  of  . Example 4.4:  Consider the  algebra  in  Example  3.2.  We  can  see that  2 is an  associator  of  1 and 1 is an associator of  2. Whereas,  3 has no associator. Hence,  the  given  algebra is not  divisible. Remark 4.1:  The  associator  is  not  unique  in  general. Proposition  4.2: There  exist some  BCL-algebras   which are not divisible.  Proof: Let   be a  BCL-algebra  then  and  for  any  . Therefore, we might  have  the  cases  where    or    i.e.  we  might  have  .  If  this is the case  then  there  is no  associator    in    such  that    Hence   is not  always  divisible.  Theorem 4.1: Let   be a divisible BCL- algebra and define for a non-zero element    a map    by   Then   is a deformation  function  of  .  Proof:  We  will  use  the  same  strategy  used  in  the  proof  of   [3, Theorem 4.7]. ISSN 2347-1921 190 | P a g e   Let    then   Thus,  if    we have     as    is  a divisible  algebra.  If  ,  given  that , then  from  Definition 1.1  BCL-2 we  see  that  Hence,  This proves  that   is a  deformation  function  of  .  REFERENCES [1]  Liu,Y. H., 2011. A New Branch of the pure Algebra: BCL-Algebras. Advances in Pure Mathematics, 1(5) :297-299.   [2] Kim,H. S., J. Neggers and K. S. So, 2012. Some Aspects of d-Units in d/BCK-Algebras: Journal of Applied Mathematics,(2012) :10 pages.   [3]  Allen,P. J., H. S. kim and J. Nggers, 2011. Deformations of d/BCK-Algebras.Bull. Korean Math. Soc.,  48(2): 315-324.    [4]  Huang,Y. S., 2006. BCI-algebra: Science press, China. [5]  MengJ. and Y. B. Jun, 1994. BCK-algebras: Kyung Moon Sa Co., Seoul, Korea.  

On BCL-Algebra

https://rajpub.com/index.php/jam/article/download/2455/pdf_21

On BCL-Algebra

Abstract

Similar works

Full text

Available Versions

KHALSA PUBLICATIONS