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G. C. RAO AND P. SUNDARAYYA

Received 12 April 2005; Revised 12 December 2005; Accepted 18 December 2005

We prove that ifA is aC-algebra, then for eacha∈A, Aa= {x∈A/x≤a} is itself a C-algebra and is isomorphic to the quotient algebra A/θa of A where θa= {(x,y) A×A/a∧x=a∧y}. IfAisC-algebra withT, we prove that for everya∈B(A), the centre ofA,Ais isomorphic toAa×Aa and that ifAis isomorphicA1×A2, then there

existsa∈B(A) such thatA1is isomorphicAaandA2is isomorphic toAa. Using this

de-composition theorem, we prove that ifa,b∈B(A) witha∧b=F, thenAais isomorphic toAbif and only if there exists an isomorphismφonAsuch thatφ(a)=b.

Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.

Introduction

In [1], Guzm´an and Squier introduced the variety ofC-algebras as a class of algebras of type (2, 2, 1) satisfying certain identities and proved that this variety is generated by the 3-element algebraC= {T,F,U}which is the algebraic semantic of the three valued conditional logic. In [3] Swamy et al. introduced the concept of the centreB(A)= {x∈ A/x∨x=T}of aC-algebraAwithTand proved thatB(A) is a Boolean algebra with induced operations and is equivalent to the Boolean Centre ofA. In [2], Rao and Sun-darayya defined a partial ordering on aC-algebraAand the properties ofAas a poset are studied.

In this paper, we prove that ifAis aC-algebra, then for eachx∈A,Ax= {s∈A/s≤x} is itself aC-algebra and is isomorphic to the quotient algebraA/θx, whereθx= {(s,t) A×A/x∧s=x∧t}. IfAis aC-algebra withTthen, for everya∈B(A),Ais isomorphic toAa×Aa and conversely ifAis isomorphic toA1×A2, then there exists an element

a∈B(A) such thatA1is isomorphic toAaandA2is isomorphic toAa. Using the above

decomposition theorem we prove that for anya,b∈B(A) witha∧b=F,Aais isomor-phic toAbif and only if there exists an isomorphism onAwhich sendsatob.

1. Preliminaries

First, we recall the definition of aC-algebra and some results, which will be used in the later text.

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 78981, Pages1–8

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By aC-algebra we mean an algebra of type (2, 2, 1) with operations,, and satisfy-ing the followsatisfy-ing properties:

(a)x=x;

(b) (x∧y)=xy; (c) (x∧y)∧z=x∧(y∧z); (d)x∧(y∨z)=(x∧y)(x∧z);

(e) (x∨y)∧z=(x∧z)(x∧y∧z); (f)x∨(x∧y)=x;

(g) (x∧y)(y∧x)=(y∧x)(x∧y).

Clearly, every Boolean algebra is aC-algebra. The set{T,F,U}is aC-algebra with oper-ations,, andgiven by

T F U

T T F U

F F F F

U U U U

T F U

T T T T

F T F U

U U U U

X X

T F

F T

U U

We denote this three-elementC-algebra byCand the two-elementC-algebra (Boolean algebra){T,F}by B. It can be observed that the identities (a), (b) imply that the variety of allC-algebras satisfies the dual statements of (b) to (g). In generalandare not commutative inCand the ordinary right distributive law ofoverfails inC.

The following properties of aC-algebra can be verified directly [1,3]: (i)x∧x=x;

(ii)x∧y=x∧(x∨y)=(x∨y)∧x; (iii)x∨(x∧x)=(x∧x)∨x=x; (iv) (x∨x)y=(xy)(xy);

(v)x∨x=x∨x; (vi)x∨y∨x=x∨y; (vii)x∧xy=xx.

If aC-algebraAhas an identity for, then it is unique and we denote it byT. In this case, we say thatAis aC-algebra withT. If we write F forT, then F is the identity for. In aC-algebra, we have the following [1,3]:

(viii)x∨y=Fif and only ifx=y=F; (ix) ifx∨y=T, thenx∨x=T;

(x)x∨T=x∨x;

(xi)T∨x=TandF∧x=F;

(xii) fora∈A,a=aif and only ifais left zero of bothand.

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Forx∈Adefine the relationθxonAbyθx= {(p,q)∈A×A/x∧p=x∧q}thenθxis a congruence relation onAandθx∩θx=θxx[1].

The relationdefined on aC-algebraAbyx≤yify∧x=xis a partial ordering onA in which, for everyx∈A, the supremum of{x,x} =x∨x, and the infimum of{x,x} = x∧x[2]. IfAis aC-algebra withT,xB(A) and yAare such that xy=yx, thenx∨yis the lub of{x,y}and in this case y∨xneed not be the lub ofxand y. For example, in the algebraC,T∈B(C) andT∧U=U∧TbutU∨T=Uis not the lub of

{U,T}. Ifx≤y, theny∧x=xand hencex∧y=x∧y∧x=x∧x=x. Thereforex≤y if and only ify∧x=x=x∧y.

2. TheC-algebraAx

Recall that for every Boolean algebraBanda∈Bthe set (a]= {x∈B/x≤a}([a)= {x∈

B/a≤x}) is a Boolean algebra under the induced operationsandwhere complemen-tation is defined byx∗=a∧x(x=ax).

In this section, we prove that ifAis aC-algebra andx∈A, thenAx= {s∈A/s≤x} is aC-algebra withT(=x) under the induced operations and thatAxis isomorphic to a quotient algebra ofA.

Theorem 2.1. LetAbe aC-algebra,x∈A, andAx= {s∈A/s≤x}. Then Ax,,,∗is aC-algebra withTwhere∧and∨are the operations inArestricted toAx,s∗is defined by x∧s, and “x” is the identity for.

Proof. ClearlyAx is closed underand. Ifs∈Ax, thenx∧s∗=x∧(x∧s)=(x∧ x)∧s=xs=s. So thatsAx ands∗∗=(s)=(xs)=x(xs)=x (x∨s)=x∧s=s(sinces≤x).

Now, fors,t∈Ax, (s∧t)∗=x∧(s∧t)=x∧(s∨t)=(x∧s)(x∧t)=s∗∨t∗. Finally, fors,t,u∈Ax,

(s∨t)∧u=x∧(s∨t)∧u=x∧(s∧u)(s∧t∧u)

=(x∧s)(x∧u)(x∧stu)

=(s∧u)∨s∗tu.

(2.1)

The remaining identities hold inAxsince they hold inA.

Hence Ax,,,is aC-algebra with “x” as the identity for. Observe thatAxis itself aC-algebra but it is not a subalgebra ofAbecause the unary operationis not the restriction oftoAx. Now, we give some properties ofAx.

Theorem 2.2. LetAbe aC-algebra. Then the following hold: (i)Ax= {x∧s/s∈A};

(ii)Ax=Ayif and only ifx=y; (iii)Ax∩Ay⊆Ax∧y;

(iv)Ax∩Ax=Ax∧x;

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Proof. (i), (ii), and (iii) can be verified routinely. We prove (iv) as follows. Lets∈Ax∧x,

then (x∧x)s=s and hence xs=x(xxs)=xxs=s. Also we have xs=x(xxs)=s, sincexx=xx. Now we prove (v),

Axx∧y=x∧y∧t/t∈Ax by (i)

=x∧y∧x∧s /s∈A

=x∧y∧s /s∈A=Ax∧y.

(2.2)

LetA1,A2be twoC-algebras withT1andT2. Then a mapping f :A1→A2that

pre-serves,,and carriesT1toT2is called aT-preservingC-algebra homomorphism. In

future, we deal withC-algebras withT only and hence by aC-algebra homomorphism we mean aT-preservingC-algebra homomorphism. The following lemma can be verified routinely.

Lemma 2.3. Let f :A1→A2be aC-algebra homomorphism whereA1,A2 areC-algebras

withT1andT2. Then

(i) ifA1has the uncertain elementU, then f (U) is the uncertain element ofA2;

(ii) ifa∈B(A1), then f(a)∈B(A2). The converse holds if f is one-one.

Now we prove the following.

Theorem 2.4. Let A be a C-algebra withT and x∈A, then the mappingαx:A→Ax defined byαx(s)=x∧sfor alls∈Ais a homomorphism ofAontoAxwith kernelθxand henceA/θx∼=Ax.

Proof. Fors∈A,x∧s≤xand hencex∧s∈Ax. Lets,t∈A, then

αx(s∧t)=x∧s∧t=x∧s∧x∧t=αx(s)∧αx(t),

αx(s)=x∧s=x∧(x∨s) by (ii) in the preliminaries

=x∧(x∧s)=(x∧s)∗=αx(s)∗.

(2.3)

Clearly,αx(s∨t)=αx(s)∨αx(t) andαx(T)=a. Henceαxis aC-algebra homomorphism. Now, fors∈Ax, we haveαx(s)=s. Thereforeαx is onto homomorphism. Hence by the fundamental theorem of homomorphismA/Kerαx∼=Axand Kerαx= {(s,t)∈A×A/αx(s)

=αx(t)} = {(s,t)∈A×A/x∧s=x∧t} =θx. ThusA/θx∼=Ax.

3. Decompositions ofA

IfBis a Boolean algebra anda∈B, then we know thatBis isomorphic to (a]×[a). In this section we prove similar decompositions for aC-algebra. IfAis aC-algebra withT and a∈B(A), then we prove thatAis isomorphic toAaxAa and conversely. We also prove

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Lemma 3.1. LetAbe aC-algebra withT,a∈B(A) andx,y∈A. Then

a∨x=a∨y, a∨x=ay⇐⇒x=y. (3.1)

Proof. Leta∈B(A) andx,y∈A. Assume thata∨x=a∨yanda∨x=ay. Then

x=F∨x=(a∧a)x=(ax)(ax)

=(a∨y)(a∨y)=(a∧a)y=Fy=y. (3.2)

The converse is trivial

Note thatLemma 3.1fails ifa /∈B(A). For example, in theC-algebraC, we haveU /∈

B(C),U∨T=U∨F=U, andU∨T=UF=U, butT=F. Now we prove the following decomposition theorem.

Theorem 3.2. IfAis aC-algebra withTanda∈B(A), thenA∼=Aa×Aa. Proof. Defineα:A→Aa×Aaby

α(x)=αa(x),αa(x) ∀x∈A. (3.3)

Then, byTheorem 2.4,αis well defined andαis a homomorphism.

Now, α(x)(y)⇒a∧x=a∧y and a∧x=ay. Hence x=y (by the dual ofLemma 3.1). Finally, we proveαis onto. Let (x,y)∈Aa×Aa. Thenx≤aandy≤a.

So thata∧x=xanda∧y=y.

Thus,a∧x=aax=Fanday=aay=F. Now,

x∨y∈A, α(x∨y)=αa(x∨y),αa(x∨y) =a∧(x∨y),a∧(x∨y)

=(a∧x)(a∧y), (a∧x)(a∧y)

=(x∨F,F∨y)=(x,y).

(3.4)

Henceαis an isomorphism.

Now we prove the converse of the above theorem in the following sense.

Theorem 3.3. LetA,A1,A2beC-algebras withTsuch thatA∼=A1×A2. Then there exists

an elementa∈B(A) such that

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Proof. Letφ:A→A1×A2be an isomorphism anda=φ−1(T1,F2) (whenT1,T2denote

the-identities ofA1,A2, resp.)

Now (T1,F2)∈B(A1)×B(A2)=B(A1×A2) and hencea∈B(A).

Define f :A1→Aaby f(x1)=φ−1(x1,F2) for allx1∈A1.

Now

a∧φ−1x 1,F2

=φ−1T 1,F2

∧φ−1x 1,F2

=φ−1x 1,F2

(sinceφ−1is a homomorphism). (3.6)

Thereforeφ−1(x

1,F2)∈Aa. Thus f is well defined.

It can be routinely verified that f preserves,and that f is one-one. Now we prove that f preserves the unary operation.

Letx1∈A1, then

fx

1

=φ−1x

1,F2

=φ−1T

1∧x1,F2∧T2

=φ−1T 1,F2

∧φ−1x

1,T2 sinceφ−1is homomorphism

=a∧φ−1x 1,F2

=a∧fx1

=fx1 .

(3.7)

Finally, we prove f is onto.

Letx∈Aa. Thenφ(x)=(x1,x2) for somex1∈A1,x2∈A2.

Now

x1,x2

(x)(a∧x)(a)∧φ(x)

=T1,F2

∧x1,x2

=x1,F2

. (3.8)

Thusx2=F2and f(x1)=φ−1(x1,F2)=φ−1(x1,x2)=x.

Hence f is onto. ThusA1∼=Aa. SimilarlyA2∼=Aa.

Finally, fora,b∈B(A) witha∧b=F, we derive a necessary and sufficient condition forAato be isomorphic toAb. First we prove the following lemmas.

Lemma 3.4. IfAis aC-algebra withT,a∈B(A),x∈Aa, andy∈Aa, thenx∨y=y∨x. Proof. Letx∈Aa,y∈Aa. Thenx≤aandy≤a. Hencea∧y=F=a∧x. Now

a∧(x∨y)=(a∧x)(a∧y)=x∨F=x,

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Therefore,a∧(x∨y)=a∧(y∨x). Similarlya∧(x∨y)=a(yx). By the dual ofLemma 3.1,

x∨y=y∨x. (3.10)

Lemma 3.5. LetAbe aC-algebra withT. Then, fora,b∈B(A),a∧b∈B(Aa). Proof. Clearlya∧b≤a. Now

(a∧b)(a∧b)∗=(a∧b)∨a∧(a∧b)

=(a∧b)∨a∧(a∨b)=(a∧b)(a∧b)

=a∧(b∨b)=aT=a.

(3.11)

Hence,a∧b∈B(Aa).

Now, we prove the theorem.

Theorem 3.6. LetAbe aC-algebra withTanda,b∈B(A) such thata∧b=F. ThenAa is isomorphic toAbif and only if there exists an isomorphismα:A→Asuch thatα(a)=b. Proof. Leta,b∈B(A) witha∧b=F. Letφ:Aa→Abbe an isomorphism.

Nowa∧b=(a∧b)∨F=(a∧b)(a∧b)=(a∨a)∧b=b becauseB(A) is a Boolean algebra. So thatb∈Aaandb∗=a∧b. Similarly,b∧a=a. Now by Theorems

2.2,3.2, andLemma 3.5, we have

(i)A∼= Aa×Aa∼=Aa×Aa∧b×A(a∧b)∗=Aa×Ab×Aa∧b

under the isomorphismx→β (a∧x,b∧x, (a∧b)∧x); (ii)A∼=Ab×Ab∼=Ab×Ab∧a×A(b∧a)∗∼=Ab×Aa×Aa∧b

under the isomorphismx→γ (b∧x,a∧x, (a∧b)x); (iii)Aa×Ab×Aa∧b∼=Ab×Aa×Aa∧b

under the isomorphism (x,y,z)→δ (φ(x),φ−1(y),z).

Now defineα:A→Abyα=γ−1δβ. Thenαis an isomorphism ofAontoAand

α(a)=γ−1δβ(a)=γ1δ(a,F,F) (sinceba=F=aa)

=γ−1(b,F,F) sinceφ(a)=b,φ(F)=F =b sinceγ(b)=(b,F,F).

(3.12)

Henceαis an isomorphism ofAsuch thatα(a)=b.

Conversely, suppose thatα:A→Ais an isomorphism such thatα(a)=b.

Letλbe the restriction ofαtoAa. Now we prove thatλis an isomorphism ofAaonto Ab. Forx∈Aa,

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So thatλ(x)∈Ab. Henceλis well defined. Clearlyλis a homomorphism and one-one. Letx∈Ab. Sinceαis onto, there existsy∈Asuch thatα(y)=x. Nowa∧y∈Aa and λ(a∧y)(a∧y)(a)∧α(y)=b∧x=x(sincex≤b).

Henceλis an isomorphism ofAaontoAb.

Acknowledgments

The authors thank Professor U. M. Swamy for his support and guidance in the prepara-tion of this paper. The second author gratefully acknowledges University Grants Com-mission for their financial support in the form of UGC-JRF. We thank referees for their helpful suggestions.

References

[1] F. Guzm´an and C. C. Squier, The algebra of conditional logic, Algebra Universalis 27 (1990), no. 1, 88–110.

[2] G. C. Rao and P. Sundarayya,C-algebra as a poset, International Journal of Mathematical Sci-ences, 4 (2005), no. 2, 225–236.

[3] U. M. Swamy, G. C. Rao, and R. V. G. Ravi Kumar, Centre of aC-algebra, Southeast Asian Bulletin of Mathematics 27 (2003), no. 2, 357–368.

G. C. Rao: Department of Mathematics, Andhra University, Visakhapatnam 530 003, India E-mail address:gcraomaths@mail.yahoo.co.in

10.1155/IJMMS/2006/78981

References

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