Article Welcome To ITB Journal A11157
ITB J. Sci., Vol. 44 A, No. 3, 2012, 204-216
204
Boolean Algebra of C-Algebras
G.C. Rao1 & P. Sundarayya2
1
2
Department of Mathematics, Andhra University, Visakhapatnam-530 003, India
Department of Mathematics, GIT, GITAM University, Visakhapatnam-530 045, India
E-mails: gcraomaths@yahoo.co.in; psundarayya@yahoo.co.in
Abstract. A C- algebra is the algebraic form of the 3-valued conditional logic,
which was introduced by F. Guzman and C.C. Squier in 1990. In this paper,
some equivalent conditions for a C- algebra to become a boolean algebra in
terms of congruences are given. It is proved that the set of all central elements
B(A) is isomorphic to the Boolean algebra BS ( A) of all C-algebras Sa, where a
B(A). It is also proved that B(A) is isomorphic to the Boolean algebra BR ( A)
of all C-algebras Aa, where a
B(A).
Keywords: Boolean algebra; C-algebra; central element; permutable congruences.
1
Introduction
The concept of C-algebra was introduced by Guzman and Squier as the variety
generated by the 3-element algebra C={T,F,U}. They proved that the only
subdirectly irreducible C-algebras are either C or the 2-element Boolean algebra
B ={T,F} [1,2].
For any universal algebra A, the set of all congruences on A (denoted by Con A)
is a lattice with respect to set inclusion. We say that the congruences , are
permutable if = . We say that Con A is permutable if =
for all ,
Con A . It is known that Con A need not be permutable for any Calgebra A.
In this paper, we give sufficient conditions for congruences on a C-algebra A to
be permutable. Also we derive necessary and sufficient conditions for a Calgebra A to become a Boolean algebra in terms of the congruences on A. We
also prove that the three Boolean algebras B(A), the set of C-algebras BS ( A) and
the set of C-algebras BR ( A) are isomorphic to each other.
Received November 1st, 2011, Revised April 5th, 2012, Accepted for publication July 7th, 2012
Copyright © 2012 Published by LPPM ITB, ISSN: 1978-3043, DOI: 10.5614/itbj.sci.2012.44.3.1
Boolean Algebra of C-Algebras
2
205
Preliminaries
In this section we recall the definition of a C-algebra and some results from
[1,3,5] which will be required later.
Definition 2.1. By a C -algebra we mean an algebra < A, , , > of type
(2,2,1) satisfying the following identities [1].
(a)
(b)
(c)
(d)
(e)
(f)
(g)
x = x;
( x y)
x (y
x (y
( x y)
x (x
( x y)
= x y;
z ) = ( x y ) z;
z ) = ( x y) ( x z );
z = ( x z) ( x y z);
y ) = x;
( y x) = ( y x) ( x y).
Example 2.2. [1]:
The 3- element algebra C={T, F, U} is a C-algebra with the operations
and defined as in the following tables.
x
x
T
F
U
F
T
U
T
T
F
U
F
U
T F
F F
U U
U
F
U
T
T
F
U
F
T T T
T F U
U U U
Every Boolean algebra is a C-algebra.
Lemma 2.3. Every C-algebra satisfies the following laws [1,3,5].
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
x x = x;
x y = x ( x y) = ( x y) x;
x ( x x) = x;
( x x ) y = ( x y) ( x y);
( x x ) x = x;
x x = x x;
x y x = x y;
x x y=x x;
x ( y x) = ( x y )
x.
The duals of all above statements are also true.
U
,
206
G.C. Rao & P. Sundarayya
Definition 2.4. If A has identity T for
( that is, x T = T x = x for all
x A ), then it is unique and in this case, we say that A is a C -algebra with T.
[1].
We denote T ' by F and this F is the identity for
Lemma 2.5 [1]: Let A be a C -algebra with T and x, y
(i)
(ii)
(iii)
(iv)
A. Then
x y = F if and only if x = y = F
if x y = T then x x = T .
x T= x x
x F=x x .
Theorem 2.6. Let < A, , , > be a C-algebra. Then the following are
equivalent [6]:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
A is a Boolean algebra.
x ( y x) = x, for all x, y A .
x y = y x, for all x, y A .
( x y ) y = y , for all x, y A .
x x is the identity for , for every x A .
x x = y y , for all x, y A .
A has a right zero for .
for any x, y A , there exists a A such that x a = y
for any x, y A , if x y = y , then y x = x.
a = a.
Definition 2.7. Let A be a C-algebra with T. An element x A is called a
central element of A if x x = T . The set of all central elements of A is called
the Centre of A and is denoted by B(A) [5].
Theorem 2.8. Let A be a C-algebra with T .Then < B ( A), , , > is a Boolean
Algebra [5].
3
Some Properties of a C -algebra and Its Congruences
In this section we prove some important properties of a C-algebra and we give
sufficient conditions for two congruences on a C-algebra A to be permutable.
Also we derive necessary and sufficient conditions for a C-algebra A to become
a Boolean algebra in terms of the congruences on A.
Lemma 3.1. Every C-algebra satisfies the following identities:
Boolean Algebra of C-Algebras
(i) x
(ii) x
y = x (y
y = x (y
207
x );
x ).
Proof. Let A be a C-algebra and x, y
A. Now,
x y = x ( x y) = x ( x y x ) = [ x ( x x )] [ x y ( x ( x x ))] =
[ x ( x x )] [ x y x ( x x )] = [ x ( x x )] [ x y ( x x )] = ( x y )
( x x ) = x ( y x ). Similarly x y = x ( y x ).
Lemma 3.2. Let A be a C-algebra and x, y A. Then x y x = x y y .
Proof. Let A be a C-algebra and x, y A. By Lemma 2.3[b],[f] and Lemma
3.1, we have x y x = x (( y x ) y ) = [ x ( y x )] y = ( x y )
y= x y y= x y y.
Lemma 3.3. Let A be a C-algebra with T , x, y
x y = y x.
A and x
y = F . Then
Proof. Suppose that x y = F. Then F = x y = x ( x y) = ( x x )
( x y) = ( x x ) F = x x . now x y = F ( x y) = ( x y) ( x y)
= ( x x y) ( x y x y ) (By Def 2.1) = ( x y) ( x y x) (By
2.3[g]) = ( x x ) ( y x) = F ( y x) = y x.
In [6], it is proved that if A is a C-algebra with T then B( A) = {a A |
a a = T } is a Boolean algebra under the same operations , , in the Calgebra A. Now we prove the following.
Theorem 3.4. Let A be a C-algebra with T and a, b
a b B(A) . Then a B(A).
A such that
Proof. Let A be a C-algebra with T and a, b A such that a b B(A). Then
T = (a b) (a b) = (a b) (a b )
= (a b a ) (a b b ) = (a b a ) (a b a ) (By Theorem 3.2)
=a b a .
Therefore, T = a b a ...(I)
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G.C. Rao & P. Sundarayya
Now, a a
Hence a
=
=
=
=
(a
(a
(a
a
a) T
a ) (a b a ) (by(I))
(a b a )) (a (a b a ))
(a (a b a ))
= a (a b a ) (By 2.3[b])
= a b a = T.
B( A).
The converse of the above theorem need not be true. For example, in the Calgebra C, F B(C ) but F U = U B(C). We have the following
consequence of the above theorem.
A and a b
Corollary 3.5. Let A be a C-algebra with T, a, b
a B( A).
Proof. Let a b B( A). Then we have, (a b)
a B( A) a B( A).
In [1], it is proved that if A is a C-algebra, then
a congruence on A and
x
=
x
x x
x
B( A)
= {( p, q) | x
B( A) . Then
a
b
p = x q} is
. In [6], if A is C-algebra with T then
is a factor congruence if and only if x B( A). They also proved that x ,
permutable congruences whenever both x, y
important properties of these congruences.
a b
Proof. (i) ( x, y)
a
=
; (ii)
a
x=a
b
b a
b
a b
.
y
b a b x=b a b x
b a x=b a x
( x, y)
Therefore
a b
b a
b a
. Similarly,
b a
a b.
are
A . Then we have the
a b
b
y
x
B( A) . Now we prove some
Theorem 3.6. Let A be a C-algebra with T and a, b
following (i)
B( A)
Hence
a b
=
b a
.
Boolean Algebra of C-Algebras
(ii) Let ( x, y)
( z, y )
a
a
b . Then there exists z
. Thus b x = b z and a z = a
z=a b a z=a b a
a
b
a b
209
A such that ( x, z)
and
x=a b
y. Now, a b
y. Therefore, ( x, y )
y=a b
b
a b
. Thus
.
In the following we give an example of a C-algebra G without T in which the
Con A is not permute.
Example 3.7. Consider the C-algebra G = {a1, a2 , a3 , a4 , a5} where a1 = (T ,U ) ,
a2 = ( F , U ), a3 = (U , T ), a4 = (U , F ), a5 = (U ,U ) under pointwise operations
in C .
x
x
a1
a2
a2
a1
a2
a3
a4
a5
a1
a1
a2
a5
a5
a5
a1
a2
a2
a2
a2
a2
a2
a3
a4
a3
a5
a5
a3
a4
a5
a4
a3
a4
a4
a4
a4
a4
a4
a5
a5
a5
a5
a5
a5
a5
a5
a2
a3
a4
a5
a1
a2
a1
a1
a1
a1
a2
a1
a5
a1
a5
a1
a5
a3
a3
a3
a3
a3
a3
a4
a5
a5
a3
a4
a5
a5
a5
a5
a5
a5
a5
This algebra (G, , , ) is a C-algebra with out T.
Let = diagonal of A. Then we have the following:
a1
a3
= {( x, y ) | a1
x = a1
y}
=
{(a3 , a4 ), (a4 , a5 ), ( a5 , a3 ), ( a4 , a3 )( a5 , a4 ), ( a3 , a5 )}
=
{(a1 , a2 ), (a2 , a5 ), (a5 , a1 ), ( a2 , a1 )( a5 , a2 ), ( a1, a5 )}
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G.C. Rao & P. Sundarayya
Now,
a1
a3
Therefore
a3
a1
a1
=
a1
=
a1
a3
a3
{(a4 , a1 ), (a4 , a2 )( a3 , a1 ), (a3 , a2 )}
a3
{(a2 , a4 ),(a2 , a3 )(a1 , a3 ),(a1 , a4 )}
a3
a1 .
Theorem 3.8. Let A a C-algebra with T and a
b A, a ,
b
permute and
a
b
=
a b
B( A) . Then for any
.
Proof. Let A be a C-algebra with T and a B(A) . By Theorem 3.6,
Now let ( p, q) a b . Then a b p = a b q b a b
a b
a b.
p = b a b q b a p = b a q. Consider, r = (a p) (a q) .
Now a r = a [(a p) (a q)] = (a p ) (a a q ) = (a p ) ( F q )
= (a p) F a p. Therefore (r, p)
b [(a p) (a q)] = [b a p] [b a
=b
((a
=b
q . Therefore (q, r)
b
a
=
q)
a b
(a
q)) = b
. Thus
b
(( a
Now, b r =
q] = (b a q) (b a q)
q) = b (T q) (since a B( A))
a)
(r, q)
b
( p, r)
a
b
a.
. Thus ( p, q)
is a congruence on A and hence
a
permutable congruences and hence
a
b
=
b
a
=
a b
a
b
=
Proof. i) We know that if a
theorem
a
b
=
b
a
=
a b
; ii) a b B( A)
b
a b.
B(A) then a
a
a.
a
,
Hence
b
are
.
Corollary 3.9. Let A be a C-algebra with T and a, b
a b B( A)
b
b
=
A . Then i)
a b.
B(A) and hence by the above
Similarly, we can prove ii).
Let A be a C-algebra. If Con(A) is permutable, then A need not be a Boolean
algebra. For example, in the C-algebra C, the only congruences are , and
they are permutable. But C is not a Boolean algebra. Now we give equivalent
conditions for a C-algebra to become a Boolean algebra in terms of congruence
relations.
Theorem 3.10. Let ( A, , , ) be a C-algebra with T. Then the following are
equivalent. (i) Let ( A, , , ) be a Boolean algebra. (ii)
x
A. (iii)
x x
=
for all x
A.
x
x
=
for all
Boolean Algebra of C-Algebras
211
(2): Let A be a Boolean algebra and x
A . Let ( p, q) x
x.
Then x p x q and x p = x q. Now, p = ( x x ) p = ( x p)
. Therefore
( x q) = ( x q) ( x q) = ( x x ) q = q. Thus x
x
(iii). (iii) (i): Suppose
x
x = . Since
x
x = x x , we get (ii)
Proof. (1)
A. We prove that x x = A A. Let ( p, q) A A .
Write t = ( x p) ( x q). Now, x t = x (( x p) ( x q)) = ( x p)
( x x q ) = ( x p) ( x x ) = x ( p x ) = x p. Also, x t = x
x x
(( x
=
for all x
p) ( x
q)) = x
q)) = ( x
x
p) ( x
q. Therefore ( p, t )
x
x
q)) = ( x
and (t , q )
x
x) ( x
q)) = ( x
. Thus ( p, q)
x
(x
x
.
Hence we get x x = A A. Also x
x
x and
x = x x = . That is
are permutable factor congruences. Therefore, by Theorem 2.6, we have
x B( A). Thus A = B( A) and hence A is a Boolean algebra.
4
The C-algebra S x
We prove that, for each x
A , S x = {x t | t
A} is itself a C-algebra under
induced operations , and the unary operation is defines by ( x t )
x t'
. We observe that Sx need not be a subalgebra of A because the unary operation
in Sx is not the restriction of the unary operation on A. Also for each x A , the
set Ax = {x t | t A} is a C-algebra in which the unary operation is given by
( x t )* = x t . We prove that the B(A) is isomorphic to the Boolean algebra
BS ( A) of all C-algebras Sa where a
B(A) . Also, we prove that B(A) is
isomorphic to the Boolean algebra BR ( A) of all C-algebras Aa , a
Theorem 4.1. Let < A, , , > be a C-algebra, x
B( A) .
A and S x = {x t | t
Then < S x , , ,* > is a C-algebra with x as the identity for
are the operations in A restricted to Sx and for any x t
, where
A}.
and
S x , here ( x t )*
is x t .
Proof. Let t , r , s
A . Then ( x t ) ( x r ) = x (t
( x r) = x (t r) Sx . Thus
**
,
r ) S x and ( x t )
are closed in Sx. Also
Consider ( x t ) = x ( x t ) = x ( x
*
is closed in Sx.
t ) = x t . Now [( x t ) ( x r )]*
212
G.C. Rao & P. Sundarayya
= [ x (t
r )]* = x (t
consider
(x t) (x r)
x (t s) x (t
x r = ( x t )* ( x r )*.
r)=x t
( x s) x [(t r) s] = x
r s)
( x t ) ( x s)
Now,
(t s) (t
r s)
( x t ') ( x r) ( x s)
( x t ) ( x s ) ( x t ) ( x r ) ( x s ) . The remaining identities of a
C-algebra also hold in Sx because they hold in A. Hence, Sx is itself a C-algebra.
Also x is the identity for because x x t = x t = x t x . Here x x is
the identity for .
Theorem 4.2. Let A be a C-algebra. Then the following holds.
(i) S x = S y if and only if x = y ;
(ii) S x
Sy
Sx y ;
Sx = Sx x ;
(iv) ( S x ) x y = S x y .
(iii) S x
Proof. (i) Suppose S x = S y . Since x = x
x S x = S y and y = y
y
S y = S x . Therefore x = y t and y = x r for some t, r A . Now,
x = y t = ( y t y ) ( y y t ) = ( x y ) ( y x) = ( y x) ( x y ) =
( x r x) ( x x r ) = x r = y. The converse is trivial. (ii) Suppose
t S x S y . Then t = x s = y r for some s, r A. Now, t = x x s
=x t=x
we have Sx
y r
Sx
Sx
x
y x r|r
S x defined by
with kernel
x
Proof. Let t , r
and
x
x
S x = S x x . (iv) (S x ) x y = {x y t |
A} = {x y r | r A} = S x y
.
Theorem 4.3. Let A be a C-algebra with T and x
A
x =x
by (ii). Since x
Sx . Hence S x
Sx
x
t S x } = {x
S x y . (iii) S x
x
(t ) = x t for all t
and hence A/
A. Then
x
x
Sx.
(t
r) = x t
(t ) = x t = ( x t )* = (
x
A, then the mapping
x
:
A is a homomorphism of A to S x
r=x t
(t ))* . Clearly,
x
x r=
(t r ) =
x
x
(t )
(t )
x
x
(r )
(r ).
in S x. Therefore x
Also x (T ) = x T = x x , which is the identity for
is a homomorphism. Hence by the fundamental theorem of homomorphism
A/Ker x S x and Ker x = {(t , r ) A A | x (t ) = x (r )} = {(t , r ) A A |
Boolean Algebra of C-Algebras
x t = x r} = {(t , r )
hence A/
A A| x
t=x
r}
x
213
(by Lemma 2.3 [b]) = and
Sx.
x
Theorem 4.4. Let A be a C-algebra with T and a
B(A), then A
Sa S a .
Proof. Define
: A Sa Sa by ( x) = ( a ( x), a ( x)) for all x A.
Then, by Theorem 4.3,
is well-defined and
is a homomorphism. Now, we
( a ( x), a ( x))
prove that
is one-one. Let x, y A . Then ( x) = ( y)
=(
a
( y),
a
x=a
( y))
(a y ) =
x=a t,
a y=a
( x y) = (
=
=
=
=
Therefore,
(a x, a
x) = ( a
y, a
y)
a x=a
y
and
y. Now x = F x = (a a ) x = (a x) (a x) = (a y )
is onto. Let ( x, y) Sa Sa . Then
y. Finally, we prove that
and y = a r for some t, r A . Therefore, a x = x,
and
Now,
a y =T y =T
a x = T , a y = y.
a
( x y ), a ( x y ))
(a ( x y ), a ( x y ))
((a x) (a y ), (a x) (a
( x T , T y)
( x, y ).
a
is onto and hence
is an isomorphism. Therefore A
Lemma 4.5. Let A be a C-algebra. Then for a, b
(i) a b = b a if and only if Sa
y ))
b
= Sa
Sa S a .
A:
Sb
(ii) Sa b = Sup{Sa , Sb } in the poset ( {S x | x
The converse is not true.
A}, ), then a b = b a.
b = b a. Then clearly Sa b Sa Sb . By
Theorem 4.2(ii) Sa Sb Sa b . Hence Sa b = Sa Sb . Conversely assume
that Sa b = Sa Sb . Clearly a b Sa b = Sa Sb . Therefore a b Sb
a b = b t for some t A. Now b a = b a b = b b t = b t =
a b. (ii) Assume that a, b A and Sa b = Sup{Sa , Sb }. Then Sa b = Sb a
and hence a b Sa b = Sb a . Therefore a b = (b a) t for some t A.
Proof. (i) Suppose that a
214
G.C. Rao & P. Sundarayya
Now (b a) (a b) = (b a ) ((b a ) t ) = (b a ) t = a b. Similarly
we can prove that (a b) (b a ) = b a. Hence a b = b a. The
converse need not be true, for example for the C-algebra C ,
SU = {U }, ST = {T } and U T = T U . But SU T (= SU ) is not an upper
bound of {SU , ST }.
Now we prove BS ( A)
{Sa | a B( A)} is a Boolean algebra under set
inclusion.
Theorem
Let
4.6.
BS ( A) ={Sa | a
be
< A, , , >
a
C-algebra
with
T.
Then
B( A)} is a Boolean algebra under set inclusion.
Proof. Clearly (BS ( A) , ) is a partially ordered set under inclusion. First we
show for a, b
B( A) , S a
b
is the infimum of {Sa , Sb } and S a
supremum of {Sa , Sb } for all
B( A) . Let
a, b
B( A) . Then
a, b
a b = b a and a b = b a. Hence by the above Lemma 4.5, S a
infimum of {Sa , Sb } . Let t
t = a x = (a (a b)) x = (a
Sb
Sb
a
= S a b . Therefore S a
Sa . Then t = a x for some x
b
is the
A . Now
Similarly
(b a) x = (a b) a x Sa b .
b
is the
b
is an upper bound of S a , Sb . Suppose Sc is
an upper bound of S a , Sb . t
Sa b . Then t = (a b) x for some x A .
Now t = (a b) x = (a x) (a b x) = (a x) (b a x) Sc (since
a x Sa Sc , b a x Sb Sc and Sc is closed under ). Therefore
S a b is the supremum of {S a , Sb }. Denote the supremum of {Sa , Sb } by
S a S b and the infimum of {Sa , Sb } by S a S b . Now ST Sa = ST a = ST
and S F
greatest
Sa = S F
= S F . Therefore ST is the least element and S F is the
(BS ( A) , ) .
Now
for
any
element
of
a
a, b, c B( A),(Sa Sb ) Sc = S(a
Sc ) (Sb
Sc ).
Also,
Sa
b) c
= S( a
Sa = Sa
a
c ) (b c )
= ST
= S( a
c)
and
Sa
S( b
c)
= ( Sa
Sa = Sa
a
= SF .
Therefore (BS ( A) , ) is a complimented distributive lattice and hence it is a
Boolean algebra.
Boolean Algebra of C-Algebras
Theorem 4.7. Let A be a C-algebra with T Define
(a) = sa for all a B(A). Then
Proof. Let a, b
(a b) = S(a
Clearly
b)
= Sa
: B ( A)
BS ( A ) by
is an isomorphism.
( a b) = S ( a
B( A) . Then
215
Sb = (a)
(b) ,
is both one-one and onto. Hence B( A)
b)
= Sa
Sb = ( a )
(b).
(a ) = Sa = ( Sa ) = ( (a)) .
BS ( A ) .
In [3] we defined a partial ordering on a C-algebra by x y if and only if
y x = x and we studied the properties of this partial ordering. We gave a
number of equivalent conditions in terms of this partial ordering for a C-algebra
to become a Boolean algebra. In [4] we proved that, for each x A ,
Ax = {s A | s x} is itself a C-algebra under induced operations , and
the unary operation is defined by s* = x
s we also observed that Ax need not
be an algebra of A because the unary operation in Ax is not the restriction of
A, we proved that Ax is isomorphic to the
the unary operation. For each x
p = x q} . We can
easily see that the C-algebras S x , Ax are different in general where x A.
quotient algebra A/
x
where
x
= {( p, q) A A | x
Now, we prove that the set of all Aa 's where a B(A) is a Boolean algebra
under set inclusion. The following theorem can be proved analogous to
Theorem 4.6.
Theorem 4.8. Let A be a C-algebra with T. Then BR( A) :={Aa | a
Boolean Algebra under set inclusion in which
{Aa , Ab } = Aa b and the infimum of {Aa , Ab} = Aa b .
the
B( A)} is a
supremum of
The proof of the following theorem is analogous to that of Theorem 4.7.
Theorem 4.9. Let A be a C-algebra with T Define f : B( A)
BR ( A) by
f (a ) = Aa for all a B(A). Then f is an isomorphism.
The following corollary can be proved directly from Theorems 4.7 and 4.9.
216
G.C. Rao & P. Sundarayya
Corollary 4.10. Let A be a C-algebra with T. Then BR( A) , B( A) and BS ( A)
are isomorphic to each other.
References
[1]
[2]
[3]
[4]
[5]
[6]
Guzman, F. & Squier, C., The Algebra of Conditional Logic, Algebra
Universalis, 27, pp. 88-110, 1990.
Swamy, U.M., Rao, G.C., Sundarayya, P. & Kalesha Vali, S., Semilattice
Structures on a C-algebra, Southeast Asian Bulletin of Mathematics, 33,
pp. 551-561, 2009.
Rao, G.C., Sundarayya, P., C-algebra as a Poset, International Journal of
Mathematical Sciences, Serials Pub., New Delhi, 4(2), pp. 225-236, Dec.
2005.
Rao, G.C., Sundarayya, P., Decompositions of a C-algebra, International
Journal of Mathematics and Mathematical Sciences, Hindawi Pub. Cor.
U.S.A., 2006(3), Article ID 78981, pp. 1-8, 2006.
Swamy, U.M., Rao,G.C. & Ravi Kumar, R.V.G., Centre of a C-algebra,
Southeast Asian Bulletin of Mathematics, 27, pp. 357-368, 2003.
Stanely, B. & Sankappanavar, H.P., A Course in Universal Algebra,
Springer Verlag, 1981.
204
Boolean Algebra of C-Algebras
G.C. Rao1 & P. Sundarayya2
1
2
Department of Mathematics, Andhra University, Visakhapatnam-530 003, India
Department of Mathematics, GIT, GITAM University, Visakhapatnam-530 045, India
E-mails: gcraomaths@yahoo.co.in; psundarayya@yahoo.co.in
Abstract. A C- algebra is the algebraic form of the 3-valued conditional logic,
which was introduced by F. Guzman and C.C. Squier in 1990. In this paper,
some equivalent conditions for a C- algebra to become a boolean algebra in
terms of congruences are given. It is proved that the set of all central elements
B(A) is isomorphic to the Boolean algebra BS ( A) of all C-algebras Sa, where a
B(A). It is also proved that B(A) is isomorphic to the Boolean algebra BR ( A)
of all C-algebras Aa, where a
B(A).
Keywords: Boolean algebra; C-algebra; central element; permutable congruences.
1
Introduction
The concept of C-algebra was introduced by Guzman and Squier as the variety
generated by the 3-element algebra C={T,F,U}. They proved that the only
subdirectly irreducible C-algebras are either C or the 2-element Boolean algebra
B ={T,F} [1,2].
For any universal algebra A, the set of all congruences on A (denoted by Con A)
is a lattice with respect to set inclusion. We say that the congruences , are
permutable if = . We say that Con A is permutable if =
for all ,
Con A . It is known that Con A need not be permutable for any Calgebra A.
In this paper, we give sufficient conditions for congruences on a C-algebra A to
be permutable. Also we derive necessary and sufficient conditions for a Calgebra A to become a Boolean algebra in terms of the congruences on A. We
also prove that the three Boolean algebras B(A), the set of C-algebras BS ( A) and
the set of C-algebras BR ( A) are isomorphic to each other.
Received November 1st, 2011, Revised April 5th, 2012, Accepted for publication July 7th, 2012
Copyright © 2012 Published by LPPM ITB, ISSN: 1978-3043, DOI: 10.5614/itbj.sci.2012.44.3.1
Boolean Algebra of C-Algebras
2
205
Preliminaries
In this section we recall the definition of a C-algebra and some results from
[1,3,5] which will be required later.
Definition 2.1. By a C -algebra we mean an algebra < A, , , > of type
(2,2,1) satisfying the following identities [1].
(a)
(b)
(c)
(d)
(e)
(f)
(g)
x = x;
( x y)
x (y
x (y
( x y)
x (x
( x y)
= x y;
z ) = ( x y ) z;
z ) = ( x y) ( x z );
z = ( x z) ( x y z);
y ) = x;
( y x) = ( y x) ( x y).
Example 2.2. [1]:
The 3- element algebra C={T, F, U} is a C-algebra with the operations
and defined as in the following tables.
x
x
T
F
U
F
T
U
T
T
F
U
F
U
T F
F F
U U
U
F
U
T
T
F
U
F
T T T
T F U
U U U
Every Boolean algebra is a C-algebra.
Lemma 2.3. Every C-algebra satisfies the following laws [1,3,5].
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
x x = x;
x y = x ( x y) = ( x y) x;
x ( x x) = x;
( x x ) y = ( x y) ( x y);
( x x ) x = x;
x x = x x;
x y x = x y;
x x y=x x;
x ( y x) = ( x y )
x.
The duals of all above statements are also true.
U
,
206
G.C. Rao & P. Sundarayya
Definition 2.4. If A has identity T for
( that is, x T = T x = x for all
x A ), then it is unique and in this case, we say that A is a C -algebra with T.
[1].
We denote T ' by F and this F is the identity for
Lemma 2.5 [1]: Let A be a C -algebra with T and x, y
(i)
(ii)
(iii)
(iv)
A. Then
x y = F if and only if x = y = F
if x y = T then x x = T .
x T= x x
x F=x x .
Theorem 2.6. Let < A, , , > be a C-algebra. Then the following are
equivalent [6]:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
A is a Boolean algebra.
x ( y x) = x, for all x, y A .
x y = y x, for all x, y A .
( x y ) y = y , for all x, y A .
x x is the identity for , for every x A .
x x = y y , for all x, y A .
A has a right zero for .
for any x, y A , there exists a A such that x a = y
for any x, y A , if x y = y , then y x = x.
a = a.
Definition 2.7. Let A be a C-algebra with T. An element x A is called a
central element of A if x x = T . The set of all central elements of A is called
the Centre of A and is denoted by B(A) [5].
Theorem 2.8. Let A be a C-algebra with T .Then < B ( A), , , > is a Boolean
Algebra [5].
3
Some Properties of a C -algebra and Its Congruences
In this section we prove some important properties of a C-algebra and we give
sufficient conditions for two congruences on a C-algebra A to be permutable.
Also we derive necessary and sufficient conditions for a C-algebra A to become
a Boolean algebra in terms of the congruences on A.
Lemma 3.1. Every C-algebra satisfies the following identities:
Boolean Algebra of C-Algebras
(i) x
(ii) x
y = x (y
y = x (y
207
x );
x ).
Proof. Let A be a C-algebra and x, y
A. Now,
x y = x ( x y) = x ( x y x ) = [ x ( x x )] [ x y ( x ( x x ))] =
[ x ( x x )] [ x y x ( x x )] = [ x ( x x )] [ x y ( x x )] = ( x y )
( x x ) = x ( y x ). Similarly x y = x ( y x ).
Lemma 3.2. Let A be a C-algebra and x, y A. Then x y x = x y y .
Proof. Let A be a C-algebra and x, y A. By Lemma 2.3[b],[f] and Lemma
3.1, we have x y x = x (( y x ) y ) = [ x ( y x )] y = ( x y )
y= x y y= x y y.
Lemma 3.3. Let A be a C-algebra with T , x, y
x y = y x.
A and x
y = F . Then
Proof. Suppose that x y = F. Then F = x y = x ( x y) = ( x x )
( x y) = ( x x ) F = x x . now x y = F ( x y) = ( x y) ( x y)
= ( x x y) ( x y x y ) (By Def 2.1) = ( x y) ( x y x) (By
2.3[g]) = ( x x ) ( y x) = F ( y x) = y x.
In [6], it is proved that if A is a C-algebra with T then B( A) = {a A |
a a = T } is a Boolean algebra under the same operations , , in the Calgebra A. Now we prove the following.
Theorem 3.4. Let A be a C-algebra with T and a, b
a b B(A) . Then a B(A).
A such that
Proof. Let A be a C-algebra with T and a, b A such that a b B(A). Then
T = (a b) (a b) = (a b) (a b )
= (a b a ) (a b b ) = (a b a ) (a b a ) (By Theorem 3.2)
=a b a .
Therefore, T = a b a ...(I)
208
G.C. Rao & P. Sundarayya
Now, a a
Hence a
=
=
=
=
(a
(a
(a
a
a) T
a ) (a b a ) (by(I))
(a b a )) (a (a b a ))
(a (a b a ))
= a (a b a ) (By 2.3[b])
= a b a = T.
B( A).
The converse of the above theorem need not be true. For example, in the Calgebra C, F B(C ) but F U = U B(C). We have the following
consequence of the above theorem.
A and a b
Corollary 3.5. Let A be a C-algebra with T, a, b
a B( A).
Proof. Let a b B( A). Then we have, (a b)
a B( A) a B( A).
In [1], it is proved that if A is a C-algebra, then
a congruence on A and
x
=
x
x x
x
B( A)
= {( p, q) | x
B( A) . Then
a
b
p = x q} is
. In [6], if A is C-algebra with T then
is a factor congruence if and only if x B( A). They also proved that x ,
permutable congruences whenever both x, y
important properties of these congruences.
a b
Proof. (i) ( x, y)
a
=
; (ii)
a
x=a
b
b a
b
a b
.
y
b a b x=b a b x
b a x=b a x
( x, y)
Therefore
a b
b a
b a
. Similarly,
b a
a b.
are
A . Then we have the
a b
b
y
x
B( A) . Now we prove some
Theorem 3.6. Let A be a C-algebra with T and a, b
following (i)
B( A)
Hence
a b
=
b a
.
Boolean Algebra of C-Algebras
(ii) Let ( x, y)
( z, y )
a
a
b . Then there exists z
. Thus b x = b z and a z = a
z=a b a z=a b a
a
b
a b
209
A such that ( x, z)
and
x=a b
y. Now, a b
y. Therefore, ( x, y )
y=a b
b
a b
. Thus
.
In the following we give an example of a C-algebra G without T in which the
Con A is not permute.
Example 3.7. Consider the C-algebra G = {a1, a2 , a3 , a4 , a5} where a1 = (T ,U ) ,
a2 = ( F , U ), a3 = (U , T ), a4 = (U , F ), a5 = (U ,U ) under pointwise operations
in C .
x
x
a1
a2
a2
a1
a2
a3
a4
a5
a1
a1
a2
a5
a5
a5
a1
a2
a2
a2
a2
a2
a2
a3
a4
a3
a5
a5
a3
a4
a5
a4
a3
a4
a4
a4
a4
a4
a4
a5
a5
a5
a5
a5
a5
a5
a5
a2
a3
a4
a5
a1
a2
a1
a1
a1
a1
a2
a1
a5
a1
a5
a1
a5
a3
a3
a3
a3
a3
a3
a4
a5
a5
a3
a4
a5
a5
a5
a5
a5
a5
a5
This algebra (G, , , ) is a C-algebra with out T.
Let = diagonal of A. Then we have the following:
a1
a3
= {( x, y ) | a1
x = a1
y}
=
{(a3 , a4 ), (a4 , a5 ), ( a5 , a3 ), ( a4 , a3 )( a5 , a4 ), ( a3 , a5 )}
=
{(a1 , a2 ), (a2 , a5 ), (a5 , a1 ), ( a2 , a1 )( a5 , a2 ), ( a1, a5 )}
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G.C. Rao & P. Sundarayya
Now,
a1
a3
Therefore
a3
a1
a1
=
a1
=
a1
a3
a3
{(a4 , a1 ), (a4 , a2 )( a3 , a1 ), (a3 , a2 )}
a3
{(a2 , a4 ),(a2 , a3 )(a1 , a3 ),(a1 , a4 )}
a3
a1 .
Theorem 3.8. Let A a C-algebra with T and a
b A, a ,
b
permute and
a
b
=
a b
B( A) . Then for any
.
Proof. Let A be a C-algebra with T and a B(A) . By Theorem 3.6,
Now let ( p, q) a b . Then a b p = a b q b a b
a b
a b.
p = b a b q b a p = b a q. Consider, r = (a p) (a q) .
Now a r = a [(a p) (a q)] = (a p ) (a a q ) = (a p ) ( F q )
= (a p) F a p. Therefore (r, p)
b [(a p) (a q)] = [b a p] [b a
=b
((a
=b
q . Therefore (q, r)
b
a
=
q)
a b
(a
q)) = b
. Thus
b
(( a
Now, b r =
q] = (b a q) (b a q)
q) = b (T q) (since a B( A))
a)
(r, q)
b
( p, r)
a
b
a.
. Thus ( p, q)
is a congruence on A and hence
a
permutable congruences and hence
a
b
=
b
a
=
a b
a
b
=
Proof. i) We know that if a
theorem
a
b
=
b
a
=
a b
; ii) a b B( A)
b
a b.
B(A) then a
a
a.
a
,
Hence
b
are
.
Corollary 3.9. Let A be a C-algebra with T and a, b
a b B( A)
b
b
=
A . Then i)
a b.
B(A) and hence by the above
Similarly, we can prove ii).
Let A be a C-algebra. If Con(A) is permutable, then A need not be a Boolean
algebra. For example, in the C-algebra C, the only congruences are , and
they are permutable. But C is not a Boolean algebra. Now we give equivalent
conditions for a C-algebra to become a Boolean algebra in terms of congruence
relations.
Theorem 3.10. Let ( A, , , ) be a C-algebra with T. Then the following are
equivalent. (i) Let ( A, , , ) be a Boolean algebra. (ii)
x
A. (iii)
x x
=
for all x
A.
x
x
=
for all
Boolean Algebra of C-Algebras
211
(2): Let A be a Boolean algebra and x
A . Let ( p, q) x
x.
Then x p x q and x p = x q. Now, p = ( x x ) p = ( x p)
. Therefore
( x q) = ( x q) ( x q) = ( x x ) q = q. Thus x
x
(iii). (iii) (i): Suppose
x
x = . Since
x
x = x x , we get (ii)
Proof. (1)
A. We prove that x x = A A. Let ( p, q) A A .
Write t = ( x p) ( x q). Now, x t = x (( x p) ( x q)) = ( x p)
( x x q ) = ( x p) ( x x ) = x ( p x ) = x p. Also, x t = x
x x
(( x
=
for all x
p) ( x
q)) = x
q)) = ( x
x
p) ( x
q. Therefore ( p, t )
x
x
q)) = ( x
and (t , q )
x
x) ( x
q)) = ( x
. Thus ( p, q)
x
(x
x
.
Hence we get x x = A A. Also x
x
x and
x = x x = . That is
are permutable factor congruences. Therefore, by Theorem 2.6, we have
x B( A). Thus A = B( A) and hence A is a Boolean algebra.
4
The C-algebra S x
We prove that, for each x
A , S x = {x t | t
A} is itself a C-algebra under
induced operations , and the unary operation is defines by ( x t )
x t'
. We observe that Sx need not be a subalgebra of A because the unary operation
in Sx is not the restriction of the unary operation on A. Also for each x A , the
set Ax = {x t | t A} is a C-algebra in which the unary operation is given by
( x t )* = x t . We prove that the B(A) is isomorphic to the Boolean algebra
BS ( A) of all C-algebras Sa where a
B(A) . Also, we prove that B(A) is
isomorphic to the Boolean algebra BR ( A) of all C-algebras Aa , a
Theorem 4.1. Let < A, , , > be a C-algebra, x
B( A) .
A and S x = {x t | t
Then < S x , , ,* > is a C-algebra with x as the identity for
are the operations in A restricted to Sx and for any x t
, where
A}.
and
S x , here ( x t )*
is x t .
Proof. Let t , r , s
A . Then ( x t ) ( x r ) = x (t
( x r) = x (t r) Sx . Thus
**
,
r ) S x and ( x t )
are closed in Sx. Also
Consider ( x t ) = x ( x t ) = x ( x
*
is closed in Sx.
t ) = x t . Now [( x t ) ( x r )]*
212
G.C. Rao & P. Sundarayya
= [ x (t
r )]* = x (t
consider
(x t) (x r)
x (t s) x (t
x r = ( x t )* ( x r )*.
r)=x t
( x s) x [(t r) s] = x
r s)
( x t ) ( x s)
Now,
(t s) (t
r s)
( x t ') ( x r) ( x s)
( x t ) ( x s ) ( x t ) ( x r ) ( x s ) . The remaining identities of a
C-algebra also hold in Sx because they hold in A. Hence, Sx is itself a C-algebra.
Also x is the identity for because x x t = x t = x t x . Here x x is
the identity for .
Theorem 4.2. Let A be a C-algebra. Then the following holds.
(i) S x = S y if and only if x = y ;
(ii) S x
Sy
Sx y ;
Sx = Sx x ;
(iv) ( S x ) x y = S x y .
(iii) S x
Proof. (i) Suppose S x = S y . Since x = x
x S x = S y and y = y
y
S y = S x . Therefore x = y t and y = x r for some t, r A . Now,
x = y t = ( y t y ) ( y y t ) = ( x y ) ( y x) = ( y x) ( x y ) =
( x r x) ( x x r ) = x r = y. The converse is trivial. (ii) Suppose
t S x S y . Then t = x s = y r for some s, r A. Now, t = x x s
=x t=x
we have Sx
y r
Sx
Sx
x
y x r|r
S x defined by
with kernel
x
Proof. Let t , r
and
x
x
S x = S x x . (iv) (S x ) x y = {x y t |
A} = {x y r | r A} = S x y
.
Theorem 4.3. Let A be a C-algebra with T and x
A
x =x
by (ii). Since x
Sx . Hence S x
Sx
x
t S x } = {x
S x y . (iii) S x
x
(t ) = x t for all t
and hence A/
A. Then
x
x
Sx.
(t
r) = x t
(t ) = x t = ( x t )* = (
x
A, then the mapping
x
:
A is a homomorphism of A to S x
r=x t
(t ))* . Clearly,
x
x r=
(t r ) =
x
x
(t )
(t )
x
x
(r )
(r ).
in S x. Therefore x
Also x (T ) = x T = x x , which is the identity for
is a homomorphism. Hence by the fundamental theorem of homomorphism
A/Ker x S x and Ker x = {(t , r ) A A | x (t ) = x (r )} = {(t , r ) A A |
Boolean Algebra of C-Algebras
x t = x r} = {(t , r )
hence A/
A A| x
t=x
r}
x
213
(by Lemma 2.3 [b]) = and
Sx.
x
Theorem 4.4. Let A be a C-algebra with T and a
B(A), then A
Sa S a .
Proof. Define
: A Sa Sa by ( x) = ( a ( x), a ( x)) for all x A.
Then, by Theorem 4.3,
is well-defined and
is a homomorphism. Now, we
( a ( x), a ( x))
prove that
is one-one. Let x, y A . Then ( x) = ( y)
=(
a
( y),
a
x=a
( y))
(a y ) =
x=a t,
a y=a
( x y) = (
=
=
=
=
Therefore,
(a x, a
x) = ( a
y, a
y)
a x=a
y
and
y. Now x = F x = (a a ) x = (a x) (a x) = (a y )
is onto. Let ( x, y) Sa Sa . Then
y. Finally, we prove that
and y = a r for some t, r A . Therefore, a x = x,
and
Now,
a y =T y =T
a x = T , a y = y.
a
( x y ), a ( x y ))
(a ( x y ), a ( x y ))
((a x) (a y ), (a x) (a
( x T , T y)
( x, y ).
a
is onto and hence
is an isomorphism. Therefore A
Lemma 4.5. Let A be a C-algebra. Then for a, b
(i) a b = b a if and only if Sa
y ))
b
= Sa
Sa S a .
A:
Sb
(ii) Sa b = Sup{Sa , Sb } in the poset ( {S x | x
The converse is not true.
A}, ), then a b = b a.
b = b a. Then clearly Sa b Sa Sb . By
Theorem 4.2(ii) Sa Sb Sa b . Hence Sa b = Sa Sb . Conversely assume
that Sa b = Sa Sb . Clearly a b Sa b = Sa Sb . Therefore a b Sb
a b = b t for some t A. Now b a = b a b = b b t = b t =
a b. (ii) Assume that a, b A and Sa b = Sup{Sa , Sb }. Then Sa b = Sb a
and hence a b Sa b = Sb a . Therefore a b = (b a) t for some t A.
Proof. (i) Suppose that a
214
G.C. Rao & P. Sundarayya
Now (b a) (a b) = (b a ) ((b a ) t ) = (b a ) t = a b. Similarly
we can prove that (a b) (b a ) = b a. Hence a b = b a. The
converse need not be true, for example for the C-algebra C ,
SU = {U }, ST = {T } and U T = T U . But SU T (= SU ) is not an upper
bound of {SU , ST }.
Now we prove BS ( A)
{Sa | a B( A)} is a Boolean algebra under set
inclusion.
Theorem
Let
4.6.
BS ( A) ={Sa | a
be
< A, , , >
a
C-algebra
with
T.
Then
B( A)} is a Boolean algebra under set inclusion.
Proof. Clearly (BS ( A) , ) is a partially ordered set under inclusion. First we
show for a, b
B( A) , S a
b
is the infimum of {Sa , Sb } and S a
supremum of {Sa , Sb } for all
B( A) . Let
a, b
B( A) . Then
a, b
a b = b a and a b = b a. Hence by the above Lemma 4.5, S a
infimum of {Sa , Sb } . Let t
t = a x = (a (a b)) x = (a
Sb
Sb
a
= S a b . Therefore S a
Sa . Then t = a x for some x
b
is the
A . Now
Similarly
(b a) x = (a b) a x Sa b .
b
is the
b
is an upper bound of S a , Sb . Suppose Sc is
an upper bound of S a , Sb . t
Sa b . Then t = (a b) x for some x A .
Now t = (a b) x = (a x) (a b x) = (a x) (b a x) Sc (since
a x Sa Sc , b a x Sb Sc and Sc is closed under ). Therefore
S a b is the supremum of {S a , Sb }. Denote the supremum of {Sa , Sb } by
S a S b and the infimum of {Sa , Sb } by S a S b . Now ST Sa = ST a = ST
and S F
greatest
Sa = S F
= S F . Therefore ST is the least element and S F is the
(BS ( A) , ) .
Now
for
any
element
of
a
a, b, c B( A),(Sa Sb ) Sc = S(a
Sc ) (Sb
Sc ).
Also,
Sa
b) c
= S( a
Sa = Sa
a
c ) (b c )
= ST
= S( a
c)
and
Sa
S( b
c)
= ( Sa
Sa = Sa
a
= SF .
Therefore (BS ( A) , ) is a complimented distributive lattice and hence it is a
Boolean algebra.
Boolean Algebra of C-Algebras
Theorem 4.7. Let A be a C-algebra with T Define
(a) = sa for all a B(A). Then
Proof. Let a, b
(a b) = S(a
Clearly
b)
= Sa
: B ( A)
BS ( A ) by
is an isomorphism.
( a b) = S ( a
B( A) . Then
215
Sb = (a)
(b) ,
is both one-one and onto. Hence B( A)
b)
= Sa
Sb = ( a )
(b).
(a ) = Sa = ( Sa ) = ( (a)) .
BS ( A ) .
In [3] we defined a partial ordering on a C-algebra by x y if and only if
y x = x and we studied the properties of this partial ordering. We gave a
number of equivalent conditions in terms of this partial ordering for a C-algebra
to become a Boolean algebra. In [4] we proved that, for each x A ,
Ax = {s A | s x} is itself a C-algebra under induced operations , and
the unary operation is defined by s* = x
s we also observed that Ax need not
be an algebra of A because the unary operation in Ax is not the restriction of
A, we proved that Ax is isomorphic to the
the unary operation. For each x
p = x q} . We can
easily see that the C-algebras S x , Ax are different in general where x A.
quotient algebra A/
x
where
x
= {( p, q) A A | x
Now, we prove that the set of all Aa 's where a B(A) is a Boolean algebra
under set inclusion. The following theorem can be proved analogous to
Theorem 4.6.
Theorem 4.8. Let A be a C-algebra with T. Then BR( A) :={Aa | a
Boolean Algebra under set inclusion in which
{Aa , Ab } = Aa b and the infimum of {Aa , Ab} = Aa b .
the
B( A)} is a
supremum of
The proof of the following theorem is analogous to that of Theorem 4.7.
Theorem 4.9. Let A be a C-algebra with T Define f : B( A)
BR ( A) by
f (a ) = Aa for all a B(A). Then f is an isomorphism.
The following corollary can be proved directly from Theorems 4.7 and 4.9.
216
G.C. Rao & P. Sundarayya
Corollary 4.10. Let A be a C-algebra with T. Then BR( A) , B( A) and BS ( A)
are isomorphic to each other.
References
[1]
[2]
[3]
[4]
[5]
[6]
Guzman, F. & Squier, C., The Algebra of Conditional Logic, Algebra
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Swamy, U.M., Rao, G.C., Sundarayya, P. & Kalesha Vali, S., Semilattice
Structures on a C-algebra, Southeast Asian Bulletin of Mathematics, 33,
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Rao, G.C., Sundarayya, P., C-algebra as a Poset, International Journal of
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2005.
Rao, G.C., Sundarayya, P., Decompositions of a C-algebra, International
Journal of Mathematics and Mathematical Sciences, Hindawi Pub. Cor.
U.S.A., 2006(3), Article ID 78981, pp. 1-8, 2006.
Swamy, U.M., Rao,G.C. & Ravi Kumar, R.V.G., Centre of a C-algebra,
Southeast Asian Bulletin of Mathematics, 27, pp. 357-368, 2003.
Stanely, B. & Sankappanavar, H.P., A Course in Universal Algebra,
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