Abstract: We define a binary operation * on Pre A*-algebra and show that <A, *> is a semilattice. We also prove some results on the partial ordering ≤* induced by semilattice structure <A, *> and a number of equivalent conditions for A to become a Boolean algebra interms of this partial ordering and binary operation and also find the necessary conditions for (A, ≤*) is a lattice.
INTRODUCTION
The study of lattice theory had been made by Birkhoff (1948). In a draft paper The Equational Theory of Disjoint Alternatives, Manes (1989), introduced the concept of Ada (Algebra of disjoint alternatives) (A,∧, ∨, (-)', (-)π, 0, 1, 2) which is however differ from the definition of the Ada of Manes (1993) later paper Adas and the equational theory of if-then-else. While the Ada of the earlier draft seems to be based on extending the If-Then-Else concept more on the basis of Boolean Algebra and the later concept is based on C-algebra (A,∧, ∨,') introduced by Guzman and Squier (1990).
Koteswara Rao (1994) first introduced the concept A*-Algebra (A,∧, ∨, (-)∼, (-)π, 0, 1, 2) not only studied the equivalence with Ada, C-algebra, Ada’s connection with 3-Ring, Stone type representation but also introduced the concept of A*-clone, the If-Then-Else structure over A*-algebra and Ideal of A*-algebra. Koteswara Rao and Venkateswara Rao (2003) Studied about Boolean Algebras and A*-algebras and the methods of generating A*-algebras from Boolean algebras and vice-versa. Koteswara Rao and Venkateswara Rao. (2004) were introduced the concept of Prime Ideals and Congruences in A*-Algebras. Koteswara Rao and Venkateswara Rao (2005) also obtained a Cayley theorem for A*-Algebras. Koteswara Rao and Venkateswara Rao (2008) were developed the concept of A*-Modules and If-Then-Else Algebras over A*-algebras. Recently Satyanarayana et al. (2009) discussed some Structural Compatibilities of Pre A*-Algebra,Venkateswara Rao (2000) introduced the concept of Pre A*-algebra (A, ∧, ∨, (-)∼) as the variety generated by the 3-element algebra A = {0, 1, 2} which is an algebraic form of three valued conditional logic. It was proved that the only sub directly irreducible Pre A*-algebra are either A or two element Boolean algebra B = {0, 1}. Venkateswara Rao et al. (2009) generated Pre A*-algebras from Boolean algebras and defined congruence relation and Ternary operation on A.
In this study, we define a binary operation on Pre A*-algebra A and show that <A, *> is a semilattice. We also prove some properties on the partial ordering ≤* induced from semilattice structure <A, *>. We also give number of equivalent conditions for A to become a Boolean algebra in terms of this partial ordering * and partial ordering ≤*.
PRELIMINARIES
Definition 1
Boolean algebra is an algebra (B, ∨, ∧. (-)', 0, 1) with two binary
operations, one unary operation (called complementation) and two nullary operations
which satisfies:
| (i) | (B, ∨, ∧) is a distributive lattice |
| (ii) | x∧0 = 0, x∨1 = 1 |
| (iii) | x∧x' = 0, x∨x' = 1 |
We can prove that x'' = x, (x∨y)' = x'∧y', (x∧y)' = x'∨y' for all x, yεB
Definition 2
An algebra (A, ∧, ∨, (-)~) satisfying
| (a) x~~ = x, ∀ x∈A, (b) x ∧ x = x, ∀ x∈A, | (c) x ∧ y = y ∧ x, ∀ x, y∈A, |
| (d) (x ∧ y)~ = x~ ∨ y~, ∀ x, y ∈A, | (e) x ∧ (y ∧ z) = (x ∧ y) ∧ z, ∀x, y, z ∈A |
| (f) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), ∀x, y, z ∈A, | (g) x ∧ y = x ∧ (x~ ∨ y), ∀x, y, z ∈A, |
is called a Pre A*-algebra.
Example
3 = {0, 1, 2} with operations ∧, ∨, (-)~ defined below
is a Pre A*-algebra.
Note 1
The elements 0, 1, 2 in the above example satisfy the following laws:
| (a) 2~ = 2 | (b) 1 ∧ x = x for all x ∈ 3 |
| (c) 0 ∨ x = x for all x ∈ 3 | (d) 2 ∧ x = 2 ∨ x = 2 for all x ∈ 3. |
Example
2 = {0, 1} with operations ∧, ∨, (-)~ defined below is
a Pre A*-algebra.
Note 2
| (i) | (2, ∨, ∧, (-)~) is a Boolean algebra. So every Boolean algebra is a Pre A* algebra. |
| (ii) | The identities 1.2(a) and 1.2(d) imply that the varieties of Pre A*-algebras satisfies all the dual statements of 1.2(b) to 1.2(g). |
Note 3
Let A be a Pre A*-algebra then A is Boolean algebra iff x∨(x∧y) =
x, x∧ (x∨ y) = x (absorption laws holds)
Lemma 1
Every Pre A*-algebra with 1 satisfies the following laws
(a) x∨1 = x∨x~ |
(b) x∧0 = x∧x~ |
Proof
(a) x∨1 = x∨(x~∧1) = x∨x~ (since x∧1,
∀x∈A)
(b) x∧0 = x∨(x~∨0) = x∨x~ (since x∨0 =
x, ∀x∈A)
Lemma 2
Every Pre A*-algebra satisfies the following laws.
(a) x∨ (x~∨ x) = x(b) (x∨x~)∧y = (x∧y)
∨ (x~∧y)
(c) (x∨x~) ∧x = x(d) (x∨y)∧z = (x∧z)∨(x~∧y∧z)
Proof
Definition 3
Let A be a Pre A*-algebra. An element x ∈A is called central element
of A if x∨x~ = 1 and the set {x ∈A/x∨x~ = 1}
of all central elements of A is called the centre of A and it is denoted by
B (A). Note that if A is a Pre A*-algebra with 1then 1, 0∈B (A). If the
centre of Pre A*-algebra coincides with {0, 1} then we say that A has trivial
centre.
Theorem 1
Let A be a Pre A*-algebra with 1, then B (A) is a Boolean algebra with the
induced operations ∧, ∨, (-)~
Proof
First we prove that B (A) is a sub algebra of A. Suppose x, y∈B(A)
then x∨x~ = y∨y~ = 1.
If x∨x~ = 1 then x~∨(x~)~ = 1 ⇒ x~∈B(A) (x∨y)∨(x∨y)~ = (x∨y)∨1 = x∨(y∨1) = x∨(y∨y~) = x∨1 = x∨x~ = 1
Therefore, x∨y∈B(A) and hence, B (A) is sub algebra of A. Also since 1∨0 = 1, 1∈B(A) and each x∈B(A), x∧1 = x∧ (x∨x~) = x. Hence, 1 (= x∨x~), which is an identity for ∧ in B (A). Then by lemma 2(a), x∨(x~∧x) = x ⇒ x∨0 = x (since x∈B(A), x∨x~∧x = 0). This shows that 0 (= x~∧x) is an identity for ∨ in B (A). Now we prove that x∨y~ = 1. Suppose that x∨y~. By dual of 1.2(g), we have x∨y = x∨(x~∧y) = x∨0 = x (since x∨y~ = 1 ⇒ x~∧y = 0).
Suppose that x∨y = x. Then x∨y~ = x∨y∨y~ = x∨1 = 1. Hence, B(A) is a Boolean algebra.
Lemma 3
Let A be a Pre A*-algebra with 1,
(a) If y ∈B(A) then x∧x~∧y = x∧x~, ∀
x∈A
(b) x∧(x∨ y) = x∨(x∧y) = x if and only if x, yB∈(A)
Proof
If x = 2 then result is clear.
If x, y∈B(A), then x∨x~ (x∧x~ = 0) and y∨y~
= 1(y∧y~ = 0)
Now x∧x∧y = 0∧y = 0 (since y∈B(A)) = x∈x
(b) x∧(x∨y) = (x∧x)∨(x∧y) = x∨(x∧(x~∨y))
= (x∨x)∧ (x∨(x~∨y)) = x∧(x∨x~∨y)
= x∧(x∨x~) = x
Definition 4
A non -empty set A equipped with a binary operation * is called a semilattice
if it satisfies the following properties:
| (i) | x*x = x for all x∈A |
| (ii) | x*y = y*x for all x,y∈A |
| (iii) | x*(y*z) = (x*y) *z, for all x, y, z∈A |
Note 4
For our purpose we use binary operation * in a more generalized form to
represent the given operation meet.
SEMILATTICE STRUCTURE ON PRE A*-ALGEBRA
Theorem 2
Let A b e a Pre A*-algebra define a binary operation * on A by x*y = x∧y,
for all x, y∈A then <A, *> is a semilattice.
Proof
x*x = x∧x = x for all x∈
| For x, y∈A we have x*y = x∧y = y∧x = y*x |
x* (y*z) = x* (y∧z) = x∧(y∧z) = (x∧y) ∧z = (x*y) *z, for all x, y, z∈A
Hence, <A, *> is a semilattice.
Definition 5
Let A be a Pre A*-algebra define a relation ≤* on A by x≤*
y iff x*y = x
Lemma 4
Let, A be a Pre A*-algebra then (A, ≤*) is a poset
Proof
Since x*x = x∧x = x, x≤* x, for all x∈A. Therefore
≤* is reflexive. Suppose ≤*y, y≤*z,
for all x, y, z∈A then x*y = x and y*z = y. Now x*z = (x*y) *z = x* (y*z)
= x*y = x. That is x≤*z,this shows that ≤* is Transitive.
Let, x≤*and y≤*x for all x, y∈A then x*y =
x and y*x = y ⇒ x = y. This shows that ≤* is anti symmetric.
Therefore (A, ≤*) is a poset.
Note 5
Let A be a Pre A*-algebra then (A, ≤*) is a poset. We have
x≤*y iff x*y = x, so x*y≤*y for all y∈A this
shows that x*y is the infimum of {x, y}.
Note 6
Let A be a Pre A*-algebra with 0, 1, 2 then x≤*1 (x∧1
= x for all x∈A) and 2≤*x (2∧x = 2 for all x∈A).This
gives that 1 is the greatest element and 2 is the least element of the poset
(A, ≤*).The Hasse diagram of the poset (A, ≤*) is:
Note 7
We have AxA = {a1 = (1, 1), a2 (1, 0), a3
(1, 2), a4 (0, 1), a5 (0, 0), a6 (0, 2), a7
(2, 1), a8 = (2, 0), a9 (2, 2)} is a Pre A*-algebra under
point wise operation and AxA is having four central elements and remaining are
non central elements, among that a9 (2, 2) is satisfying the property
that a9~ = a9. The Hasse diagram is of the
poset (AxA, ≤) given below:
Observe that, x≤* a1 (x∧a1 = x) and a9≤*x (x∧a9 = a9) for all x∈AxA. This shows that a1 is the greatest element and a9 is the least element of AxA.
Note 8
The poset (A, ≤*) need not be a join semilattice. For example
L = {(2, 1), (1, 2), (2, 0) (0, 2), (2, 2)}is a Pre A*-algebra(sub algebra of
AxA) and the poset (L, ≤*) is not a semilattice because Sup {(2,
1) ), (1, 2)}does not exist.
Lemma 5
The following conditions hold for any elements x and y in a Pre A*-algebra
A
| (i) | x∧y≤* x |
| (ii) | x∨y≤* x∨ x~ |
Proof
| (i) | Consider (x∧y) *x = (x∧y) *x = (x∧y)∧x = x∧y. Therefore, x∧y≤*x |
| (ii) | consider (x∨y) * (x∨x~) = (x∨y)∧(x∨x~) = x∨(y∨x~) = x∨y (by dual of 1.1g) |
Therefore x∨y≤*x∨x~
Lemma 6
Let A be a Pre A*-algebra then * is distributive over ∨ and ∧ i.e.,
(i) x*(y∨z) = (x*y)∨(x*z)
(ii) x*(y∧z) = (x*y)∧(x*z)
Proof
(i) (x*y)∨(x*z) = (x∧y)∨(x∧z) = x∧(y ∨ z) = x* (y∨z)
(ii) (x*y)∨(x*z) = (x∧y)∧(x∧z) = x∧(y∧z) = x*(y∨z)
Theorem 3
Let A be a Pre A*-algebra for any x∈A then x∨x~ is the
supremum of {x, x~}the semilattice <A, *>.
Proof
Lemma 7
In the poset (A, ≤*) and x, y∈A. If x≤*
then for a∈A
(i) a∧x≤* a≤y
(ii) a∨x≤*a∨y
Proof
Lemma 8
Let A be a Pre A*-algebra for any x, y∈A, x∧y≤*y
then, x∧y is the lower bound of {x, y}
Proof
Suppose x ∧y≤*y then x∧y is lower bound of y. Now
(x∧y) *x = (x∧y)∧x = x∧y⇒ x∧y≤*x.
Therefore, x∧y is lower bound of x
� x∧y is the lower bound of {x, y}
Theorem 4
Let A be a Pre A*-algebra for any x, y∈A then Inf {x, y} = x∧y
in the semilattice (A, *)
Proof
Note 9
In general for a Pre A*-algebra A with 1, x∨y need not be the least upper
bound of {x, y} in (A, ≤*). For example 2∨x = 2∧x = 2,
∀x∈A is not a least upper bound. However we have the following theorem.
Theorem 5
In a semilattice (A, *) with 1, for any x, y∈B(A) then sup {x, y} =
x∨y
Proof
Theorem 6
If A is a Pre A*-algebra and x∧ (x∨y) = x, for all x, y∈A then
(A, ≤*) is a lattice.
Proof
By Theorem 4 every pair of elements have infimum.
If x∧(x∨y) = x, for all x, y∈A then by Theorem 5 every pair of elements have supremum. Hence, (A, ≤*) is a lattice.
Lemma 9
Let A be a Pre A*-algebra then
(i) x∧(x*y) = x ∧y
(ii) (x*y)∧x = x*y
Proof
(i) x∧(x*y) = x∧(x∧y) = x∧y
(ii) (x*y)∧x = (x∧y)∧x = x∧y = x*y
Now we present a number of equivalent conditions for a Pre A*-algebra become a Boolean algebra.
Theorem 7
The following conditions are equivalent for any Pre A*-algebra (A, ∧,
∨, (-)~)
(1) A is Boolean Algebra
(2) x≤* x∨y for all x, y∈A
(3) y≤* x∨y for all x, y∈A
(4) x∨y is an upper bound of {x, y} in (A, ≤*) for all x, y∈A
(5) x∨y is an supremum of {x, y} in (A, ≤*) for all x, y∈A
(6) x∨x~ is the greatest element in (A, ≤*) for every x∈A
Proof
(2)⇒(3)
Suppose x≤*x∨y then x∧(x∨y) = x therefore, x∧(x∨
y) = x
Now y*(x∨y) = y∧(x∨y) = y. Therefore y≤* x∨y
(3)⇒(4)
Suppose that y≤*x∨y⇒y*(x∨y) = y therefore, y∧(x∨y)
= y
Since y≤* x∨y then x∨y is upper bound of y
Now x*(x∨y) = x≤ (x∨y) = x (by supposition)
x≤* x∨y ⇒ x∨y is upper bound of x
x∨y is an upper bound of {x, y}
(4)⇒(5)
Suppose x∨y is an upper bound of {x, y}
Suppose z is an upper bound of {x, y}, then x≤*z, y≤*z that is x*z = x, y*z = y
Y x∧z = x, y∧z = y
Now z*(x∨y) = z∧(x∨y) = (z∧x)∨(z∧y) = x∨y
Therefore, x∨y≤*z.
Hence, sup {x, y} = x∨y
(5)⇒(6)
Suppose sup {x, y} = x∨y then x, y∈B(A)
Now sup{x∨x~, y} = x∨x~∨y = x∨x~
(by definition 3)
Y y≤*x∨x~ therefore, x∨x~ is the greatest
element in (A, ≤*)
(6)⇒(1)
Suppose x∨x~ is the greatest element in A then y≤*x∨x~
Y (x∨x~)*y = y ⇒ (x∨x~)∧y = y
Now y∨(x∧y) = [(x∨x~ )∧y]∨(x∧y) = [(x∨x~)∨x]∧y
= (x∨x~)∨y = y (by supposition)
Therefore, by note 3 we have B is Boolean algebra.
Theorem 8
Let A be a pre A*-algebra x∧x~ is the least element in (A,
≤*) for every x∈A then A is Boolean algebra
Proof
By Note 3 we have B is Boolean algebra.