INTRODUCTION
In an outline study, Manes (1989) introduced the concept
of Ada (Algebra of disjoint alternatives) (A, ∧, ∨, ()'()_{π},
0, 1, 2), (where A is a nonempty set; ∧ and ∨ are two binary operations
on A; ()' and ()_{π} are unary operations on A and 0, 1, 2 are
distinguished elements in A). Which is however different from the definition
of the Ada of his later paper (Manes, 1993). While the
Ada of the earlier draft seems to be based on extending the IfThen Else concept
more on the basis of Boolean algebra and the later concept is based on Calgebra
(A, ∧, ∨, ') introduced by Guzman and Squier (1990).
Koteswara Rao (1994) firstly introduced the concept A*Algebra
(A, ∧, ∨, ()^{∼} ()_{π}, 0, 1, 2) and studied
its equivalence with Ada, Calgebra and Ada’s connection with 3 Ring.
Further he made an effort on the IfThenElse structure over A*algebra and
introduced the concept of Ideal of A*algebra. Venkateswara
Rao (2000) introduced the concept of Pre A*algebra (A, ∧, ∨, ')
as the variety generated by the 3element algebra A = {0, 1, 2} which is an
algebraic form of three valued conditional logic. Satyanarayana
et al. (2010) generated semilattice structure on Pre A*Algebras.
Venkateswara Rao and Srinivasa Rao (2009) defined a
partial ordering on a Pre A*algebra A and studied its properties. Satyanarayana
et al. (2010) derive necessary and sufficient conditions for pre
A*algebra A to become a Boolean algebra in terms of the partial ordering. Srinivasa
Rao (2009) studied the structural compatibility of Pre A*algebra with Boolean
algebra.
This study perceives a fundamental congruence θ_{x} on a Pre A*algebra
and confer its various properties. Also it establishes certain properties of
operations Γ_{x} (p, q) and Φ_{x}. It has been proved
that θ is a factor congruence on A if and only if θ = θ_{x}
for some xεB(A). Further it has been derived that the centre B(A) of a
Pre A*algebra A with 1 is isomorphic with the Boolean centre
(A) of A.
PRELIMINARIES
Definition 1: Boolean algebra is an algebra (B, ∧, ∨, ()', 0,
1, 2) 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.
Here, we concentrate on the algebraic structure of Pre A*algebra and state
some results which will be used in the later text.
Definition 2: An algebra (A, ∧, ∨, ()^{~}) where A is
nonempty set with 1, ∧, ∨ are binary operations and ()^{∼}
is a unary operation satisfying:
(a) 

(b) 

(c) 

(d) 

(e) 

(f) 

(g) 
is called a Pre A*algebra 
Example 1: 2 = {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 ε 3 
(c) 
0 ∨ x = x for all x ∈ 3 
(d) 
2 ∧ x = 2 ∨ x = 2 for all x ε 3. 
Example 2: 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 (a) and (d) imply that the varieties of Pre A*algebras
satisfies all the dual statements of (b) to (g) of definition 2 
Note 3: Let A be a Pre A*algebra then A is Boolean algebra if and only if x∨(x∧y) = x, x∧(x∨y) = x (absorption laws holds).
Lemma 1: Every Pre A*algebra satisfies the following laws.
Definition 3: Let A be a Pre A*algebra. An element xεA 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 1 then 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: Venkateswara Rao and Srinivasa Rao (2009):
Let A be a Pre A*algebra with 1, then B (A) is a Boolean algebra with the induced
operations ∧, ∨, ()^{∼}.
Lemma 2: Venkateswara Rao and Srinivasa Rao (2009):
Let A be a Pre A*algebra with 1:
(a) 
If y ε B(A) then 
(b) 
x∧(x∨y) = x∨(x∧y) = x if and only if x, y εB(A) 
Congruences on Pre A*algebra
Definition 4: Srinivasa Rao (2009): Let A be a Pre
A*algebra and θ be binary relation on A. Then θ is said to be an
equivalence relation on A if θ satisfies the following:
(i) 
Reflexive: (x, x) ε θ, for all xεA 
(ii) 
Symmetric: (x, y) ε θ ⇒(y, x) ε θ, for all x,
y ε A 
(iii) 
Transitive: (x, y) ε θ and (y, z) ε θ⇒(x, z)
ε θ, f or all x, y, z ε A 
We write x θ y to indicate (x, y)εθ.
Definition 5: Srinivasa Rao (2009): A relation
θ on a Pre A* algebra (A, ∧, ∨, ()^{∼}) is called
congruence relation if:
(i) 
θ is an equivalence relation 
(ii) 
θ is closed under ∧, ∨, ()^{~} 
Lemma 3: Srinivasa Rao (2009): Let (A, ∧,
∨, ()^{~}) be a Pre A*algebra and let a ε A. Then the relation
θ_{a} = {(x, y)ε AxA/a∧x = a∧y} is:
(a) 
a congruence relation 
(b) 

(c) 

(d) 

We will write x θ_{a}y to indicate (x, y)εθ_{a}.
Lemma 4: Srinivasa Rao (2009): Let A be a Pre
A*algebra with 1 and x, yεB(A). Then
(1) 
θ_{x} ∩ θ_{y} = θ_{x∨y} 
(2) 
θ_{x} o θ_{y} ⊆ θ_{x∧y} 
(3) 
θ_{x} ∨ θ_{y} ⊆ θ_{x∧y} 
Theorem 2: Srinivasa Rao (2009): Let Abe a Pre
A*algebra, then A ={θa/aεA} is a Pre A*algebra, is called quotient
Pre A*Algebra, whose operations are defined as:
(i) 
θ_{a}∧θ_{b} = θ_{a∧b} 
(ii) 
θ_{a}∨θ_{b} = θ_{a∨b} 
(iii) 
(θ_{a})^{~} = θ_{a~} 
Lemma 5: Let A be a Pre A*algebra and x, y ε A. Then, θ_{x} ⊆ θ_{y} if and only if y = x ∧ y.
Proof: Suppose that θ_{x} ⊆ θ_{y}.
Since x∧y = x∧x∧y, we have (y, x∧y) ε θ_{x }and
therefore, (y, x∧y) εθ_{y}.
(Supposition)
⇒ y∧y = y∧x∧y
⇒ y = x∧y
Conversely suppose that y = x∧y
Let (p, q)ε θ_{x} ⇒x∧p = x∧q
Now y∧p = x∧ y∧p (Supposition)
= y∧x∧p
= y∧x∧q
= y∧q (Supposition)
Therefore, (p, q)εθ_{y} and hence, θ_{x}⊆θ_{y}.
Lemma 6: Let A be a Pre A*algebra and x,yεA. Then θ_{x}⊆θ_{y} and θ_{y}⊆θ_{x∧y}.
Proof: We know that x∧(x∧ y) = x∧y, by above Lemma we have
θ_{x}⊆θ_{x∧y}.
Also, we know that y∧(x∧y) = x∧y, by above Lemma we have θ_{y}⊆θ_{x∧y}.
Lemma 7: Let A be a Pre A*algebra and x, yεB(A). Then θ_{x∨y} ⊆θ_{x}.
Proof: Let (p, q) ε θ_{x∨y}. Then (x∨y)∧p
= (x∨y)∧q
Now x∧p = x∧(x∨y)∧p (By Lemma 2)
= x∧(x∨y)∧q
= x∧q
Therefore (p, q) εθ_{x} and hence, θ_{x∨y}⊆θ_{x}.
Note that it is similar way θ_{x∨y}⊆θ_{y}.
Theorem 3: Let A be a Pre A*algebra with 1 and xεA. Then θ_{x} is the smallest congruence on A containing (1, x).
Proof: We know that θ_{x} is a congruence of A and clearly
(1,x) ε θ_{x}.
Let θ be a congruence on A and (1,x) εθ.
Suppose that (p, q)ε θ_{x} ⇒x∧p = x∧q
Since (1,x)ε θ, we have (1∧p, x∧p) and (1∧q, x∧q)
εθ; that is (p, x∧p) and (q, x∧q) εθ. Therefore
(p, q) εθ and hence θ_{x} ⊆ θ.
Theorem 4: Let A be a Pre A*algebra with 1 and x, yεA then the following are equivalent:
(1) 
x, yεB(A) 
(2) 
θ_{x∨y}⊆θ_{y} 
(3) 
θ_{x∨y}⊆θ_{x}∩θ_{y} 
(4) 
θ_{x∨y} = θ_{x}∩θ_{y} 
Proof: (1)⇒(2) Suppose that x, yεB(A)
Let (p ,q) εθ. Then (x∨y) ∧p = (x∨y)∧q
Now y∧p= y∧(y∨x) ∧p (Since x, yεB(A) by Lemma 2)
= y∧(x∨y) ∧q = y∧q
Therefore, (p, q) εθ_{y} and hence, θ_{x∨y}⊆θ_{y}.
(2)⇒(3) Suppose that θ_{x∨y}⊆θ_{x}.
By symmetry we have, θ_{x∨y}⊆θ_{x}.
Therefore, θ_{x∨y}⊆θ_{x}∩θ_{y}.
(3)⇒(4) Suppose that θ_{x∨y}⊆θ_{x}∩θ_{y}.
We know that, θ_{x}∩θ_{y}⊆θ_{x∨y}
(By Lemma 3(c))
Therefore, θ_{x∨y}=θ_{x}⊆θ_{y
}(4)⇒(1) Suppose that θ_{x∨y}=θ_{x}∩θ_{y
}Then (1, x∨y) ε θ_{x∨y} = θ_{x}∩θ_{y}
(Supposition)
Therefore (1, x∨y) ε θ_{x} and (1, x∨y) εθ_{y}
⇒ x∧1= x∧(x∨y) and y∧1 = y∧(x∨y)
⇒ x = x∧(x∨y) and y= y∧(x∨y)
⇒ x, yεB(A) (By Lemma 2)
Definition 6: Srinivasa Rao (2009): Let A be
a Pre A* algebra with 1. For any x, p, q ε A, define Γ_{x}
(p, q) = (x∧p)∨(x^{~}∧q).
Now we prove certain properties of Γ_{x}(p, q) which will be used
later text.
Lemma 8: Let A be a Pre A* algebra and x, p, q εB(A). Then:
(1) 
Γ_{x}(p, q) = (x^{~}∨p)∧(x∨q) 
(2) 
Γ_{x}(p, q)^{~} = Γ_{x}(p^{~}
,q^{~}) 
Proof:
(1) 
Γx(p, q) = (x∧p)∨(x^{~}∧q) 

= [(x∧p)∨x^{~}]∧[(x∧p)∨q] (By distributive law) 

= [x^{~}∨p]∧[(x∨q)∧(x^{~}∨ p∨q)] (Definition
2 (g) and Lemma 1(d)) 

= [(x^{~}∨p)∧(x∨q)∧(x^{~}∨ p)] ∧[(x^{~}∨p)∧(x∨q)∧q)]
(By distributive law) 

= [(x^{~}∨p)∧(x∨q)]∧[(x^{~}∨p)∧ (x∨q)
∧q)] 

= (x^{~}∨p)∧(x∨q) (Lemma 2 (b)) 
(2) 
By (1) we have Γ_{x} (p, q) = (x^{~}∨p)
∧(x∨q). 

Now Γ_{x}(p, q)^{~} = [(x^{~}∨p) ∧(x∨q)]^{~} 

= (x^{~}∨p)^{ ~}∨(x∨q)^{~} 

= (x ∧ p^{~})∨(x^{~}∧q^{~}) 

= Γ_{x}(p^{~}, q^{~}) 
Lemma 9: Let A be a Pre A* algebra and x, p, q ε A. Then Γ_{x} (p, q) = Γ_{x~} (q, p).
Proof: Γ_{x} (p, q) = (x∧p)∨(x^{~}∧q)
= (x^{~~}∧p)∨(x^{~}∧q)
= (x^{~}∧q) ∨(x^{~~}∧p)
= Γ_{x~}(q, p).
Definition 7: Let A be a Pre A* algebra and xεA. Define:
Φ_{x} = {(p, q) ε AxA / Γ_{x}(p, q) = p}.
Theorem 5: Let A be a Pre A* algebra and xεB(A). Then:
(1) 
Φ_{x}⊆θ_{x~} 
(2) 
Φ_{x} is transitive relation on A. 
Proof:
(1) 
Let p, q ε A and (p, q) εΦ_{x}, that
is Γ_{x}(p, q)= p 

⇒(x∧p)∨(x^{~}∧q) = p 

Now x^{~}∧p = x^{~}∧{(x∧p∨(x^{~}∧q))} 

= {x^{~}∧x∧p} ∨{x^{~}∨x^{~}∧q} 

= (0∧p) ∨(x^{~}∧q) 

= 0∨(x^{~}∧q) (provided p ≠ 2) 

= x^{~}∧q 

If p=2 then x^{~}∧p =2 to get the required result q should
be 2. 

Therefore x^{~}∧p = x^{~}∧q , hence (p, q) ε
θ_{x~} 

Thus, Φ_{x}⊆θ_{x~} 
(2) 
Let p, q, r εA and (p, q), (q, r) ε Φ_{x},
that is Γ_{x}(p, q) = p and Γ_{x}(q, r) = q. 

Then by (1) we have (p, q), (q, r)ε θ_{x~} 

⇒ x^{~}∧p = x^{~}∧q and x^{~}∧q =
x^{~}∧r 

Now Γ_{x}(p, r) = (x∧p)∨(x^{~}∧r) 

= (x∧p)∨(x^{~}∧q) 

= Γx(p, q) 

= p 
Therefore (p, r) ε Φ_{x} and hence Φ_{x} is transitive relation on A.
Theorem 6: Let A be a Pre A*algebra induced by a Boolean algebra and θ be congruence on A. Then θ_{x} is a factor congruence on A if and only if θ = θ_{x} for some xεA.
Proof: Suppose that θ = θ_{x} for some xεB(A).
Then x^{~}εB(A) and θ_{x}∩θ_{x~}
=θ_{x∨x~} = θ_{1} = Δ_{A
}and θ_{x} o θ _{x~} = θ_{x∧x~}
= θ_{0} = AxA
Thus, θ_{x} is a factor congruence on A
Conversely suppose that θ is a factor congruence on A
Then there exist a congruence β on A such that θ ∩ β = Δ_{A}
and θ o β = AxA.
Now we show that θ = θ_{x
}Suppose that (p,q) εθ_{x} then x∧p= x∧ q.
Since (x,1) εθ we have (x∧p, 1∧p), (x∧q, 1∧ q) ε
that is (x∧p, p), (x∧q, q) εθ which imply that (p,q) εθ.
Hence, θ_{x}⊆θ
Suppose (p,q) εθ. Then (x∧p, x∧q) εθ.
Since (0,x) εβ we have (0∧p, x∧ p), (0∧q, x∧q) εβ
that is (0, x∧p), (0, x∧q) εβ which implies that (x∧p,
x∧q) εβ.
Therefore (x∧p, x∧q) εβ∩ θ = Δ_{A} and
hence x∧p = x∧q ⇒(p, q)εθ_{x
}Hence, θ⊆θ_{x
}Thus θ = θ_{x}.
Hence, θ is a factor congruence on A if and only if θ = θ_{x}
for some xεB(A).
Definition 8: Let A be a Pre A*algebra and αεCon(A). Then α is called factor congruence if there exist βεCon(A) such that α∩β = Δ_{A} and αoβ= AxA. In this case β is called direct complement of α.
Definition 9: A congruence β on Pre A*algebra A is called balanced if (β∨θ)∩(β∨θ^{~}) = β for any direct factor congruences θ and any of its direct complement θ^{~} on A.
Theorem 7: Let A be a Pre A*algebra with 1 and xεB(A). Then θ_{x} is balanced.
Proof: Let θ_{x} is a congruence on Pre A*algebra A. Let
Ψ be a factor congruence A and Ψ^{~ }be its complement. Then
by theorem (b) there exist y, zεB(A) such that Ψ = θ_{y}
and Ψ^{~} = θ_{z}.
Now (θ_{x}∨Ψ)∩(θ_{x}∨Ψ^{~})
= (θ_{x}∨θ_{y})∩ (θ_{x}∨θ_{z})
=θ_{x∨y}∩θ_{x∨z
}=θ_{(x∨y)∨(x∨z)
}=θ_{x∨(y∨z)
}= θ_{x}∨θ_{y∨z
}= θ_{x} ∨ (θ_{y}∩θ_{z})
= θ_{x} ∨(Ψ∩Ψ^{~})
= θ_{x}∨Δ_{A} (Since Ψ and Ψ^{~}
are complements)
= θ_{x
}Hence, θ_{x} is balanced.
Therefore, the set of balanced congruence which admit a balanced complement
is precisely the set
(A) = {θ_{x}/xεB(A)} and hence,
(A) is the Boolean centre of A.
Theorem 8: Let A be a Pre A*algebra with 1. Then the Boolean centre
(A) = {θ_{x} / xεB(A)} is a Boolean algebra and the map x→θ_{x~}
is an isomorphism of B(A) into
(A).
Proof: It follows from Lemma 4, theorem 6 and theorem 7.
CONCLUSION
This manuscript point ups the essential congruence θ_{x} on a
Pre A*algebra and reach your destination at a variety of properties of these.
Also it corroborate certain properties of the operations Γ_{x}(p,
q) and Φ_{x}. It has been longestablished that θ is a factor
congruence on A if and only if θ = θ_{x} for some x εB(A).
If A is a Pre A*algebra with 1 and xεA, then obtained that θ_{x}
is the smallest congruence on A containing the ordered pair (1, x). Additionally
it was ensuing that the centre B(A) of a Pre A*algebra A with 1 is isomorphic
with the Boolean centre
(A) of A.