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.

2 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) x x = ;
• (b) ( ) = ; x y x y
• (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

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.

Every Boolean algebra is a C-algebra.

Lemma 2.3. Every C-algebra satisfies the following laws [1,3,5].

• (a) x x x = ;
• (b) x y x x y = ( ) = ( ) ; x y x
• (c) x x x x ( ) = ;
• (d) ( ) = ( ) ( ); x x y x y x y
• (e) ( ) = ; x x x x
• (f) x x x x = ;
• (g) x y x x y = ;
• (h) x x y x x = ;
• (i) x y x x y x ( ) = ( ) .

The duals of all above statements are also true.

Definition 2.4. If A has identity T for \(\wedge\) ( that is, \(x \wedge T = T \wedge x = x\) for all \(x \in A\)), then it is unique and in this case, we say that A is a C-algebra with T. We denote T by F and this F is the identity for \(\vee\) [1].

Lemma 2.5 [1]: Let A be a C-algebra with T and \(x, y \in A\). Then

• (i) \(x \lor y = F\) if and only if x = y = F
• (ii) if \(x \lor y = T\) then \(x \lor x' = T\).
• (iii) \(x \lor T = x \lor x'\)
• (iv) \(x \wedge F = x \wedge x'\).

Theorem 2.6. Let \(\langle A, \wedge, \vee, ' \rangle\) be a C-algebra. Then the following are equivalent [6]:

• (i) A is a Boolean algebra.
• (ii) \(x \lor (y \land x) = x\), for all \(x, y \in A\).
• (iii) \(x \wedge y = y \wedge x\), for all \(x, y \in A\).
• (iv) \((x \lor y) \land y = y\), for all \(x, y \in A\).
• (v) \(x \vee x'\) is the identity for \(\wedge\), for every \(x \in A\).
• (vi) \(x \lor x' = y \lor y'\), for all \(x, y \in A\).
• (vii) A has a right zero for \(\wedge\).
• (viii) for any \(x, y \in A\), there exists \(a \in A\) such that \(x \lor a = y \lor a = a\).
• (ix) for any \(x, y \in A\), if \(x \lor y = y\), then \(y \land x = x\).

Definition 2.7. Let A be a C-algebra with T. An element \(x \in A\) is called a central element of A if \(x \lor x' = T\). 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 \(\langle B(A), \wedge, \vee, ' \rangle\) 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:

• (i) \(x \lor y = x \lor (y \land x')\);
• (ii) \(x \wedge y = x \wedge (y \vee x')\).

Proof. Let A be a C-algebra and \(x, y \in A\). Now,

\[x \lor y = x \lor (x' \land y) = x \lor (x' \land y \land x') = [x \land (x \lor x')] \lor [x' \land y \land (x' \land (x \lor x'))] = [x \land (x \lor x')] \lor [x' \land y \land x' \land (x \lor x')] = [x \land (x \lor x')] \lor [x' \land y \land (x \lor x')] = (x \lor y) \land (x \lor x') = x \lor (y \land x').\] Similarly \(x \land y = x \land (y \lor x')\).

Lemma 3.2. Let A be a C-algebra and \(x, y \in A\). Then \(x \lor y \lor x' = x \lor y \lor y'\). Proof. Let A be a C-algebra and \(x, y \in A\). By Lemma 2.3[b],[f] and Lemma 3.1, we have \(x \lor y \lor x' = x \lor ((y' \land x') \lor y) = [x \lor (y' \land x')] \lor y = (x \lor y') \lor y = x \lor y' \lor y = x \lor y \lor y'\).

Lemma 3.3. Let A be a C-algebra with \(T, x, y \in A\) and \(x \wedge y = F\). Then \(x \vee y = y \vee x\).

Proof. Suppose that \(x \wedge y = F\). Then \(F = x \wedge y = x \wedge (x' \vee y) = (x \wedge x') \vee (x \wedge y) = (x \wedge x') \vee F = x \wedge x'\). now \(x \vee y = F \vee (x \vee y) = (x \wedge y) \vee (x \vee y) = (x \vee x \vee y) \wedge (x' \vee y \vee x \vee y)\) (By Def 2.1) \(= (x \vee y) \wedge (x' \vee y \vee x)\) (By Lemma 2.3[g]) \(= (x \wedge x') \vee (y \vee x) = F \vee (y \vee x) = y \vee x\).

In [6], it is proved that if A is a C-algebra with T then \(B(A) = \{a \in A \mid a \lor a' = T\}\) is a Boolean algebra under the same operations \(\land, \lor, '\) in the C-algebra A. Now we prove the following.

Theorem 3.4. Let A be a C-algebra with T and \(a,b \in A\) such that \(a \lor b \in B(A)\). Then \(a \in B(A)\).

Proof. Let A be a C-algebra with T and \(a,b \in A\) such that \(a \lor b \in B(A)\). Then \(T = (a \lor b) \lor (a \lor b)' = (a \lor b) \lor (a' \land b')\) \(= (a \lor b \lor a') \land (a \lor b \lor b') = (a \lor b \lor a') \land (a \lor b \lor a')\) (By Theorem 3.2) \(= a \lor b \lor a'\).

Therefore,

\[T = a \lor b \lor a' \tag{1}\]

Now, \[a \lor a' = (a \lor a') \land T\]
\[= (a \lor a') \land (a \lor b \lor a') \quad \text{(by (1))}\] \[= (a \land (a \lor b \lor a')) \lor (a' \land (a \lor b \lor a'))\] \[= a \lor (a' \land (a \lor b \lor a'))\] \[= a \lor (a \lor b \lor a') \quad \text{(By Lemma 2.3[b])}\] \[= a \lor b \lor a' = T.\]

Hence \(a \in B(A)\).

The converse of the above theorem need not be true. For example, in the Calgebra \(C, F \in B(C)\) but \(F \vee U = U \notin B(C)\). We have the following consequence of the above theorem.

Corollary 3.5. Let A be a C-algebra with \(T, a, b \in A\) and \(a \land b \in B(A)\). Then \(a \in B(A)\).

Proof. Let \(a \land b \in B(A)\). Then we have, \((a \land b)' \in B(A) \Rightarrow a' \lor b' \in B(A)\)\(\Rightarrow a' \in B(A) \Rightarrow a \in B(A)\).

In [1], it is proved that if A is a C-algebra, then \(\theta_x = \{(p,q) \mid x \land p = x \land q\}\) is a congruence on A and \(\theta_x \cap \theta_{x'} = \theta_{x \lor x'}\). In [6], if A is C-algebra with T then \(\theta_x\) is a factor congruence if and only if \(x \in B(A)\). They also proved that \(\theta_x, \theta_y\) are permutable congruences whenever both \(x, y \in B(A)\). Now we prove some important properties of these congruences.

Theorem 3.6. Let A be a C-algebra with T and \(a,b \in A\). Then we have the following (i) \(\theta_{a \wedge b} = \theta_{b \wedge a}\); (ii) \(\theta_a \circ \theta_b \subseteq \theta_{a \wedge b}\).

Proof. (i) \[(x, y) \in \theta_{a \wedge b}\]
\(\Rightarrow a \wedge b \wedge x = a \wedge b \wedge y\)
\(\Rightarrow b \wedge a \wedge b \wedge x = b \wedge a \wedge b \wedge x\)
\(\Rightarrow b \wedge a \wedge x = b \wedge a \wedge x\)
\(\Rightarrow (x, y) \in \theta_{b \wedge a}\)

Therefore \(\theta_{a \wedge b} \subseteq \theta_{b \wedge a}\). Similarly, \(\theta_{b \wedge a} \subseteq \theta_{a \wedge b}\). Hence \(\theta_{a \wedge b} = \theta_{b \wedge a}\).

(ii) Let \((x,y) \in \theta_a \circ \theta_b\). Then there exists \(z \in A\) such that \((x,z) \in \theta_b\) and \((z,y) \in \theta_a\). Thus \(b \wedge x = b \wedge z\) and \(a \wedge z = a \wedge y\). Now, \(a \wedge b \wedge x = a \wedge b \wedge z = a \wedge b \wedge a \wedge z = a \wedge b \wedge a \wedge y = a \wedge b \wedge y\). Therefore, \((x,y) \in \theta_{a \wedge b}\). Thus \(\theta_a \circ \theta_b \subseteq \theta_{a \wedge b}\).

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 \(G = \{a_1, a_2, a_3, a_4, a_5\}\) where \(a_1 = (T, U)\), \(a_2 = (F, U)\), \(a_3 = (U, T)\), \(a_4 = (U, F)\), \(a_5 = (U, U)\) under pointwise operations in C.

This algebra \((G, \vee, \wedge,')\) is a C-algebra with out T. Let \(\Delta =\) diagonal of A. Then we have the following:

\[\theta_{a_1} = \{(x, y) \mid a_1 \land x = a_1 \land y\}\] \[= \Delta \cup \{(a_3, a_4), (a_4, a_5), (a_5, a_3), (a_4, a_3)(a_5, a_4), (a_3, a_5)\}\] \[\theta_{a_3} = \Delta \cup \{(a_1, a_2), (a_2, a_5), (a_5, a_1), (a_2, a_1)(a_5, a_2), (a_1, a_5)\}\]

Now, \[\theta_{a_1} \circ \theta_{a_3} = \Delta \cup \theta_{a_1} \cup \theta_{a_3} \cup \{(a_4, a_1), (a_4, a_2)(a_3, a_1), (a_3, a_2)\}\]
\(\theta_{a_3} \circ \theta_{a_1} = \Delta \cup \theta_{a_1} \cup \theta_{a_3} \cup \{(a_2, a_4), (a_2, a_3)(a_1, a_3), (a_1, a_4)\}\)

Therefore \(\theta_{a_1} \circ \theta_{a_3} \neq \theta_{a_3} \circ \theta_{a_1}\).

Theorem 3.8. Let A a C-algebra with T and \(a \in B(A)\). Then for any \(b \in A\), \(\theta_a\), \(\theta_b\) permute and \(\theta_a \circ \theta_b = \theta_{a \wedge b}\).

Proof. Let A be a C-algebra with T and \(a \in B(A)\). By Theorem 3.6, \(\theta_a \circ \theta_b \subseteq \theta_{a \wedge b}\). Now let \((p,q) \in \theta_{a \wedge b}\). Then \(a \wedge b \wedge p = a \wedge b \wedge q \Longrightarrow b \wedge a \wedge b \wedge p = b \wedge a \wedge b \wedge q \Longrightarrow b \wedge a \wedge p = b \wedge a \wedge q\). Consider, \(r = (a \wedge p) \vee (a' \wedge q)\). Now \(a \wedge r = a \wedge [(a \wedge p) \vee (a' \wedge q)] = (a \wedge p) \vee (a \wedge a' \wedge q) = (a \wedge p) \vee (F \wedge q) = (a \wedge p) \vee F = a \wedge p\). Therefore \((r,p) \in \theta_a \Longrightarrow (p,r) \in \theta_a\). Now, \(b \wedge r = b \wedge [(a \wedge p) \vee (a' \wedge q)] = [b \wedge a \wedge p] \vee [b \wedge a' \wedge q] = (b \wedge a \wedge q) \vee (b \wedge a' \wedge q) = b \wedge ((a \wedge q) \vee (a' \wedge q)) = b \wedge ((a \vee a') \wedge q) = b \wedge (T \wedge q)\) (since \(a \in B(A)\)) \(= b \wedge q\). Therefore \((q,r) \in \theta_b \Longrightarrow (r,q) \in \theta_b\). Thus \((p,q) \in \theta_b \circ \theta_a\). Hence \(\theta_b \circ \theta_a = \theta_{a \wedge b}\). Thus \(\theta_b \circ \theta_a\) is a congruence on A and hence \(\theta_a, \theta_b\) are permutable congruences and hence \(\theta_a \circ \theta_b = \theta_b \circ \theta_a = \theta_{a \wedge b}\).

Corollary 3.9. Let A be a C-algebra with T and \(a,b \in A\). Then i) \(a \lor b \in B(A) \Longrightarrow \theta_a \circ \theta_b = \theta_{a \lor b}\); ii) \(a \land b \in B(A) \Longrightarrow \theta_a \circ \theta_b = \theta_{a \land b}\).

Proof. i) We know that if \(a \lor b \in B(A)\) then \(a \in B(A)\) and hence by the above theorem \(\theta_a \circ \theta_b = \theta_b \circ \theta_a = \theta_{a \diamond b}\). 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 \(\Delta, \nabla\) 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 \((A, \vee, \wedge,')\) be a C-algebra with T. Then the following are equivalent. (i) Let \((A, \vee, \wedge,')\) be a Boolean algebra. (ii) \(\theta_x \cap \theta_{x'} = \Delta\) for all \(x \in A\). (iii) \(\theta_{x \vee x'} = \Delta\) for all \(x \in A\).

Proof. (1) \(\Rightarrow\)(2): Let A be a Boolean algebra and \(x \in A\). Let \((p,q) \in \theta_x \cap \theta_{x'}\). Then \(x \wedge p = x \wedge q\) and \(x' \wedge p = x' \wedge q\). Now, \(p = (x \vee x') \wedge p = (x \wedge p) \vee (x' \wedge q) = (x \wedge q) \vee (x' \wedge q) = (x \vee x') \wedge q = q\). Thus \(\theta_x \cap \theta_{x'} \subseteq \Delta\). Therefore \(\theta_x \cap \theta_{x'} = \Delta\). Since \(\theta_x \cap \theta_{x'} = \theta_{x \vee x'}\), we get (ii) \(\Rightarrow\)(iii). (iii) \(\Rightarrow\)(i): Suppose \(\theta_{x \vee x'} = \Delta\) for all \(x \in A\). We prove that \(\theta_{x'} \circ \theta_x = A \times A\). Let \((p,q) \in A \times A\). Write \(t = (x \wedge p) \vee (x' \wedge q)\). Now, \(x \wedge t = x \wedge ((x \wedge p) \vee (x' \wedge q)) = (x \wedge p) \vee (x \wedge x' \wedge q) = (x \wedge p) \vee (x \wedge x') = x \wedge (p \vee x') = x \wedge p\). Also, \(x' \wedge t = x' \wedge ((x \wedge p) \vee (x' \wedge q)) = (x' \wedge x \wedge p) \vee (x' \wedge x' \wedge q)) = (x' \wedge x) \vee (x' \wedge q) = (x' \wedge x \wedge p) \vee (x' \wedge x' \wedge q) = (x' \wedge x) \vee (x' \wedge q) = (x' \wedge x) \vee (x' \wedge q)\). Therefore \((p,t) \in \theta_x\) and \((t,q) \in \theta_{x'}\). Thus \((p,q) \in \theta_{x'} \circ \theta_x\). Hence we get \(\theta_{x'} \circ \theta_x = A \times A\). Also \(\theta_x \cap \theta_{x'} = \theta_{x \vee x'} = \Delta\). That is \(\theta_x\) and \(\theta_{x'}\) are permutable factor congruences. Therefore, by Theorem 2.6, we have \(x \in B(A)\). Thus A = B(A) and hence A is a Boolean algebra.

4 The C-algebra \(S_{x}\)

We prove that, for each \(x \in A\), \(S_x = \{x \lor t \mid t \in A\}\) is itself a C-algebra under induced operations \(\land\), \(\lor\) and the unary operation is defines by \((x \lor t)^* = x \land t'\). We observe that \(S_x\) need not be a subalgebra of A because the unary operation in \(S_x\) is not the restriction of the unary operation on A. Also for each \(x \in A\), the set \(A_x = \{x \land t \mid t \in A\}\) is a C-algebra in which the unary operation is given by \((x \land t)^* = x \land t'\). We prove that the B(A) is isomorphic to the Boolean algebra \(\mathfrak{B}_{S(A)}\) of all C-algebras \(S_a\) where \(a \in B(A)\). Also, we prove that B(A) is isomorphic to the Boolean algebra \(\mathfrak{B}_{S(A)}\) of all C-algebras \(A_x\), \(a \in B(A)\).

Theorem 4.1. Let \(\langle A, \wedge, \vee,' \rangle\) be a C-algebra, \(x \in A\) and \(S_x = \{x \vee t \mid t \in A\}\). Then \(\langle S_x, \wedge, \vee, * \rangle\) is a C-algebra with x as the identity for \(\vee\), where \(\wedge\) and \(\vee\) are the operations in A restricted to \(S_x\) and for any \(x \vee t \in S_x\), here \((x \vee t)^*\) is \(x \vee t'\).

Proof. Let \(t, r, s \in A\). Then \((x \lor t) \lor (x \lor r) = x \lor (t \lor r) \in S_x\) and \((x \lor t) \land (x \lor r) = x \land (t \lor r) \in S_x\). Thus \(\lor, \land\) are closed in \(S_x\). Also * is closed in \(S_x\). Consider \((x \lor t)^{***} = x \lor (x \lor t')' = x \lor (x' \land t) = x \lor t\). Now \([(x \lor t) \land (x \lor r)]^* = [x \lor (t \land r)]^* = x \lor (t' \lor r') = x \lor t' \lor x \lor r' = (x \lor t)^* \lor (x \lor r)^*\). Now, consider \([(x \lor t) \lor (x \lor r)] \land (x \lor s) = x \land [(t \land r) \land s] = x \lor [(t \land s) \lor (t' \land r \land s)] = x \lor (t \land s) \lor x \lor (t' \land r \land s) = [(x \lor t) \land (x \lor s)] \lor [(x \lor t') \land (x \lor r) \land (x \lor s)] = [(x \lor t) \land (x \lor s)] \lor [(x \lor t)^* \land (x \lor r) \land (x \lor s)]\) The remaining identities of a C-algebra also hold in \(S_x\) because they hold in A. Hence, \(S_x\) is itself a C-algebra. Also x is the identity for \(\checkmark\) because \(x \lor x \lor t = x \lor t \lor x\). Here \(x \lor x'\) is the identity for \(\land\).

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) \(S_x \cap S_y \subseteq S_{x \vee y}\);
• (iii) \(S_{x} \cap S_{x'} = S_{x_{x'}x'}\);
• (iv) \((S_x)_{x \vee y} = S_{x \vee y}\).

Proof. (i) Suppose \(S_x = S_y\). Since \(x = x \lor x \in S_x = S_y\) and \(y = y \lor y \in S_y = S_x\). Therefore \(x = y \lor t\) and \(y = x \lor r\) for some \(t, r \in A\). Now, \(x = y \lor t = (y \lor t \lor y) \land (y \lor y \lor t) = (x \lor y) \land (y \lor x) = (y \lor x) \land (x \lor y) = (x \lor r \lor x) \land (x \lor x \lor r) = x \lor r = y\). The converse is trivial. (ii) Suppose \(t \in S_x \cap S_y\). Then \(t = x \lor s = y \lor r\) for some \(s, r \in A\). Now, \(t = x \lor x \lor s = x \lor t = x \lor y \lor r \in S_{x \lor y}\). (iii) \(S_x \cap S_{x'} \subseteq S_{x \lor x'}\) by (ii). Since \(x \lor x' = x' \lor x\) we have \(S_{x \lor x'} \subseteq S_x \cap S_x'\). Hence \(S_x \cap S_{x'} = S_{x \lor x'}\). (iv) \((S_x)_{x \lor y} = \{x \lor y \lor t \mid t \in S_x\} = \{x \lor y \lor x \lor r \mid r \in A\} = \{x \lor y \lor r \mid r \in A\} = S_{x \lor y}\)

Theorem 4.3. Let A be a C-algebra with T and \(x \in A\), then the mapping \(\alpha_x\): \(A \to S_x\) defined by \(\alpha_x(t) = x \lor t\) for all \(t \in A\) is a homomorphism of A to \(S_x\) with kernel \(\theta_{x'}\) and hence \(A/\theta_{x'} \cong S_x\).

Proof. Let \(t,r \in A\). Then \(\alpha_x(t \vee r) = x \vee t \vee r = x \vee t \vee x \vee r = \alpha_x(t) \vee \alpha_x(r)\) and \(\alpha_x(t') = x \vee t' = (x \vee t)^* = (\alpha_x(t))^*\). Clearly, \(\alpha_x(t \wedge r) = \alpha_x(t) \wedge \alpha_x(r)\). Also \(\alpha_x(T) = x \vee T = x \vee x'\), which is the identity for \(\wedge\) in \(S_x\). Therefore \(\alpha_x\) is a homomorphism. Hence by the fundamental theorem of homomorphism \(A/Ker\alpha_x \cong S_x\) and \(Ker\alpha_x = \{(t,r) \in A \times A \mid \alpha_x(t) = \alpha_x(r)\} = \{(t,r) \in A \times A \mid x \vee t = x \vee r\} = \{(t,r) \in A \times A \mid x' \wedge t = x' \wedge r\} \theta_{x'}\) (by Lemma 2.3 [b]) = and hence \(A/\theta_{x'} \cong S_x\).

Theorem 4.4. Let A be a C-algebra with T and \(a \in B(A)\), then \(A \cong S_a \times S_{a'}\).

Proof. Define \(\alpha: A \to S_a \times S_{a'}\) by \(\alpha(x) = (\alpha_a(x), \alpha_{a'}(x))\) for all \(x \in A\). Then, by Theorem 4.3, \(\alpha\) is well-defined and \(\alpha\) is a homomorphism. Now, we prove that \(\alpha\) is one-one. Let \(x, y \in A\). Then \(\alpha(x) = \alpha(y) \Rightarrow (\alpha_a(x), \alpha_{a'}(x)) = (\alpha_a(y), \alpha_{a'}(y)) \Rightarrow (a \vee x, a' \vee x) = (a \vee y, a' \vee y) \Rightarrow a \vee x = a \vee y\) and \(a' \vee x = a' \vee y\). Now \(x = F \vee x = (a \wedge a') \vee x = (a \vee x) \wedge (a' \vee x) = (a \vee y) \wedge (a' \vee y) = y\). Finally, we prove that \(\alpha\) is onto. Let \((x, y) \in S_a \times S_{a'}\). Then \(x = a \vee t\), and \(y = a' \vee r\) for some \(t, r \in A\). Therefore, \(a \vee x = x\), \(a \vee y = a \vee a' \vee y = T \vee y = T\) and \(a' \vee x = T, a' \vee y = y\). Now, \(\alpha(x \wedge y) = (\alpha_a(x \wedge y), \alpha_{a'}(x \wedge y))\) \(= (a \vee (x \wedge y), a' \vee (x \wedge y))\) \(= (a \vee x) \wedge (a \vee y), (a' \vee x) \wedge (a' \vee y)\) \(= (x \wedge T, T \wedge y)\) = (x, y).

Therefore, \(\alpha\) is onto and hence \(\alpha\) is an isomorphism. Therefore \(A \cong S_a \times S_{a'}\).

Lemma 4.5. Let A be a C-algebra. Then for \(a, b \in A\):

• (i) \(a \lor b = b \lor a\) if and only if \(S_{a \lor b} = S_a \cap S_b\)
• (ii) \(S_{a \wedge b} = Sup\{S_a, S_b\}\) in the poset (\(\{S_x \mid x \in A\}, \subseteq\)), then \(a \wedge b = b \wedge a\). The converse is not true.

Proof. (i) Suppose that \(a \lor b = b \lor a\). Then clearly \(S_{a \lor b} \subseteq S_a \cap S_b\). By Theorem 4.2(ii) \(S_a \cap S_b \subseteq S_{a \lor b}\). Hence \(S_{a \lor b} = S_a \cap S_b\). Conversely assume that \(S_{a \lor b} = S_a \cap S_b\). Clearly \(a \lor b \in S_{a \lor b} = S_a \cap S_b\). Therefore \(a \lor b \in S_b \Rightarrow a \lor b = b \lor t\) for some \(t \in A\). Now \(b \lor a = b \lor a \lor b = b \lor b \lor t = b \lor t = a \lor b\). (ii) Assume that \(a, b \in A\) and \(S_{a \land b} = Sup\{S_a, S_b\}\). Then \(S_{a \land b} = S_{b \land a}\) and hence \(a \land b \in S_{a \land b} = S_{b \land a}\). Therefore \(a \land b = (b \land a) \lor t\) for some \(t \in A\). Now \((b \land a) \lor (a \land b) = (b \land a) \lor ((b \land a) \lor t) = (b \land a) \lor t = a \land b\). Similarly we can prove that \((a \land b) \lor (b \land a) = b \land a\). Hence \(a \land b = b \land a\). The converse need not be true, for example for the C-algebra C, \(S_U = \{U\}, S_T = \{T\}\) and \(U \land T = T \land U\). But \(S_{U \land T} (= S_U)\) is not an upper bound of \(\{S_U, S_T\}\).

Now we prove \(\mathfrak{B}_{S(A)} = \{S_a \mid a \in B(A)\}\) is a Boolean algebra under set inclusion.

Theorem 4.6. Let \(\langle A, \wedge, \vee,' \rangle\) be a C-algebra with T. Then \(\mathfrak{B}_{S(A)} = \{S_a \mid a \in B(A)\}\) is a Boolean algebra under set inclusion.

Proof. Clearly \((\mathfrak{B}_{S(A)},\subseteq)\) is a partially ordered set under inclusion. First we show for \(a,b\in B(A)\), \(S_{a\vee b}\) is the infimum of \(\{S_a,S_b\}\) and \(S_{a\wedge b}\) is the supremum of \(\{S_a,S_b\}\) for all \(a,b\in B(A)\). Let \(a,b\in B(A)\). Then \(a\wedge b=b\wedge a\) and \(a\vee b=b\vee a\). Hence by the above Lemma 4.5, \(S_{a\vee b}\) is the infimum of \(\{S_a,S_b\}\). Let \(t\in S_a\). Then \(t=a\vee x\) for some \(x\in A\). Now \(t=a\vee x=(a\wedge (a\vee b))\vee x=(a\wedge (b\vee a)\vee x=(a\wedge b)\vee a\vee x\in S_{a\wedge b}\). Similarly \(S_b\subseteq S_{b\wedge a}=S_{a\wedge b}\). Therefore \(S_{a\wedge b}\) is an upper bound of \(S_a,S_b\). Suppose \(S_c\) is an upper bound of \(S_a,S_b\). Then \(t=(a\wedge b)\vee x\) for some \(x\in A\). Now \(t=(a\wedge b)\vee x=(a\vee x)\wedge (a'\vee b\vee x)=(a\vee x)\wedge (b\vee a'\vee x)\in S_c\) (since \(a\vee x\in S_a\subseteq S_c\), \(b\vee a'\vee x\in S_b\subseteq S_c\) and \(S_c\) is closed under A). Therefore \(S_a\wedge b\) is the supremum of \(\{S_a,S_b\}\) by \(S_a\wedge S_b\) and the infimum of \(\{S_a,S_b\}\) by \(S_a\wedge S_b\). Now \(S_T\wedge S_a=S_{T\vee a}=S_T\) and \(S_F\vee S_a=S_{F\wedge a}=S_F\). Therefore \(S_T\) is the least element and \(S_F\) is the greatest element of \((\mathfrak{B}_{S(A)},\subseteq)\). Now for any \(a,b,c\in B(A)\), \((S_a\vee S_b)\wedge S_c=S_{(a\wedge b)\vee c}=S_{(a\vee c)\wedge (b\vee c)}=S_{(a\vee c)}\vee S_{(b\vee c)}=(S_a\wedge S_c)\vee (S_b\wedge S_c)\). Also \(S_a\wedge S_{a'}=S_{a\vee a'}=S_T\) and \(S_a\vee S_{a'}=S_{a\wedge a'}=S_F\). Therefore \((\mathfrak{B}_{S(A)},\subseteq)\) is a complimented distributive lattice and hence it is a Boolean algebra.

Theorem 4.7. Let A be a C-algebra with T Define \(\varphi: B(A) \to \mathfrak{B}_{S(A)}\) by \(\phi(a) = s_{a'}\) for all \(a \in B(A)\). Then \(\phi\) is an isomorphism.

Proof. Let \[a,b \in B(A)\]. Then \(\varphi(a \wedge b) = S_{(a \wedge b)'} = S_{a'} \wedge S_{b'} = \varphi(a) \wedge \varphi(b)\). \(\varphi(a \vee b) = S_{(a \vee b)'} = S_{a'} \vee S_{b'} = \varphi(a) \vee \varphi(b)\), \(\varphi(a') = S_{a'} = (S_a)' = (\varphi(a))'\). Clearly \(\phi\) is both one-one and onto. Hence \(B(A) \cong \mathfrak{B}_{S(A)}\).

In [3] we defined a partial ordering on a C-algebra by \(x \le y\) if and only if \(y \land x = x\) 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 \(x \in A\), \(A_x = \{s \in A \mid s \le x\}\) is itself a C-algebra under induced operations \(\land, \lor\) and the unary operation is defined by \(s^* = x \land s'\) we also observed that \(A_x\) need not be an algebra of A because the unary operation in \(A_x\) is not the restriction of the unary operation. For each \(x \in A\), we proved that \(A_x\) is isomorphic to the quotient algebra \(A/\theta_x\) where \(\theta_x = \{(p,q) \in A \times A \mid x \land p = x \land q\}\). We can easily see that the C-algebras \(S_x, A_x\) are different in general where \(x \in A\).

Now, we prove that the set of all \(A_a\)'s where \(a \in B(A)\) 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 \(\mathfrak{B}_{R(A)}:=\{A_a\mid a\in B(A)\}\) is a Boolean Algebra under set inclusion in which the supremum of \(\{A_a,A_b\}=A_{a\vee b}\) and the infimum of \(\{A_a,A_b\}=A_{a\wedge b}\).

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 ( ) : ( ) R A f B A B by a A^a f ( ) = for all aB(A). Then f is an isomorphism.

The following corollary can be proved directly from Theorems 4.7 and 4.9.

Corollary 4.10. Let A be a C-algebra with T. Then ( ) ( ) , ( ) and B B R A S A B A are isomorphic to each other.