Research Article
Pre A*-Algebra as a Semilattice
Andhra Loyola Institute of Engineering and Technology, Vijayawada-8.A.P., India
J. Venkateswara Rao
Mekelle University, Main Campus, P.O. Box No. 231, Mekelle, Tigrai, Ethiopia
The study lattice theory had been made by Birkhoff (1948). In a drafted paper The Equational theory of Disjoint Alternatives, around 1989, Manes (1989) introduced the concept of Ada (A, ∧, V) (-)I, (-)π, 0, 1, 2) which however differs from the definition of the Ada. While the Ada of the earlier draft seems to be based on Boolean algebras, the latter concept is based on C-algebras (A, ∧, V) (-)~ introduced by Guzman and Squier (1990).
In 1994, Koteswara Rao (1994) firstly introduced the concept of A* algebra (A, ∧, V, *, (-) ~, (-)π, 0, 1, 2) and studied the equivalence with Ada, C-algebra, and Ada and its connection with 3-ring, Stone type representation and introduced the concept of A*-Clone and the if -then-else structure over A*-algebra and ideal of A*-algebra. We introduce Pre A*-algebra (A, ∧, V, (-) ~) analogous to C-algebra as a product of A*-algebra. Recently Pre A*-algebra had been studied by Chandrasekhara Rao et al. (2007), Rao and Satyanarayana (2010), Rao and Rao (2010) and Rao and Praroopa (2011).
Definition 1: 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∈A |
is called a Pre A*-algebra.
Example: 3 = {0, 1, 2} with ∨, ∧ (-)~ defined below is a Pre A*-algebra.
Example of a Pre A*-algebra:
Example: 2 = {0, 1} with ∨, ∧ (-)~ defined below is a Pre A*-algebra.
Example of a Pre A*-algebra:
Note: The elements 0, 1, 2 in examples satisfy the following laws:
(a) | 2~ = 2 |
(b) | 1 ∧ x = x, ∀ x ∈ 3 (1 the identity for ∧) |
(c) | 1~ = 0 |
(d) | 2 ∧ x = 2, ∀ x∈3 |
(e) | 0 ∨ x = x, ∀ x∈3 (0 is the identity for ∨) |
Note:
(i) | (2, ∧, ∧ (-)~) is a Boolean algebra. So every Boolean algebra is a Pre A* algebra |
(ii) | The identities Def. 1a and Def. 1d imply that the varieties of Pre A*-algebras satisfies all the dual statements of Def. 1a to Def. 1g |
PRE A*-ALGEBRA AS A SEMILATTICE
Definition 2: Let A be a Pre A*-algebra. Define≤on A by x≤y if and only if x∧y = y∧x = x, ∀ x, y∈A.
The defined≤is said to be partial ordering on Pre A*-algebra A.
Lemma 1: If A is a Pre A*-algebra then (A, ≤) is a Poset.
Proof: Since x∧x = x, x≤x for all xεA.
Semi lattice in a Pre A*-algebra
Definition 3: A non-empty subset S of a Pre A*-algebra A equipped with a binary operation ∧(∨) is said to be a semi lattice, if the following semi lattice axioms are satisfied:
(i) | ∧(∨) is associative i. e., a ∧ (b ∧ c) = (a ∧ b) ∧ c, ∀ a, b, c∈S |
(ii) | ∧(∨) is commutative i.e., a ∧ b = b ∧ a, ∀ a, b∈S |
(iii) | ∧(∨) is idempotent i.e., a ∧ a = a, ∀ a∈S |
Theorem 1: In Pre A*-algebra A (S, ∧) and (S, ∨) are semi lattices.
Proof: In Pre A*-algebra, a ∧ (b ∧ c) = (a ∧ b) ∧ c, ∀ a, b, c∈A (Def. 1e) | |
a ∧ b = b ∧ a, ∀ a, b ∈A (Def. 1c) and a ∧ a = a, ∀ a∈A (Def. 1b) Hence, (S, ∧) is a semi lattice. By the duality in A (S, ∨) is a semi lattice. |
Theorem 2: In Pre A*-algebra A, the class of semi lattices can be equationally defined as the class of all semi group satisfying the commutative and idempotent laws.
Proof: Let (S, ∧, ∨) be a semi lattice in a Pre A*-algebra A. By the definition of semi lattice we have ∧(∨) is associative. i.e., a ∧ (b ∧ c) = (a ∧ b) ∧ c, ∀ a, b, c∈S
(i) | ∧(∨) is commutative i.e., a ∧ b = b ∧ a, ∀ a, b∈S |
(ii) | ∧(∨) is idempotent i.e., a ∧ a = a, ∀ a∈S |
Hence, (S, ∧) as well as (S, ∨) is a semi group satisfying commutative and idempotent laws.
Therefore, (S, ∧ (∨)) is a semigroup satisfying the commutative and idempotent laws.
Converse: (S, ∧) as well as (S, ∨) is a semi-group satisfying commutative and idempotent laws.
By the definition of Pre A*-algebra A:
• | a ∧ (b ∧ c) = (a ∧ b) ∧ c, ∀ a, b, c∈A (Def. 1e) |
• | a ∧ b = b ∧ a, ∀ a, b ∈A (Def. 1c) |
• | a ∧ a = a, ∀ a∈A (Def. 1b) |
Hence, ∧(∨) is associative, commutative and idempotent.
Hence, (S, ∧(∨)) is a Semilattice in a Pre A*-algebra A.
Pre A*-algebra as a semilattice
Theorem 3: Let A be a Pre A*-algebra. Then A is a semilattice.
Proof: Since A is a Pre A*-algebra:
• | Then a ∧ (b ∧ c) = (a ∧ b) ∧ c, ∀ a, b, c∈A by (Def. 1e) |
• | a ∧ b = b ∧ a, ∀ a, b ∈A by (Def. 1c) |
• | a ∧ a = a, ∀ a∈A by (Def. 1b) |
Hence, A is a semilattice
Note: We can also define Pre A*-algebra as follows:
An algebra (A, ∧, ∨, ~ ) is said to be Pre A*-algebra where A is non-empty set with 1 and ∧, ∨ are binary operations ~ is a unary operation satisfying:
• | (A, ∧) is a semilattice |
• | x~ ~ = x, ∀, x A, |
• | (x∧y)~ = x~ v y~, ∀ x, yA |
• | x ∧ (y∨z) = (x ∧ y) ∨ (x ∧ z), ∀ x, y, z∈A |
• | x ∧y = x ∧ (x~ v y), ∀ x, y, z A |
Theorem 4: Let A be a Pre A*-algebra. Then in a semilattice A, define x≤y if and only if x∧y = x. Then (A, ≤) is an ordered set in which every pair of elements has a greatest lower bound.
Conversely, given an ordered set P with that property, define x ∧ y = g.l.b. (x, y).
Then (P, ∧) is a semilattice.Proof: Let (A, ∧) be a semilattice and define≤ as above. First we check that ≤ is a partial order:
(i) | x∧x = x implies x≤x |
(ii) | If x≤y and y≤x, then x = x∧y = y∧x = y |
(iii) | If x≤y≤z, then x∧z = (x∧y) ∧ z = x ∧ (y∧z) = x∧y = x, so x≤z |
Since (x∧y) ∧ x = x ∧ (x∧y) = (x∧x) ∧ y = x∧y), we have x∧y≤x similarly x∧y≤y. Thus x ∧ y is a lower bound for {x, y}.
To see that it is the greatest lower bound, suppose z≤x and z≤y. Then z ∧ (x∧y) = (z∧x) ∧ y = z∧y = z, soz≤x ∧ y.
Converse: suppose(P, ≤) is an ordered set. | |
define x∧y = g.l.b. (x, y). Since (P, ≤) is an ordered set: |
(i) | x≤x implies x∧x = x |
(ii) | x≤y and y≤x, then x∧y = y∧x |
(iii) | z≤x and z≤y implies z ∧ (x∧y) = (z∧x) ∧ y |
Hence (P,∧) is a semilattice.
Distributive semilattice in a Pre A*-algebra
Definition 4: Let A be a Pre A*-algebra Then L is said to be distributive semilattice if any elements a, b, c in L we have the distributive law
Example: 2 = { 0, 1} is the distributive semilattice in a Pre A*-algebra.
Theorem 5: Let A be a Pre A*-algebra. Then A is a distributive semilattice.
Proof: Since A is a Pre A*-algebra, for any elements a, b, c in A we have the distributive law:
Lemma 2: Let A be a Pre A*-algebra. Then in the Poset (A, ≤):
Proof: Define≤in A as a≤b a ∧ b = a (i.e., a∨b = b) | |
Suppose a≤b then b ∧ a = a Now b ∧ (a ∨ c) = (b ∧ a) ∨ (b ∧ c) = a ∨ (b ∧ c) (by Def. 1f) |
Modular semilattice in a Pre A*-algebra
Definition 5: Let A be a Pre A*-algebra. Then L is said to be a modular semilattice if:
Example: 3 = { 0, 1, 2 } is the modular semilattice in a Pre A*-algebra.
Theorem 6: Let A be a Pre A*-algebra. Then A is a modular semilattice.
Proof: Since A is a Pre A*-algebra,
By lemma 2:
Hence, A is a modular semilattice.
Complement of an element in a Pre A*-algebra
Definition 6: Let A be a Pre A*-algebra with least element α, greatest element β. Then a∈A is said to be complement if there exists x∈A:
Note: Since ∧, ∨ are commutative in a Pre A*-algebra, we have if x is a complement of a then a is also a complement of x.
Hence, a, x are complements to one another.
Complemented semilattice in a Pre A*-algebra
Definition 7: Let A be a Pre A*-algebra then A is said to be Complemented semilattice if each element has a complement in it.
Example: The semilattice shown in this Fig. 1 is a complemented semilattice:
• | Here every element has a complement but these are not unique |
• | Here b, c are complements of a |
• | Here a, c are complements of b |
• | Here a, b are complements of c |
Example: This is example of a semilattice which is not complemented.
In Fig. 2, the elements a, e c, d have complements but the element b has no complement.
Fig. 1: | Complemented semilattice |
Fig. 2: | Example of a semilattice which is not complemented |
Theorem 7: Let A be a Pre A*-algebra. Then A is a Complemented semilattice.
Proof: Since A is a Pre A*-algebra, A is a semilattice.
Since each element in A has a complement in it, hence A is a Complemented semilattice.
Theorem 8: Let A be a Pre A*-algebra. Then A is a Complemented distributive semilattice
Proof: Since A is a Pre A*-algebra, Then A is a Complemented semilattice also distributive semilattice.
Unique complement of an element in a Pre A*-algebra
Definition 8: Let A be a Pre A*-algebra Then a∈A is said to be unique complement if a has exactly one complement in A.
Uniquely complemented semilattice in a Pre A*-algebra
Definition 9: Let A be a Pre A*-algebra. Then A is said to be uniquely complemented semilattice if each element in A has unique complement in A.
Relative complement in a Pre A*-algebra
Definition 10: Let A be a Pre A*-algebra. Let [a, b]∈A and u is an element of [a, b]. An element x of A is said to be relative complement of u in [a, b] if u∧x = a, u ∨ x = b.
Note: If x is a relative complement of u in [a, b] then we have:
Relatively complemented semilattice in a Pre A*-algebra
Definition 11: Let A be a Pre A*-algebra. Then A is said to be relatively complemented semilattice, if for any triplet of elements a, b, u such that a≤u≤b there exists at least one complement of u in [a, b] i.e., every interval of A is a complemented semilattice of A.
Example: The semilattice shown in Fig. 1 is an example of a semilattice which is complemented as well as relatively complemented.
Note: Let A be a Pre A*-algebra hen every bounded relatively complemented semilattice in A is complemented but converse is not true i.e., a complemented semilattice in a Pre A*-algebra A may or may not be relatively complemented semilattice.
Example: The semilattice shown in Fig. 3 is complemented but not relatively complemented.
Fig. 3: | Example of semilattice which is not relatively complemented |
Fig. 4: | Example of semilattice which is not section complemented |
Fig. 5: | Section complemented semilattice |
Since [α, b], [a, β] are not complemented semilattices since a has no complements in [α, b].
Example: The semilattice shown in Fig. 3 is not relatively complemented.
Since [a, β] ={a, d, β} is not a complemented semilattice since a has no complement hence it is not relatively complemented.
Section complemented semilattice in a Pre A*-algebra
Definition 12: Let A be a Pre A*-algebra with least element α. Then A is said to be section complemented semilattice if every interval of the form [α, a] is a complemented semilattice of A.
i.e., for each pair of elements a, u with u≤a there exists an element x∈A such that u∧x = α, u∨x = a.
Example: The semilattice shown in this Fig. 4 is not section complemented because [α, b] is a not complemented semilattice.
Example: Figure 5 example of a semilattice which is section complemented.
Theorem 9: Let A be a Pre A*-algebra. Then every relatively complemented semilattice in A is section complemented.
Proof: Since A is a Pre A*-algebra, if L is a relatively complemented semilattice then by the definition L every interval of A is a complemented semilattice of A.
Hence, L is section complemented semilattice.
Note: Let A be a Pre A*-algebra every relatively complemented semilattice in A is Section complemented but converse is not true.
Semi-complement of an element in a Pre A*-algebra
Definition 13: Let A be a Pre A*-algebra with least element α and u∈A. An element x∈A is said to be semi-complement of u if u ∧ x = α (x is not equal to α).
Definition 14: Let A be a Pre A*-algebra with least element α and u∈A. Then all Semi-complements of an element u forms a poset U. If this poset has a maximal element x0 then x0 is called maximal semicomplement of u i.e., if there exists x∈A such that u ∧ x = α and x0≤x implies x = x0.
Definition 15: Let A be a Pre A*-algebra with least element α. The semi-complements of an element other than the least element α is called a proper semi-complement in A. If in addition the proper semi-complement is maximal then it is called maximal proper semi-complement in A.
Semi-complemented semilattice in a Pre A*-algebra
Definition 16: Let A be a Pre A*-algebra with least element α. Then A is said to be Semi-complemented semilattice if every inner element (other than least and greatest elements in A) has at least one proper semi-complement.
Weakly complemented semilattice in a Pre A*-algebra
Definition 17: Let A be a Pre A*-algebra with least element α. Then A is said to be weakly complemented semilattice if any pair of elements a, b (a<b) of A a has semi complement, that is however, not a semi complement of b i.e., x is semi complement of a but not semi complement of b.
Example: The semilattice shown in Fig. 6 is example of a semilattice which is not weakly complemented.
This is not weakly complemented since a<b and c is semi complement of both a and b.
Theorem 10: Let A be a Pre A*-algebra. Then every weakly complemented semilattice in A is semi-complemented.
Proof: Let A be a Pre A*-algebra with least element α, greatest element β.
Let L be any weakly complemented semilattice in A
Claim: Lis semi-complemented
Let a∈L be an inner element i.e.:
• | ⇒ a is not a maximal element |
• | ⇒ ∃b∈L such that a<b |
Fig. 6: | Example of a semilattice which is not weakly complemented |
Since L is weakly complemented semilattice in A, we have that there exists semi-complement x of a which is not a semi-complement of b i.e.:
then x is proper semi-complement of a and hence L is semi-complemented. Hence every weakly complemented semilattice in A is semi-complemented.
Absorption law in Pre A*-algebra
Theorem 11: Let A be a Pre A*-algebra with least element α. Then every section complemented semilattice in A is weakly complemented.
Proof: Let A be a Pre A*-algebra with least element α. | |
And L be section complemented semilattice in A. |
Claim: L is weakly complemented. | |
Let a, b∈L such that a<b Now [α, b] is a complemented semilattice of L Since L is section complemented and a∈[α, b]⇒∃x∈L such that: |
Consider b ∧ x = (a ∨ x) ∧ x | |
= x (by absorption law in A) |
When:
Therefore, b = a which is a contradiction to a<b
Thus x≠α
Hence L is weakly complemented semilattice.
Theorem 12: Let A be a Pre A*-algebra with least element α, greatest element β. Then every uniquely complemented semilattice in A is weakly complemented.
Proof: Let A be a Pre A*-algebra with least element α, greatest element β and L be any uniquely complemented semilattice in A.
Claim: L is weakly complemented.
Let a, b∈L such that there exists unique complement a~ of a such that:
i.e., a~ is semi-complement of a
s ince a<b we have b∨a~>a∨a~ = β:
since β is greatest element in A, β> b∨a~:
b ∧ a~≠ α (suppose if b ∧ a~ =α then a~ is complement of both a and b which is a contradiction to our assumption that L is uniquely complemented semilattice in a Pre A*-algebra).
Therefore, a∧a~ = α, b∧a~≠α
Hence L is weakly complemented semilattice in A.