Abstract Algebra Theory and Applications - Computer Science ...

17.2 BOOLEAN ALGEBRAS 301Observe thatI ∨ b = (I ∨ b) ∧ I = (I ∧ I) ∨ (b ∧ I) = I ∨ I = I.Consequently, the first of the two absorption laws holds, sincea ∨ (a ∧ b) = (a ∧ I) ∨ (a ∧ b)= a ∧ (I ∨ b)= a ∧ I= a.The other idempotent and absorption laws are proven similarly. Since Balso satisfies (1)–(3), the conditions of Theorem 17.3 are met; therefore, Bmust be a lattice. Condition (4) tells us that B is a distributive lattice.For a ∈ B, O ∨ a = a; hence, O ≼ a and O is the smallest element in B.To show that I is the largest element in B, we will first show that a ∨ b = bis equivalent to a ∧ b = a. Since a ∨ I = a for all a ∈ B, using the absorptionlaws we can determine thata ∨ I = (a ∧ I) ∨ I = I ∨ (I ∧ a) = Ior a ≼ I for all a in B. Finally, since we know that B is complemented by(5), B must be a Boolean algebra.Conversely, suppose that B is a Boolean algebra. Let I and O be thegreatest and least elements in B, respectively. If we define a ∨ b and a ∧ b asleast upper and greatest lower bounds of {a, b}, then B is a Boolean algebraby Theorem 17.3, Theorem 17.4, and our hypothesis.□Many other identities hold in Boolean algebras. Some of these identitiesare listed in the following theorem.Theorem 17.6 Let B be a Boolean algebra. Then1. a ∨ I = I and a ∧ O = O for all a ∈ B.2. If a ∨ b = a ∨ c and a ∧ b = a ∧ c for a, b, c ∈ B, then b = c.3. If a ∨ b = I and a ∧ b = O, then b = a ′ .4. (a ′ ) ′ = a for all a ∈ B.5. I ′ = O and O ′ = I.6. (a ∨ b) ′ = a ′ ∧ b ′ and (a ∧ b) ′ = a ′ ∨ b ′ (De Morgan’s Laws).