Abstract Algebra Theory and Applications - Computer Science ...

17.2 BOOLEAN ALGEBRAS 303Proof. If b is an atom, let a = b. Otherwise, choose an element b 1 , notequal to O or b, such that b 1 ≼ b. We are guaranteed that this is possiblesince b is not an atom. If b 1 is an atom, then we are done. If not, choose b 2 ,not equal to O or b 1 , such that b 2 ≼ b 1 . Again, if b 2 is an atom, let a = b 2 .Continuing this process, we can obtain a chainO ≼ · · · ≼ b 3 ≼ b 2 ≼ b 1 ≼ b.Since B is a finite Boolean algebra, this chain must be finite. That is, forsome k, b k is an atom. Let a = b k .□Lemma 17.8 Let a and b be atoms in a finite Boolean algebra B such thata ≠ b. Then a ∧ b = O.Proof. Since a ∧ b is the greatest lower bound of a and b, we know thata ∧ b ≼ a. Hence, either a ∧ b = a or a ∧ b = O. However, if a ∧ b = a, theneither a ≼ b or a = O. In either case we have a contradiction because a andb are both atoms; therefore, a ∧ b = O.□Lemma 17.9 Let B be a Boolean algebra and a, b ∈ B.statements are equivalent.1. a ≼ b.2. a ∧ b ′ = O.3. a ′ ∨ b = I.The followingProof. (1) ⇒ (2). If a ≼ b, then a ∨ b = b. Therefore,a ∧ b ′ = a ∧ (a ∨ b) ′= a ∧ (a ′ ∧ b ′ )= (a ∧ a ′ ) ∧ b ′= O ∧ b ′= O.(2) ⇒ (3). If a ∧ b ′ = O, then a ′ ∨ b = (a ∧ b ′ ) ′ = O ′ = I.(3) ⇒ (1). If a ′ ∨ b = I, thenThus, a ≼ b.a = a ∧ (a ′ ∨ b)= (a ∧ a ′ ) ∨ (a ∧ b)= O ∨ (a ∧ b)= a ∧ b.□