Figure 3: P(A), where A=a, b Example 5.< P(A), S> is a Boolean algebra. Consider a special set A =a, b, its Hasse diagram is Figure 3. According to Example 2 and Example 4, it is a Boolean algebra Example 6. <(1, 2, 3, 6,> is a boolean algebra First, we can verify that it is distributive and complemented. But it is tedious. In Example ??, we have proved <(1, 2, 3, 6],> is isomorphic to P(a, bn),C> We know the mapping keep the properties of operations n, U So <(1, 2, 3, 6],> is also a Boolean Generally, we have the following theorem Theorem 9(Stone, 1936). Every finite Boolean algebra is isomorphic to the Boolean algebra of subsets of some finite set s n order to prove it, Boolean algebra is to be represented in another form, which is out of the scope of this course. It is hard and the proof is not easy to be understood. However, the result is elegant 10. Every finite Boolean algebra has 2n elements for some n According to this Corollary, if a finite lattice has not 2"elements, it must not be a boolean algebra Theorem 11. The complement a' of any element a of a Boolean algebra b is uniquely determined The mapping'is a one-to-one mapping of B onto itself. It satisfies the conditions (aUb)=anb,(a∩b)y=a′Ub Proof. We prove the theorem one by one. 1. Suppose a has two complement aand al. We have a Ua=l,ana=0, aUal= l and ana1=0.Soa1=a1∩1=a1n(aUa)=(a1∩a)∪(a1∩a)=a1∩a Similarly, we can also prove a=ar na. And it means a=a1 2. For every a E B, a is a complement of a according to definition, so(a=a. It means that is a bijecti∅ {a} {b} {a, b} Figure 3: P(A), where A = {a, b}. Example 5. < P(A), ⊆> is a Boolean algebra. Consider a special set A = {a, b}, its Hasse diagram is Figure 3. According to Example 2 and Example 4, it is a Boolean algebra. Example 6. < {1, 2, 3, 6}, | > is a Boolean algebra. First, we can verify that it is distributive and complemented. But it is tedious. In Example ??, we have proved < {1, 2, 3, 6}, | > is isomorphic to < P({a, b}), ⊆>. We know the mapping keep the properties of operations ∩, ∪. So < {1, 2, 3, 6}, | > is also a Boolean algebra. Generally, we have the following theorem. Theorem 9 (Stone, 1936). Every finite Boolean algebra is isomorphic to the Boolean algebra of subsets of some finite set S. In order to prove it, Boolean algebra is to be represented in another form, which is out of the scope of this course. It is hard and the proof is not easy to be understood. However, the result is elegant. Corollary 10. Every finite Boolean algebra has 2 n elements for some n. According to this Corollary, if a finite lattice has not 2n elements, it must not be a Boolean algebra . Theorem 11. The complement a 0 of any element a of a Boolean algebra B is uniquely determined. The mapping 0 is a one-to-one mapping of B onto itself. It satisfies the conditions. (a ∪ b) 0 = a 0 ∩ b 0 , (a ∩ b) 0 = a 0 ∪ b 0 Proof. We prove the theorem one by one. 1. Suppose a has two complement a 0 and a1. We have a ∪ a 0 = 1, a ∩ a 0 = 0, a ∪ a1 = 1 and a ∩ a1 = 0. So a1 = a1 ∩ 1 = a1 ∩ (a ∪ a 0 ) = (a1 ∩ a) ∪ (a1 ∩ a 0 ) = a1 ∩ a 0 . Similarly, we can also prove a 0 = a1 ∩ a 0 . And it means a 0 = a1. 2. For every a ∈ B, a is a complement of a 0 according to definition, so (a 0 ) 0 = a. It means that 0 is a bijection. 4