用中文解答下题Let A = {a, b, c} and B = P(A). 1) Prove that [B; ∪ , ∩ ,~] is a Boolean algebra. 2) Write out the operation tables for the Boolean algebra.

1) 首先需要证明[B; ∪ , ∩ ,~]满足Boolean代数的四个公理：结合律、交换律、分配律和互补律。

结合律：对于任意的A,B,C∈B，有(A∪B)∪C=A∪(B∪C)和(A∩B)∩C=A∩(B∩C)。这是因为集合的并、交运算符都满足结合律。

交换律：对于任意的A,B∈B，有A∪B=B∪A和A∩B=B∩A。这是因为集合的并、交运算符都满足交换律。

分配律：对于任意的A,B,C∈B，有A∪(B∩C)=(A∪B)∩(A∪C)和A∩(B∪C)=(A∩B)∪(A∩C)。这是因为集合的并、交运算符都满足分配律。

互补律：对于任意的A∈B，有A∪~A=A和A∩~A=∅。这是因为集合的补集~A定义为A的所有元素的补集，即~A={x∣x∉A}，A∪~A=A∩~A=A的所有元素的集合，因此满足互补律。

由此可知[B; ∪ , ∩ ,~]是一个Boolean代数。

2) 接下来是[B; ∪ , ∩ ,~]的操作表：

并集 ∪：

| | a | b | c |
|---|---|---|---|
| a | a | a | a |
| b | a | b | b |
| c | a | b | c |

交集 ∩：

| | a | b | c |
|---|---|---|---|
| a | a | ∅ | ∅ |
| b | ∅ | b | ∅ |
| c | ∅ | ∅ | c |

补集 ~：

| | a | b | c |
|---|---|---|---|
| | b,c | a,c | a,b |

其中，∅代表空集。

相关问题

Let A, B,C and D be sets. Prove that (A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D).

答案：我们可以证明两个集合相等的方法是证明它们互相包含。首先，假设 (x, y) 属于 (A × B) ∩ (C × D)，那么 x 属于 A 且 y 属于 B，同时 x 属于 C 且 y 属于 D。因此，x 属于 A ∩ C 且 y 属于 B ∩ D，即 (x, y) 属于 (A ∩ C) × (B ∩ D)。因此，(A × B) ∩ (C × D) 包含于 (A ∩ C) × (B ∩ D)。

反过来，假设 (x, y) 属于 (A ∩ C) × (B ∩ D)，那么 x 属于 A 且 x 属于 C，同时 y 属于 B 且 y 属于 D。因此，(x, y) 属于 A × B 且 (x, y) 属于 C × D，即 (x, y) 属于 (A × B) ∩ (C × D)。因此，(A ∩ C) × (B ∩ D) 包含于 (A × B) ∩ (C × D)。

因此，我们证明了两个集合互相包含，从而证明了它们相等。

Exercise 4. Let � = {�, �, �} and � = P(�). 1) Prove that [�; ∪ , ∩ ,�] is a Boolean algebra. 2) Write out the operation tables for the

Boolean algebra [�; ∪ , ∩ ,�].

1) To prove that [�; ∪ , ∩ ,�] is a Boolean algebra, we need to show that it satisfies the following properties:

i) Closure under complementation: For any element � in �, there exists an element � in � such that � ∪ � = � and � ∩ � = �.

ii) Closure under union and intersection: For any elements �, � in �, there exists an element � in � such that � ∪ � and � ∩ � are in �.

iii) Associativity of union and intersection: For any elements �, �, � in �, (� ∪ �) ∪ � = � ∪ (� ∪ �) and (� ∩ �) ∩ � = � ∩ (� ∩ �).

iv) Commutativity of union and intersection: For any elements �, � in �, � ∪ � = � ∪ � and � ∩ � = � ∩ �.

v) Distributivity of union over intersection and intersection over union: For any elements �, �, � in �, � ∪ (� ∩ �) = (� ∪ �) ∩ (� ∪ �) and � ∩ (� ∪ �) = (� ∩ �) ∪ (� ∩ �).

vi) Identity elements: There exist two elements �, � in � such that for any element � in �, � ∪ � = � and � ∩ � = �.

vii) Complement elements: For any element � in �, there exists an element � in � such that � ∪ � = � and � ∩ � = �.

All of these properties are satisfied by [�; ∪ , ∩ ,�], so it is indeed a Boolean algebra.

2) Here are the operation tables for [�; ∪ , ∩ ,�]:

∪   0  1  2
  _____
0|  0  1  2
1|  1  1  2
2|  2  2  2

∩   0  1  2
  _____
0|  0  0  0
1|  0  1  1
2|  0  1  2

�   0  1  2
  _____
 |  2  1  0