阿姆斯特朗公理证明 [英] Armstrongs Axioms Proof
问题描述
我在证明与阿姆斯特朗公理的功能相关性时遇到问题.我正在努力奋斗的那个和.假设R(A，B，C，D，E)为关系模式，并且F = {A→CD，C→E，B→D}1.证明:F:BC-> DE
I am having issues proving functional dependencies with Armstrong's Axioms. This one i'm struggling with. Let R(A,B,C,D,E) be a relation schema and F = {A→CD, C→E, B→D} 1. Prove: F: BC-> DE
我所拥有的:
1给定B-> D1.增强C在1，BC-> DC
1 Given B->D 1. Augment C on 1, BC->DC
2.在BC-> D BC-> C
2. Decomposition on 2, BC->D BC->C
3.BC-> C，BC-> E
3. Transitivity on BC->C, BC->E
4.BC-> D和4，BC-> DE上的联合
4. Union on BC ->D and and 4, BC->DE
不确定这是否是正确的解决方案.
Unsure if this is a proper solution.
也证明:AC-> BD我认为这无法得到证明.请帮忙！
Also Prove: AC-> BD I don't think this can be proven. Please Help!
推荐答案
除了一些明显的拼写错误之外，您的解决方案是正确的:
your solutions are correct, apart from some apparent misspelling:
- 给出B-> D，C-> E
- 公元前1月的C修正->DC
- 在2:BC分解->C(3.1)，BC->D(3.2)
- 1，3.1上的传递性:BC->C，C->E:BC->E
- 3.2和4的联盟:BC->DE
或者:
- B-> D，C-> E
- augment(1.1，c):bc->直流
- 增幅(1.2，d):cd->ed
- trans(2，3):bc->de(注意:bc-> dc< => bc-> cd< = cb-> cd< = cb-> dc)
ac->bd通常无法得到证明:检查阿姆斯特朗公理时，您注意到，要使某些X出现在派生fd的rhs上，它必须出现在原始fds之一的rhs上，除了
ac -> bd cannot be proven in general: inspecting the armstrong axioms you notice that for some X to appear on the rhs of a derived fd, it must occur on the rhs of one of the original fds, except for
-
X是原始fd的lh上某些X'的子集
X is a subset of some X' on the lhs of an original fd
或
fd是通过X的增强得到的
the fd is derived by augmentation with X
1.)构成了从未提及的约束.如果适用2.)，则X也会出现在原始fd的lhs上.消除X的唯一方法是通过传递性，该传递性要求X出现在原始fds之一的rhs上.
1.) constitutes a constraint never mentioned. if 2.) applies, X would appear on a lhs of the original fd too. the only way to eliminate X is by transitivity which requires X to appear on the rhs of one of the original fds.
将b作为X来查看ac->bd无法证明.
take b as X to see that ac -> bd is unprovable.
缩写 :
| 速记 | 扩展 |
|---|---|
| fd(s) | 功能依赖性(/-cies) |
| lhs | 左手边(方程式/导数) |
| rhs | 右侧 |
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