1. Reflexivity Property
2. Transitive property
3. Augmentation Property
4. Union
5. Splitting Property
6. Another Property Composition
Example 1:
A → BCD then we can drive more FDs using the splitting property:
A → B, A → C, A → D, A → BC, A → CD, A → BD
Example 2:
If A → BCD so, can we say AB → CD is also FD because we can augment with B on both side
AB → BCD then using splitting property we can drive AB → CD.
So, we can drive all FDs from AB → BCD
Similarly, we can get if,
UGC NET 2013 (Dec)Q. 59.
Armstrong (1974) proposed a systematic approach to derive functional dependencies. Match the following with respect to functional dependencies:
List – I
a. Decomposition rule
b. Union rule
c. Composition rule
d. Pseudo transitivity rule
List – II
i. If X → Y and Z → W then {X, Z} → {Y, W}
ii. If X → Y and {Y, W} → Z then {X, W} → Z
iii. If X → Y and X → Z then X → {Y, Z}
iv. If X → {Y, Z} then X → Y and X → Z
Solution:
(A) Decomposition Rules or Splitting Rule:
(B) Union Rule:
(C) Composition Rule:
(D) Pseudo Transitivity Rule:
So, option (D) is the correct answer.