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Inference Rules For Functional DependenciesInference Rules For Functional Dependencies Let S be the set of functional dependencies that are specified on relation schema R. Numerous other dependencies can be inferred or deduced from the functional dependencies in S. Example : Let S = {A → B, B → C} We can infer the following functional dependency from S: A → CArmstrong's Inference Rules Let A, B and C and D be arbitrary subsets of the set of attributes of the giver relation R, and let AB be the union of A and B. Then,⇒→
Proof : Step 1 : A → BC (GIVEN) Step 2 : BC → B (Using Rule 1, since B ⊆ BC) Step 3 : A → B (Using Rule 3, on step 1 and step 2)
Proof : Step 1 : A → B (GIVEN) Step 2 : A → C (given) Step 3 : A → AB (using Rule 2 on step 1, since AA=A) Step 4 : AB → BC (using rule 2 on step 2) Step 5 : A → BC (using rule 3 on step 3 and step 4)
Proof : Step 1 : A → B (Given) Step 2 : DB → C (Given) Step 3 : DA → DB (Rule 2 on step 1) Step 4 : DA → C (Rule 3 on step 3 and step 2)
Question 1:
Prove or disprove the following inference rules for functional dependencies.
Note: Read "⇒" as implies
a. {X → Y, Z → W} ⇒ XZ → YW ??
b. {X → Y, XY → Z} ⇒ X → Z
c. {XY → Z, Y → W} ⇒ XW → Z
Solution : Method : Use Armstrong's Axioms or Attribute closure to prove or disprove. a. {X → Y, Z → W} ⇒ XZ → YW ?? XZ → XZ XZ → XW (Z > W) XZ → W (decomposition rule) XZ → XZ XZ → YZ (X > Y) XZ → Y (decomposition rule) ⇒ XZ → YW (union rule) Hence True. b. {X → Y, XY → Z} ⇒ X → Z ?? XY→Z XX → Z (pseudotransitivity rule as X → Y) ⇒ X → Z Hence True. c. {XY → Z, Y → W} ⇒ XW → Z ?? W → W X → X Y → YW Z → Z WX → WX WY → WY WZ → WZ XY → WXYZ XZ → XZ YZ → WYZ Therefore WX → Z is not true You can also find the attribute closure for WX and show that closure set does not contain Z. Question 2:
Consider a relational scheme R with attributes A,B,C,D,F and the FDs
A → BC
B → E
CD → EF
Prove that functional dependency AD → F holds in R.
Step 1 : A → BC (Given)
Step 2 : A → C (Decomposition Rule applied on step 1)
Step 3 : AD → CD (Augmentation Rule applied on step 2)
Step 4 : CD → EF (Given)
Step 5 : AD → EF (transivity Rule applied on step 3 and 4)
Step 6 : AD → F (Decomposition Rule applied on step 5)


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