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Inference Rules For Functional Dependencies



Inference 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 → C

Armstrong'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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Post date: 2015-06-26 06:41:31
Post date GMT: 2015-06-26 06:41:31
Post modified date: 2015-12-16 13:10:12
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