Only IR1 - IR3 can be used IR1 for part a (Reflexive) If Y subset X, then X ?> Y

Question

Only IR1 - IR3 can be used IR1 for part a

(Reflexive) If Y subset X, then X ?> Y IR2.

(Augmentation) If X ?> Y, then XZ ?> YZ (Notation: XZ stands for X U Z) IR3.

(Transitive) If X ?> Y and Y ?> Z, then X ?> Z

b. Find the candidate keys and explain

Consider the schema R=(A, B, C, D, E) and lei the following set F of functional dependencies hold for R: F={ A rightarrow BC,CD rightarrow E, C rightarrow A B rightarrow D} Prove or disprove AD rightarrow E is in the closure of F. A proof can be made by using inference rules IR1 through IR3. A disproof should be done by showing a relational instance (counter example) that refutes the rule

Explanation / Answer

a)

Given, A -> BC

Then, A -> B, A -> C       by using Decomposition rule(If X -> YZ then X -> Y and X -> Z)

Now, consider the FD’s:

A -> C and CD -> E

AD -> E        by Pseudotransitivity rule (If X Y and WY Z, then WX Z)

Therefore, the FD AD -> E is in closure of F

b)

Attribute closure:

A -> ABCDE

B -> BD

C -> ABCDE

D -> D

E -> E

AB -> ABCDE

AC -> ABCDE

AD -> ABCDE

AE -> ABCDE

BC -> ABCDE

BD -> BD

BE -> BDE

CD -> ABCDE

CE -> ABCDE

DE -> DE

ABC -> ABCDE

ABD -> ABCDE

ABE -> ABCDE

ACD -> ABCDE

ACE -> ABCDE

ADE -> ABCDE

BCD -> ABCDE

BDE -> BDE

CDE -> ABCDE

ABCD -> ABCDE

ABCE -> ABCDE

ABDE -> ABCDE

ACDE -> ABCDE

BCDE -> ABCDE

Therefore the candidate keys are: A, C, AB, AC, AD,AE, BC, CD, CE, ABC, ABD, ABE, ACD,ACE, ADE, BCD, CDE, ABCD,ABCE, ACDE,BCDE.

The candidate keys with minimum fields are: A and C

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