explain why they are candiate keys. Only IR1 - IR3 can be used IR1. (Reflexive)

Question

explain why they are candiate keys.

Only IR1 - IR3 can be used

IR1. (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

Consider the schema R = (A, B, C, D, E) and let 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 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. Also what are the Candidate keys of R?|

Explanation / Answer

a)

Given A -> BC, B -> D

=> A -> D                          --- (1)

=> A -> DC -> E

Therefore, A -> ABCDE               --- (2)

Given C -> A, and from equation (2) A -> ABCDE,

=> C -> ABCDE

Given B -> D, BC -> CD,

=> BC -> ABCDE    Since, C->ABCDE

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 -> ABCDE

CD -> ABCDE

CE -> ABCDE

DE -> ABCDE

ABC -> ABCDE

ABD -> ABCDE

ABE -> ABCDE

ACD -> ABCDE

ACE -> ABCDE

ADE -> ABCDE

BCD -> ABCDE

BDE -> ABCDE

CDE -> ABCDE

ABCD -> ABCDE

ABCE -> ABCDE

ABDE -> ABCDE

ACDE -> ABCDE

BCDE -> ABCDE

Since,

C->A and CD->E     by functional dependency

Since, C-> A

=>AD->E

Thus, AD->E is closure to Functional dependency.

From the above statement, it proves that AD->E is closure to F.

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b)

Given A -> BC, B -> D

=> A -> D                          --- (1)

=> A -> DC -> E                     from functional dependency

Therefore, A -> ABCDE               --- (2)

Given C -> A, and from equation (2) A -> ABCDE,

=> C -> ABCDE                       --- (3)

Given B -> D,

BC -> CD,                    by Augmentation

=> BC -> ABCDE               Since, C->ABCDE by equation (3)

Given C->A                   by functional dependency

CD->AD                       by Augmentation

CD->ABCDE                    by equation (1)

Thus, the candidate keys are A, C, CD, and BC

Any combination of attributes that includes those is a super key.

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