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Z is a subset of Y so we have: Y â†’ Z

X â†’ Y and Y â†’ Z (transitivity) ïƒ¨ X â†’ Z

X â†’ Z and Z â†’ W (transitivity) ïƒ¨ X â†’ W.

Question2:

Proving completeness:

- Proving Reflexivity: Let Y âŠ† X. Then for some Z, we have YZ=X. Using B1 and B2, we

have (Yïƒ Y) implies that (YZïƒ Y). Then, we get Xïƒ Y. We thus conclude that Y âŠ† X

implies that Xïƒ Y.

- Proving Augmentation: Let Xïƒ Y. From B1, we have Xïƒ X. Then, from B3, we get

XZïƒ YZ. We thus conclude that Xïƒ Y implies that XZïƒ YZ.

- Proving Transitivity: Let Xïƒ Y and Yïƒ Z. Then from B3, we get Xïƒ Z (by assuming

W=âˆ…). We thus conclude that Xïƒ Y, Yïƒ Z implies that Xïƒ Z.

Proving soundness:

- Proving B1: Since X âŠ† X, we get from reflexivity that Xïƒ X, which proves B1.

- Proving B2: Let Xïƒ Y. Then by augmentation XZïƒ YZ. By reflexivity, we have YZïƒ Y.

Then, by transitivity we have XZïƒ Y, which proves B2.

- Proving B3: Let Xïƒ Y and Yïƒ Z. Then, by transitivity we get Xïƒ Z. By augmentation we

get XWïƒ ZW, which proves B3.

Question3:

We compute the closure of B as follows:

B+ = {B}

Bïƒ CD, then B+ = {BCD}

Dïƒ E, then B+ = {BCDE}

Bïƒ A, then B+ = {BCDEA}

Eïƒ F, then B+ = {BCDEAF}

Fïƒ G, then B+ = {BCDEAFG}

Hence, B is a primary key

ComputeX+ ({B},F) = {BCDEAFG}. We conclude that B is a super key. Since B is minimal so it

is also a candidate key.

Question4:

R={S,P,N,C,X,Y,Q}

FDs={Sïƒ NC, Pïƒ XY, SPïƒ Q}

SP is a super key. Then, Sïƒ NC and Pïƒ XY are BCNF violations.

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