# EQUIVALENCE OF FUNCTIONAL DEPENDENCY

Definition: Two or more than two sets of functional dependencies are called equivalence if the right-hand side of one set of functional dependency can be determined using the second FD set, similarly the right-hand side of the second FD set can be determined using the first FD set.

### Q 1: Given a relational schema R( X, Y, Z, W, V ) set of functional dependencies P and Q such that:

P = { X → Y, XY → Z, W → XZ, W → V} and Q = { X → YZ, W → XV } using FD sets P and Q which of the following options are correct?

1. P is a subset of Q
2. Q is a subset of P
3. P = Q
4. P ≠ Q

→Using definition of equivalence of FD set, let us determine the right-hand side of the FD set of P using FD set Q.

Given P = { X → Y, XY → Z, W → XZ, W → V} and Q = { X → YZ, W → XV }

Let's find closure of the left side of each FD of P using FD Q.

1. X+ = XYZ (using X → YZ)
2. XY+ = XYZ (using X → YZ)
3. W+ = WXVYZ (using W → XV and X → YZ)
4. W+ = WXVYZ (using W → XV and X → YZ)

Now compare closure of each X, XY, W and W calculated using FD Q with the right-hand side of FD P. Closure of each X, XY, W and W has all the attributes which are on the right-hand side of each FD of P. Hence, we can say P is a subset of Q----------1

→Using definition of equivalence of FD set, let us determine the right-hand side of the FD set of Q using FD set P.

Given P = { X → Y, XY → Z, W → XZ, W → V} and Q = { X → YZ, W → XV }

Let us find closure of the left side of each FD of Q using FD P.

1. X+ = XYZ (using X → Y and XY → Z)
2. W+ = WXZVY (using W → XZ, W → V, and X → Y)

Now compare closure of each X, W calculated using FD P with the right-hand side of FD Q. Closure of each X and W has all the attributes which are on the right-hand side of each FD of Q. Hence, we can say Q is a subset of P-----------2

From 1 and 2 we can say that P = Q, hence option C is correct.

### Q 2: Given a relational schema R( A, B, C, D ) set of functional dependencies P and Q such that:

P = { A → B, B → C, C → D } and Q = { A → BC, C → D } using FD sets P and Q which of the following options are correct?

1. a) P is a subset of Q
2. b) Q is a subset of P
3. c) P = Q
4. d) P ≠ Q

→Using definition of equivalence of FD set, let us determine the right-hand side of the FD set of P using FD set Q.

Given P = { A → B, B → C, C → D } and Q = { A → BC, C → D }

Let us find closure of the left side of each FD of P using FD Q.

1. A+ = ABCD (using A → BC, C → D)
2. B+ = B (no FD of Q has B on its left)
3. C+ = CD (using C → D)

Now compare closure of each A, B and C calculated using FD Q with the right-hand side of FD P. Closure B is B while in FD set P, B → C (B determines C), since the closure of B determined using FD Q has no C. Hence, we can say P is not a subset of Q----------1

→Using definition of equivalence of FD set, let us determine the right-hand side of the FD set of Q using FD set P.

Given P = { A → B, B → C, C → D } and Q = { A → BC, C → D }

Let us find closure of the left side of each FD of Q using FD P.

1. A+ = ABCD (using A → B, B → C, C → D)
2. C+ = CD (using C → D)

Now compare closure of each A and C calculated using FD P with the right-hand side of FD Q. Closure of each A and C has all the attributes which are on the right-hand side of each FD of Q. Hence, we can say Q is a subset of P-----------2

From 1 and 2 we can say that only Q is a subset of P since from 1 condition violated, hence option B is correct.

### Q 3: Given a relational schema R( X, Y, Z ) set of functional dependencies P and Q such that:

P = { X → Y, Y → Z, Z → X } and Q = { X → YZ, Y → X, Z → X } using FD sets P and Q which of the following options are correct?

1. P is a subset of Q
2. Q is a subset of P
3. P = Q
4. P ≠ Q

→Using definition of equivalence of FD set, let us determine the right-hand side of the FD set of P using FD set Q.

Given P = {X → Y, Y → Z, Z → X} and Q = {X → YZ, Y → X, Z → X}

Let us find closure of the left side of each FD of P using FD Q.

1. X+ = XYZ (using X → YZ)
2. Y+ = YZX (using X → YZ, Y → X)
3. Z+ = ZXY (using X → YZ, Y → X, Z → X)

Now compare closure of each X, Y, Z calculated using FD Q with the right-hand side of FD P. Closure of each X, Y, Z has all the attributes which are on the right-hand side of each FD of P. Hence, we can say P is a subset of Q----------1

→Using definition of equivalence of FD set, let us determine the right-hand side of the FD set of Q using FD set P.

Given P = { X → Y, Y → Z, Z → X } and Q = { X → YZ, Y → X, Z → X }

Let us find closure of the left side of each FD of Q using FD P.

1. X+ = XYZ (using X → Y, Y → Z, Z → X)
2. Y+ = YZX (using Y → Z, Z → X)
3. Z+ = ZXY (using X → Y, Y → Z, Z → X)

Now compare closure of each X, Y and Z calculated using FD P with the right-hand side of FD Q. Closure of each X, Y and Z has all the attributes which are on the right-hand side of each FD of Q. Hence, we can say Q is a subset of P-----------2

From 1 and 2 we can say that P = Q, hence option C is correct.

Conclusion: From the above three questions solved for equivalence of FD, we noticed the following steps which are as follows:

STEP 1: Suppose two FD sets P and Q are given, write FD of each set P and Q separately.

STEP 2: Take FD set P first and then find closure of each attribute of the left side of FD in P using FD set Q.

STEP 3: Now compare the closure of each attribute of P with the right-hand side of FD of P.

STEP 4: If closure of each attribute is equal to the right-hand side of FD of P, we say P is a subset of Q

STEP 5: Take FD set Q next and then find closure of each attribute of the left side of FD in Q using FD set P.

STEP 6: Now compare the closure of each attribute of Q with the right-hand side of FD of Q.

STEP 7: If closure of each attribute is equal to right-hand side of FD of Q, we say Q is a subset of P

STEP 8: IF P is a subset of Q and Q is a subset of P, we can say P = Q.

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