Canonical Cover in DBMS

Solution-

STEP 1: Write all the functional dependencies such that each dependency contains only one attribute on its right side i.e-

 X → W

WZ → X

WZ → Y

Y → W

Y → X

Y → Z

STEP 2: Check each functional dependency one by one whether it is essential or not

• For X → W:

Including X → W, (X) + = {X, W}

Ignoring X → W, (X) + = {X}

Now, as the the two results are different, it is concluded that X→W is essential and cannot be eliminated.

• For WZ → X:

Including WZ → X, (WZ) + = {W, X, Y, Z}

Ignoring WZ → X, (WZ) + = {W, X, Y, Z}

Now, as the the two results are same, it is concluded that WZ→X is not essential and can be eliminated.

After eliminating WZ→X ,the resultant set of functional dependencies is :

X → W

WZ → Y

Y → W

Y → X

Y → Z

Now, we will consider this reduced set in further checks. 

• For WZ → Y:

Including WZ → Y, (WZ) + = {W, X, Y, Z}

Ignoring WZ → Y, (WZ) + = {W, Z}

 Now, as the the two results are different, it is concluded that WZ→Y is essential and cannot be eliminated.

• For Y → W:

Including Y → W, (Y) + = {W, X, Y, Z}

Ignoring Y → W, (Y) + = {W, X, Y, Z}

Now, as the the two results are same, it is concluded that Y→W is not essential and can be eliminated.

After eliminating Y→W,the resultant set of functional dependencies is :

X → W

WZ → Y

Y → X

Y → Z

• For Y → X:

Including Y → X, (Y) + = {W, X, Y, Z}

Ignoring Y → X, (Y) + = {Y, Z}

 Now, as the the two results are different, it is concluded that Y→X is essential and cannot be eliminated.

•  For Y → Z:

Including Y → Z, (Y) + = {W, X, Y, Z}

Ignoring Y → Z, (Y) + = {W, X, Y}

 Now, as the the two results are different, it is concluded that Y→Z is essential and cannot be eliminated.

• From here, our essential functional dependencies are-

X → W

WZ → Y

Y → X

Y → Z