Please solve this Q

Question

Question: 6  

Given the following functional dependencies
{X → W, X → Y, Y → Z, Z → PQ}
Consider the FD set implied using above FD set.
Implied FD set = {X → Z, X → WY, X → WZ, X → YQ, Y → P}
The number of functional dependencies in implied FD set are invalid are _________.

Comments

X → Z, X → WY, X → WZ, X → YQ, Y → P  no one is invalid

take LHS of every functional dependency and take the closure of it, if you get RHS then valid otherwise invalid.

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why Y → P invalid???

If we closure of Y then we will get (Z, P,Q),...

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yeah no one is invalid.

Answer 1

Given : {X → W, X → Y, Y → Z, Z → PQ}

Implied FD set = {X → Z, X → WY, X → WZ, X → YQ, Y → P}

X -> Y and Y -> Z implies X-> Z. Hence this is valid.

X -> W and X -> Y implies X -> WY. Hence this is valid.

X ->W ; X ->Y and X -> Z implies X ->WZ . Hence this is valid.

Closure of X in original FD : {XWYZPQ} Hence X -> YQ is valid.

Closure of Y in original FD : {YZPQ} Hence Y->P is valid.

Therefore there no invalid FDs in implied FD set.