Redundancy

Question

how decomposition of a->cd  to a->c and a->d remove redundancy in right side?

Comments

its a-> cd , not a->

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how decomposition of a->cd  to a->c and a->d remove redundancy in right side?

Decomposing a->cd to a->c and a->d will not do anything this is a trivial expansion.

if some relation satisfies the dependency a->cd then it will definitely satisfies a->c and a->d.

In real if any relation r satisfies  a->c and a->d both of them then we can combine them into one dependency i.e.

a->cd.

Answer 1

First step is to split right side of FD to single attribute.

Next find closure of complete LHS. No remove the FD and find closure of same LHS using other FD given. If both closure is same, this FD is redundant. Why? Even if we remove it our relation isn't affected.

Continue this for all FD to remove right redundancy.