Prove that X X A relation R whose schema is the set of A B

Prove that {X^+}^+ = {X}^+ A relation R whose schema is the set of {A, B, C, D} attributes with FD\'s A rightarrow B and A rightarrow C. Either is a BCNF violation, because the only key for R is {A, D}. Suppose we start by decomposing R according to A rightarrow B. Do we ultimately get the same result as if we first expand BCNF violation to A rightarrow BC? Why or Why not?

Solution

Q6)

We will do this by using a proof by contradiction.

Assume that (X⁺)⁺ X⁺. Then for (X⁺)⁺, it must be that some Functional Dependency allowed additional attributes to be added to the original set X⁺.

For instance, X⁺ A where A is some attribute not in X⁺.

If this was the case, then X⁺ wouldn’t be the closure of X. The closure of X should have to include A as well.

This contradicts the fact that we were given the closure of X, X⁺.

Hence, it must be that (X⁺)⁺ = X⁺ or else X⁺ is not the closure of X.

Q7)