Show the following rule for MVDs The attributes are arbitrar

Show the following rule for MVD\'s. The attributes are arbitrary sets X, Y, Z and the other unnamed attributes of the relation in which these dependencies hold. The Union Rule. If X, Ys and Z are sets of attributes, X - > - > Y, and X - > - > Z, then X - > - > (Y U Z) Show the following rule for MVD\'s. The attributes are arbitrary sets X, Y, Z and the other unnamed attributes of the relation in which these dependencies hold. The Intersection Rule. If X, Y, and Z are sets of attributes, X - > - > Y and X - > - > Z, then X - > - > (Y Z). Show the following rule for MVD\'s. The attributes are arbitrary sets X, Y, Z and the other unnamed attributes of the relation in which these dependencies hold. The Difference Rule. If X, Y, Z and are sets of attributes, X - > - > Y, and X - > - > Z, then X - > - > (Y-Z). Show the following rule for MVD\'s. The attributes are arbitrary sets X, Y, Z and the other unnamed attributes of the relation in which these dependencies hold. Removing attributes shared by left and right side. If X - > - > Y holds, then X - > - > (Y-X) holds.

Solution

Q9 Answer:

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.

Q12 Answer: