Consider the relation T1(A,B,C,D,E) with the dependencies {EB -> C, D -> E, EA…

Answer: T1 is not in 3NF.

Find a candidate key for T1:

• Start with {A,D}.

• D -> E gives E, so we have {A,D,E}.

• EA -> B gives B, so we have {A,B,D,E}.

• EB -> C gives C, so {A,D}+ = {A,B,C,D,E}.

Therefore a candidate key is {A,D}. Prime attributes are A and D.

Check each functional dependency against 3NF:

• EB -> C: EB is not a superkey, and C is not a prime attribute. This violates 3NF.

• D -> E: D is not a superkey (it is only part of the composite key), and E is not prime. This violates 3NF.

• EA -> B: EA is not a superkey, and B is not prime. This also violates 3NF.

Since there are dependencies whose left sides are not superkeys and whose right sides are non-prime attributes, T1 is not in Third Normal Form.

Note: T2 is also not in 3NF. In T2, C is a candidate key, but the dependency A -> B has a left side that is not a superkey and a right side B that is not prime, so T2 violates 3NF as well.