4 Answers
New AnswerYes, you are getting the correct answer: DH -> A,B,C,D,E,F,G,H,I,J Thus, the relation is in the Second Normal form, because NOT-key attributes are dependent on other NOT-key attributes. P.S. You have forgotten the A in the brackets ;)
But the relation should be in First Normal Form because H->I,J and for 2NF,subset of the candidate key should not be determining non-prime attributes.
You are right about the second normal form. And I'm sorry, but I only know until the 3.normal form. Can't help you with Boyce-Codd.
Panayot Zhaltov please check once whether I am correct or not. And also how to convert the relation to BCNF
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