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SQL show that R is not in normal form Boyce-Codd

R = (J,K,L,M,N)with a set of functional dependencies {J->KL,LM->N,K->M,N->J}.

I understand the definition of BCNF. I believe that there are no trivial functional dependencies and there cannot be a super key. I am not sure about the second part. How would you define a super-letter key? Appreciate some data on this subject.

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A relationship will be in normal Boyce-Codd (BCNF) form if closing the left attributes for all functional dependencies contains all the attributes of the relationship (J, K, L, M, N). In other words, the left attributes of each functional dependency contain a key.

:

  • J -> KL. K -> M, LM -> N N -> J. , J -> KL BCNF.
  • LM -> N. N -> J, J -> KL , .
  • K -> M. , , BCNF, .
  • N -> J. J -> KL K -> M. BCNF.

, BCNF, K .

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