![]() If it was not in 2NF, then h,i or j would have to have a function dependency (either all together or some subset) which relates from a subset of e,f,g. Here, f,g->j, and so this cannot be in 2NF. It therefore follows it cannot be in 3NF or BCNF, so it must be in either 0NF or 1NF. ![]() 3NF involves partial key dependencies. In this case, to NOT be in 3NF, e,f,g would have to be involved in selecting a subset of h,i,j. There are dependencies which produce this selection ( f,g => j ). Thus this relation IS NOT in 3NF, and thus cannot be in 4NF or BCNF. ![]() 3NF involves partial key dependencies. In this case, to NOT be in 3NF, e,f,g would have to be involved in selecting a subset of h,i,j. There are dependencies which produce this selection ( f,g => j ). Thus this relation IS NOT in 3NF, and thus cannot be in 4NF or BCNF. ![]() 3NF involves partial key dependencies. In this case, to NOT be in 3NF, e,f,g would have to be involved in selecting a subset of h,i,j. There are dependencies which produce this selection ( f,g => j ). Thus this relation IS NOT in 3NF, and thus cannot be in 4NF or BCNF. Consider the following functional dependencies
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