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If you bring the functional dependencies of R in a canonical form, you can have only dependencies in which the right part is constituted only by a single attribute, which is prime, so we are sure that the relation is in 3NF. More, we can divide those dependencies into three sets:

  1. The dependencies where the LHS is a single attribute candidate key;

  2. Those where the LHS is the (only) compound candidate key;

  3. Those where the LHS is a strict subset of the compound candidate key.

The first two kinds of dependencies are compatible with the BCNF, while the third one nois not.

It is easy to show that the last set ifis empty. In fact, if there were a dependency X -> A, with X a strict subset of the compound candidate key, since A is a candidate key, that is determines all the other attributes, then also X determines all the other attributes, and so it is a candidate key. But this is a contradiction with the hypothesis, so no dependency X -> A where is X is not a superkey can exists, and the BCNF is surely satisfied.

If you bring the functional dependencies of R in a canonical form, you can have only dependencies in which the right part is constituted only by a single attribute, which is prime, so we are sure that the relation is in 3NF. More, we can divide those dependencies into three sets:

  1. The dependencies where the LHS is a single attribute candidate key;

  2. Those where the LHS is the (only) compound candidate key;

  3. Those where the LHS is a strict subset of the compound candidate key.

The first two kinds of dependencies are compatible with the BCNF, while the third one no.

It is easy to show that the last set if empty. In fact, if there were a dependency X -> A, with X a strict subset of the compound candidate key, since A is a candidate key, that is determines all the other attributes, then also X determines all the other attributes, and so it is a candidate key. But this is a contradiction with the hypothesis, so no dependency X -> A where is X is not a superkey can exists, and the BCNF is surely satisfied.

If you bring the functional dependencies of R in a canonical form, you can have only dependencies in which the right part is constituted only by a single attribute, which is prime, so we are sure that the relation is in 3NF. More, we can divide those dependencies into three sets:

  1. The dependencies where the LHS is a single attribute candidate key;

  2. Those where the LHS is the (only) compound candidate key;

  3. Those where the LHS is a strict subset of the compound candidate key.

The first two kinds of dependencies are compatible with the BCNF, while the third one is not.

It is easy to show that the last set is empty. In fact, if there were a dependency X -> A, with X a strict subset of the compound candidate key, since A is a candidate key, that is determines all the other attributes, then also X determines all the other attributes, and so it is a candidate key. But this is a contradiction with the hypothesis, so no dependency X -> A where is X is not a superkey can exists, and the BCNF is surely satisfied.

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source | link

If you bring the functional dependencies of R in a canonical form, you can have only dependencies in which the right part is constituted only by a single attribute, which is prime, so we are sure that the relation is in 3NF. More, we can divide those dependencies into three sets:

  1. The dependencies where the LHS is a single attribute candidate key;

  2. Those where the LHS is the (only) compound candidate key;

  3. Those where the LHS is a strict subset of the compound candidate key.

The first two kinds of dependencies are compatible with the BCNF, while the third one no.

It is easy to show that the last set if empty. In fact, if there were a dependency X -> A, with X a strict subset of the compound candidate key, since A is a candidate key, that is determines all the other attributes, then also X determines all the other attributes, and so it is a candidate key. But this is a contradiction with the hypothesis, so no dependency X -> A where is X is not a superkey can exists, and the BCNF is surely satisfied.