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I know that this is not a question answer site but for sake of explaining my doubt I have to post the entire question..

Consider the following statements.

If relation R is in 3NF and every key is simple, then R is in BCNF
If relation R is in 3NF and R has only one key, then R is in BCNF

Both 1 and 2 are true
1 is true but 2 is false
1 is false and 2 is true
Both 1 and 2 are false

Ans given is $a$, but how can it be so?

I agree with the first statement but for the second one consider that we have a relation $R={\{A,B,C,D\}}$ where $AB$ is the key, and say $C->B$ then it satisfies $3nf$ right? But it is $NOT$ in $BCNF$ right? as here we have $non-prime$ $deriving$ $a$ $prime$ $attribute$

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$S_1:$ If relation R is in 3NF and every key is simple, then R is in BCNF.

$S_2:$ If relation R is in 3NF and R has only one key, then R is in BCNF

Both statements are correct.

Your counterexample is as follow:

$R(A,B,C,D)$ where fds that hold are $FD = \{AB \rightarrow CD, C\rightarrow B\}$.

Now, this example does not satisfy premice of any of the statement because there are two candidate keys i.e $\{AB\}, \{AC\}$.

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  • $\begingroup$ Thanks brother.. :) $\endgroup$ – Turing101 Jan 5 at 8:43

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