Third Normal Form and Boyce Code normal form 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 $C-&gt;B$ then it satisfies $3nf$ right? But  it is $NOT$ in $BCNF$ right? as here we have $non-prime$  $deriving$  $a$  $prime$  $attribute$
$S_1:$ 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 $\{AB\}, \{AC\}$.