# 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 $$C->B$$ 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.

$$R(A,B,C,D)$$ 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 $$\{AB\}, \{AC\}$$.