Third Normal Form and Boyce Code normal form
$S_1:$ 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 $\{AB\}, \{AC\}$.