If relation R is in 3NF and every key is simple, then R is in BCNF
given that R is 3 NF.
the FD's possible is X ---> Y
i) either X is Super Key
ii) or Y should be Prime Attribute
if really X is a SuperKey, then it is also BCNF
if Y is a Prime attribute without X is a superkey then it may lead to doesn't satisfy BCNF.
But they given that every Key is simple ===> Y is prime attribute means Y is a Key.
( Key is simple means, it have only one attribute. ex:- 'AB' is not simple key because it contain morethan one attribute.
Every attribute in the Key is Prime attribute, there is only one attribute in the key means, the prime attribute itself is a Key )
if Y is a key then according to the FD X ----> Y, X is also a SuperKey. ===> it is in BCNF.
If relation R is in 3NF and R has only one key, then R is in BCNF
for understand this, let AB is the only key of R(A,B,C,D).
then can A --> B FD is exist ? No, if it exist, we have to say A is the Key...
From these we can understand that, there is No Dependency between the attributes of Same Key.
Given that we have only one key ===> Prime Attr ---> Prime Attr FD's can't possible
then only choice is to exist a dependency, X --->Y is X should be Key ====> it is in BCNF.
check https://gateoverflow.in/218551/normal-forms question for more understanding the line from Arjun Sir's explanation
Actually only when a 3NF relation contains overlapping candidate keys, it cannot be in BCNF. In all other cases, 3NF implies BCNF.