Thank you for very clear video. Just a question: why are we considering not only A and E but also BC and CD as candidate keys? By definition of candidate key, the set with minimum number of attributes are just A and E, both with 1 attribute, while BC and CD have both two attributes.
Bro the size of the key doesn't matter because the statement is- minimal set of super key not minimum so any size if its the minimal version and u can't remove anything from it then it can become the candidate key
Thank you for the help. Quick question, finding the candidate keys with 2 combinations of all attributes will always work regardless of how many Attributes and functional dependencies right? that is, there wont ever be a need to use 3 or more combinations of attributes correct?
