I am working through an example problem in which we are trying to identify which of the following relations is in Third Normal Form (3NF). Here are the relations we are given:>
In simple english, here are the 3 Normal Forms:
1NF: The existence of "the key" ensures that the table is in 1NF(Key must be there).
2NF:It is required that "each" non-key attributes be dependent on "the whole key" to ensure 2NF.
3NF: further requiring that "each" non-key attributes be dependent on "nothing but the key" ensures 3NF.
Now, for this:
R1(ABCD) ACD -> B AC -> D D -> C AC -> B
Look at these ACD -> B and AC -> B: clearly violates the 2NF condition. Forget 3NF, this relation is not even in 2NF. "the whole key"-->concept does not hold.
I think, you have proved the same using SET.