Write the complete proof.
Steps in determining the highest normal form:
step 1: finding all the possible candidate keys of the relation
step 2: dividing all the attributes into prime and non-prime attributes.
step 3:checking for the 1st normal form and then 2nd normal form. If it fails in nth normal form, it is n-1 is the highest normal form.
(1)
step1:
so they are candidate keys.
step2: all are prime attributes
step3: The relation is in 1NF as given relational DBMS does not allow multiple values.
But, it is not in 2NF since A derives B being it is a candidate key.
So the highest normal is 1NF.
(2)
step1:
so ABD is the candidate key.
step2: {A,B,D} ------prime attributes
{C}--------------non-prime
step3: It is in 1NF.
but not in 2NF since A derives B being it is a part of candidate key.
So the highest normal is 1NF.
(3)
step1:
so AD is the candidate key.
step2: AD -----prime attribues
BC--non-prime attributes
step3: It is in 1NF.
But not in 2NF. AD is a candidate key. Part of AD, A derives C.
So the highest normal is 1NF.
(4)
step1:
so it is candidate key.
step2 : AB------------prime attributes
CD----------------non-prime attributes.
step3: It is in 1NF.
It is in 2NF.
But not in 3NF because there is transitive depandency.
If the relation R needs to be in BCNF it needs to satisfy the following
If is a relation
then X is a super key or Y is a prime attribute.
If there exists partial dependancy it is not in 2NF.
and ZX is a candidate key then if
it is not in 2NF.
We have to separate the tables
If there is a relation like this it is in transitive depandacy.
It is not in 3NF.
Then we have to separate them.
Consider the relational schemas given below and the respective sets of functional dependencies va...
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