Since only AB+ = AB, ABCD+ = ABCD
We can conclude following points-
1. A+ != A, so, A+ has something extra,
2. B+ != B , so B+ has something extra,
now AB+ = AB so no new attributes can be added to A+ or B+,
which implies – A+ = AB, B+ = AB.
3. Now, ABC+ != ABC,
so ABC+ = ABCD, similarly ABD+ = ABCD
4. Now A brings B and B brings A =, so ACD+ = ABCD, BCD+ = ABCD
5. Now when X+ = R ( X is a set of attributes and R is the full relation), we know that X is superkey.
6. In this way, ABC, ADC, ABD, and BCD are superkeys.
7. From point 3 we also conclude that C brings D and D brings C.
8. From 4, 7 we can easily conclude the following –
a. AB+ = AB
b. AC+ = ABCD
c. AD+ = ABCD
d. BC+ = ABCD
e. BD+ = ABCD
f. CD+ = ABCD
which implies AC, AD, BC, BD, CD ALL ARE SUPER KEYS.
9. From 7 and 8, C+ has to contain either A or B, but any of them brings the other,
so, C+ = ABCD, similarly, D+ = ABCD.
Here, we find the Candidate keys to be C and D, which implies A and B are non-prime attributes.
10. Since the CKs are atomic hence the relation R is in 2NF,
also, A+ = AB and B+ = AB implies there are transitive dependencies of the type Non-prime implies Non-Prime.
Hence the relation R is not in 3NF.
PS – There can be a more technical direct way to solve this question but it is one of a kind, new question, not seen yet, so this approach can be followed!