r(abcde) ab->cde a->c d->e solve:
1nf presume in 1nf.
2nf ab candidate key here. a->c violation.
so, decompose them following:
r1 = (ac) + (ab) = (abc)
r2 = r - (ac) + (ab) = (bde) + (ab) = (abde)
3nf
???
the third normal form of relational schema following:
r1 (a b d) r2 (a c) r3 (d e) you can verify finding canonical cover of set of dependencies, is:
a b → d → c d → e 
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