Solution

Answer: Option 3 — T1 is not in 3NF.

Key Points

Given Relations & Functional Dependencies (FDs)

T1(A, B, C, D, E) with FDs: { EB → C, D → E, EA → B }

T2(A, B, C, D) with FDs: { C → A, A → B, A → D }

3NF : A relation is in 3NF iff for every FD X → Y, either (i) X is a superkey, or (ii) every attribute in Y is prime (i.e., part of some candidate key).

Step-1: Candidate Keys of T1

Try AD:

• D → E
• With A and E, EA → B
• With E and B, EB → C

Thus (AD)⁺ = {A, B, C, D, E} ⇒ AD is a (minimal) candidate key.

No single attribute (A or D alone) closes to all attributes; other pairs fail similarly. Hence the only candidate key is AD. Therefore, the prime attributes are A and D; B, C, E are non-prime.

3NF Check for T1

• EB → C: LHS EB is not a superkey; RHS C is non-prime ⇒ violates 3NF.
• D → E: LHS D is not a superkey; RHS E is non-prime ⇒ violates 3NF.
• EA → B: LHS EA is not a superkey; RHS B is non-prime ⇒ violates 3NF.

Conclusion for T1: Not in 3NF. (In fact, it also violates 2NF since D → E is a partial dependency of a non-prime attribute on part of the composite key AD.)

Step-2: Candidate Keys of T2

Compute closure of C:

• C → A, and A → B, A → D ⇒ C⁺ = {A, B, C, D} ⇒ C is a (minimal) candidate key.

A alone is not a key (A⁺ = {A, B, D}). Therefore the only candidate key is C, and C is the only prime attribute.

3NF Check for T2

• C → A: LHS is a superkey ⇒ OK.
• A → B: LHS A is not a superkey; RHS B is non-prime ⇒ violates 3NF.
• A → D: LHS not a superkey; RHS D is non-prime ⇒ violates 3NF.

Conclusion for T2: Not in 3NF.

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Final Verdict: Only the statement “T1 is not in 3NF” is true ⇒ Option 3.