Understanding 3NF Requirements

A relation is in 3rd Normal Form (3NF) if, for every non-trivial functional dependency (FD) of the form X → Y, one of the following is true:

• X is a superkey of the relation.
• Y is a prime attribute (meaning Y is part of at least one candidate key of the relation).

If an FD exists in a relation where neither of these conditions holds, the relation is not in 3NF.

Analyzing Relation T1 for 3NF

We are given relation T1(A,B,C,D,E) with the following functional dependencies (FDs):

• {EB → C}
• {D → E}
• {EA → B}

Step 1: Determine Candidate Keys and Prime Attributes for T1

First, let's find the attributes that are not on the right-hand side of any FD: A, B, D, E. These are potential parts of candidate keys.

We calculate the closure of attribute sets:

• {A, D, E}+ = {A, D, E} ∪ {E} (from D→E) ∪ {B} (from EA→B) ∪ {C} (from EB→C) = {A, B, C, D, E}. Thus, {A, D, E} is a superkey.
• {A, B, D}+ = {A, B, D} ∪ {E} (from D→E) ∪ {C} (from EB→C) = {A, B, C, D, E}. Thus, {A, B, D} is a superkey.

Checking for minimality, we find that both {A, D, E} and {A, B, D} are candidate keys for T1.

Prime attributes are those part of any candidate key: {A, B, D, E}. The only non-prime attribute is {C}.

Step 2: Check Functional Dependencies against 3NF Rules for T1

Let's examine each FD:

• FD: EB → C
  • Is EB a superkey? No. ({EB}+ = {E, B, C}, which does not include A or D).
  • Is C a prime attribute? No.
  • Result: Violation. Since neither condition is met, this FD violates 3NF.
• FD: D → E
  • Is D a superkey? No. ({D}+ = {D, E}).
  • Is E a prime attribute? Yes (it's part of {A, D, E} and {A, B, D}).
  • Result: Satisfies 3NF.
• FD: EA → B
  • Is EA a superkey? No. ({EA}+ = {A, E, B}).
  • Is B a prime attribute? Yes (it's part of {A, B, D}).
  • Result: Satisfies 3NF.

Conclusion for T1: Because the FD EB → C violates the 3NF conditions, relation T1 is not in 3NF.

Analyzing Relation T2 for 3NF

We are given relation T2(A,B,C,D) with the following functional dependencies (FDs):

• {C → A}
• {A → B}
• {A → D}

Step 1: Determine Candidate Keys and Prime Attributes for T2

Attributes not on the right-hand side are B, C, D.

Calculating the closure:

• {C, D}+ = {C, D} ∪ {A} (from C→A) ∪ {B, D} (from A→B, A→D) = {A, B, C, D}. Thus, {C, D} is a superkey.

Checking for minimality, {C, D} is the sole candidate key for T2.

Prime attributes: {C, D}. Non-Prime attributes: {A, B}.

Step 2: Check Functional Dependencies against 3NF Rules for T2

Let's examine each FD:

• FD: C → A
  • Is C a superkey? No. ({C}+ = {C, A}).
  • Is A a prime attribute? No.
  • Result: Violation. This FD violates 3NF.
• FD: A → B
  • Is A a superkey? No. ({A}+ = {A, B, D}).
  • Is B a prime attribute? No.
  • Result: Violation. This FD violates 3NF.
• FD: A → D
  • Is A a superkey? No.
  • Is D a prime attribute? Yes (it's part of candidate key {C, D}).
  • Result: Satisfies 3NF.

Conclusion for T2: Because the FDs C → A and A → B violate the 3NF conditions, relation T2 is also not in 3NF.

Overall Conclusion

Our analysis shows that T1 is not in 3NF due to the FD EB → C. T2 is also not in 3NF due to FDs C → A and A → B.

Therefore, the statement "T1 is not in 3NF" is TRUE.