Equivalence of Functional Dependencies (original) (raw)

Last Updated : 24 Apr, 2026

Equivalence of functional dependencies means two sets of functional dependencies (FDs) are considered equivalent if they enforce the same constraints on a relation. This happens when every FD in one set can be derived from the other set and vice versa using inference rules like Armstrong's axioms.

• Equivalent FDs result in the same set of valid relations and preserve the same data integrity.
• It helps in normalization and optimizing database design without losing any constraints.

**Find the Relationship Between Two Functional Dependency Sets

Let FD1 and FD2 be two FD sets for a relation R.

1. If all FDs of FD1 can be derived from FDs present in FD2, we can say that FD2 ⊃ FD1.
2. If all FDs of FD2 can be derived from FDs present in FD1, we can say that FD1 ⊃ FD2.
3. If 1 and 2 both are true, FD1=FD2.

All these three cases can be shown using the Venn diagram:

Equivalence of Functional Dependency

**Need to Compare Functional Dependencies

Suppose in the designing process we convert the ER diagram to a relational model and this task is given to two different engineers. Now those two engineers give two different sets of functional dependencies.

• Being an administrator we need to ensure that we must have a good set of Functional Dependencies.
• To ensure this we require to study the equivalence of Functional Dependencies.

Advantages

• It can help to identify redundant functional dependencies, which can be eliminated to reduce data redundancy and improve database performance.
• Helps to optimize database design by identifying equivalent sets of functional dependencies that can be used interchangeably.
• Ensures data consistency by identifying all possible combinations of attributes that can exist in the database.

Disadvantages

• The process of determining the equivalence of functional dependencies can be computationally expensive, especially for large datasets.
• It may require testing multiple candidate sets of functional dependencies, which can be time-consuming and complex.
• The equivalence of functional dependencies may not always accurately reflect the semantic meaning of data, and may not always reflect the true relationships between data elements.

Sample Questions

**Q.1 Let us take an example to show the relationship between two FD sets. A relation R(A,B,C,D) having two FD sets FD1 = {A->B, B->C, AB->D} and FD2 = {A->B, B->C, A->C, A->D}

**Step 1: Checking whether all FDs of FD1 are present in FD2

• A->B in set FD1 is present in set FD2.
• B->C in set FD1 is also present in set FD2.
• AB->D is present in set FD1 but not directly in FD2 but we will check whether we can derive it or not. For set FD2, (AB)+ = {A, B, C, D}. It means that AB can functionally determine A, B, C, and D. So AB->D will also hold in set FD2.

As all FDs in set FD1 also hold in set FD2, **FD2 ⊃ FD1 is true.

**Step 2: Checking whether all FDs of FD2 are present in FD1

• A->B in set FD2 is present in set FD1.
• B->C in set FD2 is also present in set FD1.
• A->C is present in FD2 but not directly in FD1 but we will check whether we can derive it or not. For set FD1, (A)+ = {A, B, C, D}. It means that A can functionally determine A, B, C, and D. SO A->C will also hold in set FD1.
• A->D is present in FD2 but not directly in FD1 but we will check whether we can derive it or not. For set FD1, (A)+ = {A, B, C, D}. It means that A can functionally determine A, B, C, and D. SO A->D will also hold in set FD1.

As all FDs in set FD2 also hold in set FD1, **FD1 ⊃ FD2 is true.

**Step 3: As FD2 ⊃ FD1 and FD1 ⊃ FD2 both are true **FD2 =FD1 is true. These two FD sets are semantically equivalent.

**Q.2 Let us take another example to show the relationship between two FD sets. A relation R2(A,B,C,D) having two FD sets FD1 = {A->B, B->C,A->C} and FD2 = {A->B, B->C, A->D}

**Step 1: Checking whether all FDs of FD1 are present in FD2

• A->B in set FD1 is present in set FD2.
• B->C in set FD1 is also present in set FD2.
• A->C is present in FD1 but not directly in FD2 but we will check whether we can derive it or not. For set FD2, (A)+ = {A, B, C, D}. It means that A can functionally determine A, B, C, and D. SO A->C will also hold in set FD2.

As all FDs in set FD1 also hold in set FD2, FD2 ⊃ FD1 is true.

**Step 2: Checking whether all FDs of FD2 are present in FD1

• A->B in set FD2 is present in set FD1.,
• B->C in set FD2 is also present in set FD1.
• A->D is present in FD2 but not directly in FD1 but we will check whether we can derive it or not. For set FD1, (A)+ = {A,B,C}. It means that A can’t functionally determine D.
• So A->D will not hold in FD1.

As all FDs in set FD2 do not hold in set FD1, FD2 ⊄ FD1.

**Step 3: In this case, FD1 ⊆ FD2 and FD2 ⊄ FD1, these two FD sets are not semantically equivalent.