The document Equivalence of Functional Dependencies Notes | EduRev is a part of the GATE Course Database Management System (DBMS).

All you need of GATE at this link: GATE

**Introduction**

Given a Relation with different FD sets for that relation, we have to find out whether one FD set is subset of other or both are equal.

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

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

**All these three cases can be shown using Venn diagram as:**

**Q. 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 in 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. 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, FD2 ⊃ FD1 and FD2 ⊄ FD1, these two FD sets are not semantically equivalent.

The term Armstrong axioms refer to the sound and complete set of inference rules or axioms, introduced by William W. Armstrong, that is used to test the logical implication of **functional dependencies.** If F is a set of functional dependencies then the closure of F, denoted as F^+, is the set of all functional dependencies logically implied by F. Armstrong’s Axioms are a set of rules, that when applied repeatedly, generates a closure of functional dependencies.

**Axiom of reflexivity:**If**A**is a set of attributes and**B**is subset of**C**, then**C**holds**B**. If**B ⊆ A**then**A → B**This property is trivial property.**Axiom of augmentation:**If**A → B**holds and**Y**is attribute set, then**AY → BY**also holds. That is adding attributes in dependencies, does not change the basic dependencies. If**A → B**, then**AC → BC**for any**C.****Axiom of transitivity:**Same as the transitive rule in algebra, if**A → B**holds and**B****→ C**holds, then**A → C**also holds.**A → B**is called as**A**functionally that determines**B**. If**X****→ Y**and**Y → Z**, then**X****→****Z.**

These rules can be derived from the above axioms.

**Union:**If**A → B**holds and**A → C**holds, then**A → BC**holds. If**X → Y**and**X****→****Z**then**X****→ Y****Z****Composition:**If**A****→ B**and**X****→ Y**holds, then**A****X****→ BY**holds.**Decomposition:**If**A****→ BC**holds then**A****→ B**and**A****→ C**hold. If**X****→ Y****Z**then**X****→ Y**and**X****→****Z****Pseudo Transitivity****:**If**A → B**holds and**BC → D**holds, then**AC****→ D**holds. If**X****→ Y**and**YZ****→ W**then**XZ****→ W**.

**Why armstrong axioms refer to the Sound and Complete?**

By sound, we mean that given a set of functional dependencies F specified on a relation schema R, any dependency that we can infer from F by using the primry rules of amrmstrong axioms holds in every relation state r of R that satisfies the dependencies in F.

By complete, we mean that using primary rules of amrstrong axioms repeatedly to infer dependencies until no more dependencies can be inferred results in the complete set of all possible dependencies that can be inferred from F.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

49 docs|15 tests