---
title: "Lossless and dependency preserving check 5"
subject: "Database Management System"
chapter: 4
key: "semester_6/database_management_system:decomp-abcd-abc-cd"
source: https://www.bctengineer.com/q/semester_6/database_management_system/decomp-abcd-abc-cd
site: BCT Engineering
---

# Lossless and dependency preserving check 5

Database Management System · Chapter 4: Database Constraints & Normalization

- Asked in: year not given
- Video: [Illustration on Lossless Join Decomposition and Dependency Preserving Decomposition. | DBMS](https://www.youtube.com/watch?v=iAarosX8ees) — Edredo for Learners
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/database_management_system/decomp-abcd-abc-cd
- Whole chapter: https://www.bctengineer.com/md/semester_6/database_management_system/chapter-4

## Question

R (A, B, C, D), R1 = (A, B, C), R2 = (C, D). F = {A → B, B → C, C → D}. Check whether the decomposition is lossless and dependency preserving.

## Short answer

Lossless (common C is key for R2) and dependency preserving: all FDs A→B, B→C, C→D are retained in R1 and R2.

## Answer

R = (A, B, C, D), R1 = (A, B, C), R2 = (C, D), F = {A → B, B → C, C → D}.

#### Lossless join test
R1 ∩ R2 = {C}. C⁺ = {C, D} = R2, so C → R2 holds: C is a key of R2. The condition (R1 ∩ R2) → R1 or R2 is satisfied.

Chase table:

| | A | B | C | D |
|---|---|---|---|---|
| R1 | a | a | a | b |
| R2 | b | b | a | a |

C → D: both rows have C = a, so D in R1 becomes a. R1 is now all "a".

**The decomposition is lossless.**

#### Dependency preservation test
Projections:
- F1 on (A, B, C): A → B, B → C (and A → C by transitivity)
- F2 on (C, D): C → D

F1 ∪ F2 contains A → B, B → C and C → D, which are exactly the three FDs of F.

**The decomposition is dependency preserving.**

**Answer: lossless and dependency preserving.** Each FD can be checked inside one relation, without a join.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is R1 = (A,B,C), R2 = (C,D) lossless?
   A. C+ = {C,D} = R2, so C is a key of R2
   B. A+ = R
   C. D is a key
   D. R1 contains all attributes
   Answer: A. The common attribute C determines R2 through C -> D, satisfying the lossless condition. (Key point: Lossless test)

2. In the chase, which FD makes R1 all 'a'?
   A. A -> B
   B. B -> C
   C. C -> D
   D. None is needed
   Answer: C. Both rows have C = a, so C -> D sets D = a in R1. (Key point: Chase step)

3. Why is the decomposition dependency preserving?
   A. A -> B and B -> C lie in R1, and C -> D lies in R2, covering all of F
   B. Because it is lossless
   C. Because R2 is small
   D. Because A is a key
   Answer: A. F1 ∪ F2 contains A->B, B->C and C->D, exactly the FDs of F, so each can be checked in one relation. (Key point: Dependency preservation)

## Close variants set in other exams

- [Lossless and dependency preserving check 2](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Adecomp-abcdeg-four-relations) — Database Management System, chapter 4
- [Lossless and dependency preserving check 3](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Adecomp-abc-ab-ac) — Database Management System, chapter 4
- [Lossless and dependency preserving check 4](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Adecomp-abcde-abc-ade) — Database Management System, chapter 4

## Related in Database Management System

- [Lossless and dependency preserving check 1](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Adecomp-abcde-r1-r2-cde) — Database Management System, chapter 4

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