Let U be a relation schema. A set of relation schemas {R1 , R2 , … , Rn} is a decomposition of U if and only if
U = R1 𝖴 R2 𝖴 …𝖴 Rn
Lossless-Join Decomposition
A decomposition {R, T} of U is a lossless-join decomposition (with respect to a set of constraints) if the constraints imply that u = r 
t for all possible instances of R, T, and U.
The decomposition is said to be lossy otherwise.
It is always the case for any decomposition {R, T} of U that u⊆ r t.
1. Properties of a Good Decomposition
A relational schema R with a set of functional dependencies F is decomposed into R1 and R2.
1. Lossless-join decomposition
Test to see if at least one of the following dependencies are in F+
R1 ∩ R2 → R1
R1 ∩ R2 → R2
If not, decomposition may be lossy
2. Dependency preservation
Let Fi be the set of dependencies in F+ that include only attributes in Ri (Notation: 
))
Test to see if (F1 𝖴 F2)+= F+
When a relation is modified, no other relations need to be checked to preserve dependencies.
3. No redundancy Example.
- R1 = (A, B, C)
- F = {A → B, B → C}
- R1 = (A, B), R2 = (B, C)
Lossless-join decomposition?
R1 ∩ R2 = {B} and B → BC ∈ F+
Dependency preserving?
F+ = {A → B, A → C, A → AB, A → BC, A → AC, A → ABC, AB → C, AB → AC, AB → BC, AB → ABC, AC → BC,
AC → {B, B → C, B → BC} ∪{many trivial dependencies}
F1 = {A → B, A → AB} ∪ (many trivial dependencies}
F2 = {B → C, B → BC} ∪ (many trivial dependencies}
(F1 ∪ F2)+ = F+
Example.
- R = (A, B, C)
- F = {A → B, B → C}
- R1 = (A, B), R2 = (A, C)
Lossless-join decomposition: yes
R1 ∩ R2 = {A} and A → AB
Dependency preserving: no
F+ = {A → B, A → C, A → AB, A → BC, A → AC, A → ABC, AB → C, AB → AC, AB → BC, AB → ABC, AC → BC,
{AC → B, B → C, B → BC} ∪ (many trivial dependencies} F1 = {A ® B, A ® AB} (many trivial dependencies}
F2 = {A → C, A → AC} ∪ (many trivial dependencies} (F1 ∪ F2)+ # F+
Cannot check B ® C without computing R1 ∩ R2
- It is always possible to decompose a relation into relations in 3NF such that the decomposition is lossless, and dependencies are
- It is always possible to decompose a relation into relations in BCNF such that the decomposition is
- It may not be possible to preserve dependencies and BCNF.
Source: Gupta Satinder Bal, Mittal Aditya (2017), Introduction to Basic Database Management System, 2nd Edition-University Science Press (2017)