**Decomposition Without Dependency Preserving**

If the decomposition of relation R with FDs (F), into R1 and R2 with their FDs (F1) and (F2), respectively, will be decomposition without dependency preservation if.

Closure of F (F+) ≠ Closure of F1 (F1+) U Closure of F2 (F2+)

In simple words, we can also say,

**After decomposition,** IF all FDs of the original table are not preserved through FDs of decomposed tables, then it is called decomposition of the table without dependency preservation.

**Dependency Preserving Example**

Suppose a relation R with A, B, C, and D, and its FDs are AB→CD, D→A. This Relation R is decomposed into tables R1 with A and D attributes and R2 with B, C, and D attributes.

**Solution**

**Step 01:** Find the closure of each attribute given in left hand side of given FD’s of Relation R(ABCD). As given in the following diagram.

**Step 02: **Find all Non-Trivial FDs of Decomposed Relations (R1 and R2) as given below.

**Step 03:** Find all those Non-trivial FDs which are not determined from the given Relation R(ABCD)

Let’s check one by one all Non-Trivial FD’s of all decomposed relations R1 and R2.

**i. Check (A→D):**

As A→D of Relation R1(AD) cannot determined from the original table R(ABCD) Because Closure of A in the Original Table cannot determine “D,” this is an in-valid Dependency.

**ii. Check (D→A):**

As D→A of Relation R1 (AD) is directly given in the FDs of Relation R(ABCD). So, This Dependency can determined from the original table R(ABCD) Because the Closure of D in the Original Table can determine “A”. So, this is a valid Dependency.

**iii. Check (B→CD):**

As B→CD of Relation R2(BCD) cannot determined from original table R(ABCD) Because Closure of B in the Original Table cannot determine “CD”. So, this is an in-valid Dependency.

**iv. Check (C→BD):**

As C→BD of Relation R2(BCD) cannot determined from original table R(ABCD) Because Closure of C in the Original Table cannot determine “BD”. So, this is an in-valid Dependency.

**v. Check (D→CB):**

As D→CB of Relation R2(BCD) cannot determined from original table R(ABCD) Because Closure of D in the Original Table cannot determine “CB”. So, this is an in-valid Dependency.

**vi. Check (BC→D):**

As BC→D of Relation R2(BCD) cannot determined from the original table R(ABCD), Because Closure of BC in the Original Table cannot determine “D,” this is an in-valid Dependency.

**vii. Check (BD→C):**

As BD→C of Relation R2(BCD) can be determined from original table R(ABCD) Because Closure of BD in the Original Table can determine “C”. So, this is a valid Dependency.

Note: As BD determines itself (BD), D determines A, and AB can determine C in Original table R (ABCD).

**viii. Check (CD→B):**

As CD→B of Relation R2(BCD) cannot determined from original table R(ABCD) Because the Closure of CD in the Original Table cannot determine “B”. So, this is an in-valid Dependency.

So, just two non-trivial FDs are valid, which are given below.

**Step 04: **Find the closure of all valid Non-trivial FD’s of Decomposed Relations (R1 and R2) as given below

**Step 05: **If all dependencies of a given relation are preserved through all valid non-trivial dependencies of its decomposed tables, then the original table will be preserved.

**Explanation of the Above diagram**

We will check all valid Non-trivial FDs of decomposed tables (one by one) and see whether these FDs can preserve all FDs of original Relation. If it preserves, then the decomposition is dependency-preserving; otherwise, it will not be dependency-preserving decomposition.

All two FDs of original Relation are not preserved through valid Non-trivial FDs of decompose tables. So, it is an example of the decomposition of tables without dependency preserving.