Consider a relation R(A, B, C, D, E, F) with the following functional…

Step 1: Find Minimal Cover

Given FDs: A→C, BD→F, EF→A, D→B

Check BD→F:

Compute D⁺: D → B (from D→B), so D⁺ = {D, B}. Then BD→F: since D⁺ already includes B, B is extraneous in BD→F.

Thus, BD→F reduces to D → F.

Minimal cover becomes:

• A → C

• D → B

• D → F

• EF → A

Note: D → B and D → F can be combined into D → BF, but this is optional for synthesis.

Step 2: 3NF Synthesis (create relation for each FD)

• From A→C → R1(A, C)

• From D→BF → R2(D, B, F)

• From EF→A → R3(E, F, A)

Step 3: Ensure a key appears in at least one relation

Find a key of R: Compute closure of DE.

D⁺ = {D, B, F} (from D→B, D→F), so DE⁺ = {D, E, B, F, A, C} = all attributes.

Thus, DE is a key.

No relation contains both D and E.

So add R4(D, E) to ensure lossless join and key presence.

Step 4: Verify dependency preservation

Check if all original FDs are preserved in the decomposition:

A→C: preserved in R1

BD→F: B and D are in R2, F is in R2 → preserved

EF→A: E and F in R3, A in R3 → preserved

D→B: D and B in R2 → preserved

All FDs are preserved.

Final Answer: Minimum number of relations = 4

Relations:

• R1(A, C)

• R2(D, B, F)

• R3(E, F, A)

• R4(D, E)