- Year not given
List all possible F+
Short answer
F+ includes all trivial and given FDs, plus derived ones like ACD → E, ACDF → G. The only candidate key is ACD.
If R = (A, B, C, D, E, F, G), F = {A → B, ABCD → E, EF → G}. List all possible F+.
The closure F⁺ is the set of all functional dependencies logically implied by F. It is built by applying Armstrong's axioms again and again.
- Reflexivity: if Y ⊆ X then X → Y
- Augmentation: if X → Y then XZ → YZ
- Transitivity: if X → Y and Y → Z then X → Z
- Derived: union (X→Y, X→Z ⇒ X→YZ), decomposition (X→YZ ⇒ X→Y, X→Z), pseudotransitivity (X→Y, YW→Z ⇒ XW→Z)
Given R = (A, B, C, D, E, F, G) and F = {A → B, ABCD → E, EF → G}.
A complete listing is not practical: R has 7 attributes, and F⁺ contains 3357 dependencies X → Y with Y non-empty (counting the trivial ones). So F⁺ is written by its structure, with all the important members listed.
1. Members of F itself
A → B, ABCD → E, EF → G
2. Trivial dependencies (reflexivity)
X → Y for every Y ⊆ X, for example A → A, AB → A, ABCDEFG → any subset of R.
3. Dependencies from the rules
- Augmentation of A → B: AC → BC, AD → BD, AF → BF, ACD → BCD, ACDF → BCDF, and so on with any Z.
- Pseudotransitivity: A → B and ABCD → E give ACD → E. Check: A → B augmented with CD gives ACD → BCD, so ACD → ABCD (add the trivial ACD → ACD and take the union); then ABCD → E gives ACD → E by transitivity.
- From EF → G and ACD → E: ACDF → EF (union/augment), then ACDF → G by transitivity.
- Union of these: ACDF → ABCDEFG.
- Augmentation of derived ones: ACDE → E (trivial), ACDF → any subset of {A,B,C,D,E,F,G}, for example ACDF → BE, ACDF → EG.
4. Attribute closures that generate F⁺
The closure X⁺ gives every right side for a left side X:
| X | X⁺ |
|---|---|
| A | AB |
| ACD | ABCDE |
| EF | EFG |
| ACDF | ABCDEFG |
| any X without A | X only (no new attributes unless X ⊇ EF) |
So F⁺ = { X → Y : Y ⊆ X⁺ }. For example, ACD → AB, ACD → BE, ACD → ABCDE, A → B, EF → G, EF → E, BEF → G.
Candidate key
A, C, D and F never appear on the right side of any FD, so they must be in every key. ACDF⁺ = ABCDEFG = R, so ACDF is the only candidate key.
✓ Answer checked by Claude Opus 5.5 (high effort)
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