- Year not given
Find closure (AG)+
Short answer
(AG)+ = {A, B, C, G, H, I}, which is all of R, making AG a superkey and candidate key (since neither A nor G alone is a key).
Find (AG)+ if R = (A, B, C, G, H, I) and F = {A → B, A → C, CG → H, G → I, B → H, CG → I}.
The attribute closure (AG)⁺ is the set of all attributes functionally determined by A and G. Start with result = AG and keep adding right sides of FDs whose left sides are already inside the result.
F = {A → B, A → C, CG → H, G → I, B → H, CG → I}
| Step | FD used | Reason | result |
|---|---|---|---|
| 0 | start | A G | |
| 1 | A → B | A in result | A B G |
| 2 | A → C | A in result | A B C G |
| 3 | CG → H | C, G in result | A B C G H |
| 4 | G → I | G in result | A B C G H I |
| 5 | B → H, CG → I | nothing new | A B C G H I |
(AG)⁺ = {A, B, C, G, H, I}, which is all of R.
Conclusions
- AG is a superkey of R, because (AG)⁺ = R.
- It is also a candidate key: A⁺ = {A, B, C, H} lacks G and I, and G⁺ = {G, I} lacks the rest, so neither A nor G alone is a key. AG is minimal.
✓ Answer checked by Claude Opus 5.5 (high effort)
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