Consider a relation R(A, B, C, D, E, F, G) with the functional dependencies: A → BC, BC → DE, D → F, CF → G. Now, find {A}+, {D}+, {BC}+. Is {B,C} a superkey?
Understand the Problem
The question is asking to analyze a relational database with given functional dependencies and determine the closure of certain attribute sets. Specifically, it focuses on finding the closures of {A}, {D}, and {BC} and whether the set {B,C} is a superkey.
Answer
Closures: {A}+ = {A, B, C, D, E, F}, {D}+ = {D, F}, {BC}+ = {B, C, D, E, F, G}. {B,C} is not a superkey.
The closures are: {A}+ = {A, B, C, D, E, F}, {D}+ = {D, F}, {BC}+ = {B, C, D, E, F, G}. {B,C} is not a superkey.
Answer for screen readers
The closures are: {A}+ = {A, B, C, D, E, F}, {D}+ = {D, F}, {BC}+ = {B, C, D, E, F, G}. {B,C} is not a superkey.
More Information
A closure of a set is the set of attributes that can be functionally determined by it. A superkey is a set of attributes that can uniquely identify tuples in a relation.
Tips
Ensure you consistently apply functional dependencies and check if the closure includes all original attributes for superkey testing.
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