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Consider the relation scheme $R = (E, F, G, H, I, J, K, L, M, N)$ and the set of functional dependencies $$\left\{ \{E, F \} \to \{G\}, \{F\} \to \{I, J\}, \{E, H\} \to \{K, L\}, \\ \{K\} \to \{M\}, \{L\} \to \{N\}\right\}$$ on $R$. What is the key for $R$?

1. $\{E, F\}$
2. $\{E, F, H\}$
3. $\{E, F, H, K, L\}$
4. $\{E\}$

Since $E,F,H$ cannot be derived from anything else $E,F,H$ should be there in key.

Using Find $\{EFH\}^+,$ it contains all the attributes of the relation.

Hence, it is key.
edited
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Question is about KEY. Its not mentioned that it should be a candidate key or a super key. Then how to determine what is the Answer, because option C is also a KEY coz it contains EFH too.
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if it is mentioned as key then its only candidate keys / primary key

a super key is any set of attribute that uniquely determines a tuple

whereas a key or candidate key is minimal super key

so option C is superkey. but set of attributes is not minimal

hence not a key
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Thanks a lot... :)
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With H even E & F  cannot be derived from anything else so E,F,H need to be there in key!!
+1
Really awesome way of finding keys. Must watch. Check this

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Sir EFH must be in the key

### D) {E}+ = {E} ≠ R

+1 vote

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m confused between commas in LHS part .{E,F,H} why we are taking as whole {EFH} ?
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CAN WE APPLY UNION RULE FOR COMBINING {E,F,H}
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my solution is based on given option that's why I take EFH as the combination. take a closer set of EFH you get all the attribute.