Question 31

The relation scheme is $R = (E, F, G, H, I, J, K, L, M, N)$.
The given functional dependencies are:

1. $\{E, F\} \to \{G\}$
2. $\{F\} \to \{I, J\}$
3. $\{E, H\} \to \{K, L\}$
4. $\{K\} \to \{M\}$
5. $\{L\} \to \{N\}$

Check attributes that appear on the right-hand side (RHS) of any FD:

• RHS attributes: $\{G, I, J, K, L, M, N\}$.
• Attributes never appearing on any RHS: $\{E, F, H\}$.

Since $E, F, H$ cannot be determined by any other attributes, every candidate key of $R$ must contain $\{E, F, H\}$.

Now compute the closure $\{E, F, H\}^+$:

• Initial: $\{E, F, H\}$
• Using $\{E, F\} \to \{G\}$: $\{E, F, G, H\}$
• Using $\{F\} \to \{I, J\}$: $\{E, F, G, H, I, J\}$
• Using $\{E, H\} \to \{K, L\}$: $\{E, F, G, H, I, J, K, L\}$
• Using $\{K\} \to \{M\}$: $\{E, F, G, H, I, J, K, L, M\}$
• Using $\{L\} \to \{N\}$: $\{E, F, G, H, I, J, K, L, M, N\} = R$

Since $\{E, F, H\}^+ = R$ and no proper subset of $\{E, F, H\}$ can be a key, $\{E, F, H\}$ is the unique minimal key for $R$.